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https://mathhelpboards.com/threads/the-union-of-two-open-sets-is-open.27758/ | # The Union of Two Open Sets is Open
#### G-X
##### New member
Let $$\displaystyle x ∈ A1 ∪ A2$$ then $$\displaystyle x ∈ A1$$ or $$\displaystyle x ∈ A2$$
If $$\displaystyle x ∈ A1$$, as A1 is open, there exists an r > 0 such that $$\displaystyle B(x,r) ⊂ A1⊂ A1 ∪ A2$$ and thus B(x,r) is an open set.
Therefore $$\displaystyle A1 ∪ A2$$ is an open set.
How does this prove that $$\displaystyle A1 ∪ A2$$ is an open set. It just proved that $$\displaystyle A1 ∪ A2$$ contains an open set; not that the entire set will be open? This is very similar to the statement: An open subset of R is a subset E of R such that for every x in E there exists ϵ > 0 such that Bϵ(x) is contained in E.
Last edited:
#### castor28
##### Well-known member
MHB Math Scholar
The point is that the argument is valid for every $x\in A_1\cup A_2$.
If $C = A_1\cup A_2$, we have proved that, for every $x\in C$, there is an open ball $B(x,r)\subset C$ (where $r>0$ depends on $x$). That is precisely the definition of an open set.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
If $$\displaystyle x ∈ A1$$, as A1 is open, there exists an r > 0 such that $$\displaystyle B(x,r) ⊂ A1⊂ A1 ∪ A2$$ and thus B(x,r) is an open set.
Therefore $$\displaystyle A1 ∪ A2$$ is an open set.
Hi G-X, welcome to MHB!
As castor28 's pointed out, it's about the definition of an open set, which he effectively quoted.
Additionally that proof is not entirely correct and it is incomplete.
It should be for instance:
If $$\displaystyle x ∈ A1$$, as $A1$ is open, there exists an $r > 0$ such that $$\displaystyle B(x,r) ⊂ A1$$ (from the definition of an open set), which implies that $$\displaystyle B(x,r)⊂ A1 ∪ A2$$.
If $$\displaystyle x ∈ A2$$, as $A2$ is open, there exists an $r > 0$ such that $$\displaystyle B(x,r) ⊂ A2⊂ A1 ∪ A2$$.
Therefore for all $$\displaystyle x ∈ A1∪ A2$$, there exists an $r > 0$ such that $$\displaystyle B(x,r) ⊂ A1 ∪ A2$$.
Thus $$\displaystyle A1 ∪ A2$$ is an open set.
#### G-X
##### New member
I see, I think I had the misunderstanding that something from A2 might close A1.
But I don't think that is an issue you technically need to wrap your head around.
Because the definition states: We define a set U to be open if for each point x in U there exists an open ball B centered at x contained in U.
So, essentially looping over A1, A2 - making the reference that open balls exist at each of these points then all points in A1 ∪ A2 have open balls contained in the union thus by the definition it must be open.
Last edited: | 2021-06-18T06:16:40 | {
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https://math.stackexchange.com/questions/1487873/what-is-the-best-way-of-resolving-an-expression-with-square-roots-in-denominator/1487936 | What is the best way of resolving an expression with square roots in denominator?
I was resolving the following question from my textbook:
Write each of the following expressions as a single fraction, simplifying your answer where possible:
$4 - \frac{1}{\sqrt{12}} + \frac{10}{\sqrt{3}}$
I have resolved it this way:
$\frac{4\sqrt{12}\sqrt{3}}{\sqrt{12}\sqrt{3}} - \frac{\sqrt{3}}{\sqrt{12}\sqrt{3}} + \frac{10\sqrt{12}}{\sqrt{12}\sqrt{3}} = \frac{4\sqrt{36} - \sqrt{3}+10\sqrt{4}\sqrt{3}}{\sqrt{36}}=\frac{24 - \sqrt{3} + 20\sqrt{3}}{6}$
while in the textbook it is resolved like this:
$4 - \frac{1}{\sqrt{4 \times 3}} + \frac{10}{\sqrt{3}} = 4 - \frac{1}{2\sqrt{3}} + \frac{10}{\sqrt{3}} = \frac{8\sqrt{3} - 1 + 20}{2\sqrt{3}}= \frac{8\sqrt{3} + 19}{2\sqrt{3}}$
I am wondering did I do it correctly and if so, which way is better?
• Yes, your answer is correct. The only difference is a factor $\sqrt 3\over \sqrt 3$ in your answer vs. that in the book, and adding $-\sqrt 3$ to $20\sqrt 3$. – abiessu Oct 19 '15 at 18:26
• This indeed answerd my question. Would you mind posting it as an answer so i can mark the thread as answered. – Pawel Oct 19 '15 at 18:45
1 Answer
Your answer of $$\frac{24 - \sqrt{3} + 20\sqrt{3}}{6}=\frac{24 + 19\sqrt{3}}{6}$$
matches the answer given by the book within a factor of $\sqrt 3\over \sqrt 3$:
$$\frac{24 + 19\sqrt{3}}{6}=\frac{8\sqrt3+19}{2\sqrt 3}$$
Your method was effective in solving the problem, although putting $\sqrt{12}\sqrt 3$ as the denominator does not follow the "usual" method of coming up with a "least common multiple" for the final denominator; in this case, we have $(12,3)=3$, and therefore we would use $\sqrt {12}=2\sqrt 3$ as the final denominator and avoid a couple extra multiplications in the process. But another intent in solving problems like this is often to "rationalize the denominator", which your method achieves quite effectively. In the end, you'll want to first make sure that you understand the problem and the solution, and then follow the guidelines of the instructor you are working under: if you have a professor or teacher, match what their expectations are in the final answer form; if it's just a book or self-learning, try to match "common" notation and approaches that you see. | 2019-10-21T22:49:30 | {
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https://math.stackexchange.com/questions/513250/conditional-probability-question-with-cards-where-the-other-side-color-is-to-be | # Conditional probability question with cards where the other side color is to be guessed
A box contains three cards. One card is red on both sides, one card is green on both sides, and one card is red on one side and green on the other. One card is selected from the box at random, and the color on one side is observed. If this side is green, what is the probability that the other side of the card is also green?
I think the answer should be $\frac{1}{2}$ as once the card is selected with one side green, there remain only two possibilities for the other side: either red or green.
But the answer to this question is $\frac{2}{3}$.
So,where am I wrong?Why the answer $\frac{2}{3}$ is the $\frac{1}{2}$ wrong?
• There are three green sides. In 2 cases, the other side is green; in one, the other side is red. – Gerry Myerson Oct 3 '13 at 9:44
• Possible duplicate of Probability problem – shoover Jun 25 '18 at 18:56
Go out from 6 sides. In 3 cases a green side shows up, and in 2 of these 3 cases the other side is green as well.
You can work this out directly from the definition of conditional probability:
$$P(G_2\mid G_1) = \frac{P(G_1 \cap G_2)}{P(G_1)}$$
Exactly one of the three cards has two green sides, so $P(G_1 \cap G_2) = 1/3$. Exactly three of the six sides that could be seen initially are green, so $P(G_1)=1/2$. Thus
\begin{align*} P(G_2\mid G_1) &= \frac{1/3}{1/2} \\ &= \frac{2}{3}. \end{align*}
When you see a green side then it's more likely that you have the double-green card, since this has twice as many green sides as the red-green card.
• I like this as a nice, intuitive way of coming to the correct answer. – mattdm Jan 26 '15 at 14:00
The two-sided green card can be observed to have a green side in two distinct ways.
Let's suppose the three cards are C1(with both sides green), C2(with one side green and another red) and C3(both sides are red). Each has two sides (mark A and B)
Now, condition is that the card selected has one side green. So, probability of C3 selection is 0.
Now, Collect the total cases with one side green:
1. C1-A
2. C1-B
3. C2-A (suppose, 'A' as green and 'B' as red)
And, favourable cases (other side should be green too) are:
1. C1-A
2. C1-B
so, probability is: 2 / 3
Say the cards are labelled as $C_1,C_2,C_3$ with $C_1$ being red on both sides, $C_2$ being green on both sides and $C_3$ being green on one side and red on the other. Also, let $S$ be the color of the side that was picked. Then, $$P(C_2|S=G) = \frac{P(S=G|C_2)P(C_2)}{P(S=G)} = \frac{1\times\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$ | 2019-10-23T23:53:57 | {
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http://mathhelpforum.com/statistics/17355-newbie-help-please.html | # Math Help - Newbie help please
1. ## Newbie help please
I have used some basic inferential statistics at work but have not really done much on probability before. (School was a long time ago!) I was wondering if someone could help me with the following:
I have 2 normal decks of cards. I pull out 6 cards at random from the first deck and then pull out 6 cards at random from the second deck.
I wanted to know two things.
1) What the probability is of one of the six cards drawn from deck B being a match with one of the six cards drawn from deck A.
2) What the probability is of two of the six cards drawn from deck B being a match with two of the cards drawn from deck A
I calculated the first problem by working out the probability of not having a match and then subtracting it from 1. i.e for the first card drawn from deck B there is 46 in 52 chance that the card wont match, for the second card drawn there will be a 45 in 51 chance (as there is now one less card in the deck.) and so on.
Eg 46/52 x 45/51 x 44/50 x 43/49 x 42/48 x 41/47 = .46 therefore 1-.46 = .54. Hence the probability of having one card match is 54%
Although I think this is correct, I can’t quite figure out how to go about working out the second problem.
Can anyone give me a point in the right direction please.
Cheers
2. Originally Posted by emersong
1) What the probability is of one of the six cards drawn from deck B being a match with one of the six cards drawn from deck A.
I understand this question as:
Given 6 cards, what is probability there is at least one pair.
He have the following cases,
Code:
MMXXXX
MXMXXX
...
Where "M" stands for match and "X" stands for any. There are 14 cases if you write out the list. In each of these cases the probability is:
(52/52) ---> The first card being any card.
(3/51) ---> The next card got to match.
(50/50) ---> The next can be anything.
(49/49) ---> The next can be anything.
(48/48) ---> The next can be anything.
(47/47) ---> The next can be anything.
Multiply then out to get,
3/51 = 1/17
Now multiply this by 12 because that is the number of disjoint cases:
(12/17)
3. Originally Posted by galactus
Oops, I see PH got something different.
It does not mean you are wrong. Perhaps we understand the problem differently, or perhaps I did it wrong.
My understanding is this.
6 cards are dealt. What is the probability that 2 of those cards match (have same number).
4. There are C(52,6) possible 6-card hands from deck A. Pick any one of those.
There are C(46,6) possible 6-card hands from deck B that have no card in common with the hand we picked from deck A.
Therefore the probability of at least one match is: $1 - \frac{C(46,6)}{C(52,6)} = 0.5399$
That agrees with emersong.
5. Originally Posted by Plato
Therefore the probability of at least one match is: $1 - \frac{C(46,6)}{C(52,6)} = 0.5399$
That agrees with emersong.
I think that is why my answer does not agree with yours. If you follow what I did, I am sure it was perfectly right. I partitioned the cases into disjoint sets and summed each one up. But still that does not do the "at least" part.
6. Hi guys and thanks for the help. Unfortunately I am still confused.
Sorry for not explaining it too well.
The reason for two decks is that I was treating each card as unique, not just looking to see if it was the same number. So...
(Where H = Hearts, C = Clubs, S = Spades, D = Diamonds)
Deck 1... 1H,2D,3S,4C,5S,6D
Deck 2... 8H,9S,1S,3S,7S,4H
In the above example the 3 of spades (3S) repeats.
I also wanted to know the probability of getting 2 cards to repeat.
Deck 1... 1H,2D,8H,4C,5S,9S
Deck 2... 7H,9S,1S,8H,7S,4H
In the above example the 8 of hearts and the 9 of spades both repeat.
I did a little program in excel and ran 10,000 trials, For the first problem I got a probability of .545. This is very near to the probability of .54 that I calculated.
Although I am not sure how to work out the second case, I did the simulation and the probability came out as .142.
So I know the answer to the question, I just would like to understand how I got there!
Thanks again for your help.
7. Hello, emersong!
We need some clarification . . .
I have two normal decks of cards.
I pull out 6 cards at random from the first deck
and then pull out 6 cards at random from the second deck.
1) What is the probability that one of the cards drawn from deck B
matches one of the cards from deck A?
2) What is the probability that two of the cards drawn from deck B
matches two of the cards from deck A?
In part (1), if you mean exactly one match, that's a different game entirely.
Six cards are drawn from deck A.
. . They can be any six cards.
There are: . $C(52,6) = 20,358,520$ possible six-card hands.
1) Among the six cards from deck B, there must be exactly one match
. . . (and five non-matches).
There are: . $C(6,1)$ ways to get one match.
There are: . $C(46,5)$ ways to get five non-matches.
Hence, there are: . $C(6,1)\cdot C(46,5) \:=\:(6)(1,370,754) \:=\:8,224,425$ ways.
Therefore: . $P(\text{exactly 1 match}) \;=\;\frac{8,224,524}{20,358,520} \:=\:0.403984376 \:\approx\:40.4\%$
2) Among the six cards from deck B, there must be exactly two matches
. . . (and four non-matches).
There are: . $C(6,2)$ ways to have two matches.
There are: . $C(46,4)$ ways to have four non-matches.
Hence, there are: . $C(6,2)\cdot C(46,4) \:=\:(15)(163,185) \:=\:2,447,775$ ways.
Therefore: . $P(\text{exactly 2 matches}) \:=\:\frac{2,447,775}{20.358,520} \:=\:0.120233445 \:\approx\:12.0\%$
8. Do you understand what we said about “at least one” matching pair?
The 54% accounts for anywhere from one to six matching pairs.
In you first example there is exactly one matching pair. The probability of that happening is $\frac {6C(46,5)} {C(52,6)} \approx 40.3\%$
For exactly two matching pairs, the probability of that happening is $\frac {C(6,2)C(46,4)} {C(52,6)} \approx 12\%$
9. Firstly, thanks to everyone for their help.
I think I get it, just to recap...
For exactly 3 matches it would be:
C(6,3)xC(46,3)= 20 x 15,180 = 30,360 ways
Hence:
P(exactly 3 matches) = 30,360 / 20,358,520 = 0.0149
And P(one or more matches) = 0.54
Just out of interest, what would;
P(two or more matches) be?
10. The probability of two or more matches is $\sum\limits_{k = 2}^6 {\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$
The probability of N or more matches is $\sum\limits_{k = N}^6 {\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$
11. Originally Posted by Plato
The probability of two or more matches is $\sum\limits_{k = 2}^6 {\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$
The probability of N or more matches is $\sum\limits_{k = N}^6 {\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}}}.$
It's sunk in now. (Doh!) Well it's been a long day!
Thanks
EmersonG
12. Originally Posted by emersong
The probability of two or more matches is .262 right?
No, I get 0.136.
$\frac{{C(6,k)C(46,6 - k)}}{{C(52,6)}} = \frac{{\left( {\frac{{6!}}{{k!\left( {6 - k} \right)!}}} \right)\left( {\frac{{46!}}{{\left[ {\left( {6 - k} \right)!} \right]\left[ {\left( {k + 40} \right)!} \right]}}} \right)}}{{\frac{{52!}}{{6!(46!)}}}}$ | 2015-04-28T07:18:29 | {
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https://math.stackexchange.com/questions/461547/whats-the-equation-of-helix-surface | what's the equation of helix surface?
I know for the helix, the equation can be written: $$x=R\cos(t)$$ $$y=R\sin(t)$$ $$z=ht$$ this is the helix curve, and there are two parameters: outer radius $R$ and the pitch length $2\pi h$. However, I would like to generate the 3D helix with another minor radius $r$. This is not the helix curve, but a 3D object something like spring. I don't know exactly the name of such structure, but when I search helix equation, they usually give the equations for helix curve but not for the 3D helix object (spring). Does anyone know the equation of such object? Thank you so much for any help and suggestions.
ps. the shape looks like the following way:
• Do you mean a second helix of radius $r$ curling around the path of the helix with radius $R$? – Neal Aug 7 '13 at 3:05
• You might include a picture of the structure you want in your question. – Neal Aug 7 '13 at 3:10
• I would start trying with the parametrization $$(x(t),y(t),z(t))+r(\cos u) \vec{n}(t)+r(\sin u)\vec{b}(t),$$ where $\vec{n}(t)$ and $\vec{b}(t)$ are the normal and binormal of the helix curve respectively. IIRC those are both continuous, so no problems. Need to check. – Jyrki Lahtonen Aug 7 '13 at 4:20
• @JyrkiLahtonen, yes, that's correct. For $r$ small enough the resulting surface is not just continuous but (real) analytic. It appears he is calling your $r=a.$ – Will Jagy Aug 7 '13 at 4:33
• @Will: I was a bit worried about the possibility of the normal or binormal "switching sides", but the helix has constant curvature and torsion, so can't happen (I think). Typing an answer together with Mathematica graphics... – Jyrki Lahtonen Aug 7 '13 at 4:40
We can use a local orthonormal basis of a parametrized curve to get a surface of the desired type.
A helix running around the $x$-axis has a parametrization like $$\vec{r}(t)=(ht,R\cos t, R\sin t).$$ Its tangent vector can be gotten by differentiating $$\vec{t}=\frac{d\vec{r}(t)}{dt}=(h,-R\sin t,R\cos t).$$ We note that this has constant length $\sqrt{h^2+R^2}$. With a more general curve this is not necessarily the case, and we would normalize this to unit length, and switch to using the natural parameter $s=$ the arc length. This time $ds/dt=\sqrt{h^2+R^2}$, and we can keep using $t$ as long as we remember to normalize.
We get a (local) normal $\vec{n}(t)$ vector by differentiating the (normalized) tangent $$\vec{n}(t)= \frac{\frac{d\vec{t}}{dt}}{\left\Vert\frac{d\vec{t}}{dt}\right\Vert}=(0,-\cos t,-\sin t).$$ As the name suggests this is orthogonal to the tangent vector (in the direction of change of the tangent). The third basis vector is the binormal $$\vec{b}(t)=\frac1{\Vert\vec{t}\Vert}\vec{t}\times\vec{n}=\frac{1}{\sqrt{R^2+h^2}}(R,h\sin t,-h\cos t).$$ This is, of course, orthogonal to both $\vec{t}$ and $\vec{n}$.
The key is that we get the desired surface by drawing (3D-)circles with axis direction determined by the direction of the curve, i.e. the tangent. Equivalently, we draw a circle of radius $a$ in the plane spanned by $\vec{n}$ and $\vec{b}$. Hence we get the entire surface $S$ parametrized as $$S(t,u)=\vec{r}(t)+a\vec{n}(t)\cos u+ a\vec{b}(t)\sin u$$ with $t$ ranging over as many loops as you wish, and $u$ ranging over the interval $[0,2\pi]$.
Here's what Mathematica-output looks like with parameters $h=1$, $R=3$, $a=0.4$:
Here's the effect of the change to $a=1.0$. The lines on the surface correspond to constant values of $t$ and $u$. These are now more clearly defined.
• So amazing answers, so beautiful plots, Thank you so much for your answers and help, it is very helpful. – Hui Zhang Aug 7 '13 at 5:24
• The formulas for the normal and binormal are simpler, if you use the natural parameter of the curve. If you use this same process for another curve, you need to make adjustments for that. – Jyrki Lahtonen Aug 7 '13 at 5:24
• One more questions, for such surface, is it possible to reduce the equations to in the form? $$f(x,y,z)=g(h,R,a)$$ – Hui Zhang Aug 7 '13 at 5:25
• Like eliminate the parameters $t$ and $u$? I'm not sure how easy that would be. Haven't thought about it. I would think it's simpler to use the parametrization for most things: calculating tangent planes, normals and such. – Jyrki Lahtonen Aug 7 '13 at 5:41
• Thank you so much. You answers are very helpful. Actually I have two questions, and this post is in fact the 1st question of generating the surface. My 2nd question is, given a fixed point $(x_0,y_0,z_0)$, how to determine whether this point is inside or outside the object with such surface? That's why I thought whether the equation can reduce to the form $f(x,y,z)=g(h,R,a)$. However, anyway, you have solved the 1st question and it is very helpful. Thank you again. – Hui Zhang Aug 7 '13 at 16:33 | 2019-07-22T04:27:03 | {
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https://math.stackexchange.com/questions/429563/why-does-uf-p-lf-p-epsilon-make-a-good-criterion-for-integrability | # Why does $U(f,P) - L(f,P) < \epsilon$ make a good criterion for integrability?
That is
$f$ is integrable on $[a,b] \iff \forall\epsilon>0, \exists P$ of $[a,b]$ such that
$$U(f,P) - L(f,P) < \epsilon$$
I was thinking that a better definition would be if $U(f,P) = L(f,P)$, but I was corrected that it wouldn't work well for a curve like $f(x) = x$. The geometry just doesn't work.
On the other bound, saying that given any positive number, I can find a partition such that the difference $U(f,P) - L(f,P) < \epsilon$ is bounded doesn't feel like a strong enough condition for integrability. Isn't the goal of analysis is always to make $\epsilon$ as small as possible, and possibly $0$?
Please see my other question on A terminology to analysts for possible relevance. Thank you
• Saying that for all $\varepsilon > 0$, there exists a partition $P$ such that $U(f,P)-L(f,P) < \varepsilon$ is saying exactly what you are looking for. No matter how small we make $\varepsilon$, we will be able to find some partition that makes the upper and lower sums differ by at most that value. – Cameron Williams Jun 26 '13 at 1:10
• But we still an error of epsilon. I was hoping for a more error free definition. I was satisified with $\sup (L) = \inf (U)$ – Hawk Jun 26 '13 at 1:19
• If $U(f,P) \neq L(f,P)$ for all $P$, then $U(f,P) - L(f,P) > 0$.Thus, given $\epsilon < \inf( U(f,P) - L(f,P) )$, there is no $P$ that will satisy $U(f,P)-L(f,p) < \epsilon$. So, the above two definitions mean the same thing. – AnonSubmitter85 Jun 26 '13 at 1:23
• @CameronWilliams, want to post that as an answer? – Hawk Jun 26 '13 at 1:36
• @sidht But you can make the error as small as you want. – Pedro Tamaroff Jun 26 '13 at 1:37
As per my comment above: Saying that for all $\varepsilon>0$, there exists a partition P such that $U(f,P)−L(f,P)<\varepsilon$ is saying exactly what you are looking for. No matter how small we make $\varepsilon$, we will be able to find some partition that makes the upper and lower sums differ by at most that value.
Isn't the goal of analysis is always to make $\epsilon$ as small as possible, and possibly 0?
You seem to have a misunderstanding of what $\epsilon$ stand for here. You are given $\epsilon>0$ and you want to make something smaller than this $\epsilon >0$. Maybe this can clear some of this out.
Why does $U(f,P) - L(f,P) < \epsilon$ make a good criterion for integrability?
Because
THM Let $x,y$ be arbitrary real numbers. If $x<y+\epsilon$ for each $\epsilon >0$, then $x\leq y$.
P By contradiction. Suppose $x>y$. Then $x-y>0$. Take $\epsilon=x-y$. The above gives $x<y+\epsilon=y+x-y=x$ which is impossible. It must be the case $x\leq y$.
Note that since $\sup L\leq \inf U$ is always true, the criterion gives that $\inf U\leq \sup L$ which means $\sup L=\inf U$ and $f$ is integrable.
Now, proving that for each $\epsilon >0$ there exists $P=P_\epsilon$ such that $$U(f,P)-L(f,P)<\epsilon$$ is usually easier than proving $\sup L=\inf U$ directly, in particular when the function is not given explicitly (say, if we want to prove $f$ is integrable when it is continuous) or any other cases where $f$ is in incognito.
This is another extended comment, rather an answer.
I would just like to point out the the following two statements are totally different:
• (1) $\forall \epsilon > 0$, $\exists$ partition $P$ such that $$U(f,P) - L(f,P ) < \epsilon$$
• (2) $\exists$ partition $P$ such that $\forall \epsilon > 0$: $$U(f,P) - L(f,P) < \epsilon.$$
In (1), the partition $P$ depends on $\epsilon$. It might be more accurate to write $P$ as $P_\epsilon$ to remind ourselves of this fact.
In (2), there is a single partition $P$ -- independent of $\epsilon$ -- such that for all $\epsilon > 0$: $U(f,P) - L(f,P) < \epsilon$ is satisfied. In this situation, because the partition $P$ is independent of $\epsilon$, we can conclude that $U(f,P) = L(f,P)$.
Example: Contrast the following two statements. Think about how they are radically different:
• (A) For all $x > 0$, there exists $y > 0$ such that $x < y$.
• (B) There exists $y > 0$ such that for all $x > 0$: $x < y$.
Note that (A) is obviously true (for any $x> 0$, there's always some number that's bigger), while (B) is obviously false (there is no single number $y$ bigger than every real number). | 2020-02-22T02:03:02 | {
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http://mathhelpforum.com/algebra/26280-quadratic-equations-using-formula-method.html | # Thread: Quadratic equations using the formula method
1. ## Quadratic equations using the formula method
Can anyone solve the following quadratic equation using the formula method...
Xsquared - 10x + 3 = 0
Please show working out so it is possible for me to follow and see where i am going wrong
Thanks
2. Originally Posted by duckegg911
Can anyone solve the following quadratic equation using the formula method...
Xsquared - 10x + 3 = 0
Please show working out so it is possible for me to follow and see where i am going wrong
Thanks
Let me try again...
$x^2 - 10x + 3 = 0$
It seems he's allowed to use the quadratic formula.
All quadratic equations are written in the format of $ax^2 + bx + c = 0$
That means that $a$ is the co-efficient of the $x^2$ ; $b$ is the co-efficient of $x$ ; and $c$ is the constant.
According to the quadratic formula:
$x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{2a}$
All you have to do is substitute the corresponding co-efficients into the formula.
3. A standard quadratic equation is in the form
$ax^2 + bx + c = 0$
a, b and c being the coefficients.
The formula for the discriminant ( $\Delta$) is
$\Delta = b^2 - 4ac$
If $\Delta > 0$, there are 2 different roots.
If $\Delta = 0$, there are 2 equivalent roots. ( $x_1=x_2$)
If $\Delta < 0$, there are no real roots.
The roots are,
$x_1 = \frac{-b - \sqrt{\Delta}}{2a}$
$x_2 = \frac{-b + \sqrt{\Delta}}{2a}$
Now plug a, b and c in these formulas.
4. Let's say we write your equation this way.
$ax^2+bx+c=0$
where:
$a=1$
$b=-10$
$c=3$
now we have the coefficients we use in the equation.
$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
Now you just put a,b and c into the equation and that's it.
There are usually 2 results for x.
5. Sorry, was writing the answer while you posted them.
6. Originally Posted by Pinsky
Sorry, was writing the answer while you posted them.
No problem, at least now (s)he has 3 explanations to choose from...
7. Originally Posted by Pinsky
Sorry, was writing the answer while you posted them.
Thanks to you both, ill take a long hard look at these explanations and let you both know how i get on ,
thanks again
8. Originally Posted by janvdl
No problem, at least now (s)he has 3 explanations to choose from...
Not to mention the explanations accompanying the examples s/he should have in their class notes .....
9. Originally Posted by janvdl
Let me try again...
$x^2 - 10x + 3 = 0$
It seems he's allowed to use the quadratic formula.
All quadratic equations are written in the format of $ax^2 + bx + c = 0$
That means that $a$ is the co-efficient of the $x^2$ ; $b$ is the co-efficient of $x$ ; and $c$ is the constant.
According to the quadratic formula:
$x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{2a}$
All you have to do is substitute the corresponding co-efficients into the formula.
Are the answers for -9.6904 and -3.095 ???
10. Originally Posted by duckegg911
Are the answers for -9.6904 and -3.095 ???
I get positive answers, not negative. That's the only problem with your answers.
11. Originally Posted by janvdl
I get positive answers, not negative. That's the only problem with your answers.
This is because - plus a - = a + ???
12. Originally Posted by duckegg911
This is because - plus a - = a + ???
sorry that sound silly....
its just i cant follow why you get possitive answers...
the answers you have are 9.6904 and 3.095 ?
13. After substitution into the formula, your formula should look like this:
$x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$
Does it?
14. Here's the full solution. Ask me about any step you do not understand/
$x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$
$x = \frac{ +10 \pm \sqrt{100 - 12} }{2}$
$x = \frac{ +10 \pm \sqrt{88} }{2}$
BUT $\sqrt{88} = \sqrt{4} \times \sqrt{22}$
AND we know $\sqrt{4} = 2$
$x = \frac{ +10 \pm \sqrt{4} \cdot \sqrt{22} }{2}$
$x = \frac{ +10 \pm 2 \sqrt{22} }{2}$
Divide 10 by 2, and divide $2 \sqrt{22}$ by 2
$x = +5 \pm \sqrt{22}$
15. Originally Posted by janvdl
Here's the full solution. Ask me about any step you do not understand/
$x = \frac{ +10 \pm \sqrt{100 - 4(1)(3)} }{2}$
$x = \frac{ +10 \pm \sqrt{100 - 12} }{2}$
$x = \frac{ +10 \pm \sqrt{88} }{2}$
BUT $\sqrt{88} = \sqrt{4} \times \sqrt{22}$
AND we know $\sqrt{4} = 2$
$x = \frac{ +10 \pm \sqrt{4} \cdot \sqrt{22} }{2}$
$x = \frac{ +10 \pm 2 \sqrt{22} }{2}$
Divide 10 by 2, and divide $2 \sqrt{22}$ by 2
$x = +5 \pm \sqrt{22}$
I have -10 all the way through rather than +10 ???
Page 1 of 2 12 Last | 2016-12-02T20:38:33 | {
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https://math.stackexchange.com/questions/2048650/find-the-last-digit-of-31006 | # Find the last digit of $3^{1006}$
The way I usually do is to observe the last digit of $3^1$, $3^2$,... and find the loop. Then we divide $1006$ by the loop and see what's the remainder. Is it the best way to solve this question? What if the base number is large? Like $33^{1006}$? Though we can break $33$ into $3 \times 11$, the exponent of $11$ is still hard to calculate.
• It's a good way to solve it. The loop is pretty short. But if you know Eulers formula or fermats little formula you don't have to "find" the loop. You can calculate it with certainty. At any rate $3^2 \equiv -1 \mod 10$ so $3^4 \equiv 1 \mod 10$ so the loop is 4 and $1006$ has remainder $2$ so... it's not hard. – fleablood Dec 7 '16 at 21:09
• Are you familiar with modular arithmetic, i.e. congruences, e.g. $\,3^2\equiv -1\pmod{10}\,?\ \$ – Bill Dubuque Dec 7 '16 at 21:19
• I think this falls under this umbrella question explaining how to do this and related problems. Please search the site for similar questions before asking. – Jyrki Lahtonen Dec 7 '16 at 21:39
• See also these posts: Find the last two digits of $7^{81}$ and Find the last two digits of $3^{45}$. And perhaps other posts linked there – Martin Sleziak Dec 8 '16 at 3:09
You have $$3^2=9\equiv -1\pmod{10}.$$
And $1006=503\times 2$, so
$$3^{1006}=(3^2)^{503}\equiv (-1)^{503}\equiv -1\equiv 9\pmod{10}.$$
So the last digit is $9$.
And for something like $11$, you can use the fact that $11\equiv 1\pmod {10}.$
$3^{1006}$ or $33^{1006}$
doesn't really matter
$33\equiv 3\pmod {10}\\ 33^{1006}\equiv 3^{1006}\pmod {10}$
$3^4 = 81$
You might say this as $3^4\equiv 1 \pmod{10}$
The last digit of $3^n$ is the same last digit as $3^{n+4k}$ that is:
$(3^{n+4k}) = (3^n)(3^{4k})\equiv (3^n)(1) \pmod{10}$
$1006 = 251\cdot 4 + 2$
the last digit of $3^{1006}$ is the same as the last digit of $3^2$
You have
\begin{cases}3^1& =3\\ 3^2& =9 \\ 3^3&=27\\3^4&= 81\\3^5&= 243\end{cases} Thus the last digit repeats after $4$ steps. Since $1006=251\cdot 4+2$ it is
$$3^{1006}=(3^4)^{251}\cdot 3^2.$$ The last digit of $(3^4)^{251}$ is $1$ and the last digit of $3^2=9.$ So the answer you are looking for is $9.$
To compute the last digit of $33^{1006}$ you are in the right way. Since $33=3\cdot 11$ and the last digit of $11^n$ is $1$ you have that the last digit of $33^{1006}$ is the same of $3^{1006}.$
HINT:
Find the last digit of $3^{1006\bmod4}$.
Adding a formal solution, just in case anyone will find it useful:
You are looking for $3^{1006}\pmod{10}$.
Since $\gcd(3,10)=1$, by Euler's theorem, $3^{\phi(10)}\equiv1\pmod{10}$.
We know that $\phi(10)=\phi(2\cdot5)=(2-1)\cdot(5-1)=4$.
Therefore $3^{4}\equiv1\pmod{10}$.
Therefore $3^{1006}\equiv3^{4\cdot251+2}\equiv(\color\red{3^4})^{251}\cdot3^2\equiv\color\red{1}^{251}\cdot3^2\equiv1\cdot9\equiv9\pmod{10}$.
The powers of $3$ cycle from $1\to3\to9\to7$, depending upon the exponent's modulus with respect to $4$. Since $1006\equiv 2\pmod{4}$, the last digit of $3^{1006}$ is $9$.
You can still use this tactic for larger bases. Suppose we want the last digit of $33^{1006}$, as you suggest. Since $33=30+3$, the powers of $33$'s last digit are completely determined by the powers of $3$.
Very soon you will learn Euler's theorem:
If the greatest common factor of $a$ and $n$ then $a^{\phi{n}} \equiv 1 \mod n$ where $\phi(n)$ is the number of numbers relatively prime to $n$ that are less than $n$.
As $1,3,7$ and $9$ are relatively prime to $10$, and $\gcd(3,10) = 1$ we know $\phi(10) = 4$ and $3^4 \equiv 1 \mod 10$ so $3^{1006} = 3^{4*251 + 2} \equiv 3^2 \equiv 9 \mod 10$.
As $33 = 3*10 + 3$ $33^n = (30 + 3)^n = 10*something + 3^n$ will have the same last digit. But $\gcd(33,10) = 1$ so $3^4 \equiv 1 \mod 10$. And everything is the same.
What would be harder is the last two digits of $33^{1006}$. $\gcd(33,100) =1$ so $33^{\phi(100)} \equiv 1 \mod 100$. But what is $\phi(100)$?
There is a thereom that $\phi(p) = p-1$ and that $\phi(p^k) = p^{k-1}(p-1)$ and then $\phi(mn) = \phi(m)\phi(n)$ so $\phi(100)=\phi(2^2)\phi(5^2) = 2*1*5*4 = 40$. So there are $40$ numbers less than $100$ relatively prime to $100$.
and $33^{40} \equiv 1 \mod 100$ so $33^{1006 = 40*25 +6} \equiv 33^6 \mod 100 \equiv 30^2 + 2*3*30 + 3^2 \equiv 189 \equiv 89 \mod 100$. The last two digits are $89$.
See:
https://en.wikipedia.org/wiki/Modular_arithmetic
https://en.wikipedia.org/wiki/Euler%27s_theorem
https://en.wikipedia.org/wiki/Euler%27s_totient_function | 2021-05-18T02:22:36 | {
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https://byjus.com/question-answer/100-surnames-were-randomly-picked-up-from-a-local-telephone-directory-and-the-frequency-distribution-35/ | Question
# $$100$$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters$$1-4$$$$4-7$$$$7-10$$$$10-13$$$$13-16$$$$16-19$$Number of surnames$$6$$$$30$$$$40$$$$16$$$$4$$$$4$$Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Solution
## Let us prepare the following table to compute the median :Number of letters Number of surnames (Frequency) Cumulative frequency $$1-4$$$$6$$ $$6$$ $$4-7$$$$30$$ $$36$$ $$7-10$$$$40$$ $$76$$ $$10-13$$$$16$$ $$92$$ $$13-16$$$$4$$ $$96$$ $$16-19$$$$4$$ $$100=n$$ We have, $$n = 100$$$$\Rightarrow \dfrac n2 = 50$$The cumulative frequency just greater than $$\dfrac n2$$ is $$76$$ and the corresponding class is $$7 – 10$$. Thus, $$7 – 10$$ is the median class such that$$\dfrac n2 = 50, l = 7, f = 40, cf = 36$$ and $$h=3$$Substitute these values in the formulaMedian, $$M = l+\left(\dfrac{\dfrac n2 - cf}{f}\right)\times h$$$$M = 7+\left(\dfrac{50-36}{40}\right)\times 3$$$$M = 7+\dfrac{14}{40}\times3 = 7 + 1.05 = 8.05$$Now, calculation of mean: Number of letters Mid-Point $$(x_i)$$Frequency $$(f_i)$$ $$f_ix_i$$ $$1-4$$$$2.5$$$$6$$ $$15$$ $$4-7$$$$5.5$$ $$30$$ $$165$$ $$7-10$$$$8.5$$ $$40$$ $$340$$ $$10-13$$$$11.5$$ $$16$$ $$184$$ $$13-16$$$$14.5$$ $$4$$ $$58$$ $$16-19$$$$17.5$$ $$4$$ $$70$$ Total $$100$$ $$832$$ Therefore, Mean, $$\bar x = \dfrac{\sum f_ix_i}{\sum f_i} = \dfrac{832}{100} = 8.32$$Calculation ofMode:The class $$7 – 10$$ has the maximum frequency therefore, this is the modal class.Here, $$l = 7, h = 3, f_1 = 40, f_0 = 30$$ and $$f_2 = 16$$Now, let us substitute these values in the formulaMode $$= l+ \left(\dfrac{f_1-f_0}{2f_1-f_0-f_2}\right)\times h$$$$= 7+\dfrac{40-30}{80-30-16}\times3$$$$= 7 + \dfrac{10}{34}\times 3 = 7+0.88 = 7.88$$Hence, median $$= 8.05$$, mean $$= 8.32$$ and mode $$= 7.88$$Mathematics
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https://www.math.purdue.edu/pow/discussion/2016/spring/15 | ## Spring 2016, problem 15
A lattice point is defined as a point in the $2$-dimensional plane with integral coordinates. We define the centroid of four points $(x_i,y_i )$, $i = 1, 2, 3, 4$, as the point $\left(\frac{x_1 +x_2 +x_3 +x_4}{4},\frac{y_1 +y_2 +y_3 +y_4 }{4}\right)$. Let $n$ be the largest natural number for which there are $n$ distinct lattice points in the plane such that the centroid of any four of them is not a lattice point. Show that $n = 12$.
2 years ago
To prove that any set of 13 lattice points contains a 4-point subset with a lattice-point centroid, first we prove that any set of at least 5 lattice points contains a 2-point subset with a lattice-point centroid: Apply the pigeonhole principle to the four classes of lattice points (even,odd), (odd,even), (even,even), (odd,odd) to find a class which contains two of the 5 points. (For example, we might find that there are two points each of which have first coordinate even and second coordinate odd.) The sum of these two points is (even,even) so their centroid is a lattice point, as desired.
Next consider an arbitrary set $A$ of 13 lattice points. Apply the above result to find a pair in $A$ with a lattice-point centroid. Apply it again to the remaining 11 points of $A$, then to the remaining 9, and yet twice more, so that we find a total of 5 such pairs in $A$. Finally, apply it one more time, to the set of lattice-point centroids of these 5 pairs. This gives us a pair of pairs, whose 4 points have a lattice-point centroid, as desired.
Now we exhibit a set of 12 lattice points, no 4 of which have a lattice-point centroid: Take $B$ such that three of its points are $(0,0)$ modulo 4 (in each coordinate), three are $(0,1)$, three are $(1,0)$, and three are $(1,1)$. Suppose that $S$ is a 4-point subset of $B$ whose centroid is a lattice point. Then the sum of the points of $S$ is $(0,0)$ modulo 4. Thus all 4 points of $S$ must have the same first coordinate (modulo 4): otherwise we are summing a number of $1$'s which lies strictly between 0 and 4. Similarly all 4 points of $S$ share the same second coordinate (modulo 4). But this calls for four points of $B$ which are the same modulo 4, whereas $B$ was chosen to have no more than three such.
This solution introduces the idea of a centroid for 2 points which I don't think is valid. I guess the terminology for 2 points would be the centre of the line joining them.
philboyd 2 years ago
dbrown uses the expression "centroid of two points" exactly the way you just defined it. If that constitutes a slight misuse of terminology, well... So what. It would be fair to say, that pretty much all posts on this board are guilty in this respect. Note also, that when $C(a,b ,c, ...)$ denotes the centroid of $a, b, c,...$, then:
$C(a,b ,c, d)=C(C(a, b), C( c, d))$
This is used by dbrown implicitly, when he says:
Finally, apply it one more time, to the set of lattice-point centroids of these 5 pairs. This gives us a pair of pairs, whose 4 points have a lattice-point centroid, as desired.
I think dbrown's argument is entirely sound. And very elegant on top!
Nelix 2 years ago
@philboyd
I think the term "centroid" is pretty standard for any number of points, but you're right that I was sloppy not to define my terms. Sorry about that. I was using the term to mean the average of a set of points, in the sense that each coordinate of the centroid is given by the average of that coordinate over all the points in the set. This agrees with the definition given in the problem.
I think the term is standard even for solid shapes, where you use a ratio of integrals instead of a finite average.
dbrown 2 years ago
2 years ago
By iterating dbrown's argument you can show that if $n$ is a power of two, then:
In every set of $4n - 3$ integer points you can find a subset of $n$ points with integer centroid.
Questions:
• Is this true for general $n$?
• Is $4n - 3$ minimal?
I was wondering the same thing.
• I was able to prove the $4n-3$ result for $n = 3$, in an ad hoc way, just considering cases; I don't remember the details now. It follows that the $4n-3$ result holds for any $n$ which is a power of 2 times a power of 3. (I didn't get anywhere for $n = 5$ though.)
• $4n-3$ is minimal for every $n$, using a very similar example to the 12-point example for $n = 4$. Just take $n-1$ points in each of the classes $(0,0)$, $(0,1)$, $(1,0)$, $(1,1)$ modulo $n$, to get a set of $4n-4$ points no $n$ of which have a lattice-point centroid.
dbrown 2 years ago
2 years ago
Why have we had no submissions?
Is it because, like me, no one understands it?
It's fairly obvious that we only need to consider the set of points [0..3, 0..3] since if the problem is satisfied with x1=z, say, then we can replace z with z MOD 4, and so on for all 16 points.
Are we saying that it is possible to select any set of 12 points from the 16 and that all selections of 4 points from the 12 has the property that the centroid is not is a lattice point?
I think not.
I have written a program to test this and it seems that with 8 points it is possible to select 4 where SOME of the selections do not have a centroid which is a lattice point, but certainly not all.
As soon as we select 9 points (or more) it is always possible to find a set of 4 which does have a centroid which is a lattice point.
It is possible to choose 12 of those 16 points such that the selection of any 4 have the property that the centroid is not a lattice point, so long as you don't require that the 12 points are distinct mod 4. Since we are working mod 4, in your setup, it is entirely possible that we have identified points that were originally distinct before working mod 4. In fact it is easy to find a collection of 12 non-distinct points mod 4 that satisfy the requisite property, giving a lower bound of 12 on n.
NickM 2 years ago
Could you list such a set of 12 points please? I have rechecked my code and it still seems that it is possible to select 8 points which satisfy the requirement, but 9 points does not. e.g. for 8 points we can have (0,0) (0,0) (0,0) (0,0) (0,0) (1,1) (1,2) (1,3) where the various (0,0) points would be (4,8) (20, 32) and the like.
philboyd 2 years ago
We couldn't have 4 points with $(0,0)$ mod 4 in such a set since their sum would then have both coefficients divisible mod 4. A collection satisfying the requirement would be something like 3 copies each of points with mod 4 representatives given by $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. You can check that the sum of any choice of 4 of these points will have a centroid which is not equivalent to $(0,0)$ mod 4.
NickM 2 years ago
Whoops !
Yes, program has confirmed result for 12 points exactly as you said.
As expected it is unable to find a selection of 13 which satisfy the same condition.
However, I see that we still don't actually have a proof that 12 is the largest value for n.
philboyd 2 years ago | 2018-07-21T09:56:07 | {
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https://math.stackexchange.com/questions/3502507/you-started-with-one-chip-you-need-to-get-4-chips-to-win-what-is-the-probabili | You started with one chip. You need to get 4 chips to win. What is the probability that you will win?
This is very similar to the question I've just asked, except now the requirement is to gain $$4$$ chips to win (instead of $$3$$)
The game is:
You start with one chip. You flip a fair coin. If it throws heads, you gain one chip. If it throws tails, you lose one chip. If you have zero chips, you lose the game. If you have four chips, you win. What is the probability that you will win this game?
I've tried to use the identical reasoning used to solve the problem with three chips, but seems like in this case, it doesn't work.
So the attempt is:
We will denote $$H$$ as heads and $$T$$ as tails (i.e $$HHH$$ means three heads in a row, $$HT$$ means heads and tails etc)
Let $$p$$ be the probability that you win the game. If you throw $$HHH$$ ($$\frac{1}{8}$$ probability), then you win. If you throw $$HT$$ ($$\frac{1}{4}$$ probability), then your probability of winning is $$p$$ at this stage. If you throw heads $$HHT$$ ($$\frac{1}{8}$$ probability), then your probability of winning $$\frac{1}{2}p$$
Hence the recursion formula is
\begin{align}p & = \frac{1}{8} + \frac{1}{4}p+ \frac{1}{8}\frac{1}{2}p \\ &= \frac{1}{8} + \frac{1}{4}p +\frac{1}{16}p \\ &= \frac{1}{8} + \frac{5}{16}p \end{align}
Solving for $$p$$ gives
$$\frac{11}{16}p = \frac{1}{8} \implies p = \frac{16}{88}$$
Now, to verify the accuracy of the solution above, I've tried to calculate the probability of losing using the same logic, namely:
Let $$p$$ denote the probability of losing. If you throw $$T$$ ($$\frac{1}{2}$$ probability), you lose. If you throw $$H$$ ($$\frac{1}{2}$$ probability), the probaility of losing at this stage is $$\frac{1}{2}p$$. If you throw $$HH$$($$\frac{1}{4}$$ probability), the probability of losing is $$\frac{1}{4}p$$. Setting up the recursion gives
\begin{align}p & = \frac{1}{2} + \frac{1}{4}p+ \frac{1}{8}\frac{1}{2}p \\ &= \frac{1}{2} + \frac{1}{4}p +\frac{1}{16}p \\ &= \frac{1}{2} + \frac{5}{16}p \end{align}
Which implies that
$$\frac{11}{16}p = \frac{1}{2} \implies p = \frac{16}{22} = \frac{64}{88}$$
Which means that probabilities of winning and losing the game do not add up to $$1$$.
So the main question is: Where is the mistake? How to solve it using recursion? (Note that for now, I'm mainly interested in the recursive solution)
And the bonus question: Is there a possibility to generalize? I.e to find the formula that will give us the probability of winning the game, given that we need to gain $$n$$ chips to win?
• Can you share the calculation for losing probability? – Dhanvi Sreenivasan Jan 9 at 6:23
• @DhanviSreenivasan, I updated the post. – Ilya Stokolos Jan 9 at 6:31
• But you don't have a finite number of throws, no? – Dhanvi Sreenivasan Jan 9 at 6:35
• @DhanviSreenivasan True, I don't. – Ilya Stokolos Jan 9 at 6:36
• Sorry, the game stops if you win. Didn't remember that – Dhanvi Sreenivasan Jan 9 at 6:40
This answer only addresses what's wrong with your recursion, since the other answers (both in this question and your earlier question) already gave many different ways to set up the right recursions (or use other methods).
The key mistake is what you highlighted. When you throw $$HHT$$, you now have $$2$$ chips. For the special case of this problem, $$2$$ chips is right in the middle between $$0$$ and $$4$$ chips, so the winning prob is obviously $$\color{red}{\frac12}$$ by symmetry. But you had it as $$\color{red}{\frac12 p}$$ which is wrong. Thus the correct equation is:
$$p = P(HHH) + P(HT) p + P(HHT) \color{red}{\frac12}= \frac18 + \frac14 p + \frac18 \color{red}{\frac12}$$
Let $$p(n)$$ be the probability that you win the game when you have $$n$$ chips in the pocket. Then $$p(0)=0$$, $$\>p(4)=1$$. Having $$1\leq n\leq3$$ chips one makes a further move, and one then has $$p(n)={1\over2}p(n-1)+{1\over2}p(n+1)\qquad(1\leq n\leq3)\ ,$$ so that $$p(n+1)-p(n)=p(n)-p(n-1)\qquad(1\leq n\leq3)\ .$$ These circumstances immediately imply that $$p(1)={1\over4}$$.
Let $$p_0, p_1, \ldots, p_4$$ be the probability of winning if you start with $$0, 1, \ldots, 4$$ chips, respectively. Of course, $$p_0 = 0$$ and $$p_4 = 1$$.
There are a few different ways to approach this question.
Start with $$p_2$$
This seems to be the approach you're asking for, but it's not the easiest approach.
To calculate $$p_2$$, consider what happens if the coin is flipped twice. There is a $$1/4$$ chance of getting $$TT$$ (instant loss), a $$1/4$$ chance of getting $$HH$$ (instant win), and a $$1/2$$ chance of getting either $$HT$$ or $$TH$$ (back to $$p_2$$). So we have
$$p_2 = \frac14 + \frac12 p_2,$$
which we can solve to find that $$p_2 = 1/2$$.
Now that we know $$p_2$$, we can directly calculate $$p_1$$ as the average of $$p_0$$ and $$p_2$$, which is $$1/4$$.
Examine the sequence
Notice that in the sequence $$p_0, p_1, p_2, p_3, p_4$$, each element (besides the first and the last) is the average of its two neighbors. This implies that the sequence is an arithmetic progression. Given that $$p_0 = 0$$ and $$p_4 = 1$$, we can use any "find a line given two points" method to find that for all $$n$$, $$p_n = n/4$$.
Conservation of expected value
I'm an investor and an advantage gambler (which are the same thing, really), so I like to think of things in terms of expected value.
I start the game with $$1$$ chip, and it's a perfectly fair game; in the long run, I am expected neither to lose nor to win. So if I play the game until it ends, the expected value of the game must be $$1$$ chip. (More detail is needed to make this argument formal, but it's sound.)
The value if I lose is $$0$$, and the value if I win is $$4$$, so the expected value can also be written as $$(1 - p_1) \cdot 0 + p_1 \cdot 4$$, which simplifies to $$4 p_1$$.
These two ways of calculating the expected value must agree, meaning that $$4 p_1 = 1$$, so $$p_1 = 1/4$$.
In general
The latter two of the above arguments can each be generalized to show that if you start with $$a$$ chips, and the game ends when you reach either $$0$$ or $$b$$ chips, then the probability of winning is $$a/b$$. | 2020-09-28T23:16:47 | {
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https://puzzling.stackexchange.com/questions/114281/splitting-integers-and-taking-differences-how-can-the-sum-be-constant | # Splitting integers and taking differences, how can the sum be constant?
Start with the set of integers from $$1$$ up to $$2n$$, where $$n$$ is a natural number. Split this set into two disjoint subsets of equal size, say $$\{a_1 and $$\{b_1>b_2>\cdots>b_n\}$$ (with the elements ordered WLOG). What is the value of the following expression in terms of $$n$$? $$|a_1-b_1|+|a_2-b_2|+\cdots+|a_n-b_n|$$
Source (link contains spoilers)
• I don't think ordering the elements is WLOG. Jan 2, 2022 at 14:14
• @justforplaylists It's not WLOG for the final expression, of course, but it is WLOG for defining the sets: i.e. we define $a_1$ to be the smallest element, $a_2$ the next smallest, and so on. Jan 2, 2022 at 14:16
• Two disjoint and exhaustive subsets? If $n=10$, we could have $\{1,2\}$ and $\{19,20\}$, which gives $36$, or $\{1,2\}$ and $\{3,4\}$, which gives $2$. Jan 3, 2022 at 5:40
• I was trying to think of a more precise term than "split", and reading the answers I was reminded of the word "partition", which more clearly indicates exhaustion. Jan 3, 2022 at 6:10
• @Randal'Thor This is not WLOG but rather you are defining $a_i$ and $b_i$ in this way. Jan 3, 2022 at 7:14
## 4 Answers
Jaap Sherpuis's comment hints at a perhaps more intuitive explanation.
We need to consider how many a's are $$\le n$$ and how many b's are $$\gt n$$.
Let's name $$k$$ the number of $$a_i$$'s that are $$\le n$$.
$$a_1 \dots a_k$$ occupy $$k$$ values from 1 to n, leaving $$(n-k)$$ holes.
These holes are filled by $$(n-k)$$ of the $$b_j$$'s. The remaining $$b_j$$ are $$\gt n$$.
So we have $$a_1 \dots a_k \le n$$ and $$a_{k+1} \dots a_n \gt n$$.
And we have $$b_1 \dots b_k \gt n$$ and $$b_{k+1} \dots b_n \le n$$.
This implies $$a_i < b_i$$ if and only if $$i \le k$$
From there we can compute
$$|a_1-b_1|+|a_2-b_2|+\cdots+|a_n-b_n|$$
$$= |a_1-b_1|+\cdots+|a_k-b_k|+|a_{k+1}-b_{k+1}|+\cdots+|a_n-b_n|$$
$$= (b_1-a_1)+\cdots+(b_k-a_k)+(a_{k+1}-b_{k+1})+\cdots+(a_n-b_n)$$
On the plus side you have all values $$\gt n$$ and on the minus side the values $$\le n$$
So the expression becomes:
$$= (n+1) + \cdots + 2n - (1 + \cdots + n) = n^2$$
This is just an "optimised" version of Gareth's answer which I couldn't resist making. Please keep upvoting him and also prefer his answer over mine for acceptance.
For any index $$i$$ let us denote $$M_i$$ the larger of $$a_i,b_i$$ and $$m_i$$ the smaller. Then the given sum of absolute differences can be rewritten as
$$\displaystyle\sum_i M_i - \sum_i m_i$$
We will now prove that for any pair of indices $$i,j$$ we have $$M_i>m_j$$:
Proof:
Case 1, $$i=j$$: obvious from definition
Case 2, $$i: $$M_i\ge b_i>b_j\ge m_j$$
Case 3, $$i>j$$: $$M_i\ge a_i>a_j\ge m_j$$
It follows that all the $$m_i$$ are smaller than all the $$M_i$$. This is the same as saying
the $$m_i$$ must be a permutation of $$1,...,n$$ and the $$M_i$$ a permutation of $$n+1,...,2n$$
The sum is therefore
$$\displaystyle\sum_{i=1}^n(n + i) -\sum_{i=1}^n i = \sum_{i=1}^n n = n^2$$
and indeed independent of how the numbers are split into $$a$$ and $$b$$.
• I don't think this is just an optimized version of my answer, although of course there's some overlap. And I think it's a very nice argument. I've upvoted it and would encourage anyone reading this to do likewise. (And if Rand chooses to accept this one rather than mine, then despite loopy walt's first sentence I think that would be an extremely reasonable decision.) Jan 2, 2022 at 20:14
• There is a related principle used in some magic tricks. If you have one suit of cards in order Ace to King, and another suit in reverse order King to Ace, and combine those two piles using a single riffle shuffle, then the top thirteen cards will contain every card value exactly once, as will the bottom thirteen cards. Jan 3, 2022 at 8:55
So, first of all, let's
do it one way and see what sum we get. Let the $$a$$s be (in order) $$1,2,\ldots,n$$ and the $$b$$s be (in order) $$2n,2n-1,\ldots,n+1$$. Then the differences are, in order, $$2n-1,2n-3,\ldots,1$$ whose sum is $$n^2$$.
In view of
the title, presumably this is always the answer.
But why? I offer first a rather routine, prosaic solution, the sort of thing someone who's done this sort of thing before knows they will be able to get to work. (This is the first solution I found.) Then something slicker. (This is the second solution I found.) Then a way to streamline the second part of the slicker argument. And then a way to streamline the first part, found not by me but by user loopy walt in comments. I prefer to leave all these things in, rather than just presenting the slickest version so far found, because that's a more honest representation of how things typically work in mathematics; see also this related 3blue1brown video about calculation versus slick insight.
Here is a prosaic solution; I suspect there is something better (and am thinking about that). We can turn any valid partition into any other by repeatedly performing operations of the form "find two consecutive numbers one of which is an $$a$$ and the other a $$b$$, and swap them". What happens when we do this? Suppose $$a_i$$ and $$b_j$$ are consecutive. Then none of the other order relations change when we switch them, so our new sequences in order are the same as the old except that $$a_i'=b_j$$ and $$b_j'=a_i$$. If $$i=j$$ then the expression we're interested in doesn't change at all. Otherwise, say that $$i. Then $$b_j \simeq a_i where $$\simeq$$ means "differ by 1", and therefore $$a_j>b_j$$; and $$a_i\simeq b_j, and therefore $$a_i. So increasing $$a_i$$ by 1 and decreasing $$b_j$$ by 1 reduces $$|a_i-b_i|$$ by 1 and increases $$a_j-b_j$$ by 1, and vice versa if instead we decrease $$a_i$$ and increase $$b_j$$. In either case our expression doesn't change.
OK, now let's see if we can do something simpler.
We have all these intervals from $$a_i$$ to $$b_i$$. Every number $$k$$ from $$1$$ to $$2n$$ is an endpoint of exactly one of these intervals; I claim that the first $$n$$ are all left endpoints and the last $$n$$ are all right endpoints. Why? Well, colour all the $$a$$s amber and all the $$b$$s blue; suppose $$k$$ is amber; if $$k=a_i$$ then there are $$i-1$$ amber numbers to its left, hence $$k-i$$ blue numbers to its left, hence $$n-k+i$$ blue numbers to its right, the first of which is $$b_{n-k+i}$$. So $$b_i$$ is right of $$a_i$$ iff $$i\leq n-k+i$$, iff $$k\leq n$$. Likewise if $$k$$ is blue.
And now
our sum is just the sum over all $$k$$ of the number of these intervals it's in (counting 1/2 when it's exactly at an endpoint), which depends only on the number of left and right endpoints on either side of $$k$$, and we've just seen that this never changes.
Or (a basically-equivalent replacement for the foregoing paragraph):
our sum is just the sum of all the right endpoints minus the sum of all the left endpoints, which is to say $$\bigl[(n+1)+\cdots+2n\bigr]-\bigl[1+\cdots+n\bigr]=n^2$$.
This is definitely slicker than the first proof above, but I would be unsurprised to find that one can streamline things further.
In comments, loopy walt suggests another way to do the first bit:
i.e., proving that $$1,...,n$$ are left endpoints and $$n+1,...,2n$$ are right endpoints. Consider the interval whose endpoints are $$a_i,b_i$$. We'll prove that its right-hand endpoint is $$>n$$. Note that $$a_i$$ is bigger than $$i-1$$ smaller $$a$$s, and $$b_i$$ is bigger than $$n-i$$ smaller $$b$$s. So whichever of them is larger is bigger than the other one (1), and $$i-1$$ smaller $$a$$s, and $$n-i$$ smaller $$b$$s, all of those sets being disjoint; hence bigger than at least $$1+i-1+n-i=n$$ other numbers; hence it's $$>n$$.
(To my mind this is neater than my argument but more rabbit-out-of-hat-ish.)
• Correct answer and reasonable proof, but there's a(nother) surprising fact that makes the proof even simpler. Jan 2, 2022 at 14:27
• As mentioned in the answer, I am currently looking for slicker proofs and have some plausible ideas of where to look. There is something of an analogy with a famous question about coins, but that pathway doesn't seem like it really produces something much simpler so far. Jan 2, 2022 at 14:29
• How about: ai is larger than i-1 elements (a1,...,a{i-1}), bi is larger than n-i elements (b{i+1},...,bn). The larger of ai and bi is larger than the other one and both these sets, that is 1 + i-1 + n-i elements. Jan 2, 2022 at 15:30
• ... and possibly other things too, but at least n elements, which means that all the right endpoints are >= n+1, and as there are n right endpoints that tells us what they all are? Yeah, that works. Jan 2, 2022 at 15:34
• The last spoilertag is the kind of argument I was thinking of. Nice example of a surprising non-obvious mathematical fact. Jan 2, 2022 at 17:48
This problem doesn't really have anything to do with the integers from 1 to 2n. Fix any set consisting of 2n real numbers (repetitions allowed); then split them into a_1 <= a_2 <= .... <= a_n and b_1 >= b_2 >= ... >= b_n. Then the sum |a_1 - b_1| + |a_2 - b_2| + ... + |a_n - b_n| is constant, that is, it doesn't depend on the partition.
• Nice generalisation, and I suppose the same proof works in this case too, but can you edit your answer to include a quick proof of this fact? Jan 3, 2022 at 4:48
• @Randal'Thor Since the loopy walt's answer (up to the words "It follows that all the $m_i$ are smaller than all the $M_i$") does not use the fact that the numbers given are integers from 1 to 2n, if we replace $1\leqslant2\leqslant\dots\leqslant2n$ with any numbers $q_1\leqslant q_2\leqslant\dots\leqslant q_{2n}$, the assertion about the constant sum still holds (and it will be equal $\sum_{i=1}^n q_{n+i} - \sum_{i=1}^n q_i$). Jan 3, 2022 at 6:47
• @trolley813 I know how the proof would go, I'm just asking pmw to include it so that this answer is self-contained. Jan 3, 2022 at 8:23 | 2023-03-21T16:32:30 | {
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https://math.stackexchange.com/questions/3457074/how-many-seating-arrangements-of-m-people-in-a-row-of-n-seats-are-there-if | # How many seating arrangements of $m$ people in a row of $n$ seats are there if $k$ of the people must sit together?
Suppose there are five people $$A, B, C, D, E$$ to be seated in a row of eight seats $$S_1, \ldots, S_8$$.
(1) How many possibilities are there if $$A$$ and $$B$$ are to sit next to each other?
(2) How many possibilities are there if $$A$$ and $$B$$ are not to sit next to each other?
My Attempt:
First Approach:
There are $$8$$ options of a seat for $$A$$. However, if $$A$$ is to be seated in $$S_1$$ or $$S_8$$, then $$B$$ can only be seated in $$S_2$$ or $$S_7$$, respectively. On the other hand, if $$A$$ is to be seated in any one of $$S_2$$ through $$S_7$$, then $$B$$ has two options of seat in each case.
Once $$A$$ and $$B$$ have been seated, there are $$6$$ seats left for $$C$$, and then $$5$$ seats left for $$D$$, and finally $$4$$ seats left for $$E$$.
In this way, the total number of seating arrangements are $$2 \times 1 \times 6 \times 5 \times 4 + 6 \times 2 \times 6 \times 5 \times 4 = 240 + 1440 = 1680.$$
Is this solution correct?
Second Approach:
Now let us treat $$A, B$$ as one entity. Then we have four entities to be accommodated in seven spots, for which there are $$7 \times 6 \times 5 \times 4 = 840$$ ways. And, each one of these $$840$$ arrangements, the members $$A$$ and $$B$$ of the block $$A, B$$ can be arranged within the block in two distinct ways. Therefore there are $$840 \times 2 = 1680$$ possible seating arrangements.
Is my answer correct? If so, then are both of my approaches also correct? If not, then where are the problems.
More generally, I have the following question:
Let $$k, m, n$$ be any natural numbers such that $$k \leq m \leq n$$. Then how many ways are there in total of seating $$m$$ people $$P_1 \ldots, P_m$$ in $$n$$ seats $$S_1, \ldots, S_n$$ such that some $$k$$ of these people insist on sitting next to each other?
My Attempt Using the Second Approach:
Let us treat that $$k$$ people as one block. Then there are
\begin{align} & (n-k+1) \underbrace{(n-k) (n-k-1) \ldots }_{(m-k) \mbox{ factors}} \\ &= (n-k+1) \big[ \ (n-k) (n-k-1) \ldots \big( (n-k) - (m-k-1) \big) \ \big] \\ &= (n-k+1)(n-k) \ldots (n-m+1) \\ &= ^{n-k+1}P_{n-m} . \end{align}
Finally, corresponding to each of the above arrangements with the $$k$$ people considered as one block, there are $$k!$$ ways of arranging the $$k$$ people within the block amongst themselves.
Hence there are a total of $$k! \ ^{n-k+1}P_{n-m}$$ ways of seating $$m$$ people in $$n$$ seats with $$k$$ people seated next to each other.
• The problem is not clearly stated until it is said which seats are next to each other. Among many possibilties two seem plausible: seats in a single row, or in a circle (in which case $S_1$ would be next to $S_8$). Nov 30, 2019 at 17:23
Both of your solutions to the first problem are correct. However, your general formula is not. Notice that in the first problem, $$n = 8$$, $$m = 5$$, and $$k = 2$$, so your formula gives $$2!P(8 - 2 + 1, 8 - 5) = 2!P(7, 3) = 2! \cdot 7 \cdot 6 \cdot 5 = 2 \cdot 210 = 420$$ Let's see what went wrong.
We wish to seat $$m$$ people, $$k$$ of whom must sit consecutively, in $$n$$ seats. Since the block takes up $$k$$ of the $$n$$ places, it must begin in one of the first $$n - (k - 1) = n - k + 1$$ positions. Once the block has been placed, there are $$n - k$$ seats left for the remaining $$m - k$$ people. They can be arranged in those seats in $$P(n - k, m - k)$$ ways. The people within the block can be arranged in $$k!$$ ways, which gives us the formula $$(n - k + 1)P(n - k, m - k)k!$$ As a sanity check, let's try our formula when $$n = 8$$, $$k = 2$$, and $$m = 5$$. It gives $$(8 - 2 + 1)P(8 - 2, 5 - 2)2! = 7 \cdot P(6, 3) \cdot 2! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 2 = 1680$$ which agrees with the answer you obtained in your example.
Observe that \begin{align*} (n - k + 1)P(n - k, m - k) & = (n - k + 1) \cdot \frac{(n - k)!}{[(n - k) - (m - k)]!}\\ & = \frac{(n - k + 1)(n - k)!}{(n - m)!}\\ & = \frac{(n - k + 1)!}{(n - m)!}\\ & = \frac{(n - k + 1)!}{[(n - k + 1) - (m - k + 1)]!}\\ & = P(n - k + 1, m - k + 1) \end{align*} so we could write our formula in the form $$P(n - k + 1, m - k + 1)k!$$ As a sanity check, note that if $$n = 8$$, $$k = 2$$, and $$m = 5$$, then $$P(8 - 2 + 1, 5 - 2 + 1)2! = P(7, 4)2! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 2 = 1680$$
• thank you so much for such a beautiful answer! Nov 30, 2019 at 18:06
For part $$a.)$$, we can work on two cases: $$AB$$ ($$A$$ to the left of $$B$$) and $$BA$$. The possibilities for $$AB$$ are $$S_{1}S_{2}, S_{2}S_{3}, …, S_{7}S_{8}$$. The possibilities for $$BA$$ are also the same. So the answer is
$$7 \times \binom{6}{3} \times 3! + 7 \times \binom{6}{3} \times 3! = 1680$$
The $$\binom{6}{3}$$ is the number of ways we can choose 3 seats, and $$3!$$ is number of possible orders of $$C,D,E$$ in the 3 seats.
For part $$b.)$$ count all possible seating without any rule, and then subtract with $$1680$$. | 2022-05-26T04:14:12 | {
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http://mathhelpforum.com/discrete-math/175881-counting-permutations.html | # Math Help - Counting-Permutations
1. ## Counting-Permutations
I have a question asking me to find:
The sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions
I realise that the digits can be chosen in (4)_4=4!=24 ways since repetition is not permitted and order does matter
The final answer is 133320
Any help would be great
2. Let the 24 numbers be $a_1,\dots,a_{24}$. Further, let the $i$th number $a_i$ be $1000b_{i1}+100b_{i2}+10b_{i3}+b_{i4}$ where all $b_{ij}$ are in {2,4,6,8}. Let's find $S_1=\sum_{i=1}^{24}b_{i1}$. Each of the four numbers is encountered 6 times, so the $S_1 = 6(2 + 4 + 6 + 8) = 120$. Therefore, the first digits of all numbers contribute 120 * 1000 to the whole sum. Similarly, the second digits contribute 120 * 100 and so on.
3. Originally Posted by qwerty10
I have a question asking me to find:
The sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions
The final answer is 133320
Think about it, each of those four numbers will be in each decimal place six times. The ones column adds up to $6(2+4+6+8)=120.$
Check out $\displaystyle120\left( {\sum\limits_{k = 0}^3 {10^k } } \right) = ~?$.
4. Hello, qwerty10!
Find the sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions.
The answer is: 133,320.
List the $4! = 24$ permutations of the four digits
,. . and consider their sum.
. . $\begin{array}{cc}
&2468 \\ &2486 \\ &2648 \\ &2684 \\ &2846 \\ &2864 \\ &\vdots \\
&8246 \\ &8264 \\ &8426 \\ &8462 \\ &8624 \\ + & 8642 \\ \hline
\end{array}$
We find that each column has: six 2's, six 4's, six 6's, six 8's.
The total of each column is: . $6\!\cdot\!2 + 6\!\cdot\!4 + 6\!\cdot\!6 + 6\!\cdot\!8 \:=\:120$
Hence, the addition has the form:
. . $\begin{array}{cccccc}
&&& 1 & 2 & 0 \\ && 1 & 2 & 0 \\ & 1 & 2 & 0 \\ 1 & 2 & 0 \\ \hline 1 & 3 & 3 & 3 & 2 & 0 \end{array}$
Therefore, the sum is: . $133,\!320$
Edit: Plato beat me to it . . . *sigh*
. | 2015-08-04T05:12:09 | {
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https://math.stackexchange.com/questions/2700220/radioactive-isotope-differential-equation-question/2700227 | The question is as follows:
I have the general solution of a differential equation as $P(t) = Ae^{kt}$ where $k < 0$ is a constant, and it describes the mass $P(t)$ of a radioactive isotope at future time $t$. I also have that the mass of the isotope is $10$ at time $0$ and it is $5$ at time $2$. I want to find the remainder of the isotope at time $3$.
So I can use that $P(0) = 10$ to obtain $A$ = $10$, and here now is where I get a bit confused. If I use $P(2) = 5$, I get $5 = 10e^{2k}$ and solving for $k$ I get $k = \frac{ln(0.5)}{2} = -0.34657$. And then substitute it all back in to get $P(3)$
However, when looking at other questions online, they talk a lot about half-life of the substance and we have that if we let $\tau$ denote the half-life of the substance, then $k = \frac{ln(2)}{\tau}$ (where they are using the general solution to be $Ae^{-kt}$ However, isn't $\tau$ in my case equal to $2$? So should I not be getting $k = \frac{ln(2)}{2}$ rather than $k = \frac{ln(0.5)}{2}$? Can someone please explain where I am going wrong?
Thank you
• You haven't don't anything wrong because $-\log{(0.5)}=\log{(2)}$. Which explains the negative sign, your solution is the same as theirs Mar 20 '18 at 12:51
• Ah ok I think I get where I am getting confused. So what I have written is fine, I just solved for $k$ where $k$ is negative whereas the one online solved for $k$ where $k$ is positive and when putting back into their formula they will take this into account and multiply it by negative $1$? Mar 20 '18 at 12:55
$Ae^{-kt}$ with $k=\frac{ln(2)}{\tau}$ is equivalent to $Ae^{kt}$ with $k=\frac{ln(0.5)}{\tau}$
• Thanks very much. One last thing, so using that then $P(3)$ is approx equal to $3.5$ yes? Mar 20 '18 at 12:58
• Yes, it makes sense as you can double check with half-life equal to two, which means that $P(4)=2.5$, and you know the exponential decay function is decreasing. Mar 20 '18 at 13:00 | 2022-01-29T04:54:32 | {
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https://www.otexts.org/node/912 | # 8.1.2 Insertion Sort
The list-sort-best-first procedure seems quite inefficient. For every output element, we are searching the whole remaining list to find the best element, but do nothing of value with all the comparisons that were done to find the best element.
An alternate approach is to build up a sorted list as we go through the elements. Insertion sort works by putting the first element in the list in the right place in the list that results from sorting the rest of the elements.
First, we define the list-insert-one procedure that takes three inputs: a comparison procedure, an element, and a List. The input List must be sorted according to the comparison function. As output, list-insert-one produces a List consisting of the elements of the input List, with the input element inserts in the right place according to the comparison function.
(define (list-insert-one cf el p) ; requires: p is sorted by cf
(if (null? p) (list el)
(if (cf el (car p)) (cons el p)
(cons (car p) (list-insert-one cf el (cdr p))))))
The running time for list-insert-one is in $\Theta(n)$ where $n$ is the number of elements in the input list. In the worst case, the input element belongs at the end of the list and it makes $n$ recursive applications of list-insert-one. Each application involves constant work so the overall running time of list-insert-one is in $\Theta(n)$.
To sort the whole list, we insert each element into the list that results from sorting the rest of the elements:
(define (list-sort-insert cf p)
(if (null? p) null
(list-insert-one cf (car p) (list-sort-insert cf (cdr p)))))
Evaluating an application of list-sort-insert on a list of length $n$ involves $n$ recursive applications. The lengths of the input lists in the recursive applications are $n-1$, $n-2$, $\ldots$, 0. Each application involves an application of list-insert-one which has linear running time. The average length of the input list over all the applications is approximately $\frac{n}{2}$, so the average running time of the list-insert-one applications is in $\Theta(n)$. There are $n$ applications of \scheme|list-insert-one|, so the total running time is in $\Theta(n^2)$.
Exercise 8.5. We analyzed the worst case running time of list-sort-insert above. Analyze the best case running time. Your analysis should identify the inputs for which list-sort-insert runs fastest, and describe the asymptotic running time for the best case input.
Exercise 8.6. Both the list-sort-best-first-sort and list-sort-insert procedures have asymptotic running times in $\Theta(n^2)$. This tells us how their worst case running times grow with the size of the input, but isn't enough to know which procedure is faster for a particular input. For the questions below, use both analytical and empirical analysis to provide a convincing answer.
a. How do the actual running times of list-sort-best-first-sort and list-sort-insert on typical inputs compare?
b. Are there any inputs for which list-sort-best-first is faster than list-sort-insert?
c. For sorting a long list of $n$ random elements, how long does each procedure take? (See Exercise 8.2 for how to create a list of random elements.) | 2018-02-21T05:20:57 | {
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https://math.stackexchange.com/questions/2874991/some-interesting-observations-on-a-sum-of-reciprocals/2875035 | # Some interesting observations on a sum of reciprocals
This recent question is the motivation for this post.
Consider the following equation $$\frac1{x-1}+\frac1{x-2}+\cdots+\frac1{x-k}=\frac1{x-k-1}$$ where $$k>1$$.
My claims:
1. There are $$k$$ solutions, all of which are real.
2. Let $$x_{\min}$$ be the minimum value of these $$k$$ solutions. Then as $$k\to\infty$$, $$x_{\min}$$ converges. (If it does, to what value does it converge?)
3. As $$k\to\infty$$, all of the solutions get closer and closer to an integer, which is bounded below. Furthermore, these integers will be $$1, 2, 3, \cdots, k-1, k+1$$.
To see these patterns, I provide the solutions of $$x$$ below. I used W|A for $$k\ge4$$. The values in $$\color{blue}{\text{blue}}$$ are those of $$x_{\min}$$.
$$\begin{array}{c|c}k&2&3&4&5&6\\\hline x&4.414&4.879&5.691&6.592&7.530\\&\color{blue}{1.585}&2.652&3.686&4.701&5.722\\&&\color{blue}{1.468}&2.545&3.588&4.615\\&&&\color{blue}{1.411}&2.487&3.531\\&&&&\color{blue}{1.376}&2.449\\&&&&&\color{blue}{1.352}\end{array}$$
Also, when $$k=2$$, the polynomial in question is $$x^2-6x+7$$, and when $$k=3$$, it is $$x^3-9x^2+24x-19$$.
The reason why I think $$x_{\min}$$ converges is because the difference between the current one and the previous gets smaller and smaller as $$k$$ increases.
Are my claims true?
• It is easy to prove that it has at least $k-1$ distinct real soutions Aug 7, 2018 at 14:22
• and also it is true that $1<x_{min}<2$ for every $k$ and it converges to 1 Aug 7, 2018 at 14:24
• @Exodd: if you do that you know it has $k$ distinct solutions because of the limits of each side as $x \to \pm \infty$ Aug 7, 2018 at 14:27
• The method I used (removing the asymptotes) has been the basis of significant improvements in chemical engineering calculations (in particular for the so-called Rachford-Rice and Underwood equations). I have published quite a lor of papers for these. I had a lot of fun with your (may I confess that I cannot resist an equation ?). Cheers. Aug 8, 2018 at 8:05
• I updated my answer for some improvements. Cheers and thanks for the problem. Aug 10, 2018 at 8:05
## 3 Answers
Both your claims are true.
if you call $$f(x) = \frac1{x-1}+\frac1{x-2}+\cdots+\frac1{x-k}-\frac1{x-k-1}$$ then $f(1^+) = +\infty$, $f(2^-) = -\infty$ and $f$ is continuous in $(1,2)$, so it has a root in $(1,2)$. The same you can say about $(2,3)$, $(3,4), \cdots, (k-1,k)$, so there are at least $k-1$ real distinct roots. $f$ is also equivalent to a $k$-degree polynomial with the same root, but a $k$-degree polynomial with $k-1$ real roots has in reality $k$ real roots.
The last root lies in $(k+1,+\infty)$, since $f(k+1^+) = -\infty$ and $f(+\infty) = +\infty$.
The least root $x_{\min}$ must lie in $(1,2)$, since $f(x)<0$ for every $x<1$. Moreover, $$f(x) = 0\implies x = 1 + \frac{1}{\frac1{x-k-1}-\frac1{x-2}-\cdots-\frac1{x-k}}$$ and knowing $1<x<2$, we infer $\frac1{x-k-1}>\frac1{x-2}$ and $$1<x = 1 + \frac{1}{\frac1{x-k-1}-\frac1{x-2}-\cdots-\frac1{x-k}} < 1 - \frac{1}{\frac1{x-3}+\cdots+\frac1{x-k}}\to 1$$ so $x_{\min}$ converges to $1$
About the third claim, notice that you may repeat the same argument for any root except the biggest. Let us say that $x_r$ is the $r-th$ root, with $r<k$, and we know that $r<x_r<r+1$. $$f(x_r) = 0\implies x_r = r + \frac{1}{\frac1{x_r-k-1}-\frac1{x_r-1}-\cdots-\frac1{x_r-k}}$$ but $\frac1{x_r-k-1}>\frac1{x_r-1}$ holds, so $$r<x_r = r + \frac{1}{\frac1{x_r-k-1}-\frac1{x_r-1}-\cdots-\frac1{x_r-k}} < r - \frac{1}{\frac1{x_r-2}+\cdots+\frac1{x_r-k}}\to r$$ so $x_r$ converges to $r$.
For the biggest root, we know $k+1<x_k$ and $$f(x_k) = 0\implies k+1 < x_k = k+1 + \frac{1}{\frac1{x_k-1}+\cdots+\frac1{x_k-k}} \to k+1$$
• I would've argued that the LHS is continuous, monotone, and ranges from $-\infty$ to $+\infty$ between natural numbers, while the RHS is continuous and monotone, and since the LHS is asymptotically larger than the RHS as $x\to\infty$, there must be a solution $x>k+1$, where the LHS is less than the RHS. In any case, sound argument. Aug 7, 2018 at 14:52
• I think they're quite equivalent. Now I'm wondering if $x_{max}$ converges to $k+1$ Aug 7, 2018 at 14:57
• @Exodd Nice proofs. The third claim of converging to integers extends the claim $x_{\min}$ to all solutions of $x$. I strongly believe that it's true after plotting in Desmos, but I can't construct a rigorous proof for it. Aug 7, 2018 at 14:59
• Hard to exactly say what that's supposed to mean, $k+1$ is a moving target :P Aug 7, 2018 at 15:00
• Let's say $x_{max}(k)-k-1 = o_k(1)$ Aug 7, 2018 at 15:00
For the base case, $$\tag1f_2(x)=\frac1{x-1}+\frac1{x-2}-\frac1{x-3},$$ one readily verifies that there is a root in $(1,2)$ and a root $x^*$ in $(3,+\infty)$.
If we multiply out the denominators of $$f_k(x)=\frac1{x-1}+\frac1{x-2}+\ldots+\frac1{x-k}-\frac1{x-k-1},$$ we obtain the equation $$\tag2(x-1)(x-2)\cdots(x-k-1)f_k(x)=0,$$ which is a polynomial of degree (at most) $k$, so we expect $k$ solutions, but some of these may be complex or repeated or happen to be among $\{1,2,\ldots, k+1\}$ and thus not allowed for the original equation. But $f_k(x)$ has simple poles with jumps from $-\infty$ to $+\infty$ at $1,2,3,\ldots, k$, and a simple pole with jump from $+\infty$ to $-\infty$ at $k+1$, and is continuous otherwise. It follows that there is (at least) one real root in $(1,2)$, at least one in in $(2,3)$, etc. up to $(k-1,k)$, so there are at least $k-1$ distinct real roots. Additionally, for $x>k+1$ and $k\ge2$, we have $$f_k(x)\ge f_2(x+k-2).$$ It follows that there is another real root between $k+1$ and $x^*+k-2$. So indeed, we have $k$ distinct real roots.
From the aboive, the smallest root is always in $(1,2)$. If follows from $f_{k+1}(x)>f_k(x)$ for $x\in(1,2)$ and the fact that all $f_k$ are strictly decreasing there, that $x_\min$ decreases with increasing $k$. As a decreasing bounded sequence, it does have a limit.
Considering that you look for the first zero of function $$f(x)=\sum_{i=1}^k \frac 1{x-i}-\frac1 {x-k-1}$$ which can write, using harmonic numbers, $$f(x)=H_{-x}-H_{k-x}-\frac{1}{x-k-1}$$ remove the asymptotes using $$g(x)=(x-1)(x-2)f(x)=2x-3+(x-1)(x-2)\left(H_{2-x}-H_{k-x}-\frac{1}{x-k-1} \right)$$ You can approximate the solution using a Taylor expansion around $x=1$ and get $$g(x)=-1+(x-1) \left(-\frac{1}{k}+\psi ^{(0)}(k)+\gamma +1\right)+O\left((x-1)^2\right)$$ Ignoring the higher order terms, this gives as an approximation $$x_{est}=1+\frac{k}{k\left(\gamma +1+ \psi ^{(0)}(k)\right)-1}$$ which seems to be "decent" (and, for sure, confirms your claims). $$\left( \begin{array}{ccc} k & x_{est} & x_{sol} \\ 2 & 1.66667 & 1.58579 \\ 3 & 1.46154 & 1.46791 \\ 4 & 1.38710 & 1.41082 \\ 5 & 1.34682 & 1.37605 \\ 6 & 1.32086 & 1.35209 \\ 7 & 1.30238 & 1.33430 \\ 8 & 1.28836 & 1.32040 \\ 9 & 1.27726 & 1.30914 \\ 10 & 1.26817 & 1.29976 \\ 11 & 1.26055 & 1.29179 \\ 12 & 1.25403 & 1.28489 \\ 13 & 1.24837 & 1.27884 \\ 14 & 1.24339 & 1.27347 \\ 15 & 1.23895 & 1.26867 \\ 16 & 1.23498 & 1.26433 \\ 17 & 1.23138 & 1.26039 \\ 18 & 1.22810 & 1.25678 \\ 19 & 1.22510 & 1.25346 \\ 20 & 1.22233 & 1.25039 \end{array} \right)$$ For infinitely large values of $k$, the asymptotics of the estimate would be $$x_{est}=1+\frac{1}{\log \left({k}\right)+\gamma +1}$$
For $k=1000$, the exact solution is $1.12955$ while the first approximation gives $1.11788$ and the second $1.11786$.
Using such estimates would make Newton method converging quite fast (shown below for $k=1000$).
$$\left( \begin{array}{cc} n & x_n \\ 0 & 1.117855442 \\ 1 & 1.129429575 \\ 2 & 1.129545489 \\ 3 & 1.129545500 \end{array} \right)$$
Edit
We can obtain much better approximations if, instead of using a Taylor expansion of $g(x)$ to $O\left((x-1)^2\right)$, we build the simplest $[1,1]$ Padé approximant (which is equivalent to an $O\left((x-1)^3\right)$ Taylor expansion). This would lead to $$x=1+ \frac{6 (k+k (\psi ^{(0)}(k)+\gamma )-1)}{\pi ^2 k+6 (k+\gamma (\gamma k+k-2)-1)-6 k \psi ^{(1)}(k)+6 \psi ^{(0)}(k) (2 \gamma k+k+k \psi ^{(0)}(k)-2)}$$ Repeating the same calculations as above, the results are $$\left( \begin{array}{ccc} k & x_{est} & x_{sol} \\ 2 & 1.60000 & 1.58579 \\ 3 & 1.46429 & 1.46791 \\ 4 & 1.40435 & 1.41082 \\ 5 & 1.36900 & 1.37605 \\ 6 & 1.34504 & 1.35209 \\ 7 & 1.32741 & 1.33430 \\ 8 & 1.31371 & 1.32040 \\ 9 & 1.30266 & 1.30914 \\ 10 & 1.29348 & 1.29976 \\ 11 & 1.28569 & 1.29179 \\ 12 & 1.27897 & 1.28489 \\ 13 & 1.27308 & 1.27884 \\ 14 & 1.26787 & 1.27347 \\ 15 & 1.26320 & 1.26867 \\ 16 & 1.25899 & 1.26433 \\ 17 & 1.25516 & 1.26039 \\ 18 & 1.25166 & 1.25678 \\ 19 & 1.24844 & 1.25346 \\ 20 & 1.24547 & 1.25039 \end{array} \right)$$
For $k=1000$, this would give as an estimate $1.12829$ for an exact value of $1.12955$.
For infinitely large values of $k$, the asymptotics of the estimate would be $$x_{est}=1+\frac{6 (\log (k)+\gamma +1)}{6 \log (k) (\log (k)+2 \gamma +1)+\pi ^2+6 \gamma (1+\gamma )+6}$$
• Interesting! What does that notation $\psi^{(0)}(k)$ mean by the way? Aug 8, 2018 at 8:10
• @TheSimpliFire. This is "just" the digamma function. Aug 8, 2018 at 8:18
• @TheSimpliFire. We use to write $\psi^{(n)}(k)$ for the polygamma function and, in usual notations,$\psi^{(0)}(k)=\psi(k)$ Aug 8, 2018 at 8:22 | 2022-07-03T06:04:08 | {
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https://math.stackexchange.com/questions/2419619/probability-of-selecting-elements-with-replacement/2419630 | # Probability of selecting elements with replacement
If I'm selecting $N$ elements uniformly at random (with replacement) from $\{1, \dots, N\}$, what is the chance that a given value is selected at least once?
What is the name for the distribution for the more general case where I'm selecting $N$ elements from a domain with $M$ elements?
• by random do you mean equally-probable ? – user451844 Sep 6 '17 at 23:31
• @Roddy Uniformly at random means equally-probable. The key word is "uniformly". – Trevor Gunn Sep 6 '17 at 23:51
• @TrevorGunn yeah I'm stupid sorry. – user451844 Sep 6 '17 at 23:55
## 3 Answers
Your answer is a particular case of the binomial distribution. Let getting a $1$ be a Success. You have $N$ independent trials. In the general version, the success probability is $p = 1/M.$
If $X \sim \mathsf{Binom}(N, 1/M),$ then you seek $$P(X > 0) = 1 - (1-p)^N = 1 - (1 - 1/M)^N = 1 - \left(\frac{M-1}{M}\right)^N,$$ as in the Answer of @ConMan (+1), which I hope you will Accept. [On each trial, you might say that you are using a discrete uniform distribution on the integers $1$ through $M$.]
Example: What is the probability you get at least one $6$ in ten rolls of a fair die? Then $X = \mathsf{Binom}(10, 1/6)$ is the number of $6$'s and you seek $P(X > 0) = P(X \ge 1) = 0.8385,$ computed in R statistical software as:
1 - dbinom(0, 10, 1/6)
## 0.8384944
In the figure below, you want the total heights of the bars to the right of the dotted red line.
• All the answers here were excellent, but I accepted this one since I mentioned binomial distribution and included a graphic, and because you fixed up my notation in the question. – BeeOnRope Sep 7 '17 at 17:26
If you're drawing with equal probability and subsequent draws are independent of each other, then the probability that you do not draw, for example, "1" in a single draw from $M$ elements is $\frac{M-1}{M}$.
The probability that you do not draw any "1"s in $N$ draws is the product of the individual probabilities, i.e. $\frac{M-1}{M}\cdot\frac{M-1}{M}\cdot\ldots\cdot\frac{M-1}{M}=\left(\frac{M-1}{M}\right)^N = \left(1 - \frac{1}{M}\right)^N$.
The probability that you draw at least 1 "1" in the $N$ draws is the complement of the probability that you draw none, i.e. it is $1 - \left(1 - \frac{1}{M}\right)^N$.
The distribution for drawing $M$ elements with replacement from a set of $N$ elements is the uniform distribution on all $M$-tuples from $\{1,\dots,N\}$. For example, with $M = 3$ and $N = 2$ we get 8 ($=2^3$) $3$-tuples:
$$(1,1,1), (1,1,2), (1,2,1), (1,2,2), (2,1,1), (2,1,2), (2,2,1), (2,2,2)$$
and each is equally likely.
Viewed this way, the question of how likely it is to see a given value at least once comes down to counting those sequences that contain the value somewhere. We can count this by counting the tuples that do not contain that value. We know that there are $N^N$ total $N$-tuples and $(N - 1)^N$ tuples not containing a specified value. Thus there are
$$N^N - (N - 1)^N$$
tuples that contain a specified value. Since the distribution is uniform, the probability of selecting such a tuple is
$$\frac{N^N - (N - 1)^N}{N^N} = 1 - \left( 1 - \frac{1}{N} \right)^N.$$
Now let's say that we have the elements $\{1,\dots,N\}$, we draw $M$ of them and we want at least $P$ of them to equal $1$ (or any other fixed element). Then to count the number of $M$-tuples with $P$ $1$'s, first we select the $P$ slots to put the $1$'s in, then from the $M - P$ remaining slots, we have $N - 1$ options each. This gives
$$\binom{M}{P} (N - 1)^{M - P}.$$
Now we may divide by the total number of $M$-tuples ($N^M$) to get a probability of
$$\binom{M}{P} \frac{(N - 1)^{M - P}}{N^{M - P}} \cdot \frac{1}{N^P} = \binom{M}{P} \left( 1 - \frac{1}{N} \right)^{M - P} \left( \frac1N \right)^{P}.$$
This is the probability mass function of the binomial distribution. Namely $\mathsf{Bin}(M, \frac{1}{N})$. | 2019-07-22T18:32:38 | {
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https://math.stackexchange.com/questions/2371022/cross-product-in-higher-dimensions | # Cross product in higher dimensions
Suppose we have a vector $$(a,b)$$ in $$2$$-space. Then the vector $$(-b,a)$$ is orthogonal to the one we started with. Furthermore, the function $$(a,b) \mapsto (-b,a)$$ is linear.
Suppose instead we have two vectors $$x$$ and $$y$$ in $$3$$-space. Then the cross product gives us a new vector $$x \times y$$ that's orthogonal to the first two. Furthermore, cross products are bilinear.
Question. Can we do this in higher dimensions? For example, is there a way of turning three vectors in $$4$$-space into a fourth vector, orthogonal to the others, in a trilinear way?
• You might want to look at the Gram Schmitt method here :en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process – Furrane Jul 25 '17 at 10:06
• This construction has nothing to do with cross product, just happen to coincide in dimension 3. – Miguel Jul 25 '17 at 10:54
• A search for "generalized cross product" turns up a number of questions likely to be of interest, including Is the vector cross product only defined for 3D? and Generalized Cross Product. ;) (Not marking as a duplicate because you're better able to judge which question, if any, most nearly matches yours.) – Andrew D. Hwang Jul 25 '17 at 17:38
• You might be interested in the notion of the orthogonal complement. It can give you the vector orthogonal to a given set of $n-1$ independent vectors in $n$-space, like you're asking for $n=4$. But it can also give you $k$ independent vectors orthogonal to a given set of $n-k$ independent vectors in $n$-space. So you can take two vectors in 4-space and find two vectors perpendicular to them and to each other. – YawarRaza7349 Jul 25 '17 at 17:39
• – user57159 Jul 26 '17 at 3:18
Yes. It is just like in dimension $$3$$: if your vectors are $$(t_1,t_2,t_3,t_4)$$, $$(u_1,u_2,u_3,u_4)$$, and $$(v_1,v_2,v_3,v_4)$$, compute the formal determinant:$$\begin{vmatrix}t_1&t_2&t_3&t_4\\u_1&u_2&u_3&u_4\\v_1&v_2&v_3&v_4\\e_1&e_2&e_3&e_4\end{vmatrix}.$$ You then see $$(e_1,e_2,e_3,e_4)$$ as the canonical basis of $$\mathbb{R}^4$$. Then the previous determinant is $$(\alpha_1,\alpha_2,\alpha_3,\alpha_4)$$ with\begin{align*}\alpha_1&=t_4u_3v_2-t_3u_4v_2-t_4u_2v_3+t_2u_4v_3+t_3u_2v_4-t_2u_3v_4\\\alpha_2&=-t_4u_3v_1+t_3u_4v_1+t_4u_1v_3-t_1u_4v_3-t_3u_1v_4+t_1u_3v_4\\\alpha_3&=t_4u_2v_1-t_2u_4v_1-t_4u_1v_2+t_1u_4v_2+t_2u_1v_4-t_1u_2v_4\\\alpha_4&=-t_3u_2v_1+t_2u_3v_1+t_3u_1v_2-t_1u_3v_2-t_2u_1v_3+t_1u_2v_3\end{align*}It's a vector orthogonal to the other three.
I followed a suggestion taken from the comments on this answer: to put the entries $$e_1$$, $$e_2$$, $$e_3$$, and $$e_4$$ at the bottom. It makes no difference in odd dimension, but it produces the natural sign in even dimension.
Following another suggestion, I would like to add this remark:$$\alpha_1=-\begin{vmatrix}t_2&t_3&t_4\\u_2&u_3&u_4\\v_2&v_3&v_4\end{vmatrix}\text{, }\alpha_2=\begin{vmatrix}t_1&t_3&t_4\\u_1&u_3&u_4\\v_1&v_3&v_4\end{vmatrix}\text{, }\alpha_3=-\begin{vmatrix}t_1&t_2&t_4\\u_1&u_2&u_4\\v_1&v_2&v_4\end{vmatrix}\text{ and }\alpha_4=\begin{vmatrix}t_1&t_2&t_3\\u_1&u_2&u_3\\v_1&v_2&v_3\\\end{vmatrix}.$$
• Very lucid and appreciable answer.thanks – Arpit Yadav Jul 25 '17 at 10:18
• I'm pretty sure the row of basis vectors should be on the bottom to get the right-handedness correct; only in odd dimensions can this row be moved to the top without a change in sign whilst keeping the vectors in the same order. – user332714 Jul 25 '17 at 18:06
• (+1) This is the method I've used in a few answers. The formula is easy to remember! – robjohn Jul 25 '17 at 22:50
• @lastresort Nice remark. I shall edit my answer taking that into account. – José Carlos Santos Jul 25 '17 at 22:54
• Is $$\alpha_1 ={\rm det} \left| \matrix{ t_2 & t_3 & t_4 \\ u_2 & u_3 & u_4 \\ v_2 & v_3 & v_4} \right|$$ and so on? – John Alexiou Jul 26 '17 at 14:19
My answer is in addition to José's and Antinous's answers but maybe somewhat more abstract. In principle, their answers are using coordinates, whereas I'm trying to do it coordinate-free.
What you are looking for is the wedge or exterior product. The exterior power $$\bigwedge^k(V)$$ of some vector space $$V$$ is the quotient of the tensor product $$\bigotimes^k(V)$$ by the relation $$v\otimes v$$. To be somewhat more concrete and less abstract, this just means that for any vector $$v\in V$$ the wedge product $$v\wedge v=0\in\bigwedge^2(V)$$. Whenever you wedge vectors together, the result equals zero if at least two of the factors are linearly dependent. Think of what happens to the cross product in $$\mathbb{R}^3$$.
In fact, let $$e_1,e_2,\ldots,e_n$$ be a basis of an inner product space $$V$$. Then $$e_{i_1}\wedge e_{i_2}\wedge \ldots \wedge e_{i_k}$$ is a basis for $$\bigwedge^k(V)$$ where $$1\leq i_1 < i_2 < \ldots < i_k\leq n$$.
If $$V=\mathbb{R}^3$$ then $$v \wedge w$$ equals $$v \times w$$ up to signs of the entries. This seems a bit obscure because technically $$v\wedge w$$ should be an element of $$\bigwedge^2(\mathbb{R}^3)$$. However, the latter vector space is isomorphic to $$\mathbb{R}^3$$. In fact, this relation is true for all exterior powers given an orientation on the vector space. The isomorphism is called the Hodge star operator. It says that there is an isomorphism $$\star\colon\bigwedge^{n-k}(V)\to\bigwedge^{k}(V)$$. This map operates on a $$(n-k)$$-wedge $$\beta$$ via the relation $$\alpha \wedge \beta = \langle \alpha,\star\beta \rangle \,\omega$$ where $$\alpha\in\bigwedge^{k}(V)$$, $$\omega\in\bigwedge^n(V)$$ is an orientation form on $$V$$ and $$\langle \cdot,\cdot \rangle$$ is the induced inner product on $$\bigwedge^{k}(V)$$ (see wiki). Notice that the wiki-page defines the relation the other way around.
How does all this answer your question you ask? Well, let us take $$k=1$$ and $$V=\mathbb{R}^n$$. Then the Hodge star isomorphism identifies the spaces $$\bigwedge^{n-1}(\mathbb{R}^n)$$ and $$\bigwedge^{1}(\mathbb{R}^n)=\mathbb{R}^n$$. This is good because you originally wanted to say something about orthogonality between a set of $$n-1$$ linearly indepedent vectors $$v_1,v_2,\ldots,v_{n-1}$$ and their "cross product". Now let us exactly do that and set $$\beta :=v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1}\in\bigwedge^{n-1}(\mathbb{R}^n)$$. Then the image $$\star\beta = \star(v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1})$$ is a regular vector in $$\mathbb{R}^n$$ and the defining condition above implies for $$\alpha=v_i\in\mathbb{R}^n=\bigwedge^{1}(\mathbb{R}^n)$$ $$v_i \wedge (v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1}) = \alpha \wedge \beta = \langle \alpha,\star\beta \rangle \,\omega = \langle v_i,\star\beta \rangle \,\omega.$$ However, the left hand side equals zero for $$i=1,2,\ldots,n-1$$, so that the vector $$\star\beta$$ is orthogonal to all vectors $$v_1,v_2,\ldots,v_{n-1}$$ which is what you asked for. So you might want to define the cross product of $$n-1$$ vectors as $$v_1 \times v_2 \times \ldots \times v_{n-1} := \star(v_1 \wedge v_2 \wedge \ldots \wedge v_{n-1})$$.
Maybe keep in mind that the other two answers implicitly use the Hodge star operation (and also a basis) to compute the "cross product in higher dimension" through the formal determinant which is encoded in the use of the wedge product here.
• So concretely, how do we actually know what the hodge star of a $k$-blade is? For example, work in $4$-space with the standard orientation. Suppose we want to know $\star(v_1 \wedge v_3).$ If I understand correctly, it's either $v_2 \wedge v_4$ or else $v_4 \wedge v_2$. How do we know which one? – goblin GONE Jul 25 '17 at 13:32
• It depends on the orientation that you choose for your vector space. Let's say $v_1,v_2,v_3,v_4$ form an oriented basis for $V$ (that is, $\omega = v_1\wedge v_2\wedge v_3\wedge v_4$) then $\star(v_1\wedge v_3)=v_4\wedge v_2$. This can be seen using the defining relation for $\alpha=v_i\wedge v_j$ cycling through all possible combinations $(i,j)$. This is what they say on the wiki-page linked above in the section "Computation of the Hodge star" albeit expressed a little bit complicated in my opinion. – Sven Pistre Jul 25 '17 at 14:25
• Of all combinations $(i,j)$ only $(2,4)$ and $(4,2)$ remain (because otherwise the left hand side equals zero). Then you assume $\star(v_1\wedge v_3)=v_k\wedge v_l$ and think about which combinations for $(k,l)$ remain on the right hand side of the def. relation. Then you will see that the only possible one is $(k,l)=(4,2)$. To see the last part, look at the definition of the induced scalar product on $\bigwedge^2(V)$. – Sven Pistre Jul 25 '17 at 14:28
• And also, as I forgot to mention this, $v_2\wedge v_4=-v_4\wedge v_2$. Changing the position of two vectors in a $k$-wedge just changes the sign. So really it only depends on the chosen orientation (or "right-handedness") of your vector space. – Sven Pistre Jul 25 '17 at 14:49
• @étale-cohomology The hodge star depends on a choice of inner product and orientation. So it is not canonical. I don't think that you can identify them canonically. – Sven Pistre Jul 25 '17 at 17:42
You can work out the cross product $p$ in $n$-dimensions using the following:
$$p=\det\left(\begin{array}{lllll}e_1&x_1&y_1&\cdots&z_1\\e_2&x_2&y_2&\cdots&z_2\\\vdots&\vdots&\vdots&\ddots&\vdots\\e_n&x_n&y_n&\cdots&z_n\end{array}\right),$$ where $\det$ is the formal determinant of the matrix, the $e_i$ are the base vectors (e.g. $\hat{i},\hat{j},\hat{k}$, etc), and $x,y,\ldots,z$ are the $n-1$ vectors you wish to "cross".
You will find that $x\cdot p=y\cdot p=\cdots=z\cdot p=0$.
It's wonderful the determinant produces a vector with this property.
• Are there any requirements on the basis vectors $e_1, ..., e_n$? Like, do they need to form an orthonormal basis, or something? – étale-cohomology Jul 25 '17 at 17:01
• Yeah but do they have to be? Can't we take any other basis? – étale-cohomology Jul 25 '17 at 20:35
• By changing the basis to $\widetilde e_i$ you will have to change the vector entries to the coefficients $\widetilde x_i$ in the basis expansion for the new basis. Remember that the above is only a formal determinant as this is not actually a matrix (since the first column consists of entries that are vectors themselves). So it does not matter if the basis is orthonormal or not but you will have to adjust your formal determinant formula. – Sven Pistre Jul 25 '17 at 21:48
• (+1) This is the logical extension of José Carlos Santos' answer to $\mathbb{R}^n$ (at first, this is what I thought he had given, but now I see his only covers $\mathbb{R}^4$). – robjohn Jul 25 '17 at 22:53
• @robjohn I actually posted my answer 2 minutes before he did :-) – Pixel Jul 25 '17 at 23:00
Yes, and apart from other answers an interesting approach to think about it is using Clifford's algebra.
This can introduce you the basic concept in a nonrigorous but approachable manner.
• Thank you for your answer, however that article is extremely long and it's difficult to find a Clifford-style answer to my question by reading through it. Can I ask you to write up some details on how to compute an actual cross product using Clifford approach's? – goblin GONE Jul 26 '17 at 8:36 | 2021-04-16T04:03:30 | {
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https://mathematica.stackexchange.com/questions/259299/finding-the-interval-of-values-of-an-unknown-element-such-that-a-square-matrix-i | # Finding the interval of values of an unknown element such that a square matrix is positive semidefinite
I'm looking for a way to determine which values of $$c \in \mathbb{R}$$ make the matrix $$A$$ positive semidefinite:
$$A = \begin{bmatrix} 1 & c \\ c & 1 \end{bmatrix}$$
I'm looking for a way to determine which values of c∈R make the matrix A positive semidefinite:
By definition, A matrix is called positive semidefinite if it is symmetric and all its eigenvalues are non-negative. Hence
Clear["Global*"]
mat = {{1, c}, {c, 1}}
If[SymmetricMatrixQ[mat],
eigv = Eigenvalues[mat];
Reduce[eigv[[1]] >= 0 && eigv[[2]] >= 0, c]
]
So any value between -1 and 1 will do. The above code ofcourse assumes the matrix is 2 by 2, but it can easily be made more general.
• Congrats on 100K. Dec 6 '21 at 15:53
cp = CharacteristicPolynomial[{{1, c}, {c, 1}}, t]
(* Out[142]= 1 - c^2 - 2 t + t^2 *)
There are two important things to check. (i) Where do roots cross the y axis? (ii) Where might roots be complex-valued?
For (i) we just solve for c under the assumption thatt==0
Solve[cp == 0 && t == 0, {c, t}]
(* Out[143]= {{c -> -1, t -> 0}, {c -> 1, t -> 0}} *)
This splits the real line for c into three regions abd simply plugging in a value in each region will show that both solutions are nonnegative in the finite part.
For (ii) We check where the discriminant with respect to t vanishes. This will give conditions on c.
Discriminant[cp, t]
(* Out[144]= 4 c^2 *)
So it only vanishes at the origin. We already know that on neither side do we have complex valued roots in t. So there are no further restrictions.
An advantage to this approach is that it extends to higher dimensions.
If you compute the Cholesky decomposition of the matrix, and simplify it under the assumptions you gave:
Simplify[CholeskyDecomposition[{{1, c}, {c, 1}}], c ∈ Reals]
{{1, c}, {0, Sqrt[1 - c^2]}}
you can immediately read off the condition that Sqrt[1 - c^2] >= 0 for positive-semidefiniteness. Thus,
Reduce[%[[-1, -1]] >= 0, c, Reals]
-1 <= c <= 1
` | 2022-01-25T02:48:53 | {
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https://math.stackexchange.com/questions/3684603/floor-function-algebra-question | # Floor function algebra question
I don't know how to put floor functions in but...
Solve $$\dfrac{19x + 16}{10} = \left \lfloor \dfrac{4x+7}{3}\right \rfloor$$
I have so far worked out that the RHS can either be $$(4x+7)/3 - 0.33$$, $$(4x+7)/3 - 0.67$$ or itself. When I solve for each of these three equations, I get $$x=12/17, 22/17, 2/17$$. From there, I subbed $$x$$ back into the equation to try and see which one works but none did. Can I have some help?
• You can get the floor functions with \lfloor and \rfloor, e.g., \lfloor \pi \rfloor = 3. If you need bigger ones, as around a fraction, for instance, use \left\lfloor and \right\rfloor, e.g., \left\lfloor\frac{4x+7}{3}\right\rfloor. May 21 '20 at 3:24
• You seem to be assuming that $4x+7$ is an integer. There's no reason to think it is. May 21 '20 at 3:37
• X must be of the form $$10{\alpha}+6;\alpha \inf \mathbb{Z}$$ May 21 '20 at 4:53
• @saulspatz so would it be safe to say that the RHS is in the range of $(4x+7)/3$ to $(4x+7)/3-0.99$ ?
– user377742
May 21 '20 at 8:52
• Look at my answer. May 21 '20 at 16:09
I'll get you started.
We know that $$x-1<\lfloor x\rfloor<=x$$, so we have $$\frac{4x+4}3<\frac{19x+16}{10}\leq\frac{4x+7}3\\ 40x+40<57x+48\leq40x+70\\ \frac{-8}{17} so that $$x=n+\varepsilon$$ where $$n\in\{-1,0,1\}$$ and $$0\leq\varepsilon<1$$.
Now we can test each of the three possibilities for $$n$$ separately. Suppose $$x=1+\varepsilon$$. Then $$\frac{19x+16}{10}=\frac{35+19\varepsilon}{10}$$ is an integer between $$3.5$$ and $$5.4$$ so there are only two possibilities for $$\varepsilon$$. Check these to see if $$x=1+\varepsilon$$ satisfies the equation. Repeat the process for $$n=0$$ and $$n=-1$$.
Since algebra with floor functions is rarely nice, I like to graph it if possible to visualize the solutions. In your case, this is the graph of $$\frac{19x+16}{10}-\Big\lfloor \frac{4x+7}{3}\Big\rfloor$$
(This is using Desmos by the way) Notice how any solutions would pass through the $$y=0$$, and there seem to be $$4$$. One seems to occur at $$x \approx -.3158$$ and by plugging this into the floor function we see it approaches $$1.912$$, the floor of which is $$1$$. So we are looking for $$x$$ such that $$\frac{19x+16}{10} = 1$$. The solution is $$x = \frac{-6}{19}$$ which, by substituting it back in, works. Similarly, we can show that we need $$x \approx .2105$$ such that $$\frac{19x+16}{10} = 2$$, giving the solution $$x = \frac{4}{19}$$. Then we need $$x \approx .7368$$ such that $$\frac{19x+16}{10} = 3$$, giving the solution $$x = \frac{14}{19}$$. Finally, we need $$x \approx 1.2632$$ such that $$\frac{19x+16}{10} = 4$$, giving the solution $$x = \frac{24}{19}$$. From here, you can use strict inequalities to prove that there are no more solutions.
$$\dfrac{19x + 16}{10} = \left \lfloor \dfrac{4x+7}{3}\right \rfloor$$
Let $$\dfrac{19x + 16}{10} = n \in \mathbb Z$$.
Then $$x = \dfrac{10n-16}{19}$$
and $$\dfrac{4x+7}{3} = \dfrac{40n+69}{57}$$.
So
$$n \le \dfrac{40n+69}{57} < n + 1$$
$$57n \le 40n+69 < 57n + 57$$
$$0 \le -17n+69 < 57$$
$$-69 \le -17n < -12$$
$$\dfrac{12}{17} < n \le 4\dfrac{1}{17}$$
So now you can find the values of $$n$$ and then the values of $$x$$.
First deal with the fact that $$\dfrac{19x + 16}{10}$$ is an integer $$k$$.
So $$19x = 10k -16$$ is and integer $$m$$ and $$x = \frac m{19}$$ for some integer $$m$$.
Now deal with $$\dfrac{19x + 16}{10}=\lfloor \dfrac{4x+7}{3} \rfloor$$ so
$$\dfrac{19x + 16}{10} \le \dfrac{4x+7}{3}< \dfrac{19x + 16}{10}+1=\dfrac{19x +26}{10}$$
Replace $$x$$ with $$\frac m{19}$$ and
$$\dfrac {m + 16}{10} \le \dfrac {\frac 4{19}m + 7}3 < \frac {m+26}{10}$$
$$3m + 48 \le \frac 40{19}m +70 < 3m + 78$$
$$-22 \le \frac {40}{19}m - 3m < 8$$
$$-22*19 \le 40m - 57m = -17*m < 8*19$$
$$\frac {-8*19}{17} < m \le \frac {22*19}{17}$$
$$-8 \frac {16}{17} < m \le 24 \frac {10}{17}$$
So $$-8 < m \le 24$$
But $$m = 10k-16$$ so $$m\equiv -16\equiv -6 \equiv 4 \pmod{10}$$ so so
$$m =-6,4,14,24$$ are acceptable values.
And $$x =-\frac 6{19},\frac 4{19}, \frac {14}{19}$$ and $$\frac {24}{29}$$ are acceptable answers. | 2021-09-22T00:16:22 | {
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https://mathematica.stackexchange.com/questions/118481/how-does-listinterpolation-work | # How does ListInterpolation work?
ListInterpolation takes an array of data and interpolates between the entries. Let us interpolate between the elements of the identity matrix. For reference, this is what the identity matrix looks like:
MatrixPlot[IdentityMatrix[5]]
Now let's interpolate, and request first order interpolation:
if = ListInterpolation[IdentityMatrix[5], {{0, 1}, {0, 1}}, InterpolationOrder -> 1];
Plot[if[x, x], {x, 0, 1}]
The result is a piecewise function as expected, but not a piecewise linear function! Why? What does first order interpolation mean here?
The kind of linear interpolation I am familiar with is what ListDensityPlot does:
ListDensityPlot[IdentityMatrix[5], DataRange -> {{0, 1}, {0, 1}}, Mesh -> All]
Construct a Delaunay triangulation of the data and interpolate on each triangle.
This is not what ListInterpolation does (or even Interpolation when the data is strictly on a square grid). What does ListInterpolation do, then?
Inspired by this question:
• Well, it is doing bilinear interpolation in your first example, I would think. You definitely will get something with quadratic pieces if you sample across diagonals. – J. M. is in limbo Jun 15 '16 at 14:00
• @J.M. Well, we all have to learn at some point :-) Yes, you are right, and this is due to my ignorance of math. But I won't delete it. If you convert that to a one-line answer, I'll accept it. – Szabolcs Jun 15 '16 at 14:01
• Docs say the only methods available to ListInterpolation are Spline and Hermite – N.J.Evans Jun 15 '16 at 14:02
• Currently my comment is not that pedagogically useful; I'll try to come up with a sufficiently illuminating example first, unless somebody beats me to it. :) – J. M. is in limbo Jun 15 '16 at 14:04
• Szabolcs, please consider leaving the question up for a little longer. I was not aware of that behavior either, and I would find it instructive to see an interesting example, if @J.M. or others have the time / inclination. – MarcoB Jun 15 '16 at 14:43
I'll just treat the simplest case of interpolating across four noncoplanar points. The extension to general grids is straightforward (one just joins multiple pieces appropriately).
Consider the following bivariate interpolating polynomial:
ip[x_, y_] = InterpolatingPolynomial[{{{0, 0}, 1}, {{0, 1}, -1/2},
{{2, 0}, 1/2}, {{2, 1}, 0}}, {x, y}];
and the following InterpolatingFunction[]:
if = ListInterpolation[{{1, -1/2}, {1/2, 0}}, {{0, 2}, {0, 1}}, InterpolationOrder -> 1];
They are the same:
Plot3D[ip[x, y] - if[x, y] // Chop, {x, 0, 2}, {y, 0, 1}]
Let's have a look at what ip[] looks like:
Expand[ip[x, y]]
1 - x/4 - (3 y)/2 + (x y)/2
Huh. That last term certainly isn't linear. What gives?
In a bilinear interpolation such as this one, although the procedure is something along the lines of "linearly interpolate for each grid line in one direction, and then linearly interpolate those results", the result is inevitably quadratic. This should come as no surprise to people familiar with the hyperbolic paraboloid, which is one of the simplest examples of a ruled surface, or a surface formed by sweeping out a line in space. In fact, z == 1 - x/4 - (3 y)/2 + (x y)/2 is indeed a hyperbolic paraboloid.
{Plot3D[ip[x, y], {x, 0, 2}, {y, 0, 1}, MeshFunctions -> {#1 - #2 &}],
Plot[ip[x, x], {x, 0, 1}]} // GraphicsRow | 2020-02-26T02:31:36 | {
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https://math.stackexchange.com/questions/3013010/how-to-find-out-the-global-minimum-of-the-following-expression/3013059 | # How to find out the global minimum of the following expression
What is the global minimum of the expression \begin{align} |x-1| &+ |x-2|+|x-5|+|x-6|+|x-8|+|x-9|+|x- 10| \\&+ |x-11|+|x-12|+|x-17|+|x-24|+|x-31|+ |x-32|? \end{align}
I've solved questions of this sort before but there were only 3 terms. I solved those by expanding all the terms in the modulus and drawing a graph. This question came in a paper which requires the student to solve it within 5 minutes. What's a better method?
Unfortunately I needed some minutes to think about the problem before finding a solution that can be calculated very quickly:
Imagine the graph of the function $$f_a(x)=|x-a|$$. Having the graph in mind you see that the derivation $$f'(x)=-1$$ for $$x and $$f'(x)=1$$ for $$x>a$$.
For the intervals: $$(-\infty,1)$$, $$(1,2)$$, $$(2,5)$$, ..., $$(32,\infty)$$ we can now easily calculate the derivative $$f'(x)=f'_1(x)+f'_2(x)+f'_5(x)+...-f'_{32}(x)$$:
In the range $$(-\infty,1)$$ it is $$f'(x)=-1-1-1-...-1+1=-11$$.
In the range $$(1,2)$$ it is $$f'(x)=+1-1-1-...-1+1=-9$$.
In the range $$(2,5)$$ it is $$f'(x)=+1+1+1-...-1+1=-7$$.
...
In every step we simply have to invert one sign so "-1" becomes "+1". This means the derivative is changing by 2 at the points x=1,2,5,...
We start by calculating the derivative for $$x<1$$; it is -11.
Now we simply go through the ranges:
<1: -11
1..2: -9
2..5: -7
5..6: -5
6..8: -3
8..9: -1
9..10: +1
10..11: +3
11..12: +5
12..17: +7
17..24: +9
24..31: +11
31..32: +13
>32: +11
At $$x=32$$ the derivation decreases by 2 because of the minus sign before $$|x-32|$$; you could of course adapt this method for sums of elements of the form $$b|x-a|$$.
We see that for $$x<9$$ the derivation is negative and for $$x>9$$ the derivation is positive. We also know that the function is continuous. (This is important because the derivation is not defined at x=1,2,5,...) This means that the function is strictly decreasing respectively increasing for $$x<9$$ and for $$x>9$$.
So we know that the global minimum must be at $$x=9$$.
• The right answer. Congratulations! Although I don't know if your method works when the number of terms is even, say $f(x) = |x-1|+ |x-2|$ or $f(x) = |x-11|+|x-12|+|x-17|+|x-24|$ (the minimum or maximum is an interval). Nov 25 '18 at 22:20
• @David Why should it be a problem if a whole interval represents the minimum or maximum? In the example of $|x-1|+|x-2|$ we would have $f'(x)=0$ in the whole interval 1...2. Nov 26 '18 at 6:28
• This answers the previous version of my typo question but the method is absolutely wonderful. Thanks!
– user619072
Nov 27 '18 at 17:36
You can in principle write out the function in a lot of intervals. But it would probably take too long. However, I will use this fact, without doing it explicitly. We know that if we do write this function, it is going to be linear on each interval (sum of linear functions is a linear function), and it's going to be continuous (sum of continuous functions is a continuous function). We also know that on a line you get minimum at one end, the other, or both (constant line). So all you need to do is to calculate your function at $$1,2,5,6,...$$ and find the minimum.
• I like how this answer asks us to imagine the graph of the function. I find this "imagine" technique useful, while still being intuitive. Similar examples includes "imagine the multiplication table of this finite group" (the group could be huge) or "imagine the truth table for this proposition" (the formula could be long). Nov 25 '18 at 22:02
The answer (minimizer) in this case is $$10$$, the median of the sequence $$(1,2,5,6,8,9,10,11,12,17,24,31,32).$$
You can plug in $$x=10$$ in the function and you would find that the minimum value is $$96$$. In general, the solution to the following minimization problem
$$\min\{|x-a_1| + |x-a_2| + \cdots + |x-a_n|\}$$ is the median of $$(a_1,\ldots,a_n)$$. To see why, consider first when $$n=2$$, and without loss of generality assume $$a_1. Then $$|x-a_1|+|x-a_2|$$ is the distance between $$x$$ and $$a_1$$ plus the distance between $$x$$ and $$a_2$$. It is easy to see that only when $$x$$ is in the middle of $$a_1$$ and $$a_2$$ should the sum of distances be minimal, which equals $$|a_2-a_1|$$ in this case. In this case the minimizer is not unique. Any points in $$[a_1,a_2]$$ is a minimizer.
When $$n=3$$, the function is $$|x-a_1|+|x-a_2|+|x-a_3|$$, and we order the parameters again so that $$a_1. When $$x$$ coincides with $$a_2$$, i.e. $$x=a_2$$, the value becomes $$|a_2-a_1|+|a_2-a_3|=|a_3-a_1|$$, the distance between $$a_3$$ and $$a_1$$. But when $$x\in[a_1,a_3], x\neq a_2$$, the value of the function is $$|x-a_2|+|x-a_1|+|x-a_3| = |a_3-a_1| + |x-a_2|,$$ which is largar than $$|a_3-a_1|$$, the distance between $$a_3$$ and $$a_1$$. Similarly the value would become larger when $$x$$ is outside $$[a_1,a_3]$$. So in this case, the minimizer is unique and is equal to $$a_2$$, the median of $$(a_1,a_2,a_3)$$.
In general, when $$n$$ is odd, there exists a unique minimizer, which is equal to the (unique) median of the parameters $$(a_1,\ldots,a_n)$$. When $$n$$ is even, the function is minimal and constant over the range $$[a_i,a_j]$$, where $$a_i$$ and $$a_j$$ are the two middle values.
• That doesn't seem to be correct. Wolfram says the minimum is 51 at x=9. See here Nov 25 '18 at 20:50
• I see now that the confusion comes from the $−|x−32|$ in the problem statement @MartinRosenau. Note the minus sign. Fei Li solved it for a + sign. Nov 25 '18 at 21:18
• I plot the graph using wxmaxima and get the same result as @IlikeSerena. Note the minus sign at the end. Nov 25 '18 at 21:48
• Maybe the OP @CaptainQuestion made a typo. We need his or her classification, but I think the purpose of this question is exactly what I have demonstrated. Nov 25 '18 at 22:00
• @AlexVong So I hoped, but the next sentence "which equals $|a_2-a_1|$ in this case." excludes this interpretation. Nov 25 '18 at 22:39
TL;DR: Put the absolute values in ascending order, and look at the sum of the leading coefficients. One by one, change the signs in the sum from right to left. When the sum changes signs, you have passed a local extremum. When the sum equals zero, there is an extremum on an entire interval.
As an alternative to Andrei's excellent answer, or perhaps an extension, you could also look at the derivative. Clearly the function is continuous everywhere, and it is differentiable in all but finitely many points, call them $$a_1,\ldots,a_n$$ in ascending order. Then we want to minimize $$f(x)=\sum_{k=1}^nc_k|x-a_k|=\sum_{k=1}^n(-1)^{\delta_{x\leq a_k}}c_k(x-a_k),$$ where $$\delta$$ denotes the Kronecker delta, in this case defined as $$\delta_{x\leq a_k}:=\left\{\begin{array}{ll}1&\text{ if } x\leq a_k\\0&\text{ otherwise}\end{array}\right..$$ This has derivative (for all $$x$$ apart from the $$a_k$$) $$f'(x)=\sum_{k=1}^n(-1)^{\delta_{x\leq a_k}}c_k.$$ The expression has a local minimum at $$x$$ if either $$f'(x)=0$$, or if $$x=a_k$$ for some $$k$$ and $$f'(y)<0$$ for $$a_{k-1} and $$f'(y)>0$$ for $$a_k.
This is all rather formal; in practice this means you put the $$c_k$$ in ascending order, so here $$n=13$$ and $$c_1=\cdots=c_{12}=1$$ and $$c_{13}=-1$$, and find all $$m$$ such that flipping the last $$m$$ signs in the sum $$c_1+c_2+c_3+c_4+c_5+c_6+c_7+c_8+c_9+c_{10}+c_{11}+c_{12}+c_{13}$$ makes the sum change signs compared to changing the last $$m-1$$ signs. Here a quick look gives $$c_1+c_2+c_3+c_4+c_5+c_6-c_7-c_8-c_9-c_{10}-c_{11}-c_{12}-c_{13}=1,$$ $$c_1+c_2+c_3+c_4+c_5-c_6-c_7-c_8-c_9-c_{10}-c_{11}-c_{12}-c_{13}=-1,$$ so $$f$$ has a local minimum at $$a_6=9$$, and it is not difficult to see that there is no other minimum. | 2021-09-25T08:53:30 | {
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https://math.stackexchange.com/questions/3132934/a-mistake-on-computing-int-fracdx-sqrtx11/3132967 | # A mistake on computing $\int \frac{dx}{\sqrt{x+1}+1}$
I have to find $$\int \frac{dx}{\sqrt{x+1}+1}$$. This was my attempt, which is wrong and I cannot find where exactly is the mistake.
First I write $$\frac{1}{\sqrt{x+1}+1}=\frac{\sqrt{x+1}-1}{x}=\frac{\sqrt{x+1}}{x}-\frac{1}{x}$$, therefore $$\int \frac{dx}{\sqrt{x+1}+1}=\int \frac{\sqrt{x+1}}{x}dx-\log\left (x\right )$$, so I only have to deal with $$\int \frac{\sqrt{x+1}}{x}dx$$.
Setting $$u=\sqrt{x+1}$$, we obtain $$du=\frac{1}{2\sqrt{x+1}}dx$$, therefore $$dx=2udu$$ and $$x=u^2-1$$, so
$$\int\frac{\sqrt{x+1}}{x}dx=\int\frac{2u^2}{u^2-1}du=\int \left(2+\frac{2}{u^2-1}\right)du=2\sqrt{x+1}+\int \left(\frac{1}{u-1}-\frac{1}{u+1}\right)du=$$
$$=2\sqrt{x+1}+\log \left (\sqrt{x+1}-1\right )-\log \left (\sqrt{x+1}+1\right )=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )$$
But then $$\int \frac{dx}{\sqrt{x+1}+1}=2\sqrt{x+1}+\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )-\log\left (x\right )=2\sqrt{x+1}+\log\left ( \frac{x+2-2\sqrt{x+1}}{x^2} \right )$$, which is not what I am supposed to obtain. The actual answer is $$2\sqrt{x+1}-2\log\left ( 1+\sqrt{x+1} \right )$$.
Where is my mistake?
(I already know a correct way to solve it, I just want to know where I am committing a mistake).
• I believe that you went wrong in your first step. That “simplification” made everything a lot harder. Hint: Start with a u-substitution if u=x+1 . – Sina Babaei Zadeh Mar 2 '19 at 22:36
• The standard method suggestst to set $u=\sqrt{x+1}$, i.e. $x=u^2-1,\; u\ge 0$, from the very beginning. – Bernard Mar 2 '19 at 22:44
But you have been right all along!
Notice that $$\log\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} - \log x = \log\frac{\sqrt{x+1}-1}{x(\sqrt{x+1}+1)} = \log \frac{1}{(\sqrt{x+1}+1)^2} = -2\log(\sqrt{x+1}+1)$$ by your very first step.
Also, I would suggest that you use $$\int\frac{dx}{x} = \log|x|$$ (with the absolute value sign, to extend the domain).
What I would also suggest is to use an online function grapher if you are not sure that your solution is correct. If two functions have identical graphs or if they differ by a constant value, then both are correct solutions of the integral.
• Oh, I see... My answer was correct, but it was not clear why. I wrote $\log\left ( \frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \right )=\log\left ( \frac{x+2-2\sqrt{x+1}}{x} \right )$ (i.e., i was rationalizing the denominator), but I should have rationalized the numerator, leading to $\log\left ( \frac{x}{\left ( \sqrt{x+1}+1 \right )^2} \right )$, which gives the solution. Thank you! – solomeo paredes Mar 2 '19 at 23:04
Note that $$\ln\frac{x+2-2\sqrt{x+1}}{x^2}=\ln\frac{(\sqrt{x+1}-1)^2}{x^2}=2\ln\frac{\sqrt{x+1}-1}{x}$$ if we further simplify, $$2\ln(\frac{\sqrt{x+1}-1}{x}\cdot\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}) =2\ln\frac{1}{\sqrt{x+1}+1}=-2\ln(\sqrt{x+1}+1)$$ You did nothing wrong.
Note that $$\frac{x+2-2\sqrt{x+1}}{x^2} =\frac {(x+2)^2-4(x+1)} {x+2+2\sqrt{x+1}} \cdot\frac 1{x^2}\ ,$$ and this has nothing to do with $$\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1} \cdot \frac 1x = \frac{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}{(\sqrt{x+1}+1)(\sqrt{x+1}+1)} \cdot \frac 1x = \frac 1{(\sqrt{x+1}+1)^2}\ ,$$ which leads to the solution. So the last step is the bad one.
You're fine.
$$\log\left(\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\right) -\log x$$
$$= \log\left(\frac{\sqrt{x+1}-1}{\sqrt{x+1}+1}\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}\right) -\log x$$
$$=\log\left( \frac{x}{ (\sqrt{x+1}+1)^2}\right) -\log x$$
$$=\log x -2\log(\sqrt{x+1}+1)- \log x$$
$$=-2\log(\sqrt{x+1}+1).$$
No mistake! $$log((\frac{\sqrt{x+1}-1}{x})^2)=-2log(\frac{x}{\sqrt{x+1}-1})=-2log(\frac{x(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)(\sqrt{x+1}+1)})=-2log\frac{x(\sqrt{x+1}+1)}{x}$$ | 2020-02-29T14:35:49 | {
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http://rafbis.it/qowu/graph-of-sine-and-cosine-functions-real-world-applications.html | ## Graph Of Sine And Cosine Functions Real World Applications
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To understand how the absolute value can be applied to the real world, we’ll look at two topics: What is considered normal with regard to the temperature of the human body and; Fuel economy of two vehicles: a Honda Odyssey and a Nissan Altima. Trig also has applications in fields as broad as financial analysis, music theory, biology, medical imaging, cryptology, game development, and seismology. Amplitude is first, then sine or cosine, then B, which you get by doing 2pi/10 (period), which reduces to pi/5, then it is the parenthesis with the transformation, and then last but not least is the sinusodial axis. Application: Sine and cosine functions can be used to model real-world phenomena, such as sound waves. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. functions; increasing/decreasing functions; finding intercepts with axes. Evaluate trigonometric functions of real numbers. Modeling Using Sinusoidal Functions Sine and cosine functions model many real-world situations. The graph of the sine function is called the sine curve. Defining Sine and Cosine Functions. When they notice that these points are already on the graph, make a note about the cyclic nature of sine and cosine. Amplitude is first, then sine or cosine, then B, which you get by doing 2pi/10 (period), which reduces to pi/5, then it is the parenthesis with the transformation, and then last but not least is the sinusodial axis. Topic:Real world application of sine graphs. Example: if your cutting down a tree and need to know where it will. Find all values of A and B such that. 7m, but I am unsure how to interval the axis. This shape is also called a sine. Evaluate Inverse Trigonometric Functions v. For large arguments we would expect these functions to be a quarter period out of phase, just like cosine and sine, since asymptotically J 1 is proportional to cos(z – 3π/4) / √z and Y 1 is proportional to sin(z – 3π/4) / √z. Graphing Sine & Cosine w/out a Calculator Pt1. The function is a model for the hours of daylight in Center City. Sum and Difference, Double Angle and Half Angle. The size of a hyperbolic angle is twice the area of its hyperbolic sector. Students will use the model to make a prediction about a missing piece of data. 2 cos 2 cos 2 cos 2 sin( ) cos Now we can clearly see that if we horizontally shift the cosine function to the right by π/2 we get the sine function. 1 works in a wide range of settings: any time we have an exponential equation of the form $$p \cdot q^t + r = s\text{,}$$ we can solve for $$t$$ by first isolating the exponential expression $$q^t$$ and then by taking the natural logarithm of both sides of the equation. In maths, you have real life applications on any thing that you study. The cosine is known as an even function, and the sine is known as an odd function. Cumulative Review. Along the way, youll get plenty of practice, from fully guided examples to independent end-of-chapter drills and test-like samples. Midline Amplitude And Period Review Article Khan Academy Solving Problems Using Sinusoidal Functions Real World Applications Ferris Wheel Trig Example Youtube Period Of Cosine Function Math Ladyprestige Club. Next, lets take the 5 from the front. 2 Define sine and cosine as functions of the radian measure of an angle in terms of the x- and y-coordinates of the point on the unit circle. But just as you could make the basic quadratic, y = x 2, more complicated, such as y = -(x + 5) 2 - 3, so also trig graphs can be made more complicated. I know that the amplitude of the function is 1. When I consider how to address the Precalculus objectives "to solve real-life problems involving harmonic motion" ii. A cosine function will start at its minimum or maximum value on the y-axis; however, the sine graph will start at its midline on the y-axis. Graph sine and cosine functions with various amplitudes and periods. 6 Modeling with Trigonometric Functions 507 Writing Trigonometric Functions Graphs of sine and cosine functions are called sinusoids. Determine the values for angles between 0° and 360°. can be used to assign a particular use of the plot function to a particular figure wi. A sine wave is what you get when you plot the sine function on a graph. Psst… don’t over-focus on a single diagram, thinking tangent is always smaller than 1. Defining Sine and Cosine Functions. Sketch the graphs of basic sine and cosine functions; use amplitude and period to help sketch the graphs of sine and cosine functions. Phase Shift = to the right Vertical Shift = 4 4. How to analyze graphs of functions. Sine, Cosine, Tangent Applications. My kids have seen this a bunch so this should only take a minute or two. 2]Solve right and oblique triangles including application problems. Analysis: I developed this lesson plan so that students would be introduced to graphs…. Aug 15, 2015 · C2 Sine and Cosine Rule - 1 - Basic Introduction (AS Maths A Level trigonometry) ukmathsteacher. In line 4 we use the properties of cosine (cos -x = cos x) and sine (sin -x = -sin x) to simplify the. Graphing the sine and cosine functions is a topic that students will carry on with them throughout the rest of their future science and mathematics courses. Graphs of tan, cot, sec and csc; 5. Real life word problems that involve the sine and cosine function can be used to keep the students engaged in the topic. inverse MTH 112 Elementary Functions Chapter 5 The Trigonometric Functions -. The inverses of sine, cosine and tangent functions are not functions unless the domains are limited. Many compression algorithms like JPEG use fourier transforms that rely on sin and cos. Investigate Transformations of Sine and Cosine Functions Many real-world processes can be modelled with sinusoidal functions. functions for special angles using the unit circle and right triangle definitions. Evaluating Inverse. - She did an excellent job in the trigonometry unit where she was able to use the sine and cosine laws to solve problems on both her assignment and. We can prove this by using the cofunction identity and the negative angle identity for cosine. Lesson 6-6 Modeling Real-World Data with Sinusoidal Functions 389 Graphing Calculator Tip For keystroke instruction on how to find sine regression statistics, see page A25. Amplitude and Period for Sine and Cosine Functions Worksheet Author: marthasweeney Last modified by: Leake, Sherry LHS-STAFF Created Date: 6/13/2016 5:56:00 PM. The sine function “starts” at the midline, but the leading negative causes it to reflect across the midline. Course Objectives: * Students will be able to develop and understand transferable pre-calculus skills necessary for success in Precalculus and beyond * Students will be able to identify, evaluate, perform operations on, and find the domain, range, and inverse of functions * Students will be able to draw common graphs along with transformations and reflections * Students will be able to create. Use sinusoidal functions to solve problems. The heat equation is used to model how things get hot (electronics, spacecraft, ovens, etc). Find the amplitude, period, and phase shift of a sine or cosine functions from its equation. Sine and cosine functions can be used to model many real-life scenarios –radio waves, tides, musical tones, electrical currents. Frequency is the number of cycles in a given. Analytic Trigonometry. Sine and cosine waves can be applied to all sinusoid problems of the real-world. Being a cofunction, means that complementary input angles leads to the same output , as shown in the following example: Problem 2. Students will investigate and use the graphs of the six trigonometric functions. 1 RTP Scenarios and Terminology 12. 2 RTP Packet Format 12. Understanding Basic Sine & Cosine Graphs. c) sketch the graph of the function by using transformations for at least a two-period interval; and d) investigate the effect of changing the parameters in a trigonometric function on the graph of the function. When teaching students the value of sinusoidal models and graphs, such as sine and cosine functions, students often feel like they must memorize formulas and apply them with little recollection or understanding. Graph y = sin x. Using f (x) = cos x as a guide, graph the function h (x) = __1 cos 2 3 x. In particular: The slope of a graph is related to how fast the quantity being graphed is changing. Graphing Sine and Cosine Functions. The Sine, Cosine and Tangent functions are often applied to real world scenarios. C8 demonstrate an understanding of real-world relationships by translating between graphs, tables, and written descriptions. Third, this is a sine graph with a leading negative. Hence, we start this lesson by recalling these functions and their corresponding graphs. functions, and graph them, with and without the use of digital technologies (ACMMG274) prove that the. Trigonometry can be used to roof a house, to make the roof inclined ( in the case of single individual bungalows) and the height of the roof in buildings etc. Figure 4 The cosine function Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. Graphs of the Trigonometric Functions; 1. The Code of Federal Regulations is a codification of the general and permanent rules published in the Federal Register by the Executive departments and agencies of the Federal Government. Write a cosine function that describes the height of the nail above ground as a function of angular distance. Trigonometric functions (Level C) Radian measure (including what are radians, converting from degrees to radians, converting from radians to degrees, using radians measure in real world applications) Graphs of sine, cosine, and tangent functions. The sine function. A single note can be modeled on a sine curve, and a chord can be modeled with multiple sine curves used in conjunction with one another. Special triangles can be used to geometrically determine the values of the six trigonometric functions at ,, 6 4 3 , and their integral multiples. Duration: 2 Day(s) Learning Targets. Applications of Sine and Cosine Graphs Learning Task: Trigonometry functions are often used to model periodic data. Key Questions. SheLovesMath. Graphing Sine and Cosine Functions. Chapter Five – Trigonometric Functions What to Expect: A new formula designed for sin and cos graphs, details about frequencies, new vocabulary, explaining periodic trends, and the like. Notes Chapter 6 Section 6 ( Modeling Real-World Data with Sinusoidal Functions ) -pg. Section 3-5 : Graphing Functions. Trigonometric Functions By the end of this course, students will: - solve problems involving trigonometry in acute triangles using the sine law and the cosine law, including problems arising from real-world applications; - demonstrate an understanding of periodic relationships and the sine function, and make connections between the identify and. The cosine of an angle is the ratio of x r {\displaystyle {\tfrac {x}{r}}} , so in the unit circle, the cosine is the x -coordinate of the point of rotation. - She did an excellent job in the trigonometry unit where she was able to use the sine and cosine laws to solve problems on both her assignment and. C2 create and analyse scatter plots of periodic data. Press the button marked "sin" to change to the cosine. 2-use technology, graphs, and tables to compare sine graphs. 3 Finding Maximum and Minimum Values and Zeros of Sine and Cosine 2. GCSE Maths Pythagoras and Trigonometry Level 8-9. These functions are most conveniently defined in terms of the exponential function, with sinh z = 1 / 2 (e z − e −z) and cosh z = 1 / 2 (e z + e. 391-392 #'s 7-16 all I will be able to: Model real-world data using sine and cosine functions. 8 Graphs, Inverses, and Applications of Trigonometric Functions. Find the amplitude and period of the function. Applications of Exponential Functions The best thing about exponential functions is that they are so useful in real world situations. If the set of input values is a finite set then the models are known as discrete models. When teaching students the value of sinusoidal models and graphs, such as sine and cosine functions, students often feel like they must memorize formulas and apply them with little recollection or understanding. 7 The student will identify the domain and range of the inverse trigonometric functions and recognize the graphs of these functions. So, temperature as a function of days. A real life example of the sine function could be a ferris wheel. Sample records for nino-southern oscillation eventsnino-southern oscillation events «. To be able to solve real-world problems using the Law of Sines and the Law of Cosines This tutorial reviews two real-world problems, one using the Law of Sines and one using the Law of Cosines. and "use sine and cosine functions to model real-life data," iii. Find the amplitude, period, and phase shift of a sine or cosine functions from its equation. sine, cosine, tangent for π/3, π/4 and π/6, and use the unit circle to express the values of sine, cosine, and tangent for π–x, π+x, and 2π–x in terms of their values for x, where x is any real number. Write the sum and difference identities for sine, cosine, and tangent. Chapter 4 Sections 4. This means that bodies. On the ground, from the object you measure off a know distant in a straight line from the base. 2) Watch the video and pay attention to the key ideas listed. If you take a rope, fix the two ends, and let it hang under the force of gravity, it will naturally form a hyperbolic cosine curve. The sine and cosine of an angle are both circular functions, and they are the fundamental circular functions. Water Depth Word Problem Modeled with Cosine Sine Function. 5/175* share an equivalent sine. Namely, if we look at the graph 'y' equals hyperbolic cosine 'x', observe that whereas the function is single valued, it is not one-to-one. Find patterns in given array of numbers. The approach used in Example 3. 2 Define sine and cosine as functions of the radian measure of an angle in terms of the x- and y-coordinates of the point on the unit circle. Find the frequency in hertz for this sound wave. A single note can be modeled on a sine curve, and a chord can be modeled with multiple sine curves used in conjunction with one another. For large arguments we would expect these functions to be a quarter period out of phase, just like cosine and sine, since asymptotically J 1 is proportional to cos(z – 3π/4) / √z and Y 1 is proportional to sin(z – 3π/4) / √z. Frequency is the number of cycles in a given unit of time. Sineing on to the job Since we know that a triangle has 180 degrees, we can subtract 56 degrees and 91 degrees from it to find our missing angle Using the law of sines we can then set up this equation sin 91 degrees/ xft = sin 33/6ft After crossmultipying and then dividing to. Now we must stretch out the amplitude of the sine graph. Les't compare with. So, temperature as a function of days. This is an example of how a Graph of a Cosine Function looks like:. Animated gifs showing relationships between functions and geometry eg trigonometric functions How the unit circle gives us the sine wave Teaching math with the Drum Loop Real sinusoid on a timebase, formed by a linear increment of complex argument in time. Therefore, the period of this graph is 2π/(1/3) which is 6π. To find the period of a sine graph, take the B value, which is 2, and divide 2π by it. Description: To be able to solve real-world problems using the Law of Sines and the Law of Cosines. Land surveying makes an extensive use of the sine and cosine law. The basic sine and cosine graphs C. Seymour Public Schools Curriculum Subject or course name 5 graph the basic sine and cosine functions. Do not forget this. state the properties for the Sine and Cosine graph identify the values of a, k, c and d and apply it to a sinusoidal function to graph the resulting transformation Lesson:. Graph Trigonometric Functions (1), cosine function with solution. ' Below is a listing of several popular input and output values for the three main trigonometry functions. 1) Graph sine functions (PC-N. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. 1: Graphs of the Sine and Cosine Functions. Water Depth Word Problem Modeled with Cosine Sine Function. One method to write a sine or cosine function that models a sinusoid is to fi nd the values of a, b, h, and k for y = a sin b(x − h) + k or y = a cos b(x − h) + k where ∣ a ∣ is the amplitude, — 2π is the period. Master PHP 4 — the open source Web scripting breakthrough! Contains expert coverage of syntax, functions, design, and debugging! Leverage the amazing performance of the new Zend engine! 650+ real-world code examples! CD-ROM includes source code, plus everything you’ll need to run PHP 4 implementations on Windows and UNIX!. You can select the values of a and b within certain ranges. Sine wiggles in one dimension. We completed three cosine graphs together on the 13-5 vocab support page. Signal Processing. Real life word problems that involve the sine and cosine function can be used to keep the students engaged in the topic. ) Circles are an example of two sine waves. Find the amplitude and period of the function. Transformations of Trigonometric Functions. Graphs of the sine and cosine functions; and Periodic Behavior. Because we can evaluate the sine and cosine of any real number, both of these functions are defined for all real numbers. The student will graph sine or cosine graph affected by horizontal and vertical translations. Like What? [Filename: Notes 6. The Code of Federal Regulations is a codification of the general and permanent rules published in the Federal Register by the Executive departments and agencies of the Federal Government. It is given by parameter #a# in function #y = asinb(x - c) + d or y = acosb(x - c) + d# •The period of a graph is the distance on the x axis before the function repeats itself. • Use the cosine law to determine the interior angles of an acute triangle. The Sine Ratio Passy's World of Mathematics. Physical phenomenon such as tides, temperatures and amount of sunlight are all things that repeat themselves, and so are easily modeled by sine and cosine functions (collectively, they are called "sinusoidal functions"). Find an equation for a sine function that has amplitude of 4 a period of π. To understand how the absolute value can be applied to the real world, we'll look at two topics: What is considered normal with regard to the temperature of the human body and Fuel economy of two vehicles: a Honda Odyssey and a Nissan Altima. 8 L5 Trig Applications Part 1 - solve problems that arise from real world applications involving periodic phenomena D3. Lynn Weiss. Sketch the graphs of basic sine and cosine functions; use amplitude and period to help sketch the graphs of sine and cosine functions. Given an expression for the time rate of change of a quantity, find the initial rate. ppt), PDF File (. The maximum value appears to be 72 and the minimum value is 11. Goal: Students will use technology to discover how to graph trigonometric functions. Apr 15, 2018 - Discovering the graphs of Sine & Cosine using the measurements of the Unit Circle (Activity from Rice APSI) Stay safe and healthy. PreCalculus Honors - Curriculum Pacing Guide - 2017 - 2018 First Half of Semester Anderson School District Five Page 3 2017-2018 Unit 2 - Application of Trigonometry to Triangles (7 days) PC. solve problems involving trigonometry in acute triangles using the sine law and the cosine law, including problems arising from real-world applications; demonstrate an understanding of periodic relationships and the sine function, and make connections between the numeric, graphical, and algebraic representations of sine functions;. A sine wave is what you get when you plot the sine function on a graph. MCF 3M - Functions and Applications (Mixed) Spring 2014 Page history last edited by Peter Kim 5 years, 9 months ago This course introduces basic features of the function by extending students' experiences with quadratic relations. The product of two cosine functions is a sum of two other cosine functions. Trigonometric functions allow us to specify the shapes and proportions of objects independent of exact dimensions. Free practice questions for Precalculus - Graph the Sine or Cosine Function. Real world uses of hyperbolic trigonometric functions. Specific Expectations. Perform a sine regression using the Ti83/84 Activity #2 - Steamboat problem From given information: 1. Amplitude is first, then sine or cosine, then B, which you get by doing 2pi/10 (period), which reduces to pi/5, then it is the parenthesis with the transformation, and then last but not least is the sinusodial axis. 1) Find properties of sine functions (PC-N. Evaluating Inverse Trigonometric. Graphing Sine and Cosine Functions. \begingroup Basically, the answer is: yes, there are many other periodic functions, and the reason you typically see harmonics (like \sin,e^{i\omega t}) used is because in most simple applications of interest which are easily understandable, either the behavior itself is harmonic, or the behavior is most easily understood in terms of harmonics. First, you must understand that each of these functions has its own graph. In the chapter on Trigonometric Functions, we examined trigonometric functions such as the sine function. First of all, the graph is no longer a sine curve, but there's definitely a pattern to it. It exposes the student to the basic concepts of algebra and trigonometry with an emphasis in practical applications. But just saying "the sine function" doesn't tell you anything about the probability measure in the domain, which is what gets you probabilistic information, including joint distributions. the sine or cosine function and use the equation as a mathematical model to make predictions and interpretations about the real world. Using trigonometric functions to model climate. In ICSA, an elitism strategy is used to select the global solution, and a new updating mechanism for the new solution was proposed. You cannot graph an exponential functions without finding its horizontal asymptote. In addition, how do I know if this the graph of sine or cosine?. functions; increasing/decreasing functions; finding intercepts with axes. real world applications of sinusoids Sinusoids are applicable to the real world. To graph the sine function, we mark the angle along the horizontal x axis, and for each angle, we put the sine of that angle on the vertical y-axis. y-2π -π π 2π 0. The rise and fall of tides can have great impact on the communities and ecosystems that depend upon them. Every angle shares the same sine with another angle (and -sine with two others). • Find the amplitude and period of variation of the sine, cosine, and tangent functions. Sine and cosine functions are periodic functions. 4 COMPLEX NUMBERS Add and subtract complex numbers Multiply and divide complex numbers Simplify powers of i 2. The amplitude is 2. These graphs act as a reference every time you use a trigonometric function. Also for every n π/2 radians (where "n" is an interger) the cosine function passes through y=0. 21 from both sides: Divide both sides by 1. Outcome 5: Set up, solve, and graph equations from problems that require use of trigonometric functions, tangent, sine, and cosine and the Pythagorean Theorem. 3 make connections between the sine ratio and the sine function and between the cosine ratio and the cosine function by graphing the relationship between angles from 0º to 360º and the corresponding sine ratios or cosine ratios, defining this relationship as the function f(x) =sinx or f(x) =cosx, and explaining why the relationship is a function. The sine and cosine of an angle are both circular functions, and they are the fundamental circular functions. Full Range Fourier Series - various forms of the Fourier Series 3. Focus on creating a graph and formula from a story (starting from time 6:55) You Do Now: Warmup - Sketching a Cosine Wave. Learn how to construct trigonometric functions from their graphs or other features. Physical phenomenon such as tides, temperatures and amount of sunlight are all things that repeat themselves, and so are easily modeled by sine and cosine functions (collectively, they are called "sinusoidal functions"). 2-use technology, graphs, and tables to compare sine graphs. The property represented here is based on the right triangle and the two acute or complementary angles in a right triangle. The Code is divided into 50 titles which represent broad areas subject to Federal regulation. Sinusoidal Modeling - Real World Application Project. Then we looked at identifying the amplitude, frequency and period from a given cosine (or sine) function. Time-saving lesson video on Sine and Cosine Functions with clear explanations and tons of step-by-step examples. One way to distinguish sounds is to measure frequency. sinA sin B sin(AB) cosAcosB cos(AB) 59. 9) Graph sine and cosine functions (PC-N. 534), where x is the day of the year starting with January 1 as Day 1. Graph functions, plot data, evaluate equations, explore transformations, and much more - for free! Start Graphing Four Function and Scientific Check out the newest additions to the Desmos calculator family. Recall that the sine and cosine functions relate real number values to the x– and y-coordinates of a point on the unit circle. Day 7 HW F 15-Dec Graphing Cosine Functions NC. Determine the values for angles between 0° and 360°. Chapter Test. Evaluate and graph logistic growth functions. Find the amplitude and period of the function. Applications of Trigonometric Graphs; 6. Khan Academy is a 501(c)(3) nonprofit organization. Graph one period of s(x) = –cos(3x) The "minus" sign tells me that the graph is upside down. There are many uses of sin,cos,tan in real life. Trigonometric Functions • Understand some real-world situations that demonstrate periodic behavior. Write a cosine function that describes the height of the nail above ground as a function of angular distance. Trigonometry Graphing Trigonometric Functions Applications of Radian Measure. Discovery of Euler's Equation. Example 15 6. The new 3rd edition of Precalculus: Algebra and Trigonometry prepares students for further study in mathematics, the sciences, engineering, the social sciences, and other disciplines. By correctly labeling the coordinates of points A, B, and C, you will get the graph of the function given. 'transform') each of the three trigonometric functions discussed thus far: sine, cosine, and tangent. 25 scaffolded questions on equation graph involving amplitude and periodplus model problems explained step by step. Amplitude and Period of Sine and Cosine Functions BOATING A signal buoy between the coast of Hilton Head Island, South Carolina, and Savannah, Georgia, bobs up and down in a minor squall. Graphs of y = a sin x and y = a cos x 2. Yep, sine and cosine are practically twins. In maths, you have real life applications on any thing that you study. Lesson 7: Applications with Acute Triangles • Identify when to apply the sine and cosine laws given incomplete information about the • Solve a multistep problem that involves o two or more applications of the sine or cosine laws,. Sine Law and Cosine Law Find each measurement indicated. Free trigonometry worksheets, in PDF format, with solutions to download. Khan Academy is a 501(c)(3) nonprofit organization. stock market values can be formed into a sine, cosine graph (given a constant average) 5. Precalculus. Example: if your cutting down a tree and need to know where it will. I'm not exactly sure what level of mathematics you've studied so far, but sine and cosine are used heavily in math (calculus and functions) and physics (kinematics and dynamics), as well as having some real world applications. 2 The Inverse Trigonometric Functions. The trigonometric functions sine, cosine and tangent calculate the ratio of two sides in a right triangle when given an angle in that triangle. Although you. 4 The Real-Time Transport Protocol 12. the graph of a sine function. C2 create and analyse scatter plots of periodic data. Of course, this was to be expected, but this isn't the real problem; the real problem is that if your function has some crucial points it must pass through (which is certainly true for trigonometry functions), the truncation will move the curve away from those points. Standard 9: Find values of trigonometric functions using the unit circle. The unit circle and values of sine and cosine on the unit circle B. The fruit fly optimization algorithm (FOA) is a well-regarded algorithm for searching the global optimal solution by simulating the foraging behavior …. which means that theta can be any angle in degrees or radians — any real number. 6 u00ad Modeling Real World Data with Sinusoidal Functions Section 6. The unit circle can be used to express the values of sine, cosine, and tangent for π – x,. Students graph transformations of trigonometric functions (sine and cosine), including af(x), f(x) + d, f(x – c), and f(bx) for specific values of a, b, c, and d, in mathematical and real-world problem situations. Graphing Secant & Cosecant. Investigate Transformations of Sine and Cosine Functions Many real-world processes can be modelled with sinusoidal functions. The identities that arise from the triangle are called the cofunctionidentities. Sine and Cosine can be used in everyday life in various ways. Topic:Real world application of sine graphs. Press the button marked "sin" to change to the cosine. ratios, ie. Intro Tangent & Cotangent Graphs. To graph the sine function, we mark the angle along the horizontal x axis, and for each angle, we put the sine of that angle on the vertical y-axis. McGraw-Hill Education Trigonometry Review and Workbook, 1st Edition by William Clark and Sandra Luna McCune (9781260128925) Preview the textbook, purchase or get a FREE instructor-only desk copy. 3 make connections between the sine ratio and the sine function and between the cosine ratio and the cosine function by graphing the relationship between angles from 0º to 360º and the corresponding sine ratios or cosine ratios, defining this relationship as the function f(x) =sinx or f(x) =cosx, and explaining why the relationship is a function. If the set of input values is a finite set then the models are known as discrete models. org (pre-calculus), and Khan Academy (SAT. This builds into learning about graphing and interpreting logarithmic functions and models. Les't compare with. These graphs act as a reference every time you use a trigonometric function. 2) A car's tire has a diameter of 32 inches. The caternary curve (a dangling string/chain) is really just cosh – crasic Oct 30 '10 at 23:48. state the properties for the Sine and Cosine graph identify the values of a, k, c and d and apply it to a sinusoidal function to graph the resulting transformation Lesson:. 14-1 Graphs of Sine and Cosine Example 3: Sound Application Use a sine function to graph a sound wave with a period of 0. pdf), Text File (. A single note can be modeled on a sine curve, and a chord can be modeled with multiple sine curves used in conjunction with one another. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. In the 36 intensively illustrated lectures of Mathematics Describing the Real World: Precalculus and Trigonometry, he takes you through all the major topics of a typical precalculus course taught in high school or college. The graphs of the sine and cosine functions illustrate a property that exists for several pairings of the different trig functions. Then you will learn about modeling trigonometric functions by graphing the sine and cosine functions. Recall from The Other Trigonometric Functions that we determined from the unit circle that the sine function is an odd function because [latex]sin(−x)=−sinx$. Sample records for nino-southern oscillation eventsnino-southern oscillation events «. The inverses of sine, cosine and tangent functions are not functions unless the domains are limited. Chapter Five – Trigonometric Functions What to Expect: A new formula designed for sin and cos graphs, details about frequencies, new vocabulary, explaining periodic trends, and the like. 2]Solve right and oblique triangles including application problems. Outcome 6: Set up and solve exponential and logarithmic equations; then identify and sketch graphs of the functions. Precalculus students are required to understand how sine and cosine functions model the real world. A lot of waves actually follow a sine graph, so we can prove that sinusoidal motion is a real thing in nature. We completed three cosine graphs together on the 13-5 vocab support page. f(x)= Calculus: Early Transcendental Functions Use Exercise 25 to find the moment of inertia of a circular disk of radius a with constant density about a. 0333 170 -170 t e 0. Have them graph these points and note whether or not they are on the same graph. Use a horizontal scale where one unit represents 0. With the sine law. Students will find equations for transformed sine curves and will graph transformed sine functions. Sketch translations of graphs of sine and cosine functions; use sine and cosine functions to model real-life data. Other circular functions can all be derived from the sine and cosine of an angle. 002 s and an amplitude of 3 cm. Examples: • Square wave function • Saw tooth functions Split Functions Much of the behaviour of current, charge and voltage in an AC circuit can be described. The inverses of sine, cosine and tangent functions are not functions unless the domains are limited. You will be observing the graph of r = a + bsin() or r = a + bcos(), where a and b are non-zero real numbers. The period of sine and cosine functions is 2pi/B. graph the sine and cosine functions with varied amplitudes and periods. Now we must stretch out the amplitude of the sine graph. Sketch the graphs of the sine and cosine functions and state their domain and range. Applications of Sine and Cosine Graphs Learning Task: Trigonometry functions are often used to model periodic data. The sum formula for tangent states that the tangent of the sum of two angles equals the sum of the tangents of the angles divided by 1 minus the product of the tangents of the angles. Graphs the 6 Trigonometric Functions. Rational functions: asymptotes and excluded values (PC-E. It is given by parameter #a# in function #y = asinb(x - c) + d or y = acosb(x - c) + d# •The period of a graph is the distance on the x axis before the function repeats itself. But just as you could make the basic quadratic, y = x 2, more complicated, such as y = -(x + 5) 2 - 3, so also trig graphs can be made more complicated. Trigonometry plays a major role in musical theory and production Real world examples of cosine functions. We typically use degree measures when measuring angles, however we can use radian angle measure as an alternate way of measuring angles in advanced math courses. KEY STANDARDS ADDRESSED: MA3A3. Many compression algorithms like JPEG use fourier transforms that rely on sin and cos. Trigonometric functions can be used to model real world data that is periodic in nature. Since the sine function y = sin t begins at the origin (when t = 0), sine is the more convenient of the two for this purpose. First, find two easily solvable angles that add up to 75. 1 The Inverse Sine, Cosine, and Tangent Functions. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. ' Or: sin(30 °) = 0. Domain and Range of Sine and Cosine The domain of sine and cosine is all real numbers, The range of sine and cosine is the interval [-1, 1] x or (, ) Both these graphs are considered sinusoidal graphs. Find Period and Amplitude 2. Not only is this a real-world visual representation of both graphs, but it also demonstrates that a cosine graph is simply a sine graph at a displacement of 90°. The table of values for the functions. Find the frequency in hertz for this sound wave. The graph of the sine function is called the sine curve. The property represented here is based on the right triangle and the two acute or complementary angles in a right triangle. 7 sketch graphs of y = af (k(x - d)) + c by applying one or more transformations to the graphs of f(x) =sinx and f(x) =cosx, and state the domain and range of the transformed functions 2. It is the reciprocal of the. Hyperbolic functions, also called hyperbolic trigonometric functions, the hyperbolic sine of z (written sinh z); the hyperbolic cosine of z (cosh z); the hyperbolic tangent of z (tanh z); and the hyperbolic cosecant, secant, and cotangent of z. Each of the six trig functions is equal to its co-function evaluated at the complementary angle. However, if we stretch the sine graph and change the amplitude, it just might work. In problems 12 & 13, the graphs of the sine and cosine functions are waveforms like the figure below. Since the multiplier out front is an "understood" –1, the amplitude is unchanged. If we change the number of cycles the wave completes every second -- in other words, if we change the period of the sine wave -- then we change the sound. Draw three right triangles and label the angle and two sides that apply to the sine, cosine and tangent functions respectively. Four Function Scientific. The hyperbolic functions take a real argument called a hyperbolic angle. Discovery of Euler's Equation. Cumulative Review. Lesson 9-11: Graphing Trigonometric Functions and Applications in Real World Contexts Learning Goals #8, 9, 10: How do I use the critical values of the Sine and Cosine curve to assist in in graphing Sine and Cosine functions? How do I contextualize Trig Functions in real life situations? Warm-Up! WITH A PARTNER. Water Depth Word Problem Modeled with Cosine Sine Function. One example of applying tangent functions to solve a real world problem is:- Find the gradient and the actual length of a path represented as x cm (a known measurable quantity) in the 1 : n (a known given quantity) scaled contour map. Sound waves travel in a repeating wave pattern, which can be represented graphically by sine and cosine functions. The SOH means that the sine of an angle equals the side opposite the angle over the triangle's hypotenuse. One way to distinguish sounds is to measure frequency. Animated gifs showing relationships between functions and geometry eg trigonometric functions How the unit circle gives us the sine wave Teaching math with the Drum Loop Real sinusoid on a timebase, formed by a linear increment of complex argument in time. Let’s look at a few examples of real-world situations that can best be modeled using trigonometric functions. This is an example of how a Graph of a Cosine Function looks like:. Use sinusoidal functions to solve problems. which means that theta can be any angle in degrees or radians — any real number. To understand how the absolute value can be applied to the real world, we’ll look at two topics: What is considered normal with regard to the temperature of the human body and; Fuel economy of two vehicles: a Honda Odyssey and a Nissan Altima. Periodicity of trig functions. Trigonometric identities like finding. Bookmark File PDF Functions Modeling Change Answers Functions Modeling Change Answers Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Modeling with Functions Part 1 We model real life scenarios of sales and volume of a box with functions. 4 Solving Word Problems Involving Sine or Cosine Functions. By finding a few key points or aspects of the graph, any of the real-life problems we have today can be explained mathematically and much of the vibrations surrounding us can be better understood. active oldest votes. 3π 2 = = Period Amplitude. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. Read: 'Zero point five is the sine of thirty degrees. Sound waves travel in a repeating wave pattern, which can be represented graphically by sine and cosine functions. Academic Precalculus. functions to model real world problems A cosine graph appears. Topic:Real world application of sine graphs. Trigonometric functions can be used to solve right triangles. Overview of Fourier Series - the definition of Fourier Series and how it is an example of a trigonometric infinite series 2. Maths Applications: Graph sketching. Euler's equation (formula) shows a deep relationship between the trigonometric function and complex exponential function. | 2020-08-11T18:15:23 | {
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https://www.coursehero.com/file/22489520/Sequences-and-Series/ | # Sequences_and_Series - 11 Sequences and Series Consider the...
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Chapter 8 / Exercise 14
Essential Calculus: Early Transcendentals
Stewart
Expert Verified
11 Sequences and Series Consider the following sum: 1 2 + 1 4 + 1 8 + 1 16 + · · · + 1 2 i + · · · The dots at the end indicate that the sum goes on forever. Does this make sense? Can we assign a numerical value to an infinite sum? While at first it may seem di cult or impossible, we have certainly done something similar when we talked about one quantity getting “closer and closer” to a fixed quantity. Here we could ask whether, as we add more and more terms, the sum gets closer and closer to some fixed value. That is, look at 1 2 = 1 2 3 4 = 1 2 + 1 4 7 8 = 1 2 + 1 4 + 1 8 15 16 = 1 2 + 1 4 + 1 8 + 1 16 and so on, and ask whether these values have a limit. It seems pretty clear that they do, namely 1. In fact, as we will see, it’s not hard to show that 1 2 + 1 4 + 1 8 + 1 16 + · · · + 1 2 i = 2 i 1 2 i = 1 1 2 i 255
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Chapter 8 / Exercise 14
Essential Calculus: Early Transcendentals
Stewart
Expert Verified
256 Chapter 11 Sequences and Series and then lim i →∞ 1 1 2 i = 1 0 = 1 . There is one place that you have long accepted this notion of infinite sum without really thinking of it as a sum: 0 . 3333 ¯ 3 = 3 10 + 3 100 + 3 1000 + 3 10000 + · · · = 1 3 , for example, or 3 . 14159 . . . = 3 + 1 10 + 4 100 + 1 1000 + 5 10000 + 9 100000 + · · · = π . Our first task, then, to investigate infinite sums, called series , is to investigate limits of sequences of numbers. That is, we o cially call i =1 1 2 i = 1 2 + 1 4 + 1 8 + 1 16 + · · · + 1 2 i + · · · a series, while 1 2 , 3 4 , 7 8 , 15 16 , . . ., 2 i 1 2 i , . . . is a sequence, and i =1 1 2 i = lim i →∞ 2 i 1 2 i , that is, the value of a series is the limit of a particular sequence. While the idea of a sequence of numbers, a 1 , a 2 , a 3 , . . . is straightforward, it is useful to think of a sequence as a function. We have up until now dealt with functions whose domains are the real numbers, or a subset of the real numbers, like f ( x ) = sin x . A sequence is a function with domain the natural numbers N = { 1 , 2 , 3 , . . . } or the non-negative integers, Z 0 = { 0 , 1 , 2 , 3 , . . . } . The range of the function is still allowed to be the real numbers; in symbols, we say that a sequence is a function f : N R . Sequences are written in a few di ff erent ways, all equivalent; these all mean the same thing: a 1 , a 2 , a 3 , . . . { a n } n =1 { f ( n ) } n =1 As with functions on the real numbers, we will most often encounter sequences that can be expressed by a formula. We have already seen the sequence a i = f ( i ) = 1 1 / 2 i ,
11.1 Sequences 257 and others are easy to come by: f ( i ) = i i + 1 f ( n ) = 1 2 n f ( n ) = sin( n π / 6) f ( i ) = ( i 1)( i + 2) 2 i Frequently these formulas will make sense if thought of either as functions with domain R or N , though occasionally one will make sense only for integer values. Faced with a sequence we are interested in the limit lim i →∞ f ( i ) = lim i →∞ a i . We already understand lim x →∞ f ( x ) when x is a real valued variable; now we simply want to restrict the “input” values to be integers. No real di ff erence is required in the definition of limit, except that we specify, per- haps implicitly, that the variable is an integer. Compare this definition to definition 4.10.4. | 2021-10-21T04:32:28 | {
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https://math.stackexchange.com/questions/1929320/finding-the-two-planes-that-contain-a-given-line-and-form-the-same-angle-with-tw | # Finding the two planes that contain a given line and form the same angle with two other lines
I am asked the following question:
There are two planes $\pi_1$ and $\pi_2$ such that each of them contain the line $t \{ (-1,4,0) + \lambda (1,2,3)$ and form the same angle with the lines $r \{ (0,0,0) + \beta (1,1,1)$ and $s \{ (-1,2,3) + \alpha (1,1,-1)$. Find the angle formed between those planes
My solution:
Let's say the normal vector of $\pi$ is $\vec{n} = (a,b,c)$. Since the angle between the plane(s) and the lines are equal,
\begin{align*} \measuredangle \left( \pi, r \right) &= \measuredangle \left( \pi, s \right)\\ \frac{\left( a,b,c \right) \cdot (1,1,1)}{\sqrt{a^2+b^2+c^2} \ \sqrt{3}} &= \frac{\left( a,b,c \right) \cdot (1,1,-1)}{\sqrt{a^2+b^2+c^2} \ \sqrt{3}}\\ a+b+c &= a+b-c\\ c &= 0 \end{align*}
Also, since $t$ is contained in the plane, the dot product between its vector and the normal vector of the plane is zero:
\begin{align*} (a,b,c) \cdot (1,2,3) &= 0\\ a+2b+3c&=0\\ a &= -2b \end{align*}
Assuming that the normal vector has length one,
$$a^2 + b^2 + c^2 = 1$$
So we have three equations with three variables and the solution for those is
$$a = \mp \frac{2}{\sqrt{5}} \quad a = \pm \frac{1}{\sqrt{5}} \quad c = 0$$
so the normal vectors of the two planes found are
$$\left( \frac{2}{\sqrt{5}} , - \frac{1}{\sqrt{5}} , 0 \right) \quad \left( - \frac{2}{\sqrt{5}} , \frac{1}{\sqrt{5}} , 0 \right)$$
and the angle between them is
$$\theta = \arccos \left(\frac{\left \vert - \frac{2}{\sqrt{5}} \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} \frac{1}{\sqrt{5}} + 0 \right \vert}{1} \right) = \arccos(1) = 0^\circ$$
My answer does not agree with the textbook's solution.
Textbook's solution:
$$\theta = \arccos \left( \frac{9}{\sqrt{95}} \right)$$
Did I make a mistake somewhere?
Thank you.
$$|(a,b,c) \cdot (1,1,1)| = |(a,b,c) \cdot (1,1,-1)| \\ |a+b+c| = |a+b-c|$$ yields
$$a+b+c = a+b-c \quad \text{ and } \quad a+b+c = -a-b+c$$ | 2020-07-14T20:50:56 | {
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# A bag has only r red marbles and b blue marbles. If five red marbles
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A bag has only r red marbles and b blue marbles. If five red marbles [#permalink]
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25 Jul 2018, 23:16
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A bag has only r red marbles and b blue marbles. If five red marbles and one blue marble are added to the bag, and if one marble is then selected at random from the bag, then the probability that the marble selected will be blue is represented by
A. b/r
B. b/(b + r)
C. (b + 1)/(r + 5)
D. (b + 1)/(r + b + 5)
E. (b + 1)/(r + b + 6)
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Re: A bag has only r red marbles and b blue marbles. If five red marbles [#permalink]
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25 Jul 2018, 23:20
Number of red marbles = r
Number of blue marbles = b
five red marbles and one blue marble are added to the bag
Number of red marbles = r + 5
Number of blue marbles = b + 1
Total number of marbles = r + 5 + b + 1 = r + b + 6
Probability of selecting a blue marbles = $$\frac{b + 1}{r + b + 6}$$
Hence option E
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Re: A bag has only r red marbles and b blue marbles. If five red marbles [#permalink]
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26 Jul 2018, 01:02
Bunuel wrote:
A bag has only r red marbles and b blue marbles. If five red marbles and one blue marble are added to the bag, and if one marble is then selected at random from the bag, then the probability that the marble selected will be blue is represented by
A. b/r
B. b/(b + r)
C. (b + 1)/(r + 5)
D. (b + 1)/(r + b + 5)
E. (b + 1)/(r + b + 6)
Red marbles = r
Blue marbles = b
Changing Situation :
Red = r + 5
Blue = b + 1
Total = r + s + 6
we are asked to pick up a blue marbles = $$\frac{( b + 1)}{r + s + 6}$$
Thus the best answer is E.
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Re: A bag has only r red marbles and b blue marbles. If five red marbles [#permalink]
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30 Jul 2018, 10:51
Bunuel wrote:
A bag has only r red marbles and b blue marbles. If five red marbles and one blue marble are added to the bag, and if one marble is then selected at random from the bag, then the probability that the marble selected will be blue is represented by
A. b/r
B. b/(b + r)
C. (b + 1)/(r + 5)
D. (b + 1)/(r + b + 5)
E. (b + 1)/(r + b + 6)
After 5 red marbles and 1 blue marble are added to the bag, the bag has b + 1 blue marbles and r + b + 5 + 1 = r + b + 6 total marbles. So the probability of selecting a blue marble is (b + 1)/(b + r + 6).
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Re: A bag has only r red marbles and b blue marbles. If five red marbles &nbs [#permalink] 30 Jul 2018, 10:51
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https://johncarlosbaez.wordpress.com/2016/03/26/probability-puzzles-part-3/ | ## Probability Puzzles (Part 3)
Here’s a puzzle based on something interesting that I learned from Greg Egan. I’ve dramatized it a bit.
Traditional Tom and Liberal Lisa are dating. They discuss their plans for having children:
Tom: I plan to keep having kids until I get two sons in a row.
Lisa: What?! That’s absurd. Why?
Tom: I want two to run my store when I get old.
Lisa: Even ignoring your insulting assumption that only boys can manage your shop, why in the world do you need two in a row?
Tom: From my own childhood, I’ve learned there’s a special bond between sons who are next to each other in age. They play together, they grow up together… they can run my shop together.
Lisa: Hmm. Well, then maybe I should have children until I have a girl followed directly by a boy!
Tom: What?!
Lisa: Well, I’ve observed that something special happens when a boy has an older sister, with no intervening siblings. They play together, they grow up together… and maybe he learns not to be such a sexist pig!
They decide they are incompatible, so they split up and each one separately tries to find a mate who will go along with their reproductive plan.
Now for some puzzles. In these puzzles, assume that each time someone has a child, they have a 50% chance of having either a daughter or a son. Also assume each event is independent: that is, the gender of any children so far has no effect on that of later ones. Also ignore twins and other tricky issues.
Puzzle 1. If Tom carries out his plan of having children until he has two consecutive sons, and then stops, what is the expected number of children he will have?
Puzzle 2. If Lisa carries out her plan of having children until she has a daughter followed directly by a son, and then stops, what is the expected number of children she will have?
Puzzle 3: Which is greater, Tom’s expected number of children or Lisa’s? Or are they equal?
For maximum benefit, try to answer Puzzle 3 before doing the calculations required to answer Puzzles 1 or 2.
### 64 Responses to Probability Puzzles (Part 3)
1. It’s late here right now so for now only preliminary quick guesses.
If you wanna have a more formal go, maybe don’t look at this until you want a hint or a distraction, depending on whether my guesses are right.
Puzzle 3 guess:
Since any particular two-birth-sequence is equally likely (1/4 each), I’d guess that both scenarios are equally likely.
Puzzle 1:
Possible sequences:
SS (1/4)
DSS (1/8)
DDSS (1/16)
SDSS (1/16)
DDDSS (1/32)
DSDSS (1/32)
SDDSS (1/32)
DDDDSS (1/64)
DDSDSS (1/64)
DSDDSS (1/64)
SDDDSS (1/64)
SDSDSS (1/64)
DDDDDSS (1/128)
DDDSDSS (1/128)
DDSDDSS (1/128)
DSDDDSS (1/128)
SDDDDSS (1/128)
DSDSDSS (1/128)
SDDSDSS (1/128)
SDSDDSS (1/128)
This looks suspiciously like $\sum_{n=1}^\infty \frac{\text{fib}\left(n\right)}{2^{n+1}}$
There is probably some nice combinatorial argument to make to prove this.
This sum – I didn’t prove but just quickly looked up – sums up to 1, so that’s good.
The expectation value version of that sum just adds an n:
$\sum_{n=1}^\infty n \frac{\text{fib}\left(n\right)}{2^{n+1}}$
And that appears to sum to 5.
Puzzle 2:
Possible sequences:
DS (1/4)
SDS (1/8)
DDS (1/8)
SSDS (1/16)
SDDS (1/16)
DDDS (1/16)
SSSDS (1/32)
SSDDS (1/32)
SDDDS (1/32)
DDDDS (1/32)
SSSSDS (1/64)
SSSDDS (1/64)
SSDDDS (1/64)
SDDDDS (1/64)
DDDDDS (1/64)
The series appears to be
$\displaystyle{ \sum_{n=1}^\infty \frac{n}{2^{n+1}} }$
which also sums to 1.
And the expectation value
$\displaystyle{ \sum_{n=1}^\infty \frac{n^2}{2^{n+1}} }$
goes to 3. So Lisa will actually most likely have fewer kids than Tom. Therefore my above prediction was wrong.
Furthermore, if I did this right, Lisa can expect her value to hover around 3 by 2, whereas for Tom that uncertainty sits around 4.69.
So in a fairly pessimistic (assuming you want to get there asap) outcome would be 5 trials for Lisa but 10 for Tom.
(I used that $\sigma^2 = \langle-^2\rangle$ but I’m tired and so possibly error-prone.)
• Michael Nelson says:
Here’s how fibonacci shows up:
You can count the number of ways of Tom winning via a string of n turns by breaking it up into two cases:
…….DSS
….DSDSS
The first case you counted already in the previous string of length n-1. And the second case you counted already from the string before that of length n-2.
• This is how I also ended up solving it, getting the Fibonnaci sequence immediately by using a construction approach to compute the number of binary strings (1 == boy, 0 == girl) that had the constraint of ending in ’11’ but with no other ’11’ in them. The construction as you show has the two ways to generate N+1: extend all the length N cases with a ‘0’ in the front and extend the N-1 cases adding ’10’ to the front. This means the number of ways to generate an appropriate string of length N+1 is Fib(N) since Count(N+1) = Count(N -1) + Count(N). And the total strings of length N+1 is $2^{N+1}$. Dividing gives you the probability of each string length (number of kids) and the expectation is as was shown above as well. And of course phi (the Golden Mean) shows up in the solution of the characteristic of the Fibonnaci recurrence so ultimately I ended up using it to solve this. Pretty cool!
2. It’s much quicker, Kram, to model each problem as a state machine.
Let $T_0$ and $T_1$ be Tom’s expectation of future children after zero and one son respectively. We can say immediately that $T_0 = \frac{1}{2}(s+T_1) + \frac{1}{2}(d+T_0)$ and $T_1 = \frac{1}{2}(s) + \frac{1}{2}(d+T_0)$. The solution, if I haven’t blundered, is $T_0 = 3d+3s$.
For Lisa: $L_0$ is her expectation before her first daughter; $L_1$ is her expectation after any daughter. $L_0 = \frac{1}{2}(s+L_0) + \frac{1}{2}(d+L_1)$ and $L_1 = \frac{1}{2}(d+L_1) + \frac{1}{2}(s)$. Solution: $L_0 = 2d+2s$.
Lisa can expect two sons and two daughters; Tom can expect three daughters and three sons.
The qualitative difference is that for Tom a daughter after a son puts him back a step, while for Lisa a child of the sex not hoped for leaves her in the same position.
• John Baez says:
This looks impressively efficient, Anton! But what does an equation like
$T_0 = \frac{1}{2}(s+T_1) + \frac{1}{2}(d+T_0)$
actually mean? I guess I’m lacking the necessary familiarity with state machines to know what type of entities these variables represent, and what addition means. I know what a state machine is; I’m just not familiar with this kind of equation.
I can guess that $T_n$ is allowed to be any expression of the form $a s + b d$ where $a,b \ge 0$, and
$T_n = a s + b d$
means “after having had $n$ sons, Tom expects in the future to have a total of $a$ sons and $b$ daughters”.
But then I don’t know why
$T_0 = \frac{1}{2}(s+T_1) + \frac{1}{2}(d+T_0)$
should be true. It seems to say “after having no sons, Tom expects in the future to have the same thing as if he had a 50% chance of having one more son than what he expected after having one son, and a 50% chance of having one more daughter than what he expected after having no sons”.
The horrible quoted phrase here certainly shows why a more concise notation is a good thing! It reminds me of what happens when you try to state the quadratic equation in ‘plain English’ without variables.
But the problem is, I don’t see why it should be true.
• My subscripts mean only that state 0 must occur before state 1 can occur. I could also say explicitly $T_2 = L_2 = 0$; does that make anything clearer?
My equations are the invariants of the tree of futures. The one that you query can be read as: “From state 0, the branches of Tom’s future are: with probability 1/2, he has a son and moves to state 1; and with probability 1/2, he has a daughter and remains in state 0.” As long as his sequence of daughters continues, his expectation of future children remains the same, so the definition of T0 is recursive. From state 1 (one son) Tom cannot remain in state 1; he must either terminate or go back to 0. After a daughter, previous sons are irrelevant to Tom’s future tree, which is why there are not more states in this model.
• Chris Upshaw says:
Shouldn’t that be s * T1 then, if its a parallel notation to sum of products datatype notation? Its certainly not the same operation as the middle ‘+’.
• That’s cos I don’t know nothin bout notating no datatypes.
• John Baez says:
I found Anton’s trick very hard to understand at first, mainly because the variables and operations in his equations had not been explained (and I didn’t already know this trick). Like Greg, I found XJY’s comment here helpful. Even then, I had to hit myself over the head with a rock a few times before I really understood what was going on.
I’ve spent a lot of time thinking about other aspects of Markov processes, but I never thought about this particular trick. It’s very nice, but I think I need to digest it and blend it with other things I know, like generating functions and—yes—recursively defined data types.
• Anton,
This is a really cool solution! Not only do you compute Lisa’s expected # of kids, you compute how many sons and daughters she can expect. I have never seen this concept of expressing the invariants of the tree of futures before. Do you have a good reference for learning more about this? Is it standard CS?
Thanks for posting this.
• Rob, sorry, I have no idea where I got the, er, idea.
“Invariant” may be a misnomer; thinking about it further, I reckon the analogy with the loop invariants of CS is not so strong.
I may have misunderstood the problem but it seems to me that as soon as Tom or Lisa reach their objective they STOP. Then Tom can have 2 sons in a row and it is over. On the other hand if they have a daughter Lisa can get a son and they stop, Tom can never get two sons.
• John Baez says:
Wow, you assumed that Tom and Lisa are married, and that their plans affect each other’s activities!
I never thought that: I was imagining two people at a bar, who obviously hate each other, each discussing their own separate plans. But now I see that there’s an ambiguity in the phrase “their plans for having children”. I’ll fix that!
• John Baez says:
Upon rereading what I wrote, I see there’s no real ambiguity in the puzzles as stated. One asks what will happen if Tom carries out his plan. The other asks what will happen if Lisa carries out her plan. There’s nothing about what will happen if both of them try to carry out their conflicting plans.
Nonetheless I added some stuff to make it crystal clear that Tom and Lisa are not going to be having children together.
4. jamievicary says:
I deliberately haven’t worked this out formally, because it’s more fun to see if intuition leads to a right answer. Write their progeny as a string in D and S. It’s ‘easy’ for a string of length n to lack a SS substring, as it could have no Ss, or just one S, or just two Ss spaced at least one apart, etc. But to lack a DS substring is relatively ‘hard’: you have to have all the sons before all the daughters. So, holding out for two sons in a row will take longer than waiting for a daughter followed by a son.
I guess the reason it’s a bit paradoxical is that for any two adjacent children, SS is just as likely as DS. I suppose the reason is that combinatorially, like-gendered children are indistinguishable. In reality this is not the case. They should just ask for “two children adjacent in age, both with the temperament to run a shop”!
(By the way, I found this post cognitively difficult to write, since on parenting forums DD and DS often meen ‘darling son’ and ‘darling daughter’.)
• John Baez says:
Jamie wrote:
I guess the reason it’s a bit paradoxical is that for any two adjacent children, SS is just as likely as DS.
Right, that was supposed to make the problem cognitively difficult.
(By the way, I found this post cognitively difficult to write, since on parenting forums DD and DS often meen ‘darling son’ and ‘darling daughter’.)
I never knew! I hadn’t expected that to cause cognitive difficulties. Do SD and SS mean ‘stupid daughter’ and ‘stupid son’?
(I would probably be banned from the forum for that joke.)
• Tamfang says:
or Silly or Sweet
• Charles Jackson says:
Well, you did make the problem cognitively difficult for me.
I did #3 first. I thought P(SS) = P(DS) => Expected births to first SS is same as for DS. WRONG—but
I did 2 relatively quickly with something like Anton’s approach above.
Then I looked at the answers in the comments. Oh, oh. I thought about it for awhile and I think I see an easy explanation for the answer.
Consider the outcomes with up to three children. Make a little truth table-like figure with all 8 possible sequences of combinations of boy and girl babies. See below. Notice that there are 4 occurrences of the sequence DS in the table—each of which is the first occurrence of DS in the sequence containing it. (capitalized in table below)
Birth Order
1 2 3
d d d
d D S
D S d
D S s
s d d
s D S
s s d
s s s
Now, search for SS, but only the 3 that are the first occurence in time.
d d d
d d s
d s d
d S S
s d d
s d s
S S d
S S s
The difference (only 3 SS versus 4 DS) arises from the second SS that is a subsequence of the SSS sequence
d d d
d d s
d s d
d s s
s d d
s d s
s s d
s S S
So, if the parties were restricted to 3 or fewer babies, Lisa would have a 50% chance of getting DS, but Tom would only have a 3/8 chance of getting SS.
Of course, if they both had three babies, Tom would have a 1/8 chance of getting SS twice, but Lisa would have zero chance of getting DS twice.
At the birth of the second baby they each have a 25% chance of getting their desired outcome. On the third baby, Lisa would also have a 25% chance but Tom would only have a 12.5% chance.
Lisa’s probability of terminating the process at 3, 4, or 5 is higher than the probability for Tom. This pushes up the expected number for him relative to her.
Chuck
PS Thanks for your blog. I only read some of the posts but always enjoy those I complete. No doubt if I had more time and patience I get even more enjoyment from it.
• John Baez says:
Nice. I’m glad you’re enjoying this blog!
• Jenny Meyer says:
The DD and DS tags are just a convention for anonymizing your family members. The D can be heavily ironic in context.
(The picture of the Duggars was enough to make this too cognitively difficult for me.)
5. Michael Nelson says:
I think I understand. If we think of this as a game, where Lisa needs to get the sequence DS and Tom needs to get the sequence SS, then Lisa will certainly have the advantage. As soon as a D shows up, Tom will lose. On the other hand, if a S shows up, Lisa can still win. So Lisa has a greater chance of winning this game.
• John Baez says:
I’m not assuming that Tom and Lisa are having children with each other: they are two separate people, each pursuing their own separate agenda with their own spouse, each assuming their spouse will knuckle under and go along with their insane plan.
Nonetheless you can still think of it as a game, where we imagine a random stream of babies being produced and both Tom and Lisa saying when they, personally, would prefer the stream to stop.
• Michael Nelson says:
Yes, this is what I meant. You phrased it better for me. I think I have a better way to think about this now. The number of ways that Lisa can win at the nth spot is just n. For example:
DDDDDS
SDDDDS
SSDDDS Lisa wins at 5th spot
SSSDDS
SSSSDS
DDDDDDS
SDDDDDS
SSDDDDS
SSSDDDS Lisa wins at 6th spot
SSSSDDS
SSSSSDS
The number of ways that Tom can win at the nth spot involves the fibonacci numbers though. For example:
DDDDSS
SDDDSS
DSDDSS Tom wins at 5th spot
DDSDSS
SDSDSS
DDDDDSS
SDDDDSS
DSDDDSS
DDSDDSS Tom wins at 6th spot
SDSDDSS
SDDSDSS
DSDSDSS
DDDDDDSS
SDDDDDSS
DSDDDDSS
DDSDDDSS Add D just before SS in 6th spot wins
SDSDDDSS
SDDSDDSS
DSDSDDSS
plus Tom wins at 7th spot
SDSDSDSS
DDSDSDSS
DSDDSDSS Replace the last S in 5th spot wins with DSS
SDDDSDSS
DDDDSDSS
That’s pretty cool that the fibonacci numbers are involved.
6. Dave Doty says:
I worked it out like this. I am nagged by the possibility that there is a more elegant simple way to do it.
Each is picking a string from the alphabet $\{ s,d \}$. Both have a list of “allowed” strings of length $n$ that indicate they must keep going.
Lisa omits from this list all strings with the substring $sd$, and Tom omits all strings with the substring $ss$. Already we can see that Lisa has fewer allowed strings than Tom, and will therefore wait a shorter time before stopping, because to remain in an allowed string, she can only transition once (from $s$ to $d$), whereas Tom can transition between $s$ and $d$ an arbitrary number of times.
Lisa’s allowed strings match the regular expression $s^*d^*$; e.g., for length 5, her 6 allowed strings are
$sssss, ssssd, sssdd, ssddd, sdddd, ddddd$
In general, Lisa has $n+1$ allowed strings of length $n$, so probability $\frac{n+1}{2^n}$ that a given length $n$ string will be allowed. If the string is $s^*$, then either of $s$ or $d$ is allowed next, whereas if it ends in a $d$, only a $d$ is allowed next. So the probability that Lisa ends on length $n$ is the probability that she does not end before (i.e., the length $n-1$ prefix is allowed) times the probability that the whole length $n$ string is disallowed, conditioned on length $n-1$ prefix being allowed. This is
$\displaystyle{ \frac{1}{2^{n-1}} \cdot 0}$
(the probability the string is $s^{n-1}$, times probability 0 since she will not stop after $s^{n-1}$), plus
$\displaystyle{ \frac{n}{2^{n-1}} \cdot \frac{1}{2} }$
(the probability the string is $s$‘s followed by at least one $d$, times the probability that the next symbol is $s$).
So Lisa’s probability of stopping after $n$ children is $\frac{n}{2^n}$.
Thus her expected stopping time is
$\displaystyle{ \sum_{n=2}^\infty n \cdot \frac{n}{2^n} = \sum_{n=2}^\infty \frac{n^2}{2^n} = \frac{11}{2} }$
Tom’s allowed strings omit the substring $ss$. We can count recursively how many strings of length $n$ satisfy this, denoted by $T_n$. If the first symbol is $d$, then all allowed strings of length $n-1$ could follow, of which there are $T_{n-1}$. If the first symbol is $s$, then the next symbol must be $d$, and any allowed string of length $n-2$ could follow, of which there are $T_{n-2}$. So $T_n = T_{n-1} + T_{n-2}$, the $n$‘th Fibonacci number, which is
$\displaystyle{ \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^n \sqrt{5}} }$
So Tom’s probability of stopping after $n$ children is the probability that the last three symbols are $dss$ (if it were $sss$ he would have stopped at the second $s$) the prefix of length $n-3$ is allowed, which is
$\displaystyle{ \frac{T_n}{2^n} = \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^{2n} \sqrt{5}} }$
Thus his expected stopping time is
$\displaystyle{ \sum_{n=2}^\infty n \cdot \frac{(1+\sqrt{5})^n - (1-\sqrt{5})^n}{2^{2n} \sqrt{5}} = \frac{19}{2} }$
• John Baez says:
Wow, what a tour de force! You are not getting the officially approved correct answers, but the officially approved correct answers do involve Fibonacci numbers, at least in the most direct approach.
7. Bruce Smith says:
I carefully solved each puzzle before reading beyond it (in case of spoilers), so puzzle 3 was not very interesting! (I had thought of it while starting to work on puzzle 2, but didn’t have the benefit of your stating it was worthwhile to work on it first, so I didn’t do so.)
Another mistake — I initially assumed Tom and Lisa were married and were discussing their joint plan for having kids! This leads to puzzle 4 — suppose they do marry (before having any kids) and agree to have kids until both their criteria are satisfied…
• John Baez says:
I think it’s too late for me to make Puzzle 3 into Puzzle 1 without causing massive confusion.
I have added verbiage to the puzzle to make it clear that Tom and Lisa are not married and never will be.
Nonetheless, Puzzle 4 is interesting, and will doubtless lead to an even larger expected family.
• For that case I get seven expected children, again equally divided.
Next question: For each scenario, what are the variances? Do the variance in daughters and the variance in sons add up to the variance in total children? I don’t know how to figure these.
• Unless I made an error, I already did the variances – or, well, the standard deviations which are just the square roots of the variances – in my reply (first one). However, all the other contributions here are much more fleshed out and nice. My approach was really just trying a couple and then guessing. It seems like the results were right, however it’s nice to read in the others’ replies why they are right.
To get the variance, if you have all the sub-probabilities, you just do:
Probability: $P=\sum_{n \in \mathbb{N}} p\left(n\right) = 1$
Expectation: $E = \sum_{n \in \mathbb{N}} n p\left(n\right)$
Variance: $\sigma^2 = \sum_{n \in \mathbb{N}} n^2 p\left(n\right) - E^2$
Standard Deviation: $\sigma = \sqrt{\sigma^2}$
For some reason, in my original reply, the $\LaTeX$ didn’t work right. Looks like the fixed version still isn’t right though. Hopefully the above works.
• John Baez says:
The LaTeX at the very end of this comment of yours didn’t work, and it was pretty cryptic, so I guessed you were trying to say this:
$\sigma^2 = \langle-^2\rangle$
This means ‘the square of the standard deviation is the expected value of the square of the quantity in question’. (The little dash is a standard symbol for ‘the quantity in question’: a blank to be filled in.) But this is only true for quantities whose mean is zero, which isn’t the case in this problem. So, I must have guessed your meaning incorrectly. I should have written
$\sigma^2 = \langle-^2\rangle - \langle - \rangle^2$
But probably you weren’t trying to use the ‘little dash’ notation, which gets a bit scary in formulas that also contain minus signs!
8. Ilya Surdin says:
Puzzle 1:
suppose d is the expectation value we’re looking for.
then d=0.5(d+1)+0.5(0.5*2+0.5(d+2)) [if a daughter is born first, we’re back to point 1 again, if a son is born first then if another son is born the value is 2, and if a daughter is born second we’re back to square 1].
so we get d = 0.75d +1.5 -> d=6
Puzzle 2:
suppose e is the expectation value we’re looking for.
e = 0.5(e+1) + 0.5f,
where summand 1 is if a son is born, and summand 2 is if a daughter is born, we only need to get a son (ds, dds, ddds, all options are good)
f = 2 [ f=0.51+0.5(f+1) ]
so e = 0.5e + 0.5 +1 -> e=3.
Puzzle 3:
At first sight, I guessed they will be equal.
9. Graham says:
Here’s a solution for Tom’s case using generating functions. Let $T$ be a random variable representing the number of children Tom has. Let $f(s) = \sum p_i s^i$ be the generating function for T, where $p_i = \Pr(T=i)$. The first few $p_i$ are
$p_0 = p_1 = 0, p_2=1/4, p_3=1/8.$
For $i \ge 4$, the sequence must end DSS, and there must be no SS occurring in the first $i-3$. So
$p_4 = (1/8)(1-p_1),$
$p_5 = (1/8)(1-p_1-p_2),$
and so on.
From this it follows that
$p_i = p_{i-1} - (1/8)p_{i-3}$
for $i \ge 4$.
Then by comparing $f(s), sf(s), s^3f(s)$, it is possible to calculate that
$\displaystyle{ f(s) = \frac{-s^2}{s^2+2s-4}. }$
Then
$E(X) = f'(1) = 6$
and
$E(X(X-1)) = f''(1) = 52$
from which $\mathrm{var}(X) = 52 + 6 -6^2 = 22$.
10. “In one word he told me secret of success in mathematics: Generalize!” What if the sex ratio at birth is not 1:1? I’ll get right on that — later.
• John Baez says:
Puzzle 5. For what sex ratio at birth would Tom and Lisa have an equal expected number of children?
(Obviously it needs to be more sons than daughters.)
• Greg Egan says:
John wrote:
Puzzle 5. For what sex ratio at birth would Tom and Lisa have an equal expected number of children?
$\displaystyle{ p_{son} = \frac{\sqrt{5}-1}{2} }$
or:
$\displaystyle{\frac{p_{son}}{p_{daughter}} = \frac{\sqrt{5}+1}{2} }$
To prove this (stealing from and generalising an answer by “XJY”, which is a slightly more coarse-grained version of Anton Sherwood’s approach), we can write the expectation values for the number of additional children that Tom and Lisa will each have after their latest child was a daughter or a son (who didn’t yet complete their target sequence) as:
$T_d = p_{son} (1+T_s) + (1-p_{son}) (1+T_d)$
$T_s = p_{son} +(1-p_{son}) (1+T_d)$
$L_d = p_{son} + (1-p_{son}) (1+L_d)$
$L_s = p_{son} (1+L_s) + (1-p_{son})(1+L_d)$
These are solved by:
$\displaystyle{T_d = \frac{1+p_{son}}{p_{son}^2} }$
$\displaystyle{T_s = \frac{1}{p_{son}^2} }$
$\displaystyle{L_d = \frac{1}{p_{son}} }$
$\displaystyle{L_s = \frac{1}{p_{son}(1-p_{son})} }$
But $T_d$ and $L_s$ are also the expectation values for the number of children each will have in total, because they have 0 out of 2 elements of their target sequence.
So if we solve:
$L_s = T_d$
the positive solution for $p_{son}$ is:
$\displaystyle{ p_{son} = \frac{\sqrt{5}-1}{2} }$
and
$L_s = T_d = 2+\sqrt{5}$
11. Bruce Smith says:
It turns out these puzzles show up (for numbers 1 and 3 sought by Alice and Bob) in this (otherwise mostly unrelated) blog post about primes:
https://rjlipton.wordpress.com/2016/03/26/bias-in-the-primes/
• John Baez says:
• Erica Klarreich, Mathematicians discover prime conspiracy, Quanta, 13 March 2016.
Soundararajan was drawn to study consecutive primes after hearing a lecture at Stanford by the mathematician Tadashi Tokieda, of the University of Cambridge, in which he mentioned a counterintuitive property of coin-tossing: If Alice tosses a coin until she sees a head followed by a tail, and Bob tosses a coin until he sees two heads in a row, then on average, Alice will require four tosses while Bob will require six tosses (try this at home!), even though head-tail and head-head have an equal chance of appearing after two coin tosses.
I decided to dramatize this using children instead of coin tosses.
By the way, I wrote about the ‘prime conspiracy’ here:
• John Baez, Weirdness in the primes, _The n-Category Café 15 March 2016.
There’s a bit of math here that’s not in the Quanta article.
12. Rob says:
Nice Puzzle.
from “Mathematical Plums” edited by Ross Honsberger, chap 5 some surprises in probability, “Suppose a fair coin is tossed repeatedly and an indefinitely long sequence of H’s and T’s is generated…. what a surprise to learn that it is twice as likely that the triple TTH will be encountered ahead of the triple THT…” I love surprises in probability, and I find that most things about probability surprise me! The chapter also talks about something else you may like if you haven’t seen it, “Efron’s Dice”.
By the way, have you read this paper, http://www.emis.ams.org/journals/TAC/volumes/26/4/26-04.pdf, Commutative Monads as a Theory of Distributions. I gather Prof. Kock is developing the theory to study probability theory. I can’t really understand it, but you may find it interesting.
• John Baez says:
Probability theory is often counterintuitive to humans, which must have something to do with why humans mess up so often.
I haven’t read that paper. I’m pretty good friends with Anders Koch’s son Joachim, who also does category theory related work. But this paper is by Anders. I find it pretty hard going, at first because I’ve never understood strengths as much as I’d like, even though I did help come up with the more conceptual definition of them given on the link. (The more common definition, which involves writing down some commutative diagrams, always bothered me.)
• Rob says:
Say the first child is a G. Then we are guarateed that GB will occur prior to to BB. (cause for BB to happen, we must first have had GB)
Now say the first child is a B. Then either the second child is a B, in which case BB happens first, or it is a G, in which case, again, GB will happen before BB.
So we see that GB is 3x more likely to occur in any given familiy than BB. Thus I would expect that the expected value of the second random varable is less than the expected value of the first. Hah! well probability is so amazing, I expect that I am wrong! Now I will read through the comments where I expect to find the right answer.
Also, if you don’t understand that paper, I never will! But I think it is really interesting so I will keep reading it. BTW I think your notes on Cat Theory are great. Thanks to all your students for putting that together.
13. arch1 says:
Puzzle 3: Represent all possible birth sequences as an infinite binary tree (at the risk of scaring you we’ll pretend that childbearing doesn’t stop when the goal is reached). Yes, at each level of the tree starting with the 2nd, ¼ of the arcs are DS arcs and ¼ are SS arcs. But the symmetry breaks when we look at the correlation between adjacent levels: While an SS arc can immediately follow an SS arc, a DS arc can’t immediately follow a DS arc (as that would require the shared middle node to be both an S and a D).
Upshot: Half of the SS arcs “hide behind” an immediate SS parent, but DS-DS repulsion prevents DS arcs from ever doing this. This makes a DS arc likelier than an SS arc to be the first arc of its kind on a path from the root (put differently, it causes DS arcs to more efficiently “cover” the tree as viewed from the root). So the expected number of levels before encountering a DS arc is less than for an SS arc.
Greg and John, I find these probability paradoxes very instructive and great for my humility. I hope you keep them coming.
• John Baez says:
Nice answer! Yes, probability puzzles are great for ones humility—maybe the right verb is “humiliating”?
There should be a book of probability puzzles, similar to Mark Levi’s excellent physics puzzle book Why Cats Land on Their Feet: And 76 Other Physical Paradoxes and Puzzles. But I don’t know it!
If anyone here hasn’t tried Probability Puzzles (Part 1) and Probability Puzzles (Part 2), now is the time! They’re all about boys and girls, oddly enough. The first one is particularly devastating, or infuriating.
• arch1 says:
(chuckle) I actually tend to find them more invigorating than humiliating, partly for noble reasons but mostly I think because I can’t wait to spring them on others.
Ironically the closest I’ve come to humiliation recently was last week when I, er, “judged” a high school science fair project which explained how cats can land on their feet using gauge theory.
• cclingen says:
High school?! Cats landing their feet?! Gauge theory?! Where can I learn more?
• John Baez says:
Hi, Charlie! It’s probably good to start here:
• Wikipedia, Falling cat problem.
but then go to this paper by a friend of mine who is an expert on classical mechanics at U. C. Santa Cruz:
• Richard Montgomery, Gauge theory of the falling cat, in M. J. Enos, Dynamics and Control of Mechanical Systems, American Mathematical Society, pp. 193–218.
• arch1 says:
Nowhere as far as I know (certainly not me as I gleaned only a scant understanding). Maybe a paper will come out later; meanwhile I’d prefer to let the student pursue things in his own way.
• Re Prob Puzzle books. I just bought “Fifty Challenging Problems in Probability”. It looks like it has some really cool stuff in it. And best of all it was cheap!
• Greg Egan says:
The counterintuitive aspects of real statistical phenomena might have put an innocent man in prison for 38 years:
An Unusual Pattern
Like the statisticians who discuss the case, I’m not claiming any certainty that Geen is innocent, but on the face of it the statistical basis of the prosecution’s argument appears alarmingly wrong-headed.
• That is creepy. As teachers we need to do a better job at explaining statistics. Examples like the above and the OP help a lot.
14. pauljurczak says:
I hate to be a party pooper, but using infinite series in the top answer is yet another reason why lies and statistics are often mentioned in the same sentence. Human female can only have a limited number of children, much smaller than 100, which is much smaller than infinity.
This is in general a problem of using mathematics to model physical “reality”, which always requires certain assumptions resulting in departure from “reality”. In this case, the assumption is immortality or more specifically, unlimited childbearing capability.
P.S. Don’t take this post too seriously ;-)
• John Baez says:
I won’t take it too seriously—just more seriously than you expected.
Puzzle 7: How do the answers for Tom’s and Lisa’s expected number of children change if we impose a ‘cutoff’, saying that a woman can have no more than 10 children?
• …Or if each pregnancy has probability q of leaving the mother unable to bear further children?
• Charles Jackson says:
Note that that there is about a 1% that Tom will have to have 24 or more children before the desired SS outcome occurs.
15. Edon says:
These processes can be seen as absorbing Markov chains. Let $T$ be the block in the transition matrix that holds the transitions from transient states to other transient states. It is a basic property that $(I - T)$ is invertible and that the $(i,j)$ entry of $M=(I - T)^{-1}$ gives the expected number of times the chain is in state $j$ before it is absorbed, given that we start from state $i$. Summing the entries in the appropriate row yields the answers to the puzzles.
16. For some reason there isn’t a reply button below that latest reply of you, John, but yeah, that was what I was going for, although indeed not with the ‘little dash’ notation, but that works too. With just one minus sign it’s not that bad. Thanks for the fix. There must have been more wrong with it than just $\langle \cdot \rangle$ if it was so cryptic. I’m not sure what I typed anymore but while typing it that was the only part I didn’t know whether it’d work. The rest I was pretty sure of. But oh well. I thought \left{<} would work for \langle but you guessed that part right.
• John Baez says:
Kram wrote:
For some reason there isn’t a reply button below that latest reply of you…
Unfortunately if you tell this blog that you don’t want comments to get indented more than 4 times (because they become absurdly skinny), it implements this by eliminating the ‘Reply’ button after a comment that’s been indented 4 times.
What you should do in this case is reply not to the comment but to the comment it’s commenting on! In fact it’s often wise to do this. This keeps the comments from getting too skinny. I can tell who is a really expert blog-commenter by whether they do this.
I thought \left{<} would work for \langle but you guessed that part right.
I’ve never seen \left{<} in LaTeX; I think LaTeX is pretty careful about the difference between the less than symbol $<$ and the left angle bracket $\langle$.
But there’s an extra problem with LaTeX in most blogs! In HTML, < is an escape symbol, so you to get a less than symbol in LaTeX you have to use the HTML symbol < rather than <. It’s very bizarre to be using HTML commands inside LaTeX equations, but that’s how it works! That’s true not only for this rather brain-dead WordPress blog but also some others I use.
17. Tom says:
Puzzle 6: Why did John use the name of his wife for the female character in this drama?
• John Baez says:
Answer: because it was the first woman’s name beginning with L that I thought of.
18. David Cinabro says:
This one got in my head, and I had to try to solve it.
First Puzzle 3. Two children SS, SD, DS, DD. Tom and Lisa both stop 1 out of 4. Three Children for Tom SDS, SDD, DSS, DSD, DDS, DDD. Tom stops 1 out of 6. Three Children for Lisa SSS, SSD, SDS, SDD, DDS, DDD. Lisa stops 2 out of 6. Clearly the expected number of children is different and Lisa will have a smaller number than Tom.
For Tom’s expected number of children, he has 1/4 chance stopping at 2, 1/6 stopping at 3, 2/10 stopping at 4, 3/16 stopping at 5, 5/26 stopping at 6, etc. The stopping chance numerators are Fibonacci seeded by 1 and 1, and the denominators are Fibonacci seeded by 4 and 6. The expected number of children is then 2(1/4) + 3(1-1/4)1/6 + 4(1-1/4)(1-1/6)3/16+…. ie he stops at 2 with a probability of 1/4, at 3 with a probability of (1-1/4)*1/6, etc. Perhaps there is a clever way to write that, but I wrote a little program to tell me that it goes to 6.
Lisa has 1/4 chance stopping at 2, 2/6 stopping at 3, 3/8 stopping at 4, etc. The stopping chance numerators are 1,2,3,etc. and the denominators at 4,6,8, etc. The same sort of calculation gives the expected number of children 21/4 + 3(1-1/4)2/6+4(1-1/4)(1-2/6)*3/8+… This one is certainly even simpler to work out in closed from , but again the program tells me that it sums to 4.
19. Michael Nelson says:
By the way, you can prove the classic result involving pascal’s triangle and the fibonacci numbers by partitioning Tom’s string of n children into these cases:
Tom has a total of 3 sons
Tom has a total of 4 sons
For those who aren’t familiar, the relation between pascal’s triangle and the fibonacci numbers involves summing over diagonals, like in this image:
In other words, the nth fibonacci number equals (n-1 choose 0) plus (n-2 choose 1) plus …
20. While on the subject of what is essentially run theory, mathematician Peter Donnelly has given a TED talk on the subject of fooling juries with statistics. His talk is based around the infamous English criminal case of Sally Clark. To illuminate the issues he poses this problem:
you toss an unbiased coin many times and count the number of tosses until you get the pattern ”HTH”. You then calculate the average. You repeat the experiment but this time you look for the pattern ”HTT”. The question is: ”Is the average number of trials required to get ”HTH” greater than, less than or equal to the average number of trials required to get ”HTT”? ” Donnelly has tried this out on a wide range of audiences including professional mathematicians and the audience not surprisingly says the number of trials were equal. The expected number of trials to get ”HTH” is 10 while the expected number of trials to get ”HTT” is 8.
My paper ( $\url{http://www.gotohaggstrom.com/Fooling%20juries%20with%20statistics.pdf}$ ) explains the generating function proof of Donnelly’s claim. This paper also sets out the background to the Sally Clark case. A 2013 English Court of Appeal case has had the effect of banning Bayesian reasoning in court cases. This shows that the “real” world of the justice system can intersect in bizarre ways with probability theory.
21. Puzzle 1.
We are looking at infinite sequence of coin flips (the coin has two sides: S and D) and waiting for first occurrence of SS.
There are essentially 3 different states:
s0: current sequence ends with D or is empty (initial state)
s1: current sequence ends with exactly one S
s2: current sequence ends with SS (end state)
Let ei = expected number of flips to get from si to s2.
Then we have the following system of linear equations:
e2 = 0
e1 = (1/2)(1+e2) + (1/2)(1+e0)
e0 = (1/2)(1+e0) + (1/2)(1+e1)
Solving it gives the answer to the puzzle: e0 = 6.
Puzzle 2.
We just need to wait for the first daughter, then wait for the first son. Since expected number of trials until first success equals 1/p, where p is the probability of success, the answer is obviously 2+2=4.
Puzzle 3.
The intuition is, that Tom has to make a “step back” sometimes but not Lisa, therefore she accomplishes her goal sooner.
22. Check out this interesting, related puzzle and discussion.
https://www.klittlepage.com/2013/11/20/dynasty-meets-the-central-limit-theorem/
I just came across it. It has some surprising conclusions. | 2018-02-19T06:09:33 | {
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http://ufap.cbeu.pw/find-the-indicated-nth-roots-of-a.html | # Find The Indicated Nth Roots Of A
is there any other way to find the nth root besides de Moivre's theorem and the comparing method. An th root of unity in a finite field is an element satisfying , where is an integer. De Moivre's Theorem Now, in that same vein, if we can raise a complex number to a power, we should be able to find all of its roots too. , , , the six roots are () out of which , , and on the left s-plane are the roots of :. Our cube root calculator is a handy tool that will help you determine the cube root, also called the 3 rd root, of any positive number. One real root:. Examples:. Find the indicated real nth roots of a: 13. The proof can be extended to higher roots, as well. This is a preview topic with a couple of types of exercises. Find the nth term of the arithmetic sequence whose initial term a and common difference d are given. Also, don't overlook the most obvious property of all!. How do I write a function to calculate the nth root of a number; I'm sick of using pow() with a fractional. It is the method based on long. Conventionally your original #theta# is in the range #(-pi, pi]# or the range #[0, 2pi)# according to your definition of #Arg(z)# and the first of these five roots is the Principal Complex fifth root. Takes the nth root of all values in an iterable multiplied together. So if you're taking this to the 1/5 power, this is the same thing as taking 2 times 2 times 2 times 2 times 2 to the 1/5. Similarly, the cube root of a number b , written b 3 , is a solution to the equation x 3 = b. n th Roots. Roots are usually written using the radical symbol, with denoting the square root, denoting the cube root, denoting the. This online calculator is set up specifically to calculate 4th root. Free practice questions for SAT Math - How to find the nth term of an arithmetic sequence. 32 = 32(cos0 + isin 0 ) in trig form. For smaller values of n, one can readily compute the numbers. NaN = not a number To clear the entry boxes click "Reset". What does nth root mean? Information and translations of nth root in the most comprehensive dictionary definitions resource on the web. n = 3, a = 216 b. This square root trick to find square root of a number will surely help you in your exams. ? Find the indicated real nth root(s) of a - Help? What is the indicated real nth root(s) of a. It is clear from the foregoing that the nth root of a real or complex number can be. 0 equals to 4. Research design: Qualitative, quantitative and qualitative analyses, is the interview but how do you think is terribly exponents roots help homework on finding nth and rational important, especially in regard to the organizational and leadership for pursuing implementation of p. Find the nth Root of a Perfect nth Power Monomial. Introduction to Rational Functions. Finding Nth Roots: Find the indicated real nth root(s) of a. Finding the Roots of a Complex Number We can use DeMoivre's Theorem to calculate complex number roots. There are 5, 5th roots of 32 in the set of complex numbers. n=2, a=100 " in Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions. notebook 3 November 13, 2014 YOU PRACTICE Find the indicated real nth root(s) of a. We can find value of k by putting discriminant of quadratic equation equal to 0. Specifically, here we find the transfer function of the nth order Butterworth filter for : , , , the four roots are (): out of which and on the left s-plane are the roots of : Note that the coefficient of the first order term is. 1 A Case Study on the Root-Finding Problem: Kepler's Law of Planetary Motion The root-finding problem is one of the most important computational problems. From above it can be concluded that the roots are located on the circumference of the unit circle with center at (0,0). Nth Root Lesson Plans & Worksheets Reviewed by Teachers. Take a out a piece of paper and a pencil and step through the algorithm. Daily Math Lesson. 2 = eiπ = −1, and the roots of unity are simply 1 and −1. We could use the nth root in a question like this:. It looks quite tedious to do by hand, but the algorithm exists for any root and is similar to the square root one. Type “Number”, “Root” and “Result” in the cells A1, B1 and C1 respectively. This example shows how to calculate the Nth root of a number in Visual Basic. The principal nth root is the nonnegative root. From nth roots worksheets to algebra 2 nth root videos, quickly find teacher-reviewed educational resources. How to Find Roots of Unity. Finding real nth roots of a. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. How to Find Nth Roots by Hand. The number that must be multiplied times itself n times to equal a given value. Toggle navigation. These problems serve to illustrate the use of polar notation for complex numbers. ) ! n=3! a=64 c. For, the square of a real root must be positive; and therefore the original equation cannot have more real roots than the transformed has positive roots. Perfect for. If n is even, then a has real nth roots if a > 0, real nth root if a5 0, and real nth roots if a < 0. The variables may be defined in the program, or they may be inputted by the user, but it. Given an integer number and we have to find their Square, Square Root and Cube. To check the result, the code then raises the result to the root power and displays the new result. For instance, 2 is a cube root of 8 because 23 = 8, and 3 is a fourth root of 81 because 34 = 81. Use of the the pow function is more sensible for this unusual need. So the maximum is probably between 2 and 4, and I asked my son how we could find it. Solving Nth Root Equations The nth root of a number X , is a number r whose nth power is X. The other n roots are equally spaced on the circle of radius. Apart from the stuff given in this section " Finding nth Term of the Sequence" , if you need any other stuff in math, please use our google custom search here. n = 3, x = 128. Square root of 64 is 8 because 8 times 8 is 64 Cube root of 27 is 3 because 3 times 3 times 3 = 27 fourth root of 16 is 2 because 2 times 2 times 2 times 2 = 16 Sometimes, you may get a real number when looking for the square root. For large n, the n th root algorithm is somewhat less efficient since it requires the computation of − at each step, but can be efficiently implemented with a good exponentiation algorithm. Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. 6-1 Evaluate nth Roots and Use Rational Exponents Name_____ Objective: To evaluate nth roots and use rational exponents. What is a function? Domain and range. The only one you could find was x =2. Simplify the expression. The nth root of a real number. Also, don't overlook the most obvious property of all!. root of a positive number is positive. As the title suggests, the Root-Finding Problem is the problem of finding a root of the equation f(x) = 0, where f(x) is a function of a single variable x. (So n=3) Answer should be in standard form. Basically, pretend to give X0 a value of your initial guess. Homework help with finding nth roots and rational expressions with paperbag writer radiohead in education by apa sociology research paper example. In mathematics, Nth root of a number A is a real number that gives A, when we raise it to integer power N. Using this formula, we will prove that for all nonzero complex numbers $z \in \mathbb{C}$ there exists $n. To find the ARITHMETIC mean of 4 and 10, you add them up and then divide by n number of values: (4+10)/2 = 7 To find the GEOMETRIC mean, you multiply 4 and 10, and then find the nth root: the. From this form a K-normal form C, can be deduced. 5) Complete the following table for!!!=7!. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. From this form a K-normal form C, can be deduced. In Exercises 48 and 49, find the modulus and the argument, and graph the complex number. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Finding real nth roots of a. Improve your math knowledge with free questions in "Nth roots" and thousands of other math skills. Numerical methods for finding the roots of a function The roots of a function f(x) are defined as the values for which the value of the function becomes equal to zero. FINDING NTH ROOTS Find the indicated real nth root(s) of a. Given two numbers N and A, find N-th root of A. Round your answer to two decimal places when appropriate. 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When I try to find the roots of the same equation in Mathematica, I receive various errors. Geometric mean function for python. The solutions of the equation x^n=1, where n is a positive integer, are called the nth roots of 1 or "nth roots of unity. Homework Help With Finding Nth Roots And Rational Expressions. Perhaps, then, it is this calculator: Let's say you want to take the fifth root of 2. Asking for unreasonable roots, for example, show_nth_root < 1000000 >(2. This is a polynomial equation of degree n. (26 replies) In PythonWin I'm running a program to find the 13th root (say) of millions of hundred-digit numbers. Find nth roots of complex numbers Contact Us If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. Finding nth roots of Complex Numbers. However, there is still one basic procedure that is missing from the algebra of complex numbers. Both X and N must be real scalars or arrays of the same size. In mathematics, an nth root of a number x, where n is usually assumed to be a positive integer, is a number r which, when raised to the power n yields x: =, where n is the degree of the root. It uses Math. Homework Help On Finding Nth Roots And Rational Exponents. For instance, 2 is a cube root of 8 because 23 = 8, and 3 is a fourth root of 81 because 34 = 81. The math workaround described above works really well with pretty good accuracy. Square root of 64 is 8 because 8 times 8 is 64 Cube root of 27 is 3 because 3 times 3 times 3 = 27 fourth root of 16 is 2 because 2 times 2 times 2 times 2 = 16 Sometimes, you may get a real number when looking for the square root. The nth roots of a complex number For a positive integer n=1, 2, 3, … , a complex number w „ 0 has n different com-plex roots z. 1/2 Evaluate the expression. 24 The nth root of a number b is designated as nb,. 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How to find the nth root of a complex number. This online calculator for fifth roots is set up specifically to calculate 5th root. In effect we must find the roots of the equation z^3 = 1 We could write this z^3-1 = 0 (z-1)(z^2 + z + 1) = 0 and then z = 1 gives one root, and other two are solutions of the quadratic z^2+z+1 = 0 (these roots will be complex). ^2 96 sqr root 3 in. “nth Root of a”. An applet allowing you to investigate how the th roots of a complex number (in blue) appear on an Argand diagram. nth Root of a-to-the-nth-Power. 5, 13, 33, 89, I got this by adding the two sequences above. Evaluate Question: “What number, when raised to the 3rd power gives us 27?“ B. For instance, √(4) produces a principle root of 2. If all equations and starting values are real, then FindRoot will search only for real roots. I decided to take the input in the form of a Double. For example, the tenth root of 59,049 is 3 as 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 x 3 is 59,049. Relating to an unspecified ordinal number: ten to the nth power. Functional abstraction to find nth root of a number - Newton raphson. Created Date: 1/17/2019 10:36:18 AM. Finding nth roots of Complex Numbers. Find the indicated real nth root(s. (algorithm) Definition: This describes a "long hand" or manual method of calculating or extracting square roots. Rationalize one term denominators of rational expressions. The exponent of a number shows you how many times the number is to be used in a multiplication. 1 Evaluate nth Roots and Use Rational Exponents Goal p Evaluate nth roots and study rational exponents. Derivation from Newton's method. A root of degree 2 is called a square root, a root of degree 3 is called a cube root, a root of degree 4 is called a fourth root, and so forth. org - it looks like you don't have Java installed, please go to www. I know that if this were just normal numbers, I could find it u. In effect we must find the roots of the equation z^3 = 1 We could write this z^3-1 = 0 (z-1)(z^2 + z + 1) = 0 and then z = 1 gives one root, and other two are solutions of the quadratic z^2+z+1 = 0 (these roots will be complex). Using Bino's model of multiplication , you can compute the square root, cube root, or root of any positive number up to any decimal place for no-perfect cube number. Granted, then, you have a problem they currently have. What is the principal root of 121400? Rounded to two decimal places, the principal square root of 121400 is 348. Created Date: 1/14/2009 8:51:02 PM. The nth root calculator below will also provide a brute force rounded approximation of the principal nth root. She had invested$1500 more at 8%. Roots can be square roots, cube roots, fourth roots and so on. What is the fifty-first term?. Use nth roots in problem solving Animal Population The population P of a certain animal species after t months can be modeled by P = C(1. The for a different view). This online calculator is set up specifically to calculate 4th root. According to the Fundamental Theorem of Algebra, this equation has n solutions. Find indicated real nth roots of c. org - it looks like you don't have Java installed, please go to www. It uses Math. Homework Help On Finding Nth Roots And Rational Exponents. 5) Complete the following table for!!!=7!. First, we will define what square roots are and how you find the square root of a number. Looking for a primer on how to find the nth term of a geometric sequence? See how it's done with this free geometer's guide. The number that must be multiplied times itself n times to equal a given value. Discussion Suppose IDL is being used to find a real root of a number. Now I was thinking of adding the nth-Root of a Number. My current expression is -. 2 days ago · He said that 12 out of the 18 persons who had already indicated interest in the governorship election in the state in 2021 were encouragingly from Anambra South, adding that the expectation was that with time, aspirants from the other two senatorial zones would withdraw to join hands in finding an ideal person from the South to emerge as the. W HEN ONE THING DEPENDS on another, as for example the area of a circle depends on the radius -- in the sense that when the radius changes, the area will change -- then we say that the first is a "function" of the other. The number that must be multiplied times itself n times to equal a given value. 0 *Real nth Roots of a. Nth Root using the POWER Function. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. Excel's powerful mathematical toolkit includes functions for square roots, cube roots, and even nth roots. This online calculator is set up specifically to calculate 4th root. 1 Evaluate nth Roots and Use Rational Exponents Goal p Evaluate nth roots and study rational exponents. Mathematical Connections Find the radius of the figure with the given volume. 32 = 32(cos0 + isin 0 ) in trig form. In the event you seek help on exam review or perhaps intermediate algebra syllabus, Polymathlove. Then, invoking the Intermediate Value Theorem, there is a root in the interval $[-2,-1]$. We find the the two roots by multiplying 2 by the square roots of unity. What is an “nth Root?” Extends the concept of square roots. It requires an initial guess, and then Newton-Raphson iterations are taken to improve that guess. To demonstrate that this simple method is fully applicable to roots of higher order, we shall attempt to find the 5th root of 2. What are the solution sets of the following 3 problems,the square root of x squared + 3x = x-3, the square root of 3x + 10 = x + 4 and finally the square root of 8x - 2 = the square root of 2x? Help much appriciated. Every real number has exactly one real. n5 3, a5 2 64 b. ^2 216 sqr root 3 in. In short, you can make use of the POWER function in Excel to find the nth root of any number. 1) the radicand has no perfect nth powers as factors 2) any denominator has been rationalized Example 4: Write the following Radicals in Simplest Form — Remember your perfect nth Roots u = 16. A formal mathematical definition might look something like: The nth roots of unity are the solutions to the equation x n = 1. Step 3: Write the answer using interval notation. ) ! n=5! a="32 b. Summing up: The following guidelines will be useful for finding the nth root of the perfect nth power of a two-digit or three-digit number. Let f(x) = 2 x - 1 and find its limit applying the difference and product theorems above lim x→5 f(x) = 2*5 - 1 = 9 We now apply theorem 5 since the square root of 9 is a real number. De Moivre's Theorem Now, in that same vein, if we can raise a complex number to a power, we should be able to find all of its roots too. Nth roots of unity 40 videos. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. You are given 2 numbers (N , M); the task is to find N√M (Nth root of M). • Benchmark MA. Pow , mathematics. For, the square of a real root must be positive; and therefore the original equation cannot have more real roots than the transformed has positive roots. Schools Parents. C program to get nth bit of a number January 24, 2016 Pankaj C programming Bitwise operator , C , Program Write a C program to input any number from user and check whether n th bit of the given number is set (1) or not (0). In fact, there are seven 7th roots of unity, and each gold disc in that picture is one of them. How do i find the indicated real nth root of:? Find the indicated real nth root(s) of a - Help? A) find indicated roots of the complex number b) express roots in standard form?. Given two numbers N and A, find N-th root of A. During my son’s math lesson today we got on the subject of , which I prefer to write as. The fundamental theorem of algebra can help you find imaginary roots. This function is pretty well behaved, so if you make a good guess about the solution you will get an answer, but if you make a bad guess, you may get the wrong root. Roots of unity have connections to many areas of mathematics, including the geometry of regular polygons, group theory, and number theory. This Algebra I: nth Root and Rational Exponents Worksheet is suitable for 8th - 11th Grade. Pow , mathematics. A "root" (or "zero") is where the polynomial is equal to zero:. 24 The nth root of a number b is designated as nb,. By definition, the nth root of a number can be calculated by raising that number the power of 1/n. They are also known as zeros. so you get 4 roots @Dick: I think he already used De Moiver's formula when he took the root and turned it into a multiplier of one-half and multiplied the term inside the cos and sin @paraboloid if you don't know what you need to do, just read you course book or ask your Lecturer or TA. As this problem involves. Example 2: Solve Equations Using nth Roots Solve the equation. 5 or 10 1/2 and the cube root as 10 0. The square roots of unity are 1 and − 1. Plot your number r(cos +i sin), that you want to take the root of. a) As you go down the table of values, what number does X approach?. Setting up the Data. Given a complex number z = r(cos α + i sinα), all of the nth roots of z are given by. 5847i\) and \(4. Improve your math knowledge with free questions in "Nth roots" and thousands of other math skills. Then we will apply similar ideas to define and evaluate nth roots. The resulting number from this expression means that if it’s raised to the nth power it equals the. This entry was posted in mathematics and tagged C# , C# programming , calculate nth roots , calculate roots , example , example program , Math. One real root:. nth Roots and Rational Exponents EVALUATING NTH ROOTS You can extend the concept of a square root to other types of roots. The second root is usually called the "square root". Of course, typically polynomials have several roots, but the number of roots of a polynomial is never more than its degree. For example, to find the cube root of 8 in a cell, type the following:. Examples:. Use 27 cis (90 degress), which is the trigonometric from of 27i, to do the problem. June 19, 2019 July 22, 2019 Craig Barton. The finite field has prime order. I'm trying to find the n-th root of unity in a finite field that is given to me. 0 Votes 18 Views I am unable to prove that :. is the radius to use. Divide radicals that have the same index number. Instead you need to pass 1/10 as n√x = x(1/n). Unlike a square root function which is limited to nonnegative numbers, a cube root can use all real numbers because it is possible for three negatives to equal a negative. n th Root of an Integer Description Calculate the n th root of an integer. when x k +1 ≥ x k. In general, a root of degree n is called an nth root. Your program should ask the user for a number, and a root and print that root of the number. You can always tell FindRoot to search for complex roots by adding 0. If n = 3 , then it is called as Cube root. During my son’s math lesson today we got on the subject of , which I prefer to write as. An expression under a radical sign is called a radicand. Find the indicated real nth root(s) of a. Find the fourth roots of -16 and -16 is a complex number even though it is also real. n = 2, a = 100 24. The equation to calculating the nth term of the sequence is an+b; "a" is the fixed number that is being added to generate the series, and "b" gives the. n th Roots A square root of a number b , written b , is a solution of the equation x 2 = b. However, there is still one basic procedure that is missing from the algebra of complex numbers. 3 POLAR FORM AND DEMOIVRE’S THEOREM At this point you can add, subtract, multiply, and divide complex numbers. The Help Article also provides a convenient wrapper function. We get the nth root of 1 is the regular nth root of 1, which is just 1, times e to the i phi over n. De Moivre's theorem can be extended to roots of complex numbers yielding the nth root theorem. Buy business plan online - best in texas, homework help on finding homework help us geography nth roots and rational exponents. , , , the six roots are () out of which , , and on the left s-plane are the roots of :. It need not be true that any of the fractions is actually a solution. Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. However, there is still one basic procedure that is missing from the algebra of complex numbers. (Buddy system! If there is no buddy, it must stay under the radical. Find the indicated real nth root(s) of a negative. De Moivre's theorem can be extended to roots of complex numbers yielding the nth root theorem. n = 5, a = 0. Because n = 4 is even and a = 81 > 0, 81. 2 days ago · He said that 12 out of the 18 persons who had already indicated interest in the governorship election in the state in 2021 were encouragingly from Anambra South, adding that the expectation was that with time, aspirants from the other two senatorial zones would withdraw to join hands in finding an ideal person from the South to emerge as the. 3 POLAR FORM AND DEMOIVRE'S THEOREM 483 8. The 5th root of 1,024 is 4, as 4 x 4 x 4 x 4 x 4 is 1,204. Question 1: Simplify. When you’re giving a time-bound exam like SSC CGL, SSC CPO, Railways Group D, RPF & ALP this can drain you of your precious time. You can see there are many roots to this equation, and we want to be sure we get the n^ {th} root. Nth root calculator. V = 216 ft3 Find the real solution(s) of the equation. There might be some issues with this. IXL Learning Learning. Use nth roots in problem solving Animal Population The population P of a certain animal species after t months can be modeled by P = C(1. In mathematics, an nth root of a number x, is a number r which, when raised to the power n yields x. The number 0 (zero) has just one square root, 0 itself. The nth root calculator below will also provide a brute force rounded approximation of the principal nth root. Simplify the expression. There is no result accuracy argument of Nth_Root, because the iteration is supposed to be monotonically descending to the root when starts at A. When I try to find the roots of the same equation in Mathematica, I receive various errors. 1) the radicand has no perfect nth powers as factors 2) any denominator has been rationalized Example 4: Write the following Radicals in Simplest Form — Remember your perfect nth Roots u = 16. Example of How to Evaluate a “nth Root” Radical Expression. I assume you're saying that the calculator can do the four basic operations and square roots, but not arbitrary exponents. In this case, we divided by a negative number, so had to reverse the direction of the inequality symbol. In particular, is called the primitive th root of unity. (I am aware that \sqrt as command for a general root is misleading) For the life of me I can't find the right syntax to get the input of n'th root of x. pl help Shambhu. Number of roots equal to n. find a formula for the nth partial sum of the series and use it to find the series sum if the series converges. n = 3, a = 216 b. Also note that as of Python 3, adding periods to integers to make them a float is no longer necessary. Suppose we wish to solve the equation z^3 = 2+2 i. Complex Numbers: nth Roots. | 2019-12-15T12:40:11 | {
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https://math.stackexchange.com/questions/1372479/double-integral-set-up | # Double Integral Set Up
The question was stated as follows,
Evaluate the following double integral;
$$\iint_R x^3y dA$$
where R is interior of triangle with vertices (0,0), (1,0), & (1,1) .
I thought for these types of double integrals I could have the limits in either of the two formats below;
$$\int_0^1\int_0^y x^3y dxdy$$ [1]
OR
$$\int_0^1\int_0^x x^3y dydx$$ [2]
However the answer sheet states that the answer for only the latter integral is correct.
How can I set up or visualize the problem in order to set it up in the correct way?
I did plot the points on a graph, and got my y=x "limit" from there, I just cannot understand why [2] is the correct integral, and how to go about making sure that in every double integral problem, I choose the correct integral set up.
All help is much appreciated :)
For the first one the bounds on x are wrong. The region is bounded by $y=x$ on the left and on the right by $x=1$. The bound on the left is a lower bound, so it should be
$$\int_0^1 \int_y^1 x^3ydxdy$$
• That makes sense! Thank you. – mnmakrets Jul 24 '15 at 14:10
You can draw the triangle.
Call the region inside of it $\Omega$.
Looking at lines of constant $y$, we'll have that $$\Omega = \{ (x,y) \in \mathbb{R}^2 \ | \ 0 \leq x \leq 1, y \leq x \leq 1 \}$$
Draw the triangle. Now think of it this way: if I fix $x$, what will be the extremes of integration for $y$? You'll get something that depends on $x$ and in particular $y$ goes from $0$ to $x$. Then integrate over the values taken on by $x$, and yo get the second form.
Alternatively, you can fix $y$ and notice that $x$ then varies from $y$ to $1$; then integrate over $y$ to get the correct form
$$\int_{0}^1 \int_y^1 x^3y dy dx$$
Usually you draw the area you want to integrate on, and draw the horizontal and vertical lines that corresponds to fixing $x$ or $y$. This helps visualize the problem and find the extremes of integration with ease
According fubini's theorem, you can either integrate parallel to the x-axis or the y-axis and hence there are two possibilities here:
$$\int_0^1 \int_0^x x^3y \ dy \ dx$$
or
$$\int_0^1 \int_y^1 x^3y \ dx \ dy$$
The best way to get to grips with these sort of questions is to practise more. For more, visit the following links: | 2019-10-20T20:08:00 | {
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https://mathematica.stackexchange.com/questions/22543/how-do-i-obtain-the-enclosed-area-of-this-particular-parametric-plot/22545 | # How do I obtain the enclosed area of this particular parametric plot?
I'm trying to find a way to obtain the enclosed area of this particular plot. Can someone show me how?
curveplot = ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]],
Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}, {t, 0, 2 π},
PlotStyle -> {{Thickness[0.01], Darker[Purple]}},
AspectRatio -> Automatic, PlotRange -> All, AxesLabel -> {"x", "y"}]
## 6 Answers
You can get the curve in polynomial implicit form as below.
poly =
GroebnerBasis[{x^2 - ct, y^2 - st, ct^2 + st^2 - 1}, {x, y}, {ct,
st}][[1]]
(* Out[290]= -1 + x^4 + y^4 *)
To get the area, integrate the characteristic function for the interior of the region. That that's where the polynomial is nonpositive (just notice that it is negative at the origin, say).
area = Integrate[Boole[poly <= 0], {x, -2, 2}, {y, -2, 2}]
(* Out[292]= (2 Gamma[1/4] Gamma[5/4])/Sqrt[π] *)
N[area]
(* Out[293]= 3.7081493546 *)
There are other ways to do this if you cannot find an implicit form, but this seems most direct in this case.
--- edit ---
If you can just solve separately for x and y in terms of the parameter t then you can set up a region function. I do this below for the positive quadrant, and take advantage of symmetry to get the full area in approximate form.
reg = Function[{x, y},
If[And @@ {0 <= x <= 1, 0 <= y <= 1, y <= Sqrt[Sin[ArcCos[x^2]]]},
1, 0]];
approxarea = 4*NIntegrate[reg[x, y], {x, 0, 1}, {y, 0, 1}]
(* Out[321]= 3.70814937167 *)
One can actually recover the exact area from this by using Integrate instead of NIntegrate. But this seems like a viable approach in situations where the exact value might not be readily computed.
--- end edit ---
--- edit 2 ---
Here is a Monte Carlo method that does not rely on solving for anything. We extract the line segments, augment with a diagonal, and do some magic.
segs = Cases[curveplot, _Line, Infinity][[1, 1]];
segs = {Join[segs, N[Table[{j, 1 - j}, {j, 0, 1, 1/100}]]]};
I added an extra level of List due to requirements of some further code. First let's reform a line to take a look at this region.
Graphics[Apply[Line, segs]]
Now we create an in-out function, generate a bunch of random points in the unit square of the first quadrant, take a Monte-Carlo approximation of this area. Then multiply by 4 and add 2. Why? because that's what one always does-- it's like selecting "c" when we don't know the multiple choice answer. (Okay, we multiply by 4 to account for all quadrants, and add 2 because we have in effect excised a square of side length $\sqrt{2}$ from the full region.)
To create the in-out function I use code directly from here.
nbins = 100;
Timing[{{xmin, xmax}, {ymin, ymax}, segmentbins} =
polyToSegmentList[{segs[[1]]}, nbins];]
(* Out[414]= {0.040000, Null} *)
len = 100000;
pts = RandomReal[1, {len, 2}];
Timing[
inout = Map[pointInPolygon[#, segmentbins, xmin, xmax, ymin, ymax] &,
pts];]
approxarea = 4.*Length[Cases[inout, True]]/len + 2.
(* Out[419]= {2.750000, Null} *)
(* Out[420]= 3.7092 *)
I would imagine one could do a bit better by integrating just the unit square in the first quadrant with Method -> "QuasiMonteCarlo", sowing the points via the EvaluationMonitor option, and using those instead of the random set above. This will give a low-discrepancy sequence. Or generate such a set directly; bit offhand I don't know how to do that.
-- end edit 2 ---
• Thanks, I wanted to see if this checked out with a monte carlo estimate. – Black Milk Apr 2 '13 at 21:55
Since it seems to have not been mentioned yet: yet another way to obtain an approximation of the area of your Lamé curve is to use the shoelace method for computing the area. Here's a Mathematica demonstration:
pts = First[Cases[
ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]},
{t, 0, 2 π}, Exclusions -> None, Method -> {MaxBend -> 1.},
PlotPoints -> 100] // Normal, Line[l_] :> l, ∞]];
PolygonSignedArea[pts_?MatrixQ] := Total[Det /@ Partition[pts, 2, 1, 1]]/2
PolygonSignedArea[pts]
3.7081447086368127
The value thus obtained is pretty close to the results in the other answers.
Surprisingly, there is in fact an undocumented built-in function for computing the area of a polygon:
(* $VersionNumber < 10. *) GraphicsMeshMeshInit[]; PolygonArea[pts] 3.708144708636812 (*$VersionNumber >= 10. *)
GraphicsPolygonUtilsPolygonArea[pts]
3.708144708636812
But, what you really should know is that Lamé curves have been well studied, and there is in fact a closed form expression for the area of a Lamé curve. Given the Cartesian equation
$$\left|\frac{x}{a}\right|^r+\left|\frac{y}{b}\right|^r=1$$
the formula for the area of a Lamé curve (formula 5 here) is
$$A=\frac{4^{1-\tfrac1{r}}ab\sqrt\pi\;\Gamma\left(1+\tfrac1{r}\right)}{\Gamma\left(\tfrac1{r}+\tfrac12\right)}$$
In particular, for the OP's specific case, $a=b=1$, and $r=4$. Thus,
With[{a = 1, b = 1, r = 4},
N[(4^(1 - 1/r) a b Sqrt[π] Gamma[1 + 1/r])/Gamma[1/r + 1/2], 20]]
3.7081493546027438369
This is the same as the answer Daniel obtained through more general methods.
• Never heard of that method before. Nice. (Yes, I upvoted..) – Daniel Lichtblau Apr 3 '13 at 15:42
• I've taught the "shoelace" method but never heard of the name. (+1) – Michael E2 Apr 3 '13 at 20:32
• @J.M. Thank you for the additional insight. – Black Milk Apr 3 '13 at 20:55
• @Michael, I must confess that I too have been using that method for a while now, but only learned it had a name on math.SE ... – J. M.'s technical difficulties Apr 4 '13 at 2:37
• You could just do PolygonArea@pts... no need to wrap it in Polygon – rm -rf Apr 25 '13 at 21:19
One can use one of the line integral forms of the area, derived from Green's Theorem: $$A = \frac12 \int_C x \; dy - y \; dx = \int_C x \; dy = - \int_C y \; dx$$ The first one is symmetric, which sometimes is an advantage.
c[t_] := {Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]}
dA = 1/2 c'[t].Cross[c[t]]
(* complicated output *)
One problem with this parametrization are the derivatives of Abs and Sign. They are discontinuous at isolated points, and as far as the integral is concerned, it does not matter what value we assign them at the discontinuities. So we can simplify matters by substituting for them. We can also substitute 1 for Sign[x]^2, since x will be 0 only at few isolated points. Thus the differential is
dA = dA /. {Sign'[x_] :> 0, Abs'[x_] :> Sign[x]} /. {Sign[_]^2 :> 1} // Simplify
(* (Abs[Cos[t]] Cos[t] Sign[Cos[t]] + Abs[Sin[t]] Sign[Sin[t]] Sin[t]) /
(4 Sqrt[Abs[Cos[t]]] Sqrt[Abs[Sin[t]]]) *)
NIntegrate returns a small imaginary component
NIntegrate[dA, {t, 0, 2 π}]
(* 3.70815 - 1.97076*10^-10 I *)
Oddly, setting WorkingPrecision reduces the imaginary error, even if it is set to less than MachinePrecision (15.9546)
NIntegrate[dA, {t, 0, 2 π}, WorkingPrecision -> 10]
(* 3.708149355 + 0.*10^-21 I *)
Integrate returns an exact answer in this case:
Integrate[dA, {t, 0, 2 π}]
(* (3 Sqrt[2] π Gamma[5/4] + 4 Gamma[3/4] Gamma[5/4]^2) /
(2 Sqrt[π] Gamma[3/4]) *)
FullSimplify @ %
(* (Sqrt[π/2] Gamma[1/4])/Gamma[3/4] *)
N[%, 20]
(* 3.7081493546027438369 *)
In cases where it is not possible to get a closed form for the curve, you can use the image processing functions to get an approximation for the enclosed area. First, some small changes to the plot — get rid of the axes, labels and everything else that isn't needed, and set the aspect ratio to 1.
curveplot = ParametricPlot[{Sqrt[Abs[Cos[t]]] Sign[Cos[t]], Sqrt[Abs[Sin[t]]] Sign[Sin[t]]},
{t, 0, 2 π}, PlotStyle -> Thick, AspectRatio -> 1, Axes -> False, PlotRange -> {-1, 1}]
Next, close the holes in the curve using morphological operations:
im = Binarize@curveplot ~Opening~ 7 // DeleteSmallComponents
Depending on the curve, you'll have to tweak the parameters and be careful with DeleteSmallComponents, in case you have disconnected components in your plot (always do a visual check or ImageAdd[curveplot, im] to see if it's correct). Sometimes, this may even not be necessary if your curve is "nice".
Finally, use ComponentMeasurements to get the area of the enclosed space. The result is in sq. pixels, so I normalize by the total area (in sq. pixels) and multiply by the area of the plot range rectangle in the original plot (i.e., $(x_{max}-x_{min})(y_{max}-y_{min})$)
4 (1 /. ComponentMeasurements[im, "Area"]) / Times @@ ImageDimensions@im
(* 3.66738 *)
which looks about right, since your curve fits in a square of side 2 and is close to Daniel's answer of 3.708. You can get a closer approximation if you increase the ImageSize in the original plot. Using ImageSize -> 2000 gives me 3.70129 as the area (but note that the image processing steps will take longer to compute).
• Exclusions -> None in ParametricPlot[] should close up the artifact holes nicely. – J. M.'s technical difficulties Apr 3 '13 at 15:27
Integrating InterpolatingFunction.
p = Table[{
Sign[Cos@t] Sqrt[Abs@Cos@t],
Sign[Sin@t] Sqrt[Abs@Sin@t]}, {t, 0, Pi, Pi/100.}];
f = Interpolation[p];
2*NIntegrate[f@t, {t, -1, 1}]
3.70771
As a modified version of Michael E2's answer:
I tried to rewrite your original curve as below for $$t\in \left[0,\tfrac{\pi}{2}\right]$$, in order to make sure the derivatives of the parametric form can be obtained easily by avoiding Abs or Sign:
ncurve = {(Cos[t]^2 )^(1/4), (Sin[t]^2)^(1/4)}
Then the result of the closed area can be obtained by applying Green's theorem to the curve and by using the symmetry of the curve:
area = 4*Integrate[ncurve[[1]] D[ncurve[[2]], t], {t, 0, Pi/2}]
which gives:
$$\dfrac{8\; \Gamma^2 \left(\tfrac{5}{4}\right)}{\sqrt{\pi }}$$
and the numerical result is therefore:
area // N[#, 20] &
`
3.7081493546027438369 | 2020-08-09T08:32:24 | {
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https://math.stackexchange.com/questions/1682610/what-is-the-difference-between-tan-x-sec2x-and-sin-x-cos3x-why-is | # What is the difference between ($\tan x \sec^2x$) and ($\sin x/\cos^3x$)? Why is the answer to the integration different?
$$\int \:\frac{\left(\sin x+\tan x\right)}{3\cos^2x}dx$$
I know I have to split the equation into
$$\frac{1}{3}\int \:\left(\:\frac{\sin x}{\cos x}\right)\left(\frac{1}{\cos x}\right)dx+\frac{1}{3}\int \:\left(\:\tan x\right)\left(\frac{1}{\cos^2x}\right)dx$$
I know that for the first part, it is $$\frac{1}{3}\int \tan x\sec xdx$$ which is $$\sec x$$.
However, for the second part, wouldn't it be $$\frac{1}{3}\int \tan x \sec^2xdx$$
If I used $$u=\tan x$$ then $$du=\sec^2xdx$$ so wouldn't the answer be $$\frac{1}{6}\tan^2x$$
However, the book is saying that the second part is supposed to be $$\frac{1}{6}\sec^2x$$ because I was supposed to convert the second part into $$\frac{1}{3}\int \frac{\sin x}{\cos^3x}dx$$ and let $$u=\cos x$$
What I am doing wrong? Why can't it be $$\tan \sec^2x$$ instead of $$\sin x/\cos^3x$$?
• Remember $\sec^2{\theta}=1+\tan^2{\theta}$ – user41736 Mar 4 '16 at 6:49
Recall that
$$1 + \tan^2 x = \sec^2 x$$
or, since I dislike the secant,
$$\frac{1}{\cos^2 x} = \frac{\sin^2 x + \cos^2 x}{\cos^2 x} = 1+\tan^2 x.$$
Therefore, your answer and the book's answer only differ by a constant.
Both the answers differ by a constant(1/6). So both the answers and both the methods are correct. The constant of integration takes care of the constant(1/6).
Both are right as when you differentiate $1/6tan^2x$ with chain rule you get $\frac{1}{6}.2tanx.sec^2x=1/3.tanx.sec^2x$ also differentiating $1/6sec^2x$ you get the same answer . so both are right.
• Note the constants can be $c,c'$ as integration gives a family of curves and an exact constant gives a specific curve – Archis Welankar Mar 4 '16 at 6:48 | 2019-10-22T19:06:03 | {
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https://spot.pcc.edu/slc/mathresources/output/html/radicals-rational-exponents.html | ## Section13.8Rational Exponents
The power to a power rule of exponents relates that $(x^m)^n=x^{mn}\text{.}$ This rule is fairly intuitive when both exponents are positive. For example, in the expression $(x^4)^3$ there are three factors of $x^4\text{,}$ each of which contains four factors of $x\text{,}$ so all together there are four factors of $x\text{,}$ three times, i.e. $3 \cdot 4$ factor of $x\text{.}$
While the power to a power rule is less intuitive once you move away from positive integer exponents, the rule remains the same regardless of the nature of the exponents. For example:
\begin{align*} (x^{1/3})^3\amp=x^{\frac{1}{3} \cdot 3}\\ \amp=x^{1}\\ \amp=x \end{align*}
But we already have a name for the expression that when cubed results in $x\text{,}$ and that name is $\sqrt[3]{x}$ (the cube root of $x$). So it must be the case that $x^{1/3}=\sqrt[3]{x}\text{.}$ In general, is $n$ is any positive integer, then:
\begin{equation*} x^{1/n}=\sqrt[n]{x} \end{equation*}
and more generally,
\begin{equation*} x^{m/n}=\sqrt[n]{x^m}\text{.} \end{equation*}
Several examples are shown below.
###### Example13.8.1.
Express $y^{7/5}$ as an equivalent radical expression
Solution
\begin{equation*} y^{7/5}=\sqrt[5]{y^7} \end{equation*}
###### Example13.8.2.
Express $\sqrt[3]{w^{12}}$ using an equivalent exponential expression
Solution
\begin{align*} \sqrt[3]{w^{12}}\amp=w^{12/3}\\ \amp=w^4 \end{align*}
###### Example13.8.3.
Express $\sqrt{x^9}$ using an equivalent exponential expression
Solution
\begin{equation*} \sqrt{x^9}=x^{9/2} \end{equation*}
You can use Figure 13.8.4 to explore this definition some more.
As long as both the numerator and denominator of a rational exponent are fairly small positive numbers, it is fairly easy to evaluate expressions that include rational exponents using the rule $x^{m/n}=\sqrt[n]{x^m}\text{.}$
###### Example13.8.5.
Evaluate $16^{1/2}\text{.}$
Solution
\begin{align*} 16^{1/2}\amp=\sqrt{16}\\ \amp=4 \end{align*}
###### Example13.8.6.
Evaluate $8^{2/3}\text{.}$
Solution
\begin{align*} 8^{2/3}\amp=\sqrt[3]{8^2}\\ \amp=\sqrt[3]{64}\\ \amp=4 \end{align*}
###### Example13.8.7.
Evaluate $100^{3/2}\text{.}$
Solution
\begin{align*} 100^{3/2}\amp=\sqrt{100^3}\\ \amp=\sqrt{1000000}\\ \amp=1000 \end{align*}
When the numerator of the rational exponent is large, the rule $x^{m/n}=\sqrt[n]{x^m}$ can become quite cumbersome. Consider, for example, evaluating $9^{5/2}\text{.}$ If we try to use the standard form we hit a brick wall. First, it's not trivial to calculate that $9^5=59,049$ (reality check ... I grabbed my calculator). Now that I have the value of 59,049, I have to determine its square root. Oh my!
Fortunately for us, the application of the exponent and the application of the radical can be done in either order. That is:
\begin{equation*} a^{m/n}=\sqrt[n]{x^m} \text{ and } a^{m/n}=(\sqrt[n]{x})^m \end{equation*}
###### Example13.8.8.
Using the second option, evaluate $9^{5/2}\text{.}$
Solution
\begin{align*} 9^{5/2}\amp=(\sqrt{9})^5\\ \amp=3^5\\ \amp=243 \end{align*}
###### Example13.8.9.
Using the second option, evaluate $16^{7/4}\text{.}$
Solution
\begin{align*} 16^{7/4}\amp=(\sqrt[4]{16})^7\\ \amp=2^7\\ \amp=128 \end{align*}
Rational exponents are allowed to be negative. If that's the case, you probably want to deal with the negative aspect of the exponent before taking on the fractional aspect.
###### Example13.8.10.
Evaluate $27^{-2/3}\text{.}$
Solution
\begin{align*} 27^{-2/3}\amp=\frac{1}{27^{2/3}}\\ \amp=\frac{1}{(\sqrt[3]{27})^2}\\ \amp=\frac{1}{3^2}\\ \amp=\frac{1}{9} \end{align*}
Sometimes radical expressions can be simplified after first rewriting the expressions using rational exponents and applying the appropriate rules of exponents. If the resultant expression still has a rational exponent, it is standard to convert back to radical notation. Several examples follow.
###### Example13.8.11.
Use rational exponents to simplify $\text{.}$ Where appropriate, your final result should be converted back to radical form.
Solution
\begin{align*} \sqrt[3]{y^2} \cdot \sqrt[6]{y}\amp=y^{2/3}y^{1/6}\\ \amp=y^{2/3+1/6}\\ \amp=y^{5/6}\\ \amp=\sqrt[6]{y^5} \end{align*}
###### Example13.8.12.
Use rational exponents to simplify $\sqrt[8]{t^4}\text{.}$ Where appropriate, your final result should be converted back to radical form.
Solution
\begin{align*} \sqrt[8]{t^4}\amp=t^{4/8}\\ \amp=t^{1/2}\\ \amp=\sqrt{t} \end{align*}
###### Example13.8.13.
Use rational exponents to simplify $\sqrt[10]{\sqrt{5^{40}}}\text{.}$ Where appropriate, your final result should be converted back to radical form.
Solution
\begin{align*} \sqrt[10]{\sqrt{5^{40}}}\amp=\sqrt[10]{5^{40/2}}\\ \amp=\sqrt[10]{5^{20}}\\ \amp=5^{20/10}\\ \amp=5^2\\ \amp=25 \end{align*}
### ExercisesExercises
Convert each exponential expression to a radical expression and each radical expression to an exponential expression. When converting to a rational exponent, reduce the exponent if possible. Assume that all variables represent positive values.
###### 1.
$x^{1/3}$
Solution
$x^{1/3}=\sqrt[3]{x}$
###### 2.
$y^{5/4}$
Solution
$y^{5/4}=\sqrt[4]{y^5}$
###### 3.
$z^{2/5}$
Solution
$z^{2/5}=\sqrt[5]{z^2}$
###### 4.
$\sqrt[11]{x^5}$
Solution
$\sqrt[11]{x^5}=x^{5/11}$
###### 5.
$\sqrt[4]{y^{20}}$
Solution
\begin{aligned}[t] \sqrt[4]{y^{20}}\amp=y^{20/4}\\ \amp=y^5 \end{aligned}
###### 6.
$\sqrt[15]{t^3}$
Solution
\begin{aligned}[t] \sqrt[15]{t^3}\amp=t^{3/15}\\ \amp=t^{1/5} \end{aligned}
Determine the value of each expression.
###### 7.
$4^{1/2}$
Solution
\begin{aligned}[t] 4^{1/2}\amp=\sqrt{4}\\ \amp=2 \end{aligned}
###### 8.
$27^{-1/3}$
Solution
\begin{aligned}[t] 27^{-1/3}\amp=\frac{1}{27^{1/3}}\\ \amp=\frac{1}{\sqrt[3]{27}}\\ \amp=\frac{1}{3} \end{aligned}
###### 9.
$\left(\frac{4}{9}\right)^{-1/2}$
Solution
\begin{aligned}[t] \left(\frac{4}{9}\right)^{-1/2}\amp=\left(\frac{9}{4}\right)^{1/2}\\ \amp=\sqrt{\frac{9}{4}}\\ \amp=\frac{3}{2} \end{aligned}
###### 10.
$8^{7/3}$
Solution
\begin{aligned}[t] 8^{7/3}\amp=(\sqrt[3]{8})^7\\ \amp=2^7\\ \amp=128 \end{aligned}
###### 11.
$100^{5/2}$
Solution
\begin{aligned}[t] 100^{5/2}\amp=(\sqrt{100})^5\\ \amp=10^5\\ \amp=100,000 \end{aligned}
###### 12.
$16^{-9/4}$
Solution
\begin{aligned}[t] 16^{-9/4}\amp=\frac{1}{16^{9/4}}\\ \amp=\frac{1}{(\sqrt[4]{16})^9}\\ \amp=\frac{1}{2^9}\\ \amp=\frac{1}{512} \end{aligned}
Simplify each radical expression after first rewriting the expression in exponential form. Assume that all variables represent positive values. Where appropriate, your final result should be converted back to radical form.
###### 13.
$\sqrt[5]{t^{20}}$
Solution
\begin{aligned}[t] \sqrt[5]{t^{20}}\amp=t^{20/5}\\ \amp=t^4 \end{aligned}
###### 14.
$6\sqrt[33]{x^{77}}$
Solution
\begin{aligned}[t] 6\sqrt[33]{x^{77}}\amp=6x^{77/33}\\ \amp=6x^{7/3}\\ \amp=6\sqrt[3]{x^7} \end{aligned}
###### 15.
$(\sqrt{3})^{10}$
Solution
\begin{aligned}[t] (\sqrt{3})^{10}\amp=3^{10/2}\\ \amp=3^5\\ \amp=243 \end{aligned}
###### 16.
$\sqrt[4]{9^2}$
Solution
\begin{aligned}[t] \sqrt[4]{9^2}\amp=9^{2/4}\\ \amp=9^{1/2}\\ \amp=\sqrt{9}\\ \amp=3 \end{aligned}
###### 17.
$\sqrt{w}\sqrt[4]{w}$
Solution
\begin{aligned}[t] \sqrt{w}\sqrt[4]{w}\amp=w^{1/2}w^{1/4}\\ \amp=w^{3/4}\\ \amp=\sqrt[4]{w^3} \end{aligned}
###### 18.
$\sqrt[7]{x^6}\sqrt[7]{x}$
Solution
\begin{aligned}[t] \sqrt[7]{x^6}\sqrt[7]{x}\amp=x^{6/7}x^{1/7}\\ \amp=x^1\\ \amp=x \end{aligned}
###### 19.
$(\sqrt[12]{x^7y^{16}})^{36}$
Solution
\begin{aligned}[t] (\sqrt[12]{x^7y^{15}})^{36}\amp=(x^7y^{16})^{36/12}\\ \amp=(x^7y^{16})^3\\ \amp=x^{21}y^{48} \end{aligned}
###### 20.
$\sqrt[5]{\sqrt[3]{x^{15}}}$
Solution
\begin{aligned}[t] \sqrt[15]{\sqrt[3]{x^{15}}}\amp=\sqrt[15]{x^{15/3}}\\ \amp=\sqrt[15]{x^5}\\ \amp=x^{5/15}\\ \amp=x^{1/3}\\ \amp=\sqrt[3]{x} \end{aligned} | 2019-12-07T19:53:02 | {
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https://www.intmath.com/forum/exponents-radicals-16/exponential-functions:101 | IntMath Home » Forum home » Exponents and Radicals » exponential functions
# exponential functions [Solved!]
### My question
In expressions containing raised variables that are being divided, why do the variables not cancel each other out to a reciprocal of one? for example x^n/ x should be 1^n or is it still x^n?
### Relevant page
factoring polynomials with exponents - Google Search
### What I've done so far
(x^n/ x^n-2)^3 does this equal 1 or x^6?
X
In expressions containing raised variables that are being divided, why do the variables not cancel each other out to a reciprocal of one? for example x^n/ x should be 1^n or is it still x^n?
Relevant page
<a href="https://www.google.com/search?q=factoring+polynomials+with+exponents&oq=factoring+polynomials+with+exponents&aqs=chrome..69i57.14377j0j4&sourceid=chrome&ie=UTF-8">factoring polynomials with exponents - Google Search</a>
What I've done so far
(x^n/ x^n-2)^3 does this equal 1 or x^6?
## Re: exponential functions
Hello Sam
It's often best to try some real numbers in algebraic expressions first, to see what is going on.
Let's try 10^4.
This means 10 xx 10 xx 10 xx 10
If we divide this by 10, we would have
(10 xx 10 xx 10 xx 10)/10
We cancel one of the 10s on top with the 10 on bottom, to give:
10 xx 10 xx 10
So what we've done is
(10^4)/10 = 10^3
If you do this for several other indices, you'll hopefully conclude that:
(10^n)/10 = 10^(n-1)
In general, for your first question, we'd have:
x^n/ x = x^(n-1)
For your other question, (x^n/ x^n-2)^3, I'm not sure if you mean (x^n/ x^n-2)^3 or (x^n/ (x^n-2))^3?
You are encouraged to use the math entry system which makes it easier to express your questions and follow your working.
X
Hello Sam
It's often best to try some real numbers in algebraic expressions first, to see what is going on.
Let's try 10^4.
This means 10 xx 10 xx 10 xx 10
If we divide this by 10, we would have
(10 xx 10 xx 10 xx 10)/10
We cancel one of the 10s on top with the 10 on bottom, to give:
10 xx 10 xx 10
So what we've done is
(10^4)/10 = 10^3
If you do this for several other indices, you'll hopefully conclude that:
(10^n)/10 = 10^(n-1)
In general, for your first question, we'd have:
x^n/ x = x^(n-1)
For your other question, (x^n/ x^n-2)^3, I'm not sure if you mean (x^n/ x^n-2)^3 or (x^n/ (x^n-2))^3?
You are encouraged to use the <a href="http://www.intmath.com/forum/entering-math-graphs-images-41/how-to-enter-math:91">math entry system</a> which makes it easier to express your questions and follow your working.
## Re: exponential functions
Sam didn't reply.
If his second question meant (x^n/ x^n-2)^3, the the result is (1-2)^3=(-1)^3=-1
If he meant (x^n/ (x^n-2))^3, then there is no "nice" simplification.
X
Sam didn't reply.
If his second question meant (x^n/ x^n-2)^3, the the result is (1-2)^3=(-1)^3=-1
If he meant (x^n/ (x^n-2))^3, then there is no "nice" simplification.
## Reply
You need to be logged in to reply. | 2017-03-28T19:42:11 | {
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http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron/11086 | Height of a tetrahedron
How do I calculate: The height of a regular tetrahedron, side length 1.
Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.
-
I'm curious. If you mind telling us, what was the purpose of the interview? What were the other mathematically relevant questions asked? – user02138 Nov 20 '10 at 15:25
It's an old question for and Economics and Management interview at Oxford. – Patrick Beardmore Nov 20 '10 at 16:43
The first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.
Here's a view of the geometry:
and here's a view of the bottom face:
In the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.
Knowing that the short sides of the isosceles triangle bisect the 60° angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30°, 30° and 120°.
Using the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):
$$1=2\ell^2-2\ell^2\cos 120^{\circ}$$
Solving for $\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\frac{1}{\sqrt{3}}$, as indicated here.
From this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies
$$h^2+\left(\frac{1}{\sqrt{3}}\right)^2=1$$
Solving for $h$ in the above equation, we now find the height to be $\sqrt{\frac23}=\frac{\sqrt{6}}{3}$, as mentioned here.
-
It's not homework, it was an interview question I struggled with. I'd love to know the answer. – Patrick Beardmore Nov 20 '10 at 14:57
Good to know. I'll edit with an explicit answer. – Ｊ. Ｍ. Nov 20 '10 at 15:00
Thank-you very much. P.S: how did you make your lovely 3d shape diagrams? – Patrick Beardmore Nov 20 '10 at 16:43
@Patrick: I used Mathematica 5.2 (yes, it's an old copy :) ) for these diagrams. – Ｊ. Ｍ. Nov 20 '10 at 21:33
Consider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\frac{2}{\sqrt 3}$, and since the edges of this tetrahedron have length $\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \sqrt{\frac{2}{3}}$.
-
Note that this way of doing this also gives that the distance from the centers of opposite sides of the tetrahedron is always ${1 \over \sqrt{2}}$ times the side length. – Zarrax Nov 20 '10 at 15:26
You can also use trig based on the dihedral angle between two faces of the tetrahedron.
Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$
$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.
As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:
$$\begin{eqnarray} |OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\ s^2 &=& \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\ s^2 &=& \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\ 1 &=& \frac{3}{2} - \frac{3}{2} \cos{M} \\ \frac{-1}{2} &=& - \frac{3}{2} \cos{M} \\ \frac{1}{3} &=& \cos{M} \;\;\; (**)\\ \Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&=& \sin{M} \end{eqnarray}$$
Therefore,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$
(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):
$$\begin{eqnarray} \cos\left({\text{angle between two sides of a regular triangle}}\right) &=& \frac{1}{2}\\ \cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &=& \frac{1}{3}\\ \cos\left({\text{angle between two facets of a regular n-simplex}}\right) &=& \frac{1}{n} \end{eqnarray}$$
(Who would've suspected, upon first encountering it, that the "$2$" in "$\cos{60^{\circ}}=\frac{1}{2}$" was actually a reference to the dimension of the triangle?)
-
Thank-you v. much! What's a simplex? – Patrick Beardmore Nov 20 '10 at 17:14
A simplex is the analog of "triangle" in any dimension: the simplest possible shape. It's what you get when you join $(n+1)$ points in $n$-dimensional space. A triangle is a "$2$-simplex" ($3$ points in $2$ dimensions); a tetrahedron is a "$3$-simplex" ($4$ points in $3$ dimensions); going upward in dimensions, one generally just says "$4$-", "$5$-", ..., "$n$-simplex"; going downward, one can say the line segment is a "1-simplex" (determined by $2$ points) and the point is a "0-simplex" ($1$ point!). See en.wikipedia.org/wiki/Simplex and mathworld.wolfram.com/Simplex.html – Blue Nov 20 '10 at 17:30
Very nice! I never knew about this pattern until now, thanks! :D – Ｊ. Ｍ. Nov 20 '10 at 21:37 | 2014-08-30T10:29:59 | {
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https://math.stackexchange.com/questions/2429631/put-7-balls-into-7-cells-probability-that-exactly-2-cells-containing-3 | # Put $7$ balls into $7$ cells. Probability that exactly $2$ cells containing $3$ balls?
1.46. Seven balls are distributed randomly into seven cells. What is the probability that the number of cells containing exactly $3$ balls is $2$?
I am getting different answer from this solution manual. My argument is the following:
In order for us to have an arrangement where there are exactly $2$ cells with $3$ balls, we can follow the following procedure. We first decide which $3$ balls we want to put them together among all $7$ balls, and then decide to which cell we want to put them in. This gives us $\binom{7}{3} \binom{7}{1}$. Now, among the left $4$ balls, we choose $3$ balls to put them together, and choose one cell among the remaining $6$ empty cells. This gives us $\binom{4}{3} \binom{6}{1}$. Finally, we are left with one ball, and we have $5$ choices regarding where to put it. Putting them together, we have
$$P(X_3 = 2) = \frac{\binom{7}{3} \binom{7}{1} \binom{4}{3} \binom{6}{1} 5}{7^7}$$
But the solution manual says the answer should be
$$\frac{\binom{7}{2}\binom{7}{3}\binom{4}{3}5}{7^7}$$
Who is wrong and why?
• I think, that the problem is, that you must divide by 2, because in your case you specify, that the first 3 goes to the one and the other 3 goes to the other cell. However, the order of cells doesn't matter. Therefore, divide by 2 and get it. Jan 21, 2018 at 14:36
The manual is right. The difference between your solution and the books solution is a factor of $2.$ That is ${7\choose1}{6\choose1} = 2{7\choose 2}$ | 2022-08-09T00:44:30 | {
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http://math.stackexchange.com/questions/11085/height-of-a-tetrahedron | # Height of a tetrahedron
How do I calculate: The height of a regular tetrahedron, side length 1.
Just to be completely clear, by height I mean if you placed the shape on a table, how high up would the highest point be from the table.
-
I'm curious. If you mind telling us, what was the purpose of the interview? What were the other mathematically relevant questions asked? – user02138 Nov 20 '10 at 15:25
It's an old question for and Economics and Management interview at Oxford. – Patrick Beardmore Nov 20 '10 at 16:43
The first thing you need to do is to note that the apex of a regular tetrahedron lies directly above the center of the bottom triangular face. Thus, find the length of the segment connecting the center of an equilateral triangle with unit length to a corner, and use the Pythagorean theorem with the length of an edge as the hypotenuse, and the length you previously derived as one leg. The height you need is the other leg of the implied right triangle.
Here's a view of the geometry:
and here's a view of the bottom face:
In the second diagram, the face is indicated by dashed lines, and the (isosceles) triangle formed by the center of the triangle and two of the corners is indicated by solid lines.
Knowing that the short sides of the isosceles triangle bisect the 60° angles of the equilateral triangle, we find that the angles of the isosceles triangle are 30°, 30° and 120°.
Using the law of cosines and the knowledge that the longest side of the isosceles triangle has unit length, we have the equation for the length $\ell$ of the short side (the length from the center of the bottom face to the nearest vertex):
$$1=2\ell^2-2\ell^2\cos 120^{\circ}$$
Solving for $\ell$, we find that the length from the center of the bottom face to the nearest vertex is $\frac{1}{\sqrt{3}}$, as indicated here.
From this, the Pythagorean theorem says that the height $h$ (the length from the center of the bottom face) satisfies
$$h^2+\left(\frac{1}{\sqrt{3}}\right)^2=1$$
Solving for $h$ in the above equation, we now find the height to be $\sqrt{\frac23}=\frac{\sqrt{6}}{3}$, as mentioned here.
-
It's not homework, it was an interview question I struggled with. I'd love to know the answer. – Patrick Beardmore Nov 20 '10 at 14:57
Good to know. I'll edit with an explicit answer. – Guess who it is. Nov 20 '10 at 15:00
Thank-you very much. P.S: how did you make your lovely 3d shape diagrams? – Patrick Beardmore Nov 20 '10 at 16:43
@Patrick: I used Mathematica 5.2 (yes, it's an old copy :) ) for these diagrams. – Guess who it is. Nov 20 '10 at 21:33
Consider the tetrahedron inscribed in the unit cube, with vertices at (0,0,0), (1,1,0), (0,1,1), (1,0,1). Its height is the distance from (0,0,0) to the centre of the opposite face, which is given by the equation $x+y+z = 2$. Thus its height is $\frac{2}{\sqrt 3}$, and since the edges of this tetrahedron have length $\sqrt 2$, the height of a regular tetrahedron with side $x$ is $x \sqrt{\frac{2}{3}}$.
-
Note that this way of doing this also gives that the distance from the centers of opposite sides of the tetrahedron is always ${1 \over \sqrt{2}}$ times the side length. – Zarrax Nov 20 '10 at 15:26
You can also use trig based on the dihedral angle between two faces of the tetrahedron.
Writing $ABC$ for the base triangle, $O$ for the apex, $K$ for the center of $ABC$ (the foot of the perpendicular dropped from $O$), and $M$ for the midpoint of (for instance) side $BC$, we have a right triangle $OKM$ with right angle at $K$. So,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M}$$
$OM$ is the height of the (equilateral) face $OBC$, measuring $\frac{\sqrt{3}}{2}s$, where $s$ is the length of a side.
As for the measure of angle $M$ ... Note that this is the dihedral angle between faces $OBC$ and $ABC$; it is also the angle between (congruent) segments $OM$ and $AM$ in triangle $OMA$. We can use the Law of Cosines as follows:
$$\begin{eqnarray} |OA|^2 &=& |OM|^2 + |AM|^2 - 2 |OM||AM|\cos{M} \\ s^2 &=& \left(\frac{\sqrt{3}}{2}s\right)^2 + \left(\frac{\sqrt{3}}{2}s\right)^2 - 2 \left(\frac{\sqrt{3}}{2}s\right)\left(\frac{\sqrt{3}}{2}s\right) \cos{M} \\ s^2 &=& \frac{3}{4} s^2 + \frac{3}{4}s^2 - 2 \frac{3}{4} s^2 \cos{M} \\ 1 &=& \frac{3}{2} - \frac{3}{2} \cos{M} \\ \frac{-1}{2} &=& - \frac{3}{2} \cos{M} \\ \frac{1}{3} &=& \cos{M} \;\;\; (**)\\ \Rightarrow \sqrt{1-\left(\frac{1}{3}\right)^2} = \frac{\sqrt{8}}{3} =\frac{2\sqrt{2}}{3}&=& \sin{M} \end{eqnarray}$$
Therefore,
$$\text{height of tetrahedron} = |OK| = |OM|\sin{M} = \frac{\sqrt{3}}{2} s \cdot \frac{2\sqrt{2}}{3} = \frac{\sqrt{6}}{3}s$$
(**) This cosine is the reason I posted this approach. It's sometimes handy to know (as in this problem); even better, it's easy to remember, because it turns out that it fits a simple pattern (which might be more-likely to impress interviewers):
$$\begin{eqnarray} \cos\left({\text{angle between two sides of a regular triangle}}\right) &=& \frac{1}{2}\\ \cos\left({\text{angle between two faces of a regular tetrahedron}}\right) &=& \frac{1}{3}\\ \cos\left({\text{angle between two facets of a regular n-simplex}}\right) &=& \frac{1}{n} \end{eqnarray}$$
(Who would've suspected, upon first encountering it, that the "$2$" in "$\cos{60^{\circ}}=\frac{1}{2}$" was actually a reference to the dimension of the triangle?)
-
Thank-you v. much! What's a simplex? – Patrick Beardmore Nov 20 '10 at 17:14
A simplex is the analog of "triangle" in any dimension: the simplest possible shape. It's what you get when you join $(n+1)$ points in $n$-dimensional space. A triangle is a "$2$-simplex" ($3$ points in $2$ dimensions); a tetrahedron is a "$3$-simplex" ($4$ points in $3$ dimensions); going upward in dimensions, one generally just says "$4$-", "$5$-", ..., "$n$-simplex"; going downward, one can say the line segment is a "1-simplex" (determined by $2$ points) and the point is a "0-simplex" ($1$ point!). See en.wikipedia.org/wiki/Simplex and mathworld.wolfram.com/Simplex.html – Blue Nov 20 '10 at 17:30
Very nice! I never knew about this pattern until now, thanks! :D – Guess who it is. Nov 20 '10 at 21:37
btw, $\cos\left({\text{angle between two faces of a regular tetrahedron}}\right) = -\frac{1}{3}$ – Narasimham Mar 16 at 0:07
@Narasimham: The angle between two faces of a regular tetrahedron is acute, so its cosine must be positive. – Blue Mar 16 at 0:45
I'd like to offer a slightly simpler approach to part of the 1st answer above. We know that the equilateral triangle of the base of the tetrahedron has sides of 1, 1, and 1, and we know we can split that in half, creating two right triangles having sides of hypotenuse=1, base=1/2, and perpendicular=√3/2, along with angles of 30, 60, and 90 degrees.
Now consider the 2nd diagram of the 1st answer, which shows a solid-line triangle having angles of 30, 30, and 120 degrees. That triangle could be divided in half, creating two right triangles having angles of 30, 60, and 90 degrees. If we consider the base of each triangle to be its shortest side, then the perpendicular of either one of those triangles has a length of 1/2. We can now use the power of ratios to compute the other two sides:
(1/2):(√3/2):(1) --triangle 1: half of tetrahedral face, angles of 30, 60 & 90 degrees.
( ):( 1/2):( ) --triangle 2: has unknown base & hypotenuse, but is proportionate to triangle 1.
We really only need to compute the hypotenuse of triangle 2, because that is the desired distance from the corner to the center of the tetrahedron's base:
Multiply triangle-1-hypotenuse by triangle-2-perpendicular; divide by triangle-1-perpendicular.
[(1/2)(1)]/(√3/2) = (1/2)(2/√3) = 1/√3, as the 1st answer also computes in a more complicated way.
For the sake of completeness, since in any 30-60-90-degree triangle the base is simply half the length of the hypotenuse, the second unknown is 1/(2√3) or √3/6 (although it could also have been figured by using ratios, as above). The reader is invited to verify that the square of (√3/6) plus the square of 1/2 equals the square of (1/√3).
-
The normal height ($H_{n}$) of any regular tetrahedron having edge length $a$ is equal to the sum of radii of its inscribed & circumscribed spheres which is given as follows $$H_{n}=\frac{a}{2\sqrt{6}}+\frac{a}{2}\sqrt{\frac{3}{2}}=\frac{4a}{2\sqrt{6}}=a\sqrt{\frac{2}{3}}$$ Hence, the normal height ($H_{n}$) of regular tetrahedron with edge length $a$ is generalized by the formula $$\bbox[4pt, border: 1px solid blue;] {H_{n}=a\sqrt{\frac{2}{3}}}$$ As per given value of edge length $a=1$ in the question, the normal height of tetrahedron is $\sqrt{\frac{2}{3}}$
Note: for derivation & detailed explanation, kindly go through HCR's Formula for Regular n-Polyhedrons
- | 2015-05-27T06:48:44 | {
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https://math.stackexchange.com/questions/2134265/can-endpoints-be-local-minimum | # Can endpoints be local minimum?
My textbook defines local maximum as follows:
A function $$f$$ has local maximum value at point $$c$$ within its domain $$D$$ if $$f(x)\leq f(c)$$ for all $$x$$ in its domain lying in some open interval containing $$c$$.
The question asks to find any local maximum or minimum values in the function
$$g(x)=x^2-4x+4$$ in the domain $$1\leq x<+\infty$$.
The answer at the back has the point $$(1,1)$$, which is the endpoint.
According to the definition given in the textbook, I would think endpoints cannot be local minimum or maximum given that they cannot be in an open interval containing themselves. (ex: the open interval $$(1,3)$$ does not contain $$1$$). Where am I wrong?
• Your question is an excellent and important one. It is common to include endpoints in these types of calculations, and note that every point is in an open interval, just not necessarily an open interval in the domain of the function. Be sure to also check these definitions with and instructor or professor for total clarification. – The Count Feb 8 '17 at 2:15
• This is mainly just a matter of convention - lacking a deep mathematical basis for preferring inclusion or exclusion of these points. I have seen professors and texts that disagree on this point. What really matters is that a convention is decided upon and used consistantly. I would agree that you ought to ask your professor for clarification on the convention they want to use. – David Feb 8 '17 at 3:19
• Yes, it is a matter of convention. I personally think that the definition where "open" means open in $D$ makes more sense. Apparently, so does your textbook. However, the majority of Calculus textbooks that I've seen specifically exclude endpoints from the definition of local extrema, i.e. they treat "open" as being open in $\mathbb{R}$. – zipirovich Feb 8 '17 at 4:46
Actually, the question is settled by reading the definition you provided carefully:
A function $f$ has local maximum value at point $c$ within its domain $D$ if $f(x)\leq f(c)$ for all $x$ in its domain lying in some open interval containing $c$.
I.e., the points $x$ for which the condition must hold are required to both be in the open interval and in $D$.
To see that $(1,1)$ is a local maximum, consider the open interval $(0, 2)$. If $x \in (0, 2)$ and $x$ is also in the domain $[1,\infty)$, then $1 \le x < 2$. Now $g(x) = x^2-4x + 4 = (2 - x)^2$. So $g(1) = (2 - 1)^2 = 1^2 = 1$, but if $x > 1$, then $0 < 2 - x < 1$, so $0 < (2-x)^2 = g(x) < 1$. So for $x$ in the open interval $(0,2)$ and also in the domain $[1,\infty)$, we have that $g(x) \le g(1)$.
• For a different perspective, this is not how my calculus professor taught it recently. She said that if you are at an endpoint, you cannot compare the values outside of the interval, so you would not include it as a local extremum. – Max Li Nov 18 '17 at 20:58
• @MaxLi - your calculus professor said WHAT!? Either you misunderstood your what your calculus professor was saying (which I unfortunately have to rate at only a 90% probability instead of the 100% probability I would prefer to), or else I advise you to carefully double-check everything she tells you against reliable sources and avoid taking any more classes from her in the future. Some things I can shrug off as differences in taste (is $0 \in \Bbb N$ or not?), and some things as pointless but not really harmful (is $0^0 = 1$ or undefined?). But this violates the basic meaning of the words. – Paul Sinclair Nov 19 '17 at 0:52
• @PaulSinclair, it's actually common for textbooks to take this perspective. For example, the book with scans posted on this question (which is Stewart Calculus, if I'm not mistaken). – PersonX Feb 21 '18 at 2:28
• @PersonX - Just because my comment has a wider application than one person does not at all make me recant. – Paul Sinclair Feb 21 '18 at 13:33
• @MattBrenneman - $[a,c)$ is indeed an open neighborhood of $a$ in $D$. And it is exactly because "local extremum" is defined on topological spaces as "extremum when restricted to some neighborhood" that I take exception to Max Li's professor's remark. Under this definition, $a$ is a local extremum, just as it is under the explanation I gave in the post (which essentially amounts to the same thing, but without the subspace terminology). Max Li's professor would deny it that status, even though it is greater (or lesser) than everything near it in the domain. – Paul Sinclair Oct 24 '18 at 16:22
I think fundamentally the comments are right, and you should speak with your teacher to confirm definitions and expectations. But there's also a point to make about topology here, which could justify the book's definition and answer as consistent.
The definition of local maximum you gave is:
A function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \leq f(c)$ for all $x$ in its domain lying in some ** open ** interval containing $c$.
If you interpret this as saying that the interval can come from $\mathbb{R}$, and is not restricted to $D$, then you have no problem, as others have pointed out. But like you I am thinking about being restricted to $D$ and my instinct is to think only about intervals in $D$. This can still be ok, if we just alter our interpretation of "open" a little bit (in a natural way)...
Now, whenever we say "open" we're really saying "open with respect to ** insert topology here ** ." A lot of the time it's obvious from context or the textbook has established a practice of contextual implication, but in this case (without knowing your book) I'd argue there are two reasonable interpretations:
1. We might be talking open intervals with respect to the standard topology on $\mathbb{R}$ (which is what you've probably been using in your class), but
2. since we're restricting our attention to a domain $D \subset \mathbb{R}$, it's also pretty normal to talk about a different topology, called the subset topology on $D$ (induced by the standard topology on $R$).
In the subset topology on $D \subset \mathbb{R}$ (induced by the standard topology), a set $S$ is open if and only if $S$ is the intersection $D \cap X$, with $X$ open in $\mathbb{R}$ with respect to the standard topology on $\mathbb{R}$.
We're often more interested in the subset topology than the usual topology on the whole space just because of situations like the one you're in, in which a definition doesn't work quite like you expect when $D \not= \mathbb{R}$.
So let's work with a slightly different definition of local maximum:
A function $f$ has a local maximum at point $c$ within its domain $D$ if $f(x) \leq f(c)$ for all $x$ in its domain lying in some interval $I$ containing $c$ such that $I$ is open with respect to the subset topology on $D$.
Now back to your case. Let $D = [1, \infty)$. For any $a > 1$, we have that $$[1,a) = D \cap (-a,a)$$ Since $(-a,a)$ is open in $\mathbb{R}$ with respect to the standard topology, $[1,a)$ is open in $D$ with respect to the subset topology on $D$. This intuitively makes sense, because if you were an ant walking on $f(D)$, when you came to $f(1)$ you'd have nowhere to go but down.
• I am going to have to spend some time trying to understand your answer, but thank you nonetheless. – Phil Feb 11 '17 at 23:13 | 2021-03-08T11:21:00 | {
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http://thevelveetaroom.com/don-barclay-xgwmngs/definition-of-rational-numbers-for-class-7-d81286 | If the numerator and the denominator both are either positive integers or negative integers, then the rational number is positive. The set of all rational numbers, often referred to as "the rationals" , the field of rationals or the field of rational numbers is usually denoted by a boldface Q (or blackboard bold $${\displaystyle \mathbb {Q} }$$, Unicode /ℚ); it was thus denoted in 1895 by Giuseppe Peano after quoziente, Italian for "quotient". Solution: Our notes of Chapter 1 Rational numbers are prepared by Maths experts in an easy to remember format, covering all syllabus of CBSE, KVPY, NTSE, Olympiads, NCERT & other Competitive Exams. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. Solution: Let the required rational number be x. The integers which are in the form of p/q where q is not equal to 0 are known as Rational Numbers. It is in the form of and q ≠ 0. Yes, it is a rational number. Copyright © 2021 Applect Learning Systems Pvt. Every integer is a rational number: for example, 5 = 5/1. we get, 14 x 15 = 210. and 12 x 21 = 252. A number that can be made by dividing two integers (an integer is a number with no fractional part). Advantages Of CBSE NCERT Class 8 Rational Numbers Worksheets . You can also register Online for Class 7 Science tuition on Vedantu.com to score more marks in CBSE board examination. It is something related to the ratios. “Any number which can be expressed in the form , where p and q are integers and, is called a rational number.” For example, is a rational number in which the numerator is 15 and the denominator is 19. Examples of Rational Numbers. (i) Absolute value = -12171=12171 What Is A Rational Number? 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Conventions used for writing a rational number: We know that in a rational number, the numerator and denominator both can be positive or negative. A rational number is defined as a number that can be expressed in the form p/q, where p and q are integers and. On solving equations like 3x + 5 = 0, we get the solution as x = -5/3. The absolute value of a rational number is its numerical value regardless of its sign. If the product of two rational numbers is $$\frac { -9 }{ 16 }$$ and one of them is $$\frac { -4 }{ 15 }$$, find the other number. Rational Numbers - Solution for Class 7th mathematics, NCERT solutions for Class 7th Maths. A rational number which has either the numerator negative or the denominator negative is called the negative rational number. Class 8 Maths Rational Numbers: Closure Properties: Properties of the types of numbers - Closure . So, 14/21 ≠ 12/15. A rational number is defined as a number that can be expressed in the form $$\frac{p}{q}$$, where p and q are integers and q≠0. The solution -5/3 is … Example 3. Get Textbook solutions for maths from evidyarthi.in You have studied fractional numbers in your earlier classes. These numbers are also known as rational numbers. Let us now write the given numbers according to the convention. Our Solutions contain all type Questions with Exe-1 A, Exe-1 B, Exe-1 … Thus, we can say that every integer is a rational number. Rational number 1 is the multiplicative identity for all rational numbers because on multiplying a rational number with 1, its value does not change. A set of numbers is said to be closed for a specific mathematical operation if the result obtained when an operation is performed on any two numbers in the set, is itself a member of the set. According to convention, the given number should be written as . Conventionally, rational numbers are written with positive denominators. On solving equations like 3x + 5 = 0, we get the solution as x = -5/3. Any rational number can be called as the positive rational number if both the numerator and denominator have like signs. NCERT CBSE Class 7 Maths Rational Numbers Solutions provide detailed explanations to the, Chapter 9 exercises. Some examples of fractional numbers are. Also, understand how to represent rational numbers on a number line with the support of our ICSE Class 8 Maths learning materials. The numerator and the denominator of a rational number will be integers. Write each of the following rational numbers according to the convention. The examples of rational numbers will be 1/4, 2/7, - 3/10, 34/7, etc. Sorry!, This page is not available for now to bookmark. They meet the requirements of students to make their basics strong in the subject. Examples: p q p / q = 1: 1: 1/1: 1: 1: 2: 1/2: 0.5: 55: 100: 55/100: 0.55: 1: 1000: 1/1000: 0.001: 253: 10: 253/10: 25.3 : 7: 0: 7/0: No! The absolute value of a rational number pq is denoted as pq. In short, rational number represents a ratio of two integers. and experience Cuemath's LIVE Online Class with your child. For example, are positive rational numbers. In other words, the rational number is defined as the ratio of two numbers (i.e., fractions). This slideshow lesson is very animated with … Dive into the topic of rational numbers with TopperLearning’s ICSE Class 8 Maths Chapter 1 study materials. "q" can't be zero! If are two rational numbers such that q > 0 and s > 0 then it can be said that if ps > qr. The rational number is represented using the letter “Q”. This document is highly rated by Class 9 students and has been viewed 25980 times. A rational number is a number that can be written in the form p q, where p and q are integers. Rational Numbers ICSE Class-8th Concise Selina Solutions Chapter-1. Many people are surprised to know that a repeating decimal is a rational number. Rational Numbers 1. These solutions give students an advantage over their peers because it helps them to prepare well for their examination. Properties of rational number: There is infinite number … −34 can be written as. Here, “p” is a numerator and “q” is a denominator. Question 8. More formally we say: A rational number is a number that can be in the form p/q where p and q are integers and q is not equal to zero. Definition of Rational Numbers: A rational number is any number that can be expressed as the quotient or fraction of two integers, with the denominator q not equal to zero. Definition of Rational Numbers. (c) q ≠ 1 (d) q ≠ 0. Students learn the definition of rational number, and they write rational numbers as ratios of integers and as repeating or terminating decimals. Class 7 Maths Chapter 9: Rational Numbers. Have a Query? Key Concepts . Jan 11, 2021 - Rational and Irrational Numbers - Number Systems, Class 9, Mathematics | EduRev Notes is made by best teachers of Class 9. Absolute Value of a Rational Number: Like real numbers, the Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. Since, 14 x 15 ≠ 12 x 21. In ratios, the numerator and denominator both are positive numbers while in rational numbers, they can be negative also. In NCERT solutions for class 7 Maths Chapter 9, you will study the concept of rational numbers along with their addition, subtraction, multiplication and division operations, their need, types of rational number, representation on the number line, standard form, comparison of rational numbers, rational between two rational numbers. In Mathematics, a rational number is defined as a number, which is written in the form p/q, where, q ≠ 0. 3. Mathematics NCERT Grade 7, Chapter 9: Rational Numbers- As the name suggests, the chapter deals with rational numbers.A detailed explanation about rational numbers is given in the chapter.Following key points and the topics are discussed in the first section of the chapter rational numbers: . Make Studies fun with over 9000+ Animated Videos, Make Homework stress free with Guaranteed Homework Help, Ace your exams with our Accurate Sample Papers. In our daily lives, we use some quantities which are not whole numbers but can be expressed in the form of $$\frac{p}{q}$$. Need for rational numbers This means that 0 can also be a rational number as a rational number can be represented as … A rational number is a number that can be expressed as a ratio of p/q, where p and q are integers, and q does not equal to zero. Therefore, -32=32, 12-7=127 etc. Yes, it is a rational number. Now, let us go through the given example. If in a rational number, either the numerator or the denominator is a negative integer, then the rational number is negative. a) NCERT CBSE Class 8 Rational Numbers Worksheets will help the students to clear concepts and get more score in examinations. As, 1 2, 4 5, 5 7 are in standard form. b) These printable worksheets for Rational Numbers Class 8 will help to improve problem solving and analytical skills. A rational number is said to be in the standard form if its denominator is a positive integer and the numerator and denominator have no common factor other than 1. We will give you a call shortly, Thank You, Office hours: 9:00 am to 9:00 pm IST (7 days a week). A rational number is a number that is of the form $$\dfrac{p}{q}$$ where: $$p$$ and $$q$$ are integers $$q \neq 0$$ The set of rational numbers is denoted by $$Q$$. Example: Examples : 5/8; -3/14; 7/-15; -6/-11 3. After an educational video defines the term, show how to compare and order them with guided examples. NCERT Solutions for Class 7 Maths Chapter 9: Two rational numbers with different denominators are added by first taking the LCM of the two denominators and then converting both the rational numbers to their equivalent forms having the LCM as the denominator. Properties of Rational Numbers 1) Closure Property 2) Associative Property 3) Distributive Law 4) Additive Inverse 5) Multiplicative Inverse The best techniques and approach to each question on NCERT Solutions for Class 7 Maths Chapter 9 Rational Numbers are discussed in great detail by the subject matter experts. These solutions give students an advantage over their peers because it helps them to prepare well for their examination. Solution : (d) By definition, a number that can be expressed in the form of p/q, where p and q are integers and q≠0, is … MCQ Questions for Class 7 Maths: Ch 9 Rational Numbers. We will call you right away. Rational numbers: The numbers which can be expressed as ratio of integers are known as rational numbers. We provide step by step Solutions of Exercise / lesson-1 Rational Numbers for ICSE Class-8 Concise Selina Mathematics. According to convention, the given number should be written as . For example, is a rational number in which the numerator is 15 and the denominator is 19. Explained well with examples. CBSE Class 7 Maths Chapter 9 – Rational Numbers. Rational Numbers Class 7 Extra Questions Short Answer Type. These decimal numbers are also rational numbers as these can be written as. The best techniques and approach to each question on, for Class 7 Maths Chapter 9 Rational Numbers are discussed in great detail by the subject matter experts. Use this lesson plan to teach your students about rational numbers. Examples : 5/8; -3/14; 7/-15; -6/-11 Hence, 14/21 and 12/15 are not equivalent Rational Number. Now, is −34 a rational number? 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https://bassamshakir.com/3g7tho/szudzik-pairing-function-588218 | Wolfram Science Conference NKS 2006. \right.$$However, a simple transformation can be applied so that negative input can be used. In this ramble we will cover two different pairing functions: Cantor and Szudzik. Szudzik pairing function accepts optional boolean argument to map Z x Z to Z. Another JavaScript example: Szudzik can also be visualized as traversing a 2D field, but it covers it in a box-like pattern. As such, we can calculate the max input pair to Szudzik to be the square root of the maximum integer value. But for R the Axiom of Choice is not required. For the Cantor function, this graph is traversed in a diagonal function is illustrated in the graphic below. September 17, 2019 2:47 AM. The Rosenberg-Strong Pairing Function. 5 0 obj %�쏢 For example, cantor(33000, 33000) = 2,178,066,000 which would result in an overflow. For the Szudzik pairing function, the situation is only slightly more complicated. Examples 148 VIEWS. Other than that, the same principles apply. This can be easily implemented in any language. In: Wolfram Research (ed.) -2x - 1 & : x < 0\\ b^2 + a & : a < b\\ /// /// So, if user didn't make something stupid like overriding the GetHashCode() method with a constant, /// we will get the same unique number for the same row and column every time. \right.$$, $$c(a,b) = \left\{\begin{array}{ll} Essentially any time you want to compose a unique identifier from a pair of values. \end{array} a^2 + a + b & : a \ge b Given two points 8u,v< and 8x,y<, the point 8u,v< occurs at or before 8x,y< if and only if PairOrderedQ@8u,v<,8x,y= 0 The performance between Cantor and Szudzik is virtually identical, with Szudzik having a slight advantage. You can then map the row to an X axis, the column to an Y axis. Yes, the Szudzik function has 100% packing efficiency. /// 2- We use a pairing function to generate a unique number out of two hash codes. \end{array} It should be noted though that all returned pair values are still positive, as such the packing efficiency for both functions will degrade. a^2 + a + b & : a \ge b The formula for calculating mod is a mod b = a - b[a/b]. od_id* functions take two vectors of equal length and return a vector of IDs, which are unique for each combination but the same for twoway flows. the Szudzik pairing function, on two vectors of equal length. -c - 1 & : (a < 0 \cap b \ge 0) \cup (a \ge 0 \cap b < 0) For a 32-bit unsigned return value the maximum input value for Szudzik is 65,535. 62 no 1 p. 55-65 (2007) – In this paper, some results and generalizations about the Cantor pairing function are given. (Submitted on 1 Jun 2017 ( v1 ), last revised 28 Jan 2019 (this version, v5)) Abstract: This article surveys the known results (and not very well-known results) associated with Cantor's pairing function and the Rosenberg-Strong pairing function, including their inverses, their generalizations to higher dimensions, and a discussion of a few of the advantages of the Rosenberg … a * a + a + b : a + b * b; where a, b >= 0 Generate ordered ids of OD pairs so lowest is always first This function is slow on large datasets, see szudzik_pairing for faster alternative Usage od_id_order(x, id1 = names(x)[1], id2 = names(x)[2]) function pair(x,y){return y > x ? More than 50 million people use GitHub to discover, fork, and contribute to over 100 million projects. See Also. A pairing function is a mathematical function taking two numbers as an argument and returning a third number, which uniquely identifies the pair of input arguments. x��\[�Ev���އ~�۫.�~1�Â� ^"�a؇� ڕf@B���;y=Y�53�;�ZUy9y�w��Y���"w��+����:��L�����݇�h"�N����3����V;e��������?�/��#U|kw�/��^���_w;v��Fo�;����3�=��~Q��.S)wҙ�윴�v4���Z�q*�9�����>�4hd���b�pq��^['���Lm<5D'�����"�U�'�� They may also differ in their performance. (yy+x) : (xx+x+y);} function unpair(z){var q = Math.floor(Math.sqrt(z)), l = z - … <> However, cantor(9, 9) = 200. \end{array} Matthew P. Szudzik. One nice feature about using the Szudzik pairing function is that all values below the diagonale are actually subsequent numbers. Pairing library using George Cantor (1891) and Matthew Szudzik (2006) pairing algorithms that reversibly maps Z × Z onto Z*. Simple C# class to calculate Cantor's pairing function - CantorPairUtility.cs. The full results of the performance comparison can be found on jsperf. Nothing really special about it. The cantor pairing function can prove that right? It should be noted that this article was adapted from an earlier jsfiddle of mine. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. c & : (a < 0 \cap b < 0) \cup (a \ge 0 \cap b \ge 0)\\ The algorithms have been modified to allow negative integers for tuple inputs (x, y). In mathematics, a pairing function is a process to uniquely encode two natural numbers into a single natural number.. Any pairing function can be used in set theory to prove that integers and rational numbers have the same cardinality as natural numbers. F{����+��j#,��{"1Ji��+p@{�ax�/q+M��B�H��р��� DQ�P�����K�����o��� �u��Z��x��>� �-_��2B�����;�� �u֑. \end{array}$$index = \left\{\begin{array}{ll} Comparing against Cantor we see: Yes, the Szudzik function has 100% packing efficiency. cantor pairing function inverse. An example in JavaScript: How Cantor pairing works is that you can imagine traversing a 2D field, where each real number point is given a value based on the order it which it was visited. Cantor pairing function: (a + b) * (a + b + 1) / 2 + a; where a, b >= 0 The mapping for two maximum most 16 bit integers (65535, 65535) will be 8589803520 which as you see cannot be fit into 32 bits. Ask Question Asked 1 year, 2 months ago. A quadratic bijection does exist. - pelian/pairing Different pairing functions known from the literature differ in their scrambling behavior, which may impact the hashing functionality mentioned in the question. 1. ambuj_kumar 16. Let's not fail silently! It returns a vector of ID numbers. Abstract This article surveys the known results (and not very well-known re- sults) associated with Cantor’s pairing function and the Rosenberg-Strong pairing function, including their inverses, their generalizations to higher dimensions, and a discussion of a few of the advantages of the Rosenberg- Strong pairing function over Cantor’s pairing function … Like Cantor, the Szudzik function can be easily implemented anywhere. k cursive functions as numbers, and exploits this encoding in building programs illustrating key results of computability. $$b = \left\{\begin{array}{ll} ��� ^a���0��4��q��NXk�_d��z�}k�; ����HUf A��|Pv х�Ek���RA�����@������x�� kP[Z��e �\�UW6JZi���_��D�Q;)�hI���B\��aG��K��Ӄ^dd���Z�����V�8��"( �|�N�(�����������/x�ŢU ����a����[�E�g����b�"���&�>�B�*e��X�ÏD��{pY����#�g��������V�U}���I����@���������q�PXғ�d%=�{����zp�.B{����"��Y��!���ְ����G)I�Pi��қ�XB�K(�W! , To find x and y such that π(x, y) = 1432: The graphical shape of Cantor's pairing function, a diagonal progression, is a standard trick in working with infinite sequences and countability. And as the section on the inversion ends by saying, "Since the Cantor pairing function is invertible, it must be one-to-one and onto." \right.$$, https://en.wikipedia.org/wiki/Pairing_function. function(x, y, z) { max = MAX(x, y, z) hash = max^3 + (2 * max * z) + z if (max == z) hash += MAX(x, y)^2 if (y >= x) hash += x + y else hash += y return hash} This pairing function only works with positive numbers, but if we want to be able to use negative coordinates, we can simply add this to the top of our function: x = if x >= 0 then 2 * x else -2 * x - 1 -2y - 1 & : y < 0\\ An Elegant Pairing Function Matthew Szudzik Wolfram Research Pairing functions allow two-dimensional data to be compressed into one dimension, and they play important roles in the arrangement of data for exhaustive searches and other applications. A pairing function is a function which maps two values to a single, unique value. Additional space can be saved, giving improved packing efficiency, by transferring half to the negative axis. Value. The pairing function then combines two integers in [0, 226-2] into a single integer in [0, 252). The primary downside to the Cantor function is that it is inefficient in terms of value packing. Enter Szudzik's function: a >= b ? Matthew P. Szudzik 2019-01-28. Active 1 year, 2 months ago. Pairing functions with square shells, such as the Rosenberg-Strong pairing function, are binary perfect. Two pairing functions are … I found Cantor's and Szudzik's pairing function to be very interesting and useful, however it is explicitly stated that these two functions are to be used for natural numbers. A pairing function for the non-negative integers is said to be binary perfect if the binary representation of the output is of length 2k or less whenever each input has length k or less. In[13]:= PairOrderedQ@8u_,v_<,8x_,y_= b ? It is always possible to re-compute the pair of arguments from the output value. We quickly start to brush up against the limits of 32-bit signed integers with input values that really aren’t that large. Viewed 40 times 0. Szudzik M (2006) An elegant pairing function. Source. Cantor pairing function: (a + b) * (a + b + 1) / 2 + a; where a, b >= 0 The mapping for two maximum most 16 bit integers (65535, 65535) will be 8589803520 which as you see cannot be fit into 32 bits. b^2 + a & : a < b\\ /// 3- We use the unique number as the key for the entry. 2y & : y \ge 0 The inverse function is described at the wiki page. Use a pairing function for prime factorization. \right.$$,$$a = \left\{\begin{array}{ll} %PDF-1.4 Szudzik, Matthew P. Abstract This article surveys the known results (and not very well-known results) associated with Cantor's pairing function and the Rosenberg-Strong pairing function, including their inverses, their generalizations to higher dimensions, and a discussion of a few of the advantages of the Rosenberg-Strong pairing function over Cantor's pairing function in practical applications. Java : 97% speed and 66.67% memory : using Szudzik's Pairing Function and HashSet. So for a 32-bit signed return value, we have the maximum input value without an overflow being 46,340. Proof. This relies on Cantor's pairing function being a bijection. Usage. \end{array} Szudzik, M. (2006): An Elegant Pairing Function. I used Matthew Szudzik's pairing function and got this: $(p - \lfloor\sqrt{p}\rfloor^2)\cdot\lfloor\sqrt{p}\rfloor = n$ There, we need to make a distinction between values below the diagonale and those above it. This is useful in a wide variety of applications, and have personally used pairing functions in shaders, map systems, and renderers. 2x & : x \ge 0 x^2 + x + y & : x \ge y \right.$$Trying to bump up your data type to an unsigned 32-bit integer doesn’t buy you too much more space: cantor(46500, 46500) = 4,324,593,000, another overflow. In elementary set theory, Cantor's theorem is a fundamental result which states that, for any set, the set of all subsets of (the power set of , denoted by ()) has a strictly greater cardinality than itself. stream This means that all one hundred possible variations of ([0-9], [0-9]) would be covered (keeping in mind our values are 0-indexed). \end{array} PREREQUISITES. In theoretical computer science they are used to encode a function defined on a vector of natural numbers : → into a new function : →$$index = {(x + y)(x + y + 1) \over 2} + y. 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Wolfram Science Conference, pp 1–12 in this ramble we will cover different., such as the key for the first 100 combinations, an efficiency of 50.. Which would result in an overflow being szudzik pairing function comparing against Cantor we see: Yes, the pairing... Will cover two different pairing functions known from the literature differ in scrambling! ) to be the square root of the performance between Cantor and Szudzik you can then map the row an. Value for Szudzik is virtually identical, szudzik pairing function Szudzik having a slight advantage between values the. In this ramble we will cover two different pairing functions in shaders, map,... The negative axis both functions will degrade values for the Cantor function is that all values below diagonale. A slight advantage is not required full results of the points in the plane Z x Z to Z than!, are binary perfect % packing efficiency for both functions will degrade an Elegant pairing to! Question Asked 1 year, 2 months ago 33000 ) = 200 from an jsfiddle! Row to an x axis, the column to an y axis with Szudzik having a slight advantage diagonal! To compose a unique identifier from a pair of values transformation can be applied so that negative input be... 2 } + y ) key for the Cantor function, this graph is traversed a... Are binary perfect an Elegant pairing function then combines two integers in [ 0, 252.. Be easily implemented anywhere on Cantor 's pairing function then combines two integers in [ 0, ). Scrambling behavior, which may impact the hashing functionality mentioned in the plane for visually. A box-like pattern using the Szudzik function can be saved, giving improved packing efficiency by... Differ in their scrambling behavior, which may impact the hashing functionality in! Neither szudzik pairing function nor Szudzik pairing function to generate a unique number as key. Is inefficient in terms of value packing like Cantor, the Szudzik function has 100 % packing for! With square shells, such as the key for the first 100 combinations an. Key for the first 100 combinations, an efficiency of 50 % > x ) 2... With input values, 2 months ago { �ax�/q+M��B�H��р��� D Q�P�����K�����o��� �u��Z��x�� > � �-_��2B����� ; �� �u֑ Szudzik., on two vectors of equal length Z x Z to Z $����+��j # ��..., 9 ) = 200 in the plane ) = 2,178,066,000 which would result in overflow. Is inefficient in terms of value packing of pair ( 9, 9 ) = 2,178,066,000 would! Y > x k cursive functions as numbers, and renderers b [ a/b ] should be noted though all. Like Cantor, the Szudzik function can be understood as an ordering the! Primary downside to the Cantor function is that all values below the diagonale those. In shaders, map systems, and exploits this encoding in building programs illustrating key results of computability perfect... Pair ( 9, 9 ) = 200 maximum integer value have personally used pairing functions work natively with input! On two vectors of equal length �� { 1Ji��+p @ { �ax�/q+M��B�H��р��� D �u��Z��x��. Below the diagonale and those above it it in a perfectly efficient function we would expect the of. Szudzik function can be easily implemented anywhere 1 year, 2 months ago max input to... Algorithms have been modified to allow negative integers for tuple inputs (,... Two vectors of equal length value of pair ( x, y (... And contribute to over 100 million projects in this ramble we will cover two different functions. Is traversed in a diagonal function is described at the wiki page, Z. That really aren ’ t that large of the maximum integer value a diagonal is. Cantor we see: Yes, the column to an y axis compose a unique from. @ { �ax�/q+M��B�H��р��� D Q�P�����K�����o��� �u��Z��x�� > � �-_��2B����� ; ��.... Choice is not required primary downside to the Cantor function, this graph is traversed in box-like! We use a pairing function accepts optional boolean argument to map Z x to., map systems, and renderers two vectors of equal length from a of! Is virtually identical, with Szudzik having a slight advantage an overflow being.! Efficiency for both functions will degrade of arguments from the literature differ in scrambling... Associated with that unordered pair we see: Yes, the Szudzik pairing functions work natively with input... Of pair ( 9, 9 ) = 2,178,066,000 which would result in an overflow 50 million people GitHub! Square root szudzik pairing function the points in the graphic below fork, and exploits this in! ) to be 99 in shaders, map systems, and exploits this encoding in building programs illustrating key of! > = b pair of values examples /// 2- we use a pairing function and HashSet memory! Always possible to re-compute the pair of arguments from the output value the column to an y axis a... Is always possible to re-compute the pair of values transformation can be saved, giving packing! In a diagonal function is that it is inefficient in terms of value packing the max input pair Szudzik!, a simple transformation can be easily implemented anywhere a slight advantage input pair to Szudzik to be the root... Generating visually meaningful ciphertext Image found on jsperf 200 pair values are still positive, as such we! The row to an y axis of two hash codes signed return value, we can calculate max. Diagonal function is that all returned pair values for the first 100 combinations, an efficiency of %! Combines two integers in [ 0, 226-2 ] into a single integer in [,! Their scrambling behavior, which may impact the hashing functionality mentioned in the plane we cover. But it covers it in a diagonal function is that all values below the diagonale are actually subsequent.. Return value, we can calculate the max input pair to Szudzik to the. The Question is illustrated in the plane 2006 ): an Elegant pairing function is that all values below diagonale... Contribute to over 100 million projects 252 ) function and HashSet Szudzik having slight. Without an overflow being 46,340 252 ) have personally used pairing functions with square shells, such as Rosenberg-Strong. A pairing function, on two vectors of equal length \over 2 } + y$. The square root of the maximum input value without an overflow being 46,340: Yes, Szudzik. Generate a unique identifier from a pair of values pp 1–12 noted though that all pair... Binary perfect it should be noted that this article was adapted from an earlier jsfiddle of mine, by half. Graph is traversed in a diagonal function is that all values below the diagonale those... The key for the Cantor function, on two vectors of equal length 1 \over... Function superseeds od_id_order as … Java: 97 % speed and 66.67 % memory: using Szudzik function! Would result in an overflow being 46,340 simple C # class to calculate Cantor pairing... Will cover two different pairing functions in shaders, map systems, and contribute to over 100 million.. Uniquely associated with that unordered pair input can be applied so that negative input values is a mod b a! You can then map the row to an x axis, the Szudzik can! Subsequent numbers to calculate Cantor 's pairing function for prime factorization ’ t that.. Have the maximum input value for Szudzik is 65,535 pairing functions in shaders, map systems, and renderers mine... Q�P�����K�����o��� �u��Z��x�� > � �-_��2B����� ; �� �u֑ outputs a single non-negative integer that is uniquely associated with unordered. A perfectly efficient function we would expect the value of pair ( x + y ) { return y x. Negative integers for tuple inputs ( x, y ) { return >! Outputs a single non-negative integer that is uniquely associated with that unordered pair easily implemented anywhere of! | 2021-09-23T05:29:33 | {
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https://brilliant.org/discussions/thread/how-to-count-rectangles/ | # How to Count Rectangles!
Hi there! Me again. *munches*
Sorry, I still haven't finished my hamburger since the last time we count squares, I hope you don't mind that.
Anyway, we are here to see how to count rectangles of a certain size in a grid.
Consider this question: How many $$3\times2$$ rectangles are there in the $$4\times5$$ grid below?
I'm a little bit picky here, note that dimensions here are expressed as width$$\times$$height, so in this case I only want $$3\times2$$ rectangles (that is, a rectangle of width 3 and height 2) and not the $$2\times3$$ ones.
If we mark an 'X' on every bottom-right corner of every $$3\times2$$ rectangle, we would see this:
Hey, that means counting the number of $$3\times2$$ rectangles in the original grid is the same as counting how many X's are there, or in other words, it is the same as counting the number of unit squares in an $$2\times4$$ grid.
There are $$2\times4=8$$ unit squares in a $$2\times4$$ grid, therefore, the number of $$3\times2$$ rectangles in the original grid is 8! Hooray!
*munches*
Mmm... so in general, how many $$x\times y$$ rectangles ($$x$$ is the width of the rectangle, $$y$$ is the height of the rectangle) are there in an $$a\times b$$ grid ($$a$$ is the width of the grid, $$b$$ is the height of the grid), given that $$x\leqslant a$$ and $$y\leqslant b$$?
Again imagine marking an 'X' on every bottom-right corner of every $$x\times y$$ rectangle, we will get a grid full of X's.
The width of the "X grid" is $$a-x+1$$, while the height of the "X grid" is $$b-y+1$$.
Since the number of $$x\times y$$ rectangles in an $$a\times b$$ grid equals the number of unit squares in the "X grid", hence the number of $$x\times y$$ rectangles in an $$a\times b$$ grid is $(a-x+1)(b-y+1)$
*munches*
The number of $$x\times y$$ ($$x$$ is the width of the rectangle, $$y$$ is the height of the rectangle) rectangles (the exact $$x\times y$$ rectangle, not including rotated variations such as $$y\times x$$) in an $$a\times b$$ grid ($$a$$ is the width of the grid, $$b$$ is the height of the grid) provided that $$x\leqslant a$$, $$y\leqslant b$$ is $(a-x+1)(b-y+1)$
##### This is one part of Quadrilatorics.
Note by Kenneth Tan
2 years, 10 months ago
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I think another way to do it is breaking them up. For example, if we want to count the number ot rectangles in $$10×10$$ grid, is first counting the number of rectangles in $$10×1$$ grid then xounting the number of rectangles in $$10×2$$ grid, then in $$10×3$$, and so in, in this way we would be able to establish a pattern in the number of rectangles, and by arithmetic/geometric progression, we would het the answer.
- 2 years, 10 months ago
Yes, but in this case I only want to count rectangles of a specific size.
- 2 years, 10 months ago
We can do in that case too. Your questions are awesome.
- 2 years, 10 months ago
You may see the solution of my problem to see my approach.
- 2 years, 10 months ago | 2019-01-23T06:20:32 | {
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http://pacianca.it/improper-integrals-comparison-test.html | # Improper Integrals Comparison Test
Integrals over unbounded intervals. Improper integrals (Sect. Let f and g. Show graphically that the benchmark you chose satisfies the conditions of the Comparison Test. Comparison tests for convergence. The book gives students the prerequisites and tools to understand the convergence, principal value, and evaluation of the improper/generalized Riemann integral. , find the values of p for which this integral converges, and the values for which it diverges. Calculus of parametric curves. Comparison Tests. Math 185 - Calculus II. This articles needs some examples, better references, a discussion of the relation with other convergence tests, an explanation of the comparison test for improper integrals, and so forth. We want to prove that the integral diverges so if we find a smaller function that we know diverges the area analogy tells us that there would be an infinite amount of area under the smaller function. (right) Comparison of test prediction accuracy when using ourIDKkernel to a numerical estimation of the kernel integral using random features, as a function of the number of features used for estimation. Partial Fractions. Does Z 1 0 1 p 1 x2 dxconverge? Notice that the function p1 1 x2 has a vertical asymptote at x= 1, so this is an improper integral and we will need to consider the. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. 1 x2 3 + 3 4. Problem 4 (15 points) For p >0 consider the improper integral Derive the p -test for these integrals, i. Where k is going to be some positive number. The comparison test. However, for x>1 we have x2 >x, so that 1/ex2 > 1/ex. Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Select the second example from the drop down menu, showing Use the same guidelines as before, but include the exponential term also: The limit of the ratio seems to converge to 1 (the "undefined" in the table is due to the b terms getting so small that the algorithm thinks it is dividing by 0), which we can verify: The limit comparison test says that in this. aa g x dx f x dx ³³ ff When we cannot evaluate an integral directly, we first try to determine whether it converges or diverges by comparing it to known integrals. Frequently we aren't concerned along with the actual value of these integrals. Theorem 1 (Comparison Test). We are covering these tests for definite integrals now because they serve as a model for similar tests that check for convergence of sequences and series -- topics that we will cover in the next chapter. For example, one possible answer is BF, and another one is L. Partial Fractions 32 1. Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. Page 632 Numbers 74. By definition, these integrals can only be used to compute areas of bounded regions. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. This is , ∃ a psoitive number ,independent of ,such that ∫ + < M, 0< < −. If Maple is unable to calculate the limit of the integral, use a comparison test (either by plotting for direct comparison or by limit comparison if applicable). 7 Review of limits at infinity:-2 -1 0 1 2 2 4 6 x y y ex 1 2 3-4-2 0 2 x y Use the above comparison test to determine whether 1 1 x2 5. State the Comparison Test. Solutions; List of Topics that may. That means we need to nd a function smaller than 1+e x. Give a reasonable "best" comparison function that you use in the comparison (by "best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). Consider the following These are all improper because the function being integrated is not finite at one of the limits of integration. I Examples: I = Z ∞ 1 dx xp, and I = Z 1 0 dx xp I Convergence test: Direct comparison test. 10) Activity: Choosing a Technique of Integration EWA 5. Since the integral has an infinite discontinuity, it is a Type 2 improper integral. Integral Test: If f(x) is a decreasing positive function from [1,∞) to [0,∞), then the series P f(n) converges if and only if the improper integral R ∞ 0 f(x)dx converges (i. Review Problems from your textbook: Integration plus L'Hospital's Rule page 579 #1-15, 33-37, 73-87 odds. In such cases the following test is useful. To be convincing, we need to find a series with smaller terms whose sum diverges. give me a formal proof of comparison test for improper integrals. The rst of these is the Direct Comparison Test (DCT). 7) I Integrals on infinite. Comparison Tests. The key tools are the Comparison Test and the P-Test. Comparison Test which requires the non negativity of relevant functions, we shall consider instead f(x) and g(x). Int[1 to infinity] (sin^2 x)/x^3 dx. We work through several examples for each case and provide many exercises. Improper. We can summarize this line of argumentation with the following: Theorem. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. via integration by parts with u = t/2, dv = sin(2t) dt = π/8. Improper Integrals. NEET: 45 Day Crash Course | Day 45 Last Day | Alcohols, Phenols, Ethers | Unacademy NEET | Anoop Sir Unacademy NEET 117 watching Live now. Improper Integral. The concepts used include regular inequalities as well as the. There are 2 types of comparison tests: direct comparison and limit comparison. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. One way to determine the convergence of an improper integral is by comparing it to other integrals that we do know the convergence of. Using the Comparison Test from Calculus, we determine whether several improper integrals in the question converge or diverge. Step 2: Click the blue arrow to submit. integral(1 to infinity) ln(x) dx/x^(1. This chapter has explored many integration techniques. Convergence tests for improper integrals. We have two main tests that we can use to establish convergence or divergence -- the Direct Comparison Test and the Limit Comparison Test. Although we state it for Type 1 integrals,. The comparison test Identify an improper integral. Ok so from this test i came to the conclusion that the integral of sinx/lnx converges. Here I give a quick idea of what the direct comparison test for improper integrals and use it show whether an improper integral converges or. Z 1 1 1 (2 x+ 1)3 dx 2. The Comparison Test and Limit Comparison Test also apply, modi ed as appropriate, to other types of improper integrals. A similar statement holds for type 2 integrals. Comparing Improper Integrals. 2 The Integral Test and p-series test File WS 11. integral toolbox. 14 Improper integrals While the rst integral on the right-hand side diverges, the second one converges, as can be proved by the same procedure as above. The direct comparison test is a simple, common-sense rule: If you’ve got a series that’s smaller than a convergent benchmark series, then your series must also converge. There are two we will talk about here. The Limit Comparison Theorem for Improper Integrals Limit Comparison Theorem (Type I): If f and g are continuous, positive functions for all values of x, and lim x!1 f(x) g(x) = k Then: 1. Otherwise the integrals are divergent. Here is the statement: Limit Comparison Test: Suppose f(x) and g(x) are continuous and positive for all x 2[a;1). Here I give a quick idea of what the direct comparison test for improper integrals and use it show whether an improper integral converges or. Z 1 1 1 x+ 3ln(x) dx 6. If you're behind a web filter, please make sure that the domains *. Often we aren't concerned with the actual value of these integrals. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. ∫∞ 1 1 x3+1dx Solution. improper integrals (comparison theorem) 5 $\mu$ test for convergence of improper integral of first kind. ) - Duration: 1:02:59. Smith , Founder & CEO, Direct Knowledge. integral 1/(x^2+3) from 1 to infinity 3. Comparison Test: If 0 < a n ≤ b n and P b n converges, then P a n also converges. \\int \\frac{2+ e^{-x} dx}{x} from 1 to infinity Wolframalpha tells me this integral diverges, now i just need to know what to compare it to. Give a reasonable "best" comparison function that you use in the comparison (by "best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). Here’s an example. Comparison Test for Improper Integrals In some cases, it is impossible to find the exact value of an improper integral, but it is important to determine whether the integral converges or diverges. Choose "Evaluate the Integral" from the topic selector and click to. Solutions; Summary of the convergence tests from Chapter 11 (ht Libby Runte). Review Problems from your textbook: Integration plus L'Hospital's Rule page 579 #1-15, 33-37, 73-87 odds. if 0 < k < 1, then Z 1 a g(x)dx converges Z 1 a f(x)dx converges 2. ∫∞ 4 e−y y dy Solution. Comparison Test for Integrals Comparison Test for Integrals Theorem If f and g are continuous functions with f(x) g(x) 0 for x a, then (a) If R 1 a f(x)dx is convergent, then R 1 a g(x)dx is convergent. Hence the Comparison test implies. The limit comparison test gives us another strategy for situations like Example 3. The last convergence tool we have is the comparison test. For Example, One Possible Answer Is BF, And Another One Is L. Hint: 0 < E^-x Lesserthanorequalto 1|for X Greaterthanorequalto 1. Question: For Each Of The Improper Integral Below, If The Comparison Test Applies, Enter Either A Or B Followed By One Letter From C To K That Best Applies, And If The Comparison Test Does Not Apply, Enter Only L. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t ⌠1 ⌡ ⎮ converges, so by the comparison test, this integral also converges. Improper Integrals (with Examples) May 2, 2020 January 8, 2019 Categories Formal Sciences , Mathematics , Sciences Tags Calculus 2 , Latex By David A. Read lecture notes, page 1 to page 3; An integral with an infinite upper limit of integration to be evaluated. Iff(x) is continuous on ( —x, 00), then f(x) dx + — lim 2. For x > e we surely have Then also and the Comparison test is inconclusive. I Examples: I = Z ∞ 1 dx xp, and I = Z 1 0 dx xp I Convergence test: Direct comparison test. Improper Integral: Comparison Test. If is convergent then is. On the other hand, if L 0, then we must compare f x to a suitable comparison function in order to determine the convergence or divergence of the. The last convergence tool we have is the comparison test. Active 3 years, 5 months ago. Comparison Test and Limit Comparison Test. f(x)dx = lim DEFINITION Integrals with infinite limits of integration are improper integrals of Type I. Z 4 0 1 (x 22) dx Comparison Test for Integrals Theorem If fand gare continuous functions with f(x) g(x) 0 for x a, then (a) If R 1 a f(x)dxis convergent, then R 1 a g(x)dxis convergent. And if your series is larger than a divergent benchmark series, then your series must also diverge. A similar statement holds for type 2 integrals. Improper at x = 0, where the t is much larger than the t3, so this “looks like” the p-type dt 0t ⌠1 ⌡ ⎮ which converges since p < 1. This means the limits of integration include $\infty$ or $-\infty$ or both. Arc Length: Week #5: Feb 10 - 14 The Comparison Tests. Let gðxÞ A 0 for all x A a, and suppose that. LIATE; Trig Integrals. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. Hence it is convergent by comparison test. ∫∞ 3 z2 z3−1dz Solution. In this section we use a different technique to prove the divergence of the harmonic series. The improper integral. ∫∞ 1 z−1 z4+2z2dz Solution. I have discussed about comparison test to examine the convergence of improper integral of finite range. The question is simple: Decide whether the following integral converges: There are essentially three methods to answer this question: 1. The Integral Test (14 minutes, SV3 » 41 MB, H. The comparison test. f ( x) = e - x 2 ( ln x) 2. For Example, One Possible Answer Is BF, And Another One Is L. integral e^-x/x^2 from 1 to infinity 2. If f(x) has a tail you must do further tests. integral 1/(x^2+3) from 1 to infinity 3. We have seen that the integral test allows us to determine the convergence or divergence of a series by comparing it to a related improper integral. There is a discontinuity at $$x = 0,$$ so that we must consider two improper integrals: \[{\int\limits_{ - 2}^2 {\frac{{dx}}{{{x^3}}}} } = {\int\limits. 5 Direct Comparison Test. So it seems plausible that results which tell you about the convergence of improper integrals would have. We can handle only one 'problem' per integral. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Thus the integral converges. Trig Substitution. Hence it is convergent by comparison test. Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. with bounds) integral, including improper, with steps shown. (Note this is a positive number when a is negative, so this answer makes sense. Consider the improper integral ∫ 1 1 1 xp dx: Integrate using the generic parameter p to prove the integral converges for p > 1 and. One cannot apply numerical methods like LEFT or RIGHT sums to approximate the value of such. Page 631 Number 48. Improper integrals. The Organic Chemistry Tutor 409,621 views 20:18. integral 1/(x^2+3) from 1 to infinity 3. Finally we need to discuss a second type of improper integral. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Use the Comparison test. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. Improper Integrals. {Hint: for implies that. Proof: harmonic series diverges. (b) If R 1 a g(x)dxis divergent, then R 1 a f(x)dxis divergent. The instructions were to show that the integral from 1 to infinity of sinx/sqrt(x) converges. Alternating series test for convergence. Improper Integrals ( Part 5 ) Infinite Limits. The p-Test. The improper integral Z ∞ 1 1. Here, the answer just provides the base function that the problem most closely resembles. 1 dx p 1 if p > 1 1 x p diverges if p < 1 Example 2 On the surface, the graphs of the last three examples seem very much alike and there is nothing to suggest why one of the areas should be infinite and the other two finite. The following test will be helpful. Improper Integrals: Solutions Friday, February 6 Bounding Functions More Comparison Test Pretty much the only function you care to compare things to here is 1=x (or, more generally, 1=xp). (right) Comparison of test prediction accuracy when using ourIDKkernel to a numerical estimation of the kernel integral using random features, as a function of the number of features used for estimation. 138 Improper Integrals M. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t ⌠1 ⌡ ⎮ converges, so by the comparison test, this integral also converges. Usually it's more important to know whether an improper integral converges than it is to know what it converges to. You should not extend the inequality to $\int_{9}^{\infty } \frac{1}{x}dx$ because it's divergent and a convergent integral is always less than a divergent integral, so it's of no use. if k = 0, then Z 1 a g(x)dx converges =) Z 1 a f(x)dx converges 3. Improper Integrals ( Part 5 ) Infinite Limits. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Note the analogy with the geometric series if r ¼ e t so that e tx ¼ rx. Example: Determine whether the series X∞ n=1 1 √ 1+4n2 converges or diverges. Therefore putting the two integrals together, we conclude that the improper integral is convergent. improper integral convergent. Improper Integrals Convergence and Divergence, Limits at Infinity & Vertical Asymptotes, Calculus - Duration: 20:18. Improper Integrals-Section 7. Let us evaluate the corresponding improper integral. {comparison} 6. Integrals over bounded intervals of functions that are unbounded near an endpoint. Comparison test for convergence/divergence. Limit test: Let and be two positive function defined on. Use the Comparison Theorem to decide if the following integrals are convergent or divergent. However, we know that continuity is "almost necessary" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught. Comparison Test which requires the non negativity of relevant functions, we shall consider instead f(x) and g(x). Z 1 1 1 2x2 x dx 3. 9: Feb 1: Feb 2 Week 4 Feb 3 Areas of Regions and Volumes of Solids of Revolution(6. Improper Integrals. This is because Z b · f(t)dt = F(b)−F(·) and F(b) is increasing and F(b) has a limit if and only if F(b) is bounded. In fact, 1 t+t3 < 1 (bigger denominator = smaller fraction), and the p-type integral dt 0t ⌠1 ⌡ ⎮ converges, so by the comparison test, this integral also converges. If is convergent then is. (a) integrate limit 6 to 7 x/x-6 dx Since the integral has an infinite interval of integration, it is a Type 1 improper integral. Improper Integrals. Spending a class hour to introduce students to the concepts in the flipped class lesson is found to be helpful. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The Limit Comparison Test. Comparison Test and Limit Comparison Test. Comparison, Limit comparison and Cauchy condensation tests. We already know the second integral is finite, so the first one has to be finite as well. 9: Feb 1: Feb 2 Week 4 Feb 3 Areas of Regions and Volumes of Solids of Revolution(6. And if your series is larger than a divergent benchmark series, then your series must also diverge. Usually it's more important to know whether an improper integral converges than it is to know what it converges to. Part 3: Tests for Convergence and Divergence. 16 Improper Integrals. First, try the comparison. Unit 8 (Chapter 8): Sequences, l'hospital, & improper integrals Limits, Sequences Finding Limits Analytically Comparison Test Notes Extra Practice. Both the Direct and Limit Comparison Tests were given in terms of integrals over an infinite interval. In many cases we cannot determine if an integral converges/diverges just by our use of limits. 6 Show that the improper integral R 1 1 1+x2 dxis convergent. If we look at the other one, and we decide the other one is bursting at the seems, we know it's safe to open ours up. (b) If R 1 a g(x)dxis divergent, then R 1 a f(x)dxis divergent. Select the second example from the drop down menu, showing Use the same guidelines as before, but include the exponential term also: The limit of the ratio seems to converge to 1 (the "undefined" in the table is due to the b terms getting so small that the algorithm thinks it is dividing by 0), which we can verify: The limit comparison test says that in this. Improper Riemann Integrals is the first book to collect classical and modern material on the subject for undergraduate students. You solve this type of improper integral by turning it into a limit problem where c approaches infinity or negative infinity. For Example, One Possible Answer Is BF, And Another One Is L. that the limit is the value of the improper integral. if 0 < k < 1, then Z a 0 g(x)dx converges Z a 0 f(x)dx converges 2. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Lecture 4 - Comparison tests for improper integrals Safet Penjic. Discrete calculus-How to integrate discrete functions Numerical ODEs-Using sequences to solve ODEs Numerical integration-Using sequences to solve definite integrals Series-Infinite series as improper discrete integrals Convergence tests 1-Comparison-type tests Convergence tests 2-Geometric series-type tests. UNSOLVED! Use the comparison test to determine if the following integral is convergent or divergent. We will only state these theorems for integrals that are improper at b. Suppose 0 f(x) g(x) for x aand R b a f(x)dxexists for all b>a. MATH1014 Tutorial 8 The comparison test for improper integrals Iff(x) andg(x) arecontinuousandf(x) ≥g(x) ≥0,then a f(x)dxconvergesimplies a g(x)dxconverges. If 0 f(x) g(x) on [a;1), it can be shown that the convergence of R 1 a g(x)dx implies the convergence of R 1 a f(x)dx, and the divergence of R 1 a f(x)dx implies the divergence of R 1 a g(x)dx. Integrals over unbounded intervals. Improper Integrals. Theorem 1 (Comparison Test). We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. (a) integrate limit 6 to 7 x/x-6 dx Since the integral has an infinite interval of integration, it is a Type 1 improper integral. I thought about splitting the integral and using the direct comparison test, but I'm not sure what Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Warning: Now that we have introduced discontinuous integrands, you will need to check. BC Calculus Improper Integrals Day 3 Notesheet Name: _____ Sometimes, we cannot find the antiderivative of an integrand of an improper integral. Since G(t) is an increasing function, it follows that a L G(t) L - ε y 0 FIGURE 1 If G(t) is increasing with least upper bound L, then G(t) eventually lies within of L L− < G(y 0. (a) ∫ 0 1 1 2x−1 dx (b) ∫ 1 1 xe x2 dx (c) ∫ 2 0 x−3 2x−3 dx (d) ∫ 1 0 sin d 3. The primary tool in that toolbox is the set of integrals of power functions. ∫∞ 6 w2+1 w3(cos2(w)+1)dw Solution. Frequently we aren't concerned along with the actual value of these integrals. THE INTEGRAL AND COMPARISON TESTS 93 4. (May need to break up integral into several. Express it as a limit and determine whether it converges or diverges; if it converges, find the value. There is a discontinuity at $$x = 0,$$ so that we must consider two improper integrals: \[{\int\limits_{ - 2}^2 {\frac{{dx}}{{{x^3}}}} } = {\int\limits. Trig Substitution. We will only state these theorems for integrals that are improper at b. How do you use basic comparison test to determine whether the given series converges or diverges How do you use the Comparison Test to see if #1/(4n^2-1)# converges, n is going to infinity? See all questions in Direct Comparison Test for Convergence of an Infinite Series. {comparison} 6. This chapter has explored many integration techniques. An improper integral of type 1 is an integral whose interval of integration is infinite. Here is the statement: Direct Comparison Test: Suppose that f(x) and g(x) are continuous functions on the interval [c;1). Integration using Tables and CAS 39 1. Sometimes it is impossible to find the exact value of an improper integral and yet it is important to know whether it is convergent or divergent. Remainder estimation. There are two classes of improper integrals: (1) those in which at least one of the limits of integration is infinite (the interval is not bounded); and (2) those of the type where f (x) has a point of discontinuity (becoming infinite) at x = c, a c b (the function is not bounded). 2020-04-21T09:24:18Z http://oai. The Comparison Test Suppose 0 ≤ f(x) ≤ g(x) for all x ≥ a. This technique is important because it is used to prove the divergence or convergence of many other series. Section 1-9 : Comparison Test for Improper Integrals. Solution 2 (b). Improper Integrals — One Infinite Limit of Integration. is divergent. Improper Integrals In general, To determine if an integral converges or diverges, you can use the Comparison Test described below. Mansi Vaishnani 2. In my assignment I have to evaluate the (improper) integral, by means of the "comparison theorem". Comparison Test for Improper Integrals. o If an > bn for all n and if ba diverges, then an also diverges. Express it as a limit and determine whether it converges or diverges; if it converges, find the value. Comparison Test Suppose [and. Testing Convergence of Improper Integrals. A Comparison Test Sometimes it is not possible to obtain the value of improper integrals. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. David Jerison. If 0 f(x) g(x) on [a;1), it can be shown that the convergence of R 1 a g(x)dx implies the convergence of R 1 a f(x)dx, and the divergence of R 1 a f(x)dx implies the divergence of R 1 a g(x)dx. 3 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1;1). The calculator will evaluate the definite (i. Example: Z 1 1 1 p x dx We have the following general result related to the last two examples. Improper integrals are definite integrals where one or both of the boundaries is at infinity, or where the integrand has a vertical asymptote in the interval of integration. An integral is called an improper integral if one of, or both, of the conditions hold: The interval of integration is infinite. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. Solution 2 (a). {comparison} 4. Evaluate the …. Show that the improper integral. Partial Fractions. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. On the interval [1 ;1], we split up the integral into two separate improper integrals. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Note that Z t 1 1 x dx= [lnx]t 1 = lnt!1 as t!1: Hence, R 1 1 1 x dxdiverges. and is clearly finite, so the original integral is finite as well. Warning: Now that we have introduced discontinuous integrands, you will need to check. are all improper because they have limits of integration that involve ∞. {comparison} 8. 2 (Improper Integrals with Infinite Discontinuities) Consider the following three. Convergence test: Direct comparison test Remark: Convergence tests determine whether an improper integral converges or diverges. For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. xml 03/07/2013 12:21:24 mchinn [Discussion Draft] [Discussion Draft] March 7, 2013 113th CONGRESS 1st Session Rules Committee Print 113-4 of H. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. Example: Determine whether the series X∞ n=1 1 √ 1+4n2 converges or diverges. Hint: 0 < E^-x Lesserthanorequalto 1|for X Greaterthanorequalto 1. with bounds) integral, including improper, with steps shown. In my assignment I have to evaluate the (improper) integral, by means of the "comparison theorem". Practice: Limit comparison test. Both of the limits diverge, so the integral diverges. integral e^-x/x^2 from 1 to infinity 2. Solution to this Calculus Improper Integral problem is given in the video below!. ∫∞ 1 z−1 z4+2z2dz Solution. The p-Test implies that the improper integral is convergent. ∫∞ 4 e−y y dy Solution. Logistic Model problems from Pg414-415. Even though you were instructed not to use the Comparison Test to solve this problem; you can use it to check your result. 4 Day 2 Improper Integrals and the Comparison Test Thursday, January 30, 2020 7:54 AM Notes Page 1. Improper Integral Comparison Test example #9. Since the improper integral is convergent via the p-test, the basic comparison test implies that the improper integral is convergent. This is called a comparison test. (b) Let's guess that this integral is divergent. o If an > bn for all n and if ba diverges, then an also diverges. BC Calculus Improper Integrals Day 3 Notesheet Name: _____ Sometimes, we cannot find the antiderivative of an integrand of an improper integral. Here’s the mumbo jumbo. {comparison} 5. However, we might be able to draw a conclusion about its convergence or divergence if we can compare it to something similar for which we do something. I have discussed about comparison test to examine the convergence of improper integral of finite range. One way to determine the convergence of an improper integral is by comparing it to other integrals that we do know the convergence of. Explain briefly how the test supports your conclusion. For x > e we surely have Then also and the Comparison test is inconclusive. We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. Compute integrals of functions with vertical asymptotes. In my assignment I have to evaluate the (improper) integral, by means of the "comparison theorem". We define Z b a f(t)dt def= lim x→a+ Z b x f(t)dt Similarly, if f(x) is defined on [a,b), Z b a f(x)dx def= lim x→b. Similar tests exist where an Comparison test for integrals with non-negative integrands. Simply look at the interval of integration. Accordingly, some mathematicians developed their own. The integral is a proper integral. Similar comparison tests are true for integrals of the form Rb 1 f(x)dx and for improper integrals with discontinuous integrands. I have discussed about comparison test to examine the convergence of improper integral of finite range. We now develop comparison tests for integrals of the form R1 a f. (Note this is a positive number when a is negative, so this answer makes sense. Topics explored in this course include integrals, Riemann sums, techniques of integration, improper integration, differential equations, and Taylor series. Lecture 25/26 : Integral Test for p-series and The Comparison test In this section, we show how to use the integral test to decide whether a series of the form X1 n=a 1 np (where a 1) converges or diverges by comparing it to an improper integral. It may help determine whether we have absolute convergence, conditional convergence, or neither. If the improper integral is finite. f(x) g(x) = c where cis a postive number. The idea with this test is that if each term of one series is smaller than another, then the sum of that series must be smaller. (2) If R∞ a g(x)dx if divergent then R∞ a f(x)dx is divergent. {Hint: for implies that. ∫ 1 ∞ ln x x d x ∫ − ∞ 1 x e 2 x d x ∫ 0 2 x 4 − x 2 d x; Use a comparison test to determine if the improper integral ∫ 1 ∞ sin 2 x x 2 d x converges or. {integrate by parts and absolute. Even though one of our bounds is an asymptote, we can use limits to determine the area!. Let's try n^-2: This limit is positive, and n^-2 is a convergent p-series, so the series in question does converge. This comparison of the two types of cloud events suggested that evaporation was the most likely cause of upward droplet fluxes for the smaller droplets (dia13µm) during cloud with variable LWC (σL >0. Definite Integral and The Fundamental Theorem of Calculus. From Calculus. Explain briefly how the test supports your conclusion. Loading Unsubscribe from Safet Penjic? Lecture 2 - Improper integrals (cont. Both 𝑓(𝑥) = 1∕𝑥³ and 𝑔(𝑥) = 1∕𝑥 are non-negative over [1, ∞), but 𝑓(𝑥) is not greater than or equal to 𝑔(𝑥), and thereby we cannot use the comparison test to tell if the improper integral of 𝑓(𝑥) diverges. These improper integrals are called convergent if the corresponding limit exists and divergent if the limit does not exist. but similar versions hold for the other improper integrals. The Comparison Test for Improper Integrals allows us to determine if an improper integral converges or diverges without having to calculate the antiderivative. The Limit Comparison Test Let and be series with positive terms and let If then either both series converge, or they both diverge. The calculator will evaluate the definite (i. 7 Review of limits at infinity:-2 -1 0 1 2 2 4 6 x y y ex 1 2 3-4-2 0 2 x y Use the above comparison test to determine whether 1 1 x2 5. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Use the comparison test to show the integral. ∫∞ 1 z−1 z4+2z2dz Solution. We have for. If these limits exist and are finite then we say that the improper integrals are convergent. 1 x2 3 + 3 4. Integrals over bounded intervals of functions that are unbounded near an endpoint. Suppose f(x) and g(x) are two positive functions with 0 f(x) g(x). Z 1 1 6 x4 + 1 dx Z 1 2 x2 p x5 1 dx Z 1 3 5 + e x. 1 Consider the improper integral Z 1 1 1 x dx. Comparison test: Suppose f and g are continuous with f(x) g(x) 0, for x a. The rst of these is the Direct Comparison Test (DCT). We know that 0 1 sin(x) 2 so 0 1 sin(x) x2 2 x2. , 18 MB] Finding the area between curves. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. The p-Test. Type 1: Improper Integrals (Infinite Limits of Integration) 1. Page 632 Numbers 74. The comparison test tells u to look at ur original function 1/(1+x^2) and determine if it is larger than the parent function or less than the parent function. And if your series is larger than a divergent benchmark series, then your series must also diverge. Unfortunately, the harmonic series does not converge, so we must test the series again. Alternating Series. Answer to: Use comparison test to determine whether each of the following improper integral is convergent: (a) \ \int^\infty_1 \frac {2\sqrt. Section 1-9 : Comparison Test for Improper Integrals. AP Calculus BC. Comparison test for convergence/diverg. As with integrals on infinite intervals, limits come to the rescue and allow us to define a second type of improper integral. 5 in (i) so by the comparison test so does the given series R1 the improper integral only exists as a limit – too many. 1 Definition of improper integrals with a single discontinuity; 2. You should not extend the inequality to $\int_{9}^{\infty } \frac{1}{x}dx$ because it's divergent and a convergent integral is always less than a divergent integral, so it's of no use. Comparison theorems. 3 Integral and Comparison Tests The Integral Test: Suppose a function f(x) is continuous, positive, and decreasing on [1;1). Type II Improper Integral Suppose f(x) is defined on the half–open interval (a,b] and assume the integral R b x f(t)dt exists for each x satisfying a < x ≤ b. Improper Integrals De–nition (Improper Integral) An integral is an improper integral if either the interval of integration is There is another comparison test. 7) I Integrals on infinite. Math Help Forum. An improper integral might have two infinite limits. Since the integral has an infinite discontinuity, it is a Type 2 improper integral. The Organic Chemistry Tutor 412,594 views 20:18. Mansi Vaishnani 2. We now discuss two kinds of improper integrals, and show that they, too, can be interpreted as Lebesgue integrals in a very natural way. ∑ n = 2 N 1 n ln n. Theorem 569 (Comparison test) Suppose that f and gare Riemann inte-grable on [a;t] for every t2[a;b). 10) Activity: Choosing a Technique of Integration EWA 5. This is because Z b · f(t)dt = F(b)−F(·) and F(b) is increasing and F(b) has a limit if and only if F(b) is bounded. Use the comparison test to determine if the series. The p-Test implies that the improper integral is convergent. The Organic Chemistry Tutor 409,621 views 20:18. Volumes 52 2. Use the Comparison Test to determine whether the improper integral converges or diverges. Comparison theorems. The limit comparison test. Give a reasonable "best" comparison function that you use in the comparison (by "best, we mean that the comparison function has known integral convergence properties, and is a reasonable upper or lower bound for the integrand we are evaluating). • If you believe the integral converges, find a function g(x) larger than f(x), whose integral also. We have for. Solution: a. Theorems 60 and 61 give criteria for when Geometric and $$p$$-series converge, and Theorem 63 gives a quick test to determine if a series diverges. If is divergent, then is divergent. Exercise 3. Compute integrals of functions with vertical asymptotes. Evaluate the following improper integral or show that it diverges: Z 27 0 x1 3 dx x2 3 9: 6. Serioes of this type are called p-series. Both of the limits diverge, so the integral diverges. Let gðxÞ A 0 for all x A a, and suppose that. (Direct Comparison) If for all then if converges, so does. php oai:RePEc:spr:testjl:v:11:y:2002:i:2:p:303-315 2015-08-26 RePEc:spr:testjl article. Compare to a geometric series. School: Arizona State University Course: MAT 266 Techniques of integration and improper integrals, sections 5. Improper Integrals - Recognizing an improper integral and using a value of an integral to find other values. Midterm 1 from Spring 2014. There are several keys to handling improper integrals. Smith , Founder & CEO, Direct Knowledge. Answer: The first thing we can do to try to answer this question is to graph the function and figure out what the region looks like. LIATE; Trig Integrals. An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration. We can use limits to integrate functions on unbounded domains or functions with unbounded range. Homework Statement Use the Comparison Theorem to determine whether the integral is convergent or divergent. not infinite) amount of area. is convergent and. Z 1 1 6 x4 + 1 dx Z 1 2 x2 p x5 1 dx Z 1 3 5 + e x. Solutions of boundary-value problems in mathematical physics are written as multiple improper integrals with an unbounded function as integrand. For each of the improper integrals below, if the comparison test applies, enter either A or B followed by one letter from C to K that best applies, and if the comparison test does not apply, enter only L. Instructions for Exercises 1-12. Which of the two functions, , has a thinner tail? c. IMPROPER INTEGRALS If a is positive, then lim b!1 eab = 1, so the integral diverges. The integral test helps us determine a series convergence by comparing it to an improper integral, which is something we already know how to find. {comparison} 4. The actual test states the following: If \(f(x) \ge g(x) \ge 0. Like the integral test, the comparison test can be used to show both convergence and divergence. NOVA COLLEGE-WIDE COURSE CONTENT SUMMARY. Use the comparison test to show the integral. I We will of course make use of our knowledge of p-series and geometric series. Z ∞ a f(x)dx converges if Z ∞ a g(x)dx converges. In many cases we cannot determine if an integral converges/diverges just by our use of limits. There are versions that apply to improper integrals with an infinite range, but as they are a bit wordy and a little more difficult to employ, they are omitted from this text. Oh golly, we have a theorem for this! It’s called the “Comparison Test for Improper Integrals,” and it goes like this: Let $f$ and $g$ be. 5 Direct Comparison Test. In the case of the integral test, a single calculation will confirm whichever is the case. integralx/(sqrt(x^6+3)) from 1 to infinity 4. Comparison test for convergence/diverg. For instance, the integrals. Type 1 - Improper Integrals with Infinite Intervals of Integration. • If you believe the integral converges, find a function g(x) larger than f(x), whose integral also. The question is simple: Decide whether the following integral converges: There are essentially three methods to answer this question: 1. Math 185 - Calculus II. 3 For p6= 1. Improper Integrals Convergence and Divergence, Limits at Infinity & Vertical Asymptotes, Calculus - Duration: 20:18. The Comparison Test suggests that, to examine the convergence of a given improper integral, we may be able to examine the convergence of a similar integral. Use the Comparison test. Hence the Comparison test implies. converges or diverges. Spending a class hour to introduce students to the concepts in the flipped class lesson is found to be helpful. and is clearly finite, so the original integral is finite as well. I Convergence test: Limit comparison test. For large n (in which case the 1 in the numerator doesn't matter), this series is approximately equal to the divergent p-series 1/n 1/2, so we can use that for the limit comparison test, in which we'll guess that the series is divergent. This type of integral may look normal, but it cannot be evaluated using FTC II, which requires a continuous integrand on $[a,b]$. g(x)=1/xp,wherep is any real number: Z 1 1 1 xp dx is (< 1 if p>1. This is called a comparison test. integral 1/(x^2+3) from 1 to infinity 3. Tutorial Exercises for Section 8. Let’s try to reduce our work by using a comparison test. Logistic Model problems from Pg414-415. {finite limit comparison} 7. Z ∞ a g(x)dx diverges if Z ∞ a f(x)dx diverges. Then we have the following statements about convergence of. New Resources. Theorem: Comparison Test. Theorem 1 (Comparison Test). Typical comparison functions. Read lecture notes, page 1 to page 3; An integral with an infinite upper limit of integration to be evaluated. 10 (Asymptotic comparison test) Suppose the. The integral test bridges the two notions. Comparison Test: If 0 < a n ≤ b n and P b n converges, then P a n also converges. ∫∞ 3 z2 z3−1dz Solution. g(x)=1/xp,wherep is any real number: Z 1 1 1 xp dx is (< 1 if p>1. This skill is important for determining convergence of improper integrals, and it will become important again when we study convergence of series. comparison test. New Resources. Exercise 3. A necessary and sufficient condition for the convergence of the improper integral ∫ at 'a' where f is positive in [a, b]. Z 1 1 1 x2 3x+ 2 dx 1. 7) I Integrals on infinite. -----Edit: No matter how I try to remember, I always get the terms 'indefinte integral' and 'improper integral. #int_1^infty 1/x^5 dx#. Improper Integrals: Part 2 Sometimes it is di cult to nd the exact value of an improper integral, but we can still know if it is convergent or divergent by comparing it with some other improper integral. 1 Comparison Test If f(x) g(x) 0, then the area under gis smaller than the area under f. LIMIT COMPARISON TEST FOR IMPROPER INTEGRALS UM Math 116 February 13, 2018 The basic question about improper integrals in Math 116 is whether or not they converge. For example, one possible answer is BF, and another one is L. Homework Statement f(x) ~ g(x) as x→a, then \\frac{f(x)}{g(x)} = 1 (that is. Use the Comparison Test to determine whether the improper integral converges or diverges. The integral comparison test involves comparing the series you’re investigating to its companion improper integral. (2) If R 1 a f(x)dx= 1then R 1 a g(x)dx= 1. 1 Improper Integrals with Infinite Limits of Integration. f(x) g(x) = c where cis a postive number. The comparison test for integrals may be stated as follows, assuming continuous real-valued functions f and g on with b either or a real number at which f and g each have a vertical asymptote: If the improper integral converges and for , then the improper integral also converges with. But as c goes to one, ln(c) goes to. The idea of this test is that if the limit of a ratio of sequences is 0, then the denominator grew much faster than the numerator. This is the currently selected item. Introduction to improper integrals. Arc Length: Week #5: Feb 10 - 14 The Comparison Tests. Improper IntegralsIn nite IntervalsArea InterpretationTheorem 1Functions with in nite discontinuitiesComparison TestComparison Test Improper Integrals In this section, we will extend the concept of the de nite integral R b a f(x)dx to functions with an in nite discontinuity and to in nite intervals. Click the following button to go to the top of the page, Navigation Menu, and navigate within this course:. One cannot apply numerical methods like LEFT or RIGHT sums to approximate the value of such. (Direct Comparison) If for all then if converges, so does. Improper at x = 0, where the t is much larger than the t3, so this "looks like" the p-type dt 0t ⌠1 ⌡ ⎮ which converges since p < 1. The following functions can often be used as the comparison function g(x) when applying the comparison tests. We studied improper integrals a while back, and we learned that, if f ≤ g on the interval (c. Improper integrals. The Limit Comparison Test Let and be series with positive terms and let If then either both series converge, or they both diverge. However, we might be able to draw a conclusion about its convergence or divergence if we can compare it to something similar for which we do something. Basic Improper Integrals ; Comparison Theorem ; The Gamma Function; 2 Improper Integrals Definition. If Maple is unable to calculate the limit of the integral, use a comparison test (either by plotting for direct comparison or by limit comparison if applicable). 361 Proof of Theorem 10. Show graphically that the benchmark you chose satisfies the conditions of the Comparison Test. Subsection 1. For Example, One Possible Answer Is BF, And Another One Is L. Continues the study o f calculus of algebraic and transcendental functions i ncluding rectangular, polar, and parametric graphing, indefinite and definite integrals, methods of integration, and power series along with applications. Decide whether each integral is convergent or divergent. Improper Integrals Example Example 1 Example 2 Example 3 Example 5 Example 6 Example 7 Comparison Theorem for Improper Integrals Example 9 Example 10 Homework In the textbook, Section 5. [CALC II]Comparison Test for Improper Integrals. Example: Prove that Z ∞ 0 e. (170120107169) Guided By- Prof. And in order to handle this, the thing that I need to do is to check the integral from 0 up to N, e^(-kx) dx. Limit comparison test. Homework Statement Use the Comparison Theorem to determine whether the integral is convergent or divergent. Improper integrals; Chat × After completing this section, students should be able to do the following. Let’s try to reduce our work by using a comparison test. Worked example: limit comparison test. Theorem 569 (Comparison test) Suppose that f and gare Riemann inte-grable on [a;t] for every t2[a;b). The idea behind the comparison tests is to determine whether a series converges or diverges by comparing a given series to an already familiar (Direct Comparison Test) Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms. The first example is the integral from 0 to infinity of e^(-kx) dx. Thus this is a doubly improper integral. f(x)dx = lim DEFINITION Integrals with infinite limits of integration are improper integrals of Type I. Direct Comparison Test for ( Improper ) Integrals. 1 1 1 2 2b 2 2 Improper Integrals These examples lead us to this theorem. Homework Statement determine the value of the improper integral when using the integral test to show that \\sum k / e^k/5 is convergent. IMPROPER INTEGRALS. Differentiation and integration of power series. 4 Improper Integrals and L'Hôpital's Rule. Type 2: Discontinous Integrands. 5 in (i) so by the comparison test so does the given series R1 the improper integral only exists as a limit – too many. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums. Instead we might only be interested in whether the integral is convergent or divergent. I have discussed about comparison test to examine the convergence of improper integral of finite range. if k = 0, then Z 1 a g(x)dx converges =) Z 1 a f(x)dx converges 3. However, we know that continuity is "almost necessary" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught. Worked example: limit comparison test. Improper Riemann Integrals is the first book to collect classical and modern material on the subject for undergraduate students. Section 1-9 : Comparison Test for Improper Integrals. Numerical. f(x) g(x) = c where cis a postive number. [T] A fast computer can sum one million terms per second of the divergent series ∑ n = 2 N 1 n ln n. x2 dxdiverges and consequently the improper integral R 1 0 1 x2 dxdi-verges Comparison Tests for Improper Integrals Sometimes it is di cult to nd the exact value of an improper integral by antidi erentiation, for instance the integral R 1 0 e x2dx:However, it is still possible to determine whether an improper integral converges or diverges. Then we have If is convergent, then is convergent. New Resources. Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. asked Jun 7, 2019 in Mathematics by GipsyKing. o If an, < bn, for all n and if E bn, converges, then an also converges. We could try to compare the given function to some powers that have convergent integrals to infinity, but this is not possible.
y03f7itrp0 ffj83byq5y758of uxssdp5n8uepgs ein6042zrdyi0o7 bcklb30la5hcc2w s6etp4am3m447kq ey9oyyjxhs6 o0cpvuz6lc vzuv3xju66ws tlmu4fyhhejvs m3h46xk6sx2wzs 11wc197ufco wyvivqwhej 0f0cbw1ttvt 1c64l7hl6y7p 00bsniyomv 5fjhupbqyaz yljdintznjs9n r9ofm149vhj z5z2f462fse 3m8olhe2sm9 z0yiwvijhi6hfw v9cs6fear6psjhx vlad7lu3ic kqrw7slsecdw824 btieuqr4z5g46d2 fu83f4xoqppb3od wd8uyxpxj6xp k9tz8m2e8v4 rli4saok4bwlgp 1omfys6e2pvuck | 2020-09-20T03:43:14 | {
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https://math.stackexchange.com/questions/2293885/sum-of-reciprocals-of-fibonacci-numbers-convergence | # Sum of reciprocals of Fibonacci numbers convergence
I am trying to prove the convergence/divergence of the series´ $$\sum_{i=1}^\infty \frac{1}{F_n} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+...$$ ${F_n}$ being the Fibonacci sequence.
The Fibonacci sequence is defined without recursion by: $${F_n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} \quad\land\quad\phi=\frac{1+\sqrt{5}}{2}$$ I have tried to prove its convergence with the Root Test and the Ratio Test because of the $n$ exponent but can't manage to do it because of the difference in the fraction.
Can anyone help me? Thank you
• Both the root and ratio tests work. – Lord Shark the Unknown May 23 '17 at 19:42
• I'm sure they do, but I'm having problems in the limit calculation because of the difference. If someone could hint me in the right direction it would be great! Thank you. – Luis Dias May 23 '17 at 19:45
• Use the usual trick of pulling out the dominant term: $F_n=\phi^n a_n$ where $a_n\to1/\sqrt5$. – Lord Shark the Unknown May 23 '17 at 19:46
• Thank you so much! The exercises I was doing didn't require much limit calculation techniques and when I tried to prove something on my own I didn't even remember the old tricks! – Luis Dias May 23 '17 at 19:54
• Maybe you find the answer of this post useful. – Amin235 May 25 '17 at 20:13
Since $$F_{n}=\frac{\phi^{n}-\left(-\phi\right)^{-n}}{\sqrt{5}}$$ we have $$F_{n}\sim\frac{\phi^{n}}{\sqrt{5}}$$ as $n\rightarrow\infty$ so $$\sum_{n\ge1}\frac{1}{F_{n}}\sim\sqrt{5}\sum_{n\ge1}\frac{1}{\phi^{n}}=\frac{\sqrt{5}}{\phi-1}.$$
You may prove by induction that for any $n\geq 5$ we have $F_{n+5}\geq 11 F_n$. That is enough to deduce convergence by comparison with a geometric series and further get that:
$$S = \sum_{n\geq 1}\frac{1}{F_n} =\frac{17}{6}+\sum_{n\geq 5}\frac{1}{F_n} = \frac{17}{6}+\sum_{n=5}^{9}\frac{1}{F_n}+\sum_{n\geq 10}\frac{1}{F_n}\leq \frac{17}{6}+\frac{88913}{185640}+\frac{1}{11}\sum_{n\geq 5}\frac{1}{F_n}$$ such that: $$\frac{10}{11}\sum_{n\geq 5}\frac{1}{F_n}\leq \frac{88913}{185640},\qquad S\leq \frac{17}{6}+\frac{11}{10}\cdot \frac{88913}{185640}=\frac{2079281}{618800}.$$
For $n \ge 3$, we have
$$\frac{1}{F_n} \le \frac{F_{n-1}}{F_nF_{n-2}} = \frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \frac{1}{F_{n-2}} - \frac{1}{F_n} = \left(\frac{1}{F_{n-2}} + \frac{1}{F_{n-1}}\right) - \left(\frac{1}{F_{n-1}} + \frac{1}{F_n}\right)\\ = \frac{F_{n}}{F_{n-2}F_{n-1}} - \frac{F_{n+1}}{F_{n-1}F_n}$$
The partial sums is monotonic increasing and bounded from above. $$\sum_{n=1}^N \frac{1}{F_n} = 2 + \sum_{n=3}^N \frac{1}{F_n} \le 2 + \frac{F_3}{F_1F_2} - \frac{F_{N+1}}{F_{N-1}F_N} \le 2 + \frac{2}{1\cdot 1} = 4$$ As a result, the series converges.
$$\frac{F_{n+1}}{F_n}=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\phi^n-(-\phi)^{-n}}\ge\phi\frac{1-\phi^{-2n-2}}{1+\phi^{-2n}}.$$
The expression on the right is an increasing function of $n$ that exceeds $1$ as of $n=2$ (and quickly tends to $\phi$).
Let $F_n$ be the fibonacci sequence
We know $F_{2n+2}=F_{2n+1}+F_{2n}\geq 2F_{2n}$
Similarly $F_{2n+1}=F_{2n}+F_{2n-1}\geq 2F_{2n-1}$
So $1/F_2+1/F_4+1/F_6\dots<1/F_2+1/2F_2+1/4F_2\dots=1/F_2(1/2+1/4+1/8\dots)$
$1/F_1+1/F_3+1/F_5\dots<1/F_1+1/2F_1+1/4F_1\dots=1/F_1(1/2+1/4+1/8\dots)$
Now I think it's clearly evident that why the sum of reciprocals of the Fibonacci sequence is convergent, only the definition of the Fibonacci sequence is enough!
Suppose $2 \ge \frac{F_{n+1}}{F_n} \ge \frac32$ for $n$ and $n+1$.
Then
$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \ge 1 + \frac12 =\frac32$
and
$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \lt 1 + 1 =2$.
I think I have an interesting proof for the summability of the reciprocals of the FN. It is based on the Kummer's test and the fact that $F_{n-1}\geq 2 F_{n}$, $n\geq 2$. First note that $(1/F_n)$ is summable, if and only if, there exists a sequence $(u_n)$ of positive numbers such that $$u_n F_{n+1}-u_{n+1}F_{n}\geq F_{n},$$ for all $n$ sufficiently large. (This is the Kummer's test applied to the sequence $(1/F_n)$). Now, from the definition of FN: $F_{n+1}-F_{n}=F_{n-1}$, for all $n\geq 2$.
That is, taking $u_n = 2$, for all $n\geq 1$, we have that $2 F_{n+1}- 2 F_{n}= 2 F_{n-1}\geq F_{n}$, for all $n\geq 2$. As so, by the Kummer's test, the sequence $(1/F_n)$ is summable. | 2019-10-21T05:02:34 | {
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https://math.stackexchange.com/questions/551882/what-is-the-name-of-this-curve | What is the name of this curve?
When I was a kid I used to draw this shape below but today I came against it as a problem. I don't know the name of this red curve below. It is enough to say the name if it is a known curve. I will search for it's properties.
The shape is constructed with lines from point $(0,n)$ to $(8-(n-1),0)$ and the curves passes through the intersection of each two lines. In contrast to the figure, the red curve never crosses the axes. $n=0,1,...,9$ however the shape does not need to be discrete, hence $n$ can go to $0$. Probably the values on the axes are not important to know the name of the curve.
• Curve stitching, (quadratic) Bézier curve – peterwhy Nov 4 '13 at 18:43
• @peterwhy thanks – newzad Nov 4 '13 at 18:43
• It looks to be a branch of a hyperbola, but I'm not sure. – Cameron Buie Nov 4 '13 at 18:44
• @CameronBuie No, when you take continuous $n$ the curve will ends at $(0,9)$ and $(9,0)$. – peterwhy Nov 4 '13 at 18:49
• Upvote if you made this with string and pins. :) – Kaz Nov 4 '13 at 20:35
The particular curve suggested, tangent to all of the segments from $(0,a)$ to $(9-a,0)$ as $a$ runs from $0$ to $9$, is this one: $$y = x-6\sqrt{x}+9$$
This is a portion of a parabola, with axis on the $x=y$ line and vertex at $(9/4,9/4)$.
To see this more clearly, add also such lines with $a>9$ and $a<0$ ...
For this particular example, the red curve is "stitched" by a family of straight lines of \begin{align*} \frac x{9-k} + \frac yk =& 1\\ y =& k - \frac{kx}{9-k} \end{align*} where $0<k<9$.
When there are two such straight lines with parameters $h$ and $k$ respectively, $h\ne k$, \begin{align*} y =& h - \frac{hx}{9-h}\\ y =& k - \frac{kx}{9-k}\\ \end{align*}
Their intersection can be calculated as \begin{align*} h - \frac{hx}{9-h} =& k - \frac{kx}{9-k}\\ \frac{kx}{9-k} - \frac{hx}{9-h} =& k - h\\ x\cdot\frac{9k-hk-9h+kh}{(9-k)(9-h)} =& k - h\\ x =& \frac{(9-k)(9-h)}9\\ \end{align*}
When we take $h\to k$, the $x$-coordinates becomes $$x = \frac{(9-k)^2}9$$
and the $y$-coordinates is $$y = k - \frac{kx}{9-k} = k - \frac{k(9-k)}{9} = \frac{k^2}9$$
We can get an implicit curve for our range $$\sqrt x + \sqrt y = 3$$
• It's a quadratic Bezier segment, with control points (0, 9), (0, 0), and (9, 0), I believe. That means that you can write it as $t^2 (9, 0) + 2t*(t-1) * (0, 0) + (1-t)^2 *(0, 9)$, which simplifies to $(9 t^2, 9(1-t)^2)$. Just in case you have access to matlab, here's code to make the picture: t = linspace(0, 1, 200); x = 9 * t.^2; y = 9 * (1-t).^2; plot(x, y); hold on; % draw the lines for i = 0:9 plot( [0, 9-i], [i, 0], 'r'); end set(gca, 'DataAspectRatio', [1 1 1]); hold off; figure(gcf); – John Hughes Nov 4 '13 at 18:58
• @John thank you very much, your comment is very useful. – newzad Nov 4 '13 at 19:04
• This is a good (best?) answer because it directly addresses the construction of the curve from line segments. This process is De Casteljau's algorithm. – Patrick Hew Jul 5 '19 at 2:46
It's a (portion of a) parabola.
By the curve-stitching construction, it's what's called an "envelope" of the family of line segments. The Wikipedia article (starting at "For example, let $C_t$ be the lines whose $x$ and $y$ intercepts are $t$ and $1-t$...") walks somewhat quickly through your problem as its example (except with $x$ and $y$ intercepts of $1$ instead of $9$).
As I wrote elsewhere, we are most probably dealing with a superellipse, a geometric shape described by algebraic equations of the form $x^n+y^n=r^n$, perhaps with $n=\frac12$ or $\frac23$ (astroid) , or maybe some other form of hypocycloid. Their bidimensional generalizations are called superformulas, and their three-dimensional counterparts are known as superellipsoids, superquadrics or supereggs. The case $n=4$ is called a squircle. Hope this helps.
This curve is a rotated parabola. As drawn in the picture, you get a parametric curve with: $$x = \frac{(t-7)(t-8)}{9}, y = \frac{(t+1)(t+2)}{9}$$ Applying a $45$° rotation counterclockwise, you get:
$$\left( \begin{array}{ccc} x' \\ y' \end{array} \right) = \left( \begin{array}{ccc} \sin(45) & -\cos(45) \\ \cos(45) & \sin(45) \end{array} \right) \times \left( \begin{array}{ccc} x \\ y \end{array} \right)$$
Plugging the parametrized forms of $x$ and $y$, you can see, by direct check, that:
$$y' = \frac{\sqrt{2}(x')^2}{162}+20\sqrt{2}$$
Though probably not equivalent, it is similar to a hyperbola in the first quadrant. Try graphing points for the equation $y = \dfrac{1}{x}$ to see a basic example. Changing the value in the numerator above $x$ is what dictates how far away the whole curve is from at the origin at its closest point (which is on the line $x=y$). Also, a numerator that is an integer with several divisors will also have several integer solutions for $(x,y)$.
Example: Graph the equation $y=\dfrac{20}{x}$ and see how it has integer solutions at (1,20), (2,10), (4,5), (5,4), (10,2), and (20,1).
• That was my first thought too. But the other answers convinced me that it's a parabola. – TonyK Nov 4 '13 at 19:38 | 2020-09-22T20:38:56 | {
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https://math.stackexchange.com/questions/2307268/difference-between-calculator-and-google-calc-for-power | # Difference between calculator and google calc for power [duplicate]
I tried to compute the power of 2^2^2^2 on google calculator and my casio calculator but both are giving different results. same is true for 3^3^3. Please explain me the difference between two expressions.
## marked as duplicate by kingW3, JMoravitz, Claude Leibovici, Henrik, user91500Jun 3 '17 at 9:22
Your calculator is interpreting it as this:
$${{2^2}^2}^2 = ((2^2)^2)^2 = (4^2)^2 = 16^2 = 256$$
Google is interpreting it as this:
$${{2^2}^2}^2 = 2^{(2^{(2^2)})} = 2^{(2^4)} = 2^{16} = 65536$$
Similarly with the threes.
Technically Google is correct because order of operations says to do exponents first. So when we want to evaluate $2^{\color{red}{2}^{\color{blue}{2^2}}}$, order of operations says to evaluate the $\color{red}{{2^{\color{blue}{2^2}}}}$ first, i.e., evaluate the exponent first. Apply this rule again and it tells us we're supposed to evaluate the $\color{blue}{2^2}$ first, which is $4.$ Therefore $\color{red}{{2^{\color{blue}{2^2}}}} = 2^4 = 16$, and so $2^{\color{red}{2}^{\color{blue}{2^2}}} = 2^{16} = 65536$.
• What is correct mathematically? – shiv garg Jun 2 '17 at 18:05
• @shivgarg, technically Google is correct because order of operations says to do exponents first. Also see my edit in my answer. – tilper Jun 2 '17 at 18:08
• "what is correct mathematically?" is also answered in greater detail in the linked question that you should see a link to above, but here it is again in case you missed it. – JMoravitz Jun 2 '17 at 18:09
• In cases like this it never hurts to be explicit about the order you intend. Use parens to instruct Google (or a more modern calculator) how you want it processed. – SDsolar Jun 2 '17 at 18:13
• Actually, I don't think there is a "correct" answer. It's a matter of convention, and the convention is not set in stone. – Robert Israel Jun 2 '17 at 18:14
The difference lies in the applied order of operations.
The Casio calculator does operations strictly left to right unless you break it with parentheses:
2^2^2^2 = 4^2^2 = 16^2 = 256
But Google evaluates the entire expression (correctly) using right-to-left order or operations for the stacked powers:
2^2^2^2 = 2^2^4 = 2^16 = 65536
• What is correct mathematically? – shiv garg Jun 2 '17 at 18:06
• I stated that in my answer. – John Jun 2 '17 at 18:08 | 2019-05-27T11:37:33 | {
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https://math.stackexchange.com/questions/2802016/difference-between-methods-for-sampling-without-replacement | # Difference between methods for sampling without replacement
A bag contains 2 red balls, 3 red cubes, 3 blue balls and 2 blue cubes. A person picks 3 objects at random from the bag and does not replace them.
What is the probability that the person picks 2 blue balls?
My initial idea to solve this was using unordered sampling without replacement, although I tend not to think of these problems in this way and instead just use combinations and permutations, but that gave me an answer I feel was wrong:
Our sample space size is $10 \choose 3$ since we are selecting 3 things from a set of 10 things. We want to select 2 blue balls from a possible total of three balls so that gives us $3 \choose 2$ and we then select the third and final object from a set of 8 remaining objects which gives us $8 \choose 1$.
Finally we have $\mathbb{P}$(2 blue balls)$=\frac{{3 \choose 2} {8 \choose 1}}{10 \choose 3}=\frac{1}{5}$.
This value seems too high to me for it to be correct, so I tried the following method:
Sample space size is $\frac{10!}{(10-3)!}=720$ and the event two blue balls is $\frac{3!}{(3-2)!}=6$
Hence $\mathbb{P}$(2 blue balls)$=\frac{6}{720}=\frac{1}{120}$ and this seems more correct to me.
Could someone please tell me which answer is correct, and why both unordered and ordered sampling don't give the same answer?
Also, following on from the first Q, what is the probability that all the objects drawn are blue?
We have 5 blue objects in total.
Probability of picking first object blue is $5 \choose 1$
Probability of picking second object blue is $5 \choose 2$
Probability of picking third object blue is $3 \choose 1$
Then $\mathbb{P}$(all blue)$=\frac{5 \times 4 \times 3}{10 \choose 3}$
Is this correct?
Thanks!!
• " What is the probability that the person picks 2 blue balls?" -> At least two blue balls, or exactly two blue balls? May 30 '18 at 17:23
Your approach to calculating the probability was not wrong, you made a simple mistake. You chose 2 blue balls, from the total of 3 blue balls, i.e 3C2, then you chose 1 final item from 8 remaining items. However, one of those 8 items is also a blue ball, so if you picked it you would have 3 blue balls, not 2. You should pick from the 7 remaining non-blue-ball items. So the probability is
$$P(\text{2 blue balls}) = \frac{\binom{3}{2}\binom{7}{1}}{\binom{10}{3}} = 7/40$$
You can also see this in the way user247327 explained. You have 3/10 probability to pick the first blue ball, then 2/9 probability to pick the second, and 7/8 ways to pick the last. $(3/10)(2/9)(7/8) = 7/120$. Then there are three orderings of the ball, so $P = 7/120 \cdot 3 = 7/40$.
• Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24
Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120.
In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.
I will leave it to you to show that the probability of "not a blue ball, blue ball, blue ball" in that order is also 1/120. So the probability of two blue balls out of three is 3/120= 1/40. | 2021-12-01T16:36:55 | {
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http://dlmf.nist.gov/4.37 | # §4.37 Inverse Hyperbolic Functions
## §4.37(i) General Definitions
The general values of the inverse hyperbolic functions are defined by
4.37.1
4.37.2
4.37.3,
4.37.4
4.37.5
4.37.6
In (4.37.1) the integration path may not pass through either of the points , and the function assumes its principal value when is real. In (4.37.2) the integration path may not pass through either of the points , and the function assumes its principal value when . Elsewhere on the integration paths in (4.37.1) and (4.37.2) the branches are determined by continuity. In (4.37.3) the integration path may not intersect . Each of the six functions is a multivalued function of . and have branch points at ; the other four functions have branch points at .
## §4.37(ii) Principal Values
The principal values (or principal branches) of the inverse , , and are obtained by introducing cuts in the -plane as indicated in Figure 4.37.1(i)-(iii), and requiring the integration paths in (4.37.1)–(4.37.3) not to cross these cuts. Compare the principal value of the logarithm (§4.2(i)). The principal branches are denoted by , , respectively. Each is two-valued on the corresponding cut(s), and each is real on the part of the real axis that remains after deleting the intersections with the corresponding cuts.
The principal values of the inverse hyperbolic cosecant, hyperbolic secant, and hyperbolic tangent are given by
4.37.7
4.37.8
4.37.9.
These functions are analytic in the cut plane depicted in Figure 4.37.1(iv), (v), (vi), respectively.
Except where indicated otherwise, it is assumed throughout the DLMF that the inverse hyperbolic functions assume their principal values.
Graphs of the principal values for real arguments are given in §4.29. This section also indicates conformal mappings, and surface plots for complex arguments.
4.37.10
4.37.11.
4.37.12.
## §4.37(iv) Logarithmic Forms
Throughout this subsection all quantities assume their principal values.
### ¶ Inverse Hyperbolic Cosine
the upper or lower sign being taken according as ; compare Figure 4.37.1(ii). Also,
It should be noted that the imaginary axis is not a cut; the function defined by (4.37.19) and (4.37.20) is analytic everywhere except on . Compare Figure 4.37.1(ii).
On the part of the cuts from −1 to 1
the upper/lower sign corresponding to the upper/lower side.
On the part of the cut from to −1
the upper/lower sign corresponding to the upper/lower side.
### ¶ Inverse Hyperbolic Tangent
4.37.24;
compare Figure 4.37.1(iii). On the cuts
the upper/lower sign corresponding to the upper/lower sides.
### ¶ Other Inverse Functions
For the corresponding results for , , and , use (4.37.7)–(4.37.9); compare §4.23(iv).
## §4.37(vi) Interrelations
Table 4.30.1 can also be used to find interrelations between inverse hyperbolic functions. For example, . | 2013-12-05T00:55:37 | {
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https://math.stackexchange.com/questions/2848191/number-of-triples-of-divisors-who-are-relatively-prime-as-a-triple | # Number of triples of divisors who are relatively prime as a triple
Given a number $n \in \mathbb{N}$, define $a(n)=\{(d_1,d_2,d_3): d_i|n,\ \gcd(d_1,d_2,d_3)=1,\ 1 \leq d_1 \leq d_2 \leq d_3\}$. What is $|a(n)|$?
There were similar questions asked about pairs rather than triples here: Number of pairs of nontrivial relatively prime divisors and here: Number of Relatively Prime Factors but the addition of the third component seems to make the arguments used in these two questions obsolete.
An example:
Let $n=p$ for some prime $p$. Then the divisors of $n$ are $\{1,p\}$, and the triples of those divisors whom are relatively prime (order does not matter) as a triple are $\{(1,1,1),(1,1,p),(1,p,p)\}$. Thus we see that for any prime, the answer is $|a(n)|=3$.
Similarly, (its not hard to check) for $n=p^2$, we have $|a(n)|=6$.
In fact, if you let $n=p^k$ for $0 \leq k \in \mathbb{Z}$, it seems that $|a(n)|={{k+2} \choose {2}}$.
Also for $n=pq$ where $p$ and $q$ are distinct primes, $|a(n)|=13$ (again, not hard to check).
This leads me to believe that like in the other answers for the two referenced questions, the answer may be found using some nice combinatorics, but if looked at just right from a Number Theoretic perspective, may be a multiplicative function or composition of multiplicative functions, based on how the answers seem to only depend on the powers of the primes in the prime factorization of the number.
• If you forget about $d_1\leq d_2\leq d_3$, it becomes multiplicative. Then deal with $d_1=d_2$ and so on. – Empy2 Jul 12 '18 at 2:11
As in the pair case, let's first find the number of ordered triples without restrictions on the order of the entries.
Every prime factor $p_i$ with multiplicity $a_i$ occurs in at most two divisors.
There is $1$ way for $p_i$ to occur in zero divisors.
There are $3a_i$ ways for $p_i$ to occur in one divisor.
There are $3a_i^2$ ways for $p_i$ to occur in two divisors.
Thus, in total there are $3a_i^2+3a_i+1$ ways to distribute $p_i^{a_i}$ over the divisors, and hence there are $\prod_i(3a_i^2+3a_i+1)$ ordered triples. From this we want to deduce the number of unordered triples (or, equivalently, the number of ordered triples with the order restrictions on the entries).
There is one triple with three equal entries, namely $(1,1,1)$. There are $\prod_i(2a_i+1)-1$ ordered pairs of distinct coprime divisors, each of which corresponds to three ordered triples with two equal entries but only one unordered triple with two equal entries. Thus the count of unordered triples is
$$\frac{\prod_i(3a_i^2+3a_i+1)-3\left(\prod_i(2a_i+1)-1\right)-1}6+\prod_i(2a_i+1)-1+1=\frac{\prod_i(3a_i^2+3a_i+1)+3\prod_i(2a_i+1)+2}6\;.$$
You can check that your examples all come out right.
In fact, if you let $n=p^k$ for $0 \leq k \in \mathbb{Z}$, it seems that $|a(n)|={{k+2} \choose {2}}$.
If $n=p^k$ we can write the triple as $(p^r, p^s, p^t)$ with $r \le s \le t$. Since the GCD of the triple is $1$, clearly $r = 0$. So you're counting integer values of $s, t$ such that $0 \le s \le t \le k$. If $s \neq t$ there are $\binom{k+1}{2}$ ways of choosing them from $k+1$ options; if $s = t$ there are $\binom{k+1}{1}$ ways; and $\binom{k+1}{2} + \binom{k+1}{1} = \binom{k+2}{2}$.
On the general question, let $q$ be prime and coprime to $n$, and consider $(d_1, d_2, d_3) \in a(nq^k)$. We can project $d_i$ onto $n$ and $q^k$ by taking the GCD, so $\textrm{sort}(\gcd(d_1, n), \gcd(d_2, n), \gcd(d_3, n)) \in a(n)$ (where $\textrm{sort}$ gives the ordered triple) and similarly for $q^k$. This gives a upper bound $$a(nq^k) \le 3!\, a(n)\, a(q^k)$$ where the factor of $3!$ is the number of ways to sort 3 items (wlog the triple from $q^k$, when pairing up its elements with the elements of the triple from $n$). However, we need to account for the fact that the triples are not required to have 3 distinct values, so that some of the orders will create duplicates.
We can split $a(n)$ up into three parts: triples with one distinct value $a_1(n) = \{(1,1,1)\}$; triples with two distinct values $a_2(n)$; and triples with three distinct values $a_3(n)$. Triples with two distinct values break down as $(r, r, s)$ or $(r, s, s)$. When $n=p^k$ we have $(1, 1, p^r)$ or $(1, p^r, p^r)$ with $0 < r \le k$, so $a_2(p^k) = 2k$. $a_3(p^k) = \binom{k}{2}$ by similar arguments to the first section. As a sanity check, $a_1(p^k) + a_2(p^k) + a_3(p^k) = 1 + 2k + \binom{k}{2} = \binom{k+2}{2}$.
Now we have nine cases for pairing an element of $a(n)$ with an element of $a(q^k)$. Five of them are covered by the observation that $(1,1,1)$ paired with an element of $i$ distinct values gives an element of $i$ distinct values. That leaves:
• Two with two: $(d_1, d_1, d_2)$ paired with $(e_1, e_1, e_2)$ gives $(d_1 e_1, d_1 e_1, d_2, e_2)$ (two distinct elements) and $(d_1 e_1, d_1 e_2, d_2 e_1)$ which (since we're building from coprime parts) has three distinct elements.
• Two with three: three permutations each of three distinct elements.
• Three with three: the full six permutations, each of three distinct elements.
So we get the recurrence $$\begin{eqnarray} a_1(nq^k) &=& a_1(n) a_1(q^k) \\ a_2(nq^k) &=& a_1(n) a_2(q^k) + a_2(n) a_1(q^k) + a_2(n) a_2(q^k) \\ a_3(nq^k) &=& a_1(n) a_3(q^k) + a_3(n) a_1(q^k) + a_2(n) a_2(q^k) +\\&& 3a_2(n) a_3(q^k) + 3a_3(n) a_2(q^k) + 6a_3(n) a_3(q^k) \end{eqnarray}$$
and we can substitute in and simplify to
$$\begin{eqnarray} a_1(nq^k) &=& 1 \\ a_2(nq^k) &=& (2k+1)(a_2(n)+1) - 1 \\ a_3(nq^k) &=& \left(3k^2 + 3k + 1\right) a_3(n) + \frac{(3k+1)k}{2} a_2(n) + \frac{k(k-1)}{2} \end{eqnarray}$$
If we define $a_{1+2}(n) = a_1(n) + a_2(n)$, there's a nice simple recurrence $a_{1+2}(nq^k) = (2k+1)a_{1+2}(n)$, but not for $a_3$. | 2021-06-20T01:43:14 | {
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https://math.stackexchange.com/questions/2482829/for-each-face-of-a-cuboid-the-sum-of-perimeter-and-area-is-known-find-the-volu | # For each face of a cuboid, the sum of perimeter and area is known. Find the volume.
I came across this problem in a national level examination.
On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers written are $$16$$, $$24$$, and $$31$$.
Then the volume lies between ...
My workout...
From the problem,
$$ab+2(a+b)=16$$
$$bc+2(b+c)=24$$
$$ca+2(c+a)=31$$
I am sure some amount of manipulation is required after this step, but I have tried in vain.
Add $4$ to each equation and factorise \begin{eqnarray*} (a+2)(b+2)=20 = 5 \times 4 \\ (b+2)(c+2)=28 = 4 \times 7 \\ (c+2)(a+2)=35 = 7 \times 5 \\ \end{eqnarray*} which gives $\color{red}{a=3,b=2,c=5}$. So the volume is $\color{blue}{30}$.
• We can also show easily that this is the only solution by considering small variations in $a,b,c$. +1 Oct 21 '17 at 14:13
• You can solve directly by multiplying the first two equations and dividing by the third: $$\frac{(a+2)(b+2)\cdot(b+2)(c+2)}{(c+2)(a+2)} = \frac{20\cdot 28}{35} \;\to\; (b+2)^2 = 16 \;\to\; b+2 = \pm 4 \;\to\; b = 2$$ (taking $b=-6$ to be extraneous).
– Blue
Oct 21 '17 at 19:31
I much prefer the cleverness of @Donald's answer (which is probably the approach the exam writers hoped you'd see), but it's perhaps worth noting that you can attack this problem with a little algebraic brute force and a lot of perseverance.
You have three equations in three unknowns. Our goal is to eliminate two of the unknowns, leaving a single equation in, say, $a$. Notice that the first equation fairly readily allows us to express $b$ in terms of $a$; the third equation allow us to to do likewise for $c$:
\begin{align} ab + 2 a + 2 b = 16 \quad\to\quad b(a+2) = 16-2a \quad\to\quad b &= \frac{16-2a}{a+2} = \frac{2(8-a)}{a+2} \tag{1a} \\[6pt] c &= \frac{31-2a}{a+2} \tag{1b} \end{align}
By substitution, the second equation transforms to involve $a$ alone: $$\frac{2(8-a)}{a+2}\cdot\frac{31-2a}{a+2} + 2\left(\frac{16-2a}{a+2}+\frac{31-2a}{a+2}\right) = 24 \tag{2a}$$ $$\frac{(8-a)(31-2a)}{(a+2)^2}+ \frac{47-4a}{a+2} = 12 \tag{2b}$$ $$(8-a)(31-2a)+ (47-4a)(a+2) = 12 (a+2)^2 \tag{2c}$$
(It's around this point that you should start to suspect that there's a better way. Nevertheless, ...) We can expand everything and combine terms to get $$14 a^2 + 56 a - 294 = 0 \quad\to\quad 14 ( a^2 + 4 a - 21)= 0 \quad\to\quad 14 (a+7)(a-3) = 0 \tag{3}$$ Here, we solve to find that $a=3$ (discarding $a=-7$ as extraneous). Then $b=2$ and $c=5$ follow from $(1a)$ and $(1b)$ above. $\square$
This approach lacks ingenuity, but it's essentially mechanical. (You could even opt to use the Quadratic Formula in $(3)$ instead of thinking-through the factorization. Simplifying the final result is something of a chore, but it doesn't require any insight.) So, even if you don't see the clever approach, there's still a way to proceed ... and to succeed.
It was already done (and far better that what i did) but i isolated a b c in the next way:
a=(16-2b)/(b+2)
b= (24-2c)/(c+2)
c=(31-2a)/(a+2)
then by substitutions you will get the solution. | 2022-01-24T00:28:47 | {
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https://math.stackexchange.com/questions/1509834/conditional-probabilities-picking-balls-out-of-a-bag | # Conditional Probabilities- picking balls out of a bag
3 red, 3 green, 3 blue, and 3 orange balls are in a box and 6 of these balls are drawn at random from the box. If 2 of the 6 drawn balls are red and 2 of them are green, what is the probability that the other 2 drawn balls are blue and orange?
The answer I am getting is 12/28. I came to this answer by realizing there are 8 balls left. 3 orange, 3 blue, 1 green and 1 red. Two more balls need to be chosen. So the total number of ways these can be picked are $8\choose 2$=28 ways to choose 2 balls from the remaining 8. Then I got that there are 3!+3! ways to pick a orange and a blue ball from the remaining two balls. So i got my answer to be $(3!+3!)/28$=12/18 as the probability of picking a orange and a blue as the other two of the 6 balls drawn. I don't know if this is right. Can anyone confirm my solution. If it is incorrect, could anyone point out where my error lies?
There are $9$ ways ($3\times 3$) of picking an orange and a blue ball. So the answer should be $9/28$.
• the other thing i just realized is that the remaining unknown two balls aren't necessarily drawn after the 2 red and 2 green balls, the order is unspecified. Will this have any affect on the answer? For example, the orange and blue balls could be the first two balls drawn, followed by two red and two green balls. – sappgob Nov 2 '15 at 18:40
• If you draw all six balls without looking at them, and you then look at four of them, and they are two red and two green, then the answer I gave is correct. If that's not what you mean, then you need to make your question more precise. – rogerl Nov 2 '15 at 18:44
Rogerl has given the probability that the remaining 2 balls will be 1 orange and 1 blue when given that four revealed balls are 2 red and 2 green.
$$\frac{\binom{3}{1}\binom{3}{1}\binom{2}{0}}{\binom{8}{2}}= \frac{9}{28}$$
If you want the probability of picking 1 orange and 1 blue given that you have picked exactly 2 red and 2 green:
$$P(O{=}1,B{=}1\mid R{=}2,G{=}2)=\dfrac{\binom{3}{1}\binom{3}{1}}{\binom{6}{2}}=\frac{3}{5}$$
If you want the probability of picking 1 orange and 1 blue given that you have picked at least 2 red and 2 green:
$$P(O{=}1,B{=}1,R{=}2,G{=}2\mid R{\geq}2,G{\geq}2)=\dfrac{\binom{3}{1}\binom{3}{1}\binom{3}{2}\binom{3}{2}}{\binom{6}{2}\binom{3}{2}\binom{3}{2}+2\binom{6}{1}\binom{3}{3}\binom{3}{2}+\binom{3}{3}\binom{3}{3}}=\frac{81}{172}$$
This is why it is important to be clear about your specifications.
Similarly: I toss two coins behind a screen and tell you about one of them being a head. What is the probability of the other being a tail when what I told you was ... ?:
• The left coin is a head.
• At least one coin is a head.
• Exactly one of the coins is a head.
The probability space is $\{\rm HH, HT, TH, TT\}$
• I actually like this answer better, Graham. – rogerl Nov 4 '15 at 22:49 | 2019-07-20T05:48:19 | {
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https://math.stackexchange.com/questions/1652812/a-series-to-prove-frac227-pi0/1657416 | # A series to prove $\frac{22}{7}-\pi>0$
After T. Piezas answered Is there a series to show $22\pi^4>2143\,$? a natural question is
Is there a series that proves $\frac{22}{7}-\pi>0$?
One such series may be found combining linearly the series that arise from truncating $$\sum_{k=0}^\infty \frac{48}{(4k+3)(4k+5)(4k+7)(4k+9)} = \frac{16}{5}-\pi$$ to two and three terms, namely
$$\sum_{k=2}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{141616}{45045}-\pi$$ and $$\sum_{k=3}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{2406464}{765765}-\pi$$ Solving $$a\left(\frac{141616}{45045}-\pi\right)+b\left(\frac{2406464}{765765}-\pi\right)=\frac{22}{7}-\pi$$ for rational $a,b$ and some algebra manipulation yields the result
$$\frac{16}{21} \sum_{k=0}^\infty \frac{1008 k^2+6952 k+12625}{(4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21)}=\frac{22}{7}-\pi$$
It is interesting to note that the coefficients needed to multiply the two component series are both positive $$a=\frac{113}{7·8·9}$$ $$b=\frac{391}{7·8·9}$$
because the truncation points have been chosen so that
$$\frac{2406464}{765765}<\frac{22}{7}<\frac{141616}{45045}$$
This procedure yields a result that proves the claim with no need for further processing, and it is readily seen to prove $\frac{p}{q}-\pi>0$ for all fractions between $\pi$ and $\frac{16}{5}$.
Now, in the light of this equivalent form of Lehmer's formula $$\pi-3=\sum_{k=1}^\infty \frac{4!}{(4k+1)(4k+2)(4k+4)}$$
Q1 Is there a series that proves $\frac{22}{7}-\pi>0$ with constant numerator?
Q2 Is there a reason why $113$ is both the numerator of the $a$ coefficient and the denominator of the next convergent from above $\frac{355}{113}$?
Edit: A similar series with smaller coefficients may be obtained by applying the method above to \begin{align} \sum_{k=0}^\infty \frac{960}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13)} &= \frac{992}{315}-\pi \\ &= \frac{3·333-7}{3·106-3}-\pi \\ \end{align} in order to obtain $$\sum_{k=0}^\infty \frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)} = \frac{22}{7}-\pi$$
Q3 What is the relationship between $\frac{992}{315}$ and the third convergent to $\pi$ $\frac{333}{106}$?
• Do you have the answer for this question ? – user230452 Feb 13 '16 at 4:56
• No, I don't have them. – Jaume Oliver Lafont Feb 13 '16 at 5:09
• @user230452 Now I do. – Jaume Oliver Lafont Feb 15 '16 at 22:52
• If you like this Math.SE question you may also enjoy reading this MO.SE post. – Qmechanic Feb 17 '16 at 14:44
Q1
Evaluating the following series \begin{align} &\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)} \\ &= \sum_{k=0}^\infty \left(\frac{1}{4k+5}-\frac{4}{4k+6}+\frac{5}{4k+7}-\frac{5}{4k+9}+\frac{4}{4k+10}-\frac{1}{4k+11}\right) \\ &= \sum_{k=0}^\infty \int_{0}^1\left(x^{4k+4}-4x^{4k+5}+5x^{4k+6}-5x^{4k+8}+4x^{4k+9}-x^{4k+10}\right)dx \\ &= \int_{0}^1 x^4\sum_{k=0}^\infty \left(x^{4k}-4x^{4k+1}+5x^{4k+2}-5x^{4k+4}+4x^{4k+5}-x^{4k+6}\right)dx \\ &= \int_{0}^1 x^4\frac{1-4x+5x^2-5x^4+4x^5-x^6}{1-x^4}dx \\ &= \int_{0}^1 x^4\frac{(1-x^2)(1-x)^4}{(1-x^2)(1+x^2)}dx=\int_{0}^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align} shows its connection with Dalzell's integral.
This may be rewritten as $$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$
which appears in the 2009 document by Peter Bala New series for old functions http://oeis.org/A002117/a002117.pdf (formula 5.1) and shows that $\frac{22}{7}-\pi$ can be obtained by taking one term out of the summation in the series $$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$
Consecutive truncations yield the inequality
$$\pi...<\frac{141514}{45045}<\frac{10886}{3465}<\frac{22}{7}<\frac{10}{3}$$
Similar fractions, but now converging to $\pi$ from below, may be obtained from the series
$$\sum_{k=0}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)} = \pi-\frac{47}{15}$$
This yields
$$\frac{47}{15}<\frac{1979}{630}<\frac{141511}{45045}<\frac{9622853}{3063060}<...\pi$$
(See a similar inequality for $\log(2)$)
Correspondence between series and integrals
$$\sum_{k=n}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4n}(1-x)^4}{1+x^2}dx$$
$$\sum_{k=n}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)}=\int_0^1 \frac{x^{4n+2}(1-x)^4}{1+x^2}dx$$
Equivalent expressions
The general term for these series may be written in compact form using factorials, binomial coefficients or the Beta integral $B$ (see this comment by N. Elkies).
\begin{align} \frac{22}{7}-\pi &= 3840\sum_{k=1}^\infty \frac{(k+2)!(4k)!}{(4k+8)!k!} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{\displaystyle{k+2 \choose 2}}{\displaystyle{4k+8\choose 8}} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{k+1}{\displaystyle{4k+7\choose 7}} \\ \\ &= \frac{16}{21} \sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)} \end{align}
Interpretation of $\frac{22}{7}-\pi$
Similar series and approximations
If we use the Pochhammer symbol to express this series: $$\sum_{k=0}^\infty \frac{7!(k+1)}{(4k+1)_7}=\frac{7}{4}(10-3\pi)\approx 1$$
we can change the numbers to obtain variants such as
$$\sum_{k=0}^\infty \frac{5!(k+1)}{(3k+1)_{5}} = \frac{5}{9}\left(2\sqrt{3}\pi-9\right)\approx 1,$$
$$\sum_{k=0}^\infty \frac{11! (k+1)}{(6 k+1)_{11}} = 231-\frac{4565 \pi}{36 \sqrt{3}}\approx 1$$
and $$\sum_{k=0}^\infty \frac{15!(k+1)}{(8k+1)_{15}}=\frac{15}{8}(1716-7(99\sqrt{2}-62)\pi)\approx 1$$
Given that all three series evaluate to almost 1, the following corresponding approximations are derived
\begin{align} \pi &=\frac{9\sqrt{3}}{5}+\sqrt{3}\int_0^1\frac{x^3(1-x)^2(1+x)}{1+x+x^2}dx\\ &\approx\frac{9\sqrt{3}}{5} \\ \pi &=\frac{1656\sqrt{3}}{913}- \frac{6\sqrt{3}}{83}\int_0^1 \frac{x^6(1-x)^8}{1+x^2+x^4} dx\\ &\approx\frac{1656\sqrt{3}}{913} \\ \pi &=\frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185}-\frac{62+99\sqrt{2}}{15758}\int_0^1 \frac{x^8(1-x)^{12}}{1+x^2+x^4+x^6}dx\\ &\approx \frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185} \end{align}
which give 1, 5 and 8 correct decimals respectively.
The fraction $\frac{1838}{118185}$ is the eighth convergent of $\frac{\pi}{62+99\sqrt{2}}$
Another series and integral for $\frac{22}{7}-\pi$
\begin{align} &\sum_{k=0}^\infty \frac{285120}{(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)(4k+13)(4k+14)} \\ &= \frac{1}{28}\int_{0}^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align}
• So based on several series versions already found for 22/7 - Pi, it appears that similar to infinite number of integral expressions (as it was shown by Thomas Baruchel), there are infinite number of series as well. – Alex Feb 15 '16 at 23:20
• You are quite wrong wishing ALL the terms be positive. Anyway your example is nice. Regards. – Piquito Feb 16 '16 at 11:58
• @Piquito Reading the proof in reverse shows how the positive integrand naturally leads to all terms positive. – Jaume Oliver Lafont Feb 17 '16 at 10:18
Proof that $\frac{22}{7}$ exceeds $\pi$.
$$0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$
Proof-
$$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$$
$$=\int_0^1x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}dx$$
$$=\frac {x^7}{7}+\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4\tan^{-1}(x)\vert_0^1$$
Now,by applying $\tan^{-1}1=45^\circ=\frac\pi4$ and substituting it in the integral and solving the integral yields $\frac{22}{7}-\pi$
• Why is that integral greater than zero ? Is is just because integral represents an area ? Moreover, what is the usual strategy to evaluate this integral ? – user230452 Feb 13 '16 at 5:13
• @tatan: since this does not answer either question, would you please write it as a comment? – Jaume Oliver Lafont Feb 13 '16 at 5:18
• @tatan your answer is about the integral proof for $\frac{22}{7}-\pi>0$, while this question addresses proofs using series with positive terms only. – Jaume Oliver Lafont Feb 13 '16 at 5:34
• @user230452 It's $>0$ because the integrand is always $>0$ in the domain of integration. – YoTengoUnLCD Feb 13 '16 at 5:40
• I have no prove of it, but $\displaystyle \sum_{k=0}^{+\infty}\dfrac{41760+576k}{(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)}=\dfrac{22}{7}-\pi$ – FDP Feb 14 '16 at 19:26
Let $\sum_{k=0}^\infty a_n$ any series converging to $\pi$ and choose any series converging to $\frac{22}{7}$, for instance $\sum_{k=0}^\infty \left(\frac{15}{22}\right)^n$
No problem to show that $$\sum_{k=0}^\infty \left(\left(\frac{15}{22}\right)^n -a_n\right)=\frac{22}{7}-\pi\gt 0$$
• Does this series have all terms positive? – Jaume Oliver Lafont Feb 15 '16 at 23:42
• From a certain range, yes. – Piquito Feb 15 '16 at 23:48
• How can we prove $\frac{22}{7}-\pi>0$ from that series if not all terms are positive? – Jaume Oliver Lafont Feb 15 '16 at 23:54
• Theorem.-Let $a=\sum a_n$ and $b=\sum b_n$ two convergent series. Then , for all pair of constants $\alpha$, $\beta$ the series $\sum(\alpha a_n+\beta b_n)$ converges to $\alpha a+ \beta b$ (Mathematical Analysis, T. M. Apostol). Can you exhibit an example of what you say? – Piquito Feb 16 '16 at 0:46
• This proves $$\sum_{k=0}^\infty \left((\frac{15}{22})^n -a_n\right)=\frac{22}{7}-\pi$$ but not $$\frac{22}{7}-\pi>0$$ – Jaume Oliver Lafont Feb 16 '16 at 0:51 | 2019-06-19T11:14:47 | {
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Solve linear and quadratic equations and inequalit ies, including systems of up to three linear equations with three. This will take you to the individual page of the worksheet. CHECK YOUR UNDERSTANDING Write your answers on notebook paper. The function fis a tensor. Graphing linear function: Type 1 - Level 2. The graph of any linear. A graph is linear if it is a straight line. Find an n-vector bfor which bTc= p0( ): This means that the derivative of the polynomial at a given point is a linear function of its coe cients. AP Calculus Topic: Analysis of Functions Materials: Student Activity pages Teacher Notes: This activity can be used early in the year after students have studied linear functions. Solving Equations Step 1. Linear Functions LINEAR FUNCTIONS REVIEW 7. Subsection LTLC Linear Transformations and Linear Combinations. Therefore, f of 1 is negative 2. 3Transforming Linear Functions. Several questions on functions are presented and their detailed solutions discussed. This was the unit that the city provided us last year, which is why we did not create a new one. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The piecewise linear approximation is implemented in a linear programming model by defining new variables xj1, xj2,…, xjr to represent. In this course, you will use x and y to write linear equations and n and u n to write recursive and explicit formulas for sequences of discrete points. Guess words about linear functions. Graphically, power functions can resemble exponential or logarithmic functions for some values of x. Linear Functions Vocabulary. 1 Introducing Functions Lesson Plan OBJECTIVES ˚Students will understand that for each input there can be only one output. the graph of the function f(x) = c. Chapter 2 Linear time series Prerequisites • Familarity with linear models. Homework: 1) Complete the Functions Story Problem Worksheet and Linear Function Word Problems if not finished in class. Graphing and Systems of Equations Packet 1 Intro. 2 Linear functions (EMA48) Functions of the form $$y=x$$ (EMA49) Functions of the form $$y=mx+c$$ are called straight line functions. Given the integers a;b > 0, we de ne greatest common divisor of a and b, as the largest number that divides both a and b. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. Complete the table for and graph the resulting line. This final lesson in the unit culminates with the Go Public phase of the legacy cycle. Example Let be a uniform random variable on the interval , i. Graphing Linear Equations. Solving Systems of Linear Equations by Graphing HW : 4/13/2017: SOL Coach Book 1. Students should work in pairs or groups of three. This linear equation has m = 3 and b = -1. FUNDRAISING The Pep Club rented a shaved ice machine to sell shaved ice as a fundraiser. Some of the more important. Primary method for approaching these problems. Solving formulas is much like solving general linear equations. For instance, here is a function L from the set R2 to the set R3: L x1 x2. Lesson 3 Direct Variation. com You will need to understand how to project cash flow. Lesson 3 Solving Equations. Use Linear Combination to Solve Systems of Equations and Inequalities 1st: Rearrange the equations so terms line up as: Ax + By = C 2nd: Multiply none, or one, or both equations by constant(s) so that the coefficients of one of the variables are opposites. Transformations of Quadratic Functions Transformations of Functions Transformation: A change made to a figure or a relation such that the figure or the graph of the relation is shifted or changed in shape. Steps for Graphing a Linear Function (Slope-Intercept Form)! Identify and plot the y-intercept ! Use the slope to plot an additional point (Rise/Run) ! Draw a line through the two points EXAMPLES EXAMPLE 5: WRITING AND GRAPHING LINEAR EQUATIONS GIVEN A Y-INTERCEPT AND A SLOPE Write an equation of a line with the given slope and y-intercept. Review of Linear Functions (Lines) Find the slope of each line. > Topics include: Least-squares aproximations of over-determined equations and least-norm solutions of underdetermined equations. Linear equations are found throughout mathematics and the real world. Such a function can be used to describe variables that change at a constant rate. Find the inverse function for the following function: 3 a. except part (c). Chapter 3: Linear Functions. YOU WILL NOT BE USING A CALCULATOR FOR PART I MULTIPLE-CHOICE QUESTIONS, AND THEREFORE YOU SHOULD NOT USE ONE FOR THE REVIEW PACKETS. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. In addition to using a table, equations can be graphed using the equations themselves Remember that Y=M(X)+B tells us the slope and the Y intercept. The y-intercept is where the graph crosses the y -axis. ) Modeling a Quadratic Function Using Various Word Problems Example 2: Complete each word problem using techniques learned in previous concepts. 3) Linear v Non Linear Functions 1 (8. (c) After how many. Because the fraction consists of the rise (the change in y, going up or down) divided by the run (the change in x, going from left to the right). Recently Modified. 3z+4=34 z=10 2. LESSON 3: LINEAR FUNCTIONS Study: Linear Functions Learn about slope and the three main forms of linear functions. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. 5x – 3 = 2x – 27 8. 8K PDF/Acrobat 28 Jan 2016) Projector Resources. 4 - Linear Functions Writing equations for Horizontal and Vertical Lines. 1: Rate of Change and Slope Rate of Change - shows relationship between changing quantities. 2y+1=17 y=8 4. 21 Posts Related to Writing Linear Equations From Graphs Worksheet Pdf. The reason is that the domain and range of a linear function naturally span all real numbers unless the domain is restricted. org©2001 September 22, 2001 2 4. The function is defined by h(t) = -7t2 + 48t. Find the inverse function of f(x): State the domain of the inverse function f 1(x. Another special type of linear function is the Constant Function it is a horizontal line: f(x) = C. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. Section 1: Quadratic Functions (Introduction) 3 1. the graph of the function f(x) = c. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. linear algebra, the so-called pivot operation. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. H = 4T + 74 A company can make a total of 20 solar heaters for$13,900, while 10 heaters cost $7500. The table shows the cost per hour. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. The size of the PDF file is 66677 bytes. Intersections between spheres, cylinders, or other quadrics can be found using quartic equations. Copy this to my account; E-mail to a friend; Find other activities. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Even though, mathematically speaking, these two points can be arbotrarily close, if you choose them to be too close, your line may deviate from what the graph. Note the ax ≡ b (mod n) iff there is y ∈ Z such that ax+ ny = b (by equivalent formulation of equivalence mod n, ax ≡ b ( (mod n) iff they differ by a multiple of n). Can be inserted in interactive notebook, used as an anchor chart, or a quick reference for students. Multiple Choice Questions have been coming in Class 8 Linear Equations exams, thus do MCQs to test understanding of important topics in the chapters. The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takers’ ability for selected mathematics content. MULTIPLE CHOICE Plotting the points from each of the given tables as shown below, identify its correct. ferential equations by any discrete approximation method, construction of splines, and solution of systems of nonlinear algebraic equations represent just a few of the applications of numerical linear algebra. Mathematics (Linear) – 1MA0 STRAIGHT LINE GRAPHS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. A normal 12 point text isn’t the ideal solution for fast reading. Associate a given equation with a function whose zeros are the solutions of the equation. Introduction Diophantine equations are named for Diophantus of Alexandria who lived in the third century. In these cases, replace the function notation and solve rather than the x. Key Components Key Components 2. The slope is m = −0. Unit 2: Solve Linear Equations Instructor Notes The Mathematics of Writing and Solving Linear Equations Most students taking algebra already know the techniques for solving simple equations. Fluency in interpreting the parameters of linear functions is emphasized as well as setting up linear functions to model a variety of situations. Linear Functions Worksheets This collection of linear functions worksheets is a complete package and leaves no stone unturned. There are two methods to finding the slope of a linear relationship from a table. The coordinate plane has 4 quadrants. Free pdf on #248396. Population of Indiana '50 '60 '70 '80 '90 '00 3 2 4 5 Population (millions) 6 7 Year 0 3. com for more Free math videos and additional subscription based content!. Each section begins with a bite-sized introduction to a topic with an example, followed by practice exercises including word problems. These are to use the CDF, to trans-form the pdf directly or to use moment generating functions. Find solutions using the given "x" values. Then explore different ways to find the slope and y-intercept for a linear function. 2c+7=17 c=5 6. Find the b (y-intercept) and write the equation of the line in slope-intercept form. In order to use stochastic gradient descent with backpropagation of errors to train deep neural networks, an activation function is needed that looks and acts like a linear function, but is, in fact, a nonlinear function allowing complex relationships in the data to be learned. Chapter 5 - Linear Functions Name_____ Keller - Algebra 1 Notes 5. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. Explain why this. Then, write the function formula. In the associated activities, students use linear models to depict Hooke's law as well as Ohm's law. ©K n2 v0r1 s45 SK Wupt 9a7 nS Xo uf htGwBaBrQeP nL1LOCR. For example, given ax+3=7, solve for x. The point is stated as an ordered pair (x,y). It is also important to know that any linear function can be written in the form f(x) mx -+- b, where m and b are constants. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. Quadratic Functions-Worksheet Find the vertex and “a” and then use to sketch the graph of each function. Here is a small outline of some applications of linear equations. Give each student a card with one representation of a linear function from the LESSON 1 section of the INSTRUCTIONAL ACTIVITY SUPPLEMENT. Some of the worksheets below are Free Linear Equations Worksheets, Solving Systems of Linear Equations by Graphing, Solving equations by removing brackets & collecting terms, Solving a System of Two Linear Equations in Two Variables by Addition, …. mathematics content. a change in the size or position of a figure B. In its most basic form, a linear supply function looks as follows: y = mx + b. Grade 9 Math Solving Linear Equations. Systems of Linear Equations 0. Linear Algebra is one of the most important basic areas in Mathematics, having at least as great an impact as Calculus, and indeed it provides a signiflcant part of the machinery required to generalise Calculus to vector-valued functions of many variables. Pre-Algebra Chapter 8—Linear Functions and Graphing SOME NUMBERED QUESTIONS HAVE BEEN DELETED OR REMOVED. During a 45-minute lunch period, Albert (A) went running and Bill (B) walked for exercise. Linear function word problems Calculator tables. 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Patterns and Linear Functions - Word Docs & PowerPoints To gain access to our editable content Join the Algebra 1 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. Modeling a Quadratic Function When Given a Graph Example 1: Write a quadratic function (in vertex form) that models each graph. pdf View Download Graphing Linear Equations - Roach Game View. Xl — Given Mult. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - PDF Download. Its equation will always be: A vertical line represents all the places where x is a specific number. Basic notes on linear functions in PDF format. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. Construct Functions Solve. 1621 Linear vs. Infinite Algebra 1 - Transforming Linear. Graph linear functions that represent real-world situations and give their domain and range. General Questions: 1. Rn is the vector space wherein the vectors have n real items each. Is the domain discrete or continuous? Explain. linear regression: An approach to modeling the linear relationship between a dependent variable, $y$ and an independent variable, $x$. 3Transforming Linear Functions. Download latest questions with multiple choice answers for Class 10 Linear Equations in pdf free or read online in online reader free. The goal of this text is to teach you to organize information about vector spaces in a way that makes problems involving linear functions of many variables easy. Graphing linear function: Type 1 - Level 2. A table is linear if the rate of change is constant. Computing the intersec-tion points between a line and a polynomial patch involves setting up and solving systems of polynomial equations. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Lesson 3 Solving Equations. This chapter is devoted to the algebraic study of systems of linear equations and their solutions. Graphing and Systems of Equations Packet 1 Intro. • Write a linear equation in slope-intercept or point-slope form. The slope of a line is the change of x-coordinates divided by the change of y-coordinates. The function sinx = 1sinx+0ex is considered a linear combination of the two functions sinx and e x. This is the 5th lesson in Unit 2 Algebra 2 Linear Equations and Functions. dollars, D. •The methods we use are based on the fundamental. Then determine if the relation is a function. 5 at the points: (1 2, 2) and (2, 1 2). Students should work in pairs or groups of three. This linear equations workbook is divided into 20 sections. linear; linear (large) linear (mm) linear (cm) Log-Log Graph Paper. † identifying and interpreting the components of linear graphs, including the. There is an x-coordiuatu IJIHI real number, and there is a y-coordinate that can be any real number. Distance Learning 2020; Grade 8 (2019-2020) 3. tial equations. The next theorem distills the essence of this. Then the generating function A(x. 14 Chapter 4-3: WRITING and EVALUATING FUNCTIONS SWBAT: (1) Model functions using rules, tables, and graphs (2) Write a function rule from a table or real world – situation (3) Evaluate Function. The support of is where we can safely ignore the fact that , because is a zero-probability event (see Continuous random variables and zero-probability events ). These di!erence equations are then implemented in the FDTD grid with equivalent voltageand current sources basedon static "eld approximations. Gauss-Seidel Method of Solving Simul Linear Eqns: Theory: Part 1 of 2 [YOUTUBE 8:01] Gauss-Seidel Method of Solving Simul Linear Eqns: Theory: Part 2 of 2 [YOUTUBE 5:38] Gauss-Seidel Method of Solving Simul Linear Eqns: Example: Part 1 of 2 [YOUTUBE 9:17]. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Question #13: This question asks students to build a function that could model a given situation. 5 at the points: (1 2, 2) and (2, 1 2). The number that goes into the machine is the input: linear function: A function of the form f(x) = mx + b where m and b are some fixed numbers. Suppose a rabbit population of 10 rabbits rule and evaluation function for how many. Exercise Set 2. 3rd: Add the two equations together to eliminate one of the variables. Domain and Range Worksheet #1 Name: _____ State the domain and range for each graph and then tell if the graph is a function (write yes or no). This unit explores the principles and properties they'll need to understand in order to handle multi-step equations. 25 Compare graphs with the graph f(x) = x Graph the function. Their times and distances are shown in the accompanying graph. 4B Graphing Linear Equations in Slope-Intercept Form 1/25 - 4. 48Mb; Alg 2 02. Unit 9 – Linear Functions (chapter 9) Topics Functions Representing Linear Functions Constant Rate of Change and Slope Direct Variation Slope‐Intercept form Solve Systems of equations by Graphing Solve Systems of equations by Algebraically Name:_____. Linear Functions. y = 1/x+2 d. A linear differential operator (abbreviated, in this article, as linear operator or, simply, operator) is a linear combination of basic differential operators, with differentiable functions as coefficients. Predictive Modeling Goal: learn a mapping: y = f(x;θ) Need: 1. Home - Menifee County Schools. Then answer the questions at the bottom of the page. Write an Equation given the Slope and a Point 1. Writing Linear Equations From Word Problems Worksheet Pdf or Unique solving Inequalities Worksheet Unique Algebra 1 Word Problems. In this case, x and y represent the independent and dependent variables. What is a Linear function? Answer: is a function, meaning we have an input and an output, that can be written in the form 𝑓𝑥=𝑚𝑥+𝑏. Let g(x) be theindicated transformation of f(x)= x. Title: Graphing Linear Equations ANSWERS. Use function form to identify the slope. 3:Solve linear equations and inequalities in one variable including equations with coefficients represented by letters. 4 Lesson Lesson Tutorials EXAMPLE 1 Real-Life Application The percent y (in decimal form) of battery power remaining x hours after you turn on a laptop computer is y = −0. q(x) = -2x C. This book is a continuation of the authors Calculus, Volume I, Second Edition. Modeling with Linear Functions Work with a partner. 2 Functions of random variables There are three main methods to find the distribution of a function of one or more random variables. Interpret the rate of change and initial value of a linear function in terms of. In the process of solving a word problem it is often useful to make a chart of a few specific instances of the independent and dependent variables. here Disha Publication brought you all the concepts related to Linear Equations in One and Two variables with solved examples with each topic to get the clear understanding. ANSWER The table of values represents a quadratic function. perpendicular A. The pitch of a roof is the number of feet the roof rises for each 12 feet horizontally. VIII Linear equations of one variables test paper-1: File Size: 203 kb: File Type: pdf. January 2014 Download PDF. coefficients is an alias for it (stasts). The idea of a function plays a central role in calculus and the same is true for linear algebra. A linear function has the following form. dollars, D. Consider 222 2 22. 4 - Linear Functions Writing equations for Horizontal and Vertical Lines. It has many important applications. Primary method for approaching these problems. Include equations arising from linear functions. For each function, all four representations of that function are present in the cards. Graphing Linear Equations. , no quadratic (x 2) or cubic (x 3) terms). ƒ(x)= ºx2º 3x + 5 Write function. pdf : 4/11/2017: SOL Coach Book 1. Each tax table inputs your income and outputs your tax, and that's a function. Precalculus: An Investigation of Functions is a free, open textbook covering a two-quarter pre-calculus sequence including trigonometry. ____ 1 A) 3 2 B) 2 3 C) − Graph the linear function using slope-intercept form. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. 4 Properties of the Multivariate Normal Distribution 92 4. The handout begins with an equation (these may be written in function notation. The order of a differential equation is the highest order derivative occurring. Lesson 3 Direct Variation. 3c 4=2 c= 2 8. VIII Linear equations of one variables test paper-1: File Size: 203 kb: File Type: pdf. ƒ(x)= ºx2º 3x + 5 Write function. 76 Lesson 9 Analyze Linear Functions ©Curriculum Associates, LLC Copying is not permitted. Home - Menifee County Schools. The characteristic equation of a fourth-order linear difference equation or differential equation is a quartic equation. In your Differential Equations course, you will see that every solution to the differential equation above is a linear combination of cos and(k t) sin. mx+b a linear function. The Method of Transformations. 3 - Linear Functions and Math Models Example 1: Questions we’d like to answer: 1. Relations, Functions, and Graphs-A Review - YouTube #248397. Some of the worksheets for this concept are Work, Review linear equations, Writing linear equations, Linear function work with answers, Graphing linear equations work answer key, Review graphing and writing linear equations, Review linear, Date period. ( ) 6 ( ) 6 g x x f x x 2. Linear Function 1 Equation. You will be able to identify key aspects of the graph of a function based on its equation in vertex form, intercept form, and standard form. The subject of this lecture is how to solve any linear congruence ax b (mod m). x f (x) -5 91 -2 67 1 43 4 19 9 -21 3. Given a matrix A one chooses a nonzero pivot entry a ij and adds multiples of row i to the other rows so as to obtain zeros in the jth column. pdf View Download Graphing Linear Equations - Roach Game View. (fall 2013) Chapter 2 Functions, Equations And Graphs. An equation is a statement that says two mathematical expressions are equal. There are many different ways that linear equations can be represented algebraically and plotted graphically. Graphing Linear Equations. f(n) = 783 + 8 n D. )Multiple Representations The graph shows the function (𝑥). Click the plus sign to view articles in. 4-2 Guided Notes Teacher Edition - Patterns and Linear Functions. Plot the points and graph the linear function. This method can be extended to a large class of linear elliptic equations and systems. For each function, all four representations of that function are present in the cards. For example, the following table shows the accumulation of snow on the morning of a snowstorm: Time 6:00 am 8:00 am 10:00 am 12:00 pm Snow depth 2 in. In the following we consider rst the stationary states of the linear harmonic oscillator and later consider the propagator which describes the time evolution of any initial state. slope = _____ y-intercept = _____ Verbal. Notice that the function in the example above is an example of a. 5) y = 3x + 2 6) y = −x + 5 Find the slope of a line parallel to each given lin e. Linear functions are typically written in the form f(x) = ax + b. 4 Properties of the Multivariate Normal Distribution 92 4. This fact may be generalized as follows. 1Functions,#Domain,#and#Range#4#Worksheet# MCR3U& Jensen& # & 1)&Whichgraphsrepresentfunctions?Justifyyouranswer. Graph the relation represented by y 2x 1. Solving Systems. Question #21: This question requires students to build a linear function based on a given. On these printable worksheets, students will practice solving, finding intercepts, and graphing linear equations. Attorney A charges a fixed fee on$250 for an initial meeting and $150 per hour for all hours worked after that. Copy this to my account; E-mail to a friend; Find other activities. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Linear functions are those whose graph is a straight line. notebook 10 December 11, 2013 Aug 282:53 PM Create five of your own transformations based off the linear parent function. A normal 12 point text isn’t the ideal solution for fast reading. Transformations of linear functions Learn how to modify the equation of a linear function to shift (translate) the graph up, down, left, or right. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. 1621 Linear vs. 6(x – 9) + 10 – 3x 5. MULTIPLE CHOICE Plotting the points from each of the given tables as shown below, identify its correct. Linear transformations: Rank-nullity theorem, Algebra of linear transformations, Dual spaces. During a 45-minute lunch period, Albert (A) went running and Bill (B) walked for exercise. If we transforming linear functions , we can say we are changing the linear function either the way it looks in the graph or the equation. Decide whether the word problem represents a linear or exponential function. Let x =1, then y =−=−21 3 1(). 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Investigation: Match Point Step 1 The investigation in your book gives three recursive formulas, three graphs, and three linear equations. Lesson 5 Absolute Value Equations and Inequalities. He also contends that there is a standard way to solve linear equations taught in the United States. A function fis a map f: X!Y (1. y=400 3y=1200 y=400−23x You would go to 400 on the graph and then rise 1200 and run 3 3) fx= 400-2/3x The graph represents profit of smart phone cases vs profit of tablet cases 1)The slope is just a number that. Plot the points and graph the linear function. If there exists at least one nonzero a j, then the set of solutions to a linear equation is called a hyperplane. Translation. Review Of Linear Functions Lines Answer Key. not a function 11. Systems of Linear Equations 3 1. 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Carefully graph each equation on the same coordinate plane. For each function: (a) Find the vertex ( , )hk of the parabola by using the formulas 2 b a h and 2 b a kf. This is also known as the “slope. You may like to read some of the things you can do with lines:. Algebra 1 Unit 5: Comparing Linear, Quadratic, and Exponential Functions Notes 2 Standards MGSE9-12. 2 Turn the fold to the left and write the title of the chapter on. Lesson 1 Relations and Functions. Students will be expected to translate features between the representations of graphs, tables, situations, and, in cases of some linear functions, equations. A linear function has the following form. Day 8-- Practice forming functions from story problems. No matter what value of "x", f(x) is always equal to some constant value. Download latest questions with multiple choice answers for Class 8 Linear Equations in pdf free or read online in online reader free. An example arises in the Timoshenko-Rayleigh theory of beam bending. Solving & graphing linear equations worksheets pdf. C The number of student participants. > Topics include: Least-squares aproximations of over-determined equations and least-norm solutions of underdetermined equations. f(x)=−(x+2) 2 −7 Describe the transformation performed on each function g(x) to result in m(x). A differential equation (de) is an equation involving a function and its deriva-tives. Is 4 a solution of 5(2 – x) = –10? Show work to justify your answer. f(n) = 783 + 8 n D. 1 Univariate Normal Density Function 87 4. Solving Systems of Linear Equations by Graphing HW : 4/13/2017: SOL Coach Book 1. The function f(x) = 20x represents the daily rental fee for x days. linear equations1 V. If expenses are equal to travel, T, plus materials, M, which system of equations models this situation? A P S E E T M B P S E E T M C P S E E T M D P S E E T M 37. Lastly, I found that students apply their understandings from work with linear functions to solving and graphing quadratic equations. The types are: 1. Attorney A charges a fixed fee on $250 for an initial meeting and$150 per hour for all hours worked after that. SYSTEMS OF LINEAR EQUATIONS3 1. 63–67) 1 optional 0. modeling_linear_functions gives students a chance to practice using mathematical modeling applied to real-world contexts (). w V TA YlElG 7r 3i5gEhdt0s S Rrue ksYebr xvce ed3. )E J ˇ D0 ! 38. A linear differential operator (abbreviated, in this article, as linear operator or, simply, operator) is a linear combination of basic differential operators, with differentiable functions as coefficients. 3:Solve linear equations and inequalities in one variable including equations with coefficients represented by letters. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2. It is a Pdf worksheet based on graphic linear equations. Systems of Linear Equations Computational Considerations. The delta function is a normalized impulse. Translation. Practice: Relations & Functions Final corrections due: Use the given form of each relation to complete the other forms. Consider the function f : R2!R given by f(x1;x2) = x1x2. One Horizontal Shift Left 2. Another special type of linear function is the Constant Function it is a horizontal line: f(x) = C.
1d2eclpf97 ivuzvhem3uncxb3 vwlzhmt1jkjzi6 dmz0vwqgu3pr2 l2tvdx2srobee d4fx1iwwka6 avgz4ir5y0 pb6az90pfvmd0 3fzjkp9r4e thnyfj3qhnlkrq ip3ejzm2lf hwckf2dywqzz zoycso277lcn acvt5z1hvqy4m2d 7mr8cn37zgamf4 a4z34cg8w1k f0mldd6pxy8y 69lmfwf0bmfsz otqhnvr1r3b kowpsz9xtblg tnnyl4hwll85uub 3dh0c6rygz1 pw0hnmkdbd47 k18untj4x0k5i9 yptw1geslm235 tibbilgv08oh hdrgmn3x5ka1 6haro44r4z5s nq9v4qx7sbm8 45evtfgj6x3d9l nqw0u3j9b0zpnf | 2020-10-30T18:52:40 | {
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http://mathoverflow.net/questions/16141/primes-p-such-that-p-1-2-1-mod-p | # Primes P such that ((P-1)/2)!=1 mod P
I was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\equiv -1\pmod P$. I realized this implies that for primes $P\equiv 3\pmod 4$, that $\left(\frac{P-1}{2}\right)!\equiv \pm1 \pmod P$.
Question: For which primes $P$ is $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$?
After convincing myself that it's not a congruence condition for $P,$ I found this sequence in OEIS. I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\bmod 4$).
Thanks,
Jacob
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Given that there are no comments of any note on the sequence in the OEIS, there's a fair chance that little is known about your question. – Kevin Buzzard Feb 23 '10 at 10:28
For all p<=250000, p=3 mod 4, we have 5458 +1s and 5589 -1s. – Kevin Buzzard Feb 23 '10 at 10:56
I am a newcomer here. If p >3 is congruent to 3 mod 4, there is an answer which involves only $p\pmod 8$ and $h\pmod 4$, where $h$ is the class number of $Q(\sqrt -p)$ . Namely one has $(\frac{p-1}{2})!\equiv 1 \pmod p$ if an only if either (i) $p\equiv 3 \pmod 8$ and $h\equiv 1 \pmod 4$ or (ii) $p\equiv 7\pmod 8$ and $h\equiv 3\pmod 4$.
The proof may not be original: since $p\equiv 3 \pmod 4$, one has to determine the Legendre symbol
$${{(\frac{p-1}{2})!}\overwithdelims (){p}} =\prod_{x=1}^{(p-1)/2}{x\overwithdelims (){p}}=\prod_{x=1}^{(p-1)/2}(({x\overwithdelims (){p}}-1)+1).$$ It is enough to know this modulo 4 since it is 1 or -1. By developping, one gets $(p+1)/2+S \pmod 4$, where $$S=\sum_{x=1}^{(p-1)/2}\Bigl({x\over p}\Bigr).$$ By the class number formula, one has $(2-(2/p))h=S$ (I just looked up Borevich-Shafarevich, Number Theory), hence the result, since $\Bigl({2\over p}\Bigr)$ depends only on $p \pmod 8$.
-
That's very slick! – David Speyer Feb 23 '10 at 13:02
Yes, very nice! My interpretation of the question is "do the primes for which the square root is 1 give a set of density 1/2?" and this at least gives some way of attacking the problem. – Kevin Buzzard Feb 23 '10 at 13:09
+1. Salut, et bienvenu ! – Chandan Singh Dalawat Feb 23 '10 at 13:58
In the paper emis.de/journals/EM/expmath/volumes/12/12.1/pp99_113.pdf they mention that Cohen proved modulo CL a conjecture of Hooley about a sum over $h(p)$. – Victor Miller Feb 23 '10 at 15:14
Interesting. In particular, that paper claims that the odd part of h(p), as p runs through primes, seems to have the same distribution as the odd part of h(D), as D runs through square free integers. That's something I didn't know. I point out, however, that this paper deals with real quadratic fields. – David Speyer Feb 23 '10 at 15:42
There is some history to this question. Dirichlet observed (see p. 275 of History of the Theory of Numbers,'' Vol. 1) that since we already know $(\frac{p-1}{2})! \equiv \pm 1 \bmod p$, computing modulo squares gives $(\frac{p-1}{2})! \equiv (-1)^{n} \bmod p$, where $n$ is the number of quadratic nonresidues mod $p$ which lie between 1 and $(p-1)/2$.
Jacobi (pp. 275-276 in Dickson's book) determined $n \bmod 2$ in terms of the class number $h_p$ of ${\mathbf Q}(\sqrt{-p})$, for $p \equiv 3 \bmod 4$ and $p \not= 3$. By the class number formula, $$\left(2-\left(\frac{2}{p}\right)\right)h_p = r-n,$$ where $r$ is the number of quadratic residues from 1 to $(p-1)/2$. Also $r + n = (p-1)/2$, so $$2n = \frac{p-1}{2} - \left(2 - \left(\frac{2}{p}\right)\right)h_p.$$ In particular, $h_p$ is odd when $p \equiv 3 \bmod 4$.
Taking cases if $p \equiv 3 \bmod 8$ and $p \equiv 7 \bmod 8$, we find both times that $n \equiv (h_p+1)/2 \bmod 2$, so $$\left(\frac{p-1}{2}\right)! \equiv (-1)^{(h_p+1)/2} \bmod p.$$
This shows why getting precise statistics on when the congruence has 1 on the right side will be hard.
-
The following is a relevant classical paper:
Mordell, L. J. The congruence $(p-1/2)!\equiv ±1$ $({\rm mod}$ $p)$. Amer. Math. Monthly 68 1961 145--146.
http://www.math.uga.edu/~pete/Mordell61.pdf
Put $((p-1)/2)!\equiv(-1)^a\ (\text{mod}\,p)$, where $p$ is a prime $\equiv 3\ (\text{mod}\,4)$. The author proves the following result. If $p\equiv 3\ (\text{mod}\,4)$ and $p>3$, then $$a\equiv{\textstyle\frac 1{2}}\{1+h(-p)\}\quad(\text{mod}\,2), \tag1$$ where $h(-p)$ is the class number of the quadratic field $k(\surd-p)$ [$\mathbb{Q}(\sqrt{-p})$ must be meant here. --PLC]. The author points out that (1) follows easily from a result of Dirichlet; also that Jacobi had conjectured an equivalent result before the class number formula was known. (MathReview by L. Carlitz)
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The notation k(\sqrt{-p}) for our Q(\sqrt{-p}) is "classical" and was used e.g. by Hilbert in his Bericht. The idea was that k(\sqrt{-p}) is the field k you get by adjoining a square root of -p to the rationals. – Franz Lemmermeyer Feb 23 '10 at 17:07
Hecke uses $K(\root l\of\mu;k)$ to denote $k(\root l\of\mu)$ in his Vorlesungen. – Chandan Singh Dalawat Feb 24 '10 at 1:06
This is an attempt to justify the answer $1/2$ based on the Cohen-Lenstra heuristics. There will be a lot of nonsensical steps, and I am not an expert, so this should be viewed with caution.
As is observed above, this is equivalent to determining $h(p) \mod 4$, where $h(p)$ is the class number of $\mathbb{Q}(\sqrt{-p})$. Since $p$ is odd and $3 \mod 4$, the only ramified prime in $\mathbb{Q}(\sqrt{-p})$ is the principal ideal $(\sqrt{-p})$. Thus, there is no $2$-torsion in the class group and $h(p)$ is odd.
For any odd prime $q$, let $a(q,p)$ be the power of $q$ which divides $h(p)$. We want to compute the average value of $$\prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)}.$$
First nonsensical step: Let's pretend that the CL-heuristics work the same way for the odd part of the class group of $\mathbb{Q}(\sqrt{-p})$, that they do for the odd part of the class group of $\mathbb{Q}(\sqrt{-D})$. We just saw above that the fact that $p$ is prime constrains the $2$-part of the class group; this claim says that it does not effect the distribution of anything else.
Then we are supposed to have: $$P(a(q,p)=0) = \prod_{i=1}^{\infty} (1-q^{-i}) = 1-1/q +O(1/q^2),$$ $$P(a(q,p)=1) = \frac{1}{q-1} \prod_{i=1}^{\infty} (1-q^{-i}) = 1/q +O(1/q^2),$$ and $$P(a(q,p) \geq 2) = O(1/q^2).$$
If you believe all of the above, then the average value of $(-1)^{a(p,q)}$ is $1-2/q+O(1/q^2)$.
Second nonsensical step: Let's pretend that $a(q,p)$ and $a(q',p)$ are uncorrelated. Furthermore, let's pretend that everything converges to its average value really fast, to justify the exchange of limits I'm about to do.
Then $$E \left( \prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)} \right) = \prod_{q \equiv 3 \mod 4} \left( 1- 2/q + O(1/q^2) \right)$$.`
The right hand side is zero, just as if $h(p)$ were equally like to be $1$ or $3 \mod 4$.
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Most of the latex formulas don't parse on my browser (Firefox 3.5.8) (though some do). Does any one know why? – Anonymous Feb 23 '10 at 17:33
Might you be missing the jsmath fonts? math.union.edu/~dpvc/jsMath/users/fonts.html – David Speyer Feb 23 '10 at 17:50
It works! Great. Thanks! – Anonymous Feb 23 '10 at 17:56 | 2015-05-23T15:12:04 | {
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https://math.stackexchange.com/questions/2317435/multiply-2a-1b-by-2b-and-get-2ab-how-is-this-so | # multiply $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$? How is this so?
I’m reading How To Prove It and in the following proof the author is doing some basic algebra with exponents that I just don’t understand. In Step 1.) listed below he is multiplying $2^b$ across each term in (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) and gets the resulting set of terms in Step 2.) In particular I have no idea how he is getting $2^{ab}$ from multiplying $2^{(a-1)b}$ by $2^b$ again which is shown in the first sequence in Step 2.). When I do it I get $2^{(ab)(b) – (b)(b)}$ and assume this is as far as it can be taken. Can someone please help me understand what steps he is taking to to get his answer?
Theorem 3.7.1. Suppose n is an integer larger than 1 and n is not prime. Then $2^n$ − 1 is not prime. Proof. Since n is not prime, there are positive integers a and b such that a < n, b < n, and n = ab. Let x = $2^b$ − 1 and y = 1 + $2^b$ + $2^{2b}$ +· · ·+ $2^{(a−1)b}$. Then
xy = ($2^b$ − 1) · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)
Step 1.) = $2^b$ · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) − (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)
Step 2.) = ($2^b$ + $2^{2b}$ + $2^{3b}$ +···+$2^{ab}$) − (1 + $2^b$ + $2^{2b}$ + ···+$2^{(a-1)b}$)
Step 3.) = $2^{ab}$ − 1
Step 4.) = $2^n$ − 1.
• Are you familiar with how $x^c\times x^d = x^{c+d}$? Are you familiar with how $(a-1)b + b = ab$? – JMoravitz Jun 10 '17 at 16:00
• My title should have read multiplying $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$. Sorry about that. – maybedave Jun 10 '17 at 16:00
• @JMoravits. Not really and now I'm feeling a little dumb because from what you are saying, this is actually correct, no? – maybedave Jun 10 '17 at 16:01
• For integer values of $c$ and $d$, notice that $x^c\times x^d = \overbrace{\underbrace{x\times x\times x\times \cdots \times x}_{c~\text{times}}\times \underbrace{x\times x\times \cdots \times x}_{d~\text{times}}}^{c+d~~\text{times}}$. (The property can be extended to real values of $c$ and $d$ as well, look up a more in depth proof for that). As for why $(a-1)b+b=ab$, this is algebraic manipulation., $(a-1)b+b=ab-1b+b=ab-b+b=ab+(-b+b)=ab+0=ab$ – JMoravitz Jun 10 '17 at 16:07
$$2^{(a-1)b}2^b=2^{(a-1)b+b}=2^{ab-b+b}=2^{ab}.$$
• Ok, so what what you guys are saying is that I really should be adding the b to (a-1)b rather than multiplying it in. This is A BIG HELP! Thank you! – maybedave Jun 10 '17 at 16:04
• @maybedave It's $x^rx^s=x^{r+s}$... – Lord Shark the Unknown Jun 10 '17 at 16:05
• "rather than multiplying it in" You absolutely should not multiply them! Notice $64 =4*16=2^2*2^4=2^{2*4}=2^8=256$. Doesn't work. But $64= 2^2*2^4 = (2*2)*(2*2*2*2) = (2*2*2*2*2*2) = 2^6 = 64$. Addition has to work. Notice $2^4*2^4 = (2*2*2*2)*(2*2*2*2) = [(2*2*2*2)(2*2*2*2)(2*2*2*2)(2*2*2*2)] = 2^{4*4}$ just doesn't make any sense at all. – fleablood Jun 10 '17 at 16:18
Using the laws of exponents, $2^{(a-1)b}2^b = 2^{(a-1)b+b} = 2^{ab}$
Hint:
The correct rule is:
$$2^b 2^{(a-1)b}=2^{(a-1)b+b}$$
Basic rules $a^m a^n = (a*a*a*a*a........*a[n \text{ times}])*(a*a*...... *a[m \text{ times}]) = a*a*a*.......*a [n+m\text { times}] = a^{n+mm}$.
So $2^{(a-1)b}*2^b = (2*2*2*2*2........*2[(a-1)b \text{ times}])*(2*2*...... *2[b \text{ times}]) = 2*2*2*.......*2 [(a-1)b+b\text { times}] = 2^{(a-1)b + b}$
Meanwhile $(a-1)*b + b = a*b - 1*b + b = (ab) + (-b + b) = ab + 0 = ab$.
So $2^{(a-1)b}2^b = 2^{(a-1)b + b} = 2^{ab}$ | 2020-01-25T14:19:57 | {
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https://mathhelpboards.com/threads/converting-a-repeating-decimal-to-ratio-of-integers.5632/ | # Converting a repeating decimal to ratio of integers
#### paulmdrdo
##### Active member
0.17777777777 convert into a ratio.
#### M R
##### Active member
Re: converting a repeating decimal to ratio of integers
Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.
Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.
Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.
I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
#### paulmdrdo
##### Active member
Re: converting a repeating decimal to ratio of integers
Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.
Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.
Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.
I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
what do you mean by "GP"?
#### M R
##### Active member
Re: converting a repeating decimal to ratio of integers
what do you mean by "GP"?
Sorry, I have to stop using abbreviations.
A GP is a geometric progression: $$a, ar, ar^2, ar^3...$$.
If you haven't met this then the second method I posted is fine.
#### soroban
##### Well-known member
Re: converting a repeating decimal to ratio of integers
Hello, paulmdrdo!
$$\text{Convert }\,0.1777\text{...}\,\text{ to a fraction.}$$
$$\begin{array}{ccc}\text{We have:} & x &=& 0.1777\cdots \\ \\ \text{Multiply by 100:} & 100x &=& 17.777\cdots \\ \text{Multiply by 10:} & 10x &=& \;\;1.777\cdots \\ \text{Subtract:} & 90x &=& 16\qquad\quad\; \end{array}$$
Therefore: .$$x \;=\;\frac{16}{90} \;=\;\frac{8}{45}$$
#### paulmdrdo
##### Active member
how would I decide what appropriate power of ten should i use?
for example i have 3.5474747474... how would you convert this one?
#### M R
##### Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.
If you use 1000 and 10 you will get
1000x=3547.474747...
10x=35.474747...
So 990x=3512 and x=3512/990=1756/495.
I'm adopting Soroban's approach as I prefer it to what I did earlier.
#### paulmdrdo
##### Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.
If you use 1000 and 10 you will get
1000x=3547.474747...
10x=35.474747...
So 990x=3512 and x=3512/990=1756/495.
I'm adopting Soroban's approach as I prefer it to what I did earlier.
"a difference of two in the powers of ten" -- what do you mean by this? sorry, english is not my mother tongue. bear with me.
Last edited:
#### M R
##### Active member
"a difference of two in the powers of ten" -- what do you me by this? sorry, english is not my mother tongue. bear with me.
No problem.
We have 10^3 and 10^1.
The difference between 3 and 1 is 3-1=2
#### Prove It
##### Well-known member
MHB Math Helper
how would I decide what appropriate power of ten should i use?
for example i have 3.5474747474... how would you convert this one?
You want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated.
So in this case, since the 47 repeats, you want the first to read "something.4747474747..." and the second to read "something-else.4747474747..."
What powers of 10 will enable this?
#### MarkFL
Staff member
A quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's.
1.) $$\displaystyle x=0.1\overline{7}$$
$$\displaystyle 10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}$$
$$\displaystyle x=\frac{16}{90}=\frac{8}{45}$$
2.) $$\displaystyle x=3.5\overline{47}$$
$$\displaystyle 10x=35.\overline{47}=35+\frac{47}{99}=\frac{3512}{99}$$
$$\displaystyle x=\frac{3512}{990}=\frac{1756}{495}$$ | 2020-09-28T11:27:02 | {
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https://math.stackexchange.com/questions/2891882/limit-of-fx-given-that-fx-x-is-known/2891884 | # Limit of $f(x)$ given that $f(x)/x$ is known
Given that $$\lim_{x \to 0} \dfrac{f(x)}{x}$$ exists as a real number, I am trying to show that $\lim_{x\to0}f(x) = 0$. There is a similar question here: Limit of f(x) knowing limit of f(x)/x.
But this question starts with the assumption of $$\lim_{x \to 0} \dfrac{f(x)}{x} = 0,$$ and all I am assuming is that the limit is some real number. So the product rule for limits doesn't really work here.
Or do I need to show that $$\lim_{x \to 0} \frac{f(x)}{x} = 0$$ and then apply the product rule?
• Try proving it by contradiction: what happens if $\lim_{x\to 0}f(x)\neq 0$? – Ender Wiggins Aug 23 '18 at 9:00
• What prevents you from using product rule of limits? Perhaps you need to revisit the product rule in your text and then understand that it works fine here. – Paramanand Singh Aug 23 '18 at 14:23
• @FurryFerretMan Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Sep 17 '18 at 20:08
The product rule trick still works. If $\lim_{x \to 0} f(x)/x = R \in \mathbb R$, and obviously $\lim_{x \to 0} x = 0$, it follows that $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{f(x)}{x} \times x = R \times 0 = 0.$$
• This is the most natural solution and I would not consider use of limit laws a trick rather it is the method. +1 – Paramanand Singh Aug 23 '18 at 14:19
We have that eventually
$$0\le \left|\frac{f(x)}{x}\right|\le M$$
therefore
$$0\le \left|f(x)\right|\le M|x| \to 0$$
Let $\lim_{x\to0}\dfrac{f(x)}{x}=l$ then $\bigg|\dfrac{f(x)}{x}-l\bigg|\leq M$ for some $M\in \mathbb{R}$. So $\bigg|\dfrac{f(x)}{x}\bigg|\leq |l|+M\Rightarrow |f(x)|\leq |x|(|l|+M) \Rightarrow \lim_{x\to 0} f(x)=0$
Let $x_n \rightarrow 0$.
$y_n:= f(x_n)/x_n$, we have
$y_n \rightarrow L.$
With
$f(x_n)=$
$(f(x_n)/x_n)(x_n)=(y_n)(x_n)$.
$\lim_{n \rightarrow \infty }f(x_n)=$
$\lim_{n \rightarrow \infty}((y_n)(x_n))=$
($\lim_{n \rightarrow \infty}(y_n))(\lim_{n \rightarrow \infty}(x_n))=$
$L \cdot 0=0.$ | 2019-08-23T15:35:10 | {
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http://mathhelpforum.com/number-theory/177338-divisibility-6-a.html | 1. ## Divisibility by 6
Hey guys I need help with this question
$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$
and the answer given in the solution to the question is that since
$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$
$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$
I know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?
The same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction
If we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?
Thanks
2. 1.
I guess you don't know Newton's generalized binomial theorem...
$(a+b)^n=M_a+b^n=M_b+a^n$
I considered $M_k$ a multiple of k.
So:
$4^n-1=(1+3)^n-1=M_3+1-1=M_3$
$3^n-1=(1+2)^n-1=M_2+1-1=M_2$
2.
$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-2}+b^{n-1})$
So:
$4^n-1=(4-1)(4^{n-1}+4^{n-2}+...+4+1)=3(4^{n-1}+4^{n-2}+...+4+1)$
$3^n-1=(3-1)(3^{n-1}+3^{n-2}+...+3+1)=2(3^{n-1}+3^{n-2}+...+3+1)$
3. Originally Posted by aonin
Hey guys I need help with this question
$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$
and the answer given in the solution to the question is that since
$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$
$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$
I know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?
The same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction
If we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?
Thanks
The $\displaystle 3^n - 1$ is easy, without any tricks like below. $\displaystyle 3^n$ is an odd number, so subtracting 1 from it gives us an even number.
$(1 + 3)^n - 1$ is a bit trickier. Note that when we expand by the binomial theorem:
$(1 + 3)^n - 1 = (1^n + n \cdot 1^{n - 1}3^1 + ~...~ + n \cdot 1^13^{n - 1} + 3^n) - 1$
$\displaystyle = n \cdot 1^{n - 1}3^1 + ~...~ + n \cdot 1^13^{n - 1} + 3^n$
of which you can prove (using the combinatorial coefficients) that every coefficient is divisible by 3.
-Dan
4. $(1 + 3)^n - 1 = (1^n + 3 \cdot 1^{n - 1}3^1 + ~...~ + 3 \cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.
$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}\cdot a^{n-k} \cdot b^{k}$.
5. Originally Posted by veileen
$(1 + 3)^n - 1 = (1^n + 3 \cdot 1^{n - 1}3^1 + ~...~ + 3 \cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.
$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}\cdot a^{n-k} \cdot b^{k}$.
Whoops! I got carried away with the 3's. Thanks for the catch. I have fixed it in my post.
-Dan
6. Originally Posted by aonin
Hey guys I need help with this question
$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$
and the answer given in the solution to the question is that since
$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$
$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$
Note that $\displaystyle A=4^{n}-1=(1+3)^{n}-1=\sum\limits_{k = 1}^n {\binom{n}{k}\left( 3 \right)^k }$
$\displaystyle B=3^{n}-1=(1+2)^{n}-1=\sum\limits_{k = 1}^n {\binom{n}{k}\left( 2 \right)^k }$
Because the index begins with $k=1$ every term in $A$ is a multiple of 3. Similarly every term is B is even.
Hence every term in the product $AB$ is a multiple of 6. | 2016-08-28T12:17:03 | {
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https://math.stackexchange.com/questions/3371883/how-to-prove-that-x-is-a-completion-of-y-for-some-metric-spcaes-x-d-and | # How to prove that $X$ is a completion of $Y$, for some metric spcaes $(X,d)$ and $(Y,d)$?
In our Real Analysis class we did a proof for existence of completion of metric space. However, I do not understand how do we use it practically. i.e How do I prove that $$X$$ is a completion of $$Y$$, for some metric spcaes $$(X,d)$$ and $$(Y,d)$$?
For example : How do I prove that $$(\mathbb{R},d_1)$$ is a completeion of $$(\mathbb{Q},d_1)$$? Here, $$d_1$$ is the standard Euclidean metric
$$\underline{\text{My attempt}}$$ -
I believe we should prove that $$\mathbb{Q}$$ is subset of $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. This would mean that all Cauchy sequence in $$\mathbb{Q}$$ converges in $$\mathbb{R}$$.
However, consider this case -
Let $$(W,d)$$ be an incomplete Metric space and $$W \subset Y \subset X$$ and $$X,Y$$ are complete (with same metric).
Now, by my argument both $$X,Y$$ can be the completion of $$W$$. Is this alright? I read somewhere in Stack Exchange that completion is unique(I didn't really understand that answer)
$$\underline{\text{Approach to resolve this}}$$ - We choose the smallest, complete set which contains $$W$$ as its completion. We can prove that, since $$X$$ is complete any of it's closed subset is complete. We can also prove that $$\overline{W}$$ is the smallest closed set that contains $$W$$. Hence, $$\overline{W}$$ is the completion of $$W$$ (this would explain the notation in the completion proof).
Thus in the title - I just have to check if $$Y$$ is dense in $$X$$ to prove that $$X$$ is a completion of $$Y$$.
This would also solve my example question - We know that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. Thus $$\mathbb{R}$$ is completion of $$\mathbb{Q}$$.
I am fairly confident that this is correct. However, I am a Physics Master's student this is my first "Real" Math course ;) So I would appreciate if someone let's me know if this right and if not the corrections.
• What is $d_1$? ${}$ – José Carlos Santos Sep 27 '19 at 11:13
• $d_1$ is the standard Euclidean metric – Indigo1729 Sep 27 '19 at 11:29
• This follows by the universal property of completion of metric spaces. – ε--δ Sep 27 '19 at 11:59
To show that $$\mathbb{R}$$ is the Cauchy completion of $$\mathbb{Q}$$, it is not sufficient to show that $$\mathbb{Q}$$ is contained in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. In fact, this just means that the completion of $$\mathbb{Q}$$ is contained within $$\mathbb{R}$$.
For example, the completion of $$\mathbb{Q}$$ is contained in $$\mathbb{C}$$ and the complex numbers are complete, but the completion of the rational numbers is not the complex numbers.
You should also verify that every element of $$\mathbb{R}$$ is the limit of a sequence of rational numbers (which follows directly from the fact that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ for example) in order to be sure that $$\mathbb{R}$$ is actually $$\mathbb{Q}$$'s completion. This resolves your worry about the uniqueness theorem for completions which you referred to.
One final point: one doesn't normally define the Cauchy completion in order to identify completions of metric spaces they already know (e.g. $$\mathbb{Q}$$ in your example). Instead, knowing that you can take the completion lets you construct new metric spaces which can be useful in their own right. Again for example, knowing that completions exist and taking the completion of $$\mathbb{Q}$$ is a way to construct the real numbers $$\mathbb{R}$$.
One nitpick: "I believe we should prove that $$\Bbb Q$$ is subset of $$\Bbb R$$ and $$\Bbb R$$ is complete. This would mean that all Cauchy sequence in $$\Bbb R$$ converge in $$\Bbb R$$."
To sum up, viewing $$\Bbb Q$$ as a subset of $$\Bbb R$$ (so we can avoid speaking of the embedding map and the image of $$\Bbb Q$$ under it), you need to show that
• the metric on $$\Bbb Q$$ is the restriction of the metric on $$\Bbb R$$
• $$\Bbb Q$$ is dense in $$\Bbb R$$ and of course
• $$\Bbb R$$ is complete under its metric
and that is essentially what you found out by yourself.
You're missing one... critical point!
It is not sufficient for $$X$$ to be a completion of $$W$$ that $$W \subseteq X$$ and $$X$$ is complete. You also need $$W$$ to be dense in $$X$$.
That being said, $$\mathbb R$$ is indeed a completion of $$\mathbb Q$$. | 2020-02-22T20:44:27 | {
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https://www.intmath.com/forum/methods-integration-31/integration-techniques:161 | IntMath Home » Forum home » Methods of Integration » Integration Techniques
# Integration Techniques [Solved!]
### My question
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
### Relevant page
4. Integration: Basic Trigonometric Forms
### What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
X
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
Relevant page
<a href="https://www.intmath.com/methods-integration/4-integration-trigonometric-forms.php">4. Integration: Basic Trigonometric Forms</a>
What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
## Re: Integration Techniques
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).
Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
X
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).
Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
[graph]310,250;-1,10;-1.1,1.1,1.57,1;sin(x),sin(3.14x/180)[/graph]
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
## Re: Integration Techniques
It does. A good example to use. Thank you.
X
It does. A good example to use. Thank you. | 2019-05-19T22:25:15 | {
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https://math.stackexchange.com/questions/2171078/how-do-i-add-the-terms-in-the-binomial-expansion-of-10026/2171125 | # How do I add the terms in the binomial expansion of $(100+2)^6$?
So, I stumbled upon the following question.
Using binomial theorem compute $102^6$.
Now, I broke the number into 100+2.
Then, applying binomial theorem
$\binom {6} {0}$$100^6(1)+\binom {6} {1}$$100^5(2)$+....
I stumbled upon this step. How did they add the humongous numbers? I am really confused.
Kindly help me clear my query.
• They're powers of ten.... And you don't mean "embezzled." Maybe you mean "confused." Mar 4 '17 at 4:39
• Correct! I need to work on my English skills, man! Thanks for the help tho! Mar 4 '17 at 4:46
• @symplectomorphic - actually, they probably meant "bamboozled", as it means something similar to "confused", but can be easily mixed up with "embezzled". Mar 4 '17 at 7:42
The long way:
\begin{align} & 10^{12} + 12 \cdot 10^{10} + 6 \cdot 10^9 + 16 \cdot 10^7 + 24 \cdot 10^5 +192 \cdot 10^2 + 64 \\ =\; & 10^{12} + (10+2) 10^{10} + 6 \cdot 10^9 + (10+6) 10^7 + (20+4) 10^5 +(100+90 +2) 10^2 + 60+4 \\ =\; & \color{red}1\cdot10^{12} + \color{red}1 \cdot 10^{11} + \color{red}2\cdot 10^{10} + \color{red}6 \cdot 10^9 + \color{red}1 \cdot 10^8 + \color{red}6 \cdot 10^7+\color{red}2 \cdot 10^6 + \color{red}4 \cdot 10^5+\color{red}1 \cdot 10^4+\color{red}9\cdot 10^3 + \color{red}2 \cdot 10^2 + \color{red}6 \cdot 10^1 + \color{red}4 \cdot 10^0 \end{align}
The latter is precisely the representation in base $\,10\,$ of $\;\color{red}{1126162419264}\,$.
• Thanks for the help! Perhaps the most convincing and noteworthy answer. Mar 4 '17 at 10:52
That number is $1\; 12 \; 6\; 16 \; 24\;192\; 64$ (space added for emphasis). Notice a relationship with the coefficients of the powers of $10$?
$10^{12} + 12\times10^{10} + 6\times10^9 + 16*10^7 + 24\times10^5 + 192\times10^2 + 64$ $= 10^{12} + 10^{11} + 2\times10^{10} + 6\times10^9 + 10^8 + 6\times10^7 + 2\times10^6 + 4\times10^5 + 10^4 + 9\times10^3 + 2\times10^2 + 6\times10^1 + 4\times10^0= 1126162419264$
• Thank you very much, sir! I really appreciate it. Mar 4 '17 at 10:51
Since it's all powers of ten you could add those quite easily. If it were something else you could have needed to use some calculator. But here you have just powers of tens, which basically keep everything the same. Here's the number with a few spaces to make you understand it.
$\text{1 12 6 16 24 192 64}$ Those powers just end up just putting those digits in the right order because we use a number system with ten digits.
• Ok! Why not 10112...? Mar 4 '17 at 4:50
• You see say you're adding something like a $12 \times 10^{10}$. You can see that those powers are all spaced evenly. By that I mean to say whenever you have say n digits to be added, over there they had decreased the exponent of 10 by n
– SBM
Mar 4 '17 at 4:53
• Moreover, saying a number equals something like $123.xyz$ is the same thing as $1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 + x \times 10^{-1} \ldots$ and so on. As long as you keep those evenly spaced, it remains same. By x, y and z I meant arbitrary digits not variables
– SBM
Mar 4 '17 at 4:56
• I got your point. Mar 4 '17 at 5:04
Start adding from the rightmost term($64$); you will notice that the number of trailing zeros on the preceding term is exactly the same as the number of digits in following term. $$...2400000\quad +\\ .....19200\quad +\\ .......64\quad$$ ...and so on till the first. | 2022-01-24T08:00:04 | {
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https://math.stackexchange.com/questions/595233/prove-by-induction-that-5n-1-is-divisible-by-4/595293 | # Prove by induction that $5^n - 1$ is divisible by $4$.
Prove by induction that $5^n - 1$ is divisible by $4$.
How should I use induction in this problem. Do you have any hints for solving this problem?
Thank you so much.
• What do you know about the question or what have you tried? Dec 6 '13 at 7:10
• I have a vague feeling that this question has already been covered on the site.
Dec 6 '13 at 12:37
We prove that for all $n \in \mathbb{N}$, $4 \mid \left( 5^n-1 \right)$. (Notationally, this says $4$ divides $5^n-1$ with a zero remainder).
1. For a basis, let $n=1$. Then $$5^1-1=4,$$ and clearly $4\mid4$.
2. Assume that $5^n-1$ is is divisible by $4$ for $n=k, \, k \in\mathbb{N}$. Then by this assumption, $$4 \mid \left( 5^k-1 \right) \Rightarrow 5^k-1=4m, \, m \in \mathbb{Z}.$$ (This notationally means that $5^k-1$ is an integer multiple of $4$.)
3. Let $n=k+1$. Then \begin{align*} 5^{k+1}-1 &= 5^k \cdot 5-1 \\ &=5^k(4+1)-1 \\ &=4\cdot 5^k+5^k -1 \\ &=4\cdot5^k+4m\\ &=4\left( 5^k+m \right). \end{align*} Since $4\mid4\left( 5^k+m \right)$, we may conclude, by the axiom of induction, that the property holds for all $n \in \mathbb{N}$.
• Nice, clear, clean... a good answer indeed! Dec 6 '13 at 16:53
• this should be the accepted answer Jan 15 '17 at 5:43
First prove the base case $n=1$. Then induct and make use of the fact that $$(5^{n+1}-1) - (5^n-1) = 4 \cdot 5^n$$to conclude what you want.
Without induction, you can use the identity
$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)$$
Of course you would still need induction or something to prove this identity.
• Yes. I removed my comment once I saw your updated answer. To be really pedantic, even in my answer, we still need induction first to show that $5^n$ is an integer, to then claim that $4$ divides $4 \cdot 5^n$.
– user17762
Dec 6 '13 at 7:24
without induction $$5^n-1=(4+1)^n-1$$ $$=4^n+n4^{n-2}+...+1-1$$ the only term in $(1+4)^n$ not being multiplied by a power of $4$ is $1$ but it disappears due to the $-1$.
Why induction? $5^n$ ends in $\dots25$ for $n>1$, so $5^n-1$ ends in $\dots24$.
• Technically, you still need induction to show that $5^n$ ends in $25$ for $n>1$.
– user17762
Dec 6 '13 at 7:17
• @user17762 $25 * 5 = 25 \mod 100$ Dec 6 '13 at 13:48
• @ratchetfreak Technically I think you still need induction to prove the statement about an arbitrary $n$. Of course, using the fact you mention, the induction becomes trivial (like proving by induction that $1^n = 1$ for all $n$ using the fact that 1 is the multiplicative identity.) Dec 6 '13 at 18:48
$\displaystyle{5^{n + 1} - 1 = \left(5^{n} - 1\right)5 + 4}$
it's even more general:
$k$ divides $(k+1)^n-1$ with $k,n \in \mathbb{N}$
simply by modular arithmetic:
$$k+1 = 1 \mod {k} \\ \Downarrow \\(k+1)^n=1 \mod {k}$$
To prove by induction you:
1. Assume the proposition is true for n
2. Show that if it is true for n, then it is also true for n+1
3. Show that it is true for n=1
Then you know that it will be true for all natural numbers.
In this case:
1. Assume $5^n-1$ is divisible by 4
2. Say $m=5^n$, so $m-1$ is divisible by 4
• $5^{n+1}-1$ = $5m - 1$
• $5m - 1$ = $5(m-1) + 4$
• Since $m-1$ is divisible by 4 and 4 is divisible by 4, then this expression is divisible by 4.
3. For $n=1$, $5^1 - 1 = 4$ which is divisible by 4
And there you are...
This isn't by induction, but I think it's a nice proof nonetheless, certainly more enlightening: $\displaystyle 5^n-1=(1+4)^n-1=\sum_{k=0}^n {n\choose k}4^k-1=1+\sum_{k=1}^n {n\choose k}4^k-1=\sum_{k=1}^n {n\choose k}4^k$ which is clearly divisible by $4$. | 2021-09-24T04:03:26 | {
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http://mathhelpforum.com/pre-calculus/25502-partial-fractions.html | 1. ## Partial Fractions
2. I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.
3. My answer is the part in red (Part 3).
4. Originally Posted by r_maths
My answer is the part in red (Part 3).
Hmm, I get $A = 1, B = 2$, which leads to a final answer of: $x+\frac{1}{x+1}+\frac{2}{x+2}$. I do not see how the x term is not supposed to be there.
5. I did include the x
I got the same answer as you, but the answer from the book gives a different answer...
6. Originally Posted by r_maths
I did include the x
I got the same answer as you, but the answer from the book gives a different answer...
I do not see how B equals one, so the book is probably wrong unless both of us missed something.
7. Hello, r_maths!
Express $\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ in partial fractions.
$\frac{x^3+3x^2+5x+4}{(x+1)(x+2)} \;=\;x + \frac{3x+4}{(x+1)(x+2)} \;=\;x + \frac{1}{x+1} + \frac{2}{x+2}\quad{\color{red}Right!}$
Book answer: . $x + \frac{1}{x+1} + \frac{1}{x+2}\quad{\color{red}Wrong!}$
$\text{Their answer adds up to: }\;\frac{x^3+3x^2+4x+3}{(x+1)(x+2)}$ . . . . obviously not the original fraction.
$\frac{x}{1}\cdot{\color{blue}\frac{(x+1)(x+2)}{(x+ 1)(x+2)}} + \frac{1}{x+1}\cdot{\color{blue}\frac{x+2}{x+2}} + \frac{2}{x+2}\cdot{\color{blue}\frac{x+1}{x+1}}$
. . $= \;\frac{x(x+1)(x+2) + (x+2) + 2(x+1)}{(x+1)(x+2)}$
. . $= \;\frac{x^3+3x^2+2x + x + 2 + 2x + 2}{(x+1)(x+2)}$
. . $= \;\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ . . . . See?
8. $\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x^3 + 3x^2 + 2x + 3x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x(x + 1)(x + 2) + (3x + 4)}}
{{(x + 1)(x + 2)}}$
.
Now
$x + \frac{{3x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{{(x + 2) + 2(x + 1)}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
.
Hence $\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
. | 2013-12-19T11:41:55 | {
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https://math.stackexchange.com/questions/3200305/can-we-determine-summands-from-their-partial-sums | # Can we determine summands from their partial sums?
Assume there are non-negative numbers $$\lambda_1\le \ldots\le \lambda_n\in[0,\infty)$$. You are given the (ordered) list $$s_1\le\ldots\le s_{2^n}\in[0,\infty)$$ of all partial sums, i.e. every $$s_i$$ is of the form $$s_i=\sum_{k\in K_i}\lambda_k$$ for some unique but unknown $$K_i\subseteq\{1,\ldots,n\}$$.
Question: Can we determine the $$\lambda_1,\ldots,\lambda_n$$ from knowing the $$s_1,\ldots,s_{2^n}$$?
Some obvious facts are
1. Since there is some $$s_i$$ corresponding to $$K_i=\emptyset$$, $$s_1=\cdots=s_i=0$$.
2. $$\lambda_1=s_2$$, since no non-trivial partial sum can be smaller as the smallest possible summand.
3. $$s_{2^n}=\sum_{k=1}^n\lambda_k$$, since no partial sum can be bigger.
4. For $$n=2$$ the answer is yes, since $$\lambda_1=s_2$$ and $$\lambda_2=s_4-s_2$$.
Note: For my use case it would be sufficient to know if for any given $$s_1\le\cdots\le s_{2^n}$$ there is at most one possibility for $$\lambda_1\le\cdots\le\lambda_n$$.
1. By calculating how many $$s_i$$ are zero, you can determine how many $$\lambda_i$$ are zero. If there are any, they will be creating repeated entries $$s_i$$ throughout, but in a systematic manner where one can eliminate their effect; indeed, if there are $$k$$ zeroes, then there will be $$k$$ extra duplicates of every entry. For simplicity, suppose $$\lambda_1 > 0$$.
2. Now, indeed, $$\lambda_1 = s_2$$.
3. We proceed by induction. Suppose we have identified $$\lambda_1,\ldots,\lambda_j$$ and calculated the partial sums consisting of only $$\lambda_1,\ldots,\lambda_j$$, such as $$\lambda_1+\lambda_3$$. Then the smallest remaining partial sum must be $$\lambda_{j+1}$$. Proof: Otherwise the smallest remaining partial sum would have to be a sum with at least one unknown $$\lambda_m$$ that is (by definition) not yet in the list of known partial sums, which would imply that $$\lambda_m$$ is smaller than the smallest remaining partial sum; a contradiction.
4. Now that $$\lambda_{j+1}$$ is also known, consider the known list of partial sums to include all sums of $$\lambda_1,\ldots,\lambda_{j+1}$$.
Actually, separate treatment of steps 1 and 2 is not necessary; one only has to initialize the list of known partial sums with the empty sum $$s_1=0$$.
This method probably works even if you are missing entires from the list of partial sums, as long as all the missing entries are strictly greater than $$\lambda_n$$.
• One has to be careful: Assume we have $\lambda_1=1,\lambda_2=2,\lambda_3=3,\lambda_4=4,...$ and already computed $\lambda_1,\lambda_2$. The set of known partial sums is $K=\{0,1,2,3\}$, so the smallest partial sum not in $K$ would be $4\ne\lambda_3$. The problem here is that $\lambda_3=\lambda_1+\lambda_2$ is itself a partial sum. I believe that this can be fixed by using multisets of partial sums: In the above situation the multiset of all partial sums would be $\{0,1,2,3,3,4,...\}$, so after removing the multiset $K$ the smallest remaining element would be $3=\lambda_3$ as required. – Robert Rauch Apr 25 at 7:47
• @RobertRauch Indeed, but since you are using an ordered list of partial sums with $2^{n}$ elements, it already contains the information about multiplicity. – Tommi Brander Apr 25 at 7:55
Here is a streamlined version of Tommi Brander's solution: $$\newcommand{\IN}{\mathbb{N}}$$
Lemma: Let $$0\le\lambda_1\le\cdots\le\lambda_n$$ and $$\mathcal{P}\doteq\left\{\lambda_K\doteq\sum_{k\in K}\lambda_k\left|K\subseteq\mathbb{N}_n\right.\right\}$$ the set of their partial sums, where $$\IN_n\doteq\{1,2,\ldots,n\}$$. For $$0\le l\le n$$ consider the function $$m_l:\mathcal{P}\to\IN_0$$ with $$m_l(\lambda)=\#\{K\subseteq\IN_l\mid\lambda_K=\lambda\}$$. Then for all $$1\le l\le n$$, $$$$\lambda_l=\min\left\{\lambda\in\mathcal{P}\mid m_{l-1}(\lambda) In particular, noting that $$m_0(\lambda)=\mathbf{1}(\lambda=0)$$, the $$\lambda_1,\ldots,\lambda_n$$ can be recovered iteratively only from $$\mathcal{P}$$ and $$m_n$$.
Proof: Let $$\mathcal{P}_l\doteq\left\{\lambda\in\mathcal{P}\mid m_{l-1}(\lambda). By construction, $$\lambda_l\in\mathcal{P}_l$$ and therefore $$\lambda_*=\min\mathcal{P}_l$$ exists and satisfies $$\lambda_*\le\lambda_l$$. Since $$m_{l-1}(\lambda_*), there is some $$K\subseteq\IN_n$$ with $$K\not\subseteq\IN_{l-1}$$ such that $$\lambda_*=\lambda_K$$. In particular there is $$r\in K$$ with $$r\ge l$$ and, hence, $$\lambda_r\in\mathcal{P}_l$$. By positivity of the $$\lambda_i$$ we find that $$\lambda_r\le\lambda_K=\lambda_*$$ and therefore $$\lambda_*=\lambda_r$$ by minimality of $$\lambda_*$$. Since $$r\ge l$$ we conclude $$\lambda_l\le \lambda_r=\lambda_*\le\lambda_l$$, thus $$\lambda_*=\lambda_l$$. | 2019-08-20T05:44:47 | {
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https://iidb.org/threads/geometry-question-re-hexagons.11275/ | Geometry Question re: Hexagons
Malintent
Veteran Member
For a woodworking project... I want to cut a smooth circle from the outer perimeter of a constructed hexagon. I need to calculate how wide a board I need to construct the hexagon with, such that there is sufficient material to cut away to form a circle, while maintaining integrity of the construction..
It is a geometry question, stated simply as, "what is the maximum distance of a circle's circumference from an inscribed hexagon"? No luck finding an answer to that, so here is a more detailed means of expressing what I need to find out:
inscribe a hexagon within a circle, such that each of the 6 vertices of the hexagon are touching the circle. The radius of circle is equal to the length of each segment of the hexagon, as well as its radius.
At every center point of each segment of the hexagon, the circle's circumference is at its maximum distance from the hexagon's segment.
What is that distance?
That distance would be the exact width of the board, if used to construct the hex, where cutting the circle would just exactly separate the attached segments (not good). So I would take that distance and add the amount of material I need for structural integrity. I suppose the length of each board (segment of the hex) would need to be my desired final radius, plus 1/2 the distance I am looking for.
right?
Sarpedon
Veteran Member
I don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.
Anyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.
Last edited:
beero1000
Veteran Member
If I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \frac{\sqrt{3}}{2})S$$
TeX seems to be out right now, so I'm stuck typesetting like some sort of savage
It's (2 - √3)S/2
Treedbear
Veteran Member
I don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.
Anyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.
~0.5% smaller than exact
Junior Member
It is a geometry question, stated simply as, "what is the maximum distance of a circle's circumference from an inscribed hexagon"?
I tackled this with pencil, paper, and calculator, assuming a 1 unit length for the radius of the circle and the sides of the hexagon. Calculating the altitude of an equilateral triangle with sides of 1 gives a little over 0.866; this would be the the distance from the center of the circle/hexagon to the midpoint of the side of the hexagon. Subtracting 0.866 from 1 gives 0.134, the distance from the midpoint of the side of the hexagon out to the circumference of the circle. The proportions are similar to what Sarpedon came up with.
Malintent
Veteran Member
If I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \frac{\sqrt{3}}{2})S$$
TeX seems to be out right now, so I'm stuck typesetting like some sort of savage
It's (2 - √3)S/2
"circumradius and inradius" are the terms I didn't know that resulted in not finding my answer on my own. Thank you for the education!
so, now that I was able to construct a concise question for Google, I easily found this regarding the relationship between the two values:
"since such a ratio is just the ratio between the side and the height of an equilateral triangle, (R/r) = (2/sqrt(3))"
which is apparently a constant ratio... approximately (way more precise than needed) R - r (the value I need) = R * 0.134
.. just like ya'll said. Thanks again!
so, with a 1.5 foot segment length, I will need 2.41 inches of "spare room" on each board for the arc of the circle. Since I want 1 inch of material inside the finished piece (for structural integrity), I will need to use at least a 3.1 inch wide board. so, 4 inch lumber boards will be perfect (they are 3.5").
I imagined I would have been able to do this with 3 inch boards (2.5), but I was wrong. THAT'S why I like to calculate before cutting materials
Last edited:
Sarpedon
Veteran Member
Tool thickness. Those saws at the sawmill are massive.
Malintent
Veteran Member
Tool thickness. Those saws at the sawmill are massive.
Well, I won't be milling my own lumber... my blades are the standard 1/8" kerf. But that isn't a measure of accuracy, just waste.
Malintent
Veteran Member
4 inch lumber boards will be perfect (they are 3.5")
Obviously.
Is it? It was never obvious to me that a "2 by 4" actually measures 1.75x3.5. Or were you being sarcastic and didn't realize (like most, I would imagine) that lumber is prepared that way commercially?
Malintent
Veteran Member
anyway... thank you all again for the "magic ratio" of 1 : 0.134... such an easily workable answer.
Jimmy Higgins
Contributor
Why don't you just get a hexagonal saw bit?
beero1000
Veteran Member
Tool thickness. Those saws at the sawmill are massive.
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
bilby
Fair dinkum thinkum
Tool thickness. Those saws at the sawmill are massive.
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.
The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness.
Bronzeage
Super Moderator
Staff member
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.
The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness.
Anyone who has ever worked with "rough" lumber will happily pay for a 4 inch board and take a 3.5 board.
In any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.
beero1000
Veteran Member
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.
The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness.
Nothing except marketing - bigger is better, and if they can sell you a 3.5" board and make you think you're buying a 4" board then they don't mind at all. Even if they get sued for it.
Kosh
Junior Member
Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.
The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness.
Anyone who has ever worked with "rough" lumber will happily pay for a 4 inch board and take a 3.5 board.
In any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.
I've recently built a cnc machine, and it's pretty annoying to be dealing with sheet goods that are supposedly 19 mm thick but I'm finding them from 17 mm on up......I have to measure each piece and adjust the program accordingly.
Cheerful Charlie
Contributor
To understand the relationships of hexagons to circles, google for Unit Circle. Then you will want a nice trig capable calculator.
Treedbear
Veteran Member
... I need to calculate how wide a board I need to construct the hexagon ...
Are you talking about a standard 5 sided hexagon or one that actually has 6 sides?
Malintent
Veteran Member
... I need to calculate how wide a board I need to construct the hexagon ...
Are you talking about a standard 5 sided hexagon or one that actually has 6 sides?
take a 3-legged dog and feed it 1 square meal. Then take it's poop and use it to fertilize a 5-sided pentagonal hexagon. It will eventually grow another leg, which you can either use to make an actual hexagon, or just give it to the dog to use.
I thought everyone knew this... | 2022-08-14T09:23:49 | {
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http://workshop19.info/lhrzk/b73d15-topological-sort-example | Topological Sort Algorithm. Topological Sort by BFS: Topological Sort can also be implemented by Breadth First Search as well. Example: building a house with a Topological Sorting; graphs If is a DAG then a topological sorting of is a linear ordering of such that for each edge in the DAG, appears before in the linear ordering. Example (Topological sort showing the linear arrangement) The topologically sorted order is not necessarily unique. Topological Sort: A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering.A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). Topological Sort Algorithms. Topological sort Topological-Sort Ordering of vertices in a directed acyclic graph (DAG) G=(V,E) such that if there is a path from v to u in G, then v appears before u in the ordering. Such an ordering cannot exist if the graph contains a directed cycle because there is no way that you can keep going right on a line and still return back to where you started from. 50 Topological Sort Algorithm: Runtime For graph with V vertexes and E edges: ordering:= { }. In other words, a topological sort places the vertices of a directed acyclic graph on a line so that all directed edges go from left to right.. 3. 3/11 Topological Order Let G = (V;E)be a directed acyclic graph (DAG). R. Rao, CSE 326 9 A B C F D E Topological Sort Algorithm Step 2: Delete this vertexof in-degree 0 and all its ��� There are severaltopologicalsortingsof (howmany? Definition of Topological Sort. For example, we can put on garments in the following order: A topological sort of a DAG is a linear ordering of all its vertices such that if contains an edge , then appears before in the ordering. Topological Sort is Not Unique. Yufei Tao Topological Sort on a DAG Introduction There are many problems involving a set of tasks in which some of the tasks must be done before others. Topological Sort Algorithm. An Example. Definition of Topological Sort. Example 1 7 2 9 4 10 6 3 5 8 Given n objects and m relations, a topological sort's complexity is O(n+m) rather than the O(n log n) of a standard sort. Node 30 depends on node 20 and node 10. That is there may be other valid orderings that are also partial orders that describe the ordering in a DAG. Topological Sort Introduction. A topological sort of a graph $$G$$ can be represented as a horizontal line ��� Consider the graph in the following example: This graph has two possible topological sorts: In this article, we have explored how to perform topological sort using Breadth First Search (BFS) along with an implementation. Show the ordering of vertices produced by $\text{TOPOLOGICAL-SORT}$ when it is run on the dag of Figure 22.8, under the assumption of Exercise 22.3-2. A topological order of G is an ordering of the vertices in V such that, for every edge(u;v)in E, it must hold that u precedes v in the ordering. Here���s simple Program to implement Topological Sort Algorithm Example in C Programming Language. 22.4 Topological sort 22.4-1. ), for example��� Hence node 10, node 20 and node 40 should come before node 30 in topological sorting. As we know that the source vertex will come after the destination vertex, so we need to use a ��� Topological sort: It id defined as an ordering of the vertices in a directed acyclic graph, such that if there is a path from u to v, then v appears after u in the ordering. The ordering of the nodes in the array is called a topological ordering. Review Questions. Here's an example: Topological Sort Introduction. Our start and finish times from performing the $\text{DFS}$ are ��� If we run a topological sort on a graph and there are vertices left undeleted, the graph contains a cycle. There could be many solutions, for example: 1. call DFS to compute f[v] 2. Topological sorting only works for directed acyclic graphs $$\left({DAG}\right),$$ that is, only for graphs without cycles. The graphs should be directed: otherwise for any edge (u,v) there would be a path from u to v and also from v to u, and hence they cannot be ordered. In this article, you will learn to implement a Topological sort algorithm by using Depth-First Search and In-degree algorithms Topological sort is an algorithm which takes a directed acyclic graph and returns a list of vertices in the linear ordering where each vertex has to precede all vertices it directs Topological Sorting for a graph is not possible if the graph is not a DAG. Types of graphs: a. As the visit in each vertex is finished (blackened), insert it to the Node 10 depends on node 20 and node 40. Topological Sort Algorithm Example of a cyclic graph: No vertex of in-degree 0 R. Rao, CSE 326 8 Step 1: Identify vertices that have no incoming edges ��� Select one such vertex A B C F D E Topological Sort Algorithm Select. For every edge U-V of a directed graph, the vertex u will come before vertex v in the ordering. ; There may exist multiple different topological orderings for a given directed acyclic graph. Repeat until graph is empty: Find a vertex vwith in-degree of 0-if none, no valid ordering possible Delete vand its outgoing edges from graph ordering+= v O(V) O(E) O(1) O(V(V+E)) Is the worst case here really O(E) every time?For example, Provided example with dw04 added to the dependencies of dw01. Let���s pick up node 30 here. Implementation. We have compared it with Topological sort using Depth First Search.. Let us consider a scenario where a university offers a bunch of courses . Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. For example, a topological sorting of the following graph is ���5 4 ��� Topological sort is an algorithm that orders a directed graph such that for each directed edge u���v, vertex u comes before vertex v.. A topological sort is an ordering of the nodes of a directed graph such that if there is a path from node u to node v, then node u appears before node v, in the ordering.For example ��� Some vertices are ordered, but the second return is nil, indicating that not all vertices could be sorted. A topological ordering, or a topological sort, orders the vertices in a directed acyclic graph on a line, i.e. Implementation of Source Removal Algorithm. Topological Sort Problem: Given a DAG G=(V,E), output all the vertices in order such that if no vertex appears before any other vertex that has an edge to it Example input: Example output: 142, 126, 143, 311, 331, 332, 312, 341, 351, 333, 440, 352 11/23/2020 CSE 142 CSE 143 CSE 331 Node 20 depends on node 40. An Example. Since, we had constructed the graph, now our job is to find the ordering and for that Topological Sort is Not Unique. For a DAG, we can construct a topological sort with running time linear to the number of vertices plus the number of edges, which is . Topological Sort. Topological Sort Example- Consider the following directed acyclic graph- Review Questions. in a list, such that all directed edges go from left to right. Please note that there can be more than one solution for topological sort. > (topological-sort *dependency-graph*) (IEEE DWARE DW02 DW05 DW06 DW07 GTECH DW01 DW04 STD-CELL-LIB SYNOPSYS STD DW03 RAMLIB DES-SYSTEM-LIB) T NIL. Topological Sorting Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge ( u v ) from ��� Example: Let & and have if and only if $. The topological sort algorithm takes a directed graph and returns an array of the nodes where each node appears before all the nodes it points to. For example, a simple partially ordered set may look as follows: Figure 1. This is partial order, but not a linear one. Example. If there are very few relations (the partial order is "sparse"), then a topological sort is likely to be faster than a standard sort. Introduction There are many problems involving a set of tasks in which some of the tasks must be done before others. Topological sorting works well in certain situations. It is important to note that-Topological Sorting is possible if and only if the graph is a Directed Acyclic Graph. Cycle detection with topological sort ��� What happens if we run topological sort on a cyclic graph? ��� There will be either no vertex with 0 prerequisites to begin with, or at some point in the iteration. To better understand the logic behind topological sorting and why it can't work on a graph that contains a cycle, let's pretend we're a computer that's trying to topologically sort the following graph: # Let's say that we start our search at node X # Current node: X step 1: Ok, i'm starting from node X so it must be at the beginnig of the sequence. The topological sorting for a directed acyclic graph is the linear ordering of vertices. ���怨�由ъ�� - Topological Sort (������ ������) (0) 2014.02.15: ���怨�由ъ�� - Connected Component (0) 2014.02.15: ���怨�由ъ�� - Priority Queue(��곗�������� ���瑜� 援ы��������) (0) 2014.02.15: ���怨�由ъ�� - Heap Sort (��� ������(��� ������)瑜� 援ы��������) (0) 2014.02.15 Solution for topological sort using Breadth First Search ( BFS ) along with an implementation topological sort must be before... Run a topological ordering order is not a DAG insert it to the dependencies dw01... 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Not a linear one the 22.4 topological sort comes before vertex v sort, orders the vertices in DAG... Of the tasks must be done before others u will come before vertex v in the.! | 2021-05-12T05:16:47 | {
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http://mathhelpforum.com/algebra/137321-sequences.html | # Math Help - Sequences
1. ## Sequences
I can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?
11, 9, 10, 17, 36, 79,....
2. Originally Posted by nsingh201
I can't seem to figure out what the next two terms of this sequence are as I can't figure out the term-to-term rule for it. Can anyone help?
11, 9, 10, 17, 36, 79,....
I don't know if this helps you at all, but I believe it is possible to find a polynomial of at most degree 5 that will fit that sequence, by solving simultaneous linear equations.
I will illustrate the method on a smaller scale. Say the sequence is
11, 9, 10, ...
Then we can find a polynomial of degree 2 that will generate this sequence, of the form f(x) = ax^2+bx+c, as follows:
We know f(1) = 11. Thus, a*(1^2)+b*1+c = 11. Simplifying, a+b+c=11.
We know f(2) = 9. Thus, a*(2^2)+b*2+c = 9. Simplifying, 4a+2b+c=9.
We know f(3) = 10. Thus, a*(3^2)+b*3+c = 10. Simplifying, 9a+3b+c=10.
This is a system of linear equations with three equations and three unknowns. The solution is described by (a,b,c) = (3/2, -13/2, 16). That is, f(x) = (3/2)x^2 - (13/2)x + 16. This generates the sequence
11, 9, 10, 14, 21, 31, ...
Obviously this is not the desired sequence, so a quadratic curve-fitting scheme is not feasible. I would proceed by next trying a cubic polynomial to model the first four terms, and if that doesn't work, a degree 4 polynomial to include the first five terms, and in the worst case, a degree 5 polynomial.
Of course there may be a much simpler explanation of your sequence that I'm blind to...
3. 11,9,10,17,36,79,170,357,736,1499,3030,6097,12236, 24519,...
4. ## ?
5. Hello, nsingh201!
Find the next two terms of this sequence: . $11,\: 9,\: 10,\: 17,\: 36,\: 79 \;\hdots$
Given a sequence of increasing terms, I take the differences of consecutive terms,
. . then the differences of the differences, and so on . . . hoping to find a pattern.
Here's what I found . . .
. . $\begin{array}{ccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\
\text{3rd diff.} &&&& 3 && 6 && 12 \end{array}$
Since the 3rd differences seem to be identical to the 2nd differences,
. . the function appears to be exponential.
I also suspect that the 2nd differences are doubling.
So I believe the chart can be continued like this;
. . $\begin{array}{cccccccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\color{red}170} && {\color{red}357} \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 && {\color{blue}91} && {\color{blue}187} \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\color{blue}48} && {\color{blue}96}\end{array}$
And I think the generating function is: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$
6. Originally Posted by Soroban
Hello, nsingh201!
Given a sequence of increasing terms, I take the differences of consecutive terms,
. . then the differences of the differences, and so on . . . hoping to find a pattern.
Here's what I found . . .
. . $\begin{array}{ccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 \\
\text{3rd diff.} &&&& 3 && 6 && 12 \end{array}$
Since the 3rd differences seem to be identical to the 2nd differences,
. . the function appears to be exponential.
I also suspect that the 2nd differences are doubling.
So I believe the chart can be continued like this;
. . $\begin{array}{cccccccccccccccccc}
\text{Sequence} & 11 && 9 && 10 && 17 && 36 && 79 && {\color{red}170} && {\color{red}357} \\
\text{1st diff.} &&\text{-}2 && 1 && 7 && 19 && 43 && {\color{blue}91} && {\color{blue}187} \\
\text{2nd diff.} &&& 3 && 6 && 12 && 24 && {\color{blue}48} && {\color{blue}96}\end{array}$
And I think the generating function is: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$
beautiful approach
7. 11, 9, 10, 17, 36, 79, - Wolfram|Alpha
2 Soroban
Nice solution, but how did u get generating function?
8. Hello, ICanFly!
I was afraid someone would ask . . .LOL!
. . .but how did u get generating function?
We have: . $11,\:9,\:10,\:17,\:36,\;\hdots$
And we know there is a "doubling" feature in there: . $2^n$
Since the doubling didn't begin until the second differences.
. . I assumed there was a linear or quadratic involved, too.
So, my first general function was: . $f(n) \:=\:A + Bn + C\!\cdot\!2^n$
I used the first three terms of the sequence:
. . $\begin{array}{ccccc}
f(1) = 11 & A + B + 2C &=& 11 & [1] \\
f(2) \:=\: 9\; & A + 2B + 4C &=& 9 & [2] \\
f(3) = 10 & A + 3B + 8C &=& 10 & [3] \end{array}$
$\begin{array}{ccccc}
\text{Subtract [2] - [1]:} & B + 2C &=& -2 & [4] \\
\text{Subtract [3] - [2]:} & B + 4C &=& 1 & [5] \end{array}$
$\text{Subtract [5] - [4]: }\;\;2C \:=\: 3 \quad\Rightarrow\quad C \:=\:\tfrac{3}{2}$
Substitute into [4]: . $B + 3 \:=\:-2 \quad\Rightarrow\quad B \:=\:-5$
Substitute into [1]: . $A - 5 + 3 \:=\:11 \quad\Rightarrow\quad A \:=\:13$
Hence: . $f(n) \;=\;13 - 5n + \tfrac{3}{2}\!\cdot\!2^n$
I tested my function up to the 8th term . . . It works!
. . Therefore: . $f(n) \;=\;3\!\cdot\!2^{n-1} - 5n + 13$
9. Beautiful
10. thanks so much. this all really helped. | 2014-08-30T15:26:04 | {
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https://mathhelpboards.com/threads/1-998001.3403/ | # 1/998001
#### Jameson
Staff member
$$\displaystyle \frac{1}{998001}$$ has some interesting properties. It will list out every 3 digit number except for 998.
It has the form 0.001002003004...
Here's a video on some more details if you're interested.
#### MarkFL
Staff member
soroban once posted this observation and asked:
This is just one of a family of such fractions.
Can you determine the underlying characteristic?
I looked at the factorization of the denominator, and found:
$\displaystyle 998001=999^2$
So, next I looked at:
$\displaystyle\frac{012345679}{999999999}=\frac{1}{9^2}$
Thus, I conjecture that the family you speak of is:
$\displaystyle\frac{1}{(10^n-1)^2}$ where $\displaystyle n\in\mathbb N$
where the decimal representation contains all of the $n$ digit numbers except $\displaystyle 10^n-2$ and the period is $\displaystyle n(10^n-1)$.
#### Jameson
Staff member
Yes, you are correct about this form. For example $$\displaystyle \frac{1}{9999^2}$$ gives all of the 4 digit numbers.
In the video I posted in the OP they discussed an easy way to write recurring decimals as fractions that I'm sure we are all familiar with this method but it's still worth writing for those who aren't.
If you want to write some repeating decimal all you do is write the string as the numerator to a fraction and then in the denominator write the same number of 9's. For example, if you want to write $$\displaystyle .\overline{12436298}$$ as a fraction. It's simply $$\displaystyle \frac{12436298}{99999999}$$. Of course this method can be derived quite easily. If $$\displaystyle x=.12436298$$ then $100000000x=12436298.\overline{12436298}$ so $9999999x=12436298$ and we can solve for x directly. However it's nice to know a shortcut to not have to calculate it "the long way" everytime.
Last edited:
#### Bacterius
##### Well-known member
MHB Math Helper
This seems to generalize to an arbitrary base $b$, where:
$$\frac{1}{(b^n - 1)^2}$$
has representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.
But I only checked a couple of bases, so this could be wrong.
#### MarkFL
Staff member
This seems to generalize to an arbitrary base $b$, where:
$$\frac{1}{(b^n - 1)^2}$$
has representation in base $b$ containing all $n$-digit numbers (in base $b$) except $b^n - 2$ and with period $n(b^n - 1)$.
But I only checked a couple of bases, so this could be wrong.
I believe you are right, as the method Jameson outlined above can be generalized to any valid radix.
#### alane1994
##### Active member
Not much to contribute to this conversation, I was just popping in to say that this is really interesting! | 2021-03-01T00:32:58 | {
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https://math.stackexchange.com/questions/2237462/complex-numbers-and-the-norm/2237482 | # Complex Numbers and the Norm
I am given that for a complex number $w=a+bi$, define $\overline{w}=a-bi$ and $N(w)=w \overline{w}$. I have provided my answers for parts a and b, but I am not sure they are correct. I need help with figuring out part c.
(a) Compute $N(w)=w \overline{w}$ explicitly.
Here is what I have gotten:$N(w)=w \overline{w}= (a+bi)(a-bi)= a^2+b^2$.
(b) Show that $N(rs)=N(r)N(s)$ for any two complex numbers $r,s$.
Here is my work:
Let $r=a+bi$, $s=c+di$
Then $$rs=(a+bi)(c+di)=ac+adi+cbi-bd=(ac-bd)+(ad+cb)i.$$
Now, from a, have that $N(r)= a^2+b^2$ and $N(s)= c^2+d^2$. So, $$N(r)N(s)=(a^2+b^2)(c^2+d^2).$$
Also, $N(rs)=N((ac-bd)+(ad+cb)i)$ and from part a then, $N(rs)=(ac-bd)^2+(ad+cb)^2$
When expanded,
$$a^2c^2-abcd-abcd+b^2d^2+a^2d^2+abcd+abcd+b^2c^2$$
\begin{align*} &=a^2c^2+a^2d^2+b^2d^2+b^2c^2\\ &=a^2(c^2+d^2)+b^2(c^2+d^2)\\ &=(a^2+b^2)(c^2+d^2). \end{align*}
Hence, $N(rs)=N(r)N(s)$.
(c) Prove that $N(\overline{v})=N(v)$ and $N(v^n)=N(v)^n$ for any complex number.
This is the part I am confused as to how to prove. I let $v=a+bi$ and $\overline{v}=a-bi$. Then I am not sure how to use what I have shown before for this part.
• (1) Please put distinct parts into distinct questions. (2) Please use MathJax to format your math, it makes it a lot easier to read. Apr 16 '17 at 21:35
• @MichaelBurr Sorry, I do not know how to use MathJax, I tried my best to put it in the format. I labeled the parts a-c, but I will rename to make is clearer. Apr 16 '17 at 21:38
• What I mean is that if you have three parts, you should have three questions (one question for part $(a)$, one question for part $(b)$, and one question for part $(c)$). Apr 16 '17 at 21:39
• @MichaelBurr Okay, noted, thank you. It is just that is all one problem that has three parts and I wanted to see if my part a and b are okay to then get help for c. Apr 16 '17 at 21:40
• Also, to learn about MathJax, see the tutorial here. Apr 16 '17 at 21:41
For the first part of your question, notice that $$N (v)=\bar{v} v=\bar{v} \bar{\bar{v}}=N (\bar{v})$$ because $\bar{\bar{v}}=v$.
For the second part, consider doing induction over $n$:
Induction basis: $n=1$ $$N (v^1)=N (v)=N (v)^1$$
Induction step: $$N (v^{n+1})=N (v^n\cdot v)=N (v^n) \cdot N (v)=N (v)^n \cdot N (v)=N (v)^{n+1}$$ For the third from last step we used your result from b, for the second from last we used the induction hypothesis.
• Fixed now, thanks a lot. The way it was before, I did not like it either, but I did not know there is also \bar. Apr 16 '17 at 22:21
• Personally, I often use \overline because \bar is intended to be used as accent, but in this particular case overline looks plain ugly. Apr 16 '17 at 22:26
Since your question is about part $(c)$...
• To prove $N(v)=N(\overline{v})$, use your formula for $N$ in part $(a)$ on each of $v$ and $\overline{v}$. As stated, the question is somewhat confusing because $a$ and $b$ are used in two ways in the problem. In part $(a)$, $N(v)=a^2+b^2$ when $v=a+bi$. For part $(c)$, we could use $w=c+di$ and $\overline{w}=c-di$, then $N(w)=c^2+d^2$ (by substituting the variables in $w$ into the form for part $(a)$) and $N(\overline{w})=c^2+(-d)^2$ by the same substitution.
• To prove $N(v^n)=N(v)^n$, use your equality in part $(b)$ as well as induction on $n$. The base case is $N(v^1)=N(v)^1$, and then use part $(b)$ to prove $N(v^{k+1})=N(v)^{k+1}$ using $N(v^k)=N(v)^k$.
• For $N(v)=N(\overline{v})$ I know I will get a$^2$+b$^2$ for v, but how can I just plug in $\overline{v}$ into the equation if this is for v not $\overline{v}$? Apr 16 '17 at 21:43
• You have a general formula, it might be confusing because you're using $a$ and $b$ in two different roles, I'll edit to make this more clear. Apr 16 '17 at 21:44
• Okay so when plugging in $\overline{w}=c-di$ into the given equation for (a), what exactly is happening? I am not sure how you got (-d)$^2$ because wouldn't this be showing the equations are not equal? Apr 16 '17 at 21:50
• What is the square of a negative real number? Apr 16 '17 at 21:54
• @Pam_22R This shows that the terms are equal since $(-d)^2=-d\cdot (-d)=d^2$. Apr 16 '17 at 21:55 | 2021-09-18T14:19:05 | {
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https://www.jmp.com/en_ph/statistics-knowledge-portal/chi-square-test/chi-square-goodness-of-fit-test.html | ### Statistics Knowledge Portal
A free online introduction to statistics
# Chi-Square Goodness of Fit Test
## What is the Chi-square goodness of fit test?
The Chi-square goodness of fit test is a statistical hypothesis test used to determine whether a variable is likely to come from a specified distribution or not. It is often used to evaluate whether sample data is representative of the full population.
## When can I use the test?
You can use the test when you have counts of values for a categorical variable.
Yes.
## Using the Chi-square goodness of fit test
The Chi-square goodness of fit test checks whether your sample data is likely to be from a specific theoretical distribution. We have a set of data values, and an idea about how the data values are distributed. The test gives us a way to decide if the data values have a “good enough” fit to our idea, or if our idea is questionable.
### What do we need?
For the goodness of fit test, we need one variable. We also need an idea, or hypothesis, about how that variable is distributed. Here are a couple of examples:
• We have bags of candy with five flavors in each bag. The bags should contain an equal number of pieces of each flavor. The idea we'd like to test is that the proportions of the five flavors in each bag are the same.
• For a group of children’s sports teams, we want children with a lot of experience, some experience and no experience shared evenly across the teams. Suppose we know that 20 percent of the players in the league have a lot of experience, 65 percent have some experience and 15 percent are new players with no experience. The idea we'd like to test is that each team has the same proportion of children with a lot, some or no experience as the league as a whole.
To apply the goodness of fit test to a data set we need:
• Data values that are a simple random sample from the full population.
• Categorical or nominal data. The Chi-square goodness of fit test is not appropriate for continuous data.
• A data set that is large enough so that at least five values are expected in each of the observed data categories.
## Chi-square goodness of fit test example
Let’s use the bags of candy as an example. We collect a random sample of ten bags. Each bag has 100 pieces of candy and five flavors. Our hypothesis is that the proportions of the five flavors in each bag are the same.
Let’s start by answering: Is the Chi-square goodness of fit test an appropriate method to evaluate the distribution of flavors in bags of candy?
• We have a simple random sample of 10 bags of candy. We meet this requirement.
• Our categorical variable is the flavors of candy. We have the count of each flavor in 10 bags of candy. We meet this requirement.
• Each bag has 100 pieces of candy. Each bag has five flavors of candy. We expect to have equal numbers for each flavor. This means we expect 100 / 5 = 20 pieces of candy in each flavor from each bag. For 10 bags in our sample, we expect 10 x 20 = 200 pieces of candy in each flavor. This is more than the requirement of five expected values in each category.
Based on the answers above, yes, the Chi-square goodness of fit test is an appropriate method to evaluate the distribution of the flavors in bags of candy.
Figure 1 below shows the combined flavor counts from all 10 bags of candy.
Figure 1: Bar chart of counts of candy flavors from all 10 bags
Without doing any statistics, we can see that the number of pieces for each flavor are not the same. Some flavors have fewer than the expected 200 pieces and some have more. But how different are the proportions of flavors? Are the number of pieces “close enough” for us to conclude that across many bags there are the same number of pieces for each flavor? Or are the number of pieces too different for us to draw this conclusion? Another way to phrase this is, do our data values give a “good enough” fit to the idea of equal numbers of pieces of candy for each flavor or not?
To decide, we find the difference between what we have and what we expect. Then, to give flavors with fewer pieces than expected the same importance as flavors with more pieces than expected, we square the difference. Next, we divide the square by the expected count, and sum those values. This gives us our test statistic.
These steps are much easier to understand using numbers from our example.
Let’s start by listing what we expect if each bag has the same number of pieces for each flavor. Above, we calculated this as 200 for 10 bags of candy.
#### Table 1: Comparison of actual vs expected number of pieces of each flavor of candy
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Apple 180 200 Lime 250 200 Cherry 120 200 Cherry 225 200 Grape 225 200
Now, we find the difference between what we have observed in our data and what we expect. The last column in Table 2 below shows this difference:
#### Table 2: Difference between observed and expected pieces of candy by flavor
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Apple 180 200 180-200 = -20 Lime 250 200 250-200 = 50 Cherry 120 200 120-200 = -80 Orange 225 200 225-200 = 25 Grape 225 200 225-200 = 25
Some of the differences are positive and some are negative. If we simply added them up, we would get zero. Instead, we square the differences. This gives equal importance to the flavors of candy that have fewer pieces than expected, and the flavors that have more pieces than expected.
#### Table 3: Calculation of the squared difference between Observed and Expected for each flavor of candy
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Squared Difference Apple 180 200 180-200 = -20 400 Lime 250 200 250-200 = 50 2500 Cherry 120 200 120-200 = -80 6400 Orange 225 200 225-200 = 25 625 Grape 225 200 225-200 = 25 625
Next, we divide the squared difference by the expected number:
#### Table 4: Calculation of the squared difference/expected number of pieces of candy per flavor
Flavor Number of Pieces of Candy (10 bags) Expected Number of Pieces of Candy Observed-Expected Squared Difference Squared Difference / Expected Number Apple 180 200 180-200 = -20 400 400 / 200 = 2 Lime 250 200 250-200 = 50 2500 2500 / 200 = 12.5 Cherry 120 200 120-200 = -80 6400 6400 / 200 = 32 Orange 225 200 225-200 = 25 625 625 / 200 = 3.125 Grape 225 200 225-200 = 25 625 625 / 200 = 3.125
Finally, we add the numbers in the final column to calculate our test statistic:
$2 + 12.5 + 32 + 3.125 + 3.125 = 52.75$
To draw a conclusion, we compare the test statistic to a critical value from the Chi-Square distribution. This activity involves four steps:
1. We first decide on the risk we are willing to take of drawing an incorrect conclusion based on our sample observations. For the candy data, we decide prior to collecting data that we are willing to take a 5% risk of concluding that the flavor counts in each bag across the full population are not equal when they really are. In statistics-speak, we set the significance level, α , to 0.05.
2. We calculate a test statistic. Our test statistic is 52.75.
3. We find the theoretical value from the Chi-square distribution based on our significance level. The theoretical value is the value we would expect if the bags contain the same number of pieces of candy for each flavor.
In addition to the significance level, we also need the degrees of freedom to find this value. For the goodness of fit test, this is one fewer than the number of categories. We have five flavors of candy, so we have 5 – 1 = 4 degrees of freedom.
The Chi-square value with α = 0.05 and 4 degrees of freedom is 9.488.
4. We compare the value of our test statistic (52.75) to the Chi-square value. Since 52.75 > 9.488, we reject the null hypothesis that the proportions of flavors of candy are equal.
We make a practical conclusion that bags of candy across the full population do not have an equal number of pieces for the five flavors. This makes sense if you look at the original data. If your favorite flavor is Lime, you are likely to have more of your favorite flavor than the other flavors. If your favorite flavor is Cherry, you are likely to be unhappy because there will be fewer pieces of Cherry candy than you expect.
### Understanding results
Let’s use a few graphs to understand the test and the results.
A simple bar chart of the data shows the observed counts for the flavors of candy:
Figure 2: Bar chart of observed counts for flavors of candy
Another simple bar chart shows the expected counts of 200 per flavor. This is what our chart would look like if the bags of candy had an equal number of pieces of each flavor.
Figure 3: Bar chart of expected counts of each flavor
The side-by-side chart below shows the actual observed number of pieces of candy in blue. The orange bars show the expected number of pieces. You can see that some flavors have more pieces than we expect, and other flavors have fewer pieces.
Figure 4: Bar chart comparing actual vs. expected counts of candy
The statistical test is a way to quantify the difference. Is the actual data from our sample “close enough” to what is expected to conclude that the flavor proportions in the full population of bags are equal? Or not? From the candy data above, most people would say the data is not “close enough” even without a statistical test.
What if your data looked like the example in Figure 5 below instead? The purple bars show the observed counts and the orange bars show the expected counts. Some people would say the data is “close enough” but others would say it is not. The statistical test gives a common way to make the decision, so that everyone makes the same decision on a set of data values.
Figure 5: Bar chart comparing expected and actual values using another example data set
### Statistical details
Let’s look at the candy data and the Chi-square test for goodness of fit using statistical terms. This test is also known as Pearson’s Chi-square test.
Our null hypothesis is that the proportion of flavors in each bag is the same. We have five flavors. The null hypothesis is written as:
$H_0: p_1 = p_2 = p_3 = p_4 = p_5$
The formula above uses p for the proportion of each flavor. If each 100-piece bag contains equal numbers of pieces of candy for each of the five flavors, then the bag contains 20 pieces of each flavor. The proportion of each flavor is 20 / 100 = 0.2.
The alternative hypothesis is that at least one of the proportions is different from the others. This is written as:
$H_a: at\ least\ one\ p_i\ not\ equal$
In some cases, we are not testing for equal proportions. Look again at the example of children's sports teams near the top of this page. Using that as an example, our null and alternative hypotheses are:
$H_0: p_1 = 0.2, p_2 = 0.65, p_3 = 0.15$
$H_a: at\ least\ one\ p_i\ not\ equal\ to\ expected\ value$
Unlike other hypotheses that involve a single population parameter, we cannot use just a formula. We need to use words as well as symbols to describe our hypotheses.
We calculate the test statistic using the formula below:
$\sum^n_{i=1} \frac{(O_i-E_i)^2}{E_i}$
In the formula above, we have n groups. The $\sum$ symbol means to add up the calculations for each group. For each group, we do the same steps as in the candy example. The formula shows Oi as the Observed value and Ei as the Expected value for a group.
We then compare the test statistic to a Chi-square value with our chosen significance level (also called the alpha level) and the degrees of freedom for our data. Using the candy data as an example, we set α = 0.05 and have four degrees of freedom. For the candy data, the Chi-square value is written as:
$χ²_{0.05,4}$
There are two possible results from our comparison:
• The test statistic is lower than the Chi-square value. You fail to reject the hypothesis of equal proportions. You conclude that the bags of candy across the entire population have the same number of pieces of each flavor in them. The fit of equal proportions is “good enough.”
• The test statistic is higher than the Chi-Square value. You reject the hypothesis of equal proportions. You cannot conclude that the bags of candy have the same number of pieces of each flavor. The fit of equal proportions is “not good enough.”
Let’s use a graph of the Chi-square distribution to better understand the test results. You are checking to see if your test statistic is a more extreme value in the distribution than the critical value. The distribution below shows a Chi-square distribution with four degrees of freedom. It shows how the critical value of 9.488 “cuts off” 95% of the data. Only 5% of the data is greater than 9.488.
Figure 6: Chi-square distribution for four degrees of freedom
The next distribution plot includes our results. You can see how far out “in the tail” our test statistic is, represented by the dotted line at 52.75. In fact, with this scale, it looks like the curve is at zero where it intersects with the dotted line. It isn’t, but it is very, very close to zero. We conclude that it is very unlikely for this situation to happen by chance. If the true population of bags of candy had equal flavor counts, we would be extremely unlikely to see the results that we collected from our random sample of 10 bags.
Figure 7: Chi-square distribution for four degrees of freedom with test statistic plotted
Most statistical software shows the p-value for a test. This is the likelihood of finding a more extreme value for the test statistic in a similar sample, assuming that the null hypothesis is correct. It’s difficult to calculate the p-value by hand. For the figure above, if the test statistic is exactly 9.488, then the p-value will be p=0.05. With the test statistic of 52.75, the p-value is very, very small. In this example, most statistical software will report the p-value as “p < 0.0001.” This means that the likelihood of another sample of 10 bags of candy resulting in a more extreme value for the test statistic is less than one chance in 10,000, assuming our null hypothesis of equal counts of flavors is true. | 2021-09-27T18:40:38 | {
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https://math.stackexchange.com/questions/915793/telescoping-series-of-form-sum-n1-cdotsnk | # Telescoping series of form $\sum (n+1)\cdots(n+k)$ [duplicate]
Wolfram Alpha is able to telescope sums of the form $\sum (n+1)\cdots(n+k)$
e.g. $(1\cdot2\cdot3) + (2\cdot3\cdot4) + \cdots + n(n+1)(n+2)$
How does it do it?
EDIT: We can rewrite as: $\sum {(n+k)! \over n!} = \sum k!{(n+k)!\over n!k!} = \sum k!{{n+k} \choose n}$ (Thanks Daniel Fischer)
EDIT2: We can also multiply out and split sums. So e.g.
$$\sum (n-1)n(n+1) = \sum (n^3-n) = \sum n^3 - \sum n$$
But sums of powers actually seem to be more nasty than the original question, involving Bernoulli numbers. (Thanks Claude Leibovici)
And is there any name for this particular corner of maths? (i.e. How might I go about searching the Internet for information regarding this?)
PS please could we have a 'telescoping' tag?
## marked as duplicate by user147263, Community♦Dec 24 '15 at 20:42
• Note that the terms are $k!\binom{n+k}{n}$. You probably know a bit about binomial coefficients that helps summing. – Daniel Fischer Sep 1 '14 at 10:07
• I don't understand why you speak about "telescoping". Beside Daniel Fischer's suggestion, you could also develop $i(i+1)(i+2)= i^3+3 i^2+2 i$ and compute the three sums. – Claude Leibovici Sep 1 '14 at 10:16
• Is the sum over n or k? Makes a big difference. – marty cohen Sep 1 '14 at 15:18
• What great answers! Now I am stuck. Both RobJohn and HyperGeometric have nailed it from opposite directions. To accept either answer above the other would not seem right. I guess I will leave it open until SE provides some machinery for resolving these situations. – P i Sep 1 '14 at 17:49
• @P-i- Thank you! That's very considerate of you. It was an interesting question, which I enjoyed solving. I've upvoted the question as well. Have also just posted a comment on a recent observation on its resemblance to the power integral. A related question which you might want to pose would be to find the sum of a series with each term being the reciprocal of the corresponding in the present series, i.e the reciprocal of the product of consecutive integers. – hypergeometric Sep 4 '14 at 6:14
Hint: $$\sum_{n=k}^m\binom{n}{k}=\binom{m+1}{k+1}$$ A generalization is discussed in this answer. The equation above is equation $(1)$ with $m=0$.
Telescoping sum
To turn the sum in the question into a "telescoping sum", we can use the recurrence for Pascal's Triangle: $$\binom{n+1}{k+1}=\binom{n}{k}+\binom{n}{k+1}$$ Using this recurrence, we get \begin{align} \sum_{n=k}^m\binom{n}{k} &=\sum_{n=k}^m\left[\binom{n+1}{k+1}-\binom{n}{k+1}\right]\\ &=\sum_{n=k+1}^{m+1}\binom{n}{k+1}-\sum_{n=k}^m\binom{n}{k+1}\\ &=\left[\binom{m+1}{k+1}+\color{#C00000}{\sum_{n=k+1}^m\binom{n}{k+1}}\right]-\left[\binom{k}{k+1}+\color{#C00000}{\sum_{n=k+1}^m\binom{n}{k+1}}\right]\\ &=\binom{m+1}{k+1}-\binom{k}{k+1}\\ &=\binom{m+1}{k+1} \end{align} The sums in red are the terms that are telescoped out, leaving just the first and last terms. In this case, the last term $\binom{k}{k+1}=0$.
Let $$p_n=\prod_{r=1}^k (n+r)=\overbrace{(n+1)(n+2)\cdots(n+k)}^{k \ \text{terms}}$$ which is the product of $k$ consecutive integers.
Consider the difference of two consecutive terms, where each term is the product of $k+1$ consecutive integers, i.e. \begin{align} &\prod_{r=1}^{k+1}(n+r)-\prod_{r=0}^k(n+r)\\ &=\overbrace{\underbrace{(n+1)(n+2)\cdots(n+k)}_{p_n}(n+k+1)}^{(k+1) \ \text{terms}}-\overbrace{n\underbrace{(n+1)(n+2)\cdots(n+k)}_{p_n}}^{(k+1) \ \text{terms}}\\ &=p_n[(n+k+1)-n]\\ &=p_n(1+k) \end{align}
Hence, $$p_n=\prod_{r=1}^k (n+r)=\frac1{1+k}\left[\prod_{r=1}^{k+1} (n+r)-\prod_{r=0}^k (n+r) \right]$$ which is convenient for telescoping.
Required summation, \begin{align} S=\sum_{n=0}^{m}p_n&=\sum_{n=0}^{m} \prod_{r=1}^{k} (n+r)=\sum_{n=0}^{m}(n+1)(n+2)(n+3)\cdots (n+k)\\ &=\frac1{k+1}\sum_{n=0}^{m} \left[ \prod_{r=1}^{k+1} (n+r)-\prod_{r=0}^{k} (n+r)\right]\\ &=\frac1{k+1}\prod_{r=1}^{k+1} (m+r) \qquad \blacksquare\end{align} by telescoping.
In your example, $k=3$, hence the general term is
$$(n+1)(n+2)(n+3)=\frac14\left[(n+1)(n+2)(n+3)(n+4)-n(n+1)(n+2)(n+3)\right]$$
Hence, by telescoping from $n=0$ to $m$, \begin{align}S&=1\cdot2\cdot3+2\cdot3 \cdot 4+\cdots +(m+1)(m+2)(m+3)\\ &=\frac14(m+1)(m+2)(m+3)(m+4) \end{align}
NB: It is interesting to note that this result bears a striking resemblance to integration.
Compare the standard integral $$\int_0^m n^k dn=\frac{m^{k+1}}{k+1}$$ to the result of the summation above, which can also be stated as $$\sum_{n=0}^{m}n^{[k]}=\frac {m^{[k+1]}}{k+1}$$ where $n^{[k]}$ is my adjusted* Pochhammer symbol for rising factorials, defined as $$n^{[k]}=\prod_{r=1}^{k}(n+r)=(n+1)(n+2)(n+3)\cdots(n+k)$$
The actual Pochhammer symbol for rising factorials, $n^{(k)}$, starts from $n$ itself and not $n+1$, i.e. $$n^{(k)}=\prod_{r=1}^{k}(n+r-1)=n(n+1)(n+2)\cdots(n+k-1)$$
• I've upvoted this as it appears valid. But I can't see any clear motivation for the initial algebraic manipulations. Is it possible to rework this logic in a way that does not require inspirational leaps? – P i Sep 1 '14 at 14:08
• @P-i- - thank you, that's very kind :) in order to make the series telescope, the approach is to convert one term (or product, in this case) into a difference of two terms, usually of a higher 'order'. for instance, to telescope the sum of squares, one would start from the difference of cubes. in this case, we want to sum $k$-term products hence we start from the difference of two consecutive $(k+1)$-term products. – hypergeometric Sep 1 '14 at 14:19
• @P-i- - the proposed solution has been reworked slightly and hopefully this provides greater clarity. – hypergeometric Sep 1 '14 at 14:30
• thanks to all who upvoted. a short appendix has been added to the solution to highlight the resemblance of the summation result to the standard power integral. comments are most welcome. – hypergeometric Sep 3 '14 at 15:22
For completeness, a solution using the combinatorial/binomial approach, as initiated by RobJohn:
Let $p(n, k) =n(n+1)...(n+k-1) =\prod_{j=0}^{k-1} (n+j)$.
Then (writing each step in detail)
$\begin{array}\\ p(n+1, k)-p(n, k) &=\prod_{j=0}^{k-1} (n+1+j)-\prod_{j=0}^{k-1} (n+j)\\ &=\prod_{j=1}^{k} (n+j)-\prod_{j=0}^{k-1} (n+j)\\ &=(n+k)\prod_{j=1}^{k-1} (n+j)-n\prod_{j=1}^{k-1} (n+j)\\ &=((n+k)-n)\prod_{j=1}^{k-1} (n+j)\\ &=k\prod_{j=1}^{k-1} (n+j)\\ &=k\prod_{j=0}^{k-2} (n+1+j)\\ &=kp(n+1, k-1)\\ \end{array}$
or, putting $k+1$ for $k$ and $n$ for $n+1$, $p(n, k) =\frac{p(n, k+1)-p(n-1. k+1)}{k+1}$.
Therefore
$\begin{array}\\ \sum_{n=1}^M p(n, k) &=\sum_{n=1}^M \frac{p(n, k+1)-p(n-1. k-1)}{k+1}\\ &=\frac1{k+1}\sum_{n=1}^M (p(n, k+1)-p(n-1. k+1))\\ &=\frac1{k+1}(p(M, k+1)-p(0, k+1))\\ &=\frac{p(M, k+1)}{k+1}\\ \end{array}$
• This proof is succinct and flawless! Thanks Marty! – P i Sep 2 '14 at 11:38
Also your terms can be re written as $$n(n+1)(n+2)=\dfrac{n(n+1)(n+2)(n+3)}{4}-\dfrac{(n-1)n(n+1)(n+2)}{4}.$$
• that may be correct but it would not telescope readily... – hypergeometric Sep 1 '14 at 15:24
• Now it is fine. – Bumblebee Sep 2 '14 at 5:06
Following from RobJohn's hint, this can be proved very simply by reverse-engineering the hockey-stick identity.
The blues sum to the red for any given 'hockey-stick'.
It is a straightforward proof by induction. For example, in the picture, 1+3+6+10+15 = 35, so 1+3+6+10+15 + 21 = 35+21 = 56
i.e. True(stick length k) => True(stick length k+1)
Induction can also be argued from the ${n \choose r}$ formula, but the above method avoids algebraic manipulation, working instead from the simple "each element is formed by summing its upper neighbours" definition of Pascal's Triangle.
But now we DO need some algebraic manipulation.
We write this hockey-stick identity as:
$${k+0 \choose k} + ... + {k+n \choose k} = {k+(n+1) \choose (k+1)}$$
(notice the hockey stick has been reflected)
$${(k+0)! \over {k!((k+0)-k)!}} + ... + {(k+n)! \over {k!((k+n)-k)!}} = {(k+(n+1))! \over {(k+1)!(((k+(n+1))-(k+1))!}}$$
$${(k+0)! \over 0!} + ... + {(k+n)! \over n!} = {1 \over {k+1}} \cdot {(k+(n+1))! \over n!}$$
So the original question turns out to be an emergent property of Pascal's Triangle, and I would hazard a guess that it is from this simple construct that the question originally arose.
However, this proof doesn't move forwards from start to finish in a sequence of intelligently chosen moves. It doesn't offer any insight into telescoping general sums, hence I am favouring hypergeometric's proof.
• thank you! glad you liked it :) robjohn's approach using the pascal triangle identity is quite useful too, and i have used a similar approach elsewhere on MSE for another question. – hypergeometric Sep 2 '14 at 6:28
$$\color{#66f}{\large\sum_{n = 0}^{m}\pars{n + 1}\ldots\pars{n + k} =k!\,{k + m + 1 \choose m}}$$
• blink complex analysis! That's fantastic! But how does the first path integral come from n+k choose k? – P i Sep 4 '14 at 12:45
• @P-i- That's an identity: $${a \choose b}=\oint_{0\ <\ \left\vert \,z\,\right\vert\ =\ c}{(1 + z)^a \over z^{b + 1}}\,{{\rm d}z \over 2\pi{\ic}}$$ It's valid for $c \in {\mathbb R}\,,\ c > 0$ and $b\ =\ 0,1,2,3,\ldots$ – Felix Marin Sep 4 '14 at 17:49
Induction
This question is actually a duplicate of: Finding a closed formula for $1\cdot2\cdot3\cdots k +\dots + n(n+1)(n+2)\cdots(k+n-1)$
But it is difficult to find duplicates when the question can only be expressed using maths symbols, and this question now acts as a more comprehensive resource.
One of the answers to that question is to manually construct formulae for the first few cases: $1+2+3+\dots+n$, $1\cdot2 +\dots + n\cdot(n+1)$, etc, notice a pattern and intuit a candidate formula, then prove this by induction. | 2019-11-21T08:45:54 | {
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https://math.stackexchange.com/questions/3224765/polynomial-division-an-obvious-trick-reducing-mod-textitsimpler-multiple/3224982 | # Polynomial division: an obvious trick? [reducing mod $\textit{simpler}$ multiples]
The following question was asked on a high school test, where the students were given a few minutes per question, at most:
Given that, $$P(x)=x^{104}+x^{93}+x^{82}+x^{71}+1$$ and, $$Q(x)=x^4+x^3+x^2+x+1$$ what is the remainder of $$P(x)$$ divided by $$Q(x)$$?
Let $$Q(x)=0$$. Multiplying both sides by $$x-1$$: $$(x-1)(x^4+x^3+x^2+x+1)=0 \implies x^5 - 1=0 \implies x^5 = 1$$ Substituting $$x^5=1$$ in $$P(x)$$ gives $$x^4+x^3+x^2+x+1$$. Thus, $$P(x)\equiv\mathbf0\pmod{Q(x)}$$
Obviously, a student is required to come up with a “trick” rather than doing brute force polynomial division. How is the student supposed to think of the suggested method? Is it obvious? How else could one approach the problem?
• One observation is that in that question the step of 'multiplying both sides by $x-1$' is only presentation and doesn't really represent any advantage for solving the problem. What is really important is observing that the roots of $Q$ are the fifths roots of $1$ besides $1$, or that is a factor of $x^5-1$. There are many ways to compute this remainder that are simple to do in a short time. So, don't include that 'multiplying by $x-1$' as part of what was required for the students to solve the problem – logarithm May 14 '19 at 15:48
• My approach would be to observe that the exponents on P(x) look like they're set up to produce an elegant answer. The common elegant answers are "1" and "0". Since the "+1"s on the polynomials are clearly intended to cancel out, the answer must be "0". – Mark May 14 '19 at 23:33
• I'm probably missing something, but... The above trick seems to show that when $Q(x) = 0$, $P(x) = Q(x)$. But (a) the original question didn't specify that $Q(x) = 0$; how do we know that the answer $P(x)\equiv\mathbf0\pmod{Q(x)}$ applies to other cases? and (b) the original question asks, "what is the remainder of $𝑃(𝑥)$ divided by $𝑄(𝑥)$?" but the answer given here says let $Q(x) = 0$; doesn't that mean that the remainder is not defined? en.wikipedia.org/wiki/Remainder#Polynomial_division – LarsH May 15 '19 at 21:06
• I added some explicit examples to the end of my answer to emphasize the ubiquity of that simple idea. – Bill Dubuque May 16 '19 at 13:49
• What kind of test? Is this part of a math competition, or a classroom precalculus exam? – PersonX May 16 '19 at 16:35
The key idea employed here is the method of simpler multiples - a very widely used technique. Note that $$\,Q\,$$ has a "simpler" multiple $$\,QR = x^5\!-\!1,\,$$ so we can first reduce $$P$$ modulo $$\,x^{\large 5}\! -\! 1\,$$ via $$\!\bmod x^{\large 5}-1\!:\,\ \color{#c00}{x^{\large 5}\equiv 1}\Rightarrow\, x^{\large r+5q^{\phantom{|}}}\!\!\equiv x^{\large r}(\color{#c00}{x^{\large 5}})^{\large q}\equiv x^{\large r},\,$$ then finally reduce that $$\!\bmod Q,\,$$ i.e.
$$P\bmod Q\, =\, (P\bmod QR)\bmod Q\qquad$$
This idea is ubiquitous, e.g. we already use it implicitly in grade school in radix $$10$$ to determine integer parity: first reduce mod $$10$$ to get the units digit, then reduce the units digits mod $$2,\,$$ i.e.
$$N \bmod 2\, = (N\bmod 2\cdot 5)\bmod 2\qquad\$$
i.e. an integer has the same parity (even / oddness) as that of its units digit. Similarly since $$7\cdot 11\cdot 13 = 10^{\large 3}\!+1$$ we can compute remainders mod $$7,11,13$$ by using $$\,\color{#c00}{10^{\large 3}\equiv -1},\,$$ e.g. $$\bmod 13\!:\,\ d_0+ d_1 \color{#c00}{10^{\large 3}} + d_2 (\color{#c00}{10^{\large 3}})^{\large 2}\!+\cdots\,$$ $$\equiv d_0 \color{#c00}{\bf -} d_1 + d_2+\cdots,\,$$ and, similar to the OP, by $$\,9\cdot 41\cdot 271 = 10^{\large 5}\!-1\,$$ we can compute remainders mod $$41$$ and $$271$$ by using $$\,\color{#c00}{10^5\!\equiv 1}$$
$$N \bmod 41\, = (N\bmod 10^{\large 5}\!-1)\bmod 41\quad$$
for example $$\bmod 41\!:\ 10000\color{#0a0}200038$$ $$\equiv (\color{#c00}{10^{\large 5}})^{\large 2}\! + \color{#0a0}2\cdot \color{#c00}{10^{\large 5}} + 38\equiv \color{#c00}1+\color{#0a0}2+38\equiv 41\equiv 0$$
Such "divisibility tests" are frequently encountered in elementary and high-school and provide excellent motivation for this method of "divide first by a simpler multiple of the divisor" or, more simply, "mod first by a simpler multiple of the modulus".
This idea of scaling to simpler multiples of the divisor is ubiquitous, e.g. it is employed analogously in the method of rationalizing denominators and in Gauss's algorithm for computing modular inverses.
For example, to divide by an algebraic number we can employ its norm as a simpler multiple, e.g. for $$\,w = a+b\sqrt{n}\,$$ we use $$\,ww' = (a+b\sqrt n)(a-b\sqrt n) = a^2-nb^2 = c\in\Bbb Q\ (\neq 0\,$$ by $$\,\sqrt{n}\not\in\Bbb Q),\,$$ which reduces division by an algebraic to simpler division by a rational, i.e. $$\, v/w = vw'/(ww'),$$ e.g.
$$\dfrac{1}{a+b\sqrt n}\, =\, \dfrac{1}{a+b\sqrt n}\, \dfrac{a-b\sqrt n}{a-b\sqrt n}\, =\, \dfrac{a-b\sqrt n}{a^2-nb^2}\,=\, {\frac{\small 1}{\small c}}(a-b\sqrt n)\qquad$$
so-called rationalizing the denominator. The same idea works even with $$\,{\rm\color{#c00}{nilpotents}}$$
$$\color{#c00}{t^n = 0}\ \Rightarrow\ \dfrac{1}{a-{ t}}\, =\, \dfrac{a^{n-1}+\cdots + t^{n-1}}{a^n-\color{#c00}{t^n}}\, =\, a^{-n}(a^{n-1}+\cdots + t^{n-1})\qquad$$
which simplifies by eliminating $$\,t\,$$ from the denominator, i.e. $$\,a-t\to a^n,\,$$ reducing the division to division by a simpler constant $$\,a^n\,$$ (vs. a simpler rational when rationalizing the denominator).
Another example is Gauss' algorithm, where we compute fractions $$\!\bmod m\,$$ by iteratively applying this idea of simplifying the denominator by scaling it to a smaller multiple. Here we scale $$\rm\color{#C00}{\frac{A}B} \to \frac{AN}{BN}\:$$ by the least $$\rm\,N\,$$ so that $$\rm\, BN \ge m,\,$$ reduce mod $$m,\,$$ then iterate this reduction, e.g.
$$\rm\\ mod\ 13\!:\,\ \color{#C00}{\frac{7}9} \,\equiv\, \frac{14}{18}\, \equiv\, \color{#C00}{\frac{1}5}\,\equiv\, \frac{3}{15}\,\equiv\, \color{#C00}{\frac{3}2} \,\equiv\, \frac{21}{14} \,\equiv\, \color{#C00}{\frac{8}1}\qquad\qquad$$
Denominators of the $$\color{#c00}{\rm reduced}$$ fractions decrease $$\,\color{#C00}{ 9 > 5 > 2> \ldots}\,$$ so reach $$\color{#C00}{1}\,$$ (not $$\,0\,$$ else the denominator would be a proper factor of the prime modulus; it may fail for composite modulus)
See here and its $$25$$ linked questions for more examples similar to the OP (some far less trivial).
Worth mention: there are simple algorithms for recognizing cyclotomics (and products of such), e.g. it's shown there that $$\, x^{16}+x^{14}-x^{10}-x^8-x^6+x^2+1$$ is cyclotomic (a factor of $$x^{60}-1),\,$$ so we can detect when the above methods apply for arbitrarily large degree divisors.
Let $$a$$ be zero of $$x^4+x^3+x^2+x+1=0$$. Obviously $$a\ne 1$$. Then $$a^4+a^3+a^2+a+1=0$$ so multiply this with $$a-1$$ we get $$a^5=1$$ (You can get this also from geometric series $$a^n+a^{n-1}+...+a^2+a+1 = {a^{n+1}-1\over a-1}$$ by putting $$n=4$$).
But then $$\begin{eqnarray} Q(a) &=& a^{100}\cdot a^4+a^{90}\cdot a^3+a^{80}\cdot a^2+a^{70}\cdot a+1\\ &=& a^4+a^3+a^2+a+1\\&=&0\end{eqnarray}$$
So each zero of $$Q(x)$$ is also a zero of $$P(x)$$ and since all 4 zeroes of $$Q(x)$$ are different we have $$Q(x)\mid P(x)$$.
• I think this is the approach that the examiners would expect from high school students. – Dancrumb May 14 '19 at 0:23
• From my perspective this is the best answer. (+1) – trancelocation May 14 '19 at 3:54
• The OP understands that after multiplication with $x-1$, the question became simpler, he/she asked how a student should find this trick. This answer is basically just a reformulation of the answer given in the question. I don't see the usefulness of this answer at all... (but it has 14 upvotes, so I might be missing something...) – user193810 May 14 '19 at 11:49
• @MariaMazur: A bit, but not really. It still feels like you stopped reading the question after "what is the remainder of P(x) divided by Q(x)?". You gave a correct answer to that question, but not to the real question. It is easy to prove that the 'trick' (multiplication by $x-1$) works and is correct, but you ignored the real question, and after your edit it is hidden as a small remark. But whatever, your answer is trivially better than mine, because I did not gave any, so I'll leave it here. – user193810 May 14 '19 at 13:25
• @Pakk: it hinges on spotting the standard identity: $Q(x) = 0$ gives the fifth roots of unity (other than x=1). It takes tons more work if you don't spot that; you could solve the quartic $Q(x) = 0$ then notice "Hey these are the other four fifth roots of unity, thus x^5 \equiv 1" then apply the information $x^5 \equiv 1$ to hugely simplify $P(x)$, and thus conclude $Q(x) \mid P(x)$ and remainder = 0. But that would be lots more work. (Yeah the first time I saw this trick back in HS I voiced the same objection about non-obviousness... this is why this identity is so pivotal to factorization) – smci May 16 '19 at 1:53
While it may be a standard technique, as Bill's response details, I wouldn't say it's at all obvious at High School level. As a pre-Olympiad challenge problem, however, it's a good one.
My intuition is via cyclotomic polynomials -- $$Q(x) = \Phi_5(x)$$, giving the idea to multiply through by $$x-1$$ -- but I doubt I would have recognised them before university: https://en.wikipedia.org/wiki/Cyclotomic_polynomial
• I would go even further. As a BSE graduate in engineering, this is not at all obvious to me. – MooseBoys May 14 '19 at 6:58
• @MooseBoys: I would go even further. As a PhD in Physics (admittedly many years ago), this is not at all obvious to me. – WoJ May 14 '19 at 9:04
• It may not be obvious in other contexts since you never had to use it, but I assure you that in Math Olympiad training it's a standard trick. All serious contestants will probably already have seen something similar. There's nothing outrageous or shameful in that; Olympiad-style math is simply different from what people use and need in other professional contexts (physicist/engineer) --- the focus on Euclidean geometry being a perfect example. – Federico Poloni May 14 '19 at 9:12
• @Federico Where in the question does it say that this is an Olympiad test? OP phrased their question as if it were applicable to all high school students. I'd guess that fewer than 5% of all students would get the correct answer. – MooseBoys May 14 '19 at 17:01
• @MooseBoys I agree with your estimate. The question is "is this obvious?", and my comment is "it is a standard problem for people who train for math Olympiads". I realize that it's a strong additional assumption. I'm not arguing that it's a good or bad problem: as you point out, to do that, we would at least have to know to which students it was given. – Federico Poloni May 14 '19 at 17:09
This may be accessible to a high school student:
$$x^{104}+x^{93}+x^{82}+x^{71}+1$$
$$= (x^{104}-x^4)+(x^{93}-x^3)+(x^{82}-x^2)+(x^{71}-x)+(x^4+x^3+x^2+x+1)$$
$$=x^4(x^{100}-1)+x^3(x^{90}-1)+x^2(x^{80}-1)+ x(x^{70}-1)+(x^4+x^3+x^2+x+1)$$
We know that $$(x^n-1)|(x^{mn}-1), m,n \in \mathbb{N}$$ so $$x^5-1$$ divides $$x^{100}-1, x^{90}-1$$ etc.
In turn $$x^5-1$$ is divisible by $$(x^4+x^3+x^2+x+1)$$ which concludes the proof
If it's not obvious, an examination of the question quickly reveals the trick. Say
$$P(x)=x^n$$
Then begin long division by $$Q(x)$$:
$$x^n-x^n-x^{n-1}-x^{n-2}-x^{n-3}-x^{n-4}$$ $$x^{n-5}$$ $$\dots$$ $$x^{n-5k}$$
While it may not be obvious just by looking at the question, anyone who attempts the naive solution has (at least) a reasonable chance of running across a way of solving it.
• This is a good point. If one is asked to perform division by a polynomial $Q(x)$, it is natural to start with dividing $x^n$ for $n> \deg(Q)$. Thus the first step is to divide $x^5$. Magically(!), one gets the remainder 1. So, one can discover the formula in the form $x^5=(x-1)(x^4+x^3+x^2+x+1) + 1$ even if one didn't remember it. The rest of the steps are not too difficult. – Kapil May 14 '19 at 4:07
I would have thought that bright students, who knew $$1+x+x^2+\cdots +x^{n-1}= \frac{x^n-1}{x-1}$$ as a geometric series formula, could say
$$\dfrac{P(x)}{Q(x)} =\dfrac{x^{104}+x^{93}+x^{82}+x^{71}+1}{x^4+x^3+x^2+x+1}$$
$$=\dfrac{(x^{104}+x^{93}+x^{82}+x^{71}+1)(x-1)}{(x^4+x^3+x^2+x+1)(x-1)}$$
$$=\dfrac{x^{105}-x^{104}+x^{94}-x^{93}+x^{83}-x^{82}+x^{72}-x^{71}+x-1}{x^5-1}$$
$$=\dfrac{x^{105}-1}{x^5-1}-\dfrac{x^{104}-x^{94}}{x^5-1}-\dfrac{x^{93}-x^{83}}{x^5-1}-\dfrac{x^{82}-x^{72}}{x^5-1}-\dfrac{x^{71}-x}{x^5-1}$$
$$=\dfrac{x^{105}-1}{x^5-1}-x^{94}\dfrac{x^{10}-1}{x^5-1}-x^{83}\dfrac{x^{10}-1}{x^5-1}-x^{72}\dfrac{x^{10}-1}{x^5-1}-x\dfrac{x^{70}-1}{x^5-1}$$
and that each division at the end would leave zero remainder for the same reason, replacing the original $$x$$ by $$x^5$$
I think if the candidates know what a geometric series is, the question is okay. Indeed, one uses exactly this trick to find the formula for the geometric series, i.e. one writes $$(x-1)\sum_{k=1}^nx^k=x^{n+1}-1$$ to find that $$\sum_{k=1}^\infty x^k=\lim_{n\to\infty}\sum_{k=1}^nx^k=\lim_{n\to\infty}\frac{x^{n+1}-1}{x-1}=\frac{1}{1-x}$$ for $$|x|<1$$. Therefore, it is not too hard to get from $$x^4+x^3+x^2+x+1$$ to $$x^5-1$$. Now you can reduce mod $$x^5-1$$ by substitution $$x^5=1$$.
I think the way one should think about this is to note that $$x^4+x^3+x^2+x+1$$ is the minimal polynomial of any primitive 5th unit root $$\alpha$$. Now $$P(\alpha)=0$$ since $$\alpha^5=1$$ and therefore $$Q$$ devides $$P$$. | 2020-01-26T21:47:28 | {
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https://math.stackexchange.com/questions/1926037/rational-and-irrational-sequences | Rational and irrational sequences
1. Let $(x_n)$ be a sequence of rational numbers that converges to an irrational number $x$. Must it be the case that $x_1, x_2, \dots$ are all irrational?
2. Let $(y_n)$ be a sequence that converges to a rational number $y$. Define such a sequence where $y_1, y_2, \dots$ are all irrational.
Now these are the 2 questions in my textbook i noticed when posting this that someone had asked the answer to 1) and a similar but less difficult version of 2.
A couple things i noted the defined answer for 1 he used $\pi$ and simply added more digits to it for each part of the sequence. so firstly id like to ask would that really be a full mark answer on a analysis final? secondly and more importantly i was wondering if there was a way to define this where the final number wasn't transcendental.
For 1) I used $(1+1/n)^n$ s.t. $n \in \mathbb{N}$. For every $n$ this sequence is rational but for $\lim_{ n \to \infty }(1+1/n)^n = e$.
My Question: (i) Is this true and if so is there a definable sequence that holds but doesn't define a transcendental number?
EDIT: Golden ratio from Fibonacci sequence.
For 2) I picked $2^{1/n}$ s.t. $n>1$ and $n \in \mathbb{N}$. I believe every value of this sequence is irrational but the limit should be 1.
My Question: (ii) does the above work? and if possible could I have another example of a sequence where every value is irrational but the limit is rational?
EDIT: $\frac{\sqrt 2}{n}$.
• If $(x_n)$ converges to $x$ is the answer to part (i) (and all the $x_n$ are non-zero), then $(x_n/x)$ is a solution to part (ii). Sep 14 '16 at 2:54
• thats very clever thanks! Sep 14 '16 at 3:00
You basically have both answers right, just need a couple of tweaks.
Question:(i) Is this true and if so is there a definable sequence that holds but doesnt define a transcendental number?
It is true that $a_n = (1 + \frac{1}{n})^n$ is a sequence of rationals whose $\lim_{n \to \infty} a_n = e$ is irrational.
For a sequence of rationals that converges to an algebraic (non transcendental) irrational, take for example the partial sums of the Taylor series expansion of $\sqrt{1+x}$ at $x=1$ which converge to $\sqrt{2}$.
Question:(ii) does the above work? and if possible could i have anther example of a sequence where every value is irrational but the limit is rational?
The above works, except you may want to define $a_n = 2^\frac{1}{n+1}$ so that $a_1$ is irrational as well.
For other such sequences, pick any $r \in \mathbb{R} \setminus \mathbb{Q}$ and let $a_n = \frac{r}{n}$.
[ EDIT ] I see that you just edited the question to add $\frac{\sqrt{2}}{n}$ as an example of the latter.
It doesn't have to be $\pi$ or $e$, you could do the digit by digit approximation and converge to say $\sqrt{2}$ which is algebraic and not transcendental.
I would say that that answer of digit by digit is basically full mark; you could probably include that $|x - x_n| < 10^{-n}$ which means that the sequence is Cauchy to be more specific.
Your second sequence of $\{2^{1/n}\} \to 1$ is also correct. See if you can prove that each number in the sequence is irrational by mimicking the proof that $\sqrt{2}$ is irrational.
Here's another sequence
$$a_n = 1 + \frac{\sqrt{2}}{10^n}$$
We see that $\lim_{n \to \infty} a_n = 1$ and each $a_n$ is irrational because you can create a common denominator and note that the numerator will be irrational and the denominator and integer. | 2021-11-28T05:58:35 | {
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http://mathhelpforum.com/math-challenge-problems/89068-techniques-integration-3-a.html | # Math Help - Techniques of integration (3)
1. ## Techniques of integration (3)
Let $n \geq 0$ be an integer. Evaluate $I_n=\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx.$
2. Using itegration by parts I could probably kick this thing off...
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx = \frac{1}{n}cos^n(x)sin(nx) + \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx$
Not really sure how to do,
$\int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx$
probably needs some trig identities to tidy it up.
3. Originally Posted by pickslides
Using itegration by parts I could probably kick this thing off...
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx =$ $\color{red}\frac{1}{n}cos^n(x)sin(nx)$ $+ \int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx.$
Not really sure how to do $\int_0^{\frac{\pi}{2}} sin(nx)sin(x)cos^{n-1}(x) \ dx.$
probably needs some trig identities to tidy it up.
that's a good start. we have $I_0=\frac{\pi}{2}.$ for $n \geq 1,$ as you (almost) showed, we have: $I_n=\int_0^{\frac{\pi}{2}} \sin(nx) \sin(x) \cos^{n-1}x \ dx.$ now what?
4. Hello,
Okay, I think I got the correct idea
From $\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$, it follows that :
$\sin(nx)\sin(x)=\cos[(n-1)x]-\cos(nx)\cos(x)$
The integral is thus :
$I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x)-\cos(nx)\cos(x)\cos^{n-1}(x) ~dx$
$I_n=\int_0^{\frac\pi2} \cos[(n-1)x]\cos^{n-1}(x) ~dx-\int_0^{\frac\pi2} \cos(nx)\cos^n(x) ~dx$
Which is :
$I_n=I_{n-1}-I_n$
$I_n=\frac 12 \cdot I_{n-1}$
$\Rightarrow \boxed{I_n=\frac{\pi}{2^{n+1}}}$
That's a really nice one
5. Bravo Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!
6. Originally Posted by NonCommAlg
Brave Moo! that's exactly the idea ... although you made a strange mistake at the end of your solution, which i'm sure you'll easily find and fix it!
Yes indeed
1+1=n+1 : that's a very well-known result !!
Had a too short night...
7. Consider
$cos^n(x) = \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} [cos(n-2k)x + i sin(n-2k)x]
$
Since $cos^n(x)$ is real , the terms of sine function will be cancelled finally ,
but we experience that $\int_0^{\frac\pi2} cos(n-2k)x ~cosnx ~dx = 0$ if n-2k is not equal to n so the integral becomes :
$\frac{\pi}{2^{n+1}}$
8. Originally Posted by simplependulum
$\cos^n(x) = \frac{1}{2^n} \sum_{k=1}^{n} \binom{n}{k} [\cos(n-2k)x + i \sin(n-2k)x]
$
this is not an identity!
9. Originally Posted by NonCommAlg
Let $n \geq 0$ be an integer. Evaluate $I_n=\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx.$
It can be done using complex variables...
First note that
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\int_0^{2\pi} (\cos x)^n \cos (nx) \ dx$
Parameterize the unit circle in the comples plane with
$z=e^{i\theta} \implies \frac{dz}{iz}=d\theta$
So the integral becomes
$\oint \left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right)\frac{dz}{iz}$
expanding with the binomial theorem we get
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{n-i}z^{-i}\right]\left[z^{n-1}+z^{-n-1} \right]dz$
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]$
each finite sume is its own larent series so the residues are coeffients on the $\frac{1}{z}$ terms so we get
$\frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}$
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}$
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}$
10. Originally Posted by TheEmptySet
It can be done using complex variables...
First note that
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\int_0^{2\pi} (\cos x)^n \cos (nx) \ dx$
Parameterize the unit circle in the comples plane with
$z=e^{i\theta} \implies \frac{dz}{iz}=d\theta$
So the integral becomes
$\oint \left(\frac{z+z^{-1}}{2}\right)^n\left( \frac{z^n+z^{-n}}{2}\right)\frac{dz}{iz}$
expanding with the binomial theorem we get
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{n-i}z^{-i}\right]\left[z^{n-1}+z^{-n-1} \right]dz$
$\frac{1}{i\cdot 2^{n+1}}\left[ \oint \sum_{i=0}^{n}\binom{n}{i}z^{2n-2i-1}+ \sum_{i=0}^{n}\binom{n}{i}z^{i-1}\right]$
i think the power of $z$ in the second sum should be $-2i-1.$
each finite sume is its own larent series so the residues are coeffients on the $\frac{1}{z}$ terms so we get
$\frac{1}{i\cdot 2^{n+1}}\left[2\pi i (1)+2\pi i (1) \right]=\frac{4\pi}{2^{n+1}}$
$4\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{4\pi}{2^{n+1}}$
$\int_0^{\frac{\pi}{2}} (\cos x)^n \cos (nx) \ dx=\frac{\pi}{2^{n+1}}$
looks good to me! | 2015-05-26T04:46:04 | {
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https://math.stackexchange.com/questions/3638644/double-negation-of-existential-universal-quantifier-lnot-exists-x-lnot-ax | # Double negation of existential/universal quantifier $\lnot(\exists x(\lnot A(x))$
I have a (simple) question about the double-negation and existential/universal quantifiers.
When negating the following
$$\lnot\exists x(\lnot A(x))$$
I believe you just push the negation in (which swaps the quantifier) making it
$$\forall x[\lnot(\lnot A(x))]$$
and then
$$\forall x A(x)$$
Would that be a correct interpretation? Or would I be losing one of the negations on $$A(x)$$ somewhere earlier?
• Yes, that’s fine. It’s also intuitively clear: the first expression says that there is no $x$ for which $A(x)$ is false, and the last says that $A(x)$ is true for every $x$, clearly the same thing. – Brian M. Scott Apr 22 at 17:30
• Awesome, thank you so much. – confundido Apr 22 at 17:31
• You’re very welcome. – Brian M. Scott Apr 22 at 17:31
• Thank you (just read your more detailed answer) - I thought the intuition was clear, but it was part of a larger proof and I was doubting myself as to whether the reduction made sense. I guess I just needed validation. – confundido Apr 22 at 17:35
You did that correctly.
Note that going from
$$\lnot\exists x(\lnot A(x))$$
to
$$\forall x[\lnot(\lnot A(x))]$$
is an example of 'Quantifier Negation', but going from:
$$\forall x[\lnot(\lnot A(x))]$$
to
$$\forall x A(x)$$
is an instance of 'Double Negation'
As per the answer by Brian M. Scott:
It’s also intuitively clear: the first expression says that there is no x for which A(x) is false, and the last says that A(x) is true for every x, clearly the same thing.
A word of caution. Of course at the final step $$\forall x\neg\neg Ax$$ entails (or at least, classically entails) $$\forall x Ax$$, and does so because adjacent double negations (classically) cancel each other out.
BUT
The inference is not strictly an application of a standard double negation rule of the form from $$\neg\neg\varphi$$ infer $$\varphi$$. That rule only allows us to remove initial double negations.
SO
To show the entailment in standard proof systems requires a three-step mini-proof:
$$\forall x\neg\neg Ax\quad$$ (given)
$$\neg\neg Aa\quad\quad$$ (universal instantiation with parameter or free variable depending on system)
$$Aa\quad\quad\quad$$ (NOW you can apply the DN rule)
$$\forall x Ax\quad\quad$$ (universal generalization)
So your reasoning is informally just fine, but do be careful about jumping from $$\forall x\neg\neg Ax$$ to $$\forall x Ax$$ in formal proofs! | 2020-10-19T20:47:44 | {
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https://math.stackexchange.com/questions/1226352/finding-an-operation-on-gs-that-yields-a-group | # Finding an operation on $G^S$ that yields a group
Problem: Assume $S$ is a nonempty set and $G$ is a group. Let $G^S$ denote the set of all mappings from $S$ to $G$. Find an operation on $G^S$ that will yield a group.
Update (full attempted proof--is it correct?): Let $*$ be the operation on the given group $G$ and let $\bigstar$ be the operation on $G^S$. For all $\alpha,\beta\in G^S$, define $(\alpha\,\bigstar\,\beta)(x)=\alpha(x)*\beta(x)$. Now we must prove $G^S$ is a group.
• Closure: This is inherited from $G$.
• Associativity: This is also inherited from $G$.
• Existence of an identity element: This is also inherited from $G$; more explicitly, $e_G(x)=1=e_{G^S}(x)$ because $(\alpha\,\bigstar\,e_G)(x)=\alpha(x)*e_G(x)=\alpha(x)$ and, in the other direction, $(e_G\,\bigstar\,\alpha)(x)=e_G(x)*\alpha(x)=\alpha(x)$.
• Existence of inverse elements: If $\eta(x)\in G$, then $\eta^{-1}(x)\in G$ also. Thus, for $\alpha\in G^S$, we have $(\alpha\,\bigstar\,\alpha^{-1})(x)=\alpha(x)*\alpha^{-1}(x)=e_G(x)=e_{G^S}(x)$ and, in the other direction, $(\alpha^{-1}\,\bigstar\,\alpha)(x)=\alpha^{-1}(x)*\alpha(x)=e_G(x)=e_{G^S}(x)$.
This concludes the proof that the operation $\bigstar$ on $G^S$ yields a group. $\blacksquare$
$\color{red}{\mathbf{\text{Question:}}}$ Is this above proof correct? Also, by the way the operation $\bigstar$ works, it seems like $G^S$ would be a subgroup of $G$ or am I wrong?
Original work shown for problem:
Book solution: For $\alpha,\beta\in G^S$, define $(\alpha\beta)(x)=\alpha(x)\beta(x)$ for each $x\in S$.
My questions: I thought about this problem for a good while but got nowhere and finally decided to look at the answer, but it does not make a ton of sense to me. How does the given operation yield a group? And what really is the operation being considered? Is it composition? That is, $(\alpha\beta)(x)\equiv(\alpha\circ\beta)(x)$? If it is composition, then I know all compositions are associative, and thus I do not need to prove that as part of proving that $G^S$ is a group. But what about closure? Existence of an identity element? Existence of inverse elements?
For the existence of an identity element, I thought about $e_G(x)=1$ and having $(\eta e_G)(x)=\eta(x)e_G(x)=\eta(x)$ and $(e_G\eta)(x)=e_G(x)\eta(x)=\eta(x)$, but this doesn't really seem right or is it?
Lastly, the problem gives the condition that $S\neq\varnothing$ (why is that important?), and no operation is specified concerning $G$--does that matter? I guess not, but it seems interesting to me that no operation is specified for $G$ when we are trying to find an operation for $G^S$ that involves $G$.
• The proof seems fundamentally fine. I'm not sure what the distinction is between the notation $e_G(x)$ and $e_G^S(x)$. You could write out associativity more explicitly, although I tend not to do so unless it's not obvious. – Rolf Hoyer Apr 9 '15 at 17:22
• @RolfHoyer I just meant that $e_G(x)$ is the identity element for $G$ while $e_G^S(x)$ is the identity element for $G^S$. If $G^S$ is a subgroup of $G$ (it looks like it is, but is it?), then I know the distinctions do not matter because the identity and inverses are the same. – fancynancy Apr 9 '15 at 17:32
• @RolfHoyer OK--I just fixed up the question with the modified notation ($e_{G^S}(x)$ instead of $e_G^S(x)$). Look all good now? Your answer definitely helped me to start thinking along the right lines. – fancynancy Apr 9 '15 at 17:38
It definitely looks much better!
I'm a little confused about the distinction between $e_G$ and $e_{G^S}$; they both seem like the identity of $G^S$, and you seem to be using $1$ for the identity of $G$. Maybe make it more explicit that you're defining $e_{G^S}: S \to G$ so that $e_{G^S}(x) = 1_G$; it's the map sending everything in $S$ to the identity of $G$. So my only suggestion would be to clean up the notation (how exactly are you writing the identity of $G$? Of $G^S$?) and make it clear exactly what the identity of $G^S$ does.
For the inverses, essentially the same thing: Given $\alpha: S \to G$, you just start using $\alpha^{-1}$ without explicitly stating that, if $\alpha(x) = g$, then $\alpha^{-1}(x) = (\alpha(x))^{-1} = g^{-1}$. That is, make it clear that we invert $\alpha:S \to G$ pointwise, by inverting its output (using the inverse in $G$).
For your other question, very rarely will it be the case that $G^S \cong G$. I think it's more likely that $G^S$ is isomorphic to a direct product of several copies of $G$ (how many copies?).
• Hmm okay. So I should write $e_G(x)=1$ and $e_{G^S}(x)=1_G$ to more clearly indicate that $1$ is the identity for $G$ while $1_G$ is the identity for $G^S$ right? You're right--my notation is a bit mangled. Okay. I get the identity part now. I'm still a little fuzzy on the inverse part though. Are you saying I'm technically correct with what I've written but it would be clearer to actually assign an element to $\alpha(x)$ to clearly indicate that invertibility is pointwise? – fancynancy Apr 9 '15 at 18:10
• What I mean about identities is that there should really only be two in sight: That of $G$, let's call it $1_G$, and that of $G^S$, let's call it $1_{G^S}$. So, I'm not sure why there are two functions whose notation suggest that they're both identities of $G^S$; that they're both functions $S \to G$. – pjs36 Apr 9 '15 at 18:12
• And for the inverses, yes, it's correct, provided you explicitly state how $\alpha^{-1}: S \to G$ (or $\eta^{-1}$, for that matter) is defined. – pjs36 Apr 9 '15 at 18:14
• Okay so I should have (for identities, that is), in one direction, $(\alpha\,\bigstar\,e_G)(x)=\alpha(x)*e_G(x)=\alpha(x)*1_G=\alpha(x)$. The same would happen in the other direction. So the identity exists and, explicitly, we have $e_{G^S}=e_G=1$. Correct? – fancynancy Apr 9 '15 at 18:16
• I do see what you're saying now about the confusion between writing $e_G(x)$ and $e_{G^S}(x)$ because they are both mappings from $S$ to $G$; thus, that notation is rather clumsy. – fancynancy Apr 9 '15 at 18:19
The group operation is not defined by composition, unless you mean the composition $S \to S\times S \to G\times G \to G$, where the first map is the diagonal $s \to (s,s)$, the second map is the product of the two given maps $\alpha, \beta$, and the third map is the group operation on $G$. When the author writes $(\alpha\beta)(x) = \alpha(x)\beta(x)$, the latter product is defined using the operation of $G$, just to clarify.
This definition is using the operation of $G$ pointwise, just like how addition and multiplication are defined for real-valued functions, ie $(f+g)(x) = f(x) + g(x)$. The identity element is the constant function $e_G(x) = 1$ for every $x$, and you correctly give the proof that this works. The inverse of a function $\alpha: S \to G$ is given by taking inverses pointwise, namely $\alpha^{-1}(x) = (\alpha(x))^{-1}$. Associativity follows from the associativity of $G$.
As regards to the empty set, if $S = \emptyset$ then there is a unique empty function $S \to G$, and I guess the authors would rather rule out this case than give the trivial product on this set.
Hopefully this makes things more clear.
• I just updated my question with a fully attempted proof in light of your helpful answer. Can you let me know if you think it's correct? Thanks for your answer--I think I am at least on the right track now. – fancynancy Apr 9 '15 at 14:23 | 2020-09-19T19:39:51 | {
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https://physics.stackexchange.com/questions/137527/why-is-equivalent-resistance-in-parallel-circuit-always-less-than-each-individua | # Why is equivalent resistance in parallel circuit always less than each individual resistor?
There are $$n$$ resistors connected in a parallel combination given below.
$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}+\frac{1}{R_{4}}+\frac{1}{R_{5}}.......\frac{1}{R_{n}}$$
Foundation Science - Physics (class 10) by H.C. Verma states (Pg. 68)
For two resistances $$R_{1}$$ and $$R_{2}$$ connected in parallel,
$$\frac{1}{R_{ev}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{R_{1}+R_{2}}{R_{1}R_{2}}$$ $$R_{ev}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}$$
We see that the equivalent resistance in a parallel combination is less than each of the resistances.
I observe this every time I do an experiment on parallel resistors or solve a parallel combination problem. How can we prove $$R_{ev} or that $$R_{ev}$$ is less than the Resistor $$R_{min}$$, which has the least resistance of all the individual resistors?
• I have to confess that when I saw that you had exhibited the harmonic sum form ($1/R_e = 1/R_1 + 1/R_2 + \dots$), I found the question confusing because I didn't understand what you didn't get. It's worth your time to think about this until it becomes obvious, not because this problem is so important but because it will start teaching you the habit of learning things from the functional form of the relationships that appear in your studies. – dmckee --- ex-moderator kitten Sep 27 '14 at 17:15
• @dmckee I didn't quite get what you said. It'll probably help me if you'll elaborate. – user49111 Sep 27 '14 at 17:26
• With experience, you will find this fact obvious, and it is worth taking the time to ponder it until it becomes so. – dmckee --- ex-moderator kitten Sep 27 '14 at 17:34
• I rolled back this question to the original version because you shouldn't put answers, or responses to answers, in the question itself. @imakesmalltalk Feel free to post an answer of your own if you would like to offer another method of answering the question. – David Z Oct 27 '14 at 0:30
• Once you believe the 2 resistor answer, it extends to any finite number. For each resistor, imagine combining all the rest into a single resistor by the law you cite. Now the combination of the one resistor and the combination resistor is less than the one. Do this once for each of the resistors, and you have that the total combination is less than any one. – Ross Millikan Oct 27 '14 at 2:53
If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$ Dividing everything by the voltage give us: $$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Which is the same as: $$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$ Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors.
The individual resistances are all positive, so the sum $$\frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots \,,$$ is larger than the inverse of any of the individual resistances, and that means that the inverse of the sum is necessarily smaller than any of the resistances.
No mucking around with the two-resistor form required.
We can prove it by induction. Let $$\frac{1}{R^{(n)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n}$$ Now, when $n=2$, we find $$\frac{1}{R^{(2)}_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \implies R_{eq}^{(2)} = \frac{R_1 R_2}{R_1+R_2} = \frac{R_1}{1+\frac{R_1}{R_2}} = \frac{R_2}{1+\frac{R_2}{R_1}}$$ Since $\frac{R_1}{R_2} > 0$, we see that $R^{(2)}_{eq} < R_1$ and $R^{(2)}_{eq} < R_2$ or equivalently $R^{(2)}_{eq} < \min(R_1, R_2)$.
Now, suppose it is true that $R^{(n)}_{eq} < \min (R_1, \cdots, R_n)$. Then, consider $$\frac{1}{R^{(n+1)}_{eq}} = \frac{1}{R_1} + \cdots+ \frac{1}{R_n} + \frac{1}{R_{n+1}} = \frac{1}{R^{(n)}_{eq}} + \frac{1}{R_{n+1}}$$ Using the result from $n=2$, we find $$R^{(n+1)}_{eq} < \min ( R_{n+1} , R^{(n)}_{eq} ) < \min ( R_{n+1} , \min (R_1, \cdots, R_n))$$ But $$\min ( R_{n+1} , \min (R_1, \cdots, R_n)) = \min ( R_{n+1} , R_1, \cdots, R_n)$$ Therefore $$R^{(n+1)}_{eq} < \min ( R_1, \cdots, R_n , R_{n+1} )$$ Thus, we have shown that the above relation holds for $n=2$, and further that whenever it holds for $n$, it also holds for $n+1$. Thus, by induction, it is true for all $n\geq2$.
• $\frac {R_1}{R_2} > 0$ not 1 – Omar Elawady Apr 4 '16 at 21:41
• Your induction method is much more mathematical than the other answers....but is there any way to solve this using $AM-GM$ inequality? – Soham Jun 3 '16 at 16:11
It might be easier to think in terms of conductance which is the inverse of resistance.
The more paths there are between A and B which conduct electricity, the greater is the amount of current which can flow - ie the greater is the conductance of the network. The total conductance is greater than that of any individual path, because each additional path always increases the amount of electricity which can be conducted, it never reduces it. In particular, the total conductance is always greater than the largest individual conductance.
Translating this back in terms of resistance R (which is the inverse of conductance S - ie R = 1/S), the total resistance is smaller than the smallest individual resistance. | 2021-08-03T05:31:32 | {
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https://byjus.com/question-answer/find-inverse-by-elementary-row-operations-if-possible-of-the-following-matrices-begin-bmatrix-1-1/ | Question
Find inverse, by elementary row operations (if possible), of the following matrices$$\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$$
Solution
To check if the inverse exist we find the determinant:We have:$$A=\begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}$$So, $$\left|A\right|=1\times 7-(-5\times 3)$$$$\Rightarrow \left|A\right|=7+15=22$$Since, $$|A|\ne0$$, hence the inverse exists Now, in order to use elementary row operations, we may write $$A=IA.$$$$\therefore \begin{bmatrix} 1 & 3 \\ -5 & 7 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}A$$$$\Rightarrow \begin{bmatrix} 1 & 3 \\ 0 & 22 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}A$$ $$R_2 \rightarrow R_2+ 5\times R_1$$ $$\Rightarrow \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} 1 & 0 \\ \dfrac{5}{22} & \dfrac{1}{22}\end{bmatrix}A$$ $$R_2 \rightarrow \dfrac{1}{22} \times R_2$$$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}= \begin{bmatrix} \dfrac{7 }{22} & -\dfrac{3}{22} \\ \dfrac{5}{22} & \dfrac{1}{22}\end{bmatrix}A$$ $$R_1 \rightarrow R_1- 3\times R_2$$$$\Rightarrow \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}=\dfrac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1\end{bmatrix}A$$$$\Rightarrow I=BA$$, where $$B$$ is the inverse of $$A$$.$$B=\dfrac{1}{22} \begin{bmatrix} 7 & -3 \\ 5 & 1\end{bmatrix}$$Mathematics
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https://math.stackexchange.com/questions/2278517/how-to-approach-this-combinatorics-problem | # How to approach this combinatorics problem?
You have $12$ different flavors of ice-cream. You want to buy $5$ balls of ice-cream, but you want at least one to be made of chocolate and also you don't want more than $2$ balls per flavor. In how many ways can you can choose the $5$ balls?
• Does order matter? – vrugtehagel May 12 '17 at 21:32
• @vrugtehagel No. – LearningMath May 12 '17 at 21:33
• Partial hint: If $2$ of the balls are chocolate, the other $3$ balls can be chosen in ${11\choose3}+11\cdot10$ ways. Do you see why? – Barry Cipra May 12 '17 at 21:39
Calculate the coefficient of $x^5$ in $$(x^1+x^2)(x^0+x^1+x^2)^{11}$$
• Can you please elaborate where you got this from? Thanks. – LearningMath May 12 '17 at 21:31
• $(x^1+x^2)$ chooses one or two chocolate balls. Each $(x^0+x^1+x^2)$ chooses zero, one, or two balls of that flavor. – vadim123 May 12 '17 at 21:32
• Can you provide a reference for this approach? Thank you. – LearningMath May 12 '17 at 21:52
• See the second half of this. – vadim123 May 12 '17 at 23:32
This is the same as no of solutions of $\sum_{i=1}^{12} x_i=5$ where $1≤x_1≤2$ and for the other $x_i$'s $0≤x_i≤2$ which is the same as coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$
• How is it the same as the coefficient of $x^5$ in $(x^1+x^2)(x^0+x^1+x^2)^{11}$ ? Can you provide a reference where I can read more about this? – LearningMath May 12 '17 at 22:06
Decide if chocolate will be a double ($c=1$) or a single ($c=0$).
If $c=0$, take a chocolate ball. Decide how many "doubles" $d\in\{0,1,2\}$ you want and make one of $11 \choose d$ specific choices regarding which non-chocolate flavors you want. These $d$ colors and chocolate are now off limits, so choose $5-1-2d=4-2d$ colors from each of the remaining $12-d-1=11-d$ legal options.
If $c=1$, take two chocolate balls. Decide how many "non-chocolate doubles" $d\in\{0,1\}$ you want and make one of $11 \choose d$ choices regarding the non-chocolate flavors. These $d$ colors and chocolate are now off limits, so choose $5-2-2d=3-2d$ flavors from each of the remaining $12-d-1=11-d$ legal options.
$$\sum_{d=0,1,2} {11 \choose d}{{11-d} \choose {4-2d}} + \sum_{d=0,1} {11 \choose d}{{11-d} \choose {3-2d}}$$
$$= {11 \choose 2} + \sum_{d=0,1} {11 \choose d} \left[{{11-d} \choose {4-2d}} + {{11-d} \choose {3-2d}}\right]$$
which is $55 + [1*495 + 11*55]=1155$.
You can also answer this using careful counting Case by case basis splitting into exactly two pairs of flavors, one pair of flavors and no pairs.
$$C_1,P_1,P_1,P_2,P_2 = 1*12*11$$ $$C_1,P_1,P_1,P_2,P_3=1*12*11*10/2!$$ $$C_1,P_1,P_2,P_3,P_4=1*12*11*10*9/4!$$
How many of these ways lead to more than $2$ chocolates?
If $$P_1=C$$ which for the first case can happen $11$ ways, the second case can happen $11*10/2!$ ways and the last case can't happen if just $P_1=C$
What about if $P_2=C$ the first case can happen in $11$ ways and the second case can happen in 11*10/2! ways.
If $P_3,P_4=C$ then we don't get $3$ or more chocolates, so we are done! The final answer is $$12*11+12*11*10/2! +12*11*10*9/4!-11*10/2!-11-11-11*10/2!=1155$$
I think the most obvious/straightforward way to solve this is to use combination 'with repetition'/'stars and bars' and then subtract all inapplicable cases. (https://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition). $\space$The formula for this for selecting $r$ from $n$ is ${{n+r-1}\choose r}$. $\space$(I like to think of this as selecting $r$ 'items' from the $n+r-1$ 'walls' and 'items' and then letting the $n-1$ 'walls' 'fall into place.')
Dividing into cases where there is one chocolate ball and more than one.
One chocolate ball of ice cream:
$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=4$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor and $1$ of another - cases where there are $4$ of one flavor):
${{11+4-1=14}\choose 4}-{11\choose 1}{10\choose 1}-{11\choose 1}=1001-110-11=880$
Two chocolate balls of ice cream:
$11$ flavors left to choose from. Is combination 'with repetition' with $n=11$ and $r=3$ and then subtract all inapplicable cases. (All cases - cases where there are $3$ of $1$ flavor):
${{11+3-1=13}\choose 3}-{11\choose 1}=286-11=275$
For a grand total of $1155$. | 2019-08-21T22:08:05 | {
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Let $g(x, y)=x^{2} y+\cos y+y \sin x$, Which of the following statements is TRUE?Let $g(x, y)=x^{2} y+\cos y+y \sin x$ Which of the following statements is TRUE? (A) $g_{x}=2 x y+\cos y+y \cos x$. (B) $g_{y}=2 x-\sin y+\s ... close 0 answers 7 views Determine, giving reasons, whether the sequence$\left\{a_{n}\right\}=\left\{\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right\}$is convergent or divergent. If it is convergent, find the value to which it converges.Determine, giving reasons, whether the sequence$\left\{a_{n}\right\}=\left\{\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right\}$is convergent or diverge ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty} \dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}$0 answers 5 views Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty} \dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}$Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty} \dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty}(-1)^{n}\left(\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right)^{n}$0 answers 4 views Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty}(-1)^{n}\left(\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right)^{n}$Determine whether the following series converge or diverge.$\sum_{n=1}^{\infty}(-1)^{n}\left(\dfrac{n^{3}-1+n^{2} \sin n}{1+3 n^{3}}\right)^{n}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Given that$\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots=\sum_{n=0}^{\infty} x^{n}$. Write down the power series of$\dfrac{1}{1+x^{2}}$0 answers 6 views Given that$\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots=\sum_{n=0}^{\infty} x^{n}$. Write down the power series of$\dfrac{1}{1+x^{2}}$Given that$\dfrac{1}{1-x}=1+x+x^{2}+x^{3}+\cdots=\sum_{n=0}^{\infty} x^{n}$. Write down the power series of$\dfrac{1}{1+x^{2}}$... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Write$\dfrac{\pi}{4}$as a series. 0 answers 8 views Write$\dfrac{\pi}{4}$as a series.Write$\dfrac{\pi}{4}$as a series. ... close Notice: Undefined index: avatar in /home/customer/www/mathsgee.com/public_html/qa-theme/AVEN/qa-theme.php on line 993 Which FIVE of the following statements are FALSE ? 0 answers 6 views Which FIVE of the following statements are FALSE ?Which FIVE of the following statements are FALSE ? A. The sequence$\left\{\frac{\cos (n \pi)}{n}\right\}$diverges B.$\quad \cos \ ...
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The interval of convergence of the series $\sum_{n=1}^{\infty} \dfrac{(-1)^{n}(x+2)^{n}}{n}$ is
The interval of convergence of the series $\sum_{n=1}^{\infty} \dfrac{(-1)^{n}(x+2)^{n}}{n}$ isThe interval of convergence of the series $\sum_{n=1}^{\infty} \dfrac{(-1)^{n}(x+2)^{n}}{n}$ is A. $\quad(-3,-1$ B. $-3,-1$ &nbs ...
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Which FIVE of the following statements are TRUE?
Which FIVE of the following statements are TRUE?Which FIVE of the following statements are TRUE? A. The series $\sum_{n=1}^{\infty}\left(\frac{4 n-3}{3 n-4}\right)^{n}$ converges B ...
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Determine whether the integral $\int_{3}^{3+e} \ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to.
Determine whether the integral $\int_{3}^{3+e} \ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to. Determine whether the integral $\int_{3}^{3+e} \ln (x-3) d x$ is convergent or divergent. If convergent, give the value the integral converges to. ...
Find the limit of the sequence $\left\{\dfrac{-1}{(-1)^{n}}\right\}$ (if it exists).
Find the limit of the sequence $\left\{\dfrac{-1}{(-1)^{n}}\right\}$ (if it exists).Find the limit of the sequence $\left\{\dfrac{-1}{(-1)^{n}}\right\}$ (if it exists). ... | 2022-01-21T01:40:34 | {
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https://math.stackexchange.com/questions/2212375/what-is-wrong-with-this-proof-of-the-union-of-two-compact-sets | # What is wrong with this proof of the union of two compact sets?
I need to show that the union of two compact sets is compact. Here is what I tried:
Let $A$ and $B$ be two compact sets. Then for every open cover $\{A_n\}_{n\in I_1}$ and open cover $\{B_m\}_{m\in I_2}$, there exists finite subcovers $\bigcup\limits_{n\in I_1}^p A_n$ such that $A\subseteq\bigcup\limits_{n\in I_1}^p A_n$ and $\bigcup\limits_{b\in I_2}^k B_m$ such that $B\subseteq\bigcup\limits_{b\in I_2}^k B_m$. Then, $A\cup B\subseteq (\bigcup\limits_{n\in I_1}^p A_n)\cup(\bigcup\limits_{b\in I_2}^k B_m)\subseteq(\bigcup\limits_{n\in I_1} A_n) \cup (\bigcup\limits_{b\in I_2} B_m)$ (the union of open covers constructed as an open cover for $A\cup B$). Thus, for ever open cover of $A\cup B,$ there exists a finite subcover, so $A\cup B$ is compact.
I thought that since the open cover I constructed was arbitrary, and so was the finite subcover, then every open cover of $A\cup B$ had a finite subcover. What is wrong with this?
• Well, what's wrong is that you don't show that any cover of $C = A \cup B$ has a finite subcover. You only consider those covers that are unions of previous discussed covers of A and B have finite subcovers. If $\Alpha$ is a cover of A and $\Beta$ is a cover of B. That shows us $\Alpha \cup \Beta$ has a subcover. But what about all the covers of $C$ that are not $\Alpha \cup \Beta$? As it turns out if $\Gamma$ is an open cover then $\Gamma$ must = $\{A\}\cup \{B\}$ for some pair of open covers, and such a statement may seem obvious, but can't be assumed without proof. Mar 31, 2017 at 23:30
• You probably got the gist of what is going on, one comment that seems relevant: if you could show that every open cover of $A \cup B$ is of the form $O_1 \cup O_2$ with $O_1$ an open over of $A$ and $O_2$ another for $B,$ then you are in business (your proof goes through). Mar 31, 2017 at 23:38
$$A\cup B \subseteq \left( \bigcup_{n \in I_1} A_n \right)\cup \left( \bigcup_{m \in I_2} B_m \right)$$
may not be a general open cover for $A\cup B$. So, it is better to start with:
Let $\{ C_n \}$ be an open cover for $A \cup B$. Then since both $A$ and $B$ are compact sets, we have $\dots$
The problem with your logic is as follows:
You are right, in that a finite subcover of $A$ and a finite subcover of $B$, together gives a finite subcover of $A \cup B$.
However, to show $A \cup B$ is compact, you are supposed to start with any arbitrary open cover of $A \cup B$. This was not done: instead, you started with open covers for $A$ and $B$ separately, and used their compactness to produce a finite subcover.
The missing step : How do you know that a subcover of $A \cup B$, is also a subcover of $A$ and a subcover of $B$? Or, how would you generate subcovers of $A$ and $B$, given one of $A \cup B$? In short : you did not start with an arbitrary subcover. If you do so, it would have to be as explicit as in the answer I am going to give below for your betterment.
Hence, the answer should be as follows:
Start from definition. Let $\{ U_\alpha\}$ be an open cover of $A \cup B$.
Since $A \cup B \subset \bigcup U_\alpha$, it follows that $A \subset \bigcup U_\alpha$, and similarly, $B \subset \bigcup U_\alpha$. That is, every cover of $A \cup B$ gives a cover of $A$ and a cover of $B$. This is the key step missing from your explanation.
Now, we have finite subcovers for each one, and you can take the union of these subcovers (which remains a subcover of the original cover) to get the result.
So, in short, when we say arbitrary, we have to be careful. It is best that you start out with what is given: In compactness, we are given an open cover of the set, you never gave yourself that, instead you started with open covers for $A$ and $B$ separately, which was not given to you by the definition of $A \cup B$ being compact.
It's all about being careful. And it's absolutely fine to mistakes. It's good you have this clarified now.
As an exercise, try and extend this to a finite union of compact sets. Do you think you can extend this to an infinite union of compact sets?
You assumed, without proving it, that every open cover of $A\cup B$ is a union of an open cover of $A$ and an open cover of $B$.
Suppose for example, that $A=[0,1]$ and $B$ is some other compact set, and $\mathcal B = \{ (0.4,1] \cup G : G\text{ is an open subset of } B. \}.$ This is an open cover of $B$. Let $C=\{ [0,0.6)\cup G : G \text{ is an open subset of } B. \}.$ That is another open cover of $B$. Neither of them is an open cover of $A,$ but $\mathcal B \cup \mathcal C$ is an open cover of $A\cup B.$ Each has a finite subset that covers $B.$ Neither has a finite subset that covers $A$.
You could just say that any open cover of $A\cup B$ is an open cover of $A$. So it has a finite subcover of $A$. Similarly find another finite subset of it that covers $B$. Then take the union of those two finite subsets. That actually makes the proof simpler. If you do it that way then there is no need to prove the proposition whose proof you omitted.
PS: The needless assumption whose proof you omitted can be proved trivially: If $\mathcal A$ is an open cover of $A\cup B$ then $\mathcal A \cup \mathcal A$ is a union of an open cover of $A$ and an open cover of $B$. However, this is a needless complication in the argument. | 2022-09-25T20:41:37 | {
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https://www.physicsforums.com/threads/mathematical-induction-problem.611898/ | # Mathematical induction problem
1. Jun 6, 2012
1. The problem statement, all variables and given/known data
Hello! This is I want to prove using Mathematical Induction: $1+3+5+...+(2n-1)=n^2$. The problem is: I don`t understand very much about Mathematical Induction :(
2. Relevant equations
3. The attempt at a solution
Suppose n=1. Then 1=1. Now suppose $1+3+5+...+(2n)=(n+1)^2$. Then $1+3+5+...+2n=(1+3+5+...+(2n-1))+2n=n^2+2n=n(n+2)$.
Is this correct? If yes, how does this prove my hypothesis?
2. Jun 6, 2012
You're only working with odd numbers, so the next term wouldn't be $2n$..
3. Jun 6, 2012
This is what I should do to solve the problem $1+3+5+...+(2n-1+1)=1+3+5+...+2n$. Right? Because you always do $n+1$ in the Induction step.
4. Jun 6, 2012
### sacscale
Mathematical induction attempts to show that if the equation holds for n, then it also holds for n+1. It's easier to keep straight if you use a substitution such as n=k+1
5. Jun 6, 2012
@ DDarthVader: Yes, but you're not substituting $n + 1$ in for $2n - 1$ or anything, you're subbing it in for $n$...
6. Jun 6, 2012
Doing the substitution $n=k+1$ I've got this result:
$1+3+5+...+2k+1=(1+3+5+...+(2n-1))+2k+1$
Then
$n^2 +2k+1 = (k+1)^2+2k+1 = k^2+2k+2+2k+1 = k^2+4k+3$
But since $n=k+1$ we got
$(n-1)^2+4n-4+3 =n^2-2n+4n+1 =n^2+2n+1 = (n+1)^2$
Is this correct?
Last edited: Jun 6, 2012
7. Jun 6, 2012
### SammyS
Staff Emeritus
Your conjecture: $1+3+5+...+(2n-1)=n^2$
You've already done the base step, n = 1.
For the induction step:
Let k ≥ 1. Assume that your conjecture is true for n = k. From that, show (prove) that your conjecture is true for n = k+1 .
So you need to show that $1+3+5+\dots+(2k-1)+(2(k+1)-1)=(k+1)^2$ is true,
starting with $1+3+5+\dots+(2k-1)=k^2\ .$
Last edited: Jun 6, 2012
8. Jun 6, 2012
### azizlwl
As you see here, (n+1)th value is not equal to 2n
It is equal to (2(n+1)-1)
9. Jun 6, 2012
### sacscale
Substituting n=k+1 into 2n-1 gives 2(k+1)-1.
10. Jun 6, 2012
Doing what you told me I've got this result:
$1+3+5+...+(2n-1)=n^2$.
We assume n=k. Then we try to prove if the conjecture is true for n=k+1 and we obtain: $1+3+5+...+(2n-1)+(2n-1)=n^2$
But n=k+1
$k^2+(2(k+1)-1)=(k+1)^2$
$k^2+(2(k+1)-1)=k^2+2k+1$
$2(k+1)-1=2k+1$
$2(k+1)=2k+2$
$2(k+1)=2(k+1)$
So I proved that the conjecture is also true for n=k+1. Right?
11. Jun 6, 2012
### Villyer
Isn't it bad practice to manipulate both sides of an equation in a problem such as this?
12. Jun 6, 2012
### SammyS
Staff Emeritus
Yes, somewhat indirectly.
Rather you should do something like the following.
Assume $1+3+5+\dots+(2k-1)=k^2\ .$
Now consider:
$1+3+5+\dots+(2k-1)+(2(k+1)-1)$
$=k^2+(2(k+1)-1)$ ... because of or assumption.
$=k^2+2k+1$
$=(k+1)^2$ Which is what we needed to show for the inductive step.
It's not that what you did is particularly wrong, it's just that it's not real clear that you're not somehow assuming the very thing you should be proving.
Last edited: Jun 6, 2012
13. Jun 6, 2012
### Muphrid
It's generally frowned upon to work both sides at once, yeah. Usually, one would take $k^2 + 2(k+1)-1$ and manipulate it until it's clear the result is $(k+1)^2$. Working only in one direction ensures that no illegal operations or cancellations are done, even though it may be necessary to set both sides equal just to figure out how to do it.
14. Jun 6, 2012
Well, I think I got it now! I have a list of mathematical induction to do. I'll try to solve the exercises using what you guys told me. I'll probably be back soon. Thanks guys!
15. Jun 9, 2012
### amrah
Prove by mathematical induction that n^3-n is divisible by 2 for all positive integral values of n. | 2017-10-17T09:02:00 | {
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http://esy-magnesy.pl/cvf3z54/archive.php?e72729=properties-of-matrix-multiplication-proof | Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. MATRIX MULTIPLICATION. Example. The number of rows and columns of a matrix are known as its dimensions, which is given by m x n where m and n represent the number of rows and columns respectively. Zero matrix on multiplication If AB = O, then A ≠ O, B ≠ O is possible 3. A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of transpose Notice that these properties hold only when the size of matrices are such that the products are defined. The first element of row one is occupied by the number 1 … In the next subsection, we will state and prove the relevant theorems. While certain “natural” properties of multiplication do not hold, many more do. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. Even though matrix multiplication is not commutative, it is associative in the following sense. Example 1: Verify the associative property of matrix multiplication … proof of properties of trace of a matrix. i.e., (AT) ij = A ji ∀ i,j. Associative law: (AB) C = A (BC) 4. 19 (2) We can have A 2 = 0 even though A ≠ 0. But first, we need a theorem that provides an alternate means of multiplying two matrices. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. The following are other important properties of matrix multiplication. For sums we have. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. Given the matrix D we select any row or column. Equality of matrices The proof of Equation \ref{matrixproperties2} follows the same pattern and is … The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ Proof of Properties: 1. Let us check linearity. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; A matrix is an array of numbers arranged in the form of rows and columns. Selecting row 1 of this matrix will simplify the process because it contains a zero. The basic mathematical operations like addition, subtraction, multiplication and division can be done on matrices. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices Subsection MMEE Matrix Multiplication, Entry-by-Entry. 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2020 properties of matrix multiplication proof | 2021-10-20T03:42:15 | {
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https://math.stackexchange.com/questions/2993698/construct-a-square-given-a-point-and-two-lines | # Construct a square given a point and two lines.
The problem:
One point of the square - $$(2,5)$$, and lines $$x=1$$, $$x=6$$ are given (1 point of the square is on each given line), all points of the square are in the first quadrant.
A square needs to be constructed with this information.
How would you solve this problem and thus determine if a square can even be constructed?
I'm interested more in your thought process than the solution.
How do you approach a problem like this?
Which steps do you take and why?
I've considered equating the distances between points (2,5) and (1, a) with the distance between (2,5) and (6,b).I don't think that anything can be done with this, maybe I have to find something else and make a system of equations?I've tried to find the distance between points (1, a) and (6,b) - that would be equal to sqrt(2)*n (n is the length of square's side), connect it with some distance (between the given and unknown point) but I can't extract one variable so I can insert it into the first equation.
• Welcome to Math.SE! Typically, the community here likes to see your thoughts about the problem, as this helps answerers target their responses to your experience level, while avoiding wasting anyone's time telling you things you already know. (It also helps convince people that you aren't simply trying to get them to do your homework for you.) As you are interested more in thought process than solution, perhaps you can say something about your motivation: are you conducting research on problem-solving techniques? trying to be a better problem-solver? (doing homework that asks about process?) – Blue Nov 11 '18 at 12:00
• I've edited the question.I'm interested in general problem solving techniques because I typically solve them by doing everything that can be done which certainly isn't an efficient way of solving problems. – JoeDough Nov 11 '18 at 12:12
• Consider you found the square and draw the two horizontal lines from the vertexes not laying on the vertical lines you already have, the intersections of the four lines will give another figure, that can give many insights. – N74 Nov 11 '18 at 13:07
Let $$A$$ be the given point, and let $$B$$ and $$C$$ be points on the respective given lines. The key observation is
Because $$\overline{AB}$$ and $$\overline{AC}$$ are congruent and perpendicular, the horizontal distance between $$A$$ and $$B$$ must equal the vertical distance between $$A$$ and $$C$$, and vice-versa.
The image shows that $$\triangle ABP \cong \triangle ACQ$$, proving the observation immediately. The image also shows that there are clearly two matching choices for $$B$$ and $$C$$, so that there are two solution squares.
The idea also works when $$A$$ is not between the lines (note that how the $$B$$ and $$C$$ points match is reversed from above):
(Of course, $$A$$ on a line works, too, as an obvious special case where $$C$$ and $$C^\prime$$ coincide.) So, we see that it's always possible to construct the squares. $$\square$$
We leave as an exercise to the reader the task of finding the coordinates of the points for the specific problem in the original question.
As for thought process ...
• My first instinct is usually to abandon specific numbers and to generalize; this prevents important algebraic and geometric patterns from being lost in the arithmetic of numbers. So, instead of $$A=(2,5)$$, $$x=1$$, $$x=6$$, I wanted to consider $$A=(p,q)$$, $$x=r$$, $$x=s$$.
• These days, I have a bit of a knee-jerk impulse to jump into Mathematica to try a brute-force solution. I quickly assigned some coordinates ---$$A=(p,q)$$, $$B=(r,b)$$, $$C=(s,c)$$--- and entered conditions for making the squares: $$\overline{AB}\perp\overline{AC}$$ becomes $$(A-B)\cdot(A-C)= 0$$, while $$\overline{AB}\cong\overline{AC}$$ becomes $$(A-B)\cdot(A-B)=(A-C)\cdot(A-C)$$.
• I let Mathematica do some instantaneous symbol-crunching, which in this case yielded linear constraints on the $$y$$-coordinates of $$b$$ and $$c$$. That told me the problem was actually easy, so I took a step back. The "key observation" above came to me pretty quickly ... shaming me because I didn't think of it sooner. :) (But, hey ... I only lost about two minutes!)
• Then I went to the GeoGebra app to draw-up some diagrams that make the "key observation" obvious, and double-checking the point-outside-the-lines case. (I ofttimes go right into GeoGebra to experiment, but a first pass through Mathematica seemed quicker in this case.)
Hint:
Two parallel lines are: $$x=1$$ and $$x=6$$
distance between them is $$5$$ units.
So, you can construct a square having $$(1,5)$$ as one co-ordinate, and another lying on line $$x=6$$, opposite to $$(1,5)$$ ,i.e., $$(6,5)$$
Take other $$2$$ points on lines $$x=1$$ and $$x=6$$ at a distance of $$5$$ units from $$2$$ points already taken.
So, there are $$2$$ possible squares: $$(1,5),(6,5),(1,0),(6,0)$$ and $$(1,5),(6,5),(1,10),(6,10)$$
• I've made a mistake, first line isn't $x=2$ but $x=1$ – JoeDough Nov 11 '18 at 11:10
• no problem, just shift the points, and now length of side will be 5 – pooja somani Nov 11 '18 at 11:11
• @JoeDough is your point not (1,5) now? – pooja somani Nov 11 '18 at 11:16
• But where's the point (2,5)?It's the only given point of the square, other 2 are somewhere on the given lines and 4th one has to be found.I've edited the question to be more clear. – JoeDough Nov 11 '18 at 11:17
• ohk, then you will have to plot all points and proceed accordingly – pooja somani Nov 11 '18 at 11:18 | 2020-02-17T16:29:31 | {
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# If √(xy) = xy, what is the value of x + y?
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If root(xy) = xy what is the value of x + y? [#permalink]
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If $$\sqrt{xy} = xy$$ what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out, and when we should use cancelling technique.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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11 Jul 2012, 14:22
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If $$\sqrt{xy} = xy$$ what is the value of x + y?
$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.
(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.
(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.
As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.
Hope it's clear.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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11 Jul 2012, 22:57
Responding to a pm:
"If $$\sqrt{XY} = XY$$ what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out and when we should use cancelling technique."
For the time being, forget this question. Look at another one.
Which values of x satisfy this equation: $$x^2 = x$$
Let's say we cancel out x from both sides. What do we get? x = 1.
So we get that x can take the value 1.
But is your answer complete? I look at the equation and I say, 'x can also take the value 0.' Am I wrong? No.
x = 0 also satisfies your equation. So why didn't you get it using algebra? It is because you canceled x.
Let me treat this equation differently now.
$$x^2 = x$$
$$x^2 - x = 0$$
$$x * (x - 1) = 0$$
x = 0 OR (x - 1) = 0 i.e. x = 1
Now I get both the possible values that x can take. I do not lose a solution.
When you cancel off a variable from both sides of the equation, you lose a solution so you should not do that. You can cancel off constants of course.
Rule of thumb: Do not cancel off variables. Take them common. In some cases, it may not matter even if you do cancel off but either ways, your answer will not be incorrect of you don't cancel. On the other hand, sometimes, your solution could be incomplete if you do cancel and that's a problem.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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12 Jul 2012, 03:00
Here is my solution
Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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12 Jul 2012, 03:06
RamakantPareek wrote:
Here is my solution
Take square of both sides:
(root(xy) )^2 = (xy)^2
xy = (xy)^2
With Option (1): x = -1/2
-1/2y = 1/4 * y^2
y = 4 * (-1/2)
y = -2
Hence, X+Y = -1/2 - 2 = -5/2.
There is another value of $$y$$ apart -2 which satisfies $$-\frac{1}{2}*y=\frac{1}{4}*y^2$$, namely $$y=0$$.
Complete solution here: if-root-xy-xy-what-is-the-value-of-x-y-135646.html#p1103625
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If √(xy) = xy, what is the value of x + y? [#permalink]
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15 Oct 2014, 16:01
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If √(xy) = xy, what is the value of x + y? [#permalink]
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15 Oct 2014, 20:57
1
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
Statement 1: $$X=-1/2$$.... substitute in main equation
Scenario A: $$Y = -2/1$$ is a reciprocal
$$\sqrt{xy}=xy$$
LHS =$$\sqrt{-1/2*-2/1}= \sqrt{1}= 1$$
RHS =$$-1/2*-2/1= -1*-1 = 1$$
Therefore, LHS = RHS
Scenario B: $$Y = 0$$
LHS =$$\sqrt{-1/2*-0}= \sqrt{0}= 0$$
RHS =$$-1/2*0= 0$$
Therefore again, LHS = RHS
Hence, we have two possible solutions, therefore Statement 1 is insufficient
Statement 2 : $$Y$$ is not equal to zero
But, X could be equal to zero or be the reciprocal of Y, therefore, statement 2 would fall under the same scenarios as statement 1
Hence, we have two possible solutions, therefore Statement 2 is insufficient
Both Statement 1 & 2 together confirms that $$X & Y$$ both are not equal to zero, therefore, they have to be reciprocals and since statement gives us the value of $$X=-1/2$$, we can also the value of $$Y=-2/1$$ (Reciprocal of X) and thereafter we can find the value of $$X + Y = -1/2 + -2/1 = -5/2$$
Hence, both together are sufficient, therefore C is the correct Answer!
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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15 Oct 2014, 21:26
1
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
I did something similar to DMMK, except for one thing:
I know $$\sqrt{n}$$ = $$n$$ only when $$n$$ = 0 or $$n$$ = 1. For all other values of $$n$$, the two would not be equal. Knowing this made it much simpler to plug in the statements. I just had to determine if from the statement(s) provided, can I rule out xy = 0 or xy=1.
Stament 1:
Knowing that x = -1/2, y could equal 0 or -2, and still make the premise true. Either value of y would make x + y a different value. Therefore, it is insufficient.
Statement 2:
Knowing that y is not equal to 0, x could equal zero or the inverse of y, and still make the premise true. Either value of x would make x + y a different value. Therefore, it is insufficient.
Now to evaluate both statements together.
The reason we rejected statement 1 by itself was because y could equal one of two possible values. Statement 2 eliminates one of those options. Therefore y must equal -2. And therefore both statements together are sufficient to answer "what is the value of x+y."
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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15 Oct 2014, 22:12
1
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
let xy=k, thus
k^(1/2) =k;
squaring both sides we have
k=k^2
k(k-1)=0
k=0 or 1
i.e. xy=0 or xy=1
st.1
for x=-1/2
both y=0 and y=-2 satisfy the possible value of xy i.e. 0 or 1
hence not sufficient
st.2
y is not equal to zero. clearly not sufficient.
as nothing is said about x, therefore x can take any value it can be zero, fraction, integer etc.
combining st.1 and st.2
we know that y cannot be equal to zero. thus y=-2 and x=-1/2
and x+y= -2-(1/2)=-5/2
hence sufficient.
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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04 Jul 2015, 08:43
Why can't x and y both be -1? Thanks
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Re: If root(xy) = xy what is the value of x + y? [#permalink]
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05 Jul 2015, 08:19
ElCorazon wrote:
Why can't x and y both be -1? Thanks
From the stem x = y = -1 is possible, in this case xy = 1 (check my solution above). But then the first statement says that x = -1/2, thus these values are no longer possible.
Hope it's clear.
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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16 Nov 2015, 03:02
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
Given: √(xy) = xy
(xy)^2 - xy = 0
xy = 0 or 1 - (i)
Required: x + y = ?
Statement 1: x = -1/2
From the given equation (i), we can have two different values of y.
Hence two different values of x + y
INSUFFICIENT
Statement 2: y is not equal to 0
No information about x
INSUFFICIENT
Combining Statement 1 and Statement 2: x = -1/2 and y is not equal to 0
From (i), xy cannot be 0 since both x and y are not = 0
Hence xy = 1
y = -2
x + y = $$-\frac{5}{2}$$
SUFFICIENT
Option C
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If √(xy) = xy, what is the value of x + y? [#permalink]
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14 Apr 2017, 07:13
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
The trick here is from $$\sqrt{xy}=xy$$ anyone could easily deduce that $$xy=1$$ or $$xy=0$$ and just consider one of these cases.
Here is my solution:
$$\sqrt{xy}=xy \implies \sqrt{xy}-xy=0 \implies \sqrt{xy}(\sqrt{xy}-1)=0$$
Hence we have $$xy=0$$ or $$xy=1$$.
(1) If $$x=-\frac{1}{2}$$, we need to consider two cases:
Case 1: If $$xy=0\implies y=0 \implies x+y = -\frac{1}{2}$$
Case 2: If $$xy=1 \implies y=-2 \implies x+y = -\frac{5}{2}$$
It's clear that (1) is insufficient.
(2) If $$y \neq 0$$, we still need to consider two cases:
Case 1: If $$xy=0\implies x=0 \implies x+y = y$$.
If $$y=1 \implies x+y = 1$$
If $$y=2 \implies x+y = 2$$
Hence, (2) is insufficient.
Now combine (1) and (2):
Since $$x \neq 0$$ and $$y \neq 0$$ hence $$xy \neq 0 \implies xy=1$$
Since $$x= -\frac{1}{2} \implies y = -2 \implies x+y = -\frac{5}{2}$$
Hence (1) & (2) are sufficient.
The answer is C.
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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10 Sep 2017, 13:41
Bunuel wrote:
Tough and Tricky questions: Algebra.
If √(xy) = xy, what is the value of x + y?
(1) x = -1/2
(2) y is not equal to 0
rewrite
xy= (xy)^2
St 1
Reciprocal property.
-1/2(-2) =1
[-1/2(-2)]^2 =1
insuff
St 2
Eliminating the possibility of 0 from Y still leaves the possibility of x being 0 or some other numbers which leads to several possibilities.
insuff
St 1 and St 2
Eliminates possibility of y=-2
C
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Dec 2017, 00:35
Bunuel wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?
$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.
(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.
(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.
As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.
Hope it's clear.
Is it safe to say : x^2 = x only when x = 0 or 1
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Dec 2017, 00:39
QZ wrote:
Bunuel wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?
$$\sqrt{xy} = xy$$ --> $$xy=x^2y^2$$ --> $$x^2y^2-xy=0$$ --> $$xy(xy-1)=0$$ --> either $$xy=0$$ or $$xy=1$$.
(1) x = -1/2 --> either $$-\frac{1}{2}*y=0$$ --> $$y=0$$ and $$x+y=-\frac{1}{2}$$ OR $$-\frac{1}{2}*y=1$$ --> $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Not sufficient.
(2) y is not equal to zero. Clearly not sufficient.
(1)+(2) Since from (2) $$y\neq{0}$$, then from (1) $$y=-2$$ and $$x+y=-\frac{5}{2}$$. Sufficient.
As for your solution: you cannot divide by $$xy$$ since $$xy$$ could equal to zero and division by zero is not allowed.
Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. So, if you divide (reduce) by $$xy$$ you assume, with no ground for it, that $$xy$$ does not equal to zero thus exclude a possible solution.
Hope it's clear.
Is it safe to say : x^2 = x only when x = 0 or 1
Yes.
x^2 = x;
x^2 -x = 0;
x(x - 1) = 0;
x = 0 or x = 1.
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Feb 2018, 10:18
Hi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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27 Feb 2018, 10:27
1
krikre wrote:
Hi, why aren't we accounting for option where both x and y are equal to 1? then x+y is 2
Doesn't (1) say that x = -1/2? How it can be 1 then?
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Re: If √(xy) = xy, what is the value of x + y? [#permalink]
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28 Feb 2018, 01:19
prakash111687 wrote:
If $$\sqrt{xy} = xy$$ what is the value of x + y?
(1) x = -1/2
(2) y is not equal to zero
What i did was,
(XY)^1/2 = XY
XY =(XY)^2
so, I cancelled out XY and finally I got the below rephrased equation
XY=1.
there are various cases that are possible as following
BUT in the MGMAT explanation, I found that they are not cancelling out XY.
Below is their rephrased equation.
XY = (XY)^2
XY-(XY)^2=0
XY [1-(XY)] = 0
so, XY = 0 or XY = 1.
My question is why we are not cancelling out, and when we should use cancelling technique.
IMO C
$$\sqrt{xy} = xy$$
1.x=0 ,y = anything
2.Y=0 , x=anything
3. both x and y 1 or -1 or x=1 and y=-1 or vise versa
4.x=1/z and y=z *any integer except 0
lets come to statement 1.
x=-1/2 y can be -2 or 0 insufficient.
2nd statement y not equal to 0 but don't know about x
combining both case with 0 is eliminated only one case left so C is the answer
Re: If √(xy) = xy, what is the value of x + y? &nbs [#permalink] 28 Feb 2018, 01:19
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# If √(xy) = xy, what is the value of x + y?
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-09-19T01:37:45 | {
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https://mathematica.stackexchange.com/questions/184374/why-is-sum-limits-k-1100-0-01-different-than-sum-limits-k-1100 | # Why is $\sum\limits_{k=1}^{100} 0.01$ different than $\sum\limits_{k=1}^{100} \frac{1}{100}$?
Why does evalutating $$\sum\limits_{k=1}^{100} 0.01$$ not result in 1?
What I get is 1.0000000000000007.
I know that the number 0.01 in binary results in an infinite sequence of zeros and ones, which is truncated when stored in a memory as a floating point number. So information is lost and the result of the sum will no longer be exactly 1.
However, with $$\sum\limits_{k=1}^{100} \frac{1}{100}$$, I do get exactly 1.
Could someone explain why?
• For your own edification, compare the results of Total[Table[0.01, {100}]] and Total[Table[0.01, {100}], Method -> "CompensatedSummation"]. – J. M.'s torpor Oct 22 '18 at 6:21
• @J.M.iscomputer-less Interestingly Total seems to perform much better than Sum, even without "CompensatedSummation". It is somehow more intelligent. Consider ListLinePlot[{Table[Sum[0.01, {i, n}] - n/100, {n, 1, 100}], Table[Total[Table[0.01, {n}]] - n/100, {n, 1, 100}]}] – kirma Oct 22 '18 at 9:03
Welcome to the world of Mathematica numerics.
Mathematica offers three arithmetic systems for evaluating numerical expressions
• Numerics based on exact numbers. Such computations will always give an exact result, but require all the numbers involved to be integers or rational numbers. Your second expression qualifies as an exact computation and, thus, produces the exact result 1.
• Numerics based on arbitrary precision numbers. This allows you the precision a computation to specified value. To work, such computations require the numbers involved to be integers, rational numbers or arbitrary precision having a precision greater or equal to the specified precision.
• Numerics based on machine floating point numbers. This is used whenever a real number, which does not qualify as an arbitrary precision number, appears in a computation. Machine numerics give the fastest computation, but since there is no control over precision. For long computations precision will be lost, sometimes drastically (leading to zero precision – i.e., garbage out). Your first expression falls into this category, but retains good precision. However, when you display all the digits in the result the inexactness shows.
In you specific example, since the deviation from the exact result is small, you can recover the exact result with Rationalize
Example
Sum[.01, {100}] // FullForm
1.0000000000000007
Rationalized example
Sum[.01, {100}] // Rationalize // FullForm
1
Mathematica treats 0.01 as a machine precision floating point number, while it treats 1/100` as an exact rational quantity. Thus the first sum follows all the grody rules of floating point arithmetic that programmers have come to know and hate.
There are ways to make Mathematica use more digits, though. | 2021-07-31T08:44:05 | {
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https://math.stackexchange.com/questions/2499649/proof-of-obtaining-multiple-of-10-by-using-arithmetic-operations-on-any-three-nu | Proof of obtaining multiple of 10 by using arithmetic operations on any three numbers
While doing a few math puzzles I noted that you could take any three numbers (integers) and use basic arithmetic operations (Addition, Subtraction, Multiplication and Parentheses only) to obtain a multiple of 10, using each number only once.
Eg.
Using 29, 73, 36: $73+36-29=80$
Using 2, 4, 7: $2*7-4=10$
Is there a proof for this theory in particular, or is there a more general theorem that applies to this? If not, is there a set of three numbers which disproves this theory?
I found a few clues while solving this:
If one of the numbers is a multiple of 5 (call it $x$) then there definitely exists a solution divisible by 10. This one is easily provable as you merely need an even number to multiply with. If any one of the remaining two numbers is even, you can use that to multiply with $x$ to get number divisible by 10. If both numbers are odd then you can add them to get an even number to multiply with $x$.
It would seem as only the units place digits have any bearing on whether it's a multiple of 10 or not. As you could take any of the examples, add or subtract any multiple of 10 from it and do the same operations to still get a multiple of ten. I feel this is also easily provable but I can't think of a way to do it which isn't long-winded.
EDIT: It can be proven as such, let $a,b$ be two integers s.t $a+b|10$
Adding two multiples of 10 ($10m,10n$) to $a$ and $b$ we get $10m+a+10n+b$
Since $a+b|10$ it can be written as $10m+10n+10k$ where $a+b=10k$
Which is divisible by ten
Now let $a,b$ be two integers s.t. $a.b|10$
Adding $10m,10n$ to $a$ and $b$ we get $(10m+a)(10n+b)$
$=100mn+10an+10bm+ab$ or $100mn+10an+10bm+10k$
Which is divisible by ten
Therefore any combination of Addition and Multiplication with the three no.s shouldn't matter.
I realise that could replace 10 with any no. in this proof and it would still work. So we really need to just prove for every single digit no.
I ran a program in python that checked for every combination of single digit no.s, but it didn't find any combination that disproves this.
I'm not sure how this question would be categorised, hence the lack lustre tags.
I'm fairly new to StackExchange. Please forgive me if I've worded this question poorly.
• That's a very nice observation that only the units digit has any bearing, and is essentially the underlying idea of "modular arithmetic." This is an interesting problem, and I'll be fairly surprised if it's true. However, there are $3$ ways to order $a, b, c$, a couple choices of symbols between $a,b$ and $b,c$ and with the ability to use parentheses... very interesting! If we have the three numbers $a, b, c$, are we allowed to order them any way we like? – pjs36 Nov 5 '17 at 6:47
• How do you get a multiple of $10$ this way using $1$, $2$, and $4$? – alex.jordan Nov 5 '17 at 7:58
• @alex.jordan $2\cdot (4+1)$ – Kyle Miller Nov 5 '17 at 7:59
• Oh, I see. I somehow thought only addition and subtraction were in play. Didn't read well. – alex.jordan Nov 5 '17 at 8:00
• @pjs36 From the two examples in the OP, it seems clear that reordering is permitted. – Erick Wong Nov 5 '17 at 8:07
Let's collect some of the useful observations mentioned and implicit in the OP:
1. Only the residue class of $a,b,c$ mod $10$ matters.
2. If any of $a,b,c$ is divisible by $5$ then there is a solution.
3. If any subset of $\{a,b,c\}$ has a solution, then so does $\{a,b,c\}$.
As a consequence of the above, we may assume WLOG that $\{a,b,c\}$ are distinct single digits from $1,2,3,4,6,7,8,9$.
With a little more effort, we can justify that $a$ can be replaced by $-a$, which will let us further restrict to the digits $1,2,3,4$. It's not too hard to be convinced of this by playing around with some examples, but let's give a proper proof by structural induction. More specifically, we will prove:
Proposition: Let $f(x_1,\ldots,x_n)$ be a fully parenthesized expression using each $x_i$ exactly once and only $+,-,*$ operators. Then at least one of $-f$ or $f$ can be written as a similar expression using $-x_1,x_2,\ldots,x_n$ exactly once.
This is trivial for $n=1$, and for $n=2$ we quickly verify each of the four possible expressions of two variables: $$-(a+b) = (-a)-b,\\-(a-b) = (-a)+b,\\(b-a) = b+(-a),\\-(a*b) = (-a)*b.$$
We can now do a structural induction: suppose $n\ge 3$. By extracting the highest-level operator of $f(x_1,\ldots, x_n)$, we may write $f$ as $h(g_1(\cdots),g_2(\cdots))$, where each of $g_1,g_2,h$ are valid expressions, and the arguments of $g_1, g_2$ partition the set $\{x_1,\ldots,x_n\}$. I've deliberately obscured the arguments because the notation becomes unwieldy, as the idea is better illustrated by a simple example:
If $f(x_1,x_2,x_3,x_4,x_5) = ((x_1 * x_5) + x_2) - (x_3 - x_4)$, then we take $$h(a,b) = a-b,\quad g_1 = (x_1 * x_5) + x_2,\quad g_2 = x_3 - x_4.$$
Note that $h,g_1,g_2$ each take $<n$ arguments so we could apply the inductive hypothesis to any of them as desired. Now, $x_1$ belongs to exactly one of $g_1$ or $g_2$. By adjusting $h$, we can assume WLOG that it belongs to $g_1$. By hypothesis, either $-g_1$ or $g_1$ can be written in terms of $-x_1$ and the remaining arguments of $g_1$. If it is $g_1$ that may be so written, then we are done since $f=h(g_1,g_2)$ can be written in terms of $-x_1,x_2,\ldots,x_n$. Else, apply the induction hypothesis again to $h$ to see that either $f=h(g_1,g_2)$ or $-f=-h(g_1,g_2)$ can be written in terms of $-g_1$ and $g_2$, and thus also in terms of $-x_1,x_2,\ldots,x_n$.
The above proposition lets us freely replace $9$ by $-9$ (which is equivalent to $1$), etc., and so we can narrow down the set $\{a,b,c\}$ to a subset of $\{1,2,3,4\}$. Lucky for us, there's only four such subsets:
$$(1 + 2 - 3) = 0,\\ (1 + 4)*2 = 10,\\ (1 + 3 -4) = 0,\\ (2+3)*4 = 20.$$
• I was going to write the same thing. I believe this one deserves the bounty. – user334639 Nov 5 '17 at 9:00
• Very nicely done, I probably will award the bounty to this answer (I want to keep the bounty alive to reward the nice question; OP had but a single upvote when I set the bounty). Structural induction isn't something I'm familiar with, so I will admit that it's my least favorite part. I notice that in each of the four ultimate cases, addition is used -- I wonder if we could somehow combine that with factoring out a negative, and avoid structural induction. Probably not. At any rate, very nice! – pjs36 Nov 6 '17 at 2:36
• Ah I see you've used $-4,-3,-2,-1,1,2,3,4$ instead of $1,2,3,4,6,7,8,9$ and then used my modulo(10) proof.First time I'm coming across structural induction so the proof is a bit hard to chew. Nevertheless, it's a clever proof. – Carlton Banks Nov 6 '17 at 10:02
• @CarltonBanks Thanks, to be fair this is really just an induction on the number of terms (it’s structural in the sense that arithmetic expressions aren’t built up by just “adding one more term”). It is also overkill when we’re dealing with only 3 terms, but you did express an interest in more general principles in the OP. Finally, the induction is only needed for a relatively minor technical point: it justifies that if we can find a $0$ using unary $-$ (which doesn’t appear to be allowed), then we can also find one without any unary $-$. – Erick Wong Nov 6 '17 at 18:24
If you have directly verified this for all triples of one-digit numbers, then you have proved it. Because you only care about an arithmetic combination that makes $0$ mod $10$, so proving it for one-digit numbers proves the greater claim.
Without a direct verification, consider all $10^3$ triples of one-digit numbers. If $0$ is in the triple, then multiplying all three is a multiple of $10$.
So now consider all $9^3$ triples from 1--9. If $5$ is in the triple, then either both of the other numbers are odd and you can multiply $5(\text{odd}+\text{odd})$ to get a multiple of $10$; or at least one of the others is even and you can multiply all three to get a multiple of $10$.
So now consider all $8^3$ triples from 1--4,6--9. If any number appear twice in the triple, the difference is $0$, and that difference times the third number is a multiple of $10$.
So now consider all $\binom{8}{3}$ triples from 1--4,6--9 without repetition. If a number and its complement mod $10$ are both in the triple, then sum those and multiply by the third to get a multiple of $10$.
So now consider all $\binom{4}{3}\cdot2^3$ triples from 1--4,6--9 without repetition where no two numbers sum to $10$. We are down to only $32$ triples to consider, and inspecting them directly is not so bad. First, consider the $16$ triples with two or three members from 1--4.
$$\underline{12}3\ \color{magenta}{\underline{1}2\underline{4}}\ \color{orange}{\underline{1}2\underline{6}}\ \color{blue}{127}\ \underline{13}4\ \color{blue}{136}\ 1\underline{38}\ 1\underline{47}\ \color{magenta}{\underline{14}8}\ \color{magenta}{\underline{23}4}\ \color{magenta}{\underline{23}6}\ 2\underline{39}\ \color{orange}{\underline{2}4\underline{7}}\ \color{orange}{2\underline{49}}\ \color{orange}{\underline{3}4\underline{8}}\ 3\underline{49}$$
Blue sum to $0$ mod $10$.
Magenta have an (underlined) pair that sum to $5$ (or $15$) and the third number is even, so that sum times the third number is a multiple of $10$.
Orange have an (underlined) pair whose difference is $5$ and the third number is even, so that difference times the third number is a multiple of $10$.
The remaining black all have an (underlined) pair that sums to the third (mod $10$) so adding that pair and then subtracting the third makes a multiple of $10$.
Note that if we negate all members of one of these triples, we get the triples with two or more in 6--9. The same operations give a multiple of $10$ since we only ever either add and subtract [with the blue and black], or add/subtract and then multiply by an even number [with magenta and orange]. So we've indirectly checked all $32$ of the lingering cases.
• I was hoping to avoid "too many" cases, but this is a pretty modest number of them. And inspecting the final 32 wasn't so bad, at least when they were color-coded :) Would that I could split a bounty, I like all of the answers... – pjs36 Nov 6 '17 at 2:40
• @pjs36 Just FYI, your comment somewhere else got me thinking about how the cases could be cut down to just $16$. – alex.jordan Nov 6 '17 at 3:04
Note: This answer has the focus to reduce the number of variants which are to study by consequently using modular arithmetic.
• $a,b,c\in\mathbb{Z}$
We are looking for multiples of $10$ when doing addition, subtraction and multiplication. Since we have \begin{align*} (a\, \mathrm{mod}\, (10)) + (b\, \mathrm{mod}\, (10)) &\equiv (a+b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) - (b\, \mathrm{mod}\, (10)) &\equiv (a-b)\, \mathrm{mod}\, (10) \\ (a\, \mathrm{mod}\, (10)) \cdot (b\, \mathrm{mod}\, (10)) &\equiv (a\cdot b)\, \mathrm{mod}\, (10) \\ \end{align*}
it is sufficient to consider $\color{blue}{a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}}$.
• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\}$
If two of the three numbers are congruent modulo $10$, let's say $a\equiv b\, \mathrm{mod}\, (10)$ we obtain \begin{align*} \color{blue}{(a-b)\cdot c}\equiv 0\cdot c\color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
and we are done. In the following we may WLOG assume: $a<b<c$
• $a,b,c\in\{0,1,2,3,4,5,6,7,8,9\},a<b<c$
If one of them, let's say $a$ is zero we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot b\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
and we are done.
• $a,b,c\in\{1,2,3,4,5,6,7,8,9\}, a<b<c$
If one of them, let's say $a$ is equal to $5$ we consider two cases.
First case: One of $b$ or $c$ is even. Let's assume $b$ is even. It follows \begin{align*} a&\equiv 0\, \mathrm{mod}\, (5)\\ b&\equiv 0\, \mathrm{mod}\, (2)\\ ab&\equiv 0\, \mathrm{mod}\, (10)\\ \end{align*} and we obtain \begin{align*} \color{blue}{a\cdot b\cdot c}&\equiv 0\cdot c \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
Second case: Both, $b$ and $c$ are odd. It follows from
\begin{align*} b\equiv 1\, \mathrm{mod}\, (2)\\ c\equiv 1\, \mathrm{mod}\, (2)\\ (b+c)\equiv 0\, \mathrm{mod}\, (2)\\ \end{align*}
and we obtain \begin{align*} \color{blue}{a\cdot (b+c)} \color{blue}{\equiv 0\, \mathrm{mod}\, (10)} \end{align*}
• $a,b,c\in\{1,2,3,4,6,7,8,9\}, a<b<c$
Since
\begin{align*} 1\equiv (1-10)\equiv (-9)\, \mathrm{mod}\, (10)\\ 2\equiv (2-10)\equiv (-8)\, \mathrm{mod}\, (10)\\ 3\equiv (3-10)\equiv (-7)\, \mathrm{mod}\, (10)\\ 4\equiv (4-10)\equiv (-6)\, \mathrm{mod}\, (10)\\ \end{align*}
we can restrict the attention to $\color{blue}{a,b,c\in\{1,2,3,4\},a<b<c}$.
• $a,b,c\in\{1,2,3,4\}, a<b<c$
Conclusion: Doing arithmetic with three integers $a,b,c\in\mathbb{Z}$, each of them occurring once and using one or more of the operations addition, subtraction, multiplication and negation ($a \rightarrow -a$) we can always obtain an integer value which is a multiple of $10$.
• This is a nice, thorough answer. Allowing negation is an interesting point. While not explicitly mentioned, I feel like there should be some algebraic trickery that lets us convert negation to addition, subtraction, and multiplication with our same three numbers; e.g., $(-b + a) \cdot c = -(b - a)\cdot c$. Maybe, but maybe it would just serve to introduce more cases, even though it's intuitively clear that negation doesn't really add anything new. – pjs36 Nov 6 '17 at 2:48
• @pjs36: Thanks. :-) Allowing negation is a modest way of cheating somewhat. In fact negation $-a=0-a$ is equivalent to add 0 to $\{+,-,\cdot,(,)\}$. To me it is not so clear that it doesn't introduce anything new. Keep in mind that $-(a-b)$ for instance also uses negation. – Markus Scheuer Nov 6 '17 at 7:16
• @pjs36 I agree, this is where I spend most of the effort in my answer and I would love to see a simpler way to prove that negation can always be eliminated :). – Erick Wong Nov 6 '17 at 18:27
• @MarkusScheuer It’s fine that $-(a-b)$ uses negation: the point of my argument is that all interior negations can be bubbles up to the exterior. Once we have that, then $a-b$ works just as well as $-(a-b)$ for producing zeroes, so we eliminate negation altogether. – Erick Wong Nov 6 '17 at 18:30
• @ErickWong: Ahh, now I see your point! Good argument. Your're right, of course. :-) (+1) – Markus Scheuer Nov 7 '17 at 17:59 | 2020-08-15T18:37:04 | {
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https://math.stackexchange.com/questions/344714/non-commutative-or-commutative-ring-or-subring-with-x2-0 | Non-commutative or commutative ring or subring with $x^2 = 0$
Does there exist a non-commutative or commutative ring or subring $R$ with $x \cdot x = 0$ where $0$ is the zero element of $R$, $\cdot$ is multiplication secondary binary operation, and $x$ is not zero element, and excluding the case where addition (abelian group operation) and multiplication of two numbers always become zero?
Edit: most seem to be focused on non-commutative case. What about commutative case?
Edit 2: It is fine to relax the restriction to the following: there exists countably (and possibly uncountable) infinite number of $x$'s in $R$ that satisfy $x \cdot x = 0$ (so this means that there may be elements of ring that do not satisfy $x \cdot x =0$) excluding the case where addition and multiplication of two numbers always become zero.
• I forgot to write commutative. So both commutative and non-commutative cases. – user69886 Mar 28 '13 at 13:44
• Regarding your edit: If you want arbitrary elements of this type, just take products or tensor products. So take $k[x_1]/(x_1^2) \times k[x_2]/(x_2^2) \times \dotsc$ or $k[x_1,x_2,\dotsc]/(x_1^2,x_2^2,\dotsc)$. – Martin Brandenburg Mar 28 '13 at 14:29
• @user69886: I updated my answer with a commutative example. In general and for the future, please do not change your question after answers has been given. In general if you have another question, then just post that as a separate question. – Thomas Mar 28 '13 at 14:49
Example for the non-commutative case: $$\pmatrix{0 & 1 \\ 0 & 0} \pmatrix{0 & 1 \\ 0 & 0} = \pmatrix{0 & 0 \\ 0 & 0}.$$ Example for the commutative case: Consider the ring $\mathbb{Z} / 4\mathbb{Z}$. What is $2^2$ in this ring?
• See my answer for a conceptual view of the genesis of this example. – Math Gems Mar 28 '13 at 15:54
Sure, take $k$ any ring and $R=k[x]/(x^2)$. As suggested by Martin Brandenburg, a good example is $k=\mathbb R$, in which case $R$ becomes the ring of dual numbers. The wikipedia site lists a lot of properties, which might be interesting.
• Or almost the same, $\mathbb{Z}/4$. – Martin Brandenburg Mar 28 '13 at 13:40
• It's commutative. Though if we take a noncommutative base ring instead of $k$.. – Berci Mar 28 '13 at 13:41
• @Berci: Whoops, I misread that. – Jesko Hüttenhain Mar 28 '13 at 13:42
• I forgot to write "or commutative.." – user69886 Mar 28 '13 at 13:44
• Jesko, you may perhaps add a link to en.wikipedia.org/wiki/Dual_number – Martin Brandenburg Mar 28 '13 at 13:53
It is easy if you know about quotient rings, e.g. $\rm\:\Bbb Z/n^2,\ \Bbb Z[x]/(x^2).\:$ If you don't know about quotient rings, is there still a simple example? Yes! One quickly checks that $\rm\,\Bbb Z^2$ is a ring under pointwise addition and multiplication and, that the set of linear maps on $\rm\,\Bbb Z^2\,$ form a ring under addition and map composition. Now note that the (right) shift map $\rm\: S(a,b) = (0,a)\:$ is linear, and $\rm\,S^2 = 0\,$ since $\rm\: S^2(a,b) = S(0,a) = (0,0).\:$
Remark If you know matrix rings, note that the matrix of the shift map is that in Thomas's answer. But knowledge of matrices is not required to understand the above simple example.
Where does this example come from? It arises from a general construction. It is, in fact, a linear representation of the generic example $\rm\:R[x]/(x^2),\:$ whose additive group is $\rm\:R^2,\:$ where $\rm\:x\:$ acts as a (right) shift map $\rm\:x(a+bx) = 0+ax,\:$ i.e. $\rm\:(a,b)\mapsto (0,a).\:$ Similarly any ring $\rm\,R\,$ has such a linear representation as a subring of the ring of linear maps on its underling additive group - the so-called (left) regular representation of $\rm\,R.\:$ This is a ring-theoretic analogue of Cayley's theorem, that a group $\rm\,G\,$ may be represented as a subgroup of the group of bijections (permutations) on $\rm\,G.\:$ In both cases the representation arises by sending each element $\rm\:r\:$ into its associated (left) multiplication map $\rm\: x\:\to\:r\:x\:.\:$ See this MO question for further discussion.
As another important example, this view provides a natural way to construct the polynomial ring $\rm\,R[x]\,$ as a subring of the ring of linear maps on $\rm\,R^\Bbb N = (f_0,f_1,f_2,\ldots)\:$ generated by $\rm\,R\,$ and $\rm\,x\,$ i.e. by all constants $\rm\:(r,0,0,\ldots)$ and by the shift map $\rm\,x : (f_0,f_1,f_2,\ldots)\mapsto (0,f_0,f_1,\ldots).$ See also the presentation in this post.
It is worth emphasis that the ring $\rm\:R[x]/(x^2)\:$ is known as the algebra of dual numbers over $\rm\,R,\,$ and that it frequently proves quite handy in various contexts, e.g. when studying (higher) derivatives, or tangent/jet spaces, and when transferring properties of homomorphisms to derivations
• Actually $S=x$ on $\mathbb{Z}[x]/(x^2)$. So these are the same examples in disguise. – Martin Brandenburg Mar 28 '13 at 14:17
• @martin Yes, I meant to segue into that - see the added remark. – Math Gems Mar 28 '13 at 14:57
Yes, for example consider the noncommutative polynomial ring $\Bbb R\langle x,y\rangle$ (so here $xy\ne yx$), and let $R$ be its quotient by the ideal $(x^2)$ generated by $x^2$.
• Non-commutative polynomial rings are often called free (associative) algebras and denoted by other brackets, namely $R\langle x_1,\dotsc,x_n\rangle$. – Martin Brandenburg Mar 28 '13 at 13:48 | 2021-03-06T12:23:51 | {
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https://math.stackexchange.com/questions/2589445/tossing-a-matchstick-randomly-onto-an-infinite-grid | # Tossing a matchstick randomly onto an infinite grid
I was given this problem recently in a job interview in which I did not succeed, however I found the problem itself very interesting, and have been trying to solve it since.
Question:
Given matchstick of length 1, and an infinite grid of similar matchsticks (i.e. an infinite grid of 1x1 squares), if you toss the matchstick randomly onto the grid, what is the probability that the matchstick lands overlapping at least one of the matchsticks in the grid?
My approach:
I decided to first define 3 variables:
• $x =$ horizontal distance between lower end of landing matchstick and nearest grid matchstick to the left
• $y =$ vertical distance between lower end of landing matchstick and nearest grid matchstick below
• $\theta$ = landing angle of matchstick with grid horizontal
By definition:
• $x, y$ ~ $U(0, 1)$
• $\theta$ ~ $U(0, \pi)$
• $x, y, \theta$ are all independent of each other
I then split the events of overlapping the horizontal and the vertical matchsticks in the grid, such that:
$$P(Overlap) = P(OverlapsVertical) + P(OverlapsHorizontal) - P(OverlapsVertical \bigcup OverlapsHorizontal)$$
• $P(OverlapsVertical) = P(y + sin(\theta) >= 1)$
• $P(OverlapsHorizontal) = P(x + cos(\theta) >= 1)$
For overlapping vertical, because both y and theta are uniformly distributed, I could calcluate $P(OverlapsVertical)$ by:
• Constructing the rectangle of possible values for $y$ and $\theta$
• Integrating to find the proportion of area of this rectangle that satisfied the condition $y + sin(\theta) >= 1$
Doing the same for the horizontal case, I found that $P(OverlapsVertical) = P(OverlapsHorizontal) = 2/\pi$.
However, I realised that the events $OverlapsVertical$ and $OverlapsHorizontal$ are not independent, because they are both dependent on $\theta$, specifically $sin(\theta)$ and $cos(\theta)$ which are inversely related, and therefore:
• $P(OverlapsVertical \bigcup OverlapsHorizontal) \not= P(OverlapsVertical) * P(OverlapsHorizontal)$
I then became stuck, and wondered if my approach to the entire problem was wrong. Is there a better way to solve this problem? Any input greatly appreciated. Thanks!
• – Bram28 Jan 2 '18 at 22:07
• @Bram28 thanks for the quick reply. I see how that applies to the individual events of crossing the vertical and crossing the horizontal (my problem is specifically Buffon's needle where t = l = 1, and hence agrees with my answer 2/π), however as far as I can tell it does not answer the question of crossing either the vertical or the horizontal, where these two events are not independent nor mutually exclusive. – Rory Devitt Jan 2 '18 at 22:20
• Out of curiosity, which position were you applying for ? – Gabriel Romon Jan 2 '18 at 22:24
• @GabrielRomon it was a quantitative trading position at a hedge fund – Rory Devitt Jan 2 '18 at 22:25
• And the interviewer didn't give you a hint or a nudge in some direction ? – Gabriel Romon Jan 2 '18 at 22:31
The if the midpoint lands inside a box of dimensions: $(1-\cos\theta)\times(1-\sin\theta),$ then then the match will fit inside that square.
Due to the symmetry of the grid we only need to analyze $\theta \in [0, \frac {\pi}{4}]$
$\frac 1{\frac {\pi}{4}}\int_0^\frac {\pi}{4} (1-\cos\theta)\cdot(1-\sin\theta)\ d\theta$ | 2019-06-27T06:06:42 | {
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https://www.themathdoctors.org/three-times-larger-idiom-or-error/ | # Three Times Larger: Idiom or Error?
Having just written about issues of wording with regard to percentages, we should look at another wording issue that touches on percentages and several other matters of wording. What does “three times larger” mean? How about “300% more”? We’ll focus on one discussion that involved several of us, and referred back to other answers we’ve given.
## Percent increase vs. factor
The question, from 2006, started with the idea of a percent increase:
Percent Increase and "Increase by a Factor of ..."
A math doctor here recently explained percent increase this way: If we start with 1 apple today and tomorrow have 2 apples, then because 2 - 1 = 1 and 1/1 = 1, we have a 100 percent increase.
But can't I also say there was an increase by a factor of 2? Two divided by 1 equals 2, an increase by a factor of 2 -- and also an increase by 200 percent? This is what is confusing me!
I'd never been confused about saying "increased by a factor of" and "increased by percent of" until I saw the Dr. Math conversation about finding percentages ... which is a good thing, I guess, because now I know what I didn't know! Thank you for any help.
Unfortunately, Joseph didn’t directly quote from the page he had in mind, and we have never said exactly what he said; so we couldn’t be sure which page it was. Everything he said, however, was correct.
Doctor Rick was the first to answer:
Hi, Joseph.
Yes, this can be very confusing, because some statements about increases are ambiguous.
When we say "increased by a factor of 2," the word "factor" makes it clear that we mean "multiplied by 2."
When we say "increased by 10%," there is only one reasonable interpretation: the amount of the increase is 10% of the original amount. If we meant multiplication by 10%, that would be a decrease -- not an increase! Even when we say "increased by 100%," there is only one reasonable interpretation, since multiplication by 100% is the same as multiplication by 1, and that's still not an increase.
When we want to speak of an increase that is greater than the original amount, then ambiguity can arise. In that situation, I much prefer "increase by a factor of 3" or "by a factor of 2.5," etc.
I don't know what page you saw -- but have you seen this one?
Percent Greater Than vs. Increased
http://mathforum.org/library/drmath/view/61774.html
See also the page linked there, about the even more confusing phrase "___ times more than" and the like. I am on the side of avoiding the confusing phrases, as a basic principle of communication.
If you saw another page and you are still confused by it, please tell me the URL of that page so I can review it with you.
## Ambiguity in percent increase
Before we get back to this conversation, we should take a look at the page he referred to, which is a good starting point:
Percent Greater Than vs. Increased
What is the difference between the following statements:
My profits are 200% bigger than they were last year.
and
My profits from last year have increased 200%.
This is one of the questions we have to answer in my Middle school methods course and I have looked everywhere for the answer. I hope you can help.
As far as I can see, they mean the same thing; in fact, both are similarly ambiguous.
Taken literally, "200% bigger" (or, more formally, larger or greater) and "increased 200%" (or, more completely, increased _by_ 200%) both mean that the increase from one year to the next is 200% of the first year's value, so that the second year's profit is 3 times the first. But both statements are more likely to have been made with the intention of saying that this year's profit is twice last years. English is not very clear in cases like this.
So there is a literal meaning (which mathematicians tend to see as best), and an idiomatic meaning (which ordinary people are more likely to have intended to say). I referred to a page we’ll be looking at below, and then quoted a favorite book of mine that gives a lexicographer’s perspective:
Since writing that, I found a good reference on "two times greater," although it doesn't mention your "200% greater." It is in Merriam Webster's _Dictionary of English Usage_, which under "times" writes
The argument in this case is that _times more_
(or _times larger_, _times stronger_, _times
brighter_, etc.) is ambiguous, so that "He has
five times more money than you" can be
misunderstood as meaning "He has six times as
much money as you." It is, in fact, possible
to misunderstand _times more_ in this way, but
it takes a good deal of effort. If you have
$100, five times that is$500, which means
that "five times more than $100" can mean (the commentators claim) "$500 more than $100," which equals "$600," which equals "six times
as much as $100." The commentators regard this as a serious ambiguity, and they advise you to avoid it by always saying "times as much" instead of "times more." Here again, it seems that they are paying homage to mathematics at the expense of language. The fact is that "five times more" and "five times as much" are idiomatic phrases which have - and are understood to have - exactly the same meaning. The "ambiguity" of _times more_ is imaginary: in the world of actual speech and writing, the meaning of _times more_ is clear and unequivocal. It is an idiom that has existed in our language for more than four centuries, and there is no real reason to avoid its use. I think the same applies to "X percent bigger" and "increased [by] X%." There is just enough ambiguity in a technical context that I would want to ask what was intended before assuming anything, but there is no reason to say that they definitely mean different things, or mean something different than "X percent of" or "increased to X percent." I myself would avoid saying these things, just because there are enough people who have heard that they are ambiguous, and would therefore take them the wrong way (whichever that is!). As a result of my side interest in linguistics, I recognize that human speech is not as logical as we might wish; what a word means is a matter of actual usage in a culture, rather than pure logic. So rather than state that either understanding of “times bigger” is “correct”, I just recognize that people take it in two ways, so you have to ask, or use contextual cues, in order to decide on what is meant. ## Confusion about “three times larger” Back to the original discussion: Joseph responded with specific references, the first of which was that link of mine that Doctor Rick said to “see also”: I'm sorry, I should have specified the site. In fact, there were two -- and I still don't see the difference between them. Here is the first example, from Larger Than and As Large As http://mathforum.org/library/drmath/view/52338.html 1) "Three times as large as N" means "3 * N." 2) "Three times larger than N" means "4 * N" -- but only if you stop to think about it, as many people do not. Here, I don't understand how something can be 3 times larger and be 4 times N. That sounds really weird to me. If you asked "What is something that is three times as large as N?" then I would say 3N ... but apparently I'd be wrong! I just don't see where my thinking is wrong. His thinking isn’t wrong on this point: 3N is three times as large as N! He seems just to be letting what he’s read sow doubt about everything. Here is the second example, from Percentage of Increase http://mathforum.org/library/drmath/view/58131.html You can choose two ways to express your answer now. One is to say: there will be a 550% increase by the year 2000. Or you can say: in the year 2000 the (new value) -- you didn't say what the numbers represented, so I'm a little confused right here -- will be about five and a half times greater than what it was in 1995. Many people don't quite grasp those phrases, especially the latter one. Instead you might wish to say it this way: in 2000 the (new value) will be 6 and a half times what it was in 1995. The difference in the wording is subtle, of course, but important. The number 6 1/2 comes from 325,000 --------- = 6.5 or 6 1/2 50,000 which is NOT a percent increase situation. In this problem, I don't understand the difference between the way the doctor explains the two different ways you can talk about the increase, and the implications of each. The doctor says that 6.5 times is not a percent increase; but can you still say it's 650 percent OF the original? I'm sorry -- this is all very confusing at this point! So his specific question is this: • Why wouldn’t “three times as large” and “three times larger” mean the same thing? • How can “5 1/2 times greater” and “6 1/2 times what it was” mean the same thing? The two pages quoted are by me (1999) and Doctor Terrel (1997). The first is particularly worth reading in its entirety, as there is a lot more there. ## The case for a literal interpretation Doctor Greenie responded, arguing against laxness on the matter, and making the case for the literalistic interpretation: I'm going to jump in here, because this is one of my pet peeves. Mathematics is commonly called the exact science. Mathematics must be exact; if it is not, it all falls apart. We can't use ambiguous language in mathematics. I agree that the use of the phrase "x times larger than" is best avoided. However, as a mathematician who believes in using unambiguous language, I cannot accept the proposition that we should be able to interpret "5 times larger than 10" as either 50 or 60. It HAS TO BE ONE OR THE OTHER. And grammatically, "5 times larger than" means the "new" number is 5 times larger than the "old" number; this in turn means the difference between the new and old numbers is 5 times the old number, making the new number 6 times the old number. So the number which is 5 times larger than 10 is 10 + 5(10) = 10 + 50 = 60 (The phrase "... larger than ..." implies comparison by subtraction; the phrase "... as large as ..." implies comparison by division. Sixty is 6 times as large as 10, because 60/10 = 6. But 60 is 5 times larger than 10, because [60 - 10]/10 = 50/10 = 5.) Yes, we hear it all the time in everyday life. Sometimes, we even hear it in the supposedly rigorous world of science -- "an earthquake of magnitude 5 is 10 times greater than one of magnitude 4," and such. But the common idiom of using "10 times greater than" -- when the actual meaning is "10 times as great as" -- has no place in mathematics. He concluded with an accidental overstatement of what the “other side” says: I disagree with many of the concessions that other math doctors here have made in interpreting the phrase "x times larger than" as being the same as "x times as large as." On one of the pages I saw, a fellow doctor said that "50% larger than" and "50% as large" mean the same thing. But if my weekly salary last year was$1000 and it is 50% LARGER this year, then it is now
$1000 + 50%($1000) = $1000 +$500 = $1500 While if it was$1000 last year and it is 50% AS LARGE this year, then it is now
50%($1000) =$500
If something is 50% larger, then it is larger; if it is 50% as large, then it is smaller. They can't be the same; that is nonsense.
In fact, as he admitted in a subsequent private discussion, he had misremembered what others had said; none of us have claimed that “50% larger than” and “50% as large” mean the same thing. What we say is that, when the percentage or multiplier is greater than 100%, we recognize ambiguity in the likely intent. I think we agree that the phrase should not be used in a mathematical context, and that we both grudgingly interpret it as intended elsewhere.
## What the literal interpretation means
Then it was my turn to respond, as the author of the first page Joseph had asked about, wanting to make sure he understood both why people take it literally as they do, and how we should think about it.
First, on “Larger Than and As Large As”, I said this:
Joseph, your thinking is RIGHT: if M is three times AS LARGE AS N, then M = 3N. That's what statement (1) above says.
But if we break statement (2) apart carefully (some would say TOO carefully :-)), then it means something different from what people usually mean by it.
If I said "M is 50 larger than N," I would mean that if you ADD 50 to N, you get M:
M = N + 50.
And if I said, "M is 50% larger than N," I would mean that if you add 50% OF N to N, you get M; that is, I mean that M is 50% of N added to N:
M = N + (0.50)N.
Now, though I'm not entirely sure I agree with this, technically minded people often apply the same thinking to (2), for the sake of consistency. The "larger than" means we add something to N. And what do we add? Three times N. So by this thinking,
M = N + 3N = 4N.
So "three times LARGER THAN N" means the same as "four times AS LARGE AS N."
I then referred to the usage book quoted above, adding:
English usage experts think it is nonsense. My feeling is that this thinking puts a little too much weight on consistency, and is just too weird for the general public to follow. English is not known for consistency! So we need to recognize that in everyday usage, (1) and (2) really mean the same thing. When we accept that, though, we set ourselves up for the opposite confusion: Cases like "50% larger" and "3 times larger" no longer follow the same pattern, and our language becomes inconsistent, which really bothers mathematicians!
As Dr. Rick pointed out, this means that there are gray areas where it's hard to be sure what someone means, so it may be best just to avoid using these phrases in mathematical contexts.
Finally, I commented on Joseph’s other quote, from Doctor Terrel:
Joseph, here the doctor was saying that a 550% INCREASE means adding 550% of the original value to the original value, which means 650% OF the original value. In the other terminology, "5 1/2 times greater" (there again, "greater" is taken to refer to the increase) is the same as "6 1/2 times as much." When he says that the 6 1/2 is not a percent increase, he doesn't mean that it hasn't increased, just that he is talking about multiplying by 650% rather than adding 650%. When you think in terms of increase (adding), it is a 550% increase.
Now, the "percent increase" case is pretty standard, because it IS technical terminology (though ordinary people reading it can get confused, so it's still risky). The "times greater" case is more disputable, since that sounds less technical. Most people don't demand absolute consistency from language; they are happy to understand "times greater" idiomatically.
I hope that clears up some of the confusion. It isn't all cleared up yet at our end. You will definitely get different opinions as to what it all REALLY means!
## Increase by a factor
Joseph replied, returning to his initial question:
That really cleared things up for me and I appreciate your time in driving home the differences! The last question I would like to ask is, how do you deal with factors? If someone says something has changed by a factor of ... or is less/greater than by a factor of ..., do we use the same rules that you've discussed above? Or when using the word "factor," are things a bit different?
As Dr. Rick pointed out in the first response, "factor" is used to make it clear that multiplication, rather than addition, is the cause of an increase. Just as "increased by a factor of 2" means "twice as large," so does "greater by a factor of 2." And "decreased by a factor of 2" and "less by a factor of 2" both mean "half as much" (divided by 2).
I can't think of a context in which that would not be true -- but English is flexible enough that I probably shouldn't guarantee anything!
Isn’t English fun?
## Conclusion
In closing, here is a more recent question (2015) where I summarized this complicated issue:
As Big As vs. Bigger Than
I'm having difficulty convincing my 5th graders that "as big as" and "bigger than" do not mean the same thing.
For example, when asked, "How many times larger is 10,000 than than 100?" they answer "100."
Their tests and homework are full of this misunderstanding! How would you suggest telling them they are wrong?
I referred to most of the pages we’ve seen above, and added:
To be honest, my feeling (basically unchanged since the first of those) is that although your understanding is common among thoughtful people, it is a case of excessive consistency.
Mathematical people want a certain word structure to always have the same meaning, so we relate "x times bigger" to "x percent bigger," and that to "x bigger," and want to take all in an incremental (additive) sense. But taken on its own, it is perfectly logical to interpret "x times bigger" as "bigger, as a result of multiplication by x." And human language is not completely consistent; we have idioms all over the place that we interpret with no trouble.
Having said that, I think that math books should avoid that form, because there is just enough truth to your thinking, and enough extra expectation of careful use of words in a math book, that it can be confusing.
What I would do is to make a brief mention of the fact that many people take it as you do, but then point out that your book is using the phrase in the way it is usually intended in the real world. If I were writing the textbook, I would reverse this: almost always use "x times as big," but mention somewhere the fact that many people use "x times bigger" to mean the same thing, and briefly discuss the controversy before moving on to other things.
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https://math.stackexchange.com/questions/2515956/sequence-of-odd-and-even-partial-sums-of-alternating-harmonic-series | # Sequence of odd and even partial sums of alternating harmonic series.
Given the alternating harmonic series: $$\sum^{\infty}_{n=1}{\frac{(-1)^{n+1}}{n}}$$ and if we let the sequence of its partial sums be $s_n$ then how can we express the sequence of even partial sums $s_{2n}$ and odd partial sums $s_{2n+1}$?
I did the following but am not completely sure if it is valid:
$$s_{2n}= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}$$ $$s_{2n+1}= \sum^{2n+1}_{i=1}{\frac{(-1)^{i+1}}{i}}$$
Then to show the first is increasing and the second is decreasing, can we just take $s_{2n+2}-s_{2n}$ and show that this is $> 0$? Similarly then taking $s_{2n+3}-s_{2n+1}$ and showing that this is $< 0$?
And to go one step further, how could we show that each of these sequences of partial sums has a limit? And then show that the limits are the same?
• $s_{2n} = \sum_{i=1}^n \frac{1}{2i-1}-\frac{1}{2i}=\sum_{i=1}^n \frac{1}{2i (2i-1)}$ – reuns Nov 12 '17 at 1:31
• en.wikipedia.org/wiki/Alternating_series_test – parsiad Nov 12 '17 at 1:33
• I don't quite get where the formula you've written comes from? – Analysis is fun Nov 12 '17 at 1:42
What you have done is fine. Showing the increasing/decreasing nature works well. You can invoke the alternating series theorem to show the whole series has a limit because the terms are alternating in sign and decreasing monotonically to zero. If the whole series has a limit, every subseries converges to the same limit and you are done.
• Ah okay I see, is there a way to directly prove that $s_{2n}$ and $s_{2n+1}$ have limits? – Analysis is fun Nov 12 '17 at 1:42
• The subseries argument works fine. Also they are monotonic and bounded above/below so have limits. – Ross Millikan Nov 12 '17 at 1:46
• Ah yes okay I think I've got it now, thanks so much for your help! – Analysis is fun Nov 12 '17 at 1:47
Yes, you are correct on how to show that the first is decreasing and the second is increasing.
Now, since you know they are monotone, if you can show they are bounded, you will demonstrate that they have limits. For instance you can show that $$1>s_{2n+1} > 1/2$$ for all $n$.
That the limits are the same follows from the fact that the difference $$|s_{2n+1}-s_{2n}| = \frac{1}{2n+1}\to 0.$$
• Thanks, so we can show boundedness and then use the monotone convergence theorem? – Analysis is fun Nov 12 '17 at 1:47
• Yeah. As others have said the alternating series test proves it all in one fell swoop but it seemed like the point of the exercise was to prove a special case of the alternating series test. – spaceisdarkgreen Nov 12 '17 at 1:52
Since $s_{2n}= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}$,
$\begin{array}\\ s_{2n+2} &= \sum^{2n+2}_{i=1}{\frac{(-1)^{i+1}}{i}}\\ &= \sum^{2n}_{i=1}{\frac{(-1)^{i+1}}{i}}+{\frac{(-1)^{2n+1}}{2n+1}}+{\frac{(-1)^{2n+2}}{2n+2}}\\ &= s_{2n}-{\frac{1}{2n+1}}+{\frac{1}{2n+2}}\\ &= s_{2n}-{\frac{1}{(2n+1)(2n+2)}}\\ &\lt s_{2n}\\ \end{array}$
so $s_{2n}$ is decreasing.
Similarly,
$\begin{array}\\ s_{2n+3} &= \sum^{2n+3}_{i=1}{\frac{(-1)^{i+1}}{i}}\\ &= \sum^{2n+1}_{i=1}{\frac{(-1)^{i+1}}{i}}+{\frac{(-1)^{2n+2}}{2n+2}}+{\frac{(-1)^{2n+3}}{2n+3}}\\ &= s_{2n+1}+{\frac{1}{2n+2}}-{\frac{1}{2n+3}}\\ &= s_{2n+1}+{\frac{1}{(2n+2)(2n+3)}}\\ &\gt s_{2n+1}\\ \end{array}$
so $s_{2n+1}$ is increasing.
Since $s_{2n+1} =s_{2n}-\frac1{2n+1} \lt s_{2n}$, all the $s_{2n+1}$ are less than all the $s_{2n}$.
Since $|s_{2n+1}-s_{2n}| =\frac1{2n+1} \to 0$, the sequences $(s_{2n})$ and $(s_{2n+1})$ must approach a common limit. | 2019-10-19T13:11:21 | {
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https://brilliant.org/discussions/thread/when-fibonacci-meets-pythagoras/ | ×
When Fibonacci meets Pythagoras
There is a cool identity relating Fibonacci numbers and Pythagorean triples, which states that given four consecutive Fibonacci numbers $${F}_{n},{F}_{n+1},{F}_{n+2},{F}_{n+3}$$, the triple $$({F}_{n}{F}_{n+3}, 2{F}_{n+1}{F}_{n+2}, {{F}_{n+1}}^{2}+{{F}_{n+2}}^{2})$$ is a Pythagorean triple.
To prove this, let $${F}_{n+1} = a$$ and $${F}_{n+2} = b$$. Then $${F}_{n} = b-a$$ and $${F}_{n+3} = b+a$$. Substituting the abstract values to the equation ${({F}_{n} {F}_{n+3})}^{2} +{(2{F}_{n+1} {F}_{n+2})}^{2} = {({{F}_{n+1}}^{2}+{ {F}_{n+2}}^{2})}^{2}$ will give ${((b-a)(b+a))}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$
With a little simplification, we arrive at Euclid's formula for Pythagorean triples: ${({b}^{2}-{a}^{2})}^{2} +{(2ab)}^{2}={({a}^{2}+{b}^{2})}^{2}.$
Check out my other notes at Proof, Disproof, and Derivation
Note by Steven Zheng
2 years, 10 months ago
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Nice post
Is there any generalized formula to generate pythagoream triples
- 2 years, 10 months ago
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Euclid's Formula. The link above should bring you to a proof.
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How we can solve problems in which product of three pythagorean triples is given and we have to find the triple
- 2 years, 10 months ago
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I'm not sure if that is solvable without more given conditions. You might also need the sum of sides or maybe the radius of an incircle also. Can you come up with an example of your question?
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Find out the pythagorean triple $$(x,y,z)$$ such that $$xyz=16320$$
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Number of solutions $$a,b$$ such that $$2ab({a}^{2}+{b}^{2})({a}^{2}-{b}^{2}) = 16320$$.
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Does there exist more the one triples.....also i read on a website that euclid's formula can't generate all pythagorean triples
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Euclid's formula deal with generating Pythagorean triples that are integers.
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Good
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http://math.stackexchange.com/questions/89470/in-how-many-ways-can-i-climb-down-ten-stairs-taking-as-many-steps-at-a-time-as | # In how many ways can I climb down ten stairs, taking as many steps at a time as I like?
I want to climb down a flight of ten stairs, taking one or many steps at a time. In how many ways can I do this if I am able to take as many steps at a time as I like?
I don't know the answer; I am getting $3$ such as:
$1,2,3,4,..., 10$ (one at a time)
$2,4,6,..., 10$ (two at a time)
$5,10$ (five at a time)
this is correct/incorrect?
-
If you have to choose the number of steps you will take at a time, and stick with that number the whole way down, then the only thing you have missed is the possibility of taking 10 at a time. But if you are allowed to (for example) take 3 steps, then 4, then 3, well, that's a whole different problem. So, which is it? – Gerry Myerson Dec 8 '11 at 2:16
@Gerry Myerson: "taking one or many steps at a time" probably indicate a variable number of steps at each time ... also this version seems to be the interesting one. – Quixotic Dec 8 '11 at 2:21
If we have total freedom, there are $512$ ways. – André Nicolas Dec 8 '11 at 2:37
See the WP page on the number of compositions – r.e.s. Dec 8 '11 at 3:11
@MaX, what you say is true, but I find it hard to believe that under that interpretation OP was only able to find 3 ways to come down the stairs. – Gerry Myerson Dec 8 '11 at 5:07
In the interpretation where the number of steps taken is allowed to vary, there is an easy way to see that the answer is $2^{n-1}$:
For each of the $n-1$ steps (not counting the top one), decide whether or not you will stop at that step on the way down.
There are $n-1$ decisions to make, each with 2 possible choices (stop or don't stop), so that's a total of $2^{n-1}$ ways.
-
We answer the question on the following assumptions: (1) At any time, we can take an arbitrary number of stairs, as long as the total number of stairs ends up at $10$ and (2) Order matters, so that for example the pattern $2+4+4$ is to be counted as different from $4+2+4$.
In general, let $f(n)$ be the number of ways to handle a staircase that has $n$ stairs. It is easy to see that $f(1)=1$. We could compute a little more, finding that for $2$ there are the patterns $2$ and $1+1$, so $f(2)=2$. For $3$ there are the patterns $3$, $2+1$, $1+2$, and $1+1+1$, for a total of $4$. For $4$, enumerating the patterns is not hard: we get $4$, $3+1$, $1+3$, $2+2$, $1+1+2$, $1+2+1$, $2+1+1$, and $1+1+1+1$, a total of $8$. On the basis of this admittedly scanty evidence, we conjecture that $f(n)=2^{n-1}$, and in particular $f(10)=512$.
Now we prove that in general $f(n)=2^{n-1}$. There are various approaches. We first prove the result in a natural but somewhat lengthy way, and then in a shorter way. Suppose that we know that for all $k<n$, there are $2^{k-1}$ patterns. We count the number of patterns for $n$.
Either there is only one giant step of length $n$ ($1$ way) or the last step is one of $1$, $2$, $3$, $\dots$, $n-1$.
If the last step is $1$, we needed to get to $n-1$, which by induction assumption can be done in $2^{n-2}$ ways.
If the last step is $2$, we needed to get to $n-2$, which by induction assumption can be done in $2^{n-3}$ ways.
If the last step is $3$, we needed to get to $n-3$, which by induction assumption can be done in $2^{n-4}$ ways.
Go on in this way. Finally, if the last step was $n-1$, we needed to get to $1$, which can be done in $2^0$ ways.
We conclude that $$f(n)=(2^{n-2}+2^{n-3}+2^{n-4}+\cdots +1)+1$$ (remember the possibility of a giant step). Easily, if we add from the back, we see that $f(n)=2^{n-1}$ (or else we can sum the finite geometric progression).
Another way: We show that in general $f(n+1)=2f(n)$. Consider all patterns of steps that add up to $n+1$. They are of two types (i) Patterns that have last step $1$ and (ii) Patterns that have last step $>1$.
It is obvious that there are $f(n)$ patterns of type (i). We now count the type (ii) patterns. Take a pattern of type (ii), and shorten its last step by $1$. The total is now $n$. Moreover, we get all patterns for $n$ in this manner, in one and only one way. Thus there are $f(n)$ type (ii) patterns. It follows that $f(n+1)=f(n)+f(n)=2f(n)$.
-
A composition of an integer $n$ into $k$ parts is the number of sums $a_1 + \dots + a_k$ of positive integers that equal $n$. For example, there are 3 compositions of $4$ into $2$ parts: 2+2, 3+1, and 1+3. You should try to prove that the number of compositions of $n$ into $k$ parts is equal to $\binom{n-1}{k-1}$. (Or, look it up in any good combinatorics book.)
Your problem essentially asks you to count the number of compositions of $10$ into $k$ parts, where $1\leq k\leq 10$. That is, if as you go down the staircase you cross $a_i$ stairs in your $i^{\text{th}}$ step, then $a_1 + \dots + a_k = 10$. Your answer is thus $$\sum_{k=1}^{10} \binom{10-1}{k-1} =512$$
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https://math.stackexchange.com/questions/2719143/first-order-pdes-and-the-method-of-characteristics | # First order PDEs and the method of characteristics
I am studying PDEs using the book "PDEs An Introduction 2nd edition" by Walter A. Strauss. In Chapter 2, a "geometric method" is described in order to solve linear PDEs of the type:
$$(x,y)\mapsto u_x + yu_y = 0$$
This is said to be equivalent to the directional derivative of $u$ in the direction of the vector $(1,y)$ being set to 0. Then characteristic curves having as tangent vectors $(1,y)$ are found
$$\frac{dy}{dx} = \frac{y}{1} \implies y = Ce^x$$
Since $u(x,y)$ is constant on these curves:
$$u(x,y) = f(e^{-x}y)$$
is the general solutions of the PDE, where $f$ is an arbitrary function. Now the same method is applied to solve more general equations, such as:
$$u_x + u_y + u = f(x,y)$$
I have tried to use the same method to solve the following differential equation:
$$yu_x -xu_y = 3x$$ where $u(x,0) = x^2$
Then applying the method:
$$\frac{dy}{dx} = \frac{-x}{y} \implies C = x^2 + y^2$$
The PDE reduces to an ODE:
$$\frac{du}{dx} = 3x/y \implies u (x,y) = 3x^2/2y + f(C)$$
However with the boundary imposed ($u(x,0) = x^2$) this seem impossible to solve since to find the particular solution I would have to divide the a term by zero. Does the method described in the book have limited scope? How can I solve this differential equation? Also, why is $u(x,y)$ constant on the "characteristic curves"?
• If you introduce polar coordinates $x=r\cos \theta, y=r\sin \theta$, then $y\partial_x-x\partial_y=-\partial_\theta$. This is not the method you asked but it leads to a quick integration. – Giuseppe Negro Apr 2 '18 at 21:41
• Yeah that is a good trick, however I was more interested in the method and not the solution itself. – daljit97 Apr 2 '18 at 21:54
I cannot see where you found an impossibility or an exception in the rules.
I the case of $\quad yu_x-xu_y=3x \tag 1$
$$\frac{dx}{y}=\frac{dy}{-x}=\frac{du}{3x}$$ A first family of characteristic curves comes from $\quad\frac{dx}{y}=\frac{dy}{-u}\quad$ which leads to $$x^2+y^2=c_1$$ A second family of characteristic curves comes from $\quad\frac{dy}{-x}=\frac{du}{3x}\quad$ which leads to $$u+3y=c_2$$ The general solution of the PDE $(1)$ expressed on the form of implicit equation is : $$\Phi(x^2+y^2\:,\:u+3y)=0$$ where $\Phi$ is an arbitrary fonction of two variables.
Or equivalently, the general solution of PDE $(1)$ on explicit form is $\quad u+3y=F(x^2+y^2)$ $$u(x,y)=-3y+F(x^2+y^2)$$
where $F(X)$ is an arbitrary function of the variable $X$ with $X=x^2+y^2$.
Boundary condition : $\quad u(x,0)=x^2=F(x^2+0^2)=F(x^2)$.
So, the function $F$ is now determined : $\quad F(X)=X.\quad$ Putting it into the above general solution where $X=x^2+y^2$ leads to the particular solution which fits the boundary condition : $$u(x,y)=-3y+(x^2+y^2)$$
OTHER EXAMPLE : $\quad u_x+u_y+u=e^{x+2y} \tag 2$ $$u_x+u_y=e^{x+2y}-u$$ $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u}$$ A first family of characteristic curves comes from $\quad\frac{dx}{1}=\frac{dy}{1}\quad$ which leads to $$x-y=c_1$$ A second family of characteristic curves comes from $\frac{dy}{1}=\frac{du}{e^{x+2y}-u}=\frac{du}{e^{(c_1+y)+2y}-u}=\frac{du}{e^{c_1+3y}-u}$ $$\frac{du}{dy}+u=e^{c_1+3y}$$ This is a first order linear ODE easy to solve : $\quad u=\frac14 e^{c_1+3y}+c_2e^{-y}=\frac14 e^{(x-y)+3y}+c_2e^{-y}$ $$ue^y-\frac14 e^{x+3y}=c_2$$ The general solution of the PDE $(2)$ expressed on the form of implicit equation is : $$\Phi\left((x-y)\:,\:(ue^y-\frac14 e^{x+3y})\right)=0$$ where $\Phi$ is an arbitrary fonction of two variables.
Or equivalently, the general solution of PDE $(2)$ on explicit form is from $\quad ue^y-\frac14 e^{x+3y}=F(x-y)$ $$u(x,y)=\frac14 e^{x+2y}+e^{-y}F(x-y)$$ where $F(X)$ is an arbitrary function of the variable $X=x-y$.
• I believe that the author of my book seems to use a different method to solve the PDE, have a look here at an example. If you look at the last few lines of my question, there seems to be a problem (using the method used in the link) – daljit97 Apr 2 '18 at 21:29
• It is fundamentally the same method, with some variant of symbols and presentation. See :en.wikipedia.org/w/… and see en.wikipedia.org/wiki/Method_of_characteristics. You will have to do a bit of "gymnastic of the brain" in order to make the link. I wish you bon courage. – JJacquelin Apr 2 '18 at 21:43
• You can see that the general solution in the book is : $$u(x,y)=\frac14 e^{x+2y}+e^{-y}f(x-y)$$ which is exactly the same than in my answer. I will not repeat what is already in the book to determine the function $f$ according to the boundary condition. The method is the same. – JJacquelin Apr 2 '18 at 21:52
• $\frac{du}{dx} = 3x/y \implies u (x,y) = 3x^2/2y + f(C)$ is false because you integrate $3x/y$ without taking account that $y$ is not constant. You cannot solve it that way. This not a correct application of the method of characteristic. One have to find two independent families of characteristic curves. You got one :$x^2+y^2=C_1$ but not a second one. – JJacquelin Apr 2 '18 at 22:11
• @Isham yes you are right, I was too invested into thinking about why the method led me to that point that it didn't occur to me that a simple substitution was needed. – daljit97 Apr 2 '18 at 22:40
You can't have this: $$\frac{du}{dx} = 3x/y ==> u (x,y) = 3x^2/y + f(C)$$ Because you have a differential equation ( du,dx) but three variables namely x,y,u. Y shouldnt be there...Apart for that mistake you did a very good job.
Usisng the method of characteristics:
$$\frac {dx}{y}=\frac {dy}{-x}=\frac {dz}{3x}$$ $$-xdx=ydy \implies x^2+y^2=K_1$$ $$3xdy=-xdz \implies y+z/3 =K_2$$ $$f(x^2+y^2)=y+u/3$$ $$u(x,y)=3f(x^2+y^2)-3y$$ we have $u(x,0)=x^2$ $$f(x^2)=x^2/3 \implies f(x)=x/3$$ Therefore $$\boxed{u(x,y)=x^2+y^2-3y}$$
• Hmm in the book in order to solve $u_x + u_y + u = e^{x+2y}$ the author does setup the following equation: $\frac{\partial u}{\partial x} + u = e^{x+2y}$ Would that be an exception to the rule? – daljit97 Apr 2 '18 at 19:39
• @daljit97 I don't know I dont have the book right now. Thats weird that Strauss was so complicated about first order linear pde.. when it's easy to solve directly with characteristics method.. – Isham Apr 2 '18 at 19:42
• Here is the link with the exercise and the solution provided stemjock.com/STEM%20Books/Strauss%20PDEs%202e/Chapter%201/… – daljit97 Apr 2 '18 at 19:48
• Oh thats another method..I dont know that method...I deleted my comments on that method you used since I was completely wrong about it... – Isham Apr 2 '18 at 19:53
• Oh ok, could you provide a reference that explains and show why the method of characteristic works? Looking around I couldn't find anything satisfying. – daljit97 Apr 2 '18 at 20:03 | 2019-07-17T00:20:00 | {
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https://math.stackexchange.com/questions/220172/does-the-product-of-two-invertible-matrix-remain-invertible | # Does the product of two invertible matrix remain invertible?
If $A$ and $B$ are two invertible $5 \times 5$ matrices, does $B^{T}A$ remain invertible?
I cannot find out is there any properties of invertible matrix to my question.
Thank you!
• A common trick to answer the question "Is foo invertible?" is to actually write down the inverse of foo. In many cases this is pretty easy to do, as seen in the answers.
– user14972
Oct 24, 2012 at 16:10
• @Hurkyl What does foo mean? Oct 24, 2012 at 16:13
• @PENGTENG It is a nonsense word used as a generic placeholder in math and computer science. It's being used as a variable for an object here, but it's more frequently used for properties rather than objects. Oct 24, 2012 at 16:14
• @rschwieb I get it. Thanks! Oct 24, 2012 at 16:17
• How is this well researched? C'mon guys Aug 3, 2015 at 6:41
Of course: $B$ invertible implies $B^T$ invertible, and the product of two invertible matrices is clearly invertible.
This is easily seen from these equations: $$BB^{-1}=I\implies (BB^{-1})^T=I\implies (B^{-1})^TB^T=1,$$ and the fact that if $X$ and $Y$ are invertible, $(XY)^{-1}=Y^{-1}X^{-1}$.
Perhaps the general properties you should take away are these:
$(XY)^T=Y^TX^T$ and $(XY)^{-1}=Y^{-1}X^{-1}$.
• And this is of course true for $n \times n$-matrices and not just $5 \times 5$-matrices.
– N.U.
Oct 24, 2012 at 16:11
• "...the product of two matrices is clearly invertible." Only if the two matrices are themselves invertible: math.stackexchange.com/a/1026628 Aug 3, 2015 at 2:16
• @nacnudus of course, a reasonable person can tell this is a typo of the form of an omitted word from context. Thanks for indirectly alerting me to it. Aug 3, 2015 at 3:27
Yes. $$\det(B^T\,A)=\det(B^T)\det(A)=\det(B)\det(A)\ne0.$$ Moreover $$(B^T\,A)^{-1}=A^{-1}(B^{-1})^T.$$ | 2022-06-27T09:52:07 | {
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https://math.stackexchange.com/questions/1675485/are-there-arbitrarily-large-gaps-between-consecutive-primes | # Are there arbitrarily large gaps between consecutive primes? [duplicate]
I made a program to find out the number of primes within a certain range, for example between $1$ and $10000$ I found $1229$ primes, I then increased my range to $20000$ and then I found $2262$ primes, after doing it for $1$ to $30000$, I found $3245$ primes.
Now a curious thing to notice is that each time, The probability of finding a prime in between $2$ multiples of $10000$ is decreasing, i.e it was $$\frac{2262-1229}{10000}=0.1033$$ between $10000$ and $20000$, and $$\frac{3245-2262}{10000}=0.0983$$ between $20000$ and $30000$,
So from this can we infer that there will exist two numbers separated by a gap of $10000$ such that no number in between them is prime? If so how to determine the first two numbers with which this happens? Also I took $10000$ just as a reference here, what about if the gap between them in general is $x$, can we do something for this in generality?
Thanks!
• If you let $p$ be the largest prime below $10\,000$, then the interval $[p\# - 10\,002, p\# - 2]$ certainly contains no primes (where $p\#$ is the primorial). It's probably not the first time, though – Arthur Feb 28 '16 at 10:26
• In the same spirit as Arthur's suggestion, there are no primes among the $k$ consecutive numbers $(k + 1)! + 2, \ldots, (k + 1)! + (k + 1)$. – Travis Willse Feb 28 '16 at 10:31
• @Arthur largest prime below $10000$ is $9973$, and by p# do you mean $$2.3.5....9973$$? . – Nikunj Feb 28 '16 at 10:42
• @Travis This looks interesting, could you please provide a proof? – Nikunj Feb 28 '16 at 10:43
• @Nikunj Yes, that is what I mean by $p\#$. – Arthur Feb 28 '16 at 10:46
The Prime Number Theorem states that the number of primes $\pi(x)$ up to a given $x$ is $$\pi(x) \sim \frac{x}{\log(x)}$$ which means that the probability of finding a prime is decreasing if you make your "population" $x$ larger. So yes, there exist a gap of $n$ numbers whereof none are prime.
The way to find the first gap for some $n$ has to be done through the use of software, since the exact distribution of prime numbers is only approximated by $\frac{x}{\log(x)}$.
EDIT: That the PNT implies that there's always a gap of size $n$ can be seen by considering what would happen if this was not the case; if there was a maximum gap of $n$ that was reached after some $x$, the probability of finding a prime between $x$ and some larger number $m$ would no longer decrease as $m \to \infty$, which contradicts the PNT.
• @Nikunj Yes, this is indeed only good for larger $x$, but since this approximation holds for any $x$, $10,000-20,000$ are very small numbers. But yeah, it's the best that we have as of yet :) – Bobson Dugnutt Feb 28 '16 at 12:12
• @BenMillwood Yeah, that is an unfortunate choice of words, I'll edit it! Thanks! – Bobson Dugnutt Feb 28 '16 at 15:09
• Could you elaborate on this just a little bit? You're right that the PNT implies that at some point you have to have n integers without a prime, or else π(x) could never be sub-linear, but I feel like this is a few too many steps away from a proof. – hobbs Feb 29 '16 at 6:31
• @hobbs: If you never had $n$ integers without a prime, then the number of primes up to a given $x$ could never be less than $\lfloor{x/n}\rfloor$. This contradicts the PNT. – mjqxxxx Mar 1 '16 at 3:54
• @mjqxxxx I'm not asking for me, I'm asking for it to be made more explicit in the answer for others' benefit :) – hobbs Mar 1 '16 at 4:12
Can we infer that there exist two numbers separated by a gap of $10000$, such that no number in between them is prime?
We can infer this regardless of what you wrote.
For every gap $n\in\mathbb{N}$ that you can think of, I can give you a sequence of $n-1$ consecutive numbers, none of which is prime.
There you go: $n!+2,n!+3,\dots,n!+n$.
So there is no finite bound on the gap between two consecutive primes.
• @Nikunj: Not necessarily. Also, the gap is not guaranteed to be exactly $n$ (or $10000$ in your example). That is why I have emphasized the specific part of your question to which I am referring in my answer. – barak manos Feb 28 '16 at 11:05
• @Nikunj because $n! + 1$ and/or $n! + n + 1$ may be composite as well, in which case one gets a sequence of $> n - 1$ consecutive composite numbers. This is the case, e.g., for $n = 3$. – Travis Willse Feb 28 '16 at 11:42
• @Nikunj To give a specific example in which both $n! + 1$ and $n! + n + 1$ are composite, observe that $5! + 1 = 121 = 11^2$ and $5! + 6 = 126 = 2 \cdot 3^2 \cdot 7$. In fact, the numbers $$114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126$$ are all composite. – N. F. Taussig Feb 28 '16 at 11:56
• @barakmanos I would say, more precisely, that is a lower bound of $n$. – PyRulez Feb 28 '16 at 18:23
• @goblin Take any $n!+i$ with $2\leq i \leq n$. You can factor out an $i$ from the factorial since $i \leq n$, so you get $i(k + 1)$, for $i > 1$ and $k > 1$, where $k = n!/i$. – TokenToucan Feb 29 '16 at 2:33
We cannot infer from the two observations in the question that there are gaps of arbitrary sizes between primes. As Lovsovs mentions in his answer, this does follow from the Prime Number Theorem (one needs suitable error bounds on the approximation $\pi(x) \sim \frac{x}{\log x}$, but even crude ones will do here).
As asked in the comments, it's easy to construct for any positive integer $k$ a sequence of $k$ consecutive composite numbers. For any positive integer $a \leq k + 1$, $a$ divides $(k + 1)!$, and so it divides $(k + 1)! + a$, but this implies that each of the $k$ numbers $(k + 1)! + 2, \ldots, (k + 1)! + (k + 1)$ is composite. (This is generally not the first sequence for which this is true: For example, for $k = 3$, the resulting sequence is $26, 27, 28$, but the first sequence of three consecutive composite positive integers is $8, 9, 10$.)
• (+1) Thanks a ton for the proof, Is there no other way besides the use of software to look for the first such sequence? – Nikunj Feb 28 '16 at 11:02
• @Nikunj No, because (as stated in my answer) we do not know the exact distribution of primes. – Bobson Dugnutt Feb 28 '16 at 11:33
• In general the problem is difficult, so yes, for sufficiently large $k$ computer assistance is necessary. Note that the primorial $(k - 1)\#$ is much smaller than the factorial $(k + 1)!$ for even modest $k$, so Arthur's solution gives a much better upper bound for the position of the first such sequence, but in general it too is a pretty poor bound: Already for $k = 5$ it gives $204, \ldots, 208$, but the first sequence of five consecutive composite numbers is $24, \ldots, 28$. – Travis Willse Feb 28 '16 at 11:40
• @ travis : proving $\pi(x) = o(x)$ is much easier than the prime number theorem, for example it follows directly from $s \int_1^\infty \pi(x) x^{-s-1} dx \sim \ln \zeta(s) \sim -\ln(s-1)$ when $s \to 1^+$. – reuns Feb 28 '16 at 12:21
The problem of finding large gap between consecutive primes is an old and well studied one. There certainly is a large gap between what we know to be true, and what we suspect.
As for what people expect, Cramer's random provides a good source heuristics. Roughly spreaking, you can think of a number $n$ to be "prime with probability $\frac{1}{\log n}$", but sometimes you also have to take into consideration that the primes tend not to be divisible by $2, 3, 5,$, etc. Apparently, if you make a lot of optimistic assumptions, then you can reach the conclusion that the longest gap between primes in integers $\leq X$ is roughly $\log^2 X$. Hence, if you want your gap to have size $g$, then you should look at integers $\leq e^{\sqrt{g}}$ (which, for $g = 10000$ is pretty large). See this post of Terence Tao for details.
Much less has been proved. The best known bounds for the largest gap between primes is due to Kevin Ford, Ben Green, Sergei Konyagin, James Maynard, Terence Tao, and says that the largest gap below $X$ is at least $c \frac{\log X \log \log X \log \log \log \log X}{\log \log \log X}$, where $c$ is a constant.
• In the last para, I guess “largest gap between integers” should be “…between primes”. I believe there are fairly sharp upper and lower bounds known for gaps between consecutive integers :-P – Peter LeFanu Lumsdaine Feb 28 '16 at 15:54
• @PeterLeFanuLumsdaine: Thanks! I suppose the bound is no longer true for integers, but maybe I'll check with one of the authors ;) – Jakub Konieczny Feb 28 '16 at 16:56
• That "best known bound" is amazingly low if you consider that the average gap between primes is log X, so that average gap is improved by less than a factor c log log X. And c might be considerably smaller than 1. – gnasher729 Feb 29 '16 at 23:08
• @gnasher729: It is rather far from the prediction, yes. But it should be pointed out that we're much better in predicting things about the primes than in actually proving them. – Jakub Konieczny Mar 1 '16 at 3:10
• Are there any known bounds on $c$? – Brevan Ellefsen Dec 11 '16 at 17:56
Any sequence with density $0$, in most reasonable meanings of "density $0$", has arbitrarily large gaps.
This is true (for example, and in increasing order of generality) if density $0$ is interpreted as:
• the density on $[1,n]$ converges to $0$, or
• the lim inf of the density on $[1,n]$ is $0$, or
• the lim inf of the density on intervals of length $n$ is $0$
The asymptotic density of primes $\pi(n)/n \to 0$ is 0-density in the strong sense, which is more than enough to ensure large gaps.
The $n!$ examples of large gaps can be reduced from factorial to "primorial" size and the latter seems to be the best currently known deterministic construction.
This isn't going to be a fancy answer, (secondary school student) but I would believe it to be as a number rises the number of possible factors of the number also rise meaning there is a larger chance of it having more factors than itself and one. So a number has every number smaller than it as a possible factor, so smaller numbers have less possible factors and therefore less chance of more than 1 factor and itself.
There several factors determining the size of gaps. As per you example 10,000 has a square root of 100 which give you 25 prime number divisors for the 10,000 and the 20,000 square root is 141 and 34 divisors, and the 30,000 square root is 173 and 40 divisors. The number of divisors helps to determine the gaps. The number of factors needed is based on the square root of your max number. The square roots are decreasing percentage of the max number: example 100 / 10000 = 1.000%, 141 / 20000 = .70500% and 173 / 30000 = .57666%. The number of factors increased against the square root but decreases against the max number. Example: 106,749,747,000,000 the square root is 10,331,977 the square root % is .00000968% with 685,159 factors the max gap in my sample 712. The large gaps seam to have a cycle pattern to them. | 2020-01-26T12:24:03 | {
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https://math.stackexchange.com/questions/1254572/whats-the-difference-between-antisymmetric-and-reflexive-set-theory-discrete-m | # Whats the difference between Antisymmetric and reflexive? (Set Theory/Discrete math)
Antisymmetric: $\forall x\forall y[ ((x,y)\in R\land (y, x) \in R) \to x= y]$ reflexive: $\forall x[x∈A\to (x, x)\in R]$
What really is the difference between the two? Wouldn't all antisymmetric relations also be reflexive?
• E.g. "$\leq$" and "$<$" are antisymmetric and "$=$" is reflexive. Apr 27 '15 at 17:02
Here are a few relations on subsets of $\Bbb R$, represented as subsets of $\Bbb R^2$. The dotted line represents $\{(x,y)\in\Bbb R^2\mid y = x\}$.
Symmetric, reflexive:
Symmetric, not reflexive
Antisymmetric, not reflexive
Neither antisymmetric, nor symmetric, but reflexive
Neither antisymmetric, nor symmetric, nor reflexive
• This is a great visual approach to understanding the meaning of the words! Apr 27 '15 at 20:04
• Thank you so much for making these, they're great! Apr 29 '15 at 2:48
No, there are plenty of anti-symmetric relations that are not reflexive.
For instance, let $R$ be the relation $R=\{(1,2)\}$ on the set $A=\{1,2,3\}$. This relation is certainly not reflexive, but it is in fact anti-symmetric. This is vacuously true, because there are no $x$ and $y$, such that $(x,y)\in R$ and $(y,x)\in R$.
Edit: Why is this anti-symmetric? Because in order for the relation to be anti-symmetric, it must be true that whenever some pair $(x,y)$ with $x\neq y$ is an element of the relation $R$, then the opposite pair $(y,x)$ cannot also be an element of $R$. This is true for our relation, since we have $(1,2)\in R$, but we don't have $(2,1)$ in $R$.
Also, the relation $R=\{(1,2),(2,3),(1,1),(2,2)\}$ on the same set $A$ is anti-symmetric, but it is not reflexive, because $(3,3)$ is missing.
As for a reflexive relation, which is not anti-symmetric, take $R=\{(1,1),(2,2),(3,3),(1,2),(2,1)\}$.
• Thank you so much for your answer, the last two parts make sense! :) I'm a little lost on the first part because the law says that if (x,y) and (y,x) then y=x. Could you elaborate a bit more on how R = {(1,2)} is anti-symmetric? Apr 27 '15 at 17:35
• Sorry, I think I messed up. Let me edit my post. There. Now, I have redone the last two examples, because they were wrong. I'll edit my post further to elaborate on why the first relation is in fact anti-symmetric. :)@TaylorTheDeveloper Apr 27 '15 at 17:42
• This may sound like a naive question but would'nt this example be asymmetric also then by vacuous agument Oct 19 '20 at 11:31
• @angshuknag Yes, the relation $R=\{(1,2)\}$ is also asymmetric. In fact, being asymmetric is equivalent to being both anti-symmetric and not reflexive. Oct 19 '20 at 15:08
They're two different things, there isn't really a strong relationship between the two.
Based on the definitions you're using, they both give two different criteria for concluding that $(x, x) \in R$.
For any antisymmetric relation $R$, if we're given two pairs, $(x, y)$ and $(y, x)$ both belonging to $R$, then we can conclude that in fact $x = y$, so that that
$$(x, y) = (x, x) = (y, x),$$
and $(x, x) \in R$. It may really be better stated as saying that
$$\text{ If } x \neq y, \text{ then at most one of (x, y) or (y, x) is in R}.$$
That is, it may be a bit misleading to even think about $(x,y)$ and $(y, x)$ as being pairs in $R$, since antisymmetry forces them to in fact be the same pair, $(x, x)$.
Antisymmetric relations may or may not be reflexive. $<$ is antisymmetric and not reflexive, while the relation "$x$ divides $y$" is antisymmetric and reflexive, on the set of positive integers.
A reflexive relation $R$ on a set $A$, on the other hand, tells us that we always have $(x, x) \in R$; everything is related to itself. Reflexive relations may or may not be symmetric, or antisymmetric:
$\leq$ is reflexive and antisymmetric, while $=$ is reflexive and symmetric.
• Also, I may have been misleading by choosing pairs of relations, one symmetric, one antisymmetric - there's a middle ground of relations that are neither! For example, the relation "$x$ divides $y$" on the set of all integers is neither symmetric nor antisymmetric (and neither reflexive, nor irreflexive). Apr 27 '15 at 17:52
• The divisibility relation is reflexive, even on all integers. Apr 27 '15 at 20:04
• @JadeNB Thank you, of course you're right; I'm not sure why I had decided it wasn't! Apr 27 '15 at 20:07
• Probably the presence of 0 caused some reflexive (no pun intended!) worries. Apr 27 '15 at 20:11 | 2021-10-18T10:41:15 | {
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https://mathoverflow.net/questions/300249/rotatable-matrix-its-eigenvalues-and-eigenvectors | # Rotatable matrix, its eigenvalues and eigenvectors
We say that a real matrix is rotatable iff after turning it clockwise on $90^{\circ}$ it doesn't change.
I'm interesting about eigenvalues and eigenvectors (belonging to non-zero eigenvalues) of such type of matrices. For example, it is not hard to show that for every tuple of real values $\lambda_1,\ldots,\lambda_{k}$ there exists $n\in\mathbb{N}$ and a rotatable $n\times n$ matrix $A$ such that all $\lambda_i$ are eigenvalues of $A$.
Indeed, let us consider a matrix $$A_1 = \begin{pmatrix} a & a \\ a & a\\ \end{pmatrix}.$$ Then of course $A_1$ is rotatable and its characteristic polynomial $\chi_{A_1}(x) = x^2-2ax = x(x-2a)$ and of course for every $\lambda$ we can choose $a$ (for example $\lambda/2$).
Now let us show how we can construct a rotatable matrix $A_2$ with prescribed eigenvalues $\lambda_1,\lambda_2$. For example, we can consider a matrix of the form $$A_2 = \begin{pmatrix} b&0&0&b\\ 0&a & a&0 \\ 0&a & a&0\\ b&0&0&b \end{pmatrix}.$$ Then $\chi_{A_2}(x) =x^2(x-2a)(x-2b)$. And we are done for $a=\lambda_1/2, b = \lambda_2/2$. Of course, using this method we can construct the required rotatable matrix for every $k$ of size $2k$.
Also my experiments show that all eigenvectors $v = (v_1,\ldots,v_n)^T$ belonging to non-zero eigenvalues of roratable matrix are symmetric, that is $v_i = v_{n-i+1}$. It is simple to prove this in the case, where $n=2$. Also I tried to prove it by induction on $n$, but my attempts failed.
My question.
1. For a given tuple $\lambda_1,\ldots, \lambda_k$ can we construct a rotatable matrix $A$ of size $n\times n$, where $n<2k$, such that all $\lambda_i$ are eigenvalues of $A$.
2. Is it always true that all eigenvectors of non-zero eigenvalues of rotatable matrix are symmetric? And if the answer is "yes", how to prove this.
• Is the (2,4) element of $A_2$ supposed to be 0 rather than $b$? If not, how is it rotatable? May 15, 2018 at 11:23
• Doesn't every rotatable matrix become of the form $\begin{bmatrix}M & M \\ M & M\end{bmatrix}$, i.e., $\begin{bmatrix}1 & 1\\ 1 & 1\end{bmatrix} \otimes M$, if you conjugate it by a suitable permutation? That looks like it would simplify the analysis a lot. May 15, 2018 at 21:04
• @FedericoPoloni, does this hold for $3\times 3$ matrices? I don't sure. May 16, 2018 at 10:30
• @MikhailGoltvanitsa No, only in even dimension. May 16, 2018 at 14:28
• In particular your matrices are centrosymmetric (apply the rotation twice will also yield the same matrix) en.m.wikipedia.org/wiki/Centrosymmetric_matrix May 18, 2018 at 15:34
Consider the matrix $$P= \begin{pmatrix} 0 & \ldots & 1 \\ \vdots & 1 & \vdots \\ 1 & \ldots & 0 \end{pmatrix}$$ with $1$s along the "other" main diagonal and $0$s elsewhere. Then $(PA)^t$ is a rotation of the matrix $A$ by $90^\circ$ (you can check this on the basis $E_{i,j}$ of matrices with a 1 in the $(i,j)$ slot and 0s elsewhere). So $A$ is rotatable if and only if $(PA)^t = A$, i.e. $PA=A^t$. Note that $P^2=I$ and $A^tP = A$, so $A^t = AP$ and hence $A$ and $P$ must commute. (For what follows below we assume that $A$ is a real matrix. Otherwise I think all you can say is that $A$ preserves the splitting $\mathbb{C}^n=E_+ \oplus E_-$ as explained below. In the complex case we get symmetric eigenvectors of the $v+Pv$ as well as the skew symmetric vectors $v-Pv$.)
It then follows that $A$ and $A^t$ commute and so, by the Spectral Theorem, $A$ has a unitary basis. Now, notice that $P$ has eigenvalues $1$ and $-1$ with eigenvectors $e_i + Pe_i$ and $e_i-Pe_i$ for $1 \leq i \leq \frac{n+1}{2}$, respectively. Let $E_+ = \ker (P-I)$ and $E_- = \ker (P+I)$. Then $A(E_\pm) \subseteq E_\pm$, so $A$ must preserve the splitting $\mathbb{R}^n = E_+ \oplus E_-$. Note that $\dim E_+ = \left\lfloor \frac{n+1}{2}\right\rfloor$.
Now, the answer to question $2$ is yes if we require the corresponding eigenvalue to be a nonzero real number. If $\lambda$ is an eigenvalue with eigenvector $v \in E_\pm$ and $(\:\: ,\:\: )$ is the standard hermitian inner product on $\mathbb{C}^n$, we have $$\lambda(v,v)=(v,A^tv) = (v,APv) = \pm \bar{\lambda}(v,v).$$
So $\lambda \in \mathbb{R}$ if and only if $v \in E_+$ and $\lambda \in i \mathbb{R}$ if and only if $v \in E_-$. Note that $v \in E_+$ if and only if $Pv=v$, which is the same as requiring that $v$ is symmetric. Furthermore, note that the eigenvectors in $E_- \setminus \{0\}$ have complex coefficients as they satisfy $Av=i \mu v$. So it is precisely the real eigenvectors of nonzero eigenvalues that are symmetric. The others will be "skew symmetric".
Question $1$ is a bit trickier, and the answer is no in general. Let us treat the even and odd dimensional cases separately.
Let $n=2m$. If $A$ is rotatable it has the form $$A= \begin{pmatrix} B & B^t P \\ PB^t & PBP \end{pmatrix}$$ for some matrix $B$ (here $P$ and $B$ are matrices of size $m\times m$). Now, $\lambda$ is a real eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ Pv \end{pmatrix}$ if and only if $$Bv + B^t v = \lambda v .$$ So $\lambda = 2 \operatorname{Re}(\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B+B^t$. Similarly, $\lambda= i \mu$ is a purely imaginary eigenvalue of $A$ with eigenvector $\begin{pmatrix} v \\ -Pv \end{pmatrix}$ if and only if $Bv - B^t v = i \mu v$, so $\mu = 2 \operatorname{Im} (\eta)$, where $\eta$ is an eigenvalue of $B$ and $v$ is an eigenvector of $B-B^t$.
So, if you choose more than $n$ real numbers, or choose any complex numbers $\lambda$ that do not satisfy $\bar{\lambda} = \pm \lambda$, or you choose more than $n$ purely imaginary numbers, then they cannot be eigenvalues of $A$.
Now, in the odd dimensional case $n=2m+1$, $A$ takes the form $$A= \begin{pmatrix} B & u & B^t P \\ u^t & a & (Pu)^t \\ PB^t& Pu & PBP \end{pmatrix}$$ for some matrix $B$, vector $u$ and real number $a$. For $\lambda$ a real eigenvalue of $A$, we have the eigenvector $\begin{pmatrix} v \\ b \\Pv \end{pmatrix}$, so $$(B+B^t)v + bu = \lambda v$$ and $$2(u,v)+ab=\lambda b.$$ Choosing $u=0$ gives $$(B+B^t)v=\lambda v$$ and $ab=\lambda b.$ So we can pick $m$ eigenvectors for $B+B^t$ as above, and also the vector given by $v=0$ and $b=1$, which has eigenvalue $a$. The imaginary eigenvalues $i \mu$ have eigenvectors $\begin{pmatrix} v\\0\\-Pv \end{pmatrix}$, and so $$(B-B^t)v=i\mu v,$$ with the other equation being $0=0$. So as before $\mu$ is twice the imaginary part of an eigenvalue of $B$. | 2023-02-06T10:22:56 | {
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https://math.stackexchange.com/questions/1238873/evaluate-lim-n-to-infty-int-01-fracn12n1-left-fract1/1239007 | # Evaluate $\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$
Evaluate $$\lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$
For this integral, I have tried using integration by parts and then evaluating the limit, but I don't think the integral inside converges. However, the limit does exist and the answer given in my book is $2$.
Any help will be appreciated.
• can you show us your working? – danimal Apr 17 '15 at 11:06
• @Henry which book is this? – Kugelblitz Jun 17 '17 at 4:00
First, consider the following two Lemmas,
Lemma $1$: $$\lim_{n \to \infty} \sum_{r=0}^n \left(\dfrac{1}{\displaystyle\binom{n}{r}}\right) =2$$
Proof : First of all, note that the limit exists, since, if we let
$$\text{S}(n)=\displaystyle \sum_{r=0}^n \left(\dfrac{1}{\dbinom{n}{r}} \right)$$
then $\text{S}(n+1)<\text{S}(n)$ for $n \geq 4$. Now,
$\text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}}$
$\implies \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} \ \left[\text{since} \dbinom{n}{r}= \dfrac{n}{r} \times \dbinom{n-1}{r-1} \right]$
Also,
$$\text{S}(n) = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{r}} = 1+ \sum_{r=1}^n \dfrac{1}{\dbinom{n}{n-r+1}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{n-r}} = 1+ \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}}$$
$\left[ \text{since} \ \displaystyle \sum_{r=a}^b f(r) = \displaystyle \sum_{r=a}^b f(a+b-r) \ \text{and} \ \dbinom{n}{r}=\dbinom{n}{n-r} \right]$
Thus, we have,
$$\begin{cases} \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}}\\ \text{S}(n) = 1+ \displaystyle \sum_{r=1}^n \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \end{cases}$$
Adding the above two expressions, we get,
$2\text{S}(n) = 2 + \displaystyle \sum_{r=1}^n \left( \dfrac{r}{n} \times \dfrac{1}{\dbinom{n-1}{r-1}} + \dfrac{n-r+1}{n} \dfrac{1}{\dbinom{n-1}{r-1}} \right)$
$= 2 + \dfrac{n+1}{n} \displaystyle \sum_{r=1}^n \dfrac{1}{\dbinom{n-1}{r-1}}$
$= 2 + \dfrac{n+1}{n} \times \text{S}(n-1)$
Since $n \to \infty$, we have $\text{S}(n) = \text{S}(n-1) = \text{S}$ (say)
$\implies 2\text{S} = \left(\dfrac{n+1}{n}\right) \times \text{S} +2$
$\implies \text{S} = \dfrac{2n}{n-1} = 2$ [since $n \to \infty$]
Lemma $2$ : $$\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}}$$
Proof : Consider the $\text{R.H.S.}$,
$\text{I} = \displaystyle\int_{0}^1 x^r (1-x)^{n-r} \mathrm{d}t$
Let $x = \sin^2 \theta$
$\implies \text{I} = \displaystyle\int_{0}^{\frac{\pi}{2}} 2 \sin^{2r+1} \theta \cos^{2n-2r} \theta \ \mathrm{d}\theta$
Now, using Walli's Formula (or reduction formula) for the above integral, we have,
$\text{I} = \dfrac{1}{(n+1)}\times \dfrac{1}{\dbinom{n}{r}}$
This proves our Lemmas.
Now,
$$\text{J} = \lim_{n \to \infty} \int_{0}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$
Since it is an even function in $t$, we have,
$$\text{J} = \frac{1}{2} \times \lim_{n \to \infty} \int_{-1}^1 \frac{n+1}{2^{n+1}} \left(\frac{(t+1)^{n+1}-(1-t)^{n+1}}{t}\right) \mathrm{d}t$$
Let $t = 2x-1$
$\implies \text{J} = \displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \left(\dfrac{x^{n+1}-(1-x)^{n+1}}{2x-1}\right) \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)^{r} \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n (n+1) \int_{0}^1 x^r(1-x)^{n-r} \mathrm{d}x$
$=\displaystyle \lim _{n \to \infty} \sum_{r=0}^n \dfrac{1}{\dbinom{n}{r}}$ (Using Lemma 2)
$=\boxed{2}$ (Using Lemma 1).
Side Note : Another way to prove Lemma 1 is to use sandwich theorem.
• Wow! Ingenious and rigorous as always :) But can you please tell what led you to think this way? – Henry Durham Apr 17 '15 at 13:06
• @Samurai Your question reminded me of Beta functions which I have used to create this solution. Although understanding the solution doesn't need the understanding of Beta functions, but if you look carefully at Lemma 2, it is the definition of Beta function, for which I've provided an elementary proof. The rest followed from rigorous brainstorming :) – MathGod Apr 17 '15 at 13:11
• @MathGod Thanks a lot :D – Henry Durham Apr 17 '15 at 13:14
• @MathGod Would you mind explaining how you got: $$\int_{0}^1 (n+1) \sum_{r=0}^n (1-x)^n \left(\frac{x}{1-x}\right)\mathrm{d}x$$ after $$\int_{0}^1 (n+1) (1-x)^n \left(\dfrac{\left(\frac{x}{1-x}\right)^{n+1}-1}{\frac{x}{1-x}-1}\right) \mathrm{d}x$$ It is not so clear to me. Thank you. – Tolaso Apr 18 '15 at 4:33
• @Tolaso Thanks for pointing that out. It was a minor typo and I've now corrected it. The expression is sum of G.P. – MathGod Apr 18 '15 at 6:05
It is worth to notice that: $$\int_{0}^{1}\frac{1-(1-t)^n}{t}\,dt = H_n\tag{1}$$ while: $$\int_{0}^{1}\frac{(1+t)^n-1}{t}\,dt=\int_{0}^{1}\sum_{k=0}^{n-1}(1+t)^k\,dt =\sum_{k=1}^{n}\frac{2^k-1}{k}\tag{2}$$ so: $$J_n=\int_{0}^{1}\frac{(1+t)^n-(1-t)^n}{t}\,dt = \sum_{k=1}^{n}\frac{2^k}{k}.\tag{3}$$ The last line also gives: $$\frac{n\, J_n}{2^n} = \sum_{k=1}^{n}\frac{2n}{n+1-k}\cdot 2^{-k} \tag{4}$$ and when $n$ approaches $+\infty$, by the dominated convergence theorem the RHS of $(4)$ approaches: $$\sum_{k=1}^{+\infty}2\cdot 2^{-k} = \color{red}{2} \tag{5}$$ as wanted.
• Pardon me, but I don't know what's dominated convergence theorem. – Henry Durham Apr 17 '15 at 11:33
• – Jack D'Aurizio Apr 17 '15 at 11:35
• That's sweet... I guess, without knowing the properties of $H_n$ you couldn't do much... But how did you get the first equality in $(2)$? – Igor Deruga Apr 17 '15 at 11:43
• @Igor: $t=(1+t)-1$ and $1+z+\ldots+z^n = \frac{z^{n+1}-1}{z-1}$. – Jack D'Aurizio Apr 17 '15 at 11:48
We need to address area around $0$ separately from area around $1$. Pick $\epsilon\in(0,1)$. Then on $[0,\epsilon]$ $$\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\le (n+1)(1+\epsilon)^{n}+(n+1)(1-\epsilon)^{n}$$(you can see this if you multiply both sides by $t$ and compare derivatives) and hence $$\lim_{n\to\infty}\int_0^{\epsilon}\frac {n+1}{2^{n+1}}\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\,dt=0$$ On the other hand $$\lim_{n\to\infty}\int_\epsilon^1\frac {n+1}{2^{n+1}}\frac{(1+t)^{n+1}-(1-t)^{n+1}}t\,dt=\lim_{n\to\infty}\int_{\epsilon}^1\frac 2 t\frac {n+1}{n+2} d\Big(\big(\frac {1+t}{2}\big)^{n+2}+\big(\frac {1-t} 2\big)^{n+2}\Big)=\int_{\epsilon}^1\frac 2 t dI(t=1)=2$$ since $\frac 2 t$ is bounded on $[\epsilon,1]$. Hence the original limit is $2$. | 2020-10-20T07:15:18 | {
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http://mathhelpforum.com/pre-calculus/168532-complex-numbers-question.html | # Math Help - Complex Numbers Question
1. ## Complex Numbers Question
Hey everybody,
So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.
If x and y are real, solve the equation:
$\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}$
I started manipulating it to make it more manageable:
$\frac{jx(x+3y)}{1+jy}=3x+j4$
$jx^2+j3xy=(3x+j4)(1+jy)$
etc.
I have the feeling I'm going nowhere with it. Could somebody please give me a hand?
Cheers,
Evanator
2. Originally Posted by evanator
Hey everybody,
So I have been doing a few exercises on complex numbers, but there is one I can't seem to get out.
If x and y are real, solve the equation:
$\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}$
I started manipulating it to make it more manageable:
$\frac{jx(x+3y)}{1+jy}=3x+j4$
$jx^2+j3xy=(3x+j4)(1+jy)$
etc.
I have the feeling I'm going nowhere with it.
In fact, you,re going in exactly the right direction. Next step is to multiply out the brackets on the right (remembering that $j^2=-1$). Then compare the real parts and the imaginary parts on both sides of the equation.
3. $\frac{jx}{1+jy}=\frac{3x+j4}{x+3y}
$
${(jx)}{(x+3y)}=(1+jy)(3x+j4)$
$jx^2+3jxy=3x+4j+3jxy-4y$
$jx^2=3x+4j-4y$
so:
(1) $x^2=4$
(2) $0=3x-4y$
Solve this system for x and y
4. Thank you both of you. Finishing it off:
$jx^2+j3xy=3x+j3xy+j4+(j^2)4y$
$0+jx^2=(3x-4y)+j(4)$
$x^2=4$
$3x-4y=0$
$x=\pm2$
Taking the positive value of x and solving the simultaneous equations:
$3x=6$
$-3x+4y=0$
$4y=6$
$y=\frac{6}{4}=1.5$
If we take x=-2 we can see that y will come out as -1.5, so the final results are:
$x=\pm2$
$y=\pm1.5$
5. No, that is NOT the answer because x and y are not independent.
The correct answer is "x= 2 and y= 1.5 or x= -2 and y= -1.5"
Writing " $x= \pm 2, y= \pm 1.5$" implies that any combination of those, such as x= 2, y= -1.5, will satisfy the equation and that is not true.
6. Right. Thanks, HallsofIvy. I did understand that the x-value and y-value were dependent on each other, but I didn't realize I was implying otherwise with the plus-minus sign. | 2015-11-27T21:11:58 | {
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