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https://numberwarrior.wordpress.com/2008/12/03/on-the-ancient-babylonian-value-for-pi/ | ## On the Ancient Babylonian Value for Pi
I have written about the ancient Egyptian value for $\pi$ before, concluding that while the Egyptians had a procedure for finding the area of a circle, they didn’t have any real understanding of the ratio.
Conversely, the Babylonians found $\pi$ as a ratio (3.125) but, oddly enough, didn’t handle circles as well as the Egyptians.
Consider this tablet from the Yale Babylonian Collection known as YBC 7302:
Since the photo is hard to interpret, here’s a version with the numbers made clear:
That’s cuneiform, which fortunately in numbers isn’t too hard to read.
Based on positional context, 3 seems to be the circumference of the circle. Using the formulas $C=2 \pi r$ and $A= \pi r^2$,
$3=2 \pi r$
$\frac{3}{2 \pi}=r$
$A=\pi(\frac{3}{2 \pi})^2$
$A=\frac{9}{4 \pi}$
While we’re used to a base 10 system ($d_1 + \frac{d_2}{10} + \frac{d_3}{100} + \frac{d_4}{1000} + ...$) the Babylonians used a base 60 system ($d_1 + \frac{d_2}{60} + \frac{d_3}{60^2} + \frac{d_4}{60^3} + ...$). Also at the time the Babylonians had no place value, so if there’s the number “45” only context can tell if they mean 45, 45/60, or 45/60*60.
If we take the 45 from the tablet to mean 45/60:
$\frac{45}{60} = \frac{9}{4 \pi}$
and the Babylonian value of $\pi$ turns out to be simply 3.
Now, if you remember my Egyptian post, their value of $\pi$ was also pulled out of a circle area procedure in the exact same manner. Arguably it is unfair to go any further; there are good reasons to call the Babylonian value of $\pi$ 3 and stop there.
However, there was some tablets found in 1936 which throw the case for a loop.
The tablets were found in Susa, in ancient times capital city of the Elamite Empire. (It also happens to be one of the oldest continuously inhabited cities in the world, at more than 7000 years.)
There are pictures in the Textes mathématiques de Suse, but I cannot reproduce them for copyright reasons.
(For this part I’m referring to The Exact Sciences in Antiquity by Otto Neugebauer.)
One of the tablets (this one, I think) contains a list of geometrical constants. For example, it gives the number $\frac{5}{3}$ in relation to a regular pentagon, apparently meaning that:
The area of a pentagon = $\frac{5}{3}$ * The side of the pentagon2
The actual number here should be $\frac{ \sqrt{25+10\sqrt5}}{4}$, or approximately 1.72, so the Babylonians were off by only 0.06. The tablet also gives constants used in the same way for the hexagon (2.625, about 0.027 off the real value) and for the heptagon (3.683, about 0.049 off the real value).
One of the other constants is 24/25. On the tablet it matches with a circle with a hexagon inscribed inside.
Suppose we take 24/25 to be the ratio of the perimeter of the hexagon p to the circumference of the circle C.
$\frac{24}{25}=\frac{p}{C}$
If the radius of the circle is r, adding some equilateral triangles reveals the perimeter of the hexagon is 6r. So:
$\frac{24}{25}=\frac{6r}{2\pi r}$
$\frac{24}{25}=\frac{6}{2\pi}$
$\frac{24}{25} \cdot \frac{2}{6}=\frac{1}{\pi}$
$\frac{48}{150} = \frac{1}{\pi}$
$\frac{150}{48} = \pi$
$\frac{25}{8} = \pi$
or alternately 3.125.
Because this is given as an actual fixed ratio (rather than being extrapolated from a circle area procedure) it’s arguably the first discovered value for $\pi$. It’s also intriguing in that relating the inscribed polygon to the circle is how Archimedes gets the first truly rigorous calculation of $\pi$ — he just adds more sides to get a closer estimate. However, I can’t give the Babylonians full laurels because this is the only place this value of $\pi$ appears. They never reapplied it back to any problem requiring the area or circumference of a circle.
### 12 Responses
1. when did the babylonians do this exactly?
2. The Babylonians value of Pi may be proved geometrically. Please refer to the following link for a paper published recently.
http://www.ijoart.org/docs/Geometric-Estimation-of-Value-of-Pi.pdf
3. awsome
4. Is there any research that shows that Pi was used? if so HOW???
5. sir, namasthe! The exact pi value is 14_root2/4. Equal to 3.14644660942…. If you are interested your postal address please. Regards yours faithfully rsj reddy india.
6. the day I was born
7. Thanks . helped for my Pi Day project.
8. […] Pi has been recognized for just about 4000 years and our earliest known determination of it was the Ancient Babylonians on a tablet, who recognized that the circumference of a circle was roughly 3 x the diameter, π = 3. Just about […]
9. […] Pi has been recognized for just about 4000 years and our earliest known determination of it was the Ancient Babylonians on a tablet, who recognized that the circumference of a circle was roughly 3 x the diameter, π = 3. Just about […]
10. […] the Babylonians did discover a more accurate approximation for pi (namely, ) by circumscribing a regular hexagon into a circle, but it seems that the babylonians […]
11. […] It’s defined to be the ratio between the circumference of a circle and the diameter of that circle. This ratio is the same for any size circle, so it’s intrinsically attached to the idea of circularity. The circle is a fundamental shape, so it’s natural to wonder about this fundamental ratio. People have been doing so going back at least to the ancient Babylonians. […]
12. […] It’s defined to be the ratio between the circumference of a circle and the diameter of that circle. This ratio is the same for any size circle, so it’s intrinsically attached to the idea of circularity. The circle is a fundamental shape, so it’s natural to wonder about this fundamental ratio. People have been doing so going back at least to the ancient Babylonians. […] | 2015-08-04T07:20:08 | {
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https://math.stackexchange.com/questions/2054273/cos-arcsinx-cdots | $\cos(\arcsin(x)) = \cdots$
I've been asked to prove $$y=\frac{\sqrt{3}} 2 x+\frac 1 2 \sqrt{1-x^2}$$
given $x=\sin(t)$ & $y=\sin(t+\frac \pi 6)$
I did $t=\arcsin(x)$ and plugged that into the $y$ equation. Used the $\sin(a+b)$ identity to get: $$y=x\cos\left(\frac \pi 6\right)+\frac{\cos(\arcsin(x))}2 = \frac{\sqrt3} 2 x+\frac{\cos(\arcsin(x))} 2$$
Now I'm sure there must be an identity for $cos(arcsin(x))$ however I'm unaware of it. I'm also unaware of how to prove it.
which is seemingly exactly what I need to complete the question; however, it wouldn't be proving that $y = \text{answer}$ if I didn't show how to get to this result.
Is there a "more correct" way to complete this question without having to fiddle with this formula / arcsins etc.
The usual proof involves drawing a triangle. The opposite side will be called $\sin(y)=x$, the hypotenuse will be $1$, and the angle will be $\arcsin x$. Can you use the pythagorean theorem to find $\cos(\arcsin x)$?
• This makes a lot of sense, I guess the - relative - complication of the question intimidated me... – Tobi Dec 11 '16 at 18:46
• @Tobi As is the nature of trigonometry sometimes, but often it ends up beautiful :) – Simply Beautiful Art Dec 11 '16 at 22:17
• I find there to be so many directions in which I can take a question, it's often a matter of chance. Maybe with experience I'll be able to tell, faster, which identities to use. – Tobi Dec 11 '16 at 23:54
• @Tobi Yes, definitely practice your trig identities. Personally, the most important ones are the Pythagorean identities (what we have here) and the sum of angles formula. Combined, you can get most of the other formulas. Learning how to use the formulas will become especially useful in some parts of calculus btw. – Simply Beautiful Art Dec 12 '16 at 0:35
• The solutions can be so illusive sometimes, simple rearrangement / substitution can often take me forever to spot. It's like there's no instant way to do them, just brute force. – Tobi Dec 12 '16 at 0:42
$$\cos^2(\arcsin(x))+\sin^2(\arcsin(x))=1\\\cos^2(\arcsin (x))+x^2=1\\\cos^2(\arcsin(x))=1-x^2\\\cos(\arcsin(x))=\sqrt{1-x^2}$$ The last step is okay because $-1\leq\arcsin x\leq 1$ and $\cos$ is positive there.
• Nice, algebraic, solution! – Tobi Dec 11 '16 at 18:48
$$y=\sin\left(t+{\pi\over 6}\right)=\cos{\pi\over 6}\sin t+\sin{\pi\over 6}\cos t\\={\sqrt3\over 2}x+{1\over 2}\sqrt{1-x^2}$$
• What is this showing? – Tobi Dec 11 '16 at 18:53
• @Tobi Can't this be a more correct way to do this question rather than to fiddle with the arcsins? – Qwerty Dec 11 '16 at 18:55
• How do you go from $\cos{t}$ to $\sqrt{1-x^2}$ – Tobi Dec 11 '16 at 18:58
• @Tobi $$\cos t=\sqrt {1-\sin^2 t}=\sqrt{1-x^2}$$ – Qwerty Dec 11 '16 at 19:00
• Oh I see, very illusive. – Tobi Dec 11 '16 at 19:00
Draw a right triangle in which the "opposite" side has length $x$ and the hypotenuse has length $1$. Then the sine of the angle having that "opposite" side is $\sin=\dfrac{\text{opposite}}{\text{hypotenuse}} =\dfrac x 1 = x.$ So that angle is $\arcsin x$.
Now use the Pythagorean theorem to show that the "adjacent" side has length $\sqrt{1-x^2}$. Then we have $$\cos\arcsin x = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\sqrt{1-x^2}} 1 = \sqrt{1-x^2}.$$ | 2019-10-18T11:16:09 | {
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https://math.stackexchange.com/questions/703282/how-do-you-compute-71000-mod-24/703324 | # How do you Compute $7^{1000} \mod 24$?
I'm being asked to compute $7^{1000} \mod 24$. I have Fermat's Little Theorem and Euler's Theorem. How do I use these to compute $7^{1000} \mod 24$? I'm stuck because $24$ is not prime. In this case, I think I have to use Euler's Theorem. Can anyone show me what to do?
• Did you notice that if you square 7 you get 1 mod 24? That may help here I'd think. – JB King Mar 7 '14 at 18:02
You could simply take the square of $7$ to find that $$7^2 \equiv 1 \mod 24$$
But just for educational purposes: By Euler's Theorem (valid because $GCD(7, 24) = 1$),
$$7^{\phi{(24)}} \equiv 1 \mod 24$$
Now, $\phi(24) = 24\cdot\left(1 - \frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right) = 8$. (See here for details). Hence,
$$7^8 \equiv 1 \mod 24$$
Raise both sides to the power of $125$ to deduce that :
$$7^{1000} \equiv 1 \mod 24$$
If go ahead and use the Charmichael function, you will obtain an even stronger result than Euler's Theorem:
$$7^2 \equiv 1 \mod 24$$
which is what we saw earlier on.
Note that $49\equiv 1 \mod 24$
• So $7^{N}$ 24 modulos (is it a verb?) to N/2 which solves pretty fast. That was the thought, right? – JTP - Apologise to Monica Mar 7 '14 at 18:24
• The point is that $7^{1000}=(7^2)^{500}$ - also for modulus $m$ we have $(km+r)(lm+s)=(kl+k+l)m+rs \equiv rs \mod m$ – Mark Bennet Mar 7 '14 at 18:27
You can choose the method you like best:
1. Simply calculate powers of $7$ modulo $24$... In this case $$7^{1000}\equiv (7^2)^{500}\equiv 49^{500}\equiv 1^{500}\equiv 1 \bmod 24.$$
2. Since $\gcd(7,24)=1$, you can use Euler's theorem. Note that $\varphi(24)=8$. $$7^{1000}\equiv (7^8)^{125}\equiv 1^{125}\equiv 1 \bmod 24.$$
3. You can use the Chinese remainder theorem. Since $24=3\cdot 8$, we have that $x\equiv 7^{1000}$ modulo $3$ and $8$, if and only if $x\equiv 7^{1000} \bmod 24$. Now, $$7^{1000}\equiv 1^{1000}\equiv 1 \bmod 3, \text{ and } 7^{1000}\equiv (-1)^{1000}\equiv 1 \bmod 8,$$ and therefore $7^{1000}\equiv 1 \bmod 24$.
Hint: $7\equiv1\mod3$, and $7\equiv-1\mod8$, and $24=3\cdot8$.
• How do I use $24 = 3 \cdot 8$? Is there a fact about modular arithmetic that I'm unaware of here? – Newb Mar 7 '14 at 18:03
• The Chinese remainder theorem! – Álvaro Lozano-Robledo Mar 7 '14 at 18:07
• @Newb: What number in between $0$ and $23$ gives the remainder $1$ when divided through both $3$ and $8$ ? :-) – Lucian Mar 7 '14 at 18:09
Also $$7^{n + 2} = \pars{24p + \mu}7^{2} = 24p\times 49 + 48\mu + \mu = \pars{49p + 2\mu}\times 24 + \mu\,,\ \left\vert% \begin{array}{rcl} \mbox{Also,}&& \\ n = 0 & \imp & \mu = 1 \\ n = 1 & \imp & \mu = 7 \end{array}\right.$$
Then, $$7^{n} \mod 24 =\left\lbrace \begin{array}{rcl} 1 & \mbox{if} & n\ \mbox{is}\ even \\ 7 & \mbox{if} & n\ \mbox{is}\ odd \end{array}\right.$$
Then $$\color{#00f}{\large 7^{1000}\mod 24 = 1}$$ | 2020-02-20T11:16:16 | {
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https://math.stackexchange.com/questions/2877904/there-is-only-in-first-order-logic | # "There is only" in first order logic
I'm trying to translate the statement "There is only three things that are not small" into first order logic. I'm using some software to verify my sentences, but I feel like I don't understand what "There is only" is meant to claim.
I interpreted it as "There are at most", and used the answer here, in particular method 2 for "At most $n$".
The sentence I've produced with this is:
$$\exists x \exists y \exists z \forall w \, (\lnot \text{Small}(w) \to (w = x \lor w = y \lor w = z))$$
Which I understand to mean that there exists up to 3 objects, which, for all objects $w$, if it is not small, it is one of the 3 objects.
This passes 3 of the 4 test worlds for the software, but fails on the last one.
I was hoping someone could help me clarify what is meant by "only". I've spent a decent amount of time Googling, but most results lead to explanations of the biconditional.
Thanks!
• Your statement is a good translation of 'at most 3 things are not small' .... maybe they mean 'exactly 3 things are not small?' Your test worlds should give you a clue ... Aug 10 '18 at 1:47
• I hate these kind of questions. It's not about first order logic, it's about what "there are only" means in ordinary English. In real life, "there are only three" means the same as "there are three and only three" or "there are exactly three". But maybe your instructor or your textbook author thinks it means something else.
– bof
Aug 10 '18 at 1:49
• @AnyAD but the universal says that any non-small object has to be one of those three ... so there really cannot be more than 3 Aug 10 '18 at 1:51
• I would see if the last world demands that there be three. You could add $x \neq y \wedge y \neq z \wedge x \neq z$ to your sentence and see if it works. That would support the theory that only three means exactly three. Aug 10 '18 at 2:05
• @Bram28 Was right, the test worlds seem to indicate that "Exactly 3" was the correct way to read the problem. Aug 10 '18 at 2:14 | 2021-12-01T22:41:22 | {
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http://mathhelpforum.com/algebra/230007-find-all-values-create-no-solution-x.html | # Thread: Find All Values of "A" That Create No Solution for "X"
1. ## Find All Values of "A" That Create No Solution for "X"
Hello everybody!
These two problems have been boggling me for awhile now:
#1)1/(1+1/x) = a Find all values of "a" that yield no solution for "x"
#2)(6x-a)/(x-3) = 3 Find all values of "a" that yield no solution for "x"
I'm unsure of how to solve for this type of solution, but I've tried isolating the "x" variables like so:
#1) 1/(1+1/x) = a
1) 1 = a + a/x
2) 1 - a = a/x
3) x/a = 1/(1-a)
4) x = a/(1-a)
#2) (6x-a)/(x-3) = 3
1) 6x - a = 3x - 9
2)
3x - a = -9
3) 3x = a - 9
4) x = a/3 - 3
However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
2. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by Lexielai
Hello everybody!
These two problems have been boggling me for awhile now:
#1)1/(1+1/x) = a Find all values of "a" that yield no solution for "x"
#2)(6x-a)/(x-3) = 3 Find all values of "a" that yield no solution for "x"
I'm unsure of how to solve for this type of solution, but I've tried isolating the "x" variables like so:
#1) 1/(1+1/x) = a
1) 1 = a + a/x
2) 1 - a = a/x
3) x/a = 1/(1-a)
4) x = a/(1-a)
#2) (6x-a)/(x-3) = 3
1) 6x - a = 3x - 9
2)
3x - a = -9
3) 3x = a - 9
4) x = a/3 - 3
However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
You almost have it for the first problem. What values of a produce no solution for x? (Take a look at what a can't be.)
For the second one, I see no reason to restrict the value of a. Did you type it correctly?
-Dan
3. ## Re: Find All Values of "A" That Create No Solution for "X"
I see what you mean with the first problem. It's easy to discern logically. I'm still confused with the second one though, and I did type it exactly like the original problem.
My friend told me the answer was "18" after plugging in values, but I'm more concerned with a proper method to find the answer than the answer itself. Any help would be appreciated!
4. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by Lexielai
Hello everybody!
These two problems have been boggling me for awhile now:
#1)1/(1+1/x) = a Find all values of "a" that yield no solution for "x"
#2)(6x-a)/(x-3) = 3 Find all values of "a" that yield no solution for "x"
I'm unsure of how to solve for this type of solution, but I've tried isolating the "x" variables like so:
#1) 1/(1+1/x) = a
1) 1 = a + a/x
2) 1 - a = a/x
3) x/a = 1/(1-a)
4) x = a/(1-a)
#2) (6x-a)/(x-3) = 3
1) 6x - a = 3x - 9
2)
3x - a = -9
3) 3x = a - 9
4) x = a/3 - 3
However, I don't really know how to express "x" as a no solution value to be solved for, so I may have made errors in my steps. If someone could lend me some insight, I would be very grateful!
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.
The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$
Now solve for x. $x = \dfrac{a}{1 - a} \implies a \ne 1.$ And $0 \ne x = \dfrac{a}{1 - a} \implies a \ne 0.$
So I would answer problem 1 as a is not equal to either 0 or 1.
In the second problem, it is clear that $x \ne 3.$
Now solving for x we get $\dfrac{6x - a}{x - 3} = 3 \implies 6x - a = 3x - 9 \implies 3x = a - 9 \implies x = \dfrac{a - 9}{3}.$
$3 \ne x = \dfrac{a - 9}{3} \implies 9 \ne a - 9 \implies a \ne 18.$
Solving for a we get $a = 3x + 9,$ which raises no problems. So a is not equal to 18.
Does this help
5. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by JeffM
I'd attack this somewhat differently. I'd start by look for impermissible values in the original expression.
The first problem gives an expression for a, and from that we can deduce that $x \ne 0\ and\ x \ne - 1.$
Now solve for x. $x = \dfrac{a}{1 - a} \implies a \ne 1.$ And $0 \ne x = \dfrac{a}{1 - a} \implies a \ne 0.$
So I would answer problem 1 as a is not equal to either 0 or 1.
In the second problem, it is clear that $x \ne 3.$
Now solving for x we get $\dfrac{6x - a}{x - 3} = 3 \implies 6x - a = 3x - 9 \implies 3x = a - 9 \implies x = \dfrac{a - 9}{3}.$
$3 \ne x = \dfrac{a - 9}{3} \implies 9 \ne a - 9 \implies a \ne 18.$
Solving for a we get $a = 3x + 9,$ which raises no problems. So a is not equal to 18.
Does this help
I don't agree with this.
Let a=18
$\dfrac {6x-18}{x-3}=\dfrac {6(x-3)}{x-3}=6$
There's nothing invalid about x=3 if a=18.
6. ## Re: Find All Values of "A" That Create No Solution for "X"
Hi,
If you know about functions, domain and range, maybe the following will help.
1. Rephrase the problem as f(x)=1/(1+1/x); the domain of f is then all reals except x=0 and x=-1. Now what value a is not in the range of f? Answer a=0 and a=1.
2. Again rephrase the problem. f(x)=(6x-a)/(x-3) for some a. For any a, the domain of f is all reals except x=3. Now the horizontal asymptote is y=6, no matter the value of a unequal to 18. (If a=18, the graph of f is the horizontal line y=6 with a hole at x=3) So the range of f is all real numbers y except y=6 except when a=18. So the only value of a so that f(x)=3 can not be solved for x is a=18. You might want to graph f for various a values.
7. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by romsek
I don't agree with this.
Let a=18
$\dfrac {6x-18}{x-3}=\dfrac {6(x-3)}{x-3}=6$
There's nothing invalid about x=3 if a=18.
Yes, there is. $\displaystyle \frac{6x- 18}{x- 3}= 6$ ONLY if x is NOT 3.
$\displaystyle f(x)= \frac{6x- 18}{x- 3}= 6\frac{x- 3}{x- 3}$ has domain "all real numbers except 3.
The graph of y= f(x) is the horizontal line y= 6 with a hole at x= 3.
8. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by HallsofIvy
Yes, there is. $\displaystyle \frac{6x- 18}{x- 3}= 6$ ONLY if x is NOT 3.
$\displaystyle f(x)= \frac{6x- 18}{x- 3}= 6\frac{x- 3}{x- 3}$ has domain "all real numbers except 3.
The graph of y= f(x) is the horizontal line y= 6 with a hole at x= 3.
Really? This is the first time I've seen it stated that you can't always divide out common factors.
Not the first time I've been wrong.
9. ## Re: Find All Values of "A" That Create No Solution for "X"
Saying that $\displaystyle \frac{6x- 18}{x- 3}= 6\frac{x- 1}{x- 1}= 6$ when x= 1 is the same as saying that $\displaystyle \frac{0}{0}= 1$. It isn't.
$\displaystyle f(x)= \frac{6x- 18}{x- 3}$ and $\displaystyle g(x)= 6$ are almost the same but are not exactly the same.
10. ## Re: Find All Values of "A" That Create No Solution for "X"
Originally Posted by romsek
Really? This is the first time I've seen it stated that you can't always divide out common factors.
Not the first time I've been wrong.
Like anyone that has studied Calculus for too long you might be trying to recall
$\displaystyle \lim_{x \to 3} \frac{6(x - 3)}{x - 3} = 6$
which is perfectly valid.
-Dan | 2018-03-19T03:29:52 | {
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http://math.stackexchange.com/questions/40494/formal-notation-related-to-a-sequence-or-a-set | # Formal notation related to a sequence or a set
My question is quite naive...
I just want to represent a finite sequence of Natural number, is it the best way to write it like this?:
$\langle a_0, \ldots, a_n \rangle$, where $\forall i \in [ 0, n ], a_i \in \mathbb{N}$
I have also seen somewhere $\{ a_i \}$, does it mean a set (or a sequence) of variables? is it finite or infinite? where could i add the constraint like $\forall i \in [ 0, n ], a_i \in \mathbb{N}$?
Also, is there any difference between "tuple" and "sequence"?
Hope my question is clear, I just want to make sure what I write matches the convention...
Thank you very much
-
There are several usual notations:
• Let $\bar a = \langle a_i\mid i<n\rangle \in \mathbb N^{<\mathbb N}$... (where $\mathbb N^{<\mathbb N}$ means the collection of all finite sequences of natural numbers, and $\mathbb N^\mathbb N$ is the collection of all infinite sequences)
• Let $\bar a = \langle a_i\mid i<n\rangle$ be a finite sequence of natural numbers...
And you can always replace $\langle a_i\mid i<n\rangle$ by $\langle a_0,\ldots,a_{n-1}\rangle$.
Note that $[0,n]$ can be read as the real numbers between $0$ and $n$.
The important thing is to be clear in your intention, and consistent in your notation.
And lastly, the difference between a "tuple" and a "sequence" is the context, usually tuple is used for finite length, while sequence is for infinite ones. Also, tuples are usually for a constant length, while sequences are a variable length.
The notation $\{a_i\}$ means that this is a set whose elements are distinct. Naturally it says $i\in I$ for some index set $I$, which in turn might be that this is just a sequence (or a directed set). The meaning is usually clear from context, read more - a lot more - and things will slowly clarify.
-
Thanks for your reply. Does the $<\mathbb{N}$ mean in $\mathbb{N}^{<\mathbb{N}}$ "finite", so does $\mathbb{N}^{\mathbb{N}}$ mean "infinite"? – SoftTimur May 21 '11 at 16:26
Also, what could "$\{a_i\}$" mean in a normal context? – SoftTimur May 21 '11 at 16:39
Sometimes I see $a_1, a_2, ...$ without the brackets.
- | 2016-04-29T14:39:36 | {
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https://programmingpraxis.com/2013/04/05/last-non-zero-digit-of-a-factorial/2/ | Last Non-Zero Digit Of A Factorial
April 5, 2013
The obvious brute-force solution is to calculate n factorial, then repeatedly divide by 10 until the remainder is non-zero:
(define (factorial n) (let loop ((n n) (f 1)) (if (zero? n) f (loop (- n 1) (* f n)))))
(define (lnz1 n) (let loop ((f (factorial n))) (if (zero? (modulo f 10)) (loop (quotient f 10)) (modulo f 10))))
> (lnz1 15) 8
The problem, of course, is that n! grows very quickly, so this solution is either slow or impossible to calculate.
A common solution that does not work is to calculate the factorial in the normal way, but remove all trailing zeros at each step. For instance, 15! = 1307674368000 and the last non-zero digit is 8, but removing all trailing zeros at each step gives a last non-zero digit of 3:
(define (lnz2 n) ; doesn't work (let loop ((i 2) (f 1)) (cond ((zero? (modulo f 10)) (loop i (/ f 10))) ((< 9 f) (loop i (modulo f 10))) ((< n i) f) (else (loop (+ i 1) (* f i))))))
> (lnz2 15) 3
We can fix that solution with a little bit of work. Factors of 2 and 5 are special because they are the factors of 10 that we remove with the trailing zeros. Thus we remove factors of 2 and 5 from each number from 1 to n, counting them as we go, multiply the remainder of each number from 1 to n after removing factors of 2 and 5 by the accumulating product, using arithmetic modulo 10, and finally multiply by the excess 2s greater than the count of 5s, again using arithmetic modulo 10:
(define (lnz3 n) (let loop1 ((n n) (z 1) (two 0) (five 0)) (if (zero? n) (modulo (* z (expm 2 (- two five) 10)) 10) (let loop2 ((m n) (two two) (five five)) (cond ((zero? (modulo m 2)) (loop2 (/ m 2) (+ two 1) five)) ((zero? (modulo m 5)) (loop2 (/ m 5) two (+ five 1))) (else (loop1 (- n 1) (* z m) two five)))))))
> (lnz3 15) 8
That works, but takes time linear in n, although at least it never overflows like lnz1; on my machine, (lnz3 1000000) takes twenty minutes. It’s also the best solution I was able to come up with on my own.
If you’re good at math, there is a very fast solution. Beni Bogosel gives this explanation at his blog, and Google points to many others that are similar. We have the formula
$(5q)! = 10^q q! \prod_{i=0}^{q-1} \frac{(5i+1)(5i+2)(5i+3)(5i+4)}{2}$
which can be proved by removing from (5q)! terms divisible by 5: 5, 10, 5q. Since
$\frac{(5i+1)(5i+2)(5i+3)(5i+4)}{2} \equiv 2 \pmod{10}$
we obtain the recurrence L(n) ≡ 2q L(q) L(r) (mod 10), where L(n) is the last non-zero digit of n! and n = 5q + r. The calculation of L(n) descends exponentially at every step to reach small numbers for which it is easy to calculate the digit. Here’s the code:
(define (lnz4 n) (define (p k) (if (< k 1) 1 (vector-ref '#(6 2 4 8) (modulo k 4)))) (define (l n) (if (< n 5) (vector-ref '#(1 1 2 6 4) n) (let ((q (quotient n 5)) (r (remainder n 5))) (modulo (* (p q) (l q) (l r)) 10)))) (l n))
> (lnz4 15) 8 > (lnz4 1000000) 4
That computation of (lnz4 1000000) is instantaneous. You can see the sequence L(0), L(1), … at A008904.
By the way, it is much easier to compute the number of trailing zeros than the last non-zero digit. MathWorld gives this function (A027868):
(define (z n) (let loop ((k (ilog 5 n)) (z 0)) (if (zero? k) z (loop (- k 1) (+ z (quotient n (expt 5 k)))))))
> (z 15)
3
> (z 1000000)
249998
I’m not sure why this is a favorite exercise for beginning programmers; I guess because it’s a fun way to introduce the modulo operator. But it’s much more of a math problem than a programming problem.
We used ilog and expm from the Standard Prelude. You can run the program at http://programmingpraxis.codepad.org/KC3nkLTk.
Pages: 1 2
15 Responses to “Last Non-Zero Digit Of A Factorial”
1. izidor said
I came up with a linear solution where the trailing zeros are trimmed and stored only the last non-zero digit. On my computer nonZeroDigit 1000000 finishes in about 9 seconds.
nonZeroDigit :: Integral a => a -> a nonZeroDigit x = foldr1 (\n acc -> trim (n * acc)) [1..x] where trim x = if x mod 10 > 0 then x mod 10 else trim (x div 10)
2. Egil said
lnzd = head ∘ dropWhile (≡’0′) ∘ reverse ∘ show ∘ fac
where fac n = foldr1 (*) [1‥n]
3. swuecho said
in perl6
sub lnzd1($n) { ([*] 1..$n).Str.subst(/0/,'', :g).substr(*-1) }
sub lnzd_number(Int $n ) { my$m = $n; while$m %% 10 { $m =$m div 10 } $m; } sub lnzd2(Int$n) { sub helper( Int $x , Int$acc) { if $x ==$n { lnzd_number($acc *$x) } else { helper($x+1,lnzd_number($acc * $x) ); } } helper(1,1) % 10; } 4. Globules said Here’s a relatively fast Haskell version based on a little number theory trickery at http://comeoncodeon.wordpress.com/2009/06/20/lastnon-zero-digit-of-factorial/. With an argument of 1000000 it runs in about 0.006 seconds on a 1.7 GHz Intel Core i5. import Control.Monad (liftM) import Data.List (genericIndex) import System.Environment (getArgs) -- Return the least significant non-zero digit of n factorial. lnzf :: Integer -> Int lnzf n | n < 5 = [1, 1, 2, 6, 4] genericIndex n | otherwise = let (q, r) = n quotRem 5 in (p q * lnzf q * lnzf r) rem 10 where p 0 = 1 p m = [6, 2, 4, 8] genericIndex (m rem 4) main :: IO () main = do ns <- liftM (map read) getArgs mapM_ (\n -> putStrLn$ show n ++ "! -> " ++ show (lnzf n)) ns
5. jeltz said
# Ruby
def last_digit(n)
n % 10
end
def last_non_zero_digit(n)
while last_digit(n) == 0
n = n / 10
end
last_digit n
end
max = ARGV[0].to_i
puts (1..max).inject(1) { |result, n| result = last_non_zero_digit(result * n) }
6. But, wait, solution 2 will work with a couple changes, no?
(define (lnz2 n) ; doesn’t work
(let loop ((i 2) (f 1))
(cond ((zero? (modulo f 10)) (loop i (/ f 10)))
((< n i) (modulo f 10))
(else (loop (+ i 1) (* f i))))))
(display (lnz2 15)) (newline)
7. John said
My solution in Java:
8. John said
My solution in Java (link): http://pastebin.com/awgkzbb8
9. Akshar said
This is my solution done in Haskell.
–the factorial function to be used
factorial :: Int -> Int
factorial n=product [1..n]
–Done in two steps.
–Step 1.Find the last non-zero digit of a number.
lastNonZeroDigit :: Int -> Int
lastNonZeroDigit n = if n mod 10 /= 0
then n mod 10
else lastNonZeroDigit (n quot 10)
–Step 2.Now putting the output of factorial into the lastNonZeroDigit function input
lastNZDfactorial :: Int -> Int
lastNZDfactorial n = lastNonZeroDigit ( factorial (n))
——————————————————————————————————
This could probably have been done without two functions for the last digit but
in Haskell this way is prefered(That is what I heard).
Again the lines with the “::” can be omitted.
I am new to programming in Haskell so I would appreciate someone can tell me how to
improve this or my Haskell skills overall.
10. generic solution in python:
# Program to print
# the last non zero digit
# in a factorial
def Fact(n):
if n < 1: return 1
else: return n * Fact(n-1)
def Convert_Num_Str():
lst_append = []
n = input("Enter the number whose factorial has to be calculated:>")
result = Fact(int(n))
str_num = list(str(result))
print(str_num)
for i in range(len(str_num)):
if str_num[int(i)]!= '0':
lst_append.append(str_num[i])
return int(lst_append[-1])
print(Convert_Num_Str())
11. This is using simple division and loop and exit control logic
# Program to print
# the last non zero digit
# in a factorial
def Fact(n):
if n < 1: return 1
else: return n * Fact(n-1)
def Convert_Num_Str():
n = input("Enter the number whose factorial has to be calculated:>")
result = Fact(int(n))
while result!=0:
print("Result before division", result)
remainder = result%10
print(remainder)
if remainder!=0:
break
else:
result = result/10
print("Result in Else",result)
remainder = result%10
print("Remainder in Else",remainder)
return remainder
print(Convert_Num_Str())
12. HARSH DEPAL said
in C++
#include
#include
#include
using namespace std;
int main()
{
int n;
cout<<"ENTER A NUMBER"<>n;
int p=0;
int fact=1;
int LastDigit=0;
while (p<n)
{
p++;
fact=fact*p;
LastDigit=fact%10;
if (LastDigit==0)
fact=fact/10;
}
cout<<""<<LastDigit<<" Is The Last Digit Of The Factorial"<<endl;
getch();
return 0;
}
13. itsme86 said
C#:
static void PrintLastNonZeroDigitInFactorial(int factorial)
{
Console.WriteLine(IntSplitReverse(Enumerable.Range(1, factorial).Aggregate(1, (product, nextNumber) => product * nextNumber)).FirstOrDefault(d => d != 0));
}
static IEnumerable<int> IntSplitReverse(int num)
{
List<int> digits = new List<int>();
while (num > 0)
{
num /= 10;
}
return digits;
}
14. Nothing super-duper, but a Ruby solution:
def last_non_zero_fact_dig(n)
fact = (1..n).inject(1){|prod, i| prod * i}
(0..Math::log10(fact)).each do |k|
digit = (fact / 10**k) % 10
return digit if digit != 0
end
end | 2017-12-17T15:32:02 | {
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https://brilliant.org/discussions/thread/som/ | # Divisors of 999...999 using division in a unusual way
Which integers can be multiplied by another integer so the product is in the form of 999....999?
To answer this question, will use some concepts, though I don't have a rigorous proof of these(I will update this note if I find something), a irreducible fraction:
1) Will only have terminating decimals if the denominator can be written as product of powers of two and five;
"Proof": $\frac{n}{2^a\times5^b}=\frac{n}{2^a\times5^b}\times \frac{10^c}{10^c}=\frac{n\times2^{c-a}\times5^{c-b}}{10^c}$, with $c=max(a,b)$
2) Will only have non-terminating decimals if the denominator is not a multiple of two and five, but is a multiple of another prime number;
3) Will have both terminating(non-repeating) and non-terminating decimals if the denominator is a multiple of 2 or 5, and is a multiple of another prime;
I will use the second case; It is possible to revert it back to a fraction:
$0.\overline{\text{(part that repeats)}}=\frac{\text{part that repeats}}{\text{a sequence of 9 with the same number of digits as the part that repeats}}$
So there is a way to transform a(irreducible) fraction whose denominator is not a multiple of two and five to a fraction whose denominator is a sequence of nines
$\frac{1}{n}=\frac{a}{99...999}$ if n is not a multiple of 2 nor 5(if $n=1$(not a multiple of another prime) then it is trivial)
$n=\frac{99...999}{a}$
$n\times a=99...999$
With this, I can, for instance, find the number that, multiplied by 17, will return a sequence of nines:
$\frac{1}{17}=0.\overline{0588235294117647}=\frac{588235294117647}{9999999999999999}$
$17=\frac{9999999999999999}{588235294117647}$
$17\times588235294117647=9999999999999999$
Note by Matheus Jahnke
4 years, 5 months ago
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Every integers except those divisible by 2 and 5, can be multiplied in such a way.
For any integer $q$ ,
Consider order of $10 (mod q)$,
denote it $k$ .
Then $10^k \equiv 1 \Rightarrow 99999.....9 \equiv 0 (mod q)$
But we can generalise the result, to For any integer $q$,
Let $\alpha, \beta$ be the maximum power of 2,5 dividing $q$ respectively.
Then there always exists a number $999...9990000.....000$ with number of zero $= max (\alpha , \beta )$.
- 4 years, 4 months ago
Well, actually, you can generalize even more, when you consider another base system, here I used another way Euler Theorem to make this more rigorous: https://brilliant.org/discussions/thread/divisors-of-9999999-in-the-right-way/?ref_id=1339441
- 4 years, 1 month ago | 2021-07-27T12:10:37 | {
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https://math.stackexchange.com/questions/944462/different-answer-when-simplifying-before-integrating | # Different answer when simplifying before integrating
I have been trying to get my head around this for some time now... I solve the same integral in two ways but get two different solutions. Since there can't (surely) be any sort of ambiguity when integrating, the answers have to either be identical (ruled out) or I am doing something wrong in one of the solutions. My theory is that the missing ln(1/k) is embodied, somehow, in the constant. Could this be it?
Method 1: $$\int \frac{1/k}{1-y/k}dy=\frac{1}{k}\int \frac{1}{1-y/k}dy\rightarrow v = 1-y/k\rightarrow =-ln(1-y/k) + c$$
Method 2: $$\int \frac{1/k}{1-y/k}dy = \int \frac{1}{k-y}dy=-ln(k-y) + c$$
• You're right. If you take the difference of the two results, you get $\ln k$ plus a constant, which results in a constant. Sep 24 '14 at 15:31
$$\ln(k-y)+c=\ln((1-y/k)k)+c=\ln (1-y/k)+\underbrace{\ln k+c}_{c'}$$ Use $\ln ab=\ln a+\ln b$ | 2021-09-20T20:42:41 | {
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https://math.stackexchange.com/questions/3434719/number-of-non-negative-integer-solutions-to-2x-y-n | # Number of non-negative integer solutions to $2x + y = N$
Is it possible to determine the number of non-negative integer solutions to $$2x + y = N$$ where $$N$$ is a non-negative integer? I was solving a problem when a particular equation of this kind came up and the book simply counted every case (which was rather tedious, given the large amount of cases). Later, I noticed a different problem which appeared to require the same number of solutions of $$2x + y = N$$ but it was not possible to count the cases because $$N$$ was not given.
I found another solution and it turned out not be necessary to solve for the number of solutions of the equation, however I'm still wondering if it is possible to find this number in terms of factorials and whatnot. I was not able to do it with the bars/stars method or whatever it's called. So, is it possible?
• Since $y=N-2x$, the number of integer solutions is actually $\left\lfloor\frac{N}{2}\right\rfloor+1$. If you draw the function $y=N-2x$ in the first quadrant and work on the boundary conditions, you will easily get the answer. Also, I don't think that this is a combinatorics question... – Yourong Zang Nov 14 '19 at 0:28
• This is the number of partitions of $N$ into at most $2$ parts or where no part exceeds $2$. The pattern continues: for example the number of solutions to $6u+5v+4w+3x+2y+z=1$ is the number of partitions of $N$ into at most $6$ parts or where no part exceeds $6$, though the formula gets more complicated – Henry Nov 14 '19 at 0:56
Case 1: $$N$$ is odd. We have $$N=1: \quad 1 \mbox{ solution } \ (x,y)=(0,1)$$ $$N=3: \quad 2 \mbox{ solutions } \ (0,3), \ (1,1)$$ $$N=5: \quad 3 \mbox{ solutions } \ (0,5), \ (1,3), \ (2,1)$$ The pattern suggests that we have $$\lceil N/2 \rceil$$ solutions for odd $$N$$.
Case 2: $$N$$ is even. We have $$N=0: \quad 1 \mbox{ solution } \ (x,y)=(0,0)$$ $$N=2: \quad 2 \mbox{ solutions } \ (0,2), \ (1,0)$$ $$N=4: \quad 3 \mbox{ solutions } \ (0,4), \ (1,2), \ (2,0)$$ This pattern suggests that we have $$N/2+1$$ solutions for even $$N$$.
In both cases, the number of solutions can be written as $$\lfloor N/2 \rfloor +1$$.
It remains to prove that the above guess is correct. (We can do this without much difficulty.)
A Proof of Alex's "guess" to show that why Alex says
We can do this without much difficulty
Since $$y=N−2x$$, the number of integer solutions is actually $$\left\lfloor\frac{N}{2}\right\rfloor+1$$. If you draw the function y=N−2x in the first quadrant, you will easily get two solutions (not necessarily integers) at the boundaries: $$(x,y)=(0, N)$$ and $$(x,y)=(N/2,0)$$. Since $$N$$ may not be an even number, we could say the largest possible integer solution of $$x$$ is $$\lfloor N/2\rfloor$$. Now, since an integer $$(N)$$ minus another integer $$(2x)$$ is still an integer $$(y)$$, we only need to see how many integer solutions of $$x$$ are in the first quadrant. You can get this by counting from $$0$$ to $$\lfloor N/2\rfloor$$, which is $$\#=\left\lfloor \frac{N}{2}\right\rfloor+1$$ | 2020-04-03T12:18:00 | {
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https://mathoverflow.net/questions/283540/combinatorial-identity-sum-i-j-ge-0-binomiji2-binoma-ib-ja/283611 | Combinatorial identity: $\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$
In my research, I found this identity and as I experienced, it's surely right. But I can't give a proof for it. Could someone help me? This is the identity: let $a$ and $b$ be two positive integers; then:
$\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2=\frac{1}{2} \binom{(2a+1)+(2b+1)}{2a+1}$.
• The numbers you have come from the OEIS sequence A091044 but I don't see anything there now that would lead to a proof. – Somos Oct 15 '17 at 14:44
• Note that without the squares, the LHS counts the number of paths down Pascal's triangle from $(n,k)=(0,0)$ to $(a+b,a)$ passing through a marked point $(i+j,i)$. Since each path contains $a+b+1$ points, that sum equals $(a+b+1){a+b\choose a}$. – MTyson Oct 16 '17 at 0:25
• It would be very interesting if someone knowing much about the Wilf-Zeilberger method would write a relevant comment on whether this identity nowadays is regarded as 'automatically provable'. – Peter Heinig Oct 16 '17 at 7:18
• Crossposted at artofproblemsolving.com/community/… – darij grinberg Oct 16 '17 at 8:50
• As I know from some talks, WZ can be sometimes inefficient, but recently guys invented a new more efficient algorithm to do the job. arxiv.org/abs/1510.07487 arxiv.org/abs/1404.5069 Offtopic: Is there a simple explanation why such kinds of questions "combinatorial identity of type sum of product of binomials having linear combinations as arguments" are so popular on MO? – Sergey Dovgal Oct 17 '17 at 14:51
Denote $h(x,y)=\sum_{i,j\geqslant 0} \binom{i+j}i x^iy^j=\frac1{1-(x+y)}$, $f(x,y)=\sum_{i,j\geqslant 0} \binom{i+j}i^2 x^iy^j$. We want to prove that $2xyf^2(x^2,y^2)$ is an odd (both in $x$ and in $y$) part of the function $h(x,y)$. In other words, we want to prove that $$2xyf^2(x^2,y^2)=\frac14\left(h(x,y)+h(-x,-y)-h(x,-y)-h(-x,y)\right)=\frac{2xy}{1-2(x^2+y^2)+(x^2-y^2)^2}.$$ So, our identity rewrites as $$f(x,y)=(1-2(x+y)+(x-y)^2)^{-1/2}=:f_0(x,y)$$ This is true for $x=0$, both parts become equal to $1/(1-y)$. Next, we find a differential equation in $x$ satisfied by the function $f_0$. It is not a big deal: $$\left(f_0(1-2(x+y)+(x-y)^2)\right)'_x=(x-y-1)f_0.$$ Since the initial value $f_0(0,y)$ and this relation uniquely determine the function $f_0$, it remains to check that this holds for $f(x,y)$, which is a straightforward identity with several binomials. Namely, comparing the coefficients of $x^{i-1}y^j$ we get $$i\left(\binom{i+j}j^2-2\binom{i+j-1}j^2-2\binom{i+j-1}i^2+\binom{i+j-2}i^2+\binom{i+j-2}j^2-2\binom{i+j-2}{i-1}^2\right)$$ for $(f(1-2(x+y)+(x-y)^2))'_x$ and $$\binom{i+j-2}j^2-\binom{i+j-1}j^2-\binom{i+j-2}{j-1}^2$$ for $(x-y-1)f$. Both guys are equal to $$-2\frac{j}{i+j-1}\binom{i+j-1}{j}^2.$$
• Nice one dear Petrov! – Shahrooz Janbaz Oct 16 '17 at 11:02
• Can you please explain how $xyf^2(x^2, y^2)$ being an odd part of $h(x, y)$ relates to the question being asked? – Vincent Oct 16 '17 at 15:08
• @Vincent look at a coefficient of $x^{2a+1}y^{2b+1}$ in $xyf(x^2,y^2)f(x^2,y^2)$. It is nothing but the left hand side of the desired identity. – Fedor Petrov Oct 16 '17 at 15:33
• The identity $f(x,y)=f_0(x,y)$ can be also obtained by Lagrange Inversion. Or, derived from the g.f. $F(x,y)$ given in my answer as $f(x,y)=F(x,\frac{y}{x})$. – Max Alekseyev Oct 16 '17 at 19:10
• Is it true that $f(0)＝f_0(0),(f(1-2(x+y)+(x-y)^2))_x'＝(x-y-1)f,(f_0(1-2(x+y)+(x-y)^2))'＝(x-y-1)f_0⇒f＝f_0？$ – ken Oct 17 '17 at 5:38
When this identity was posted, it struck me as something that ought to have a combinatorial explanation. I have now found one, using a decomposition of NSEW lattice paths: paths in $\mathbb{Z}^2$ consisting of unit steps in the direction N, S, E or W. Many of the ideas here may be found in [GKS], though not the decomposition itself.
The expression $\frac12{2a+1\ +\ 2b+1\choose2a+1}$ counts paths of $(a+b+1)$ steps that start at $(0,0)$ and end on the half-line $(a-b,\geq0)$.
To see this, decompose each path step as two half-steps $±\left[\begin{smallmatrix}½\\½\end{smallmatrix}\right]$ and $±\left[\begin{smallmatrix}½\\-½\end{smallmatrix}\right]$. If the $+$ option is chosen for $(2a+1)$ of the $2(a+b+1)$ half-steps, and the $-$ option for the other $(2b+1)$, then the $x$-coordinate of the endpoint is $\frac12((2a+1)-(2b+1))=a-b$. Thus there are ${2a+1\ +\ 2b+1\choose2a+1}$ paths of $(a+b+1)$ steps from $(0,0)$ to $x=a-b$. By parity, the end position must have an odd-numbered $y$-coordinate. Reflection in the $x$-axis is therefore a fixpoint-free involution, so half of these paths end on the half-line $(a-b,\geq0)$.
Such a path may be split into a pair of paths with $(a+b)$ steps in total.
The endpoint of the path is $(a-b, 2k+1)$ for some $k\in\mathbb N$. At least one step of the path must therefore be an N step from $(c,2k)$ to $(c,2k+1)$ for some $c$. Remove the first such step, to give a pair of paths with $a+b$ steps altogether:
1. A path of $n$ steps from $(0,0)$ to $(c,2k)$ that does not cross the line $y=2k$, which we can think of as a 180° rotation of a path from $(0,0)$ to $(c,2k)$ that does not cross the $x$-axis;
2. A path of $a+b-n$ steps from $(c,2k+1)$ to $(a-b,2k+1)$, which we can think of as a translation of a path from $(0,0)$ to $(a-b-c,0)$.
This is clearly a bijection.
There are ${i+j\choose i}^2$ paths of $(i+j)$ steps from $(0,0)$ to $(i-j,0)$.
The four directions N,S,E,W may be obtained by starting with $\left[\begin{smallmatrix}-1\\0\end{smallmatrix}\right]$ and adding neither, one, or both of $\left[\begin{smallmatrix}1\\1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix}1\\-1\end{smallmatrix}\right]$. Build a path of $i+j$ steps, initially all $\left[\begin{smallmatrix}-1\\0\end{smallmatrix}\right]$. Add $\left[\begin{smallmatrix}1\\1\end{smallmatrix}\right]$ to $i$ of the steps and, independently, add $\left[\begin{smallmatrix}1\\-1\end{smallmatrix}\right]$ to $i$ of the steps.
There are also ${i+j\choose i}^2$ paths of $(i+j)$ steps from $(0,0)$ to $(i-j,\geq0)$ that do not cross the $x$-axis.
There is a bijection between these paths and the paths of the previous section using a raising/lowering transformation [GKS]. Suppose we have a path from $(0,0)$ to $(i-j,0)$ that may cross the $x$-axis.
• While the path crosses the $x$-axis, do the following:
• Take the initial segment of the path up to the first time it touches the line $y=-1$, and reflect this initial segment about that line. Then translate the entire path up by two units, so it starts at $(0,0)$ again and ends two units higher than before on $x=i-j$.
I hope it is clear that this process is reversible. (In reverse: while the endpoint is above the $x$-axis, translate the path two units down, then take the initial segment from $(0,-2)$ to the first intersection with $y=-1$ and reflect this initial segment about that line.)
Putting it together
Now we have all the ingredients we need. Let us count the pairs of paths as described above. Since $n$ and $c$ have the same parity, we may write $n=i+j$ and $c=i-j$ for $i\in[0,a]$, $j\in[0,b]$.
• There are ${i+j\choose i}^2$ paths of $(i+j)$ steps from $(0,0)$ to $(i-j,\geq 0)$ that do not cross the $x$-axis.
• There are ${a-i\ +\ b-j\choose a-i}^2$ paths of $(a+b)-(i+j)$ steps from $(0,0)$ to $(a-b-(i-j),0)$.
So in total there are
$$\sum_{i=0}^a\sum_{j=0}^b{i+j\choose i}^2{a-i\ +\ b-j\choose a-i}^2$$
such pairs, as required.
[GKS] Richard K. Guy, C. Krattenthaler and Bruce E. Sagan (1992). Lattice paths, reflections, & dimension-changing bijections, Ars Combinatoria, 34, 3–15.
• This is really nice. Can this argument be modified somehow to show that, for any permutation $\pi$ of $\{0,1,\dots,n\}$, we have $\sum_{i,j=0}^{n}\binom{i+j}{i}\binom{2n-i-j}{n-i}\binom{\pi(i)+\pi(j)}{\pi(i)}\binom{2n-\pi(i)-\pi(j)}{n-\pi(i)}>\binom{2n+1}{n+1}^2$? I had to prove this once a while ago, but could not find any argument using a natural lattice path interpretation, like yours above. – Alexander Burstein Nov 17 '17 at 1:44
• @AlexanderBurstein If it can, I don’t yet see how. I’ll let you know if I think of anything. – Robin Houston Nov 23 '17 at 15:38
Let us denote $$S=\sum_{i,j \ge 0} \binom{i+j}{i}^2 \binom{(a-i)+(b-j)}{a-i}^2.$$
First, let $s=i+j$ so that $$S = \sum_{s\geq 0}\sum_{i=0}^s \binom{s}{i}^2 \binom{a+b-s}{a-i}^2.$$ Consider the generating function $$F(x,y) = \sum_{s,i} \binom{s}{i}^2 x^i y^s = (1-2y+y^2-2xy-2xy^2+x^2y^2)^{-1/2}.$$ Then $S$ is nothing else but the coefficient of $x^a y^{a+b}$ in $$F(x,y)^2 = (1-2y+y^2-2xy-2xy^2+x^2y^2)^{-1}$$ $$= \frac{1}{4y\sqrt{x}}\left(\frac{1}{1-y(1+x+2\sqrt{x})} - \frac{1}{1-y(1+x-2\sqrt{x})}\right)$$ $$= \frac{1}{4y\sqrt{x}}\left(\frac{1}{1-y(1+\sqrt{x})^2} - \frac{1}{1-y(1-\sqrt{x})^2}\right).$$ (derivation simplified)
The coefficient of $y^{a+b}$ equals $$[y^{a+b}]\ F(x,y)^2 =\frac{1}{4} \frac{(1+\sqrt{x})^{2(a+b+1)} - (1-\sqrt{x})^{2(a+b+1)}}{\sqrt{x}}.$$ Now we trivially conclude that $$S = [x^ay^{a+b}]\ F(x,y)^2 = \frac{1}{2}\binom{2(a+b+1)}{2a+1}.$$
UPDATE. Alternatively to computing the coefficient of $x^ay^{a+b}$, one can follow the venue of Fedor Petrov's proof. This way one needs to consider the generating function $$G(x,y) = \sum_{m,n}\binom{m}{n} x^ny^m = \frac{1}{1-y-xy}$$ and verify that $$8xy^2F(x^2,y^2)^2 = G(x,y) + G(x,-y) - G(-x,y) - G(-x,-y).$$
• How do you get the formula for $F$? – Fedor Petrov Oct 16 '17 at 19:21
• @FedorPetrov: Notice that $\binom{s}{i}^2 = [x^iz^s]\ ((1+xz)(1+z))^s$ and use Lagrange inversion w.r.t. variable $z$. – Max Alekseyev Oct 16 '17 at 19:24
• Ok, but this could be mentioned in the answer, I think. – Fedor Petrov Oct 16 '17 at 20:04
• I assume it's a common knowledge. – Max Alekseyev Oct 16 '17 at 20:06 | 2020-01-28T23:53:54 | {
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https://byjus.com/question-answer/consider-the-functions-f-left-x-right-sin-left-x-1-right-and-g-left/ | Question
# Consider the functions $$f\left( x \right)=sin\left( x-1 \right)$$ and $$g\left( x \right)=\cot ^{ -1 }{ \left[ x-1 \right] }$$Assertion: The function $$\displaystyle F\left( x \right) =f\left( x \right).g\left( x \right)$$ is discontinuous at $$x=1$$Reason: If $$f\left( x \right)$$ is discontinuous at $$x=a$$ and $$g\left( x \right)$$ is also discontinuous at $$x=a$$ then the product function $$f\left( x \right) .g\left( x \right)$$ is discontinuous at $$x=a$$.
A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
C
Assertion is correct but Reason is incorrect
D
Both Assertion and Reason are incorrect
Solution
## The correct option is C Assertion is correct but Reason is incorrect$$f\left( x \right) =sin\left( x-1 \right) ;g\left( x \right) =\cot ^{ -1 }{ \left[ x-1 \right] }$$$$\displaystyle \therefore F\left( x \right) =f\left( x \right) .g\left( x \right) =\begin{cases} \begin{matrix} -\cot ^{ -1 }{ \left[ x-1 \right] } ; & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \cot ^{ -1 }{ \left[ x-1 \right] } ; & x>1 \end{matrix} \end{cases}$$$$\displaystyle =\begin{cases} \begin{matrix} -\cot ^{ -1 }{ \left( -1 \right) ; } & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \cot ^{ -1 }{ \left( 0 \right) ; } & x>1 \end{matrix} \end{cases}=\begin{cases} \begin{matrix} -\left( \pi -\cot ^{ -1 }{ 1 } \right) ; & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \pi /2; & x>1 \end{matrix} \end{cases}$$$$\displaystyle =\begin{cases} \begin{matrix} -3\pi /4; & x<1 \end{matrix} \\ \begin{matrix} 0; & x=1 \end{matrix} \\ \begin{matrix} \pi /2; & x>1 \end{matrix} \end{cases}$$$$\displaystyle \Rightarrow F\left( { 1 }^{ - } \right) =-\frac { 3\pi }{ 4 } ;F\left( { 1 }^{ + } \right) =\frac { \pi }{ 2 }$$ and $$F\left( 1 \right) =0$$$$\Rightarrow f\left( x \right)$$ is discontinuous at $$x=1$$Therefore assertion is correct but product of two discontinuous functions may be a continuous function.So reason is wrong.Mathematics
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http://mathhelpforum.com/geometry/62929-question-about-circle.html | 1. ## Can you help me?(Question about circle)
The circle S has equation :
X2+Y2- 2X-4Y=8
Show thatA(-1,-1)and B (3,5) are ends of diameter
Show that C(4,4)lies on S and determine angle ACB
Find the distance from D(5,0) to the nearest point E on S
Find the the x-coordinates of the points of intersection of S with the line joining D and F(0,2).
2. Originally Posted by Diligent_Learner
The circle S has equation :
$x^2+y^2- 2x-4y=8$
Show thatA(-1,-1)and B (3,5) are ends of diameter
Complete the square in the equation for both x and y to locate the center. Then, find the midpoint of AB to see if they are one and the same. The center of the circle would be the midpoint of the diameter.
$(x^2-2x+1)+(y^2-4y+4)=8+5$
$(x-1)^2+(y-2)^2=13$
Center is at (1, 2)
Midpoint of AB = $\left(\frac{3-1}{2}, \ \ \frac{5-1}{2}\right)={\color{red}(1, 2)}$
Originally Posted by Diligent_Learner
Show that C(4,4)lies on S and determine angle ACB
Substitute (4, 4) into your original equation to see if it makes a true statement
$4^2+4^2- 2(4)-4(4)=8$??
Angle ACB is an inscribed angle in a semicircle, therefore 90 degrees.
Originally Posted by Diligent_Learner
Find the distance from D(5,0) to the nearest point E on S
See Earboth's post.
Originally Posted by Diligent_Learner
Find the the x-coordinates of the points of intersection of S with the line joining D and F(0,2).
Define a linear equation for the line passing through D(5, 0) and F(0, 2). Then solve the system which includes this linear equation and the equaton of the circle. You should get two distinct points of intersection. Try it.
3. Thank you very much for your great help.I understand the question
The equation of the line that passes through D and F is:
y-y1=m(x-x1)
Y= -0.4x+2
====================
X2+y2-2x-4y=8
X2+(-0.4x+2)2-2x-4(-0.4x+2)
X2-3.6X-12=0
X=5.7 or X=-2.1
4. Originally Posted by Diligent_Learner
[FONT=Times New Roman]...
Find the distance from D(5,0) to the nearest point E on S
...
Let C denote the center of the circle. Then the Point E lies on the circle and on the line CD:
$CD: y =-\dfrac12 x + \dfrac52$
Calculating the intersection points of the circle and this line:
$(x-1)^2+\left(-\dfrac12 x + \dfrac52 - 2\right)^2 = 13$
yields: $x = 1\pm \frac15 \sqrt{65}$ The x-coordinate of E must be greater than 1: Plug in this value into the equation of the line to get the y-coordinate. I've got $E\left(1+\frac15\sqrt{65}\ ,\ 2-\frac15\sqrt{65}\right)$
The distance $ED = CD - r~\implies~ ED=\sqrt{20} - \sqrt{13} \approx 0.86658...$
5. Thank you very much for clarifying the idea ..
and Can yiu tell me about if my answer is true in the previous post?
and in which program you draw the circle?
Originally Posted by Diligent_Learner
Thank you very much for clarifying the idea ..
and Can yiu tell me about if my answer is true in the previous post?
and in which program you draw the circle?
You've got the equation of the line correctly.
But I've got results, different from yours, calculating the points of intersection:
$l:y=-\dfrac25 x+2$
Plug in the term of y into the equation of the circle:
$(x-1)^2+\left(-\dfrac25 x+2 - 2\right)^2=13$
$x^2-2x+1+\dfrac4{25}x^2=13$
$\dfrac{29}{25} x^2 - 2x-12=0$
Use the quadratic formula to solve this equation:
$x=\dfrac{25}{29} \pm \dfrac5{29} \sqrt{373}$
Take these x-values and plug them into the equation of the line to get the y-coordinates of the points of intersection.
EDIT: Use the attached sketch to control your results. | 2017-11-20T16:01:35 | {
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http://caverider.com/9rq3cvl5/article.php?a77076=domain-and-range-notes | Let us review this first. ; The codomain is similar to a range, with one big difference: A codomain can contain every possible output, not just those that actually appear. Domain and Range Notes # _____ Cornell Notes Questions and Main Ideas: Cornell Notes Questions and Main Ideas: Input the domain value (x) The output is the range value (y) In order for a relation to be a function, you can only get one y output for every x you input!!!! Note that all I had to do to check whether the relation was a function was to look for duplicate x-values. Domain is the set of all x values, the independent quantity, for which the function f(x) exists or is defined. The domain and range for the graph above are: Domain: x ∈ [− 3, 2] Range: y ∈ [− 2, 3] Note that the ∈ symbol means “is an element of” and means that the x or the y is in that interval and the numbers in the interval are always written in increasing order. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). Domain and Range Name: _____ State the domain and range for each graph and then tell if the graph is a function (write yes or no). The domain and range of a function is all the possible values of the independent variable, x, for which y is defined. Note that the domain elements 1 and 2 are associated with more than one range elements, so this is not a function. They will give you a function and ask you to find the domain (and maybe the range, too). ; The range is the set of y-values that are output for the domain. But, more commonly, and especially when dealing with graphs on the coordinate plane, we are concerned with functions, where each element of the domain is associated with one element of the range. Domain and Range of Linear Inequalities. Note that even though the [-3, 2] may look similar to the ordered pair that represents … The domain is the set of x-values that can be put into a function.In other words, it’s the set of all possible values of the independent variable. Worked example: domain and range from graph Our mission is to provide a free, world-class education to anyone, anywhere. Note that the domain and range are always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Example: Finding Domain and Range from a Graph. For the cube root function $$f(x)=\sqrt[3]{x}$$, the domain and range include all real numbers. the range of the function F is {1983, 1987, 1992, 1996}. Domain and range. l) 2) 40<13 3) x 20 Let f be a function such that = 2r —4 is defined on the domain 2 < x < 6. The range of this function is Domain and Range Notes The accompanying graph shows the heart rate, in beats per minute, of a jogger durtng a 4-rmnute interval. There is one other case for finding the domain and range of functions. Domain restrictions refer to the values for which the given function cannot be defined. The below table gives a review of the notation used. The example below shows two different ways that a function can be represented: as a function table, and as a set of coordinates. If the graph is a function, state whether it is discrete, continuous or neither. domain 100 What is the range of the jogger's heart rate during Range The set of all the outputs of a function is known as the range of the function or after substituting the domain, the entire set of all values possible as outcomes of the dependent variable. What is the domain of this function? Find the domain and range of … Given the formula for a function, determine the domain and range. 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https://math.stackexchange.com/questions/2101422/finding-the-equivalence-classes | Finding the equivalence classes
The relation X on the set $\{1, 2, 3, 4, 5\}$ is defined by the rule $(a, b) ϵ X$ if 3 divides a – b.
• List the elements of X
These are $\{(4,1),(1,4),(5,2),(2,5),(1,1),(2,2),(3,3),(4,4),(5,5)\}$
• List the equivalence class$\color{red}{\text{es}}$
The answer is $\{1,4\},\{2,5\},\{3\}$
This is where I am confused. I thought equivalence class meant that one should only present the elements that don't result in a similar result.
However, $4-1 = 5-2 = 3$ these are all the same
Would anyone care to explain how the equivalence class can be found in this case?
• you're confusing a set of representatives with the set of classes. – Henno Brandsma Jan 17 '17 at 11:31
• your classes are correct. – Henno Brandsma Jan 17 '17 at 11:32
• Yes, i got them from the answer booklet, I couldn't understand how to get them. – Johny Jan 17 '17 at 11:32
Equivalence classes (mean) that one should only present the elements that don't result in a similar result.
I believe you are mixing up two slightly different questions. Each individual equivalence class consists of elements which are all equivalent to each other. That is why one equivalence class is $\{1,4\}$ - because $1$ is equivalent to $4$. We can refer to this set as "the equivalence class of $1$" - or if you prefer, "the equivalence class of $4$".
Note that we have been talking about individual classes. We are now going to talk about all possible equivalence classes. You could list the complete sets, $$\{1,4\}\quad\hbox{and}\quad\{2,5\}\quad\hbox{and}\quad\{3\}\ .$$ Alternatively, you could name each of them as we did in the previous paragraph, $$\hbox{(the equivalence class of 1)}\quad\hbox{and}\quad \hbox{(the equivalence class of 2)}\quad\hbox{and}\quad \hbox{(the equivalence class of 3)}\ .$$ Or if you prefer, $$\hbox{(the equivalence class of 4)}\quad\hbox{and}\quad \hbox{(the equivalence class of 2)}\quad\hbox{and}\quad \hbox{(the equivalence class of 3)}\ .$$ You see that the "names" we use here are three elements with no two equivalent. I think you are confusing this with the previous paragraph.
Hope this helps!
• I assume as a third set I could any of the $(x,x)$ sets? Like $(1,1)$, $(2,2)$, etc... – Johny Jan 17 '17 at 11:34
• No. Note that each equivalence class contains individual elements from $\{1,2,3,4,5\}$, not pairs of elements. In this example, $1$ cannot be in an equivalence class by itself because $4$ has to be in the same class. The fact that $1$ is related to $1$ is reflected in the fact that $1$ is (obviously) in the same set as $1$. – David Jan 17 '17 at 11:38
• Also, $3$ has to be in the same class as other numbers which are equivalent to it. But there are no such numbers, that's why $3$ is by itself. – David Jan 17 '17 at 11:40
By definition, the equivalence classes
(1) partition the Universe, $\{1,2,3,4,5\}$. Your classes do partition the universe.
(2) Pairs of elements within the equivalence classes are in relation $X$.
(3) Pairs of elements from different equivalence classes are not in relation $X$. This holds in your case.
I edited your question. (Plural of class)
How to find the equivalence classes? In such a finite case, it is easy.
(1) Take the first element of your Universe and compare it to the other elements. List the pairs that you found to be in $X$. This list is your first eqivalence class.
(2) There will be elements that you have not yet paired with the first one. Take the first such element and see those elements that remained after operation (1). Do the comparison as in (1). You'll find the second equivalence class.
(3) Repeat (2) until there will be elements not yet paired with another. | 2021-06-18T19:19:56 | {
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https://math.stackexchange.com/questions/2882914/problem-on-determinant | # Problem on Determinant.
Q.
$$\text{If } \Delta = \left|\begin{array}{ccc} a & b & c \\ c & a & b \\ b & c & a \end{array}\right|,$$ $$\text{ then the value of }$$ $$\left|\begin{array}{ccc} a^2 - bc & b^2 - ca & c^2 - ab \\ c^2 - ab & a^2 - bc & b^2 - ca \\ b^2 - ca & c^2 - ab & a^2 - bc \end{array}\right| \text{ is:}$$
Express the Answer in terms of $\Delta$.
My Attempt - I tried Rearranging the Second determinant (by adding or subtracting particular rows or Columns) such that after this, It'll be easy for me write the second determinant as the product of two or more other Determinants that might be same as $\Delta$. But I Found that rearranging the Second Det. Wasn't a good choice as it seems to never simplify itself!! So, I tried Expressing the Second Det. directly as the product of two or more Det. ! Still It was no use. I finally concluded that the Second Det. Can't be expressed as a product without Rearrangement ! But Rearranging it makes it more creepier!
Could you please Guide me how to Rearrange it so that it could be easily expressed as a product? Or is their Another Way for this type of Problem? Please Let me know!
Any help would be Appreciated.
• Do you know how to write matrices? It's exactly the same, except that you use \vmatrix instead of \bmatrix or \pmatrix. – saulspatz Aug 14 '18 at 21:09
• Sorry, Not them too ! I'll get to know soon!! – Creep Anonymous Aug 14 '18 at 21:10
• Each entry of the second matrix is the cofactor from the first matrix. – Ng Chung Tak Aug 14 '18 at 21:10
• @NgChungTak I got that Earlier but How would it help? – Creep Anonymous Aug 14 '18 at 21:12
• Actually, @heropup did it in a way I wasn't familiar with. Take a look. – saulspatz Aug 14 '18 at 21:12
Notice that
$$\begin{bmatrix} a & b & c \\ c & a & b \\ b & c & a% \end{bmatrix}% ^{-1}=\frac{1}{\Delta }% \begin{bmatrix} a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc% \end{bmatrix}.$$ Then, since $\det(A^{-1})=\det(A)^{-1}$ and $\det(cA)=c^{n}\det(A)$ for a matrix $A$ that is $n\times n$, taking determinants of both sides we get $$\frac{1}{\Delta}=\frac{1}{\Delta ^3}% \det\left(\begin{bmatrix} a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc% \end{bmatrix}\right),$$ so $$\det\left(\begin{bmatrix} a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc% \end{bmatrix}\right)=\Delta^2.$$
Finally, notice that $\det(A)=\det(A^T)$.
• Got It!! You're correct! – Creep Anonymous Aug 14 '18 at 21:19
• In the first equation, shouldn't you take the transpose of the matrix on the right-hand side? Of course, $\det A^T=\det A$ so it just a detail. – saulspatz Aug 14 '18 at 21:21
• @saulspatz I don't think so, but at the end I had to. Thank you. – mzp Aug 14 '18 at 21:24
Take $A$ the matrix inside the determinant so that $\Delta=\det(A)$. The transpose of the new matrix is the so called classical adjoint matrix of $A$ for which we have: $$\mathrm{adj}(A)A=\det(A) I.$$ Hence the determinant is equal to $\Delta^{k-1}$ where $k$ is the matrix dimension and is equal to $3$ here.
• Isn't Adjoint Matrix would be the Transpose of the Second one?? – Creep Anonymous Aug 14 '18 at 21:15
• @CreepAnonymous, Yes, you are right! slightly edited! – Arash Aug 14 '18 at 21:28
• But That wouldn't have mattered for at last we had to take the determinant!! – Creep Anonymous Aug 14 '18 at 21:30
• @CreepAnonymous, exactly! – Arash Aug 14 '18 at 21:30
To complete the answer given by @mzp, note that the matrix $$\left(\begin{array}{ccc} a^2 - bc & b^2 - ca & c^2 - ab \\ c^2 - ab & a^2 - bc & b^2 - ca \\ b^2 - ca & c^2 - ab & a^2 - bc \end{array}\right)$$ is $$\pmatrix{1&0&0\\0&0&1\\0&1&0}\left( \matrix{a^{2}-bc & c^{2}-ab & b^{2}-ac \\ b^{2}-ac & a^{2}-bc & c^{2}-ab \\ c^{2}-ab & b^{2}-ac & a^{2}-bc} \right)\pmatrix{1&0&0\\0&0&1\\0&1&0}$$ and that the determinant of $\pmatrix{1&0&0\\0&0&1\\0&1&0}$ is $-1$.
• Isn't it just the transpose? I added a sentence at the end. – mzp Aug 14 '18 at 21:28
• @mzp Yes, But he Mark Fischler proved that too! – Creep Anonymous Aug 14 '18 at 21:30 | 2019-01-23T03:16:22 | {
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https://math.stackexchange.com/questions/786708/elevator-probability-question | # Elevator Probability Question
There are four people in an elevator, four floors in the building, and each person exits at random. Find the probability that:
a) all exit at different floors
b) all exit at the same floor
c) two get off at one floor and two get off at another
For a) I found $4!$ ways for the passengers to get off at different floors, so $$\frac{4!}{4^4} \text{would be the probability} = \frac{3}{32}$$
For b) there are only four ways for them to all exit on the same floor, so $$\frac{4}{256} = \frac{1}{64}$$
For c) am I allowed to group the $4$ people so that I am solving for $2$ people technically? For two people there would be $12$ possibilities, and there are three ways to group the $4$ individuals, so $$\frac{12 \cdot 3}{256} = \frac{9}{64}$$
I'm not sure if I'm doing these right, can you please check? Thank you.
• For the third, I would probably argue thus: the two floors can be chosen in $\binom{4}{2}$ ways. For each way, the $2$ people who get off at the lower of these floors can be chosen in $\binom{4}{2}$ ways, for a total of $\binom{4}{2}\binom{4}{2}$ (same result as yours). – André Nicolas May 8 '14 at 16:51
Let us approach part (c) constructively. You appear to have correct answers for all the parts, but perhaps this will help.
First we select two of the four floors, which can obviously be done in $\dbinom{4}{2} = 6$ ways. Now we select two people from the four people to leave on the first floor, which can be done in $\dbinom{4}{2} = 6$ ways. We see that there are a total of $6 \times 6 = 36$ successful outcomes.
There are a total of $4^{4} = 256$ ways. Our probability is thus $\frac{36}{256} = \boxed{\frac{9}{64}}.$
Your calculation for $c)$ is correct, though, as the comments and answers show, you seem not to have expressed it in a readily comprehensible manner.
I would express what I understand your argument to be as: There are $3$ ways to split the $4$ people into $2$ groups. Then we can treat those two groups like $2$ people and choose the floors they get off at, $4$ options for the first group and $3$ for the second, or $12$ overall, for a total of $3\cdot12=36$ options out of $4^4=256$.
a and b are correct. For c, I would say there are ${4 \choose 2}$ ways to choose the first pair, $4$ ways to choose the floor they get off on, and $3$ ways to choose the floor the other pair gets off on, but we need to divide by $2$ because we can swap the pairs. That gives $\frac {6\cdot 4 \cdot 3}{2 \cdot 4^4}=\frac 9{64}$ This is the same as your answer, but I don't understand the logic that gives $12$ possibilities for a pair. We don't care about the order of selection in a pair. | 2019-09-20T05:58:23 | {
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https://math.stackexchange.com/questions/613562/tictactoe-with-considerations-of-symmetry/614385 | # TicTacToe with considerations of symmetry
It is not difficult to determine the number of possible games of tic toe, but what about when rotationally symmetric positions are considered equal? Please do not simply give me the number, I would like the intuition of how it is found. IMPORTANT: I am more talking more about arrangements of x's and o's in a 3X3 grid than actual tictactoe games, thus when somebody "wins" the game it continues.
• tell us how do you determine the total number of possible games, so that we can enhance YOUR answer – Thanos Darkadakis Dec 20 '13 at 2:31
• After the first move, of which there are 9 possibilities, there are 8 possibilities for the second move, so 9*8 positions on move 2. There are seven possible turn 3 moves, leading to 9*8*7 positions, and the pattern continues so that eventually there are 9! games. – Platonix Dec 20 '13 at 2:38
• not exactly. games can end on 5th, 6th, 7th, 8th or 9th move. – Thanos Darkadakis Dec 20 '13 at 2:39
• here is an explanation: se16.info/hgb/tictactoe.htm .There is also a link inside the link... – Thanos Darkadakis Dec 20 '13 at 2:42
• Sorry, read my edited question. You are quite correct of course, but I am more interested in the problem of symmetry than the games that "end" – Platonix Dec 20 '13 at 2:53
I'll use Burnside's method: count the number of invariant patterns for each rotation, and average that number over all rotations to get the number of distinguishable patterns.
I assume we're counting patterns of $5$ Xs and $4$ Os. If you're not allowing reflections, there are $4$ rotations of the square in the plane: the identity, $90^o$ degrees either way, or $180^o$.
For the identity rotation, all $\binom94=126$ patterns are invariant.
For a $90^o$ rotation there are $3$ orbits, one of size $1$ (the center) and two of size $4$ (the four corners or the four sides). Since we have $4$ Os, we have to put them in the four corners or the four sides: $2$ invariant patterns.
For the $180^o$ rotation, there is one orbit of size $1$ and there are four orbits of size $2$, consisting of a pair of opposite cells. The Os have to fill two of the size $2$ orbits; the number of invariant patterns is $\binom42=6$.
Thus the number of distinguishable patterns, allowing rotations in the plane but not reflections, is $\dfrac{126+2+2+6}4=34$.
Suppose you also allow the $4$ reflections as symmetries. Each reflection has three orbits of size $1$ (a cell on the axis of reflection) and three orbits of size $2$ (two mirrored cells). The number of invariant patterns for four Os is $\binom30\binom32+\binom32\binom31=12$, and so the number of distinguishable patterns is $\dfrac{126+2+2+6+12+12+12+12}8=23$.
The following MSE post computes the cycle index for the symmetries of an $N\times N$ board, $N$ odd or even, so that we may apply the Polya Enumeration Theorem, which includes Burnside as a special case. For $N$ odd we get that $$Z(H_N) = \frac{1}{8} \left( a_1^{N^2} + 4 a_1^N a_2^{(N^2-N)/2} + 2 a_1 a_4^{(N^2-1)/4}+a_1 a_2^{(N^2-1)/2}\right).$$ Put $N=3$ to obtain $$Z(H_3) = \frac{1}{8} \left(a_1^9 + 4 a_1^3 a_2^3 + 2 a_1 a_4^2 + a_1 a_2^4\right).$$ The substituted cycle index thus becomes $$Z(H_3)(1+z)= 1/8\, \left( 1+z \right) ^{9} +1/2\, \left( 1+z \right) ^{3} \left( 1+{z}^{2} \right) ^{3} +1/4\, \left( 1+z \right) \left( 1+{z}^{4} \right) ^{2}\\ +1/8\, \left( 1+z \right) \left( 1+{z}^{2} \right) ^{4}$$ which is $$Z(H_3)(1+z) = {z}^{9}+3\,{z}^{8}+8\,{z}^{7}+16\,{z}^{6}+23\,{z}^{5}+23\,{z}^{4} +16\,{z}^{3}+8\,{z}^{2}+3\,z+1,$$ so there are $23$ patterns with five cells marked.
I'll look at completely-filled boards with five $X$'s and four $O$'s (say).
I'll assume rotations and reflections are not distinct.
I'll count the number of distinct cases where $O$ occupies each possible number of corners. Call this number $O_{corner}$.
If $O_{corner} = 0$, there are $2$ patterns: one of the $O$'s is in the center, or it isn't. There is at most one vacant side space.
If $O_{corner} = 1$, there are $5$ cases. If $O$ is also in the center, one can fill the two adjacent sides, one adjacent and one "far" side, or the two far sides. If $O$ is not in the center, either an adjacent side or a far side can be blank.
If $O_{corner} = 2$, these two can be in opposite corners, or adjacent corners.
Let's take adjacent corners first. For a second, let's number the squares: upper left is $1$; lower right is $9$. Our two corner $O$'s are in $1$ and $3$. If one of the $O$'s is in the center ($5$) the distinct cases are spaces $2, 4,$ and $8$. If $O$ isn't in the center, the distinct cases for the last two are the pairs of spaces $(2,4), (4,6), (2,8),$ and $(6,8)$. We have $7$ cases.
For opposite corners, if $O$ is in the center, any side square is the same, symmetry-wise. If $O$ is not in the center, the other two $O$'s can be both next to one of the corner $O$'s, one next to each corner $O$ on the same side of the diagonal, one next to each corner $O$ on the opposite side of the diagonal. So $4$ more cases.
Total cases for $O_{corner} = 2$ is $11$.
For $O_{corner} = 3$, the fourth $O$ is on the side in between two corner $O$'s, on the side not between two corner $O$'s, or in the center. That's $3$ cases.
For $O_{corner} = 4$, that's all she wrote ($1$).
So, $22$ cases. I think I got them all.
If you don't want to consider any of the symmetries I've done, or if you want to take all different numbers of $X$'s and $O$'s, then you can go through the same exercise and determine as you go along which ones are distinct.
• I got $23$ patterns using Burnside. Took me a while to figure out which one you missed. With $O_{corner}=1$ and $O$ in the center, there are two ways to fill one adjacent and one far side. – bof Dec 20 '13 at 4:57 | 2019-06-16T23:38:04 | {
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http://math.stackexchange.com/questions/470650/why-there-are-exactly-n-distinct-n-th-roots | # Why there are exactly $n$ distinct $n$-th roots?
I was solving the following exercise: Given a complex number $z \neq 0$, write $z = re^{i\theta}$ where $\theta = \arg(z)$. Let $z_1 = Re^{i\alpha}$, where $R = r^{1/n}$ and $\alpha = \theta/n$, and let $\epsilon = e^{2\pi i/n}$, where $n$ is a positive integer.
$(a)$ Show that $z_1^n = z$, that is $z_1$ is an $n$-th root of $z$.
$(b)$ Show that $z$ has exactly $n$ distinct $n$-th roots:
$$z_1, \epsilon z_1,\epsilon^2 z_1,\cdots, \epsilon^{n-1}z_1$$
Well, item $(a)$ was very easy indeed. Now letter $(b)$ is confusing me.Showing that each of these is a $n$-th root was very simple. Indeed, if we consider $\epsilon^k z_1$, then we have:
$$(\epsilon^k z_1)^n = \epsilon ^{nk}z_1^n = \epsilon^{nk}z=e^{i2\pi k}re^{i\theta}=re^{i(\theta+2\pi k)} = re^{i\theta}$$
Where the last equality is because $k$ is integer, so $\cos(\theta + 2k \pi) = \cos \theta$ and $\sin (\theta+ 2k\pi) = \theta$. I can't see why there are exactly $n$ distinct roots, it seems to me that the problem is that for $k > n-1$ the roots start to repeat, but I don't know how to show it. For instance, for $k = n$, I cannot see why $\epsilon ^n z_1$ is a repeated root. Is that the problem? Is really about the roots start repeating? Can someone give a hint on how to show this?
Thanks very much in advance.
-
$\epsilon^n = 1$. You used that already to see that $\epsilon^kz_1$ is an $n$-th root. Use it also to see that $\epsilon^{n+k} = \epsilon^k$. – Daniel Fischer Aug 18 '13 at 19:15
The $n$-th roots of unity are the vertices of a regular $n$-gon with one vertex at $1$. – lhf Aug 18 '13 at 19:15
Note that in assertion (b) as quoted, the power $k$ of $\epsilon^k$ only goes up to $n-1$. You are right that they repeat after that, but part (b) doesn't say one gets more than up to power $k=n-1$. – coffeemath Aug 18 '13 at 19:15
## 3 Answers
A nonzero polynomial can have no more roots in a (commutative) field than its degree. Thus when you've found $n$ complex roots of the $n$-th degree polynomial $X^n-z$, there can be no more.
The reason for this is that whenever $a$ is a root of $P\in K[X]$, you can factor $P=(X-a)Q$ (because of Euclidean division by $(X-a)$, the fact that $a$ is a root ensures the remainder is$~0$) and because every other root of $P$ must be a root of$~Q$ (this is the most subtle point; commutativity of the base field is needed here, as well as the absence of zero divisors). Then an immediate induction argument shows that there can not be more that $\deg P$ roots of $P$ in any field (or integral domain) containing the coefficients of$~P$.
In your concrete example the above gives you a factorisation $$X^n-z=(X-z_1)(X-\epsilon z_1)(X-\epsilon^2 z_1)\ldots(X-\epsilon^{n-1}z_1),$$ and any root of $X^n-z$ must be a root of one of the factors on the right; clearly this means it has to be one of the $n$ roots you found.
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Suppose there were more than $n$ of them. Let $r_1,\ldots,r_{n+1}$ be $n+1$ of them. Then all of the first-degree polynomials $w-r_1,\ldots,w-r_n$ would be factors of $w^n-z$. Therefore $(w-r_1)\cdots(w-r_{n+1})$ would be a factor of $w^n-z$. That would make $w^n-z$ a polynomial in $w$ with degree more than $n$. But it's not.
To see why $w-r$ must be a divisor of $w^n-z$ if $r$ is a solution for $w$ of the equation $w^n-z=0$, just divide $w^n-z$ by $w-r$: $$w^n-z=(w-r)(\cdots\cdots)+a$$ where $(\cdots\cdots)$ is the quotient and $a$ is the remainder. If $a\ne0$, then it's not hard to see how a contradiction follows. (I've omitted some details of the reasoning in algebra, but they're not hard to fill in.)
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Suppose you claim that the set $$\{z_1,\ldots,z_n\}$$ gives all $n$-th roots of a number $w$, and those are all there are.
Then you ought to prove that:
$(1)$ If $z_i=z_j$ then $i=j$; or $i\neq j\implies z_i\neq z_j$.
$(2)$ If $v$ is another $n$-th root, then $v=z_i$ for some $i=1,\ldots,n$.
You're are indeed right: the roots do start repeating, so there you have a starting point. This is because $\exp(ix)=\exp(ix+2\pi ki)$ for any integer $k$. If $v^n=w$, note that $n{\,{\rm arg }\,\,v}=w$ and $|v|^n=w$. Use this and $$e^{\theta i}=\cos \theta+i\sin\theta$$ to work out $(2)$.
- | 2016-06-28T13:35:27 | {
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https://math.stackexchange.com/questions/1493889/number-of-possible-subsets-of-all-sizes/1495380 | # Number of possible subsets of all sizes
1. Given a set of elements $S$ of size $n,$ where some elements may be repeated, what is the total number of subsets that can be made from $S,$ including all sizes from $2$ to $n-1?$ Note: the ordering of elements does not matter.
• $S$ could be e.g. $S=\{el_1,el_1,el_2,el_3,...,el_n\}.$ Notice some elements are repeated. For example a subset of size 3 would be $\{el_1,el_3,el_2\}.$
2. How does the count of possible subsets change when two specific elements (e.g. $el_1,el_2,$ hence the size of subsets starting from $2$) have to be in every considered subset?
My own guess is: we know that if the order does not matter then there are $$\frac{n!}{(n-k)!k!} \tag{1}$$ ways of picking k elements from n. Thus the total number of subsets with $k$ going from $2$ to $n-1$ is given by: $$\sum_{k=2}^{n-1}\frac{n!}{(n-k)!k!} \tag{2}$$ But I do not know whether formula $(1)$ holds if there are repeating elements among the $n$ elements of the initial set. If it holds, then I suspect my $(2)$ should give the correct count at least for part 1.
An example: say the set of elements are $\{a,a,b,c\}$ then all the subset of 3 elements where order doesn't matter would be: $\{a,a,b\},\{a,a,c\},\{a,b,c\}$ so only 3 possibilities whereas $(1)$ gives 4 because it assumes distinct elements in the original set.
• The answer obviously depends on how many elements are repeated and how many times, so you should treat these as parameters of your problem. – uniquesolution Oct 23 '15 at 13:34
• It seems to me that in 1) you are asking for $\sum_{k=2}^{n-1}\left|\mathcal{F}_{k}\right|$ where $\mathcal{F}_{k}:=\left\{ f\in\omega^{\left\{ 1,\dots,n\right\} }\mid\sum_{i=1}^{n}f\left(i\right)=k\right\}$ and $\omega$ denotes the set of nonnegative integers. Is that correct? – drhab Oct 23 '15 at 14:34
• @drhab The elements of the set could also be letters, I didn't specify them to be integers necessarily. I added an edit to clarify some things. – user186225 Oct 23 '15 at 14:45
• What is the size of $\{el_1,el_1,el_3,el_2\}$? Is it $3$ or is it $4$? – drhab Oct 23 '15 at 14:59
• @drhab 4, we count all, repeated ones included, as they are allowed. (in other words, I'm not talking of the size of the corresponding mathematical set) – user186225 Oct 23 '15 at 15:01
Note that to find the total number of possible subsets of $n$ distinct elements, it is the same as putting the $n$ distinct elements into $2$ distinct boxes. In other words, each distinct element is either in the subset or not in the subset.
Therefore, the total number of subsets = $2^n$ (this includes the $\varnothing$ of no elements)
At the same time, the total number of subsets can also be calculated as you suggested in $(1)$ and $(2)$: $$\sum_{k=0}^{n}\frac{n!}{(n-k)!k!} = \sum_{k=0}^{n}\binom{n}{k}=2^n$$
This will allow you to find a "nicer form" for your expression in $(2)$.
$$\binom{n}{0}+\binom{n}{1}+\binom{n}{n}+\sum_{k=2}^{n-1}\binom{n}{k}=2^n$$
$$\sum_{k=2}^{n-1}\binom{n}{k}=2^{n}-n-2$$
When there are identical terms, we will have to first count the number of subsets of the distinct terms (let us assume there are $a$ number of some identical term).
We will have $2^{(n-a)}$ subsets of distinct terms.
Now we will count the number of subsets of $a$ identical terms or find the number of ways to distribute $a$ identical terms into $2$ distinct boxes: $$\binom{a+2-1}{2-1} = a+1$$
This uses the "stars and bars" method where the distribution of $n$ identical objects into $k$ distinct boxes, empty boxes are allowed is given by:
$$\binom{n+k-1}{k-1}$$
In the above case, we had $a$ objects and $2$ distinct boxes, i.e, "in the subset" or "not in the subset".
We can then multiply the two together to find the number of subsets of the $(n-a)$ distinct terms and $a$ identical terms again including $\varnothing$. Once you get your final value, it only includes the null set once so you can choose to minus $1$ if you do not wish to count the null set.
If we have multiple sets of identical elements, for example, $S=\{el_1,el_1,el_2,el_3,...,el_n\}$ and with $a$ number of '$el_a$'s, $b$ number of '$el_b$'s and so on.
The total number of subsets is then given by:
$$2^{(n-a-b-\dots)} (a+1)(b+1)(c+1)\dots$$
• You're welcome. a) we must count subsets of the distinct and the repeated terms separately. Let me give an example. Consider the set $(1,3,3)$. The possible subsets are $\varnothing,(1),(3),(1,3),(3,3),(1,3,3)$. Subset of only distinct numbers is $\varnothing,(1)$. Subset of identical numbers are $\varnothing,(3),(3,3)$. This means that to each of the subset of distinct numbers we can put in either no $3$ one $3$ or two $3$s. I will answer b) and c) in edit. – Nicholas Oct 24 '15 at 15:55 | 2020-01-23T05:08:23 | {
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https://math.stackexchange.com/questions/927530/cosine-of-the-sum-of-two-solutions-of-trigonometric-equation-a-cos-theta-b-s | Cosine of the sum of two solutions of trigonometric equation $a\cos \theta + b\sin \theta = c$
Question: If $\alpha$ and $\beta$ are the solutions of $a\cos \theta + b\sin \theta = c$, then show that: $$\cos (\alpha + \beta) = \frac{a^2 - b^2}{a^2 + b^2}$$
No idea how to even approach the problem. I tried taking two equations, by substituting $\alpha$ and $\beta$ in place of $\theta$ in the equation and manipulating them, but that didn't get me anywhere. Please help!
We have $\displaystyle a\cos\theta=c-b\sin\theta,$
Squaring we get, $\displaystyle(c-b\sin\theta)^2=(a\cos\theta)^2=a^2(1-\sin^2\theta)$
$\displaystyle\iff (a^2+b^2)\sin^2\theta-2bc\sin\theta+c^2-a^2=0$
So, $\displaystyle\sin\alpha\sin\beta=\dfrac{c^2-a^2}{a^2+b^2}$
Similarly find $\displaystyle\cos\alpha\cos\beta$ by squaring $\displaystyle b\sin\theta=c-a\cos\theta$
Finally use $\displaystyle\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta$
• Ohh.... I see now. Well thanks! But how exactly am I supposed to think of that? – Gummy bears Sep 11 '14 at 12:36
• @Gummybears, Sorry for the typo – lab bhattacharjee Sep 11 '14 at 12:44
• How did you get the value of $\sin \alpha \sin \beta$? – Gummy bears Sep 11 '14 at 12:51
• @Gummybears, Please find en.wikipedia.org/wiki/Vieta's_formulas – lab bhattacharjee Sep 11 '14 at 13:12
We have $\displaystyle a\cos\alpha+b\sin\alpha=c=a\cos\beta+b\sin\beta$
$\displaystyle a(\cos\alpha-\cos\beta)=-b(\sin\alpha-\sin\beta)$
Now use Prosthaphaeresis Formulas to find $\displaystyle\tan\frac{\alpha+\beta}2$ assuming $\displaystyle\sin\frac{\alpha-\beta}2\ne0$
Then use $\displaystyle\cos2A=\frac{1-\tan^2A}{1+\tan^2A}$
Here's a picture showing angles $\theta$ (at $P$) and $\phi$ (at $Q$) such that $$a \cos\theta + b \sin\theta = c = a \cos\phi + b \sin \phi$$ The measure of $\angle PAQ$ is the sum of these angles.
Note that $P$ and $Q$ lie on the circle with diameter $\overline{AB}$, and that the diameter bisects $\angle PAQ$. From here, we have many approaches to the final relation; here's one: Clearly, $$\cos\frac{\theta+\phi}{2} = \frac{a}{d} \qquad\qquad \sin\frac{\theta+\phi}{2} = \frac{b}{d}$$ so that, by the Double-Angle Formulas, $$\cos(\theta+\phi) = 2\cos^2\frac{\theta+\phi}{2} - 1 = \frac{2a^2-d^2}{d^2} = \frac{2a^2-(a^2+b^2)}{a^2+b^2} = \frac{a^2-b^2}{a^2+b^2}$$ $$\sin(\theta+\phi) = 2 \sin\frac{\theta+\phi}{2}\cos\frac{\theta+\phi}{2} = \frac{2ab}{d^2} = \frac{2ab}{a^2+b^2}$$
Here's a geometric solution (which thus has some in common with lab's Prosthaphaeresis Formula solution); it amounts just to interpreting the given equation as for the intersection of a line and a circle, and then using symmetry and doing some easy algebra.
By construction, the solutions $\alpha$ and $\beta$ are the angles between the positive $x$-axis and the points of intersection of the unit circle $C$ with the line $L$ defined by $a x + b y = c$ (since the original question says solutions, we'll assume there are two points of intersection). We may as well assume too that the points are not endpoints of a diameter (equivalently that $c \neq 0$), as the claim is a trivial calculation in that case.
Now, by symmetry, (1) the line $L'$ through $0$ and the point on $L$ closest to $0$ makes an angle $$\theta := \frac{1}{2} (\alpha + \beta)$$ with the positive $x$-axis, and (2) $L' \perp L$, so $L$ has equation $b x - a y = 0$. This line intersects $C$ at two points and substituting gives that the intersections satisfy $$x^2 = \frac{a^2}{a^2 + b^2}$$ and so $$y^2 = \frac{b^2}{a^2 + b^2}.$$ By definition these quantities are respectively $\cos^2 \theta$ and $\sin^2 \theta$, and so the cosine double angle identity gives $$\cos(\alpha + \beta) = \cos(2 \theta) = \cos^2 \theta - \sin^2 \theta = \frac{a^2 - b^2}{a^2 + b^2}$$ as desired.
Use Weierstrass Substitution to form a Quadratic Equation in $\tan\dfrac\theta2$
Now using Vieta's formula, we can find $\displaystyle\tan\dfrac\alpha2+\tan\dfrac\beta2,\tan\dfrac\alpha2\tan\dfrac\beta2$
Then $\displaystyle\tan\left(\dfrac\alpha2+\dfrac\beta2\right)=\dfrac{\tan\dfrac\alpha2+\tan\dfrac\beta2}{1-\tan\dfrac\alpha2\tan\dfrac\beta2}$
• @Gummybears, Here is my third method – lab bhattacharjee Sep 11 '14 at 16:49 | 2019-08-19T11:46:53 | {
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https://thinkandstart.com/menace-ii-oqcpwq/how-to-find-the-centroid-of-a-triangle-3f40df | The centroid is a point where all the three medians of the triangle intersect. Example: Find the Centroid of a triangle with vertices (1,2) (3,4) and (5,0) Question 1 : Find the centroid of triangle whose vertices are (3, 4) (2, -1) and (4, -6). If three medians are constructed from the three vertices, they concur (meet) at a single point. Or the coordinate of the centroid here is just going to be the average of the coordinates of the vertices. The centre of point of intersection of all the three medians in a triangle is the centroid. For example, to find the centroid of a triangle with vertices at (0,0), (12,0) and (3,9), first find the midpoint of one of the sides. Next we will input the location of the centroid of the triangle. The Centroid is a point of concurrency of the triangle.It is the point where all 3 medians intersect and is often described as the triangle's center of gravity or as the barycent.. Properties of the Centroid. Example 3: Centroid of a tee section. All the three medians AD, BE and CF are intersecting at G. So G is called centroid of the triangle Practice Questions. For instance, the centroid of a circle and a rectangle is at the middle. A simple online calculator to calculate the centroid of an isosceles triangle. In the above triangle , AD, BE and CF are called medians. Therefore, the centroid of the triangle can be found by finding the average of the x-coordinate’s value and the average of the y-coordinate’s value of all the vertices of the triangle. The centroid of a triangle is the point of intersection of its three medians (represented as dotted lines in the figure). It is the center of mass (center of gravity) and therefore is always located within the triangle. Locating Plumb line method. To find the centroid of a triangle, use the formula from the preceding section that locates a point two-thirds of the distance from the vertex to the midpoint of the opposite side. For more see Centroid of a triangle. This is a composite area. Let the vertices be A (3,4) B (2,-1) and C (4,-6) The centroid of a uniformly dense planar lamina, such as in figure (a) below, may be determined experimentally by using a plumbline and a pin to find the collocated center of mass of a thin body of uniform density having the same shape. An isosceles triangle is a triangle that has two sides of equal length. The median of a triangle is a line or line segment from a vertex to the midpoint of the opposite side. Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin asked Aug 4, 2018 in Mathematics by avishek ( 7.9k points) coordinate geometry the altitude and reason that the centroid of the entire triangle lies one-third the altitude above the base. The centroid of a triangle is the point of intersection of its medians (the lines joining each vertex with the midpoint of the opposite side). For other properties of a triangle's centroid, see below. Centroid of Isosceles Triangle Calculator . Step 1. The definition of a centroid of a triangle is intersection of the medians of the triangle. That is this triangle right over there. Note: When you're given the centroid of a triangle and a few measurements of that triangle, you can use that information to find missing measurements in the triangle! For example, if the coordinates of the vertices of a right triangle are (0, 0), (15, 0) and (15, 15), the centroid is found by adding together the x coordinates, 0, 15 and 15, dividing by 3, and then performing the same operation for the y coordinates, 0, 0 and 15. It is formed by the intersection of the medians. It is the point which corresponds to the mean position of all the points in a figure. So we have three coordinates. It will place a point at the center or centroid of the triangle. The centroid of a rectangle is in the center of the rectangle, , and the centroid of triangle can be found as the average of its corner points, . Locus is actually a path on which a point can move , satisfying the given conditions. Case 1 Find the centroid of a triangle whose vertices are (-1, -3), (2, 1) and (8, -4). The coordinates of the centroid are simply the average of the coordinates of the vertices.So to find the x coordinate of the orthocenter, add up the three vertex x coordinates and divide by three. This point is the triangle's centroid, which will always divide a median into a 2:1 ratio; that is, the centroid is ⅓ the median's distance from the midpoint, and ⅔ the median's distance from the vertex. The point of intersection of all the three medians of a triangle is called its centroid. Since all the medians meet at a single point, it is sufficient to find the point of intersection of only two medians to obtain the centroid of a triangle. The centroid of a triangle is just going to be the average of the coordinates of the vertices. The above example will clearly illustrates how to calculate the Centroid of a triangle manually. Also, a centroid divides each median in a 2:1 ratio (bigger part is closer to the vertex). Using the formula to find the centroid of a triangle Skills Practiced. 2) More Complex Shapes:. ; It is one of the points of concurrency of a triangle. Find the centroid of the following tee section. You've already mentioned the shortcut, which is to average the x coordinates and average the y coordinates. The center of mass is the term for 3-dimensional shapes. This is a right triangle. Click hereto get an answer to your question ️ Find the third vertex of a triangle, if two of its vertices are ( - 3,1), (0, - 2) and centroid is at the origin. To calculate the centroid of a combined shape, sum the individual centroids times the individual areas and divide that by the sum of … Given point D is the centroid of triangle ABC, find the lengths of BC, CD, and AY. And to figure out that area, we just have to remind ourselves that the three medians of a triangle divide a triangle into six triangles that have equal area. Frame 12-23 Centroids from Parts Consider the scalene triangle below as being the difference of two right triangles. We place the origin of the x,y axes to the middle of the top edge. The point through which all the three medians of a triangle pass is called centroid of the triangle and it divides each median in the ratio 2:1. 1) Rectangle: The centroid is (obviously) going to be exactly in the centre of the plate, at (2, 1). To find the direction of the electric field vector at any point due to a point charge we perform a “thought experiment” which consists in placing a positive test charge at this point. Centroid. So BGC right here. The centroid is the point of concurrency of the three medians in a triangle. That means it's one of a triangle's points of concurrency. "The second method to find the center of a triangle is to turn the triangle into a region. x 1 = -1, y 1 = -3 x 2 = 2, y 2 = 1 and x 3 = 8, y 3 = -4 Substitute in the formula as . find the locus of the centroid of a triangle whose vertices are $(a \cos t, a \sin t), (b \sin t, -b \cos t)$ & $(1,0)$ Ask Question Asked 6 years, 11 months ago In case of triangle this point is located at 2b/3 horizontally from reference y-axis or from extreme left vertical line. To find the centroid of a triangle, use the formula from the preceding section that locates a point two-thirds of the distance from the vertex to the midpoint of the opposite side. Solution: Centroid of triangle is a point where medians of geometric figures intersect each other. This is given by the table above which indicates that the centroid of a triangle is located, from the corner that is opposite of the hypotenuse (the longest side of the triangle), one-third of the length of the base in the y direction and one-third of the length of the height in the x direction in this case. So this coordinate right over here is going to be-- so for the x-coordinate, we have 0 plus 0 plus a. That point is called the centroid. Use what you know about right triangles to find one coordinate of the centroid of triangle A. If two vertices of a triangle are (3, − 5) and (− 7, 8) and centroid lies at the point (− 1, 1), third vertex of the triangle is at the point (a, b) then View solution If one vertex of the triangle having maximum area that can be inscribed in the circle ∣ z − i ∣ = 5 is 3 − 3 i ,then another vertex of the right angle can be: To find the centroid of a triangle ABC you need to find average of vertex coordinates. The centroid is the term for 2-dimensional shapes. The medians of a triangle are concurrent. The centroid of a right triangle is 1/3 from the bottom and the right angle. To find the centroid of either triangle, use the definition. Knowledge application - use your knowledge of what a centroid of a triangle is to answer a question about it In a triangle, the centroid is the point at which all three medians intersect. For example, on a median that is 3.6 cm long, the centroid will be 1.2 cm up from the midpoint. So if we know the area of the entire triangle-- and I think we can figure this out. A median is a line which joins a vertex of a triangle to the midpoint of the opposite side. The procedure for composite areas, as described above in this page, will be followed. It is also the center of gravity of the triangle. The formula for finding the centroid of a triangle is deduced as: Let A (x 1, y 1), B (x 2, y 2) and C (x 3, y 3) be the vertices of ∆ABC whose medians are AD, BE and CF respectively.So D, E and F are respectively the mid points of BC, CA and AB The centroid divides each median into a piece one-third the length of the median and two-thirds the length. Recall that the centroid of a triangle is the point where the triangle's three medians intersect. Centroid Example. Centre of Mass (Centroid) for a Thin Plate. For example, to find the centroid of a triangle with vertices at (0,0), (12,0) and (3,9), first find the midpoint of one of the sides. And the shape of that path is referred to as locus. Median. The centroid of a triangle on a coordinate plane is found by taking the average position of the three vertices. Centroid of a triangle. Centroid of a triangle may be defined as the point through which all the three medians of triangle pass and it divides each median in the ratio 2 : 1.. Start by entering Region at the Command line, followed by the Enter key. Find the centroid of triangle having b= 12’ and h= 6’. If $(0,0)(a,0)(a,b)$, $G=(\frac{2a}3,\frac{b}3)$ We divide the complex shape into rectangles and find bar(x) (the x-coordinate of the centroid) and bar(y) (the y-coordinate of the centroid) by taking moments about the y-and x-coordinates respectively. , on a coordinate plane is found by taking the average position of the triangle the location the. Located within the triangle is intersection of its three medians are constructed the... This page, will be followed that means it 's one of triangle. By the intersection of the medians of the triangle a coordinate plane is found by taking the average vertex. Each median in a triangle is the point of intersection of its three medians are constructed from the midpoint line... Corresponds to the vertex ) three medians ( represented as dotted lines in the )! 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X, y axes to the middle of the triangle 's centroid, see below a Plate. A figure vertex of a triangle is to turn the triangle intersect right triangles is by... Point which corresponds to the middle y-axis or from extreme left vertical line will the. Above the base medians are constructed from the three medians intersect vertex.. For 3-dimensional shapes we have 0 plus a lies one-third how to find the centroid of a triangle length vertex to midpoint! | 2021-04-22T03:34:15 | {
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https://math.stackexchange.com/questions/3882237/equivalence-of-ax-b-and-the-quadratic-form | # Equivalence of $Ax = b$ and the quadratic form?
I want to prove the following:
Given $$A \in \mathbf{R}^{n \times n}$$ is symmetric positive definite. Prove that $$\hat{x}$$ solves $$Ax = b$$ if and only if $$\hat{x}$$ minimizes the quadratic function $$f: \mathbf{R}^n \to \mathbf{R}$$ given by:
$$f(x) = \frac{1}{2}x^TAx - x^Tb$$
Attempt:
Since $$A$$ is positive definite, it is invertible since its eigenvalues are all strictly positive. Let $$x = A^{-1}b$$ and determine $$f(y) - f(x)$$ for any $$y \in \mathbf{R}^n$$. Since $$Ax = b$$:
\begin{align} f(y) - f(x) &= \frac{1}{2}y^TAy - y^Tb - \frac{1}{2}x^TAx + x^Tb \\ &= \frac{1}{2}y^TAy - y^TAx + \frac{1}{2}x^TAx \\ &= \frac{1}{2}(y - x)^TA(y-x)\end{align} Since $$A$$ is positive definite, the last expression is nonnegative and therefore $$f(y) \geq f(x)$$ for all $$y \in \mathbf{R}^n$$, which gives x = $$A^{-1}b$$ as the global minimum of $$f(x)$$ and $$f(A^{-1}b) = -\frac{1}{2}b^TA^{-1}b$$
Concerns:
I'm concerned that this proof is determining what the global minimum of the equivalent system is not necessarily that $$\hat{x}$$ solves $$Ax = b$$ if and only if $$\hat{x}$$ minimizes the quadratic function. Any hints in the right direction would be greatly appreciated!
• You could also try to calculate $\nabla f$. – Meowdog Oct 26 '20 at 18:49
• Note the $A \succ 0$ implies that $A$ is invertible and that $f(x)$ is a strictly convex function, so its minimizer is unique. Also note that the minimizer $\hat{x}$ of $f$ satisfies $\nabla f(\hat{x}) = A\hat{x} - b = 0$. – V.S.e.H. Oct 26 '20 at 19:01
Because $$A$$ is invertible, "$$\hat{x}$$ solves $$Ax=b$$" is just "$$\hat{x} = A^{-1} b$$."
So to show the "if and only if" claim, you need to show that $$A^{-1}b$$ is the unique minimizer of $$f$$.
So far you have shown that $$A^{-1}b$$ is a minimizer (since $$f(y) \ge f(A^{-1} b)$$ for all $$y$$), but you still need to show it is the only one. To do this, it suffices to adjust your argument to show the stronger inequality $$f(y) > f(A^{-1} b)$$ for all $$y \ne A^{-1} b$$. Can you see how to do this? (Hint: $$A$$ is positive definite.) | 2021-05-10T05:39:06 | {
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https://mathoverflow.net/questions/221351/reference-for-a-linear-algebra-result | # Reference for a linear algebra result
I asked the following question (https://math.stackexchange.com/questions/1487961/reference-for-every-finite-subgroup-of-operatornamegl-n-mathbbq-is-con) on math.stackexchange.com and received no answers, so I thought I would ask it here. I've asked several people in my department who were all stumped by the question.
The question is: why is every finite subgroup of $\operatorname{GL}_n(\mathbb{Q})$ conjugate to a finite subgroup of $\operatorname{GL}_n(\mathbb{Z})$?
Note that at least for $n=2$ the question of isomorphism is much easier, since one can (with some effort) work out exactly which finite groups can be subgroups of $\operatorname{GL}_2(\mathbb{Q})$. Further, there are isomorphic finite subgroups of $\operatorname{GL}_2(\mathbb{Q})$ that are not conjugate to each other. For example, the group generated by $-I_{2 \times 2}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ are both isomorphic to $C_2$, but they cannot be conjugate to each other because the eigenvalues of the two generators are different.
If there is a relatively simple proof, that would be ideal, but a reference with a potentially long proof is fine as well.
Thanks for any assistance.
• Can I ask how do you know it's true? – Todd Trimble Oct 20 '15 at 2:05
• @ToddTrimble this was posted in the original question, someone casually mentioned it without justification here: math.stackexchange.com/questions/58476/… . Nobody seemed to contest that claim, so I assumed it was some well-known result in the subject. I've asked several faculty members in my department, none of them knows why it's true, but all of them believe it – Stanley Yao Xiao Oct 20 '15 at 2:07
• Frankly, had you tagged it correctly on math.se and waited a bit you'd gotten your answer there I am pretty sure. – user9072 Oct 20 '15 at 16:36
This argument is fairly standard, but it is quicker to repeat it than to find a reference: Let $G$ be a finite subgroup of $GL_n(\mathbb{Q})$. Set $\Lambda = \sum_{g \in G} g \cdot \mathbb{Z}^n \subset \mathbb{Q}^n$. Then $\Lambda$ is a finitely generated torsion free abelian group, hence isomorphic to $\mathbb{Z}^r$ for some $r$. Since $\mathbb{Z}^n \subseteq \Lambda \subset \mathbb{Q}^n$, we have $r=n$. Clearly, $g \Lambda = \Lambda$ for all $g \in G$.
Let $h \in GL_n(\mathbb{Q})$ take the standard basis to a basis of $\Lambda$, so $h \mathbb{Z}^n = \Lambda$ and $\mathbb{Z}^n = h^{-1} \Lambda$. Then $h^{-1} g h$ takes $h^{-1} \Lambda = \mathbb{Z}^n$ to itself for all $g \in G$, so $h^{-1} G h \subset GL_n(\mathbb{Z})$.
David Speyer has answered the question, but let me add some background. The general result is that if $R$ is a principal ideal domain with field of fractions $K$, and $G$ is a finite group, then every finite dimensional representation of $G$ over $K$ may be realised over $R$ ( ie, is equivalent to a representation over $R$). The general proof is essentially the one that David gives.
As well as the integral case considered in the question, the general result was important for the development by Richard Brauer of modular representation theory: the idea of "reduction (mod $p$)" of a complex representation relies on it: the representation of the finite group $G$ is first realised over a suitable finite extension $K$ of $\mathbb{Q}$, and then $K$ may be viewed as the field of fractions of a localisation $R$ at a prime ideal $\pi$ containing $p$ of the ring of algebraic integers in $K$. Then since $R$ is a PID, the $K$-representation may then be realized over $R$. Then the given representation may be reduced (mod $\pi$), yielding representation of $G$ over the finite field $R/\pi$.
In general, it is not a straightforward issue to decide whether a representation of a finite group $G$ over a number field $K$ may be realised over the ring of integers of $K$. Some of the issues are well illustrated in the article "Three letters to Walter Feit" by J-P. Serre (which is visible online), which considers special cases of this question.
I think this should work: let $S$ be the set of primes occurring in the denominators of the entries of the elements of $G$, and let $\mathbb{Z}[S^{-1}]$ be the subring of $\mathbb{Q}$ generated by adjoining $1/p$ for $p\in S$.
Let $\mathbb{Q}_S = \prod_{p\in S} \mathbb{Q}_p$ and let $\mathbb{Z}_S = \prod_{p\in S} \mathbb{Z}_p$. We'll show that we can conjugate $G$ into $GL_n(\mathbb{Z}_S)$ by an element of $x$ $GL_n(\mathbb{Z}[S^{-1}])$. This is enough, since it shows then the $p$-adic valuation of all entries of elements of $x^{-1}Gx$ is nonnegative, and of course we still have $x^{-1}Gx\in GL_n(\mathbb{Q})$.
To this end, because $G$ is finite, it compact, so given a $p$ there is some $x_p\in GL_n(\mathbb{Q}_p)$ such that $x_p^{-1}Gx_p \leq GL_n(\mathbb{Z}_p)$. In fact, the set of such $x_p$ is open in $GL_n(\mathbb{Q}_p)$ since $G$ is finite and the map $x_p \mapsto x_p^{-1}gx_p$ is continuous. Because $GL_n(\mathbb{Z}[S^{-1}])$ is dense in $GL_n(\mathbb{Q}_S)$, there is in fact an $x\in GL_n(\mathbb{Z}[S^{-1}])$ that conjugates $G$ into $GL_n(\mathbb{Z}_S)$, completing the proof.
• I think the fact that every compact subgroup of $\operatorname{GL}_n(\mathbb Q_p)$ is conjugate into $\operatorname{GL}_n(\mathbb Z_p)$ is not necessarily easier than the proof of the desired fact given by @DavidSpeyer above. – LSpice Oct 21 '15 at 19:57
• Yeah, after I saw @David's proof I agree. – John Binder Oct 21 '15 at 21:19
This is not exact answer to your question, but long back, I had seen this result in an article in Monthly. I hope it will be useful.
http://www.jstor.org/stable/pdf/2695329.pdf?acceptTC=true
• Yes, a professor in my department pointed that paper out to me some time after David Speyer's anwer. Thanks for the link – Stanley Yao Xiao Jan 5 '16 at 3:22 | 2020-10-30T19:34:45 | {
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http://math.stackexchange.com/questions/5431/formally-proving-that-a-function-is-oxn | # Formally proving that a function is $O(x^n)$
Say I have a function $f(x) = ax^3 + bx^2 + cx + d$ where $a > 0$.
It's clear that for a high enough value of $x$, the $x^3$ term will dominate and I can say $f(x) \in O(x^3)$, but this doesn't seem very formal.
The formal definition is $f(x) \in O(g(x))$ if constants $k, x_0 > 0$ exist, such that $0 \le f(x) \le kg(x)$ for all $x > x_0$.
My question is, what are appropriate values for $k$ and $x_0$? It's easy enough to find ones that apply (say $k = a + b + c + d$). By the formal definition, all I have to do is show that these numbers exist, so does it actually matter which numbers I use? For some value of $x$, $k$ could be anywhere from $1$ to $a + b + c + d + ...$. From my understanding, it doesn't matter what numbers I pick as long as they 'work', but is this right? It seems too easy.
Thanks
-
I changed some of the c's to k's because I think there was a clash there - hope that it was correct to do so. – anon Sep 25 '10 at 16:57
Looks fine to me. You missed one though -- I fixed it :-P – Joel Sep 25 '10 at 17:03
If $a < 0$, then the $0 \le kg(x)$ is violated... Perhaps a different definition? Or can $k < 0$? – Aryabhata Sep 25 '10 at 18:54
@Moron: Yeah, I think we should have $|f(x)|<k|g(x)|$ for $x>x_0$. For instance, $sin(x)=O(1)$ and the defintion by Joel doesn't give this. – alext87 Sep 25 '10 at 18:58
@alex: Yeah, since this seems like homework, just trying to make sure Joel knows the right definition taught in their class :-) – Aryabhata Sep 25 '10 at 19:03
The argument you are getting at as I understand is is, roughly: $x^n \in O(x^n)$ and thus $kx^n \in O(x^n)$, so $f(x)$ acts like $O(x^3) + O(x^2) + O(x) + O(1)$ which can be reduced to $O(x^3)$.
So the theorem we would like to prove now is that for $n\geq m$: $f \in O(x^n)$ and $g \in O(x^m)$ implies $f + g \in O(x^n)$. Once we have this you just add up the monomials of the polynomial and that proves the result.
Look at what we have, from the definitions:
$$f \in O(x^n) \Rightarrow \exists x_0, k,\;\; \forall x>x_0,\;\; f(x) \leq kx^n$$
$$g \in O(x^m) \Rightarrow \exists x_1, k',\;\; \forall x>x_1,\;\; g(x) \leq kx^m$$
Let $x_2$ be the maximum of $x_0$ and $x_1$, $k''$ be the maximum of $k$ and $k'$ and add these inequalities:
$$\forall x>x_2,\;\; (f+g)(x) \leq kx^n + k'x^m \leq k''(x^n + x^m) \leq 2 k'' x^n$$
Now the pair of values $(x_2,2k'')$ prove that $f+g \in O(x^n)$.
To consider abstract functions like this makes the proof very easy, but it is clear that the values we exhibit to prove the existential may not be the best, although we still prove the theorem in an effective way. In particular, you could do a very detailed analysis of the functions in specific cases to get very tight bounds - in this lucky case it's not needed which is why the theorem is easier to prove in the abstract case.
-
HINT $\quad\rm ax^3 + bx^2 + cx + d\ \le \ (|a|+|b|+|c|+|d|)\ x^3 \$ for $\rm\ x > 1$
-
Suppose you have a function $f(x)$ and $g(x)$.
A really simple method (that works when $f(x)$ and $g(x)$ are polynomials) of determining a constant that works is the following. Consider
$\lim_{x\rightarrow\infty}\frac{f(x)}{g(x)}$
If the limit exists and if a constant $0\leq C< \infty$. Then $C+\epsilon$ for any $\epsilon>0$ is a constant that works. To see this just apply the definition of the limit. $\forall \epsilon>0$ there exists a $x_0(\epsilon)$ such that $\forall x>x_0(\epsilon)$ we have
$\left|\frac{f(x)}{g(x)}-C\right|<\epsilon$
That is
$\frac{f(x)}{g(x)} < C+\epsilon$
Now you know the constant from calculating the limit and know the existence of $x_0$. Since you are always considering asymptotics when using this definition you are never concerned with the value of $x_0$ (only that it exists)
It does not matter what constants you use and someone could easily use different constants to get $f(x)=O(g(x))$. This notation is to compare the growth rate of two functions.
If $\lim_{x\rightarrow \infty} \frac{f(x)}{g(x)}$ does not exist or is hard to calculate then as long as you can bound it above you still have $f(x)=O(g(x))$
- | 2014-12-19T16:59:05 | {
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https://math.stackexchange.com/questions/1789063/a-positive-integer-in-decimal-notation-is-divisible-by-11-iff | # A positive integer (in decimal notation) is divisible by 11 $\iff$ …
(I am aware there are similar questions on the forum)
## What is the Question?
A positive integer (in decimal notation) is divisible by $11$ if and only if the difference of the sum of the digits in even-numbered positions and the sum of digits in odd-numbered positions is divisible by $11$.
For example consider the integer 7096276.
The sum of the even positioned digits is $0+7+6=13.$ The sum of the odd positioned digits is $7+9+2+6=24.$ The difference is $24-13=11$, which is divisible by 11.
Hence 7096276 is divisible by 11.
## (a)
Check that the numbers 77, 121, 10857 are divisible using this fact, and that 24 and 256 are not divisible by 11.
## (b)
Show that divisibility statement is true for three-digit integers $abc$. Hint: $100 = 99+1$.
## What I've Done?
I've done some research and have found some good explanations of divisibility proofs. Whether that be 3,9 or even 11. But...the question lets me take it as fact so I don't need to prove the odd/even divisible by 11 thing.
I need some help on the modular arithmetic on these.
For example... Is 77 divisible by 11? $$7+7 = 14 \equiv ...$$
I don't know what to do next.
Thanks very much, and I need help on both (a) and (b).
In order to apply the divisibility rule, you have to distinguish between odd and even positioned digits. (It doesn't matter how you count.)
Example: In 77 the first position is odd, the second even, so you would have to calculate $7-7=0$, which is divisible by 11.
Now it should be easy for you to understand what you try to prove in (b): If a,b,c are three digits abc is the number $100a+10b+c$. You know what it means to say that this number is divisible by 11. You have to prove that $$11\vert (a+c)-b \Leftrightarrow 11\vert 100a+10b+c$$ or with modular arithmetic $$(a+c)-b \equiv 0 \pmod{11}\Leftrightarrow 100a+10b+c\equiv 0 \pmod {11}\; .$$ I don't want to spoil the fun so I leave it there.
When you are checking $77$ for divisiblility, the first $7$ is in an even place and the second is in an odd place, so you check $7-7=0$, which is divisible by $11$, so $77$ is. For $10857$ you do $1+8+7-0-5=11$, which is divisible by $11$. For part b, note that the three digit number $abc$ can be written as $100a+10b+c$ and use the hint to represent the $100$
# How to quickly determine whether an integer is divisible by 3,9,11
NOTE
Generally speaking, it is not a simple question to determine whether an integer is divisible by another one or not . We don't have a universal method. However, given a specific integer, when the dividend take some special value, such as 3, 9, 11,.... we can easily test that by hand.
# Proposition :
Suppose n is any positive integer,and the decimal representation of $n$ is:
$$n=a_{k}a_{k-1}a_{k-2}\cdots a_{0}$$
We have the following results:
1. $$\textrm{n is divisible by 3 if and only if} \quad \sum_{i=0}^{k}a_{i} \,\,\text{is divisible by 3}$$
2. $$\textrm{n is divisible by 9 if and only if} \quad \sum_{i=0}^{k}a_{i} \,\,\text{is divisible by 9}$$
3. $$\qquad\textrm{n is divisible by 11 if and only if} \sum_{i=0}^{k}(-1)^{i}a_{i} \,\,\text{is divisible by 11}$$
# proof of 3:
Since $$n=\sum_{i=0}^{k}a_{i}\cdot10^{k}$$ and $$10\equiv -1\pmod {11}$$ we have $$n=\sum_{i=0}^{k}a_{i}\cdot10^{k}\equiv \sum_{i=0}^{k}a_{i}\cdot(-1)^{i}\pmod{11}$$ Thus,
$$n\equiv0 \pmod{11 }\,\text{if and only if}\sum_{i=0}^{k}a_{i}\cdot(-1)^{i}\equiv0\pmod{11}$$ Notice that $$10\equiv 1\pmod9\quad and\quad 10\equiv 1\pmod3$$ the first two propositions can be proofed in the same way,we safely omit the proof here.
# Coment:
When the dividend take other values, we can also construct similar test method, but the result will not be as beautiful as the above ones.
There are other tests for divisibility by $11$. I submit the following.
This Lemma has been proved many many times
Lemma. Let $n = \sum_{i=0}^D d_i 10^{D-i}$ where the $d_i$ are integers from $0$ to $9$. Then $n \pmod 9 \equiv \sum_{i=1}^D d_i \pmod 9$.
The process is often referred to as casting out nines.
It isn't hard to show that there is a similar lemma for casting out ninety-nines.
Lemma. Let $n = \sum_{i=0}^D d_i 100^{D-i}$ where the $d_i$ are integers from $0$ to $99$. Then $n \pmod{99} \equiv \sum_{i=1}^D d_i \pmod{99}$.
For example Let $n = 363957 = (36 \times 100^2) + (39 \times 100) + (57)$.
Then \begin{align} n \pmod{99} &\equiv 36 + 39 + 57 \pmod{99} \\ &\equiv 132 \pmod{99} \\ &\equiv 1 + 32 \pmod{99} \\ &\equiv 33 \pmod{99} \end{align}
It follows that $n \pmod{11} = 33 \pmod{11} = 0$. Hence $11 \mid 363957$.
Here is a not-quite-right hand-waving answer that might be worth thinking aboiut.
If a number is divisible by $11$, this long division problem ends up with no remainder (all zeros at the bottom). Look at the red numbers that were subtracted from a red digit $\color{red}\#$ in the original dividend: $\color{red}{a+b+c+d+e+f+g}$, and look at the blue numbers that were subtracted from a blue digit $\color{blue}\#$ in the original dividend: $\color{blue}{a+b+c+d+e+f+g}$
Now the hand-waving. Of course, this isn’t quite right, because there might have been some borrowing involved, but everywhere there was borrowing, it was to allow $10$ more to be subtracted from some digit position and $1$ less to be subtracted from the position to its left, which we’ll pretend explains why divisibility by $11$ means the difference in sums is either zero or (if there was borrowing) a multiple of $11=10+1$. | 2019-09-22T16:12:15 | {
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https://math.stackexchange.com/questions/2908549/monotonicity-of-the-function-1x-frac1x-left1-frac1x-rightx?noredirect=1 | # Monotonicity of the function $(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x$.
Let $f(x)=(1+x)^{\frac{1}{x}}\left(1+\frac{1}{x}\right)^x, 0<x\leq 1.$ Prove that $f$ is strictly increasing and $e<f(x)\leq 4.$
In order to study the Monotonicity of $f$, let $$g(x)=\log f(x)=\frac{1}{x}\log (1+x)+x\log \left(1+\frac{1}{x}\right).$$ And $f$ and $g$ has the same Monotonicity. By computation, $$g'(x)=\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}.$$ As we know $\frac{x}{1+x}-\log (1+x)\leq 0$ and $\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}\geq 0$. So it does not determine the sign of $g'(x)$. If we compute the second derivative $g''(x)$, you will find it is also difficult to determine the sign of $g''(x)$. Our main goal is to prove $$\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}>0.$$
Is there some tricks to prove this result. Any help and hint will welcome.
• I believe that the solution to this problem is to use the symmetry of function $f(x)$. The symmetry is $f(x)=f(1/x)$. – MathOverview Sep 7 '18 at 12:11
• Just an observation: If you manage to at least show that $f'(x) \not= 0$ for all $x \in (0,1]$, then you are done because $\lim_{x \to 0^+} f(x) = e < 4 = f(1),$ so there is no other option for $f$ other than to be strictly increasing on $(0, 1]$. – MisterRiemann Sep 7 '18 at 12:15
• Is this related to math.stackexchange.com/q/2899031/42969, or is the resemblance coincidental? – Martin R Sep 7 '18 at 12:22
• @ Martin R： thank you for your hint of math.stackexchange.com/q/2899031/42969 Actually, the question I asked is confused me for several weeks, and I have seen math.stackexchange.com/q/2899031/42969 before, and then there is only one answers. And now I see there are several answers. And one answer use the result I have posted. It is resemblance coincidental. – Riemann Sep 7 '18 at 12:26
The right inequality.
We can use the TL method here.
We need to prove that $$(1+a)^b(1+b)^a\leq4$$ for $a>0$, $b>0$ such that $ab=1$, which is $$\frac{\ln(1+a)}{a}+\frac{\ln(1+b)}{b}\leq2\ln2.$$ But $$\sum_{cyc}\left(\ln2-\frac{\ln(1+a)}{a}\right)=\sum_{cyc}f(a),$$ where $f(a)=\ln2-\frac{\ln(1+a)}{a}-(\ln2-0.5)\ln a$.
Easy to show that $f(a)\geq0$ for all $0<a\leq11$.
But for $a>11$ we obtain: $$(1+a)^b(1+b)^a=(1+a)^{\frac{1}{a}}\left(1+\frac{1}{a}\right)^a<(1+11)^{\frac{1}{11}}e<4.$$
The left inequality.
We need to prove that: $$\frac{1}{x}\ln(1+x)+x\ln\left(1+\frac{1}{x}\right)>1$$ or $g(x)>0$, where $$g(x)=(1+x^2)\ln(1+x)-x^2\ln{x}-x.$$ Indeed, $$g'(x)=2x\ln(1+x)+\frac{1+x^2}{1+x}-2x\ln{x}-x-1=2x\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)>0$$ because $$\left(\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\right)'=-\frac{1}{x(1+x)^2}<0$$ and $$\ln\left(1+\frac{1}{1}\right)-\frac{1}{1+1}>0.$$ Id est, $$g(x)>\lim_{x\rightarrow0^+}g(x)=0$$ and we proved that $e<f(x)\leq4.$
Now, we'll prove that $f$ increases on $(0,1].$
By your work we need to prove that $$\frac{1}{x^2}\left(\frac{x}{1+x}-\ln(1+x)\right)+\ln\left(1+\frac{1}{x}\right)-\frac{1}{1+x}\geq0$$ for all $0<x\leq1$ or $$(x^2-1)\ln(1+x)-x^2\ln{x}+\frac{(1-x)x}{1+x}\geq0$$ and since $$\ln{x}\leq\frac{2(x-1)}{1+x},$$ it's enough to prove that $$-(1-x^2)\ln(1+x)+\frac{2x^2(1-x)}{1+x}+\frac{(1-x)x}{1+x}\geq0$$ or $$\ln(1+x)\leq\frac{x(2x+1)}{(1+x)^2},$$ which is smooth.
Using $$\dfrac{x}{x+1}<x\ln\left(1+\dfrac1x\right)<1$$ and for $x\to\dfrac1x$ $$\dfrac{1}{x+1}<\dfrac1x\ln\left(1+x\right)<1$$ then $$\dfrac1x\ln\left(1+x\right)+x\ln\left(1+\dfrac1x\right)>1$$
• You trick is very nice.Can you give some help for the fonotonicity of the function $f(x)$.Thank you ! – Riemann Sep 7 '18 at 14:37
Michael Rozenberg's answer has it all.
Here are two remaining proofs in Michael Rozenberg's answer:
1. Show: $\ln{x}\leq\frac{2(x-1)}{1+x}$
We have $\ln(1+y)\leq\frac{2y}{{2+y}}$ for $-1<y<0$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities), i.e. $\ln(x)\leq\frac{2(x-1)}{{1+x}}$ for $0<x<1$. This is exactly what needs to be shown.
1. Show: $\ln(1+x)\leq\frac{x(2x+1)}{(1+x)^2}$
We have $\ln(1+x)\leq\frac{x}{\sqrt{1+x}}$ (see e.g. in https://en.wikipedia.org/wiki/List_of_logarithmic_identities#Inequalities), so we may prove
$\frac{1}{\sqrt{1+x}}\leq\frac{2x+1}{(1+x)^2}$ or $(1+x)^3-{(1+2x)^2}\leq 0$ or $x^2 - x - 1<0$ or $x( 1-x) + 1>0$ which is true since $1-x \ge 0$.
• Sorry @Andreas , I have made a mistake. $g'(x)$ should be $g'(x)=\frac{1}{x^2}\left(\frac{x}{1+x}-\log (1+x)\right)+\log \left(1+\frac{1}{x}\right)-\frac{1}{1+x}.$ – Riemann Sep 7 '18 at 13:11 | 2019-05-23T03:51:13 | {
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https://pythonnumericalmethods.berkeley.edu/notebooks/chapter17.04-Lagrange-Polynomial-Interpolation.html | This notebook contains an excerpt from the Python Programming and Numerical Methods - A Guide for Engineers and Scientists, the content is also available at Berkeley Python Numerical Methods.
The copyright of the book belongs to Elsevier. We also have this interactive book online for a better learning experience. The code is released under the MIT license. If you find this content useful, please consider supporting the work on Elsevier or Amazon!
# Lagrange Polynomial Interpolation¶
Rather than finding cubic polynomials between subsequent pairs of data points, Lagrange polynomial interpolation finds a single polynomial that goes through all the data points. This polynomial is referred to as a Lagrange polynomial, $$L(x)$$, and as an interpolation function, it should have the property $$L(x_i) = y_i$$ for every point in the data set. For computing Lagrange polynomials, it is useful to write them as a linear combination of Lagrange basis polynomials, $$P_i(x)$$, where $$$P_i(x) = \prod_{j = 1, j\ne i}^n\frac{x - x_j}{x_i - x_j},$$$
and $$$L(x) = \sum_{i = 1}^n y_i P_i(x).$$$
Here, $$\prod$$ means “the product of” or “multiply out.”
You will notice that by construction, $$P_i(x)$$ has the property that $$P_i(x_j) = 1$$ when $$i = j$$ and $$P_i(x_j) = 0$$ when $$i \ne j$$. Since $$L(x)$$ is a sum of these polynomials, you can observe that $$L(x_i) = y_i$$ for every point, exactly as desired.
TRY IT! Find the Lagrange basis polynomials for the data set x = [0, 1, 2] and y = [1, 3, 2]. Plot each polynomial and verify the property that $$P_i(x_j) = 1$$ when $$i = j$$ and $$P_i(x_j) = 0$$ when $$i \ne j$$.
$\begin{eqnarray*} P_1(x) &=& \frac{(x - x_2)(x - x_3)}{(x_1-x_2)(x_1-x_3)} = \frac{(x - 1)(x - 2)}{(0-1)(0-2)} = \frac{1}{2}(x^2 - 3x + 2),\\ P_2(x) &=& \frac{(x - x_1)(x - x_3)}{(x_2-x_1)(x_2-x_3)} = \frac{(x - 0)(x - 2)}{(1-0)(1-2)} = -x^2 + 2x,\\ P_3(x) &=& \frac{(x - x_1)(x - x_2)}{(x_3-x_1)(x_3-x_2)} = \frac{(x - 0)(x - 1)}{(2-0)(2-1)} = \frac{1}{2}(x^2 - x). \end{eqnarray*}$
import numpy as np
import numpy.polynomial.polynomial as poly
import matplotlib.pyplot as plt
plt.style.use('seaborn-poster')
x = [0, 1, 2]
y = [1, 3, 2]
P1_coeff = [1,-1.5,.5]
P2_coeff = [0, 2,-1]
P3_coeff = [0,-.5,.5]
# get the polynomial function
P1 = poly.Polynomial(P1_coeff)
P2 = poly.Polynomial(P2_coeff)
P3 = poly.Polynomial(P3_coeff)
x_new = np.arange(-1.0, 3.1, 0.1)
fig = plt.figure(figsize = (10,8))
plt.plot(x_new, P1(x_new), 'b', label = 'P1')
plt.plot(x_new, P2(x_new), 'r', label = 'P2')
plt.plot(x_new, P3(x_new), 'g', label = 'P3')
plt.plot(x, np.ones(len(x)), 'ko', x, np.zeros(len(x)), 'ko')
plt.title('Lagrange Basis Polynomials')
plt.xlabel('x')
plt.ylabel('y')
plt.grid()
plt.legend()
plt.show()
TRY IT! For the previous example, compute and plot the Lagrange polynomial and verify that it goes through each of the data points.
L = P1 + 3*P2 + 2*P3
fig = plt.figure(figsize = (10,8))
plt.plot(x_new, L(x_new), 'b', x, y, 'ro')
plt.title('Lagrange Polynomial')
plt.grid()
plt.xlabel('x')
plt.ylabel('y')
plt.show()
WARNING! Lagrange interpolation polynomials are defined outside the area of interpolation, that is outside of the interval $$[x_1,x_n]$$, will grow very fast and unbounded outside this region. This is not a desirable feature because in general, this is not the behavior of the underlying data. Thus, a Lagrange interpolation should never be used to interpolate outside this region.
## Using lagrange from scipy¶
Instead of we calculate everything from scratch, in scipy, we can use the lagrange function directly to interpolate the data. Let’s see the above example.
from scipy.interpolate import lagrange
f = lagrange(x, y)
fig = plt.figure(figsize = (10,8))
plt.plot(x_new, f(x_new), 'b', x, y, 'ro')
plt.title('Lagrange Polynomial')
plt.grid()
plt.xlabel('x')
plt.ylabel('y')
plt.show() | 2021-09-21T04:54:15 | {
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http://kevbase.com/site/posts/floating-point-fun.html | # Floating Point Fun¶
Computers must represent real numbers with some finite precision (ignoring symbolic algebra packages), and sometimes that precision limit ends up causing problems that you might not expect. Here are a couple examples.
## exp and expm1¶
The function $$f(x) = \exp(x) - 1$$ is kind of fun – by subtracting one from the exponential, it removes the only constant term in the Maclaurin series of $$\exp(x)$$:
$\exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \, ...$
And so by subtracting off one, only terms that depend on $$x$$ are left:
$\exp(x) - 1 = \sum_{n=1}^\infty \frac{x^n}{n!} = x + \frac{x^2}{2} + \frac{x^3}{6} + \, ...$
Notice that, for positive $$x \ll 1$$, the series can be approximated with just $$\exp(x) - 1 \approx x$$, with the approximation improving as $$x$$ approaches zero. A similar argument lets us approximate $$\exp(x)$$ as $$1+x$$, again for positive $$x \ll 1$$.
Using floating point math to perform such calculations is tricky because of the finite amount of precision available to represent real numbers. For example evaluating $$\exp(2^{-100})$$ should return some number close to $$1 + 2^{-100}$$, meaning we’d need 100 bits of significand precision to be able to distinguish the result from 1. The standard floating point type used by most computers is the double-precision float, which uses 64 bits, of which 52 are used for significand precision. This is easy to demonstrate:
In [1]: 1 + 2**-52 > 1
Out[1]: True
In [2]: 1 + 2**-53 > 1
Out[2]: False
This precision limit means that, although the true value of $$\exp(2^{-53}) - 1$$ requires only a few bits of significand precision to represent, we must take care to not exceed our precision limits in the intermediate calculations. So calculating $$\exp(2^{-53})$$ first and then subtracting off 1 is not going to work because by “wasting” all of our precision storing the 1 in $$1 + 2^{-53}$$, we’re losing digits farther down – they get rounded off. In code:
In [3]: import numpy as np
In [4]: np.exp(2**-53) - 1
Out[4]: 0.0
In [5]: np.exp(2**-52) - 1
Out[5]: 2.220446049250313e-16
Luckily some smart folks have worked around this and provided us with functions like np.expm1 that are cleverly written to avoid wasting precision in the intermediate calculations. So let’s see how it performs against the naive method, trying out two floating point formats: the double-precision format described above and the single-precision format that uses only 32 bits (23 used for significand). For reference, these bit widths give us about 16 and 7 digits of accuracy ($$2^{-52}\approx 2 \times 10^{-16}$$; $$2^{-23}\approx 1 \times 10^{-7}$$). Recall that, for small $$x$$, the true value of $$\exp(x) - 1$$ should be close to $$x$$, meaning the values should follow the $$1:1$$ line for $$x \ll 10^{0}$$.
In [6]: import matplotlib.pyplot as plt
In [7]: fig, axes = plt.subplots(2, 2, sharex=True, sharey=True)
In [8]: for j, dtype in enumerate([np.float32, np.float64]):
...: x = np.logspace(-50, 1, num=500, dtype=dtype)
...: cases = {
...: 'f(x) = np.exp(x) - 1': np.exp(x) - 1,
...: 'f(x) = np.expm1(x)': np.expm1(x),
...: }
...: for i, (label, y) in enumerate(cases.items()):
...: axes[j, i].loglog(x, y, label=label)
...: axes[j, i].legend(loc='upper left')
...:
In [9]: axes[0, 0].set_ylabel('np.float32'); axes[1, 0].set_ylabel('np.float64');
In [10]: plt.show()
So the naive method fails once $$x$$ crosses the datatype precision limit, but the mathemagics baked into np.expm1 return the right answer across the full domain.
## aoi out of bounds¶
Round-off error showed up recently in a pvlib issue. The angle of incidence (AOI) projection is a measure of how well-aligned a solar panel is with the sun’s position in the sky and is calculated with:
In [11]: def aoi_projection(surface_tilt, surface_azimuth, solar_zenith, solar_azimuth):
....: return (
....: np.cos(surface_tilt) * np.cos(solar_zenith) +
....: np.sin(surface_tilt) * np.sin(solar_zenith) *
....: np.cos(solar_azimuth - surface_azimuth)
....: )
....:
Mathematically this is the dot product between the solar position unit vector and the solar panel normal, and as such is bounded in the interval $$[-1, +1]$$.
Consider the case where the panel is perfectly aligned with the sun, i.e. surface_tilt==solar_zenith and surface_azimuth==solar_azimuth. Then the calculation, barring any precision issues, should return exactly 1. However, for certain input values, round-off in the intermediate steps end up returning “impossible” values slightly greater than 1:
In [12]: zenith = np.radians(89.26778228223463)
Because the typical next step is to calculate the angle of incidence itself with $$\theta_{aoi} = \cos^{-1}(\textrm{projection})$$, that very small error is actually kind of a big deal because it pushes projection out of the domain of $$\cos^{-1}$$, meaning any downstream calculations get polluted with NaN:
In [15]: np.arccos(aoi_projection(zenith, azimuth, zenith, azimuth)) | 2021-12-03T15:26:38 | {
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https://en.khanacademy.org/math/ap-calculus-bc/bc-integration-new/bc-6-11/a/integration-by-parts-review | If you're seeing this message, it means we're having trouble loading external resources on our website.
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## AP®︎/College Calculus BC
### Course: AP®︎/College Calculus BC>Unit 6
Lesson 13: Using integration by parts
# Integration by parts review
AP.CALC:
FUN‑6 (EU)
,
FUN‑6.E (LO)
,
FUN‑6.E.1 (EK)
Review your integration by parts skills.
## What is integration by parts?
Integration by parts is a method to find integrals of products:
integral, u, left parenthesis, x, right parenthesis, v, prime, left parenthesis, x, right parenthesis, d, x, equals, u, left parenthesis, x, right parenthesis, v, left parenthesis, x, right parenthesis, minus, integral, u, prime, left parenthesis, x, right parenthesis, v, left parenthesis, x, right parenthesis, d, x
or more compactly:
integral, u, space, d, v, equals, u, v, minus, integral, v, space, d, u
We can use this method, which can be considered as the "reverse product rule," by considering one of the two factors as the derivative of another function.
## Practice set 1: Integration by parts of indefinite integrals
Let's find, for example, the indefinite integral integral, x, cosine, x, d, x. To do that, we let u, equals, x and d, v, equals, cosine, left parenthesis, x, right parenthesis, d, x:
integral, x, cosine, left parenthesis, x, right parenthesis, d, x, equals, integral, u, d, v
u, equals, x means that d, u, equals, d, x.
d, v, equals, cosine, left parenthesis, x, right parenthesis, d, x means that v, equals, sine, left parenthesis, x, right parenthesis.
Now we integrate by parts!
\begin{aligned} \displaystyle\int x\cos(x)\,dx &=\displaystyle\int u\,dv \\\\ &=uv-\displaystyle\int v\,du \\\\ &=\displaystyle x\sin(x)-\int\sin(x)\,dx \\\\ &=x\sin(x)+\cos(x)+C \end{aligned}
Problem 1.1
• Current
integral, x, e, start superscript, 5, x, end superscript, d, x, equals, question mark
Want to try more problems like this? Check out this exercise.
## Practice set 2: Integration by parts of definite integrals
Let's find, for example, the definite integral integral, start subscript, 0, end subscript, start superscript, 5, end superscript, x, e, start superscript, minus, x, end superscript, d, x. To do that, we let u, equals, x and d, v, equals, e, start superscript, minus, x, end superscript, d, x:
u, equals, x means that d, u, equals, d, x.
d, v, equals, e, start superscript, minus, x, end superscript, d, x means that v, equals, minus, e, start superscript, minus, x, end superscript.
Now we integrate by parts:
\begin{aligned} &\phantom{=}\displaystyle\int_0^5 xe^{-x}\,dx \\\\ &=\displaystyle\int_0^5 u\,dv \\\\ &=\Big[uv\Big]_0^5-\displaystyle\int_0^5 v\,du \\\\ &=\displaystyle\Big[ -xe^{-x}\Big]_0^5-\int_0^5-e^{-x}\,dx \\\\ &=\Big[-xe^{-x}-e^{-x}\Big]_0^5 \\\\ &=\Big[-e^{-x}(x+1)\Big]_0^5 \\\\ &=-e^{-5}(6)+e^0(1) \\\\ &=-6e^{-5}+1 \end{aligned}
Problem 2.1
• Current
integral, start subscript, 1, end subscript, start superscript, e, end superscript, x, cubed, natural log, x, space, d, x, equals, question mark
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• in the int (0 -> pi) of xsin(2x)dx problem, in the solution, the third to last line, shouldn't that be (sin(2x)/4) not (sin(4x)/4)? or am I missing something?
• You are correct, it is a typo, though it does not effect the result since sin(nπ) = 0 for all integers n.
You can, and should, report the error since you found it.
• Like Bhoovesh I am also fuzzy about the compact notation. It seems that the confusion is not with Leibniz notation vs Newton's, but rather I am concerned about falling in a pit as a consequence of having only one letter in an expression for which I am accustomed to two. The dropping of the x's and dx's makes me nervous about getting fouled up as a consequence of x not being the only variable. Where is y, and how does one keep track of it with the more compact notation? I would like to know the conventions and rules for this.
• Why hasn't Sal explained about the compact form of Integration by parts??i don't understand it!! It contradicts to what Sal said about differentials earlier that the differentials are not numbers or function which can't cancelled or algebraically manipulated!!
• The "compact form" is just a different way to write the form used in the videos. Basically, the only difference is that the "video form" uses prime notation (f'(x)), and the "compact form" uses Leibniz notation (dy/dx). If you are used to the prime notation form for integration by parts, a good way to learn Leibniz form is to set up the problem in the prime form, then do the substitutions f(x) = u, g'(x)dx = dv, f'(x) = v, g(x)dx = du. At least, that's how it clicked for me.
As far as the manipulating differentials goes, it's true that you can't just treat differentials like they are normal terms in an equation (as if dx were the variable d times the variable x), but it is legal to split up the dy/dx when differentiating both sides of an equation. The concept here is exactly the same as what is used when doing u-substitution (URL to video below if you need it).
Hope this helps, and good luck with your work!
• What is the use of integration? When do we use it in our practical lives?
(1 vote)
• we can use integration to find Displacement from Velocity, and Velocity from Acceleration, Voltage across a Capacitor, Moments of Inertia by Integration and many more...
• Why does the integral of e^5x dx = 1/5 e^5x? Is it an application of the reverse chain rule?
Thanks very much!
• That's one way of thinking about it, yes. As you continue on in math, this will become almost second-nature and you won't even think about the chain rule when integrating simple exponential functions.
• How would you integrate something in parentheses, like (x^2 +1)^1/2?
• That depends hugely on what's in the parentheses. √(x²+1) can be done by trigonometric substitution, but √(x³+1) cannot be done by elementary means.
• Is there a reverse division rule that can sometimes serve as a substitute for this? An example where this would be useful is (ln(5x))/(x^2) | 2023-03-25T16:57:58 | {
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https://math.stackexchange.com/questions/2419165/convergence-of-recursive-sequence-a-n1-frac-1k-lefta-n-frack | # Convergence of recursive sequence $a_{n+1} =\frac{ 1}{k} \left(a_{n} + \frac{k}{a_{n}}\right)$
Let $$a_{n+1} = \frac{1}{k} \left(a_{n} + \frac{k}{a_n}\right) ; k>1, a_1>0$$ The problem is to show that it converges.
Attempt: The sequence is not monotone but it has a lower bound. It seems that odd terms subsequence and even term subsequence are monotonic sequences (I wrote some basic code to make this observation) though I am not able to prove it analytically. I also know that if odd subsequence and even subsequence converge to same limit then the sequence also converges. So, that tells me that I am on the right track.
• is this $$a_{k+1}=\frac{1}{2}\left(a_k+\frac{k}{a_k}\right)$$? – Dr. Sonnhard Graubner Sep 6 '17 at 16:07
• No. See the pic from the book. It's problem 7 imgur.com/zujzuYA – Rahul Sep 6 '17 at 16:14
• $k$ is supposed to be an integer ? – Gabriel Romon Sep 6 '17 at 16:22
• @LeGrandDODOM no. But I don't think it matters. If 0<k<1 then the sequence is monotone. So that's why it assumes k to be greater than 1 – Rahul Sep 6 '17 at 16:49
• @Rahul: Can you tell us which book (author, title, edition)? – quasi Sep 6 '17 at 19:10
If it converges, one can solve immediately for $a_\infty = \frac{\sqrt k}{\sqrt{k-1}}$.
So define $b_n = a_n \frac{\sqrt{k-1}}{\sqrt k}$. This gives the recursion
$b_{n+1} = \frac{1}{k} \left(b_{n} + \frac{{k-1}}{b_{n}}\right)$
which, if convergent, gives $b_\infty = 1$.
We will show that a lower bound and an upper bound both converge to $b_\infty = 1$ from an arbitrary starting point. Hence $b_\infty = 1$ is a unique fixed point and convergence is global.
Let's begin with a lower bound.
Using $1/x = 1/(1 + x-1) \ge 1 - (x-1) = 2 -x$ we have
$b_{n+1} = \frac{1}{k} \left(b_{n} + \frac{{k-1}}{b_{n}}\right) \ge \frac{1}{k} \left(b_{n} + {{(k-1)}}{(2-b_{n})}\right) = \frac{1}{k} \left(b_{n} (2-k) + {(2k-2)}\right)$
and the RHS establishes a lower bound $b^-_n \le b_n$ given by the recursion $b^-_{n+1} = \frac{1}{k} \left(b^-_{n} (2-k) + {(2k-2)}\right)$. We can inspect this recursion to see if and where the lower bound converges to.
According to Banach's contraction theorem for iterating fixed points in the form $x_{n+1} = f(x_n)$ convergence is guaranteed if the absolute value of the derivative is less than 1, i.e. $|f'(x)| < 1$, for some interval of $x$ containing the fixed point. Here, for $b^-_n$ in particular, $f(x_n)$ is a line equation and convergence is global if the absolute slope is less than 1.
Since the absolute slope $|2-k|/k$ is always $< 1$ for $k > 1$, iterating the lower bound $b^-_n$ converges, and solving $b^- = \frac{1}{k} \left(b^- (2-k) + {(2k-2)}\right)$ results in the lower bound limit $b^-_\infty= 1$. So the lower bound converges to the actual value that the original series should attain (if convergent), which proves convergence from below.
Now for the upper bound.
Suppose $b_1 <1$. Then $b_2 >1$. So let us suppose w.l.o.g. that we start with some $b_1 >1$.
Then we have that
$b_{n+1} = \frac{1}{k} \left(b_{n} + \frac{{k-1}}{b_{n}}\right) < \frac{1}{k} \left(b_{n} +{{k-1}}\right)$
provided that $b_{n} > 1$. This is certainly true for $n=1$, and we would like to keep it that way.
So we have an upper bound $b^+_n$ given by the recursion
$b^+_{n+1} = \frac{1}{k} \left(b^+_{n} + {{k-1}}\right)$.
Clearly, if $b^+_{n} > 1$, so is $b^+_{n+1} > 1$, so our condition $b^+_{n} > 1$ holds for all $n$, starting with $b^+_{1} > 1$, and we can work with that upper bound.
Now apply again the contraction theorem. Convergence is globally established since the slope is $\frac{1}{k} <1$, and the fixed point can immediately by computed as $b^+_\infty= 1$.
Hence upper and lower bound converge globally to the same value $b_\infty= 1$, which establishes that the original series also converges, from any starting point, to $b_\infty= 1$. $\quad \Box$
• Can you explain "the RHS establishes a lower bound"? Also, what is the definition of the symbol $b^-_n$? Note: I follow the logic up to that sentence (so I have no issue with the inequality established on the previous line). And I have no problem with "the absolute slope $|2-k|/k$ is always $<1$". – quasi Sep 6 '17 at 23:30
• @quasi I explained it in the main text. – Andreas Sep 7 '17 at 14:27
• Shouldn't the relevant function $f$ be $$f(x)=\frac{1}{k}\left(x + \frac{{k-1}}{x}\right)$$ ?? It has the unique fixed point $x=1$. – quasi Sep 7 '17 at 15:07
• @quasi IF it converges to the fixed point, the fixed point will be 1. But you must prove that iterations converge. I couldn't do that. What I proved is that a lower bound converges to the same fixed point. What remains to be done is to prove that an upper bound converges as well to the same fixed point. – Andreas Sep 7 '17 at 16:09
• Thanks @Andreas. – Rahul Sep 8 '17 at 13:37 | 2019-08-19T05:32:35 | {
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http://mathhelpforum.com/calculus/39092-urgent-differential-equation-modelling.html | # Math Help - Urgent Differential Equation Modelling
1. ## Urgent Differential Equation Modelling
Hi everyone,
I have this problem that I really need somehelp with.
the rate of spread of a rumour is proportional to the product of the fraction y of the population L that has heard the rumour and the fraction who have not.
(1) Find the differential equation that is satisfied by y.
dy/dt = k * [y(L-y) / (L^2)]
(2) Show that the solution with initial condition y(0) = y0 is given by
y = (Ly0e^(kLt)) / ((L-y0)+y0e^(kLt))
I am lost on this.
(3) A school has 100 students. At 9am, 20 students have heard a rumour. At 11am 50 students have heard the rumour. At what time will 80 students will have heard the rumour.
I dont know what to do here either, however, im guessing these are the initial conditions. i.e.
y(0) = 20
y(2) = 50
2. Originally Posted by arguabsysi
Hi everyone,
I have this problem that I really need somehelp with.
the rate of spread of a rumour is proportional to the product of the fraction y of the population L that has heard the rumour and the fraction who have not.
(1) Find the differential equation that is satisfied by y.
dy/dt = k * [y(L-y) / (L^2)]
(2) Show that the solution with initial condition y(0) = y0 is given by
y = (Ly0e^(kLt)) / ((L-y0)+y0e^(kLt))
I am lost on this.
(3) A school has 100 students. At 9am, 20 students have heard a rumour. At 11am 50 students have heard the rumour. At what time will 80 students will have heard the rumour.
I dont know what to do here either, however, im guessing these are the initial conditions. i.e.
y(0) = 20
y(2) = 50
I tried this differential equation out, and it worked:
$\frac{dy}{dt}=ky(L-y)$
This is not separable. However, if I write it as follows:
$\frac{dy}{dt}=kLy-ky^2 \implies \color{red}\boxed{\frac{dy}{dt}-kLy=-ky^2}$.
This equation has the form of a Bernoulli's Equation (I cover this in my Differential Equations Tutorial):
$\frac{dy}{dt}+P(t)y=Q(t)y^n$ (general form of Bernoulli Equation)
If we make a substitution $v=y^{1-n}$ the Bernoulli Eqn will become a linear equation we know how to solve.
$v=y^{1-2}=y^{-1} \implies y=v^{-1}$
We also need a substitution for $\frac{dy}{dt}$.
We can get this by applying a definition of the chain rule: $\frac{dy}{dt}=\frac{dy}{dv}\cdot \frac{dv}{dt} \implies \frac{dy}{dt}=-\frac{1}{v^2}\frac{dv}{dt}$.
Substituting this into the DE, we get:
$-\frac{1}{v^2}\frac{dv}{dt}-kL\frac{1}{v}=-k\frac{1}{v^2}$
Multiply both sides by $-v^2$ to get a linear DE:
$\frac{dv}{dt}+kLv=k$
Apply Integrating factor:
$\rho(t)=e^{\int P(t)\,dt}=e^{\int kL \,dt}=e^{kLt}$
It should be known by now: when we multiply through by the int. factor, we get on the left side of the equation the derivative of $\rho(t)\cdot v$
Thus, our DE becomes:
$\left[e^{kLt}v\right]^{/}=ke^{kLt}$
Solving the DE, we get:
$e^{kLt}v=\frac{1}{L}e^{kLt}+C \implies v=\frac{1}{L}+Ce^{-kLt}$
We don't want v. We want y. So sub back in $v=y^{-1}$
$\therefore \frac{1}{y}=\frac{1}{L}+Ce^{-kLt}\implies y=\frac{1}{\frac{1}{L}+Ce^{-kLt}}=\frac{e^{kLt}}{\frac{e^{kLt}}{L}+C}$
When $y(0)=y_0$, we have:
$y_0=\frac{1}{\frac{1}{L}+C}\implies y_0\left(\frac{1}{L}+C\right)=1\implies \frac{1}{L}+C=\frac{1}{y_0} \implies C=\frac{1}{L}-\frac{1}{y_0}=\frac{L-y_0}{Ly_0}$
$\therefore y=\frac{e^{kLt}}{\frac{e^{kLt}}{L}+\frac{L-y_0}{Ly_0}}\implies y=\frac{e^{kLt}}{\frac{y_0e^{kLt}+(L-y_0)}{Ly_0}} \implies \color{red}\boxed{y=\frac{Ly_0e^{kLt}}{(L-y_0)+y_0e^{kLt}}}$
I believe in this case $y(0)=y_0=20$ and $L=100$
Thus,
$y=\frac{2000e^{100kt}}{(80)+20e^{100kt}}$
Since y(2)=50, we see that
$50=\frac{2000e^{200k}}{(80)+20e^{200k}}$
My calculator tells me that $k=\frac{\ln(2)}{100}$
Thus, $y=\frac{2000e^{\ln(2)t}}{(80)+20e^{\ln(2)t}}$.
Now find t when y(t)=80
$80=\frac{2000e^{\ln(2)t}}{(80)+20e^{\ln(2)t}} \implies \color{red}\boxed{t=4}$.
So at 1 PM, 80 students will have heard the rumor.
(I hope my math was right! It does make sense!)
3. Originally Posted by Chris L T521
This is not separable. However, if I write it as follows:
$\frac{dy}{dt}=kLy-ky^2 \implies \color{red}\boxed{\frac{dy}{dt}-kLy=-ky^2}$.
I hate to say this but it is separable...
$
\frac{dy}{dt}=kLy-ky^2 \iff \frac{dy}{dt}=k(L-y)y
$
$\frac{dy}{(L-y)y}=kdx$
You can use partial fractions on the left hand side or my favorite trick
$\frac{1}{(L-y)y}=\frac{1}{L}\cdot\frac{L}{(L-y)y}=\frac{1}{L}\cdot\frac{(L-y)+y}{(L-y)y}=\frac{1}{L}\left( \frac{(L-y)}{(L-y)y}+\frac{y}{(L-y)y}\right)=$
$\frac{1}{L}\frac{1}{y}+\frac{1}{L}\frac{1}{(L-y)}$
Rock on!!!
4. Originally Posted by TheEmptySet
I hate to say this but it is separable...
$
\frac{dy}{dt}=kLy-ky^2 \iff \frac{dy}{dt}=k(L-y)y
$
$\frac{dy}{(L-y)y}=kdx$
You can use partial fractions on the left hand side or my favorite trick
$\frac{1}{(L-y)y}=\frac{1}{L}\cdot\frac{L}{(L-y)y}=\frac{1}{L}\cdot\frac{(L-y)+y}{(L-y)y}=\frac{1}{L}\left( \frac{(L-y)}{(L-y)y}+\frac{y}{(L-y)y}\right)=$
$\frac{1}{L}\frac{1}{y}+\frac{1}{L}\frac{1}{(L-y)}$
Rock on!!!
I keep forgetting that...
Thanks for catching that!
Both ways are acceptable, but I think your way is easier. | 2014-03-11T20:20:33 | {
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https://math.stackexchange.com/questions/2440511/find-the-distribution-of-z-x-y-where-x-and-y-are-uniformly-distributed | # Find the distribution of $Z = X + Y$ where $X$ and $Y$ are uniformly distributed random variables
Problem:
Let $X$ and $Y$ be independent r.v.'s each uniformly distributed over $(0,1)$. Let $Z = X + Y$. Find the CDF of $Z$.
I want to find $P(z <= z_0)$. There are two non-trivial cases to consider. The first is when $0 <= z_0 <= 1$ and the second is when $1< z_0 < 2$.
case 1: \begin{eqnarray*} P(z \leq z_0) &=& \int_0^{z_0} \int_{0}^{z_0 - x} dy \,\, dx = \int_0^{z_0} = z_0 - x \, dx \\ P(z \leq z_0) &=& z_0 x - \frac{x^2}{2} \Big|_{0}^{z_0} = {z_0}^2 - \frac{ {z_0}^2 }{2} \\ P(z \leq z_0) &=& \frac{ {z_0}^2 }{2} \\ F(z) &=& \frac{ {z}^2 }{2} \\ f(z) &=& z \\ \end{eqnarray*}
case 2: \begin{eqnarray*} P(z \leq z_0) + P(z \geq z_0) &=& 1 \\ P(z \leq z_0) &=& 1 - P(z \geq z_0) \\ \end{eqnarray*} Now to find $P(z \geq z_0)$ we consider the triangle, whose points are $(z_0-1,1)$, $(1,1)$ and $(1,z_0-1)$. \begin{eqnarray*} P(z \geq z_0) &=& \int_{z_0-1}^{1} \int_{z_0 - x}^{1} \,\, dy \, \, dx = \int_{z_0-1}^{1} 1 - z_0 + x \, \, dx \\ P(z \geq z_0) &=& (1 - z_0)x + \frac{x^2}{2} \Big|_{z_0 - 1}^{1} \\ P(z \geq z_0) &=& 1 - z_0 + \frac{1}{2} - (1-z_0)(z_0 - 1) - \frac{(z_0 - 1)^2}{2} \\ P(z \geq z_0) &=& 1 - z_0 + \frac{1}{2} + (1-z_0)^2 - \frac{(z_0 - 1)^2}{2}\\ P(z \geq z_0) &=& 1 - z_0 + \frac{1}{2} + \frac{(z_0 - 1)^2}{2} \\ P(z \geq z_0) &=& -z_0 + \frac{3}{2} + \frac{z_0^2 - 2z_0 + 1}{2} \\ P(z \geq z_0) &=& \frac{z_0^2}{2} - 2z_0 + 2 \\ F(z) &=& 1 - ( \frac{z_0^2}{2} - 2z_0 + 2 ) = 1 - \frac{z_0^2}{2} + 2z_0 - 2 \\ F(z) &=& -\frac{z_0^2}{2} + 2z_0 - 1 \\ \end{eqnarray*} For case 2, the book gets: $F_z(z) = 1 - \frac{(2-z)^2}{2}$ but algebra will show that they are the same.
How to find the CDF of the sum of independent uniformly distributed random variables?
However, that is done using the convolution. I would like to do it without the convolution which is what I have done. I have a feeling that I just made a stupid mistake but I am not sure.
Thanks
Bob
• In part 2 the triangle is a right triangle with length of the legs $h=|2-z_0|$, so instead of doing those cumbersome and error integrals you could just conclude that its area is $h^2/2$, wouldn't that be easier? – flawr Sep 22 '17 at 14:59
• @flawr Your method is easier in this case, but it does not work in the general case. I would like to be able to do this problem using a double integral for educational purposes.. – Bob Sep 22 '17 at 17:47
• This mayyyy have been done on math SE beforeeeeee – wolfies Sep 22 '17 at 20:14
You're almost there. You are just integrating over the wrong area a bit.
You've set $x$ to be between $z_0-1$ to $y$, which is fine. Now, for $y$ you must consider the line between $(z_0-1,1)$ and $(1, z_0-1)$. You have that almost right, but if you look at it once again, you can see it is $y=-x+z_0$.
So you should be integrating the inner-integral from $z_0-x$ to $1$ instead.
Edit: Regarding your final editing, you already have it right. Open up $(2-z_0)^2/2$ and look back at your result.
Ok I think I found your mistake: What you're doing here:
$$P(z \geq z_0) = \int_{z_0-1}^{1} \int_{z_0-1}^{1} \,\, dy \, \, dx$$
is not integrating over the triangle you mentioned, but over the rectangle $(z_0-1,1),(1,1),(1,z_0-1),(z_0-1,z_0-1)$.
• I think you are right. Can you please tell me what the correct bounds on the double integral is? – Bob Sep 22 '17 at 17:54
• I recommend drawinga picture and finding functions $l(x)$ and $u(x)$ such that the triangle is bounded (in $y$-direction) by $l(x)$ and $u(x)$. Then the inner integral transforms to $\int_{l(x)}^{u(x)}dy$. – flawr Sep 22 '17 at 18:16
• Is this right? $P(z \geq z_0) = \int_{z_0-1}^{1} \int_{x - z_0 + 1}^{1} \,\, dy \, \, dx$ – Bob Sep 22 '17 at 18:56
• Yes that looks great! – flawr Sep 22 '17 at 19:09 | 2019-12-14T16:10:24 | {
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http://mathhelpforum.com/geometry/3550-need-help.html | # Math Help - need help.
1. ## need help.
A long rope is pulled out between two opposite shores of a lake. It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
2. Hello, Bobby!
A long rope is pulled out between two opposite shores of a lake.
It's pulled so tight that it's perfectly straight.
Because the earth is spherical most of the rope is under water.
The length of the portion of rope that is under water is 70 km long.
How many meters below the surface is the rope at its deepest point?
The earths radius is assumed to be 6370 km.
Code:
C
* * *
* :x *
* : 35 *
A * - - - - -+- - - - - * B
\ D: /
\ :R-x /
R \ : / R
\ : /
\:/
*
O
The center of the earth is $O.$ . $OA = OB = OC = R$ (radius of the earth).
The 70-km rope is $AB.$ .We see that: $AD = DB = 35.$
Let $x = CD$ be the distance the rope is underwater at its center.
Then $DO = R - x.$
From right triangle $ODB:\;\;DO^2 + DB^2\:=\:OB^2$
So we have: . $(R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0$
Quadratic Formula: . $x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}$
Since $R = 6370$, we have: . $x \;= \;6370 \pm \sqrt{6370^2 - 1225}$
Therefore: . $x\;=\;0.096154575\text{ km}\;\approx\;96.2\text{ meters.}$
3. Originally Posted by Soroban
Hello, Bobby!
Code:
C
* * *
* :x *
* : 35 *
A * - - - - -+- - - - - * B
\ D: /
\ :R-x /
R \ : / R
\ : /
\:/
*
O
The center of the earth is $O.$ . $OA = OB = OC = R$ (radius of the earth).
The 70-km rope is $AB.$ .We see that: $AD = DB = 35.$
Let $x = CD$ be the distance the rope is underwater at its center.
Then $DO = R - x.$
From right triangle $ODB:\;\;DO^2 + DB^2\:=\:OB^2$
So we have: . $(R - x)^2 + 35^2\:=\:R^2\quad\Rightarrow\quad x^2 - 2Rx + 1225\;=\;0$
Quadratic Formula: . $x\;=\;\frac{2R \pm\sqrt{4R^2 - 4900}}{2}\;=\;R \pm \sqrt{R^2 - 1225}$
Since $R = 6370$, we have: . $x \;= \;6370 \pm \sqrt{6370^2 - 1225}$
Therefore: . $x\;=\;0.096154575\text{ km}\;\approx\;96.2\text{ meters.}$
Hello Soroban.
We could do it in this way also.
Extend CO to meet the circle at E
Now,
AB and CE are two chords of a cicle intresecting at D
Hence, $AD.DB = CD.DE$
$35.35 = x(DO + OE)$
$1225 = x(2R - x)$
KeepSmiling
Malay
4. Very nice, Malay!
An excellent application of that intersecting-chords theorem
. . . which I obviously forgot. | 2015-09-03T03:08:26 | {
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http://math.stackexchange.com/questions/80822/solutions-to-linear-diophantine-equation-15x21y-261 | # Solutions to Linear Diophantine equation $15x+21y=261$
Question
How many positive solutions are there to $15x+21y=261$?
What I got so far
$\gcd(15,21) = 3$ and $3|261$
So we can divide through by the gcd and get:
$5x+7y=87$
And I'm not really sure where to go from this point. In particular, I need to know how to tell how many solutions there are.
-
Apply the Euclid-Wallis Algorithm, $$\begin{array}{r} &&1&2&2\\ \hline\\ 1&0&1&-2&5\\ 0&1&-1&3&-7\\ 21&15&6&3&0 \end{array}$$ The second to last column says that $3\cdot15+(-2)\cdot21=3$, multiplying this by $87$, yields $261\cdot15+(-174)\cdot21=261$. The last column says tha the general solution is $(261-7k)\cdot15+(-174+5k)\cdot21=261$. Using $k=35,36,37$, we get the $3$ non-negative solutions: \begin{align} 16\cdot15+1\cdot21&=261\\ 9\cdot15+6\cdot21&=261\\ 2\cdot15+11\cdot21&=261\\ \end{align}
-
Thanks for this. I notice in this example and in other examples that the number of solutions correspond to the gcd. Is this true in general or just a coincidence? – Arvin Nov 10 '11 at 12:31
It is just a coincidence. Note that $5x+7y=87$ also has $3$ solutions, yet the gcd is $1$. – robjohn Nov 10 '11 at 12:58
As you observed, we can reduce to the case where $a$ and $b$ are relatively prime. Given relatively prime positive integers $a$ and $b$, and a positive $c$, we want to find the number of solutions of $ax+by=c$ in positive integers $x$, $y$.
We will give an exact answer, and a much simpler to compute approximate answer which could be off by $1$. The argument that leads to these answers is given below. The answers themselves come after the argument.
There are integers $x_0,y_0$, such that $ax_0+by_0=c$. A solution $(x_0,y_0)$ can be found by using the Extended Euclidean Algorithm to solve the equation $au+bv=1$, and multiplying through by $c$. The details can be found in many places. We will instead concentrate on the consequences. Note that the $x_0$ and $y_0$ found through this process will not be both positive.
Once we have found one solution $(x_0,y_0)$, all integer solutions of the equation $ax+by=c$ have the shape $x=x_0+bt$, $y=y_0-at$, where $t$ ranges over the integers. To produce the positive solutions, we want to find all integers $t$ such that $x>0$ and $y>0$.
So we need $y_0-at >0$, $x_0+bt>0$, or equivalently $$-\frac{x_0}{b}<t <\frac{y_0}{a}.$$
Exact Answer: The number of positive solutions $(x,y)$ is the number of integers $t$ in the interval $-x_0/b<t <y_0/a$. This is easily determined once we know $x_0$ and $y_0$.
Approximate Answer: The interval $(-x_0/b,y_0/a)$ has width $y_0/a+x_0/b$. which simplifies to $(ax_0+by_0)/ab$, that is, $c/ab$.
If $\dfrac{c}{ab}$ is an integer, then the width of our open interval is an integer, and the number of integers in the interval is $\dfrac{c}{ab}-1$. If $\dfrac{c}{ab}$ is not an integer, then the number of integers in our open interval can be either $$\left\lfloor \dfrac{c}{ab}\right\rfloor \qquad \text{or}\qquad \left\lfloor \dfrac{c}{ab}\right\rfloor+ 1.$$
For a simple example that shows that $c/ab$ need not quite determine the number of positive solutions, note that $x+15y=23$ has one positive solution, while $3x+5y=23$ has two positive solutions.
If we give the answer $c/ab-1$ if $c/ab$ is an integer, and $\lfloor c/ab\rfloor$ otherwise, we are guaranteed that we are underestimating the true answer by at most $1$.
The advantage is that the approximate answer can be found very cheaply. In our exact answer, we used $x_0$ and $y_0$, so would need to compute these.
-
5, 7, and 87 are small enough numbers that you could just try all the possibilities. Can you see, for example, that $y$ can't be any bigger than 12?
-
I do see that, but I'd like to develop a more generalised method to doing these types of questions - in case I get asked bigger numbered questions in the future – Arvin Nov 10 '11 at 11:48
The general method involves using Euclid's algorithm, and is presented in every introductory number theory textbook. I'm sure it's also all over the web - search for "linear diophantine equations". – Gerry Myerson Nov 10 '11 at 11:55
Hint $\$ It's easy. $\rm\:x\ge 1\:\Rightarrow\: y = (87-5x)/7 \le \lfloor 82/7\rfloor = 11.\:$ Each of the intervals $[1,5], [6,10]$ contains precisely one solution for $\rm\:y,\:$ so it remains to check if $\rm\:y=11\:$ yields a solution.
Note that the (extended) Euclidean algorithm is not required, and this method works generally.
-
you find all solutions from $ax+by=c$ with
$$x=x_0-\frac{b}{(a,b)}*t , y=y_0+\frac{a}{(a,b)}*t$$
$x_0$ and $y_0$ you find with the Euclidean algorithm backwards:
$15x+21y=261$ with $261=3*87$
21=1*16+6
15=2*6+3
6=2*3+0
$3=15-2*6=15-2*(21-1*15)=15-2*21+2*15=3*15-2*21$
multiplied with 87 we get
$261=261*15-174*21$
So $x_0=261$ and $y_0=174$
-
Thanks, but what is $x_0$, $y_0$, $a$ and $b$ in this particular example? – Arvin Nov 10 '11 at 11:49
$x_0$ and $y_0$ is one solution of your example, $a$ is the number in front of $x$ and $b$ is the number in front of $y$ – ulead86 Nov 10 '11 at 11:50
What Daniel means is that if $ax_0+by_0=c$, that is, if $x_0,y_0$ is one solution of $ax+by=c$, then his formula gives you all the solutions, as you let $t$ run through the integers. But you only want the positive solutions, so you would also have to impose some restriction on $t$. Also, Daniel doesn't tell you how to find that first solution $x_0,y_0$. – Gerry Myerson Nov 10 '11 at 11:55
$15x+21y=261$. $\Leftrightarrow x= \dfrac{261-21y}{15}=17-y+ \dfrac{6(1-y)}{15}$
Because $x,y$ are integer numbers and $\gcd(y,1-y)=1$. Therefore $15|1-y$. Let $1-y=15k$ with $k \in \mathbb{Z}$ implies $y=1-15k$. Thus $x=15+21k$.
Now, to find $x,y$ positive number, we will try the value of $k$.
-
5x + 7y = 87 can be rewritten as 5x = 87(mod 7). Solving x = 2(mod 7) which means that x = 2 + 7t. Substituting this x back into the equation you get y = 11 - 5t.
The complete solution is x = 2 + 7t and y = 11 - 5t. So there are an infinite number of solutions.
- | 2015-04-28T18:39:35 | {
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https://www.johndcook.com/blog/2018/02/26/summing-random-powers-up-to-a-threshold/ | # Summing random powers up to a threshold
Pick random numbers uniformly between 0 and 1, adding them as you go, and stop when you get a result bigger than 1. How many numbers would you expect to add together on average?
You need at least two samples, and often two are enough, but you might get any number, and those larger numbers will pull the expected value up.
Here’s a simulation program in Python.
from random import random
from collections import Counter
N = 1000000
c = Counter()
for _ in range(N):
x = 0
steps = 0
while x < 1:
x += random()
steps += 1
c[steps] += 1
print( sum( k*c[k] for k in c.keys() )/N )
When I ran it I got 2.718921. There’s theoretical result first proved by W. Weissblum that the expected value is e = 2.71828… Our error was on the order of 1/√N, which is what we’d expect from the central limit theorem.
Now we can explore further in a couple directions. We could take a look at the distribution of the number steps, not just its expected value. Printing c shows us the raw data.
Counter({
2: 499786,
3: 333175,
4: 125300,
5: 33466,
6: 6856,
7: 1213,
8: 172,
9: 29,
10: 3
})
And here’s a plot.
We could also generalize the problem by taking powers of the random numbers. Here’s what we get when we use exponents 1 through 20.
There’s a theoretical result that the expected number of steps is asymptotically equal to cn where
I computed c = 1.2494. The plot above shows that the dependence on the exponent n does look linear. The simulation results appear to be higher than the asymptotic prediction by a constant, but that’s consistent with the asymptotic prediction since relative to n, a constant goes away as n increases.
Reference for theoretical results: D. J. Newman and M. S. Klamkin. Expectations for Sums of Powers. The American Mathematical Monthly, Vol. 66, No. 1 (Jan., 1959), pp. 50-51
## 3 thoughts on “Summing random powers up to a threshold”
1. Paul Fabel
Nice!
To see why e is the answer, after n rolls, the probability that the sum is less than 1 is the hyper volume of the convex hull of 0 and n standard basis vectors, a Euclidean n-simplex.
This is also an easy to write down iterated integral of 1 — the bounds will be 0 to 1-x1-x2…-xi.
Calculating the cases n=2 and n=3 makes it easy to see the pattern. we get 1/2, 1/6. In general the Euclidean hyper volume is 1/n! .
Now we are almost done — a typical term in the expected value calculation is n x ( 1/(n-1)! – 1/(n!)) and this simplifies to 1/(n-2)!
Now we sum from n=2 to infinity to get e.
2. Nathan Hannon
Another, more conceptual way to see why e is the answer:
Let x_1, x_2, … be the random numbers you generated, let y_k = x_1 + x_2 + … + x_k, and let z_k = y_k mod 1. That is, each y_k is the k-th partial sum of the x_n, and each z_k is the fractional part of y_k. Then the z_k are also independent, Uniform [0, 1] random variables. (To see this, just note that, for any given values of z_1, …, z_(k-1), the marginal distribution of z_k is still Uniform [0, 1].)
Furthermore, since the y_k are increasing in increments of at most 1, the fractional part of y_k decreases precisely when the integer part increases; that is, z_k floor(y_(k-1)). In particular, y_k < 1 iff z_1 <= z_2 <= … <= z_k.
But z_1, …, z_k are independent, identically distributed, continuous random variables, so they are almost surely distinct, and all orderings are equally likely. Hence, y_k < 1 with probability 1/k!, and taking the sum gives e.
3. What the expected value of x in this code? | 2022-05-21T16:20:55 | {
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https://math.stackexchange.com/questions/1972652/probability-an-6-sided-die-will-be-higher-than-a-8-sided-die | # Probability an 6 sided die will be higher than a 8 sided die?
Say one person rolls an 8 sided die and the other rolls a six, what is the probability that the six sided die is higher than the 8?
I know that the expected value of the eight is 4.5 and the six is 3.5 but am having trouble figuring out how to find the probability.
EDIT: Answer is 15/48 but still curious if there's a way of doing this without creating a grid.
• Draw a $6 \times 8$ array and fill it with all the elementary events that can occur (each one with probability $1/48$, assuming equidistribution) then count... – Jean Marie Oct 17 '16 at 14:32
• Ok that'll work, I am curious if there's a simpler way though. What if I'm dealing with numbers too large to draw out? I suppose I could always whip up a script to do it but curious if there's a formula. – i_am_so_stupid Oct 17 '16 at 14:42
• There are cases like this one where no "witty" way exist. One thing is sure, the fact that you know the expected values of the "6" and the "8" dice is of no help for your problem. – Jean Marie Oct 17 '16 at 14:48
• Fair enough, thank you for your help – i_am_so_stupid Oct 17 '16 at 14:57
• @Skeleton Bow You are perfectly right: $n(n+1)/2=15$ for $n=5$. – Jean Marie Oct 17 '16 at 15:14
Probability That The $\boldsymbol{6}$-Sided Die Will Be Higher (from the body of the question) $$\sum_{k=1}^6\overbrace{\frac16}^{k\text{ on d}6}\cdot\overbrace{\frac{k-1}8}^{\lt k\text{ on d}8} =\frac{5}{16}$$
Probability That The $\boldsymbol{8}$-Sided Die Will Be Higher (from the original title to the question) $$\sum_{k=1}^7\overbrace{\frac18}^{k\text{ on d}8}\cdot\overbrace{\frac{k-1}6}^{\lt k\text{ on d}6} +\overbrace{\frac18}^{8\text{ on d}8}\cdot\overbrace{1\vphantom{\frac16}}^{\lt8\text{ on d}6} =\frac{9}{16}$$
• Can you explain what you did there? Your answer does not agree with what I (and others) have found. – Thanassis Oct 17 '16 at 15:20
• It is the six sided die that is to score higher. – true blue anil Oct 17 '16 at 15:36
• Sorry. I read the title. I have amended my answer. – robjohn Oct 17 '16 at 15:36
• I made the same "mistake" and deleted, I now see that OP confused us with conflicting title and body ! – true blue anil Oct 17 '16 at 15:49
• Thanks. Sorry for the conflicting title, it's been fixed. – i_am_so_stupid Oct 17 '16 at 16:10
This is rather easy. You don't even need a grid. We just have to count the number of tuples $(x,y) ; x \subset A$ and $y\subset B$ [where A is the est of all possible numbers of the six sided dice(1-6) and B is the set of all possible numbers of the eight sided dice(1-8)] such that $x>y$(by the problem). There are 15 such tuples: 5 for x = 6, 4 for x = 5 so on till 1 for x = 2. Thus the probability is P= $\frac{15}{48}$. PS: I don't know how your edit makes sense with the probability being $\frac{120}{48}$,which i believe is greater than 1.
Answer is 5!/48? So greater than one probability?
Of course you mean the the 5th triangle number / 48.
Use theorem of total probability to reach this answer, conditioning on the value the 6 sided die takes.
That is to say if you roll x with the 6 sided die. You have an x-1/8 chance of it being higher than the 8 sided die.
The generalisation is pretty clear
\begin{align}\mathsf P(X_6>X_8) ~=~& \mathsf P(X_8<7)~\mathsf P(X_6>X_8\mid X_8<7)+\mathsf P(X_8>6)~\mathsf P(X_6>X_8\mid X_8>6) \\ ~=~& \frac 68\cdot\frac {15}{36}+\frac 28\cdot 0 \\ ~=~& \frac {5}{16} \end{align}
Let $d_8$ be the result of the 8-sided die, and $d_6$ the result of the the 6-sided die.
$$P(d_8<d_6) = \sum_{i=1}^6 P((d_8 < d_6) \cap(d_6=i)) = \sum_{i=1}^6 P(d_8 < d_6|d_6=i)\cdot P(d_6=i)$$
$$=\sum_{i=1}^6 \frac{i-1}{8} \cdot \frac{1}{6} = \frac{1}{48}\sum_{i=1}^6 (i-1) = \frac{\frac{6\cdot 7}{2}-6}{48} = \frac{15}{48} = \frac{5}{16}$$
You can generalise for any $n,m$ sided dice. Assume $n>m$ we have:
$$P(d_n<d_m) = \sum_{i=1}^m P((d_n < d_m) \cap(d_m=i)) = \sum_{i=1}^m P(d_n < d_m|d_m=i)\cdot P(d_m=i)$$
$$=\sum_{i=1}^m \frac{i-1}{k} \cdot \frac{1}{m} = \frac{1}{nm}\sum_{i=1}^m (i-1) = \frac{\frac{m\cdot (m+1)}{2}-m}{nm} = \frac{\frac{m+1}{2}-1}{n} = \frac{\frac{m-1}{2}}{n} = \frac{m-1}{2n}$$
• It could perhaps be more simply derived as $\frac12$ (total squares minus diagonal squares in smaller die)/(total squares in grid) $= \frac{(m^2-m)/2}{mn} = \frac{m-1}{2n}$ – true blue anil Oct 17 '16 at 15:44
From the OP:
1. " Say one person rolls an $8$ sided die and the other rolls a six ".
2. " What is the probability that the six sided die $\underline{is\ higher\ than}$ the $8$ ? ".
$$\bbx{\ds{\large% \mbox{Hereafter,}\ \bracks{\cdots}\ \mbox{is the}\ Iverson\ Bracket}}$$
The answer to $2.$ is given by: \begin{align} {1 \over 8}\sum_{d_{8} = 1}^{8}{1 \over 6}\sum_{d_{6} = 1}^{6} \bracks{d_{6} > d_{8}} & = {1 \over 48}\sum_{d_{8} = 1}^{8}\braces{\bracks{d_{8} \leq 5} \sum_{d_{6} = d_{8} + 1}^{6}} = {1 \over 48}\sum_{d_{8} = 1}^{8}\bracks{d_{8} \leq 5}\pars{6 - d_{8}} \\[5mm] & = {1 \over 48}\sum_{d_{8} = 1}^{5}\pars{6 - d_{8}} = {1 \over 48}\bracks{5 \times 6 - {5\pars{5 + 1} \over 2}} = \bbx{\ds{5 \over 16}} = 0.3125 \end{align}
If there is a die with $m$ sides and a die with $n$ sides, assuming that $m<n$, the probability that $m$ will be greater than $n$ assuming that those are the names given to the outcomes when both of them are rolled is
$$\frac{m-1}{2n}$$
Full disclaimer: I made this up with help from the grid. This is a formula very specific to your question.
• I don't think this is a general formula. – true blue anil Oct 17 '16 at 15:28
• @trueblueanil it is the general formula (look at my answer), I just did not see any explanation on how SkeletonBow arrived there. – Thanassis Oct 17 '16 at 15:31
• Oh, I misread the question as the $8$ sided die scoring higher. Drat, can't up vote now, so sorry ! – true blue anil Oct 17 '16 at 15:33
• @trueblueanil no problem. I saw your answer (which you deleted), and figured that you thought it was for numbers higher than or equal to the ones on the 8-die. – Skeleton Bow Oct 17 '16 at 15:37
• @Thanassis to be honest, it was quite simple: I drew it out on a grid and realized that every time you add a number to the smaller $m$, the number of possibilities grew by $m-1$. This meant that I needed to find the total sum of the first $m-1$ integers and then divide that by the total number of squares on the grid, which would be $m\cdot n$ and then got the simplified result. – Skeleton Bow Oct 17 '16 at 15:40 | 2019-06-19T09:33:13 | {
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https://math.stackexchange.com/questions/2956369/whats-wrong-in-my-solution-ways-of-choosing-5-items-from-3-catagory-with-3-6 | # What's wrong in my solution? Ways of choosing 5 items from 3 catagory with 3, 6, 14 items, while having 1 item from each catagory.
The exact question is:
b) Sandra wishes to buy some applications (apps) for her smartphone but she only has enough money for 5 apps in total. There are 3 train apps, 6 social network apps and 14 games apps available. Sandra wants to have at least 1 of each type of app. Find the number of different possible selections of 5 apps that Sandra can choose?
How I approached this question :
As we need to choose 1 from each catagory, the first three apps can be choosen in :
3 x 6 x 14 = 252 ways.
Now, for the remaining 2 apps, we can choose from the 20 apps (3 + 6 + 14 - the installed 3 apps)
So, number of combinations should be 20C2 = 190, so finally the answer I got to was 3 x 6 x 14 x 190 = 47880 which is wrong, the answer is 13839 ways. After trying another method I got 13839 but, I need to know why this method is wrong. Can anyone explain this?
• For reference, the following Mathematica code confirms by brute force that the answer is 13839: Count[Map[{Count[#, y_ /; 1 <= y <= 3], Count[#, y_ /; 4 <= y <= 9], Count[#, y_ /; 10 <= y <= 23]} &, Subsets[Range[23], {5}]], x_ /; Min[x] >= 1] – Meni Rosenfeld Oct 15 '18 at 14:52
You are overcounting many, many choices, such as the following selections:
• train app A, SNS app B, games app C, games app D, games app E
• train app A, SNS app B, games app D, games app C, games app E
Your method counts these two selections as different (the italicised part is the "at least one of each app" requirement), but they are the same.
• Ok, I see now why my answer is wrong but, can you point out where/what step in my solution this overlap occurs? Is it possible to change my solution to avoid this error? – abnas Oct 15 '18 at 11:35
• @abnas When you start saying "the remaining two apps are selected from the unselected 20", that is where you overcount. There is no easy fix to your attempt. – Parcly Taxel Oct 15 '18 at 11:36
As Parcly Taxel pointed out, you overcount many times because picking the people in two different counts(using your method) doesn't account for the ordering.
My attempt(casework) is the following:
The possible group sizes per element must be either: 1, 1, 3 or 1, 2, 2.
Case 1: 1, 1, 3
There are different combinations based on the different choices of the largest group. So we compute them separately, giving
$$\binom{3}{1}\binom{6}{1}\binom{14}{3} + \binom{3}{1}\binom{6}{3}\binom{14}{1} + \binom{3}{3}\binom{6}{1}\binom{14}{1} = 7476.$$
Case 2: 1, 2, 2
Similarly the the last one, we have $$\binom{3}{1}\binom{6}{2}\binom{14}{2}+\binom{3}{2}\binom{6}{1}\binom{14}{2}+\binom{3}{2}\binom{6}{2}\binom{14}{1} = 6363.$$
Summing the two cases gives a total of $$\boxed{13839}.$$
• Is there no easier way than to enumerate all possibilities of the number of apps from each category to install? – BallpointBen Oct 15 '18 at 18:47
• I'm not sure. I was trying to come up with a nice bijection using letters of the alphabet, but this direct casework was all I could come up with. – math783625 Oct 16 '18 at 1:05
As pointed out by @Parcly Taxel,
You are overcounting the apps; keep in mind that order is irrelevant here, so train1,train2 is same as train2,train1. This is where overccounting crept into your solution.
So, how to avoid this overcounting?
Take 3T,6S and 14G apps.
In total, you want 5 apps.
So make selections like: $$(1T,1S,3G),(1T,3S,1G),(3T,1S,1G),(2T,2S,1G),(2T,1S,2G),(1T,2S,2G)$$ Now, count for number of ways you have for each of the above selection, and add.
$$(^3C_1^6C_1^{14}C_3) + (^3C_1^6C_3^{14}C_1) + (^3C_3^6C_1^{14}C_1) + (^3C_2^6C_1^{14}C_1) + (^3C_2^6C_1^{14}C_2) + (^3C_1^6C_2^{14}C_2) = 13839$$
• This is the method I, after checking my answer used to get to the correct answer. So, is this the only method to solve this problem? @math783625 used a slightly different style but it is basically the same method as this one.But, is it possible to solve this without using the different selections you used or a more direct method ? – abnas Oct 15 '18 at 11:48
• you can compensate for overcounted apps from your solution, but it will ultimately yield the same result as we posted. – idea Oct 15 '18 at 11:52
I found another solution using the Inclusion-Exclusion Principle and Complementary Counting.
Let $$S_3, S_6, S_{14}$$ be the sets of 5 objects that do not contain a 3, 6, and 14, respectively.
The number of sets that are missing at least 1 of 3, 6, or 14 is $$|S_3 \cup S_6 \cup S_{14}| = |S_3| + |S_6| + |S_{14}| - |S_3 \cap S_6| - |S_3\cap S_{14}| -|S_6\cap S_{14}| + |S_3\cap S_6\cap S_{14}|.$$
$$|S_3| = \binom{20}{5}, |S_6| = \binom{17}{5}, |S_{14}| = \binom{9}{5},$$
$$|S_3 \cap S_6| = \binom{14}{5}, |S_3\cap S_{14}| = \binom{6}{5}, |S_6\cap S_{14}| = 0,$$
$$|S_3\cap S_6\cap S_{14}| = 0.$$
Evaluating gives 19810 as the number of sets that are missing 1 of 3, 6, or 14. The total number of sets is $$\binom{23}{5} = 33649,$$ so the number of sets that are not missing any of 3, 6, or 14 is $$33649 - 19810 = \boxed{13839}.$$ | 2019-04-25T06:34:17 | {
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https://thesacrednest.com.au/blog/5zp3nvv/viewtopic.php?557c17=solve-linear-congruence | If this condition is satisfied, then the above congruence has exactly $d$ solutions modulo $m$, and that, $$x = x_0 + \frac{m}{d} \cdot t, \quad t = 0, 1, \ldots, d-1.$$. Find all solutions to the linear congruence $5x \equiv 12 \pmod {23}$. is the solution to the initial congruence. This was really helpful. We first note that $(5, 23) = 1$, hence we this linear congruence has 1 solution (mod 23). If b is not divisible by g, there are no solutions. These cookies will be stored in your browser only with your consent. Linear Congruences ax b mod m Theorem 1. It turns out x = 9 will do, and in fact that is the only solution. Observe that Hence, (a) follows immediately from the corresponding result on linear … Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. 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If u 1 and u 2 are solutions, then au 1 b (mod m) and au 2 b (mod m) =)au 1 au 2 (mod m) =)u 1 u In case the modulus is prime, everything you know from linear algebra goes over to systems of linear congruences. Browse other questions tagged linear-algebra congruences or ask your own question. We assume a > 0. You can verify that 7*59 = 413 so 7*59 â¡ 13 (mod 100). This website uses cookies to ensure you get the best experience on our website. That help us the … Since gcd(50, 105) = 5 and 65 is divisible by 5, there are 5 solutions. Then x = (100*4 + 13)/7 = 59. Proposition 5.1.1. Hence -9 can be used as an inverse to our linear congruence $5x \equiv 12 \pmod {23}$. Adding and subtracting rational expressions, Addition and subtraction of decimal numbers, Conversion of decimals, fractions and percents, Multiplying and dividing rational expressions, Cardano’s formula for solving cubic equations, Integer solutions of a polynomial function, Inequality of arithmetic and geometric means, Mutual relations between line and ellipse, Unit circle definition of trigonometric functions, Solving word problems using integers and decimals. In this case, $\overline{v} \equiv v_k \pmod m’$ is a solution to the congruence $a’ \overline{v} \equiv 1 \pmod{m’}$, so $v \equiv b’ v_k \pmod{m’}$ is the solution to the congruence $a’v \equiv b’ \pmod{m’}$. solve the linear congruence step by step. For another example, 8x â¡ 2 (mod 10) has two solutions, x = 4 and x = 9. In the table below, I have written x k first, because its coefficient is greater than that of y. This is a linear Diophantine equation and it has a solution if and only if $d = \gcd(a, m)$ divides $b$. That is, assume g = gcd(a, m) = 1. Let d = gcd(c,m), and choose q, r 2Z such that c = dq and m = d r. If b is a solution to (1), then it is also a Our rst goal is to solve the linear congruence ax b pmod mqfor x. Unfortu-nately we cannot always divide both sides by a to solve for x. Necessary cookies are absolutely essential for the website to function properly. The result is closely related to the Euclidean algorithm. Let x 0 be any concrete solution to the above equation. This reduces to 7x= 2+15q, or 7x≡ … This problem has been solved! Solution to a linear congruence equation is equivalent to finding the value of a fractional congruence, for which a greedy-type algorithm exists. solutions of a linear congruence (1) by looking at solutions of Diophantine equation (2). Let's use the division algorithm to find the inverse of modulo : Hence we can use as our inverse. By the Euler’s theorem, $$a^{\varphi (m)} \cdot b \equiv b \pmod m.$$, By comparing the above congruence with the initial congruence, we can show that, $$x \equiv a^{\varphi (m) -1} \cdot b \pmod m$$. Update: Here are the posts I intended to write: systems of congruences, quadratic congruences. The linear congruence equation ax = b (mod n) may be rewritten as ax1 = b - nx2 where x1, x2 -E- Z. Solve the linear system sa+ tm= 1: Then sba+ tbm= b: So sba b (mod m) gives the solution x= sb. Therefore, solution to the congruence $3x \equiv 8 \pmod 2$ is, $$x = x_0 + 2t, \quad t \in \mathbb{Z},$$. 24 8 pmod 16q. The algorithm can be formalized into a procedure suitable for programming. Theorem 2. There are several methods for solving linear congruences; connection with linear Diophantine equations, the method of transformation of coefficients, the Euler’s method, and a method that uses the Euclidean algorithm…, Connection with linear Diophantine equations. The brute force solution would be to try each of the numbers 0, 1, 2, â¦, m-1 and keep track of the ones that work. Now what if the numbers a and m are not relatively prime? Since 100 â¡ 2 (mod 7) and -13 â¡ 1 (mod 7), this problem reduces to solving 2y â¡ 1 (mod 7), which is small enough to solve by simply sticking in numbers. We find y = 4. Example 1. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. In particular, (1) can be rewritten as We must now see how many distinct solutions are there. A linear congruence $ax \equiv b \pmod m$ is equivalent to. The CRT is used solve systems of congruences of the form $\rm x\equiv a_i\bmod m_{\,i}$ for distinct moduli $\rm m_{\,i}$; in our situation, there is only one variable and only one moduli, but different linear congruences, so this is not the sort of problem where CRT applies. To the above congruence we add the following congruence, By dividing the congruence by $7$, we have. The solution of a linear congruence can be found in the Wolfram Language using Reduce[a*x == b, x, Modulus -> m]. the congruences whose moduli are the larger of the two powers. This means that there are exactly $d$ distinct solutions. Solve The Linear Congruence Step By Step ; Question: Solve The Linear Congruence Step By Step . This simpli es to 5t 2 (mod 8), which we solve by multiplying both sides by We can repeat this process recursively until we get to a congruence that is trivial to solve. Construction of number systems – rational numbers. Substituting this into our equation for yields: Thus it follows that , so is the solution t… The complete set of solutions to our original congruence can be found by adding multiples of 105/5 = 21. Lemma. The notion of congruences was first introduced and used by Gauss in his Disquisitiones Arithmeticae of 1801. Let’s talk. Solve the following congruence: Since $\gcd(7, 15) = 1$, that the given congruence has a unique solution. The congruence $ax \equiv b \pmod m$ has solutions if and only if $d = \gcd(a, m)$ divides $b$. 1/15 15 22 31 47 Fermat's Little Theorem is often used in computing large powers modulo n, 1 point under some conditions. Previous question Next question Get more help from Chegg. The algorithm above says we can solve this by first solving 21y â¡ -13 (mod 10), which reduces immediately to y â¡ 7 (mod 10), and so we take y = 7. stated modulo 90, and so the most satisfying answer is given in terms of congruence classes modulo 90. Solve x^11 + x^8 + 5 mod(49) I have a lot of non-linear congruence questions, so I need an example of the procedure. Solving Linear Congruence by Finding an Inverse. Although Bill Cook's answer is completely, 100% correct (and based on the proof of the Chinese Remainder Theorem), one can also work with the congruences successively; we know from the CRT that a solution exists. For example, we may want to solve 7x â¡ 3 (mod 10). Solve the following congruence: Since $\gcd(3, 2) = 1$, that, by the theorem 1., the congruence has a unique solution. linear congruences (in one variable x). Proof. One or two coding examples would’ve been great, though =P, this really helpful for my project. Solve Linear Congruences Added May 29, 2011 by NegativeB+or- in Mathematics This widget will solve linear congruences for you. This entails that a set of remainders $\{0, 1, \ldots, p-1 \}$ by dividing by $p$, whit addition and multiplication $\pmod p$, makes the field. Since 7 and 100 are relatively prime, there is a unique solution. See the answer. Existence of solutions to a linear congruence. Example 3. The result is closely related to the Euclidean algorithm. These cookies do not store any personal information. So we first solve 10x â¡ 13 (mod 21). We can calculate this using the division algorithm. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Email: donse[email protected] Tel: 800-234-2933; Let $x_0$ be any concrete solution to the above equation. First, let’s solve 7x â¡ 13 (mod 100). most likely will be coming back here in the future, Thank you! The equation 3x==75 mod 100 (== means congruence), input 3x into Variable and Coeffecient, input 100 into modulus, and input 75 into the last box. Thus: Hence for some , . If (a;m) = 1, then the congruence ax b mod mphas exactly one solution modulo m. Constructive. The algorithm says we should solve 100y â¡ -13(mod 7). If the number $m =p$ is a prime number, and if $a$ is not divisible by $p$, then the congruence $ax \equiv b \pmod p$ always has a solution, and that solution is unique. Solve the following congruence: $$x \equiv 5^{\varphi(13) -1} \cdot 8 \pmod{13}.$$, Since $\varphi (13) =12$, that it follows, By substituting it in $x \equiv 3^{11} \cdot 8 \pmod{13}$ we obtain. A linear congruence is the problem of finding an integer x satisfying, for specified integers a, b, and m. This problem could be restated as finding x such that, Two solutions are considered the same if they differ by a multiple of m. (It’s easy to see that x is a solution if and only if x + km is a solution for all integers k.). 1 point In order to solve the linear congruence 15x = 31 (mod 47) given that the inverse of 15 modulo 47 is 22, what number should be multiplied to both sides in the given congruence? Thanks :) Then $x_0 \equiv b \pmod m$ is valid. Now substitute for x in the second congruence: 3(6+7t) 4 (mod 8). Theorem. By subtracting obtained equations we have: It follows: $x – x_0 = 2t, t \in \mathbb{Z}$. first place that I’ve understood it, after looking through my book and all over the internet Example. If b is divisible by g, there are g solutions. 10 15 20 25 30 None of the above 1 point Using the binary modular exponentiation algorithm (as shown in lecture, Algorithm 5 in Section 4.2) to … Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. If this condition is met, then all solutions are given with the formula: $$x = x_0 + \left (\frac{m}{d} \right) \cdot t, \quad y= y_0 \left (\frac{a}{d} \right) \cdot t,$$. Thanks a bunch, Your email address will not be published. Then x 0 ≡ … Solution: We have gcd(42,90) = 6, so there is a solution since 6 is a factor of 12. Find more at https://www.andyborne.com/math See how to solve Linear Congruences using modular arithmetic. We first put the congruence ax â¡ b (mod m) in a standard form. It is mandatory to procure user consent prior to running these cookies on your website. Then the solutions to ax â¡ b (mod m) are x = y + tm/g where t = 0, 1, 2, â¦, g-1. This category only includes cookies that ensures basic functionalities and security features of the website. But opting out of some of these cookies may affect your browsing experience. Therefore, $x_1$ and $x_2$ are congruent modulo $m$ if and only if $k_1$ and $k_2$ are congruent modulo $d$. The proof for r > 2 congruences consists of iterating the proof for two congruences r – 1 times (since, e.g., € ([m 1,m 2],m 3)=1). For example 25x = 15 (mod 29) may be rewritten as 25x1 = 15 - 29x2. If $d \nmid b$, then the linear congruence $ax \equiv b \pmod m$ has no solutions. Featured on Meta “Question closed” … Multiply the rst congruence by 2 1 mod 7 = 4 to get 4 2x 4 5 (mod 7). The given congruence we write in the form of a linear Diophantine equation, on the way described above. The most important fact for solving them is as follows. $3x \equiv 8 \pmod 2$ means that $3x-8$ must be divisible by $2$, that is, there must be an integer $y$ such that. Since $\gcd(6,8) = 2$ and $2 \nmid 7$, there are no solutions. Let $a$ and $m$ be natural numbers, and $b$ an integer. If y solves this new congruence, then x = (my + b)/ a solves the original congruence. This means that a linear congruence also has infinitely many solutions which are given in the form: $$x = x_0 + \left( \frac{m}{d}\right) \cdot t, \quad t \in \mathbb{Z}.$$. Menu. If not, replace ax â¡ b (mod m) with –ax â¡ –b (mod m). Solving linear congruences is analogous to solving linear equations in calculus. Solving the congruence $ax \equiv b \pmod m$ is equivalent to solving the linear Diophantine equation $ax – my = b$. Example 4. The linear congruence Solve the following congruence: We must first find $\gcd(422, 186)$ by using the Euclidean algorithm: Therefore, $\gcd(422, 186) = 2$. Recall that since $(31,24)=1$ and $1|12$ there is exactly one incongruent solution modulo $24.$ To find this solution let’s use the definition of congruence, … Suppose a solution exists. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Since $2 \mid 422$, that the given congruence has solutions ( it has exactly two solutions). Which of the following is a solution for x? Get 1:1 help now from expert Advanced Math tutors For this purpose, we take any two solutions from that set: $$x_1 = x_0 + \left( \frac{m}{d}\right) \cdot k_1,$$, $$x_2 = x_0 + \left (\frac{m}{d}\right) \cdot k_2.$$, $$x_0 + \left( \frac{m}{d} \right) \cdot k_1 \equiv x_0 + \left( \frac{m}{d} \right) \cdot k_2 \pmod m$$, $$\left( \frac{m}{d} \right) \cdot k_1 \equiv \left( \frac{m}{d} \right) \cdot k_2 \pmod m.$$. Solving the congruence 42x ≡ 12 (mod 90) is equivalent to solving the equation 42x= 12+90qfor integers xand q. Let , and consider the equation (a) If , there are no solutions. This says we can take x = (105*7 + 65)/50 = 16. Example 1. Gauss illustrates the Chinese remainder theorem on a problem involving calendars, namely, "to find the years that have a certain period number with respect to the solar and lunar cycle and the Roman indiction." Linear Congruence Calculator. Therefore, if $ax \equiv b \pmod m$ has a solution, then there is infinitely many solutions. So if g does divide b and there are solutions, how do we find them? This is progress because this new problem is solving a congruence with a smaller modulus since a < m. If y solves this new congruence, then x = (my + b)/a solves the original congruence. (b) If , there are exactly d distinct solutions mod m.. If we need to solve a system of three linear congruences with one unknown, then we need first solve a system of two linear congruences, and then see which of the obtained solutions also satisfy the third congruence. Expert Answer . My colleagues and I have decades of consulting experience helping companies solve complex problems involving data privacy, math, statistics, and computing. Then first solve the congruence (a/g)y â¡ (b/g) (mod (m/g)) using the algorithm above. So, we restrict ourselves to the context of Diophantine equations. The algorithm can be formalized into a procedure suitable for programming. 1 point Solve the linear congruence 2x = 5 (mod 9). For instance, solve the congruence $6x \equiv 7 \pmod 8$. So the solutions are 16, 37, 58, 79, and 100. The method of transformation of coefficients consist in the fact that to the given equation we add or subtract a well selected true congruence. Solving the congruence a x ≡ b (mod m) is equivalent to solving the linear Diophantine equation a x – m y = b. Linear Congruences In ordinary algebra, an equation of the form ax = b (where a and b are given real numbers) is called a linear equation, and its solution x = b=a is obtained by multiplying both sides of the equation by a1= 1=a. For daily tweets on algebra and other math, follow @AlgebraFact on Twitter. If d does divide b, and if x 0is any solution, then the general solution is given by x = x Here, "=" means the congruence symbol, i.e., the equality sign with three lines. Now let’s find all solutions to 50x â¡ 65 (mod 105). First, suppose a and m are relatively prime. Your email address will not be published. This simpli es to x 6 (mod 7), so x = [6] 7 or x = 6 + 7t, where t 2Z. How do I solve a linear congruence equation manually? // Example: To solve € … 12 ( mod 10 ) get more help from Chegg symbol, i.e. the. 2011 by NegativeB+or- in Mathematics this widget will solve linear congruences and how to solve …! Multiples of 105/5 = 21 4 and x = ( 100 * 4 13... The form of a linear congruence equation manually mod 21 ) equation of the.! Example, the process becomes more complicated congruence by $7$, that the given we. \Mid 422 $, we may want to solve use the division to... Congruences whose moduli are the larger of the y coding examples would ’ ve been great, =P... Enter a mod b statement thanks a bunch, your email address will not be published = 15 ( (... Intended to write posts in the table below, I have decades of experience. Standard form of a fractional congruence, by dividing the congruence which also specifies the class that the! How you use this website uses cookies to ensure you get the experience! Residue is 7 ( mod 105 ) the class that is the solution only solution context. Mandatory to procure user consent prior to running these cookies quadratic congruences distinct solutions b..., 79, and consider the equation 42x= 12+90qfor integers xand q solving them is follows. Modulus is prime, there is infinitely many solutions denoted by$ 7 $, that by Theorem! A congruence that is the solution = 16, let ’ s solve 7x â¡ 13 ( )! Is equivalent to ( 2 ) smaller than the coefficient of the y features of the congru- Browse other tagged... Says we should solve 100y â¡ -13 ( mod 10 ) has two solutions ) them is as follows 100! To procure user consent prior to running these cookies 1/15 15 22 31 47 's. ¡ -13 ( mod m ) = 5 ( mod 21 ), that by the 6... Mod b statement mandatory to procure user consent prior to running these cookies on your website are,... A Tutor ; Upgrade to Math Mastery: Hence our solution in least residue is 7 mod! Start Here ; our Story ; Hire a Tutor ; Upgrade to Math Mastery non-zero! Than the coefficient of the website ( 100 * 4 + 13 ) /7 = 59$ distinct.... Must now see how many distinct solutions are 16, 37, 58, 79, and are! ) -- - Enter a mod b statement, 1 point under some conditions statistics, and.. B statement congruence ( a/g ) y â¡ ( b/g ) ( 21! From Chegg two powers how do we find them 65 is divisible by g solve linear congruence is. Or subtract a well selected true congruence coefficients consist in the fact that to the Euclidean algorithm ) =! B $an integer = 4 and x = 9 will do and! Get the best experience on our website experience while you navigate through the website \equiv 7 \pmod 8.... And in fact that is trivial to solve € … linear congruences Added 29! 31 x\equiv 12 \pmod { 23 }$ running these cookies may affect browsing... Now what if the numbers a and m are relatively prime, there are 5 solutions until we get to... Know how to solve € … linear congruences of transformation of coefficients consist in the form }! So there is a factor of 12 42x ≡ 12 ( mod 7 ) 12..., quadratic congruences set of solutions to our original congruence can be as! Second example, the equality sign with three lines -13 ( mod 21 ) numbers! The class that is, assume g = gcd ( 50, )... Linear algebra goes over to systems of linear congruences the inverse of modulo: Hence our in... { Z } _p $linear Diophantine equation ( 2 ) solution for x are posts! Has no solutions for you the option to opt-out of these cookies be. To the linear congruence$ ax \equiv b \pmod m $is valid you use this website uses to. 65 ( mod ( m/g ) ) using the algorithm above solve a congruence... Theorem is often used in computing large powers modulo n, 1 under. Most satisfying answer is given in terms of congruence coding examples would ve... Thus: Hence our solution in least residue is 7 ( mod 8.. Of some of these cookies on your website previous question Next question get more help from Chegg that is to! Denoted by$ 7 $, we can use as our inverse \pmod m$ is valid should... $distinct solutions x in the future about how to solve linear don! The modulus is prime, everything you know from linear algebra goes over systems! Means that there are exactly$ d $distinct solutions are 16, 37, 58, 79 and!: solve the linear congruence equation manually divide both sides of the website to properly... Is prime, everything you know from linear algebra goes over to systems of linear congruences you... Now what if the numbers a and m are not relatively prime â¡ (... Of consulting experience helping companies solve complex problems involving data privacy, Math, statistics, and$ $! A, m ) =1$ 58, 79, and in fact is... X k first, Suppose that $\gcd ( 6,8 ) =,. Mathematics this widget will solve linear Diophantine equations in two variables, we may have to the. Involving data privacy, Math, statistics, and consider the equation 42x= 12+90qfor integers xand q Hire Tutor. 6,8 ) = 1 rather than talking about equality, it is mandatory to user! Helping companies solve complex problems involving data privacy, Math, follow solve linear congruence AlgebraFact Twitter. The division algorithm to find the inverse of modulo: Hence we can use as our.... Also use third-party cookies that help us analyze and understand how you use this website uses cookies to you. And other Math, statistics, and in fact that to the Euclidean algorithm features of the k! \Pmod { 23 }$ the most satisfying answer is given in terms of congruence now see how many solutions. ) by looking at solutions of a linear congruence 2x = 5 and 65 is divisible 5! X\Equiv 12 \pmod { 24 }. $solution have a unique solution equation manually ( 9... Really helpful for my project inverse to our linear congruence equation is equivalent to solving the congruence$ 6x 7... To running these cookies may affect your browsing experience for instance, solve the linear congruence equation manually mod ). Complete set of solutions to our original congruence can be formalized into a procedure suitable programming. Second congruence: 3 ( mod 7 = 4 and x = ( 105 * 7 + 65 ) =. Third-Party cookies that help us analyze and understand how you use this website uses to! Be rewritten as 25x1 = 15 ( mod 7 = 4 and x = 4 and x 4... As our inverse that there are no solutions 10x â¡ 13 ( 100!, 1 point solve the congruence 42x ≡ 12 ( mod 90 ) is equivalent to the... 105/5 = 21 opting out of some of these cookies may affect your experience. Algorithm exists may affect your browsing experience the posts I intended to write: systems of congruences quadratic. Most satisfying answer is given in terms of congruence the congruences whose moduli are larger! For solving them is as follows by Step ; question: solve the congruence \$ \equiv! Your website we must now see how many distinct solutions are there to speak of congruence classes modulo.... 7 + 65 ) /50 = 16 1 point solve the congruence symbol, i.e., the process more! Solution, then the congruence symbol, i.e., the process becomes more complicated b/g ) mod. Two solutions ) the best experience on our website of 12 if d! ( 105 * 7 + 65 ) /50 = 16 m/g ) ) using the can... And 65 is divisible by g, there is a unique solution m. Constructive we know... We write in the number of congruences, quadratic congruences now from expert Advanced Math tutors congruences... Address will not be published - Enter a mod b statement as 25x1 = (! Two solutions ) features of the website to function properly procure user consent prior to running these may... With your consent infinitely many solutions modulo: Hence we can repeat process! To opt-out of these cookies congruences whose moduli are the posts I intended to write posts in the fact is... Is not divisible by g, there is a solution since 6 a... May 29, 2011 by NegativeB+or- in Mathematics this widget will solve linear congruences ’!, rather than talking about equality, it is customary to speak of congruence classes modulo,! Inverse of modulo: Hence our solution in least residue is 7 ( mod m \equiv 7 \pmod 8.! Of solutions to 50x â¡ 65 ( mod ( m/g ) ) using the algorithm says we can that. Congruences don ’ t always have a unique solution which also specifies the class that is trivial to solve congruences! Are quotients in the number of congruences, the equality sign with lines! Opting out of some of these cookies may affect your browsing experience for a. Already know how to solve ) 4 ( mod 29 ) may be rewritten as 25x1 = -! | 2021-04-19T07:32:06 | {
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http://mathhelpforum.com/advanced-algebra/136387-characteristic-polynomial-print.html | Characteristic Polynomial
• March 29th 2010, 06:15 PM
joe909
Characteristic Polynomial
Given that the characteristic polynomial of $A$ is $t^4 + t + 1$. How would I go about calculating the characteristic polynomial of $A^2$?
I originally thought of using the fact that the eigenvalues of $A^2$ would simply be the square of the eigen values of $A$. But unfortunately I couldnt calculate the eigenvalues of A.
Any help would be great, thanks!
• March 30th 2010, 03:02 AM
tonio
Quote:
Originally Posted by joe909
Given that the characteristic polynomial of $A$ is $t^4 + t + 1$. How would I go about calculating the characteristic polynomial of $A^2$?
I originally thought of using the fact that the eigenvalues of $A^2$ would simply be the square of the eigen values of $A$. But unfortunately I couldnt calculate the eigenvalues of A.
Any help would be great, thanks!
Let $\lambda_i\,,\,i=1,2,3,4$ the eigenvalues of $A\Longrightarrow \lambda_i^2$ are the eigenvalues of $A^2$ . Now, applying Viete formulae we get:
Coefficient of $t^3=-\sum^4_{i=1}\lambda_i=0$
Coefficient of $t^2=\sum^4_{i,j=1,\,i>j}\lambda_i\lambda_j=0$
Coefficient of $t=-\!\!\!\!\!\sum^4_{i,j,k=1\,,\,i>j>k}\lambda_i\lamb da_j\lambda_k=1$
Free coefficient $=\prod^4_{i=1}\lambda_i=1$ .
Well, use the above relations for $\lambda_i^2$ and a little algebra to find out the coefficients of the char. pol. of $A^2$ . For example, it may interest you to use
$\sum^4_{i=1}\lambda_i^2=\left(\sum^4_{i=1}\lambda_ i\right)^2-2\sum^4_{i,j=1,\,i>j}\lambda_i\lambda_j$ ...
• March 30th 2010, 07:03 AM
Opalg
I'm not sure if this is an acceptable answer, but it's slick! By the Cayley–Hamilton theorem, $A^4+A+I=0$. Therefore $A = -(A^4+I)$ and so $A^2 = (A^4+I)^2 = A^8+2A^4+I$. Hence $A^8 +2A^4-A^2+I = 0$, which says that $A^2$ satisfies the polynomial equation $t^4+2t^2-t+1=0$. So I assume that $t^4+2t^2-t+1$ is the characteristic polynomial of $A^2$. But I don't know whether that conclusion is justified. (Wondering)
• March 30th 2010, 08:21 AM
tonio
Quote:
Originally Posted by Opalg
I'm not sure if this is an acceptable answer, but it's slick! By the Cayley–Hamilton theorem, $A^4+A+I=0$. Therefore $A = -(A^4+I)$ and so $A^2 = (A^4+I)^2 = A^8+2A^4+I$. Hence $A^8 +2A^4-A^2+I = 0$, which says that $A^2$ satisfies the polynomial equation $t^4+2t^2-t+1=0$. So I assume that $t^4+2t^2-t+1$ is the characteristic polynomial of $A^2$. But I don't know whether that conclusion is justified. (Wondering)
It looks fine to me: it is of the correct degree, monic and vanishes at $A^2$...good!
Tonio | 2014-03-10T06:31:13 | {
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https://mathhelpboards.com/threads/how-to-prove-this-logarithmic-inequality.2972/ | # How to prove this logarithmic inequality?
#### anemone
##### MHB POTW Director
Staff member
Hi all, I've been having a hard time trying to solve the following inequality:
Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$
I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.
Could I get some hints on how to tackle this problem?
Any help would be deeply appreciated.
Thanks!
P.S. This question was originally asked here (Logarithm) at MMF.
Last edited by a moderator:
#### CaptainBlack
##### Well-known member
Hi all, I've been having a hard time trying to solve the following inequality:
Prove that $\displaystyle \left(\log_{24}(48) \right)^2+\displaystyle \left(\log_{12}(54) \right)^2 >4$
I've tried to change the bases to base-10 log and relating all the figures (12, 24, 48, and 54) in terms of 2 and 3 but only to make the problem to be more confounded.
Could I get some hints on how to tackle this problem?
Any help would be deeply appreciated.
Thanks!
P.S. This question was originally asked here (Logarithm) at MMF.
$$\log_{24}(48)=1+\log_{24}(2)$$
But $$2^5 \gt 24$$ so $$\log_{24}(2) \gt 1/5$$
Also: $$\log_{12}(54)=1+\log_{12}(4.5)$$, and $$4.5^5>12^3$$ so $$\log_{12}(4.5)>3/5$$
Hence:
$(\log_{24}(48))^2 + (\log_{12}(54))^2 \gt 1.2^2+1.6^2 =4$
CB
#### anemone
##### MHB POTW Director
Staff member
Hi CB, a big thank for your help in making it so straightforward and simple for me!
Thanks. | 2020-12-05T00:09:39 | {
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https://mathematica.stackexchange.com/questions/215925/extracting-argument-of-a-specific-functions-in-a-large-expression | # Extracting argument of a specific functions in a large expression
I have a large expression, for example
Cos[x]Sin[y]Sqrt[1+z]/(1+x^2)-1/(1+y)
I wish to extract the argument inside Sqrt function, namely 1+z. I thought of using /.Sqrt[x_]->(h=x), but executing
Cos[x]Sin[y]Sqrt[1+z]/(1+x^2)-1/(1+y)/.Sqrt[x_]->(h=x);h
returns x but not 1+z, why? Is there a way to achieve my goal?
In my expressions, there will only be one Sqrt, but many other different heads as well. There is a similar question, but it is only applicable to very simple expression like Sqrt[1+z].
• You did not get what you expected because you used Rule(->) rather than RuleDelayed(:>) – Bob Hanlon Mar 7 at 20:23
• @BobHanlon I see, thank you very much! – pisco Mar 10 at 10:39
Clear["Global*"]
expr = Cos[x] Sin[y] Sqrt[1 + z]/(1 + x^2) - 1/(1 + y) Sqrt[1 - z];
Cases[expr, Sqrt[t_] :> t, Infinity]
(* {1 - z, 1 + z} *)
To ensure that all of the terms are real
And @@ Thread[% >= 0] // Simplify
(* -1 <= z <= 1 *)
Or more directly,
And @@ Cases[expr, Sqrt[t_] :> (t >= 0), Infinity] // Simplify
(* -1 <= z <= 1 *)
Better yet,
FunctionDomain[expr, z]
(* -1 <= z <= 1 *)
` | 2020-08-03T15:38:42 | {
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http://belaarteconceito.com.br/cronos-stock-vctt/definite-integral-properties-78b942 | # definite integral properties
This can be done by simple adding a minus sign on the integral. Proof of : $$\int{{k\,f\left( x \right)\,dx}} = k\int{{f\left( x \right)\,dx}}$$ where $$k$$ is any number. This video explains how to find definite integrals using properties of definite integrals. If x is restricted to lie on the real line, the definite integral is known as a Riemann integral (which is the usual definition encountered in elementary textbooks). Properties of the Definite Integral The following properties are easy to check: Theorem. . () = . () Definite integral is independent of variable od integration.iii. The properties of indefinite integrals apply to definite integrals as well. EXAMPLE PROBLEMS ON PROPERTIES OF DEFINITE INTEGRALS. If . Property 2: p∫q f(a) d(a) = – q∫p f(a) d(a), Also p∫p f(a) d(a) = 0. Some standard relations. Whereas the indefinite integral f(x) is a function and it has no upper and lower limits. We begin by reconsidering the ap-plication that motivated the definition of this mathe-matical concept- determining the area of a region in the xy-plane. These cookies do not store any personal information. Integration By Parts. Also note that the notation for the definite integral is very similar to the notation for an indefinite integral. A constant factor can be moved across the integral sign.ii. Rule: Properties of the Definite Integral. There are many definite integral formulas and properties. Properties of definite integrals. The definite integral of the function $$f\left( x \right)$$ over the interval $$\left[ {a,b} \right]$$ is defined as the limit of the integral sum (Riemann sums) as the maximum length of the subintervals approaches zero. Let a real function $$f\left( x \right)$$ be defined and bounded on the interval $$\left[ {a,b} \right]$$. PROPERTIES OF INTEGRALS For ease in using the definite integral, it is important to know its properties. Next we will look at some properties of the definite integral. Additive Properties When integrating a function over two intervals where the upper bound of the first is the same as the first, the integrands can be combined. If f (x) and g(x) are defined and continuous on [a, b], except maybe at a finite number of points, then we have the following linearity principle for the integral: (i) f (x) + g(x) dx = f (x) dx + g(x) dx; (ii) f (x) dx = f (x) dx, for any arbitrary number . These properties are used in this section to help understand functions that are defined by integrals. Properties of Definite Integrals Proofs. This category only includes cookies that ensures basic functionalities and security features of the website. If f (x) is defined and continuous on [a, b], then we have (i) Zero Integral property If the upper and lower limits of a definite integral are the same, the integral is zero. This applet explores some properties of definite integrals which can be useful in computing the value of an integral. This website uses cookies to improve your experience. An integral is known as a definite integral if and only if it has upper and lower limits. Properties of Definite Integrals - II. Also, observe that when a = -p, t = p, when a = 0, t =0. The most important basic concepts in calculus are: THE DEFINITE INTEGRAL INTRODUCTION In this chapter we discuss some of the uses for the definite integral. Properties of definite integral. Related Notes: Area Problem Revisited, Concept of Definite Integral, Type I (Infinite Intervals), Type II (Discontinuous Integrands), Area Problem, Properties of Definite Integrals… Function Definite Integral Definition. Therefore, equation (11) becomes, And, if ‘f’ is an odd function, then f(–a) = – f(a). These properties are used in this section to help understand functions that are defined by integrals. Suppose that is the velocity at time of a particle moving along the … Khan Academy is a 501(c)(3) nonprofit organization. Properties of Indefinite Integrals 4. These properties are justified using the properties of summations and the definition of a definite integral as a Riemann sum, but they also have natural interpretations as properties of areas of regions. If v(t) represents the velocity of an object as a function of time, then the area under the curve tells us how far the object is from its original position. I = 0. Integral For example, we know that integraldisplay 2 0 f ( x ) dx = 2 when f ( x ) = 1, because the value of the inte- gral is the area of a rectangle of height 1 and base length 2. A definite integral is a formal calculation of area beneath a function, using infinitesimal slivers or stripes of the region. 9. We list here six properties of double integrals. 0∫π/2 (2log sinx – log sin 2x)dx = – (π/2) log 2 is proved. The properties of indefinite integrals apply to definite integrals as well. In this section we’ve got the proof of several of the properties we saw in the Integrals Chapter as well as a couple from the Applications of Integrals Chapter. A definite integral is an integral int_a^bf(x)dx (1) with upper and lower limits. Properties of the Definite Integral. Definite integrals also have properties that relate to the limits of integration. Evaluate the following problems using properties of integration. Hence. A Definite Integral has start and end values: in other words there is an interval [a, b]. Also, note that when a = p, t = q and when a = q, t = p. So, p∫q wil be replaced by q∫p when we replace a by t. Therefore, p∫q f(a)da = –q∫p f(p+q-t)dt … from equation (4), From property 2, we know that p∫q f(a)da = – q∫p f(a)da. 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Expression in Fundamental Theorem of calculus the limit of the definite integral ( given again below ) has a at... Take the constant – log sin 2x ) dx = – ( π/2 ) log 2 is proved relate the. Of this mathe-matical concept- determining the area under a curve and the definite integral is independent change! Property 2: if definite integral properties limits of definite integrals also have properties that are defined, to generate a value... Antiderivative and indefinite integral of f over two intervals that hold the same but opting out some. On data visualizations and data analysis definite integral properties integrals may not be necessary are interchanged, then value... Curves and so they are equivalent explains how to combine integrals, Trapezoidal approximation of a sum. The question “ what function produces f ( x ) is an broad! We also use third-party cookies that help us analyze and understand definition of definite! Of integration remain the same family of curves and so they are equivalent ( 1/2 [! Opt-Out if you wish Academy is a function generally represents the area under the curve the! Data visualization, data analysis, integrals may not be necessary integral as a integral. This concept a closed interval, we will look at some properties of region! N\ ) subintervals connection between the area under the curves within the specified limits the... Has an upper limit and lower limits produces f ( x ) is a (. Use this website [ -1/3+1 ] = 1- ( 1/3 ) = 2/3 at graphs integral interchanged! Riemann sum of right rectangles this is the difference between the values of the definite integral – sin! Two or more rules in the tabular form which is easy to check: Theorem better understanding Mathematical topics this! Riemann sum of right rectangles solve integration problems based on them let divide... Are defined by integrals let us divide this interval into \ ( \Delta x_i... Integral sign.ii is known as a limit of a given function end point of each for... The list of important rules that form the basis of solving definite integral to evaluate the integral the... The constant – log sin 2x ) dx ( 1 ) x_i } \ ) integral int_a^bf x! Unique value above expression in Fundamental Theorem of calculus prior to running these cookies be! Rules for how to combine integrals, Trapezoidal approximation of a region in the form. Limits of integration: \ ( { \xi_i } \ ) compute definite integral properties areas a connection between the values the. Values of the independent variable will use definite integrals be used only lower!, t = 0, t = p, and play with the limits are defined integrals. Whereas the indefinite integral choose an arbitrary point \ ( n\ ) subintervals cookies that help analyze. Video explains how to find definite integrals as well solution, free steps and graph properties of indefinite with... Cookies will be stored in your browser only with your consent compute exact areas on data visualizations and analysis... At its heart used to compute exact areas at its heart, anywhere the,. Otherwise working with them some of these cookies integration interval chapter we discuss some of these.... Limit and lower bound value to the limits of integration remain the same no upper and limits... Using properties of definite integrals as well intervals that hold the same ) nonprofit organization on closed. ’ is an interval [ a, b, and play with the limits of.... A region in the xy-plane integral the following properties are useful in solving problems requiring the application the. Under the curve from the lower bound value the limit of the region, definite integral is closely to! On them dx ( 1 ) with upper and lower bound are the same, the lines and the! Cookies on your website graph ): breaking interval Our mission is to provide a free, world-class education anyone! Anyone, anywhere - 1 ) a definite integral calculator - solve definite integrals ; Why you Should integrals..., world-class education to anyone, anywhere, assume that f ( x ):! Mandatory to procure user consent prior to running these cookies may affect your browsing experience generate a unique...., free steps and graph properties of the integral to anyone, anywhere evaluate definite! Data modeling, and more ’ is an extremely broad term integral (. Easy to check: Theorem under a curve and the part of the definite integral that ’ s to an... Post, we will use definite integrals also have properties that are defined, to generate a value! 2 is proved the ap-plication that motivated the definition of the website very helpful computing. Summation at its heart requiring the application of the website to function.! Constant factor can be solved quite simply by using this property to running cookies. For this whole section, assume that f ( x ) is even how you this! Opt-Out if you wish with them variable od integration.iii how to find integrals. Integral with two specified limits called the upper and the desired result is obtained basis of solving integral! For x∗ i x i ∗ and play with the limits of integration or otherwise working with them, analysis... Please write: this work is mine unless otherwise cited that motivated the definition of the definite integral following! Mathematics, there are a lot of useful rules for how to find many useful such. Symmetric function the limit of the more challenging problems can be useful in solving requiring. Then the value of integral changes its sign only: \ ( n\ ) subintervals lower are. Useful quantities such as areas, volumes, displacement, etc discuss of! Values: in other words there is an interval [ a, b, and when =! = -p, t = 0, t =p and when a =2p, 0! Is odd below is a number which defines the area bounded by the -axis, the lines and. Is differentiated? ” approximation of a definite integral as a definite integral given below is 501. Question 5: the function f ( x ) is odd anyone, anywhere you wish function... Over another, related, interval cookies may affect your browsing experience -p, t =0 cookies... Difference between the values of the independent variable given a velocity function the indefinite integral integral with specified... - solve definite integrals as well integration remain the same conditions given a velocity function option... Video on definite integral properties Pre-Class Exploration Name: Pledge: Please write: this work is mine otherwise. Changes its sign only of double integrals are very helpful when computing or... Category only includes cookies that ensures basic functionalities and security features of the more challenging problems be... At the specified limits moving along the … properties of definite integral is closely linked the! Basic concepts in calculus are: function limits integral Derivatives problems can be moved across definite integral properties integral sign.ii,... Functionalities and security features of the definite integral using Simpson ’ s the “ simple ” definition of the function. Problems can be done by simple adding a minus sign on the integral choose an point. Values: in other words there is a number which defines the bounded! Improve your experience while you navigate through the website end point of each for! ( \Delta { x_i } \ ) and form the integrals, combine,... Name: Pledge: Please write: this definite integral properties is mine unless otherwise.! | 2021-04-21T19:27:16 | {
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https://mathforums.com/threads/why-cant-i-add-the-period-to-find-the-next-two-solutions.329589/ | # Why can't I add the period to find the next two solutions?
#### davedave
Consider tan(2x) = $$\displaystyle \sqrt3$$ for 0 $$\displaystyle \leq x < 2\pi$$
reference angle x = $$\displaystyle \tan^{-1}$$($$\displaystyle \sqrt{3}$$) = $$\displaystyle \frac{\pi}{3}$$
In the given domain 0 $$\displaystyle \leq x < 2\pi$$, the positive ratio of tangent is in quadrants I and III. So the two standard position angles in this interval are
2x = $$\displaystyle \frac{\pi}{3}$$ and $$\displaystyle \frac{4\pi}{3}$$
Now, divide both sides by 2. Then I get
x = $$\displaystyle \frac{\pi}{6}$$ and $$\displaystyle \frac{2\pi}{3}$$ --------- (*)
The period of tangent is: period = $$\displaystyle \frac{\pi}{b}$$
where b = 2 in tan(2x).
So the period is $$\displaystyle \frac{\pi}{2}$$.
Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle. So I get
x = $$\displaystyle \frac{2\pi}{3}$$ and $$\displaystyle \frac{7\pi}{6}$$
Therefore, the solutions are x=$$\displaystyle \frac{\pi}{6}$$, $$\displaystyle \frac{2\pi}{3}$$, $$\displaystyle \frac{2\pi}{3}$$ and $$\displaystyle \frac{7\pi}{6}$$
The two middle solutions are the same.
The answer key answer says x = $$\displaystyle \frac{\pi}{6}$$, $$\displaystyle \frac{2\pi}{3}$$, $$\displaystyle \frac{7\pi}{6}$$ and $$\displaystyle \frac{5\pi}{3}$$.
I actually found my mistake. If I added $$\displaystyle \pi$$ instead of $$\displaystyle \frac{\pi}{2}$$ in (*) above, I would get the right solutions in the second cycle. But, I don't understand why you have to do that, since the period of tan(2X) is $$\displaystyle \frac{\pi}{b}$$ = $$\displaystyle \frac{\pi}{2}$$.
I am very confused. Can someone explain this? Thanks a lot.
Last edited by a moderator:
#### skeeter
Math Team
$\tan(2x) = \sqrt{3}$
$0 \le x < 2\pi \implies 0 \le 2x < 4\pi$
$2x = \dfrac{\pi}{3} \, , \, \dfrac{4\pi}{3} \, , \, \dfrac{7\pi}{3} \, , \, \dfrac{10\pi}{3}$
$x = \dfrac{\pi}{6} \, , \, \dfrac{2\pi}{3} \, , \, \dfrac{7\pi}{6} \, , \, \dfrac{5\pi}{3}$
note ...
$\dfrac{\pi}{6} + \dfrac{\pi}{2} = \dfrac{2\pi}{3}$
$\dfrac{\pi}{6} + 2 \cdot \dfrac{\pi}{2} = \dfrac{7\pi}{6}$
$\dfrac{\pi}{6} + 3 \cdot \dfrac{\pi}{2} = \dfrac{5\pi}{3}$
... adding the modified period does work
#### greg1313
Forum Staff
Note that the difference between $\dfrac{\pi}{6}$ and $\dfrac{2\pi}{3}$ is $\dfrac{\pi}{2}$.
If you had added $\dfrac{\pi}{2}$ to $\dfrac{2\pi}{3}$ twice you would have come up with the two remaining solutions, hence the period of $\dfrac{\pi}{2}$.
$$\tan(2x)=\sqrt{3}$$
$$2x=\dfrac{\pi}{3}+k\pi$$
$$x=\dfrac{\pi}{6}+k\dfrac{\pi}{2},\quad k\in\mathbb{Z}$$
#### skipjack
Forum Staff
Then, add this period to each solution of x in the first cycle to get the next two solutions in the second cycle.
You cover two periods in each cycle, so adding the period to your first solution, $$\displaystyle \frac{\pi}{6}$$, gives your second solution in the same cycle, $$\displaystyle \frac{2\pi}{3}$$. Adding the period to your second solution gives the first solution in the second cycle, $$\displaystyle \frac{7\pi}{6}$$. Adding the period once more will then give the second solution in the second cycle, $$\displaystyle \frac{5\pi}{3}$$.
As the cycle length is a multiple of the period for this question, you can alternatively add the cycle length, $\pi$, to each of your initial solutions to get the solutions in the second cycle.
To avoid confusion, I recommend that you start by finding the solution(s) in the first period (rather than cycle) and then add the period length repeatedly until the specified domain is exhausted. | 2020-03-31T14:10:17 | {
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http://mathoverflow.net/questions/92188/system-of-generators-of-a-homogenous-ideal | # System of generators of a homogenous ideal
Let $I$ be a homogenous ideal in the ring $k[x_{1},\dots,x_{n}]$. My question is:
If $\lbrace f_{1},\dots,f_{r}\rbrace$ is a minimal system of generators of $I$, then are the integers $r$ and $\deg f_i$ determined uniquely by $I$?
More precisely:
If $\lbrace g_{1},\dots,g_{s}\rbrace$ is another minimal set of generators of $I$, then do we necessarily have $r=s$ and $\deg f_{i}=\deg g_{\sigma i}$ for some $\sigma\in S_r$.
-
I do not know why the sentences after "more precisely..." is bigger and in bold. Moderators of MO, please edit it for me. Thanks – Arsenaler Mar 26 '12 at 2:32
@msnaber: The boldface was caused by your "----" following that paragraph. I added a line break which removed it. If you intended to produce a horizontal line, then this should probably look like what you intended, otherwise, remove the "----". – Sándor Kovács Mar 26 '12 at 7:23
@SándorKovács : Thank you very much! – Arsenaler Mar 26 '12 at 14:03
Yes, they are uniquely determined by $I$. This holds more generally:
Let $R = \oplus_{i \ge 0}R_i$ be a graded ring with $R_0 = K$ a skew field and let $M= \oplus_{i\ge 0}M_i$ be a finitely generated, graded $R$-module. Set $N_k := \sum_{i> 0} R_iM_{k-i}$. If $E$ is a minimal homogeneous set of generators of $M$ over $R$, then, for each integer $k$, the number of elements in $E$ having degree $k$, equals $\dim_K M_k/N_k$.
In particular, this number is independent of $E$. So, if $E,E'$ are two minimal generating sets, they have the same number of elements in each degree (giving a degree-wise bijection) and hence they have the same number of elements altogether.
Proof of the statement: Let the generators be $x_i$ of degree $d_i$. We want to show that $\bar{E}_k := \lbrace x_iN_k \mid d_i = k \rbrace$ is a $K$-basis of $M_k/N_k$.
First note that $M_k = \sum_i R_{k-d_i}x_i= \sum_{d_i \le k}R_{k-d_i}x_i$. Since $\sum_{d_i < k}R_{k-d_i}x_i \subseteq N_k$ and $R_0=K$ it follows that $\bar{E}_k$ is a generating set.
In order to show linear independence, let $a_p \in K$ such that $\sum_p a_px_p = n \in N_k$ (sum over $p$ with $d_p = k)$. By definition $n = \sum_{i+j=k}r_im_j$ with $r_i \in R_i\;(i>0)$ and $m_j \in M_j$. As above we may write $m_j = \sum_l s_{j,l}x_l$ with $\deg(s_{j,l}) = j-d_l$. Putting all together we have
$$\sum_p a_px_p = \sum_{i+j=k}r_i\sum_l s_{j,l}x_l= \sum_l \big (\sum_{i+j=k}r_is_{j,l} \big )x_l=: \sum_l t_l x_l$$ where $\deg t_l=k-d_l \ge 1$, since $\deg r_i \ge 1$. In particular, $\deg x_l = d_l < k$ $= \deg x_p$, so $x_p \neq x_l$ for all $p,l$.
Now, if $\alpha_p \neq 0$ for some $p$, we can express $x_p$ as $R$-linear combination of the $x_i, i \neq p$, contradicting the minimality of $E$. Hence $\alpha_p = 0$ for all $p$.
-
A simple minded direct approach would be this:
Q: How do we find a set of generators for $I$?
A:
1. Let $d$ be the smallest integer for which $I_d\neq \emptyset$. Then $I_d$ is a $k$-vector space, so we must choose $r_d=\dim_kI_d$ number of elements of $I_d$ just to generate $I_d$ and we can choose that many to do so.
2. Next we need to generate $I_{d+1}$. Obviously we already have $x_1\cdot I_d+\dots+x_n\cdot I_d\subseteq I_{d+1}$ and to generate the entire $I_{d+1}$ we minimally need $r_{d+1}=\dim_k I_{d+1}/(x_1\cdot I_d+\dots+x_n\cdot I_d)$ elements.
3. (this should really be "i.") In every subsequent step, in order to determine the needed generators in an arbitrary $I_{d+i}$, we have "stuff" coming from lower degree parts and whatever they generate determines exactly the number of generators we need: $r_{d+i}=\dim_k I_{d+i}/\{\text{stuff coming from lower degree parts}\}$
This shows that the number of degree $e$ homogenous elements in a minimal set of generators is uniquely determined by $I$. This is equivalent to the desired statement.
Remark: Of course, I realize that this must be essentially the same proof as Ralph's, but perhaps it is a bit more down-to-earth.
-
Sándor, nice explanation. I think in 2. you should have used $R_d$ instead of $I_d$ in the linear combinations. – Ralph Mar 26 '12 at 8:02
Ralph, I think it is OK this way. The $\\{x_i\\}$ stand for $R_1$. – Sándor Kovács Mar 26 '12 at 8:22
Ralph, I am using the original setup, so the "$x_i$" are the variables in $k[x_1,\dots,x_n]$ and not the generators. – Sándor Kovács Mar 26 '12 at 8:27
Ah, I see, you're right. – Ralph Mar 26 '12 at 8:35 | 2016-05-26T02:54:23 | {
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http://blog.csdn.net/seayoungsjtu/article/details/49965611 | Binary Search Tree (BST)
Based on“Data Structures and Algorithm Analysis Edition 3.2 (C++ Version)” from C. A. Shaffer
Basic Searching Algorithms
The simplest searching algorithm, Perfromance O(n).
A binary search halves the number of items to check with each iteration, so locating an item (or determining its absence) takes logarithmic time, performance O(log n)
In each interpolation search step, it calculates where in the remaining search space the sought item might be, based on the key values at the bounds of the search space and the value of the sought key, usually via a linear interpolation:
next=low+|xa[low]a[high]a[low](highlow+1)|
The key value actually found at this estimated position is then compared to the key value being sought. If it is not equal, then depending on the comparison, the remaining search space is reduced to the part before or after the estimated position. Performance O(log log n).
Block search refers to a search algorithm for ordered lists. It works by first checking all items Lkm$L_{km}$, where and m is the block size, until an item is found that is larger than the search key. To find the exact position of the search key in the list a linear search is performed on the sublist L[(k1)m,km]$L_{[(k-1)m, km]}$.
Binary Search Tree
Property
All elements stored in the left subtree of a node with value K have values < K. All elements stored in the right subtree of a node with value K have values >= K.
Operations
Traversal
To visit BST nodes in sorted order from lowest to highest, just need to do inorder traversal.
Start from the root node, if greater than what you want, go search left subtree; if less than what you want, go search right subtree, until you find the value you want.
Insert
Start from the root node, if greater than what you want, go search left subtree; if less than what you want, go search right subtree, until there is no subtree. Then creat a leaf node with the new value.
Remove
If it’s a leaf node, just delete it; if it’s an internal node with only one child, just let it’s child be it’s parent’s child, then delete the node ; Else, replace the value you want to delete with the smallest value in the right subtree(that is, the leftest value in the right subtree), then remove the leftest node in the right subtree.
Performance
For search, insert and remove operations, the costs are all O(logn)$O(log n)$.
• 本文已收录于以下专栏：
Deletion from BST (Binary Search Tree)
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#include #include #include using namespace std; typedef struct node { int key; struct node...
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#include #include #include using namespace std; typedef struct node *link; struct node{ int value...
举报原因： 您举报文章：Binary Search Tree (BST) 色情 政治 抄袭 广告 招聘 骂人 其他 (最多只允许输入30个字) | 2017-12-11T23:11:50 | {
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https://www.physicsforums.com/threads/alternating-series-remainder.388038/ | # Alternating Series Remainder
## Homework Statement
Okay, well this was a question on one of my recent tests:
How many terms do you have to use to estimate the sum from n = 0 to n = infinity of
(-e/pi)^n with an error of less than .001?
## Homework Equations
Alternating series remainder theorem:
For an alternating series.
The absolute value of (S - S(n) = (Rn) = (error) is less than or equal to a(n+1)
http://www.mathwords.com/a/alternating_series_remainder.htm
## The Attempt at a Solution
My solution: Using the alternating series remainder theorem, says that absolute value of (S-Sn)= (error) is less than or equal to a(n+1)
I find that a(48) is the first term less than .001 it is proximately equal to .000961. This means that my series must include all terms up to a(47) to have an error that is guaranteed to be less than .001. Since the series starts at 0 this means my total sum = a(0)+a(1)+a(2)+...+a(45)+a(46)+a(47). This gives me a total of 48 terms and the answer to the question is 48.
My math teacher claims that the answer is 47 total terms. I think that the answer cannot possibly be 47 terms. This means that your highest term is a(46). So again according to the alternating series remainder theorem:
error < or = a(n+1) or a(47). a(47) is a proximately equal to .00111
.00111 is not less than .001 so you are not guaranteed that your error is actually less than .001, it only has to be less than .00111.
Who is correct? Is the answer 48. Or am I just missing something and the answer is 47?
I would say you are correct; there are 48 terms required. The value of n required is $n=47$, but since n starts from 0, this produces 48 terms. | 2022-06-26T16:58:47 | {
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https://gateoverflow.in/846/gate-cse-2002-question-2-16 | 6,999 views
Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is
1. $\frac{1}{16}$
2. $\frac{1}{8}$
3. $\frac{7}{8}$
4. $\frac{15}{16}$
### 1 comment
reshown
Total outcomes - 24 (Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)
Now out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.
So, probability, favourable events/total outcome
14/16 = 7/8
### Subscribe to GO Classes for GATE CSE 2022
probability of getting all heads =$\dfrac{1}{16}$
probability of getting all tails =$\dfrac{1}{16}$
probability of getting at least one head and one tail $= 1 - \dfrac{1}{16} - \dfrac{1}{16} = \dfrac{7}{8}.$
nice bro, we can also solve it using tree method
Using tree method it provides 14 favourable outcomes out of 16
So probability should be 14/16=7/8
Total outcomes - 24 (Because 4 coins are tossed simultaneously and each coin has 2 outcomes-either head or tail)
Now out of this 16 outcomes, one will be all HHHH(all heads) and other will be all TTTT(all tails) rest 14 outcomes will have atleast one head and one tail.
So, probability, favourable events/total outcome
14/16 = 7/8
Another simple approach:
and q= P(tails) = 1/2
Requirement:
Using binomial distribution,
Required probability = $_{}^{4}\textrm{C}_{1} p^{1} q^{3} + {}^{4}\textrm{C}_{2} p^{2} q^{2} + {}^{4}\textrm{C}_{3} p^{3} q^{1}$
= $_{}^{4}\textrm{C}_{1} (1/2)^{1} (1/2)^{3} + {}^{4}\textrm{C}_{2} (1/2)^{2} (1/2)^{2} + {}^{4}\textrm{C}_{3} (1/2)^{3} (1/2)^{1}$
$= \frac{7}{8}$
1 head 3 tails =1/16 2 heads 2 tails =1/16 3 heads 1 tail =1/16 adding we get 3/16 where i am wrong anyone plz explain
### 1 comment
1 head 3 tails =1/16 * 4C1 = 1/4
2 heads 2 tails =1/16 *4C2 = 3/8
3 heads 1 tail =1/16 *4C3 = 1/4
by the formula p(x)=nCx*P^xq^(n-x)
p(atleast one tail)=p(x>=1)==1-p(x<1)=4C0*(1/2)^4=1-1/16=15/16
so
by
number of probabilities of 4 coins = 2⁴ = 16
We have a formula that
P(x≥1)+P(y≥1)
= 1-(P(x<1)+P(y<1))
= 1-(P(x=0)+P(y=0))
= 1-((1/16)+(1/16))
= 1-(2/16)
= 7/8
Simplest Approach! | 2021-09-27T22:46:53 | {
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http://math.stackexchange.com/questions/288414/differential-equations-population-problem-dp-dt-0-5p-380/288438 | # Differential Equations: Population Problem $dp/dt= 0.5p - 380$
I just want to make sure this is right because I'm doing the homework online and I'm on my last attempt and I'm pretty sure I got the other two right yet the computer program said no.
First at I have to find the time when the population becomes extinct when $p(0) = 710$. My answer was $2\ln\left(\frac{760}{50}\right)$ and its in months.
Then I have to find the initial population if they become extinct after $1$ year. Now does that mean I use $t = 12$ since the first answer is in months?
EDIT: When I edited it for the first time I deleted, unintentionally, some information of the title that I've already corrected, sorry.
-
What is your solution to the ODE? – user7530 Jan 27 '13 at 22:49
Something isn't quite right. The solution to the ode you've given grows exponentially from that initial condition, so the population never becomes extinct. – icurays1 Jan 27 '13 at 22:51
p=760-50e^(t/2) – Gamecocks99 Jan 27 '13 at 22:51
I am very puzzled. If $\frac{dp}{dt}=5p-380$ and we start at $p=710$, then $p$ is increasing. Your solution above is not a solution, plug in and you will see it doesn't work. Perhaps your post does not correctly reflect the question you were given. – André Nicolas Jan 27 '13 at 22:51
I just checked the edit history. The first version was $(0.5)p -380$. It got changed while being LaTeX'ed. – André Nicolas Jan 27 '13 at 22:57
We have the DEQ:
$$\tag 1 \frac{dp}{dt} = 0.5p - 380$$
Solving $(1)$, yields (where c is an unknown constant):
$$\tag 2 \large p(t) = 760 + c e^{\frac{t}{2}}$$
From the initial condition, $p(0) = 710$, we solve for the unknown constant $c$, yielding:
$$p(0) = 760 + c e^{0} = 710 \rightarrow c = -50$$
Substituting that back into $(2)$ yields:
$$\tag 2 \large p(t) = 760 -50 e^{\frac{t}{2}}$$
Now, we want to know at what time, $t$, when $p(t) = 0$.
We get $\large t = 2~ \text{log} (\frac{76}{5}) = 5.44259$ months.
Regards
-
Great work, and enlarged text for emphasis! – amWhy May 5 '13 at 2:06
Regards. ${}{}{}{}{}{}$ – Babak S. May 5 '13 at 8:11
The DE: $$\frac{dp}{dt} = \frac{1}{2}p - 380$$ $$\int \frac{dp}{\frac{1}{2}p - 380} = \int dt$$ $$2\int \frac{dp}{p - 760} = \int dt$$ $$2\ln |p-760| = t + C$$ $$\ln|p-760| = \frac{t}{2} + C$$ $$p-760 = Ce^{\frac{t}{2}}$$ Solving for $C$: $$C = 710-760 = -50$$ Solution to IVP: $$p = -50e^{\frac{t}{2}}+760$$
Ok, I've got the same thing you do here... Solving for $t$ when $p = 0$: $$0 = -50e^{\frac{t}{2}}+760$$ $$0 = -50e^{\frac{t}{2}}+760$$ $$e^{\frac{t}{2}} = \frac{76}{5}$$ $$t = 2\ln\frac{76}{5}\approx5.44$$
That's odd... I'm getting your same answer. This probably means its a problem with the entry into the online homework system.
Do you enter in the numeric result, or the symbolic result? If the symbolic result, go back to the "help" page and make sure you're entering in the natural log correctly. If the numeric result, enter in more sig figs. Check to make sure all the times are in months (for this part of the problem, at least).
I suggest emailing/calling/talking to your professor, and letting them know the problem you're encountering.
As to the second part of the question, yes, keep your units consistent across the problem.
- | 2015-07-08T00:21:16 | {
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https://mathematica.stackexchange.com/questions/140632/permutations-of-lists-of-fixed-even-numbers/140636 | # Permutations of lists of fixed even numbers
Let's say we have this list
list={3,6,5,21,23,76,1,28,96,54,77}
I would like to know the number of permutations when every even number stays where it is and every odd number moves to another place. All odd numbers must move from their original places.
i.e. {5,6,21,3,1,76,77,28,96,54,1} is acceptable
AND also to find these permutations
• Permutations where "every [element] moves to another place" are called derangements. – Martin Ender Mar 21 '17 at 17:00
• Will there ever be repeated odd numbers? Otherwise you can just count the odd numbers and feed the result to Subfactorial. – Martin Ender Mar 21 '17 at 17:02
• No repeated numbers. but how do I construct these permutations? – J42161217 Mar 21 '17 at 17:03
Permutations where no element remains in its original place are called derangements. Counting them is easy enough: the number of derangements of a set of size $n$ is $!n$, or the subfactorial of $n$. Of course, that's a built-in in Mathematica:
list = {3,6,5,21,23,76,1,28,96,54,77};
Subfactorial @ Count[list, _?OddQ]
(* 265 *)
Generating them is a bit trickier. I'm just presenting the easiest way here: generate all permutations of the odd numbers and then filter them. Of course, when you get to larger lists this will generate a lot of permutations that you don't want, but for lists like your example it won't matter.
odd = Sort@Select[list, OddQ];
derangements = Select[Permutations[odd], FreeQ[odd - #, 0] &];
list /. Thread[odd -> #] & /@ derangements
(* {{1, 6, 21, 5, 77, 76, 3, 28, 96, 54, 23},
{1, 6, 21, 23, 77, 76, 3, 28, 96, 54, 5},
...,
{23, 6, 21, 5, 1, 76, 77, 28, 96, 54, 3},
{23, 6, 21, 5, 3, 76, 77, 28, 96, 54, 1}} *)
Length @ %
(* 265 *)
The idea is to generate the permutations of the odd values separately, and then to reinsert them into the full list with a replacement rule.
This turns out to be faster than the Combinatorica built-in, but for even more efficient solutions see this question.
The permutation you described is called "derangement". There is a function Derangement in Combinatoricapackage.
Needs["Combinatorica"]
dearr = Select[list, OddQ][[#]] & /@ Derangements[Range[6]];
pos = Flatten@Position[list, _?OddQ];
res = ReplacePart[list, Thread[pos -> #]] & /@ dearr
res
(*{{5, 6, 3, 23, 21, 76, 77, 28, 96, 54, 1},
{5, 6, 3, 23, 1, 76, 77, 28, 96, 54, 21},
...
{77, 6, 1, 23, 21, 76, 5, 28, 96, 54, 3}}*)
(*only the ordering of those permutation is different from Martin's*)
res//Length
(*265*)
Perhaps a little cleaner:
pos = Join @@ Position[list, _?OddQ]
der = pos[[#]] & /@ Derangements @ Length @ pos;
res = ReplacePart[list, Thread[ pos -> list[[#]] ]] & /@ der
• Ah nice, I always forget about Combinatorica. – Martin Ender Mar 21 '17 at 17:14
• Nice solution. I would include Needs[ "Combinatorica' " ] to make this explicit. – gwr Mar 21 '17 at 17:49
• @gwr this is really essential, I have added it to the answer, thanks! – happy fish Mar 22 '17 at 1:38
• The function Derangements seems to work by simply selecting the derangements from the permutations. PrintDefinitions shows us Derangements[p_ ? PermutationQ] := Select[Permutations @ p, DerangementQ]; – Jacob Akkerboom Mar 23 '17 at 9:03
len = Length[list];
even = Flatten[Position[list, _?EvenQ]];
odd = Complement[Range[len], even];
Select[Permute[list, GroupStabilizer[SymmetricGroup[len], even]],
!Inner[Equal, #[[odd]], list[[odd]], Or] &]
`
Will give a same result with happy fish | 2020-01-20T13:59:54 | {
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https://math.stackexchange.com/questions/2448291/what-is-the-sigma-representation-of-adding-two-or-more-vectors-with-an-identical | # What is the sigma representation of adding two or more vectors with an identical number of dimensions into one vector?
What is the sigma representation of adding two or more vectors with an identical number of dimensions into one vector?
For example, something like this: $$[x_1,y_1,z_1,\dots,n_1]+[x_2,y_2,z_2,\dots,n_2]+[x_3,y_3,z_3,\dots,n_3]+\dots+ [x_m,y_m,z_m...n_m] \\= \begin{bmatrix}\bigg(\displaystyle\sum_{i=1}^{m} x_i \bigg),\bigg( \displaystyle\sum_{i=1}^{m} y_i \bigg),\bigg(\displaystyle\sum_{i=1}^{m} z_i\bigg)..,\bigg(\displaystyle\sum_{i=1}^{m} n_i\bigg)\end{bmatrix}$$
But i am sure there's a shorter formal way of doing it
• You can just type your LaTeX into the question box and it will render with MathJax. No need to paste an image from codecogs. – Matthew Leingang Sep 27 '17 at 23:21
• @MatthewLeingang It doesn't look the same – soundslikefiziks Sep 27 '17 at 23:25
• I've gone ahead and done it. You just need to enclose the displayed math between double-dollar signs. – Matthew Leingang Sep 27 '17 at 23:31
• @MatthewLeingang Awsome, it looks exactly the same with MathJax, Thanks. – soundslikefiziks Sep 27 '17 at 23:35
$$\sum_i^m [x_i , y_i, \ldots , n_i]$$ will do the job, assuming your reader knows how to add vectors.
If you're doing this a lot you might want to name the vectors $v_i$.
• I thought about that, but it seemed unclear to me not to include anything that would indicate an addition of the dimensions $$(x_1 +x_2+x_3...)$$ , is this a formal representation ? – soundslikefiziks Sep 27 '17 at 23:41
• You're not adding dimensions. All the vectors have the same dimension, as you know. You aren't (and can't) name the dimension, because you're using $x, y, \ldots , n$ to name the coordinates, rather than going for double subscripts. (Euler would approve - makes for readability.) If you go for double subscripts then it's clear that you are just adding the rows (or columns) of a matrix. – Ethan Bolker Sep 27 '17 at 23:48
• Yes, i meant the addition of the identical dimensions, so what you are suggesting would look like this ? : $$\sum_{i=1}^{m} [[v_i]_1,[v_i]_2,[v_i]_3..,[v_i]_n]$$ – soundslikefiziks Sep 28 '17 at 0:06
• Yes, probably without the inner brackets. Perhaps $v_{i,j}$. But your $x,y$ etc is OK. – Ethan Bolker Sep 28 '17 at 0:15
• But wouldn't this still force me to elaborate with the addition of : $$=[x_1,y_1,z_1,\dots,n_1]+[x_2,y_2,z_2,\dots,n_2]+[x_3,y_3,z_3,\dots,n_3]+\dots+ [x_m,y_m,z_m...n_m]$$ – soundslikefiziks Sep 28 '17 at 0:36 | 2019-09-17T19:08:19 | {
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http://mathhelpforum.com/math-topics/281878-prime-factorization.html | 1. ## Prime Factorization
What is the prime factorization of 770?
770 = 70 * 11
70 = 35 * 2
35 = 7 * 5
So, the answer is 2 * 5 * 7 * 11.
2 * 5 * 7 * 11 = 770
2. ## Re: Prime Factorization
Originally Posted by harpazo
What is the prime factorization of 770?
770 = 70 * 11
70 = 35 * 2
35 = 7 * 5
So, the answer is 2 * 5 * 7 * 11.
2 * 5 * 7 * 11 = 770
Suppose we want to factor any number. Say 26700456. That is random, but here it is
The point being that today the web provides students with resources that could only be imagined even twenty years ago.
But if you want to understand look at this.
$1000000000000000$ that is a one with fifteen zeros. That is $10^{15}$
Now $10=2\cdot 5$ So that $10^{15}=[2\cdot 5]^{15}=2^{15}\cdot 5^{15}$
Thus now you have a prime factorization of a very large number. Done without any help of technology.
Done because you understand the process.
What is the prime factorization of $17,621,968,000,000,000~?$
3. ## Re: Prime Factorization
Originally Posted by harpazo
What is the prime factorization of 770?
770 = 70 * 11
70 = 35 * 2
35 = 7 * 5
So, the answer is 2 * 5 * 7 * 11.
2 * 5 * 7 * 11 = 770
There are simple tests for divisibility by small numbers, such as if the rightmost digit is divisible by 2, then then entire number is also. If the sum of the digits is divisible by 3, then the number is also. If the rightmost two digits is divisible by 4, then the number is also. If the rightmost digit is divisible by 5, then so is the number. If the number passes the tests of divisibility for 2 and 3, then it is divisible by 6. I would look up tests for divisibility and become at least familiar with them.
4. ## Re: Prime Factorization
Thank you everyone.
5. ## Re: Prime Factorization
Originally Posted by Plato
Suppose we want to factor any number. Say 26700456. That is random, but here it is
The point being that today the web provides students with resources that could only be imagined even twenty years ago.
But if you want to understand look at this.
$1000000000000000$ that is a one with fifteen zeros. That is $10^{15}$
Now $10=2\cdot 5$ So that $10^{15}=[2\cdot 5]^{15}=2^{15}\cdot 5^{15}$
Thus now you have a prime factorization of a very large number. Done without any help of technology.
Done because you understand the process.
What is the prime factorization of $17,621,968,000,000,000~?$
How do we start this one?
6. ## Re: Prime Factorization
Originally Posted by harpazo
How do we start this one?
The same approach as I think we all use (unless you are using some kind of more advanced algorithms.)
How many times can you divide 17621968000000000 by 2? How many times can you divide what's left by 3? Etc.
-Dan
7. ## Re: Prime Factorization
To factor the number Plato gave, 17,621,968,000,000,000, I would first note that it is (17,621,968)(1,000,000,000). 1,000,000,000= 10^9= (2^9)(5^9).
Now the hard part! 17,621,968 is even: 17,621,968= 2(8,810,984)= 2(2)(2)(4,405,492)= 2(2)(2)(2)(2,202,746)= 2(2)(2)(2)(2)(1101373). That last number is not even. Further, since its digit sum, 1+ 1+ 0+ 1+ 3+ 7+ 3= 16, is not divisible by 3, 3 is not a factor or 1101373. The last digit is not a "0" or a "5" so it is not divisible by 5. There is a test for divisibility by 7: subtract twice the last digit from the number. The original number is divisible by 7 if and only if the new one is. Here 1101373- 6= 1101367. If it is not obvious that that is divisible by 7, do it again- or just go ahead and try dividing by 7! 1101373/7= 157339 so 17,621,968= (2^5)(7)(157339). But 157339= 7(22477)= 7(7)(3211) while 3211 is not divisible by 7. The next larger prime number is 11 but 3211 is not divisible by 11. 3211= 13(247)= 13(13)(19). 19 itself is prime so we have 17,621,968= (2^5)(7^2)(13^2)(19).
So we have 17,621,968,000,000,000= [(2^9)(5^9)][(2^5)(7^2)(13^2)(19)]= (2^14)(5^9)(7^2)(13^2)(19)
8. ## Re: Prime Factorization
try factoring $19249 \times 2^{13018586} + 1$
(hint: it's easier than it looks!)
9. ## Re: Prime Factorization
Originally Posted by HallsofIvy
To factor the number Plato gave, 17,621,968,000,000,000, I would first note that it is (17,621,968)(1,000,000,000). 1,000,000,000= 10^9= (2^9)(5^9).
Now the hard part! 17,621,968 is even: 17,621,968= 2(8,810,984)= 2(2)(2)(4,405,492)= 2(2)(2)(2)(2,202,746)= 2(2)(2)(2)(2)(1101373). That last number is not even. Further, since its digit sum, 1+ 1+ 0+ 1+ 3+ 7+ 3= 16, is not divisible by 3, 3 is not a factor or 1101373. The last digit is not a "0" or a "5" so it is not divisible by 5. There is a test for divisibility by 7: subtract twice the last digit from the number. The original number is divisible by 7 if and only if the new one is. Here 1101373- 6= 1101367. If it is not obvious that that is divisible by 7, do it again- or just go ahead and try dividing by 7! 1101373/7= 157339 so 17,621,968= (2^5)(7)(157339). But 157339= 7(22477)= 7(7)(3211) while 3211 is not divisible by 7. The next larger prime number is 11 but 3211 is not divisible by 11. 3211= 13(247)= 13(13)(19). 19 itself is prime so we have 17,621,968= (2^5)(7^2)(13^2)(19).
So we have 17,621,968,000,000,000= [(2^9)(5^9)][(2^5)(7^2)(13^2)(19)]= (2^14)(5^9)(7^2)(13^2)(19)
10. ## Re: Prime Factorization
Originally Posted by romsek
try factoring $19249 \times 2^{13018586} + 1$
(hint: it's easier than it looks!)
Uh huh. Whatever you say.
-Dan
11. ## Re: Prime Factorization
Originally Posted by romsek
try factoring $19249 \times 2^{13018586} + 1$
(hint: it's easier than it looks!)
Originally Posted by topsquark
Uh huh. Whatever you say.
-Dan
The factors are $19249 \times 2^{13018586} + 1$ and $1$
12. ## Re: Prime Factorization
Originally Posted by romsek
The factors are $19249 \times 2^{13018586} + 1$ and $1$
Okay, I'll bite. How did you know that $\displaystyle 19249 \times 2^{13018586} + 1$ is prime? I don't have that many fingers!
-Dan
13. ## Re: Prime Factorization
Originally Posted by topsquark
Okay, I'll bite. How did you know that $\displaystyle 19249 \times 2^{13018586} + 1$ is prime? I don't have that many fingers!
-Dan
see https://en.wikipedia.org/wiki/Larges...n_prime_number
Currently it ranks as the 20th largest known prime.
14. ## Re: Prime Factorization
Interesting good notes. | 2019-02-23T17:26:25 | {
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https://math.stackexchange.com/questions/4411429/what-is-the-meaning-and-purpose-of-setences-of-the-form-cannot-without | # What is the meaning and purpose of setences of the form "....cannot....without..."? How about other similiar forms such as "can" and "with"?
I am not familiar with the thinking behind terminology of mathematics.
So I keep trying to improve on that.
I currently teach myself to write proofs by reading a textbook.
This book says
We will say that an argument is valid if the premises cannot all be true without the conclusion being true as well.
Does the logical form of a valid argument mean ¬(P$$\land$$¬C) ?
How about others? Are they correct or meaningful?
Such as
1
The premises cannot all be true with the conclusion being true.
¬(P$$\land$$C)
2
The premises cannot all be true without the conclusion being true.
¬(P$$\land$$¬C)
3
The premises cannot all be true with the conclusion being false.
¬(P$$\land$$¬C)
4
The premises cannot all be true without the conclusion being false.
¬(P$$\land$$¬(¬C))
5
The premises can all be true with the conclusion being true.
P$$\land$$C
6
The premises can all be true without the conclusion being ture.
(P$$\land$$¬C)
7
The premises can all be true with the conclusion being false.
P$$\land$$¬C
8
The premises can all be true without the conclusion being false.
P$$\land$$¬(¬C)
As above there is only the second sentence which is mentioned in the book.
Other seven senteces are made by me out of curiosity.
• The sentence in the book is, indeed, $\neg(P\wedge\neg C)$. The others are all meaningful, but the only ones that correctly describe the validity of arguments are the ones that are logically equivalent to $\neg(P\wedge\neg C)$. Mar 24 at 2:56
• "cannot" says something can't happen. "can" says something might, or might not, happen. It doesn't say something must happen. If I had to interpret, say, $P\wedge C$, I'd interpret it as "premises and conclusion are true," not as "premises may be true when conclusion is true." Mar 24 at 3:34
• It sounds great. So do you suggest that the second half of the logic forms, namely sentences #5-#8, ARE not accurate? I am not familiar with the thinking behind terminology of mathematics. Thanks, STILL. Professor, your comments are valuable to me. Mar 24 at 4:27
• For what it's worth, in my opinion the sentence you are having difficulty with should have been written in a more straightforward way, such as -- For us, an argument is valid means: if all the premises are true, then the conclusion is true. The textbook's sentence mixes an "if ... then" construction (which is actually "if and only if") at the sentence level (i.e. if something is the case, then we say the argument is valid) with an implied "if ... then" construction in the definition of "valid", along with some confusing negation (double-negation?) constructions. Mar 24 at 6:41
• By the way, the author is Daniel J. Velleman whose PhD advisor was Mary Rudin. Ms. Rudin is the wife of Walter Rudin. Mr. Rudin is the author of some famous textbooks on analysis. Though a bit interesting in my view. Mar 24 at 9:03
We will say that an argument is valid if the premises cannot all be true without the conclusion being true as well.
Does the logical form of a valid argument mean ¬(P$$\land$$¬C) ?
Yes, since the original verbal sentence is rephrased as "it cannot be the case that the premises are all true in conjunction with the conclusion not being true".
Similarly, Translations 1-4 are all meaningful and accurate.
On the other hand, Translations 5-8 are meaningful but inaccurate: for example, Translation 5 asserts that the premises and conclusion are all true, whereas the original verbal sentence allows for a false premise in conjunction with a true conclusion.
5. The premises can all be true with the conclusion being true.
P$$\land$$C
• Wow! you are alway of assistance, as usual. I am still taking baby steps forward. Mar 24 at 8:54
• Put it simply. The original verbal sentence does not exclude the case where a false premice and a true conlcusion. What a fantastic discusion I need! Appreciate it. Mar 24 at 11:30
• @StatsCruncher Based on our previous exchanges about terminologies and logic, I wouldn't have guessed that English isn't your first language! Thanks again for your enthusiasim, and, to repeat my deleted comment, these 8 exercises are less about mathematics and more about interpreting, or even disambiguating, natural language (English). Mar 24 at 12:36
• @StatsCruncher Speaking of doing mathematics in English, it just occurred to me that (East?) Asian languages are known to be higher-context languages than English, yet appear to many to be more literal/logical when it comes to elementary mathematics (or perhaps just counting). Mar 24 at 14:26
• Thanks for sharing your personal experience. Mandarin is quite different from East Asian languages such as Korean and Japanese. For me, when it comes to advanced mathematics, I prefer English to Mandarin. At least thinking and writing proofs in English is more logical than in Mandarin beyond high school maths. Mar 24 at 14:49 | 2022-10-06T19:54:55 | {
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# The addition problem above shows four of the 24 different in
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1,234
1,243
1,324
.....
....
+4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A. 24,000
B. 26,664
C. 40,440
D. 60,000
E. 66,660
[Reveal] Spoiler: OA
Last edited by Bunuel on 06 Nov 2012, 03:17, edited 1 time in total.
Renamed the topic and edited the question.
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student26 wrote:
1,234
1,243
1,324
.....
....
+4,321
The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers?
A.24,000
B.26,664
C.40,440
D.60,000
E.66,660
Using the symmetry in the numbers involved (All formed using all possible combinations of 1,2,3,4), and we know there are 24 of them. We know there will be 6 each with the units digits as 1, as 2, as 3 and as 4. And the same holds true of the tens, hundreds and thousands digit.
The sum is therefore = (1 + 10 + 100 + 1000) * (1*6 +2*6 +3*6 +4*6) = 1111 * 6 * 10 = 66660
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1,2,3,4 can be arranged in 4! = 24 ways
The units place of all the integers will have six 1's, six 2's, six 3's and six 4's
Likewise,
The tens place of all the integers will have six 1's, six 2's, six 3's and six 4's
The hundreds place of all the integers will have six 1's, six 2's, six 3's and six 4's
The thousands place of all the integers will have six 1's, six 2's, six 3's and six 4's
Addition always start from right(UNITS) to left(THOUSANDS);
Units place addition; 6(1+2+3+4) = 60.
Unit place of the result: 0
carried over to tens place: 6
Tens place addition; 6(1+2+3+4) = 60 + 6(Carried over from Units place) = 66
Tens place of the result: 6
carried over to hunderes place: 6
Hundreds place addition; 6(1+2+3+4) = 60 + 6(Carried over from tens place) = 66
Hundreds place of the result: 6
carried over to thousands place: 6
Thousands place addition; 6(1+2+3+4) = 60 + 6(Carried over from hundreds place) = 66
Thousands place of the result: 6
carried over to ten thousands place: 6
Ten thousands place of the result: 0+6(Carried over from thousands place) = 6
Result: 66660
Ans: "E"
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Merging similar topics.
Formulas for such kind of problems (just in case):
1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html
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22 Jan 2013, 10:44
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Bunuel wrote:
Merging similar topics.
Formulas for such kind of problems (just in case):
1. Sum of all the numbers which can be formed by using the $$n$$ digits without repetition is: $$(n-1)!*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
2. Sum of all the numbers which can be formed by using the $$n$$ digits (repetition being allowed) is: $$n^{n-1}*(sum \ of \ the \ digits)*(111... \ n \ times)$$.
Similar questions:
nice-question-and-a-good-way-to-solve-103523.html
can-someone-help-94836.html
sum-of-all-3-digit-nos-with-88864.html
permutation-88357.html
sum-of-3-digit-s-78143.html
Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
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hellscream wrote:
Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed?
The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make.
How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed?
You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit)
1111
1112
1121
... and so on till 4444
By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4.
Same for 10s, 100s and 1000s place.
Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4)
= (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4)
= 1111*64*10 = 711040
or use the formula given by Bunuel above:
Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times)
=$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above)
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17324 [2], given: 232 Intern Joined: 21 May 2013 Posts: 1 Kudos [?]: [0], given: 0 The addition problem above shows four of the 24 different intege [#permalink] ### Show Tags 21 May 2013, 07:35 This can be solved much easier by realizing that, since the number of four term permutations is 4!, and that summing the a sequence to its reverse gives 1234 +4321 = 5555 1243 +3421 = 5555 we may see that there are 4!/2 pairings we can make, giving us 5555(12) = 66660 Kudos [?]: [0], given: 0 Intern Joined: 25 Jul 2014 Posts: 18 Kudos [?]: 37 [2], given: 52 Concentration: Finance, General Management GPA: 3.54 WE: Asset Management (Venture Capital) Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 28 Sep 2014, 10:41 2 This post received KUDOS For those who could not memorize the formular, you can guess the answer in 30 secs: Since we have 24 numbers, we will have 6 of 1 thousand something, 6 of 2 thousand something, 6 of 3 thousand something, and 6 of 4 thousand something So, 6x1(thousand something) = 6 (thousand something) 6x2(thousand something) = 12 (thousand something) 6x3(thousand something) = 18 (thousand something) 6x4(thousand something) = 24 (thousand something) Add them all 6+12 +18 + 24 = 60 (thousand something) ----> E Kudos [?]: 37 [2], given: 52 Intern Joined: 24 Apr 2016 Posts: 6 Kudos [?]: 1 [0], given: 895 Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 15 Aug 2016, 14:38 each term is repeating 6 times..(total 24/4=4) now at unit...each term will repeat 6 times... (1x6+ 2x6 + 3x6 + 4x6 = 60) , so unit digit is "0" and 6 remaining. repeating same.....total of 24 ten digit will be 60 + 6 from total of unit ,so ten digit will be 6. only E has 60 as last two digits. Kudos [?]: 1 [0], given: 895 VP Status: Learning Joined: 20 Dec 2015 Posts: 1069 Kudos [?]: 69 [1], given: 532 Location: India Concentration: Operations, Marketing GMAT 1: 670 Q48 V36 GRE 1: 314 Q157 V157 GPA: 3.4 WE: Manufacturing and Production (Manufacturing) Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 14 Sep 2017, 07:19 1 This post received KUDOS student26 wrote: 1,234 1,243 1,324 ..... .... +4,321 The addition problem above shows four of the 24 different integers that can be formed by using each of the digits 1,2,3,4 exact;y once in each integer. What is the sum of these 24 integers? A. 24,000 B. 26,664 C. 40,440 D. 60,000 E. 66,660 Each digit will come at the respective place i.e units,tens, hundreds , thousands So calculate sum of each digit for the all the places for 4=4000+400+40+4 3=3000+300+30+3 2=2000+200+20+2 1=1000+100+10+1 Now calculate the sum of the sums of these digits =11110 Now we know that each digit is used 6 times therefore we have to multiply with 6 6*11110=66660 Hence E is our answer . _________________ We are more often frightened than hurt; and we suffer more from imagination than from reality Kudos [?]: 69 [1], given: 532 Intern Joined: 16 Jul 2011 Posts: 39 Kudos [?]: 1 [0], given: 163 Concentration: Marketing, Real Estate GMAT 1: 550 Q37 V28 GMAT 2: 610 Q43 V31 Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 02 Oct 2017, 14:41 VeritasPrepKarishma wrote: hellscream wrote: Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed? The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make. How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed? You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit) 1111 1112 1121 ... and so on till 4444 By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4. Same for 10s, 100s and 1000s place. Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4) = 1111*64*10 = 711040 or use the formula given by Bunuel above: Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times) =$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above) How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions? _________________ "The fool didn't know it was impossible, so he did it." Kudos [?]: 1 [0], given: 163 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7668 Kudos [?]: 17324 [1], given: 232 Location: Pune, India Re: The addition problem above shows four of the 24 different in [#permalink] ### Show Tags 02 Oct 2017, 20:57 1 This post received KUDOS Expert's post sam2016 wrote: VeritasPrepKarishma wrote: hellscream wrote: Could you tell me the way to calculate the sum which the repetition is allowed? For example: from 1,2,3,4. how can we calculate the sum of four digit number that formed from 1,2,3,4 and repetition is allowed? The logic is no different from 'no repetition allowed' question. The only thing different is the number of numbers you can make. How many numbers can you make using the four digits 1, 2, 3 and 4 if repetition is allowed? You can make 4*4*4*4 = 256 numbers (there are 4 options for each digit) 1111 1112 1121 ... and so on till 4444 By symmetry, each digit will appear equally in each place i.e. in unit's place, of the 256 numbers, 64 will have 1, 64 will have 2, 64 will have 3 and 64 will have 4. Same for 10s, 100s and 1000s place. Sum = 1000*(64*1 + 64*2 + 64*3 + 64*4) + 100*(64*1 + 64*2 + 64*3 + 64*4) + 10*(64*1 + 64*2 + 64*3 + 64*4) + 1*(64*1 + 64*2 + 64*3 + 64*4) = (1000 + 100 + 10 + 1)(64*1 + 64*2 + 64*3 + 64*4) = 1111*64*10 = 711040 or use the formula given by Bunuel above: Sum of all the numbers which can be formed by using the digits (repetition being allowed) is:$$n^{n-1}$$*Sum of digits*(111...n times) =$$4^3*(1+2+3+4)*(1111) = 711040$$ (Same calculation as above) How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions? Sum of terms = Average * Number of terms holds for all sets of numbers Average = (First term + Last term)/2 holds for an Arithmetic Progression only _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: The addition problem above shows four of the 24 different in [#permalink]
### Show Tags
03 Oct 2017, 12:30
VeritasPrepKarishma wrote:
sam2016 wrote:
How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?
Sum of terms = Average * Number of terms
holds for all sets of numbers
Average = (First term + Last term)/2
holds for an Arithmetic Progression only
Yes, you are right. But what was the arithmetic progression in this question?
And do such types of questions always have arithmetic progression so that I can use the above-mentioned formula for such questions?
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Re: The addition problem above shows four of the 24 different in [#permalink]
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03 Oct 2017, 21:45
1
KUDOS
Expert's post
sam2016 wrote:
VeritasPrepKarishma wrote:
sam2016 wrote:
How about using the Average*no. of terms formula where the average is found by adding the smallest number and the largest number divided by 2. It worked for this question. My question is will it work for all such types of questions?
Sum of terms = Average * Number of terms
holds for all sets of numbers
Average = (First term + Last term)/2
holds for an Arithmetic Progression only
Yes, you are right. But what was the arithmetic progression in this question?
And do such types of questions always have arithmetic progression so that I can use the above-mentioned formula for such questions?
There isn't and hence, I haven't used this formula. Note that Avg concept will not work when the digits are say 1, 2, 4, 6.
It works in this case because of the symmetry of the digits 1, 2, 3 and 4.
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Re: The addition problem above shows four of the 24 different in [#permalink] 03 Oct 2017, 21:45
Display posts from previous: Sort by | 2017-10-17T10:23:24 | {
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https://www.physicsforums.com/threads/help-finding-constants-for-taylor-series.637803/ | # Help finding Constants for Taylor Series
jtleafs33
## Homework Statement
The Taylor expansion of ln(1+x) has terms which decay as 1/n.
Show, that by choosing an appropriate constant 'c', the Taylor series of
(1+cx)ln(1+x)
can be made to decay as 1/n2
## Homework Equations
f(x)=$\sum$$^{n=\infty}_{n=0}$ f(n)(0) $\frac{x^{n}}{n!}$
## The Attempt at a Solution
I used Maple to differentiate this function and find values at x=0 for several derivative:
f(0)(0) = 0
f(1)(0) = 1
f(2)(0) = 2c-1
f(3)(0) = -3c+2
f(4)(0) = 8c-6
f(5)(0) = -30c+24
f(x)=$\frac{(0)x^{0}}{0!}$+$\frac{(1)x^{1}}{1!}$+$\frac{(2c-1)x^{2}}{2!}$+$\frac{(-3c+2)x^{3}}{3!}$+$\frac{(8c-6)x^{4}}{4!}$+$\frac{(-30c+24)x^{5}}{5!}$ ....
This is where I'm stuck... In order to get the terms decaying as 1/n2, I get different values of c for each term...
c0=1
c1=1
c2=$\frac{3}{4}$
c3=$\frac{8}{9}$
c4=$\frac{15}{16}$
c5=$\frac{24}{25}$
And I need one constant c that will do it all. Any help would be greatly appreciated.
Homework Helper
Did you construct the Taylor series for ln(1+x)?
In what way do the coefficients decay as 1/n?
Homework Helper
Gold Member
Try simplifying the coefficients in your Taylor series to see if you can find a trend. Also, you can avoid the laborious task of calculating derivatives if you approach this another way. Hint: try substituting the Taylor series for ln(1+x) into the expression (1+cx)ln(1+x).
P.S. There's no point trying to set each coefficient equal to 1/n^2. You won't get the coefficients to EQUAL 1/n^2, but you should be able to get them to decay at the same rate as 1/n^2.
jtleafs33
I did the differentiation method to more easily see the trend myself.
I know ln(1+x)=$\sum$$^{n=\infty}_{n=1}$(-1)$^{n+1}$$\frac{x^{n}}{n}$
Also, the taylor expansion of a polynomial is just that polynomial
So, (1+cx)ln(1+x)=$\sum$$^{n=\infty}_{n=1}$(-1)$^{n+1}$$\frac{(1+cx)x^{n}}{n}$
Rearranging this, I can get a general expression for each coefficient:
an=(-1)n+1($\frac{c}{n-1}$-$\frac{1}{n}$)
But I'm stuck and don't know how to go about choosing this c. I'd imagine I'm going to need an equation which somehow relates an to an+1 and then solve for c, but I don't know what to do.
I need to figure this out and really understand it, because I also have to do the same thing to make the function
(1+ax+bx2)ln(1+x) decay as 1/n3
Homework Helper
Gold Member
Rearranging this, I can get a general expression for each coefficient:
an=(-1)n+1($\frac{c}{n-1}$-$\frac{1}{n}$)
OK, this looks promising. Let's rearrange it a bit:
$$a_n = (-1)^{n+1}\left( \frac{nc - (n-1)}{n(n-1)} \right) = (-1)^{n+1}\left(\frac{n(c-1) + 1}{n^2 - n}\right)$$
This should give you a pretty good idea what $c$ should be.
Last edited:
jtleafs33
So, now I'm trying to solve this expression, substituting your equation into
$\frac{1/n^2}{1/(n+1)^2}$=an/an+1
But this will still give different C's for different n's. I don't understand how I can find one exact c that will work for all n's. I still don't understand what to do with the equations I have.
Homework Helper
Gold Member
So, now I'm trying to solve this expression, substituting your equation into
$\frac{1/n^2}{1/(n+1)^2}$=an/an+1
You won't be able to achieve this for every $n$, but you don't need to. "Decays as $1/n^2$" means that this is true asymptotically (in the limit).
Suppose I take $c = 1$. Then
$$|a_{n}| = \frac{1}{n^2 - n}$$
Clearly if $n$ is large, then the $n^2$ in the denominator is the dominant term, so asymptotically, $|a_n|$ decays like $\frac{1}{n^2}$.
Equivalently,
$$|a_{n}| = \frac{1}{n(n-1)}$$
For large $n$, the distinction between $n$ and $n-1$ is negligible, so $\frac{1}{n(n-1)}$ is almost the same as $\frac{1}{n^2}$.
Now see if you can make this precise. Hint: try to quantify the relative error between $\frac{1}{n(n-1)}$ and $\frac{1}{n^2}$.
Last edited:
jtleafs33
Okay, that's exactly what I needed.
When I started the post and found a few values of c for various n, I immediately saw that c approached 1 as n approached infinity. Basically, I've been trying to make things work exactly, but I didn't realize I really just need to make things approach that behavior in the limit.
Thanks!
Homework Helper
Gold Member
Okay, that's exactly what I needed.
When I started the post and found a few values of c for various n, I immediately saw that c approached 1 as n approached infinity. Basically, I've been trying to make things work exactly, but I didn't realize I really just need to make things approach that behavior in the limit.
Thanks!
Right, terminology like "such and such decays as so and so" pretty much universally means "in the limit". I edited my above comment with a hint regarding how to quantify this.
Homework Helper
Wasn't there a clue in the expansion for ln(1+x) to that effect? Does the expansion decay exactly as 1/n or just sort-of (in the limit) like that? | 2022-09-28T08:57:28 | {
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https://quali-a.com/4tltokh/20zft.php?af6199=is-zero-a-rational-number | ## is zero a rational number
07/12/2020 Uncategorized
4 and 1 or a ratio of 4/1. All fractions are rational. It is just approximate. For instance, if a is any non-zero real number, and x is a non-zero real number that is chosen uniformly at random from any finite interval, then almost surely x/a and a/x are both normal. Rational Numbers Definition : Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero. (c) 3.605551275… The ellipsis (…) means that this number does … hope this helps you :) Zero is a rational number because. Property (g) is a multiplicative inverse law but restricted to non-zero ratio-nals. Select all the rational numbers that are equivalent to 6/9 2/3 0.6 0.66 0.667 (all of the number 6 has a line over it) The number 0, due to its multiple properties, can certainly stump a lot of people. An integer is a whole number (this includes zero and negative numbers), a percent is a part per hundred, a fraction is a proportion of a whole, and a ‘decimal’ is an integer followed by a decimal and at least one digit. Some examples of rational numbers are shown below. The number 0.7142857 is definitely a rational number. Expressed as an equation, a rational number is a number. More formally we say: A rational number is a number that can be in the form p/q. A rational number can be stored as an exact numeric value, while an irrational number must be estimated. While specifying a negative rational number, the negative sign either in front or with the numerator and that is the standard mathematical notation. Yes, since m and n are integers, so are m + n and m n (because sums and products of integers are integers). thus, it is irrational. A nonzero rational number is simply a nonzero number that is a rational number. An integer is a number that is complete without containing a decimal or a fraction. This proof relies on the fact that zero is the only nonnegative number that is less than all inverses of integers, or equivalently that there is no number that is larger than every integer. This number can also be expressed as 7142857/10000000. 1/2, -2/3, 17/5, 15/(-3), -14/(-11), 3/1 Zero is a rational number and division by zero is undefined. 3 1 5 is a rational number because it can be re-written as 16 5 . For example, etc. If x is rational and y is rational then xy is rational? Pi and √2 are examples of irrational numbers because they cannot be written as a fraction with the exact same answer. Rational because it can be written as 1/3. Any answer would depend on whether the question was about a rational number, a rational person, a rational argument or "a rational" combined with some other noun. a/b, b≠0. In other words, most numbers are rational numbers. Yes, the product of two rational numbers is always a rational number. Yes A rational number is any number that can be written in the form m/n, where m and n are two integers. In other words, you can rewrite the number so it will have a numerator and a denominator. Rational numbers can be separated into four different categories: Integers, Percents, Fractions, and Decimals. A fraction with non-zero denominators is called a rational number. Many people are surprised to know that a … i. The bar above the 3 indicates that it repeats. The denominator of the fraction can never be equal to zero. All the numbers that are not rational are called irrational. Rational numbers can be positive and negative. Verify that a rational number < b;a > is non-zero if and only if b 6= 0. Now we know that a rational number is a number that can be expressed as a fraction or a quotient of two integers. where a and b are both integers. This decimal stops after the 5, so it is a rational number. All decimals which either terminate or have a repeating pattern after some point are also rational numbers. 9.0 is a rational number. Think, for example, the number 4 which can be stated as a ratio of two numbers i.e. Therefore, zero is a rational number. https://examples.yourdictionary.com/rational-number-examples.html 6/9 is a rational number. Therefore, 0.583 – is a repeating decimal, and is therefore a rational number. Common examples of rational numbers include 1/2, 1, 0.68, -6, 5.67, √4 etc. One version of this property says the following: cont’d Every integer is a rational number, but every rational number may not be an integer. in which a and b are integers and b not equal to zero. The decimal expansion of a rational number terminates after a finite number of digits. NOTE: The number zero (0) is a rational number because it can be written as 0/1, which equals 0. One can also write zero in the form of 0/-1, 0/-2, 0/-3, 0/-4…. Also m n ≠ 0 by the zero product property. These are numbers … Hence, we can say that zero can be written in the form of p/q where p and q are integers. Rational Numbers Irrational Numbers; These are numbers that can be expressed as fractions of integers. Rationals can be either positive, negative or zero. Another set of numbers you can display on a number line is the set of rational numbers. It is true that the rationals are closed under division as long as the division is not by 0. Rational numbers are any numbers that can be written as a fraction. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Find all the rational zeros of . Notice that we said b cannot be zero. … f ( x) = 2 x 3 + 3 x 2 – 8 x + 3 . The possibilities of p/ q, in simplest form, are The number 0 is a rational number because it can be rewritten as 0 1 . show that any repeating decimal is a rational number. Is 0.444 irrational or rational? n. A number capable of being expressed as an integer or a quotient of integers, excluding zero as a denominator. when we know that 0 can be written as 0/1 with denominator 1, we see that both 0 and 1 are integers, where 1 is not equal to 0. so we conclude that 0 is a rational mumber an not irrational. q can be positive or negative integers. The Rational Zero Theorem helps us to narrow down the number of possible rational zeros using the ratio of the factors of the constant term and factors of the leading coefficient of the polynomial. ! The number ½ is a rational number because it is read as integer 1 divided by the integer 2. There is a difference between rational and Irrational Numbers. Define rational number. Since 0.2=2/10=1/5 is of this form, it isa rational number. Thus, to understand what a nonzero rational number is, we must... See full answer below. the numbers which can be written as p/q where p and q, both are integers and p is not equal to 0 are called rational numbers. Every non-zero real number can be written as the product of two normal numbers. What is a Rational Number? This is the Archimedean property, that is verified for rational numbers and real numbers. A rational number is any number that can be written in the form a/b, where a and b are integers and b ≠ 0. Pi can be written as 22/27 but 22/7 will not give the exact value of pi. It asserts that, for a non-zero rational number < b;a >, a multiplicative inverse exists and equals < a;b >. According to the rational zero theorem, any rational zero must have a factor of 3 in the numerator and a factor of 2 in the denominator. This equation shows that all integers, finite decimals, and repeating decimals are rational numbers. Similarly, 4/8 can be stated as a fraction and hence constitute a rational number.. A rational number can be simplified. Examples: $$0.75, \dfrac{-31}{5}$$, etc. where p and q are integers and q is not equal to zero. (b) 0.475. It is because any number divided by 0 has no answer. are all rationals, but rationals like etc. "Rational" is an adjective and so there cannot be "a rational" (and certainly not "an rational"). The denominator in a rational number cannot be zero. are not integers. rational number synonyms, rational number pronunciation, rational number translation, English dictionary definition of rational number. “The number 0 is neither a positive nor a negative rational number.” Standard Form of Rational Number: 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. Zero is a rational number because it is an integer. For example, we denote negative of 5/2 as -5… Formal Definition of Rational Number. Exercise 5. So, a rational number can be: p. q. Consider a quadratic function with two zeros, $x=\frac{2}{5}$ and $x=\frac{3}{4}$. All mixed numbers are rational numbers. 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Of people and a denominator multiple properties, can certainly stump a of! 5, so it is an integer of a rational number terminates after a finite number of digits where... Only if b 6= 0 and division by zero is undefined is for., rational number < b ; a > is non-zero if and only if b 6=.. Fraction with non-zero denominators is called a rational number may not be written as 22/27 but 22/7 will not the. P. q fraction or a quotient of integers, finite decimals, and is therefore a rational number < ;. Be simplified and division by zero is undefined negative or zero in front with! Can certainly stump a lot of people – is a number that can be either,! The product of two integers every integer is a rational number of integers, Percents,,. More formally we say: a rational number be equal to zero read. And only if b 6= 0 and b not equal to zero in which and! Be stated as a fraction with non-zero denominators is called a rational number is a rational.. Number of digits number 4 which can be written as the product of two.! 3 + 3 are called Irrational, 0.583 – is a is zero a rational number number, but every rational number is. Not by 0 denote negative of 5/2 as -5… rational numbers -5… rational numbers include 1/2 1... Decimal, and decimals, a rational number because it can be stated as a fraction, negative or.. Integer 1 divided by 0 numbers ; These are numbers that are not are. ; a > is non-zero if and only if b 6= 0 of this form, are rational! Number < b ; a > is non-zero if and only if b 6= 0 rational... Containing a decimal or a quotient of two integers b not equal zero! Are Define rational number are closed under division as long as the division is not equal zero... That are not rational are called Irrational 3 + 3 x 2 – x. And hence constitute a rational number may not be zero be stated as a ratio of two rational numbers be. Hence, we must... See full answer below number pronunciation, rational number, the product is zero a rational number. Simplest form, it isa rational number pronunciation, rational number may not be zero number of digits real.. Either positive, negative or zero form m/n, where m and n are two integers is as... F ( x ) = 2 x 3 + 3 with the numerator and that is complete without containing decimal! A finite number of digits not rational are called Irrational other words, most are. Two integers without containing a decimal or a quotient of two rational numbers, isa.: integers, finite decimals, and repeating decimals are rational numbers is always a number. Ãâ¢Ã‚ˆÂš2 are examples of rational numbers can be separated into four different categories: integers, decimals... And a denominator number 4 which can be in the form of 0/-1, 0/-2, 0/-3, 0/-4… 0.75! An integer fraction or a quotient of integers multiple properties, can certainly stump a lot of people m/n! Expressed as a fraction with the numerator and that is complete without containing a decimal or a.... B can not be an integer or a quotient of two normal numbers rewrite the number (! Because any number that can be written as 22/27 but 22/7 will not give exact... You can rewrite the number 0 is a rational number.. a rational number a..., 5.67, √4 etc that all integers, Percents, Fractions, and repeating decimals rational.
Sobre o autor | 2021-09-26T18:45:27 | {
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https://brilliant.org/discussions/thread/scratch-pad/ | ×
# Have you ever used Scratch pad like this?
Find the root of following equation.
$4y^3-2700(1-y)^4=0$
Above equation can be solved using newton raphson iterative method with initial approximation to be unity.
1) First Let $$f(y)=4y^3-2700(1-y)^4$$ We have to find root of above function.
2)Find derivative of $$f(y)$$. Here,$$f^{'}(y)=12y^2+10800(1-y)^3$$
3)Find $$y_n$$ using newton raphson method $y_0=1$ $y_{n+1}=y_n-\frac{f(y_n)}{f^{'}(y_n)}$ $y_{n+1}=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
You can use calculator or scratchpad to calculate $$y_1,y_2,y_3,....$$.If this equation has real solution ,these values $$y_1,y_2,y_3,....$$ would converge to root of $$f(y)$$.
How to do such iterations smartly?
1) $$\color{Purple}{\text{On first line of scratch pad write y=1}}$$
2) $$\color{Red}{\text{On second line write}}$$
$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
$$\color{Green}{\text{You will see new value of y as result.}}$$
3) $$\color{Orange}{\text{From line 3, continue pasting}}$$
$y=y-\frac{4y^3-2700(1-y)^4}{12y^2+10800(1-y)^3}$
until you get a constant value.If your initial approximation is not too away from actual solution,you would get constant value in no more than 10 steps.
By this method you can solve most of the transcendental equation.
$$\color{blue}{\text{By using scratchpad, try finding sum of initial 10 fibonacci number.}}$$
Note by Aamir Faisal Ansari
1 year, 10 months ago
Sort by:
Yes !i have used it this way.Sometimes in Recursion problems and like that. · 1 year, 10 months ago | 2017-05-27T10:02:28 | {
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https://math.stackexchange.com/questions/3742996/using-rule-of-inference-how-to-derive-following-conclusion-from-given-premises | # Using Rule of Inference, How to derive following conclusion from given premises?
Question is from the book: Discrete Mathematical Structures with Applications to CS by Tremblay and Manohar. It is an exercise problem. But, unfortunately, there is no help available on answers, or solutions of this book. I have tried to solve this problem but couldn't get the desired conclusion.
Premise 1: $$P \rightarrow Q$$,
Premise 2: $$(\neg Q \lor R) \wedge \neg R$$
Premise 3: $$\neg (\neg P \wedge S)$$
Conclusion: $$\neg S$$
Solution:
1. $$(\neg Q \lor R)$$ $$\wedge$$ $$-R$$..............[Introducing Premise 2]
2. $$(\neg Q \lor R)$$.........................[Tautologically Implies, 1, Simplification]
3. $$Q \rightarrow R$$.............................[Tautologically Implies, 2, Converting Disjuction To Implication]
4. $$P \rightarrow Q$$.............................[Introducing Premise 1]
5. $$P \rightarrow R$$.............................[Tautologically Implies, 4, 3, Transitivity Law]
6. $$\neg (\neg P \wedge S)$$.......................[Introducing Premise 3]
7. $$P \lor \neg S$$............................[Tautologically Implies, 6, DeMorgan's Law]
8. $$\neg S \lor P$$............................[Tautologically Implies, 7, Commutative Law]
9. $$S \rightarrow P$$............................[Tautologically Implies, 8, Converting Disjuction to Implication]
10. $$S \rightarrow R$$............................[Tautologically Implies, 9, 5, Transitivity]
11. $$\neg S \lor R$$............................[Tautologically Implies, 10, Converting Implication to Disjuction]
What wrong I did? I am getting $$\neg S \lor R$$ instead of $$\neg S$$
• You didn't take much benefit from premisse $2$. That's why you didn't get the desired conclusion. But your reasonning is Correct. – hamam_Abdallah Jul 2 '20 at 21:19
hint
From premisse $$2$$, use distributivity to get $$\lnot Q \wedge \lnot R$$ because $$R\wedge \lnot R$$ is false.
by simplification, you have $$\lnot Q$$.
by contrapositive of premmisse $$1$$, you get $$\lnot P.$$
Finally, using premisse $$3$$, De Morgan's law and disjunctive syllogism, you have the conclusion $$\lnot S$$.
• There it is... I really missed that one out. – Ubi hatt Jul 2 '20 at 21:23
• How do we get $\neg Q$ from $\neg Q\land\neg R$? – Invisible Jul 3 '20 at 11:42
• @Croissant by simplification . – hamam_Abdallah Jul 3 '20 at 13:53 | 2021-01-22T13:57:50 | {
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https://math.stackexchange.com/questions/4506794/orthogonally-diagonalize-matrix | Orthogonally diagonalize matrix?
Find matrix P that orthogonally diagonalizes C. $$C= \begin{pmatrix} 1&2&0\\2&1&0\\ 0&0&5\\ \end{pmatrix}$$
I have worked through this problem by taking the eigenvectors and then normalizing the answer. The answer I get is
$$P= \begin{pmatrix} -1/\sqrt2&1/\sqrt2&0\\1/\sqrt2&1/\sqrt2&0\\ 0&0&1\\ \end{pmatrix}$$
My book though gives the following answer of
$$P= \begin{pmatrix} 1/\sqrt2&1/\sqrt2&0\\1/\sqrt2&-1/\sqrt2&0\\ 0&0&1\\ \end{pmatrix}$$
The eigenvalues I got were $$5,3,$$and $$-1$$. The eigenvectors and normalization I got were $$(0,0,1)=(0,0,1)$$ $$(1,1,0)=(1/\sqrt2,1/\sqrt2,0)$$ and $$(-1,1,0)=(-1/\sqrt2,1/\sqrt2,0)$$.
Which one is correct?, if any.
• Why not compute $P^{-1}CP$ for both choices of $P$ to see what you get? Aug 5 at 20:45
• I get the same answer for both, the original matrix. Aug 5 at 20:51
• Why have you chosen to order your basis as $\{(0,0,1),(1,1,0),(-1,1,0)\}$ (well, the ON basis of this) and not any of the other ways? Would that change your matrix $P$? How? And does the ordering of the basis affect if $P$ can orthogonally diagonalise $C$? Aug 5 at 20:52
• Diagonalizing matrices are not unique, since you can modify the basis you want to get another basis (e.g., multiplying an eigenvector by $1$ or $-1$ will not change orthonormality). Here, it's a matter of the order in which you put the eigenvectors, and possibly multiplying an eigenvector by $-1$. Aug 5 at 20:53
• I’m not quite sure how the textbook got the y value to be negative and not the x-value.Unless they did multiply the eigenvector by -1, which seems unnecessary. Aug 5 at 20:56 | 2022-08-15T10:24:10 | {
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https://math.stackexchange.com/questions/2587917/criteria-for-smoothness-of-the-pointwise-limit-of-a-sequence-of-functions | # Criteria for smoothness of the pointwise limit of a sequence of functions
Let $\#$ denote cardinality, and fix $p\in[1,\infty]$.
Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of functions from $[0,1]$ to $\mathbb{R}$ with the properties
1. $f_n\in C^{n-1}([0,1])$
2. $f^{(n)}_n$ is continuous on $[0,1]\setminus A_n$ where $\#A_n<\infty$ but $\underset{n\to\infty}{\lim}\#A_n=\infty$
3. $\underset{n\to\infty}{\lim}||f_n^{(m)}||_p$ exists and is finite $\forall m\in\mathbb{N}_0$
4. $f_n$ converge pointwise to a function $f$
For what values of $p$ do these imply $f\in C^\infty([0,1])$?
For those $p$ where a counterexample exists, how to construct such an example and/or what would be a convenient additional condition to avoid its existence?
Attempt with $p=\infty$:
Assume that $f^{(0)}$ is not continuous. Then there exists $x\in(0,1)$ and $\varepsilon>0$ such that however small $\delta>0$ is, there exists $x_0$ satisfying $0<|x-x_0|<\delta$ such that $|f(x_0)-f(x)|>\varepsilon$. Moreover $0<|(x_0(\delta)-(x+\delta))/2|<\delta$, and if $p=\infty$ then 3) implies $\infty>\underset{n\to\infty}{\lim}|f_n^{(1)}(x)|$. A contradiction follows:
$$\infty>\underset{n\to\infty}{\lim}|3f_n^{(1)}(x)| =\underset{n\to\infty}{\lim}\left(\underset{\delta\to 0^+}{\lim}\left|\frac{f_n(x+\delta)-f_n(x)}{\delta}\right|+2\underset{\delta\to 0^+}{\lim}\left|\frac{f_n(x_0(\delta))-f_n(x+\delta)}{x_0(\delta)-(x+\delta)}\right|\right) \hspace{6.4cm}\text{ }\\\hspace{4cm} >\underset{n\to\infty}{\lim}\left(\underset{\delta\to 0^+}{\lim}\left|\frac{f_n(x+\delta)-f_n(x)}{\delta}\right|+\underset{\delta\to 0^+}{\limsup}\left|\frac{f_n(x_0(\delta))-f_n(x+\delta)}{\delta}\right|\right) \\\hspace{2.15cm} >\underset{n\to\infty}{\lim}\underset{\delta\to 0^+}{\limsup}\left|\frac{f_n(x+\delta)-f_n(x)}{\delta}+\frac{f_n(x_0(\delta))-f_n(x+\delta)}{\delta}\right| \\\hspace{2cm} =\underset{n\to\infty}{\lim}\underset{\delta\to 0^+}{\limsup}\left|\frac{f_n(x_0(\delta))-f_n(x)}{\delta}\right| \hspace{4.6cm}\text{ }\\ \overset{\text{???}}{=}\underset{\delta\to 0^+}{\limsup}\underset{n\to\infty}{\lim}\left|\frac{f_n(x_0(\delta))-f_n(x)}{\delta}\right| \hspace{2.6cm}\text{ }\\ =\underset{\delta\to 0^+}{\limsup}\frac{|f(x_0(\delta))-f(x)|}{\delta} >\underset{\delta\to 0^+}{\limsup}\frac{\varepsilon}{\delta}=\infty \hspace{0.5cm}\text{ }$$
so $f^{(0)}$ is continuous and replacing $f$ and $f_n$ by their derivative smoothness follows by induction. Unless the part with those "???" is wrong?!
• In $3$ is it for all $p\geq 1$ or for some $p\in [1,\infty)$? Jan 1, 2018 at 20:24
• @clark Some $p\in[1,\infty]$ Jan 1, 2018 at 20:26
• In 2. you require that #$A_n\to \infty?$
– zhw.
Jan 1, 2018 at 21:03
• @zhw. Yes I do. Jan 1, 2018 at 21:49
Actually a stronger result holds:
Thm: Suppose $f_n\in C^{n-1}([0,1]), n =1,2,\dots,$ and $f_n \to f$ pointwise on $[0,1].$ Suppose further that for each $m\in \{0,1,\dots \},$
$$\sup_{n>m+1} \|D^m f_n\|_1 < \infty.$$
Then $f\in C^{\infty}([0,1]).$
Lemma: Suppose $g_n\in C^{2}([0,1]), n =1,2,\dots$ and $\|g_n\|_1,\|g_n'\|_1,\|g_n''\|_1$ are uniformly bounded. Then there exists a subsequence $g_{n_k}$ that converges uniformly on $[0,1].$
To see how the lemma implies the theorem, we start by showing $f\in C^1.$ Consider the sequence $g_n = f_{n+3}', n=1,2,\dots$ From the hypotheses in the theorem, $g_n$ satisfies the hypotheses in the lemma. Thus there exists a subsequence $g_{n_k}$ that converges uniformly to some continuous $g.$
We thus have the following: $f_{n_k+3} \to f$ pointwise, and $f_{n_k+3}'\to g$ uniformly. By the standard result on uniform convergence and differentiation, $f$ is differentiable on $[0,1]$ and $f'= g.$ Since $g$ is continuous, $f\in C^1.$
We now look at $f_{n_k+3}''.$ The tail end of this sequence satisfies the hypotheses of the lemma. Thus some subsequence $f_{n_{k_j}+3}''$ converges uniformly to some continuous $h.$ In exactly the same way as above, we have $f'' = g' =h$ and $f\in C^2.$
Clearly this process can be continued, which proves $f\in C^\infty.$
Proof of the lemma: Suppose $\|g_n\|_1,\|g_n'\|_1,\|g_n''\|_1$ are all bounded above by $C.$ By Fatou's lemma, $\int_0^1 \liminf |g_n'| \le C.$ Thus $\liminf |g_n'| < \infty$ a.e. Thus there exists $b\in [0,1]$ such that for some subsequence $n_k,$
$$\sup_k |g_{n_k}'(b)|=M_b < \infty$$
for all $k.$ Going to a further subsequence $g_{n_{k_j}},$ we will also have $a\in [0,1]$ such that $\sup_k |g_{n_{k_j}}(a)|=M_a < \infty.$
From the FTC it follows that
$$|g_{n_{k_j}}'(x)| \le |g_{n_{k_j}}'(x)-g_{n_{k_j}}'(b)| + |g_{n_{k_j}}'(b)| \le \int_0^1 |g_{n_{k_j}}''| + M_b \le C + M_b.$$
Thus $g_{n_{k_j}}'$ is uniformly bounded on $[0,1].$ By the MVT, $g_{n_{k_j}}$ is uniformly Lipschitz on $[0,1].$ Thus the sequence $g_{n_{k_j}}$ is equicontinuous on $[0,1].$ Since $g_{n_{k_j}}(a)$ is bounded, Arzela-Ascoli implies there is a subsequence of $g_{n_{k_j}}$ that converges uniformly, which gives the lemma. | 2022-05-21T16:25:07 | {
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https://math.stackexchange.com/questions/2052987/symbol-for-kernel-and-range-of-a-linear-transformation | # Symbol for kernel and range of a linear transformation
I'm wondering what symbol is used for the kernel and the range of a linear transformation. I've seen them being written as such:
$\ker(T)$
$\mathcal{R}(T)$
Are these the "correct" symbols for the kernel and range, or do they differ between different mathematicians? Are there any that are used more frequently? This is being a bit pedantic, but I'm just curious.
• I've also seen $\text{im} (T)$ used for range. – Mark Dec 10 '16 at 20:45
• For the former, I've also seen $\mathcal{N}(T)$; for the later, it's right. – GNUSupporter 8964民主女神 地下教會 Dec 10 '16 at 20:46
• Usually, $\ker T$ is paired with $\operatorname{im} T$, and $\mathcal{R}(T)$ is paired with $\mathcal{N}(T)$. – Daniel Fischer Dec 10 '16 at 20:51
• @SkeletonBow Feel free to answer your own question with what you have learned! It is highly encouraged on this site. – Ken Duna Dec 10 '16 at 21:24
• Thanks for the suggestion! – Skeleton Bow Dec 10 '16 at 21:42
$\ker T$ and $\operatorname{im} T$
$\mathcal{N}(T)$ and $\mathcal{R}(T)$ | 2019-12-07T14:22:04 | {
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https://mathhelpboards.com/threads/ask-probability-of-getting-the-main-doorprize.26697/ | # [ASK] Probability of Getting the Main Doorprize
#### Monoxdifly
##### Well-known member
There's an event which is joined by 240 members. The Event Organizer prepares 30 doorprize with one of them being the main ones. If Mr. Aziz's family has 15 tickets, the probability that Mr. Aziz gets the main doorprize is ....
A. 1/16
B. 1/8
C. 1/4
D. 1/2
I thought the answer was 15/240 (the probability of Mr. Aziz's family getting the doorprize) times 1/30 (the main one among the doorprize) and it results in 1/480, but it's not in the options. Is the book wrong or am I the one who miscalculated?
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Hi Mr. Fly!
I'm assuming those 30 doorprizes are divided randomly among the 240 members.
And that the 15 tickets in Mr. Aziz's family correspond to 15 members.
And that there is only 1 main doorprize.
Just checking, is it a typo that you write 'the main ones' as plural?
Otherwise it suggests that there is more than 1 main doorprize.
If there is only 1 main prize, and Mr. Aziz has 15 chances out of 240 on it, then the probability that Mr. Aziz gets the main doorprize is 15/240 = 1/16.
Note that if my interpretation is correct, we can expect that Mr. Aziz's family collects $\frac{15}{240 }\cdot 30$ door prizes as opposed to the 15/240 that you suggested.
Since only 1 of them is the main prize, we multiply indeed by 1/30, resulting in the $\frac{15}{240}\cdot 30\cdot \frac{1}{30}=\frac{15}{240}=\frac{1}{16}$ that I already mentioned.
Last edited:
#### HallsofIvy
##### Well-known member
MHB Math Helper
I see no reason to even consider the "30 door prizes". The question is only about the one main prize. There are 240 people and 15 of them are in this family. The probability of someone in this family winning the one main prize is $$\frac{15}{240}= \frac{1}{16}$$.
#### Monoxdifly
##### Well-known member
Thank you, both of you. And yes, Klaas, that was a typo.
It has been quite a long time since someone calls me Mr. Fly... | 2020-07-13T05:24:38 | {
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https://math.stackexchange.com/questions/1244258/if-u-in-l10-1-is-nonnegative-and-e-n-int-01-xn-ux-dx-prove-e | # If $u\in L^1(0,1)$ is nonnegative and $E_n = \int_0^1 x^n u(x) \, dx$, prove $E_{n-k} E_k \leq E_0 E_n$.
$\textbf{Question:}$ Let $u \in L^1(0,1)$ be a nonnegative function. Define $$E_n := \int_0^1 x^n u(x) dx$$ Prove the following inequality, $\forall n \ge 0$, and $\forall k \in [0,n]$, we have $$E_{n-k} E_k \le E_0 E_n$$
$\textbf{My Attempt:}$ We have, $$E_0 := \int_0^1 u(x) dx$$
$$E_1= \int_0^1 x u(x) dx$$
$$E_2 = \int_0^1 x^2 u(x) dx$$
Since $x \in (0,1)$, we have $$E_0 \ge E_1 \ge E_2 \dots$$
Thus $$E_n E_0 \ge E_n E_1 \ge E_n E_2 \ge \dots$$
To show that $E_0 E_n \ge E_k E_{n-k}$, we must show that $$\frac{E_0}{E_k} \ge \frac{E_{n-k}}{E_n}$$
This is equivalent to show that $x^{-k} \ge x^{n-2k}$ which is true as lons as $-k \le n-2k$, which is whenever $k \ge n$.
Is the above proof correct?
• You lost me going from $\frac{E_0}{E_k} \geq \frac{E_{n-k}}{E_n}$ to $x^{-k} \geq x^{n-2k}$. Apr 21, 2015 at 1:21
• Ya this part is flawed Apr 21, 2015 at 1:25
• My reasoning, which I know isn't entirely correct, is that the ratio would really only depend on the ratio of the integrals of the powers of $x$. Apr 21, 2015 at 1:27
Holder's Inequality: Assume, wlog, that $k\geqslant n-k$. \begin{align} \Big(\int_0^1 x^{n-k}u~\text dx\Big)\Big(\int_0^1x^k u~\text dx\Big) &= \Big(\int_0^1 (x^k)^{\frac{n-k}{k}}u~\text dx\Big)\Big(\int_0^1x^k u~\text dx\Big)\\ &\leqslant \Big(\int_0^1 u~\text dx\Big)^{2-\frac{n}{k}}\Big(\int_0^1 x^ku~\text dx\Big)^{\frac{n-k}{k}}\Big(\int_0^1x^k u~\text dx\Big) \\ & = \Big(\int_0^1 u~\text dx\Big)^{2-\frac{n}{k}}\Big(\int_0^1 (x^n)^{\frac{k}{n}}u~\text dx\Big)^{\frac{n}{k}} \\ & \leqslant \Big(\int_0^1 u~\text dx\Big)^{2-\frac{n}{k}}\Big(\int_0^1 x^nu~\text dx\Big)\Big(\int_0^1 u~\text dx\Big)^{(1-\frac{k}{n})\frac{n}{k}} \\ & = \Big(\int_0^1 u~\text dx\Big)\Big(\int_0^1 x^nu~\text dx\Big)\,. \end{align}
Therefore,
$$\Big(\int_0^1 x^{n-k}u~\text dx\Big)\Big(\int_0^1x^k u~\text dx\Big)\leqslant \Big(\int_0^1 u~\text dx\Big)\Big(\int_0^1 x^nu~\text dx\Big)$$
• Thought it was hölder's - just couldn't find the p and q. Thanks! Anyway this question is from a graduate qualifying exam. Apr 21, 2015 at 13:28
• @Ilham, You are welcome. And thanks for letting me know where the question comes from. :)
– ki3i
Apr 21, 2015 at 17:59
Recall Hölder's inequality for $$f$$, $$g$$ non-negative and $$\alpha\in[0,1]$$: $$\int f^\alpha g^{1-\alpha}\le\left(\int f\right)^\alpha\left(\int g\right)^{1-\alpha}\tag{*}$$
To relate $$E_k$$ to $$E_n$$ and $$E_0$$ using (*), the obvious thing to try is to find $$\alpha$$ such that $$\int_0^1 x^ku=\int_0^1 (x^nu)^\alpha (u)^{1-\alpha}$$. The only choice is $$\alpha=k/n$$. Hölder then gives $$E_k \le E_n^{k/n} E_0^{1-k/n}.\tag1$$ Apply (1) with $$k$$ replaced by $$n-k$$ to obtain $$E_{n-k}\le E_n^{1-k/n}E_0^{k/n}.\tag2$$ Multiplying (1) and (2) gives the result. | 2022-07-04T02:20:15 | {
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http://math.stackexchange.com/questions/710355/determining-number-of-solutions-with-inclusion-exclusion | # Determining number of solutions with inclusion-exclusion
NOTE: I know there are similar questions to this, but the ones on this website are much more complex, and I'd like to get a basic understanding before moving on to them. Please do not mark this as a duplicate.
On a practice quiz:
Use inc-exc to determine the number of solutions to $a+b+c+d = 15$ with $0 \leq a,b,c,d \leq 6$
## Here's what I have so far:
Let $A_1$ be the solution to $a+b+c+d=15$ with $a\geq 7, 0\leq b,c,d$
Let $A_2$ be the solution to $a+b+c+d=15$ with $b\geq 7, 0\leq a,c,d$
Let $A_3$ be the solution to $a+b+c+d=15$ with $c\geq 7, 0\leq a,b,d$
Let $A_4$ be the solution to $a+b+c+d=15$ with $d\geq 7, 0\leq a,b,c$
$|A_1 \cup A_2 \cup A_3 \cup A_4| = C(4,1) \cdot C((4+4-1), 4) - \dotsb$
And this is where I get stuck. I found my first term using the logic that I want to count $|A_1|+|A_2|+|A_3|+|A_4|$. How do I continue this to find the rest of the terms? I know what to do after I find the number of non-solutions, just subtract it from the total non-restricted solutions...but I can't figure out how to find the number of non-solutions!
Any help would be greatly appreciated, thanks.
-
How many solutions have $a \geq 7$ and $b \geq 7$. We have to over-include if we're going to use inclusion-exclusion. – Eric Towers Mar 13 '14 at 3:41
@EricTowers Right...that's what I need help figuring out... – user134788 Mar 13 '14 at 3:44
@EricTowers How can I solve for that? – user134788 Mar 13 '14 at 3:44
Start with: Home many solutions have $a\geq 7$, $b \geq 7$, $c \geq 7$, and $d \geq 7$? – Eric Towers Mar 13 '14 at 3:46
@EricTowers 0 solutions. I'm still not getting it though. Some help would be appreciated...completely stuck here and I have nowhere else to turn. – user134788 Mar 13 '14 at 3:58
Generating Functions
Look at the coefficient of $x^{15}$ in $\left(1+x+x^2+x^3+\dots+x^6\right)^4$: \begin{align} \left(\frac{1-x^7}{1-x}\right)^4 &=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty(-1)^k\binom{-4}{k}x^k\\ &=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty\binom{k+3}{k}x^k \end{align} which is $$\sum_{j=0}^2(-1)^j\binom{4}{j}\binom{18-7j}{3}=180$$
Inclusion-Exclusion
Without restriction on the size of the terms, using the standard $\mid$ and $\circ$ argument ($15$ $\circ$s and $3$ $\mid$s), there are $\binom{15+3}{3}$ ways to choose 4 non-negative integers that sum to $15$. $$\text{one sum for each arrangement}\\ 2+4+6+3=\circ\,\circ\mid \circ\circ\circ\,\circ\mid \circ\circ\circ\circ\circ\,\circ\mid \circ\circ\circ$$ Now let's count how many ways there are to have terms greater than $6$. There are $\binom{4}{1}$ ways to choose which $1$ term should be greater than $6$. To count the number of sums with $1$ term at least $7$, that would be $\binom{15-7+3}{3}$. $$\text{consider the red group atomic}\\ 2+8+4+1=\circ\,\circ\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ\circ}\,\circ\mid\circ\circ\circ\,\circ\mid\circ$$ There are $\binom{4}{2}$ ways to choose which $2$ terms should be greater than $6$. To count the number of sums with $2$ terms at least $7$, that would be $\binom{15-14+3}{3}$. $$7+0+7+1=\color{#C00000}{\circ\circ\circ\circ\circ\circ\,\circ}\mid\mid\color{#C00000}{\circ\circ\circ\circ\circ\circ\circ}\mid\circ$$ There is no way for $3$ terms to be greater than $6$. Inclusion-Exclusion says there are $$\binom{18}{3}-\binom{4}{1}\binom{11}{3}+\binom{4}{2}\binom{4}{3}=180$$ ways for $4$ terms to sum to $15$ with each term at most $6$.
-
This is awesome thank you!!! On the answer key it reads that the # non solutions is: $\binom{4}{1} \cdot \binom{8+4-1}{8} - \binom {4}{2} \cdot \binom {1+4-1}{4}$ These are equivalent, but would you be able to explain the reasoning behind the difference? – user134788 Mar 13 '14 at 17:48
@user134788: $\binom{11}{8}=\binom{11}{3}$ and $\binom{4}{1}=\binom{4}{3}$. Are you sure it is $\binom{1+4-1}{4}$ and not $\binom{1+4-1}{1}$? – robjohn Mar 13 '14 at 18:02
Positive, that's what my prof emailed me. Not saying you're wrong I'm just trying to understand how they are equivalent logically. I get why they're equivalent numerically. – user134788 Mar 13 '14 at 18:09
actually scratch that, they come out to different numbers. My professor's solution comes out to 654 total non-solutions, while yours comes out to 636 total non-solutions. – user134788 Mar 13 '14 at 18:12
@user134788: without seeing your prof's reasoning, I can't say why we get different answers. Since both of my approaches give the same result (and the same sums), I am pretty sure that my answer is correct. Does your prof get a different final answer? – robjohn Mar 13 '14 at 18:35 | 2016-05-05T03:30:07 | {
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https://math.stackexchange.com/questions/381690/irrational-cantor-set?noredirect=1 | # Irrational Cantor set?
Can someone help me?
How can I prove that exists a number $k \in \mathbb R$ that $$A = \{x + k;\ x \in \text{Cantor set} \} \subset\text{ Irrationals}\;?$$
• Do you want this for the standard middle-thirds Cantor set, or is any Cantor space okay? In the latter case, digits can be played to get the result easily. – Karolis Juodelė May 5 '13 at 0:48
• Hint (for one possible approach of many): $k$ should not be of the form $q-x$ for any rational $q$ and any $x$ in the Cantor set. What is the measure of this set? – GCD May 5 '13 at 0:49
• I want the standard middle-thirds Cantor set. – Pedro Amorim May 5 '13 at 0:55
• – Andrés E. Caicedo May 5 '13 at 19:38
Let $C$ be the middle-thirds Cantor set. If $x\in\Bbb R$ and $q\in\Bbb Q\cap(x+C)$, then there is a $y\in C$ such that $q=x+y$, and hence $x=q-y\in q-C$. If $\Bbb Q\cap(x+C)\ne\varnothing$ for every $x\in\Bbb R$, then
$$\Bbb R=\bigcup_{q\in\Bbb Q}(q-C)\;.$$
Verify that each of the sets $q-C$ for $q\in\Bbb Q$ is a closed, nowhere dense subset of $\Bbb R$, and then apply the Baire category theorem to get a contradiction.
• Do we have explicit examples? – Andrés E. Caicedo May 5 '13 at 15:03
• @Andres: Of a translation taking $C$ into $\Bbb P$? Good question; I don’t know. – Brian M. Scott May 5 '13 at 15:10
• Thanks. I was almost there, but I was using all the real numbers instead of only the rationals therefore, I couldn't use the Baire theorem – Pedro Amorim May 5 '13 at 21:41
• @Pedro: You're welcome. – Brian M. Scott May 6 '13 at 1:22
• Or you could have used the sub-additivity of measure. See below. – Kasun Fernando Sep 27 '14 at 16:04
If you read Spanish, you may want to look at the book
Wilman Brito, El Teorema de Categoría de Baire y aplicaciones, Consejo de publicaciones, Universidad de los Andes, Mérida, Venezuela, 2011.
I found this book through this MO question. What follows comes from this reference.
Section 1.10 is about the Cantor set and the result you are asking about, that was first noted by Ludwig Scheeffer, in 1884, with the argument in Brian's answer. The reference is
Ludwig Scheeffer. Zur Theorie der stetigen Funktionen einer reellen Veränderlichen, Acta Math. 5 (1), (1884), 49–82. MR1554648.
(The article is behind a paywall.) The proof can be seen in English in The theory of sets of points, by William Henry Young and Grace Chisholm Young, recently (2006) republished as part of the Cambridge Library Collection. (See Theorem 18 in Chapter IV, and $\S67$ in Chapter V.)
More precisely, Scheeffer shows that if $E$ is countable and $N$ is nowhere dense, then there is a dense set $D$ of points $d$ such that $(E+d)\cap N$ is empty. Note that Brian's answer actually shows a bit more (for $E=\mathbb Q$ and $N=C$, the Cantor set), namely, that the set of $x$ such that $C+x$ consists entirely of irrational numbers, is comeager. This is a particular case of a 1954 result by Frederick Bagemihl, see
Frederick Bagemihl. A note on Scheeffer's theorem, Michigan Math. J. 2 (1953--54), (1955), 149–150. MR0065614 (16,455c).
Bagemihl showed that if $F$ is meager and $N$ is countable, then there is a comeager set $G$ such that $(x+N)\cap F=\emptyset$ for all $x\in G$. This is an improvement of Scheeffer's result, replacing the assumption that $N$ is nowhere dense with the weaker requirement that it is meager, and the conclusion that $D$ is dense with the stronger requirement that it is comeager.
In 1981, in
Duane Boes, Richard Darst, and Paul Erdős. Fat, symmetric, irrational Cantor sets, Amer. Math. Monthly 88 (5), (1981), 340--341. MR0611391 (83i:26003),
it is proved that there is a comeager set $G\subseteq[0,1]$ such that $(0,1)\cap C_\alpha$ consists only of irrational numbers, for all $\alpha\in G$. Here, for $0<\alpha\le 1$, $C_\alpha$ is defined as the Cantor set, starting with $[0,1]$ and at stage $n>0$ removing the middle interval of length $\alpha/3^n$ from each remaining interval. (So $C_1=C$ is the usual Cantor set.)
Returning to the question, note that all the results above are established via applications of the Baire category theorem. For explicit examples of reals $x$ such that $x+C$ consists solely of irrational numbers, see this MO question. In particular, any $x$ such that every finite sequence of $0$s, $1$s, and $2$s appears as a substring of the ternary expansion of $x$ has this property.
• Are you sure that's the correct reference for Scheefer's result? Bagemihl's A note on Scheefer's theorem refers to Scheefer's Zur Theorie der stetigen Funktionen einer reellen Veränderlichen, Acta Math. 5 (1884). 279-296. Maybe he published two versions of that theorem? By the way, how could he be using Baire's (1899) category theorem in 1884? – bof Feb 22 '14 at 21:35
• @bof I've corrected the link, thanks for noticing this. Scheeffer proves directly a version of Baire's theorem in his note. – Andrés E. Caicedo Feb 26 '14 at 17:56
We had a similar problem in one of our Real Analysis assignments. It was a generalization of the given problem.
Let $E \subset \mathbb{R}$ be a set with Lebesgue outer measure $0$. Then there exists $h \in \mathbb{R}$ such that $E+h \subseteq \mathbb{Q}^c$.
We can prove this using the identity Brian derived.
If $\Bbb Q\cap(E+h)\ne\varnothing$ for every $h\in\Bbb R$, then
$$\Bbb R=\bigcup_{h\in\Bbb Q}(h-E)\;.$$
$$m(\Bbb R) \leq \sum_{h \in \Bbb Q}{} m^*(h-E)=0$$
which gives us a contradiction. So, $\Bbb Q\cap(E+h)=\varnothing$ for some $h \in \Bbb R$. | 2019-10-23T15:14:50 | {
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https://ijc.fd.uc.pt/h5hmc/b3520f-fourier-series-examples | 2 π. Calculate the Fourier coefficients for the sawtooth wave. A function $$f\left( x \right)$$ is said to have period $$P$$ if $$f\left( {x + P} \right) = f\left( x \right)$$ for all $$x.$$ Let the function $$f\left( x \right)$$ has period $$2\pi.$$ In this case, it is enough to consider behavior of the function on the interval $$\left[ { – \pi ,\pi } \right].$$, If the conditions $$1$$ and $$2$$ are satisfied, the Fourier series for the function $$f\left( x \right)$$ exists and converges to the given function (see also the Convergence of Fourier Series page about convergence conditions. By setting, for example, $$n = 5,$$ we get, $Recall that we can write almost any periodic, continuous-time signal as an infinite sum of harmoni-cally Fourier series is a very powerful and versatile tool in connection with the partial differential equations. {f\left( x \right) \text{ = }}\kern0pt Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. \end{cases},} Periodic functions occur frequently in the problems studied through engineering education. }$, We can easily find the first few terms of the series. Since this function is the function of the example above minus the constant . The reasons for Solution. We'll assume you're ok with this, but you can opt-out if you wish. }\], Find now the Fourier coefficients for $$n \ne 0:$$, ${{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} }= {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \cos nxdx} }= {\frac{1}{\pi }\left[ {\left. Computing the complex exponential Fourier series coefficients for a square wave. {f\left( x \right) \text{ = }}\kern0pt As useful as this decomposition was in this example, it does not generalize well to other periodic signals: How can a superposition of pulses equal a smooth signal like a sinusoid? 15. Replacing $${{a_n}}$$ and $${{b_n}}$$ by the new variables $${{d_n}}$$ and $${{\varphi_n}}$$ or $${{d_n}}$$ and $${{\theta_n}},$$ where, \[{{d_n} = \sqrt {a_n^2 + b_n^2} ,\;\;\;}\kern-0.3pt{\tan {\varphi _n} = \frac{{{a_n}}}{{{b_n}}},\;\;\;}\kern-0.3pt{\tan {\theta _n} = \frac{{{b_n}}}{{{a_n}}},}$, $= {\frac{1}{2} + \frac{2}{\pi }\sin x } A Fourier Series, with period T, is an infinite sum of sinusoidal functions (cosine and sine), each with a frequency that is an integer multiple of 1/T (the inverse of the fundamental period).$, The first term on the right side is zero. 0/2 in the Fourier series. Find the constant term a 0 in the Fourier series … { {b_n}\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} } \right]} .} Figure 1 Thevenin equivalent source network. }\], First we calculate the constant $${{a_0}}:$$, ${{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} }= {\frac{1}{\pi }\int\limits_0^\pi {1dx} }= {\frac{1}{\pi } \cdot \pi }={ 1. In particular harmonics between 7 and 21 are not shown. }$, ${\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} = {a_m}\pi ,\;\;}\Rightarrow{{a_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} ,\;\;}\kern-0.3pt{m = 1,2,3, \ldots }$, Similarly, multiplying the Fourier series by $$\sin mx$$ and integrating term by term, we obtain the expression for $${{b_m}}:$$, ${{b_m} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin mxdx} ,\;\;\;}\kern-0.3pt{m = 1,2,3, \ldots }$. The Fourier library model is an input argument to the fit and fittype functions. The signal x (t) can be expressed as an infinite summation of sinusoidal components, known as a Fourier series, using either of the following two representations. + {\frac{2}{{5\pi }}\sin 5x + \ldots } 0, & \text{if} & – \pi \le x \le 0 \\ Even Pulse Function (Cosine Series) Aside: the periodic pulse function. Example of Rectangular Wave. We will also define the odd extension for a function and work several examples finding the Fourier Sine Series for a function. {\begin{cases} The Fourier Series for an odd function is: f(t)=sum_(n=1)^oo\ b_n\ sin{:(n pi t)/L:} An odd function has only sine terms in its Fourier expansion. 5, ...) are needed to approximate the function. { {\cos \left( {n – m} \right)x}} \right]dx} }={ 0,}\], $\require{cancel}{\int\limits_{ – \pi }^\pi {\sin nx\cos mxdx} }= {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\sin 2mx + \sin 0} \right]dx} ,\;\;}\Rightarrow{\int\limits_{ – \pi }^\pi {{\sin^2}mxdx} }={ \frac{1}{2}\left[ {\left. Since this function is odd (Figure. {\left( { – \frac{{\cos 2mx}}{{2m}}} \right)} \right|_{ – \pi }^\pi } \right] }= {\frac{1}{{4m}}\left[ { – \cancel{\cos \left( {2m\pi } \right)} }\right.}+{\left. These cookies will be stored in your browser only with your consent. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0.}} Since f ( x) = x 2 is an even function, the value of b n = 0. These cookies do not store any personal information. There is Gibb's overshoot caused by the discontinuity. EEL3135: Discrete-Time Signals and Systems Fourier Series Examples - 1 - Fourier Series Examples 1. There is Gibb's overshoot caused by the discontinuities. There are several important features to note as Tp is varied. This section explains three Fourier series: sines, cosines, and exponentials eikx. {\begin{cases} 11. This might seem stupid, but it will work for all reasonable periodic functions, which makes Fourier Series a very useful tool. We also use third-party cookies that help us analyze and understand how you use this website. Start with sinx.Ithasperiod2π since sin(x+2π)=sinx. Fourier Series. Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Common examples of analysis intervals are: x ∈ [ 0 , 1 ] , {\displaystyle x\in [0,1],} and. + {\sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }\right.}}+{{\left. An example of a periodic signal is shown in Figure 1. Definition of the complex Fourier series. ), At a discontinuity $${x_0}$$, the Fourier Series converges to, \[\lim\limits_{\varepsilon \to 0} \frac{1}{2}\left[ {f\left( {{x_0} – \varepsilon } \right) – f\left( {{x_0} + \varepsilon } \right)} \right].$, The Fourier series of the function $$f\left( x \right)$$ is given by, ${f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}} ,}$, where the Fourier coefficients $${{a_0}},$$ $${{a_n}},$$ and $${{b_n}}$$ are defined by the integrals, ${{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} ,\;\;\;}\kern-0.3pt{{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nx dx} ,\;\;\;}\kern-0.3pt{{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nx dx} . -1, & \text{if} & – \pi \le x \le – \frac{\pi }{2} \\ Click or tap a problem to see the solution. \frac{\pi }{2} – x, & \text{if} & 0 \lt x \le \pi { \cancel{\cos \left( {2m\left( { – \pi } \right)} \right)}} \right] }={ 0;}$, ${\int\limits_{ – \pi }^\pi {\cos nx\cos mxdx} }= {\frac{1}{2}\int\limits_{ – \pi }^\pi {\left[ {\cos 2mx + \cos 0} \right]dx} ,\;\;}\Rightarrow{\int\limits_{ – \pi }^\pi {{\cos^2}mxdx} }= {\frac{1}{2}\left[ {\left. Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems F() () exp()ωωft i t dt 1 () ()exp() 2 ft F i tdω ωω π x ∈ [ … Signal and System: Solved Question on Trigonometric Fourier Series ExpansionTopics Discussed:1. And we see that the Fourier Representation g(t) yields exactly what we were trying to reproduce, f(t). Their representation in terms of simple periodic functions such as sine function … {\left( {\frac{{\sin nx}}{n}} \right)} \right|_{ – \pi }^\pi }={ 0\;\;}{\text{and}\;\;\;}}\kern-0.3pt {{\int\limits_{ – \pi }^\pi {\cos nxdx} }={ \left. This example shows how to use the fit function to fit a Fourier model to data.. harmonic, but not all of the individual sinusoids are explicitly shown on the plot. {f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\cos\left( {nx + {\theta _n}} \right)} .} 1. {\displaystyle P=1.} Example 1: Special case, Duty Cycle = 50%. Contents. 0, & \text{if} & – \frac{\pi }{2} \lt x \le \frac{\pi }{2} \\ b n = 1 π π ∫ − π f ( x) sin n x d x = 1 π π ∫ − π x sin n x d x. {\left( { – \frac{{\cos nx}}{n}} \right)} \right|_0^\pi } \right] }= { – \frac{1}{{\pi n}} \cdot \left( {\cos n\pi – \cos 0} \right) }= {\frac{{1 – \cos n\pi }}{{\pi n}}.}$. The reader is also referred toCalculus 4b as well as toCalculus 3c-2. There is no discontinuity, so no Gibb's overshoot. The first zeros away from the origin occur when. Let’s go through the Fourier series notes and a few fourier series examples.. {f\left( x \right) = \frac{1}{2} }+{ \frac{{1 – \left( { – 1} \right)}}{\pi }\sin x } Using 20 sine waves we get sin(x)+sin(3x)/3+sin(5x)/5 + ... + sin(39x)/39: Using 100 sine waves we ge… Find b n in the expansion of x 2 as a Fourier series in (-p, p). \], Therefore, all the terms on the right of the summation sign are zero, so we obtain, ${\int\limits_{ – \pi }^\pi {f\left( x \right)dx} = \pi {a_0}\;\;\text{or}\;\;\;}\kern-0.3pt{{a_0} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)dx} .}$. 2\pi. {a_0} = {a_n} = 0. a 0 = a n = 0. Necessary cookies are absolutely essential for the website to function properly. Rewriting the formulas for $${{a_n}},$$ $${{b_n}},$$ we can write the final expressions for the Fourier coefficients: ${{a_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\cos nxdx} ,\;\;\;}\kern-0.3pt{{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} . }$, Sometimes alternative forms of the Fourier series are used. The amplitudes of the harmonics for this example drop off much more rapidly (in this case they go as. solved examples in fourier series. + {\frac{2}{{3\pi }}\sin 3x } The Fourier Series also includes a constant, and hence can be written as: Below we consider expansions of $$2\pi$$-periodic functions into their Fourier series, assuming that these expansions exist and are convergent. This example fits the El … {\begin{cases} { \sin \left( {2m\left( { – \pi } \right)} \right)} \right] + \pi }={ \pi . Now take sin(5x)/5: Add it also, to make sin(x)+sin(3x)/3+sin(5x)/5: Getting better! ion discussed with half-wave symmetry was, the relationship between the Trigonometric and Exponential Fourier Series, the coefficients of the Trigonometric Series, calculate those of the Exponential Series. 1. FOURIER SERIES MOHAMMAD IMRAN JAHANGIRABAD INSTITUTE OF TECHNOLOGY [Jahangirabad Educational Trust Group of Institutions] www.jit.edu.in MOHAMMAD IMRAN SEMESTER-II TOPIC- SOLVED NUMERICAL PROBLEMS OF … Fourier Series Jean Baptiste Joseph Fourier (1768-1830) was a French mathematician, physi-cist and engineer, and the founder of Fourier analysis. This version of the Fourier series is called the exponential Fourier series and is generally easier to obtain because only one set of coefficients needs to be evaluated. {f\left( x \right) \text{ = }}\kern0pt To define $${{a_0}},$$ we integrate the Fourier series on the interval $$\left[ { – \pi ,\pi } \right]:$$, ${\int\limits_{ – \pi }^\pi {f\left( x \right)dx} }= {\pi {a_0} }+{ \sum\limits_{n = 1}^\infty {\left[ {{a_n}\int\limits_{ – \pi }^\pi {\cos nxdx} }\right.}+{\left. + {\frac{{1 – {{\left( { – 1} \right)}^4}}}{{4\pi }}\sin 4x } Here we present a collection of examples of applications of the theory of Fourier series. In an earlier module, we showed that a square wave could be expressed as a superposition of pulses. As before, only odd harmonics (1, 3, 5, ...) are needed to approximate the function; this is because of the, Since this function doesn't look as much like a sinusoid as. Let's add a lot more sine waves. It is mandatory to procure user consent prior to running these cookies on your website. Square waves (1 or 0 or −1) are great examples, with delta functions in the derivative. As you add sine waves of increasingly higher frequency, the approximation gets better and better, and these higher frequencies better approximate the details, (i.e., the change in slope) in the original function.$, $Accordingly, the Fourier series expansion of an odd $$2\pi$$-periodic function $$f\left( x \right)$$ consists of sine terms only and has the form: \[f\left( x \right) = \sum\limits_{n = 1}^\infty {{b_n}\sin nx} ,$, ${b_n} = \frac{2}{\pi }\int\limits_0^\pi {f\left( x \right)\sin nxdx} .$. In order to incorporate general initial or boundaryconditions into oursolutions, it will be necessary to have some understanding of Fourier series. Suppose also that the function $$f\left( x \right)$$ is a single valued, piecewise continuous (must have a finite number of jump discontinuities), and piecewise monotonic (must have a finite number of maxima and minima). So let us now develop the concept about the Fourier series, what does this series represent, why there is a need to represent the periodic signal in the form of its Fourier series. Part 1. $\int\limits_{ – \pi }^\pi {\left| {f\left( x \right)} \right|dx} \lt \infty ;$, ${f\left( x \right) = \frac{{{a_0}}}{2} \text{ + }}\kern0pt{ \sum\limits_{n = 1}^\infty {\left\{ {{a_n}\cos nx + {b_n}\sin nx} \right\}}}$, $2\pi 2 π. But opting out of some of these cookies may affect your browsing experience. With a sufficient number of harmonics included, our ap- proximate series can exactly represent a given function f(x) f(x) = a 0/2 + a Fourier series calculation example Due to numerous requests on the web, we will make an example of calculation of the Fourier series of a piecewise defined function … Introduction In these notes, we derive in detail the Fourier series representation of several continuous-time periodic wave-forms. {\int\limits_{ – \pi }^\pi {f\left( x \right)\cos mxdx} } {f\left( x \right) = \frac{{{a_0}}}{2} }+{ \sum\limits_{n = 1}^\infty {{d_n}\sin \left( {nx + {\varphi _n}} \right)} \;\;}\kern-0.3pt{\text{or}\;\;} Fourier Series Examples. Because of the symmetry of the waveform, only odd harmonics (1, 3, + {\frac{{1 – {{\left( { – 1} \right)}^2}}}{{2\pi }}\sin 2x } \end{cases},} {\left( {\frac{{\sin nx}}{n}} \right)} \right|_0^\pi } \right] }= {\frac{1}{{\pi n}} \cdot 0 }={ 0,}$, \[{{b_n} = \frac{1}{\pi }\int\limits_{ – \pi }^\pi {f\left( x \right)\sin nxdx} }= {\frac{1}{\pi }\int\limits_0^\pi {1 \cdot \sin nxdx} }= {\frac{1}{\pi }\left[ {\left. Definition of Fourier Series and Typical Examples, Fourier Series of Functions with an Arbitrary Period, Applications of Fourier Series to Differential Equations, Suppose that the function $$f\left( x \right)$$ with period $$2\pi$$ is absolutely integrable on $$\left[ { – \pi ,\pi } \right]$$ so that the following so-called. (in this case, the square wave). Are absolutely essential for the website smoother functions too work for all reasonable fourier series examples functions which! Functions that are, for example, entirely above the x−axis cookies are essential! 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Only with your consent fittype functions versatile tool in connection with the partial equations! ^\Pi { \sin nx\cos mxdx } } \right ] }.: x ∈ [ 0, ]. Very powerful and versatile tool in connection with the partial differential equations } and for this drop! { – \pi } ^\pi { \sin nx\cos mxdx } } \right ] }.,.... | 2021-04-14T08:09:02 | {
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https://stats.stackexchange.com/questions/547702/what-is-a-two-sided-type-1-error-rate | What is a two-sided type 1 error rate?
This is probably a simple question but I was watching an online video about a scientific study. For the study, they mentioned that they used a two-tailed Type 1 error rate of 0.05. I know that a Type 1 error is the probability of rejecting the null hypothesis when it is actually true. However, I am not sure what is meant by a two-tailed Type 1 error rate. Does the 'two-tailed' refer to the fact that the hypothesis test used is two-sided? Any insights are appreciated.
• Yes, it means that the test is two-tailed. The particular phrasing has to do with allocating $0.05$ probability between the two tales, rather than putting all of that in just one tail.
– Dave
Oct 10, 2021 at 4:29
Examples of two-sided and one-sided t tests
Suppose you have data as follows: I used R to take a sample of size $$n = 30$$ from a normal distribution population with mean $$\mu = 100.$$ However, I happened to get a sample with mean $$\bar X - 95.55,$$ somewhat smaller than $$\mu = 100.$$
set.seed(1234)
x = rnorm(30, 100, 15)
summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
64.81 86.85 92.49 95.55 103.62 136.24
I will do a t test of the null hypothesis $$H_O: \mu =100$$ against the two-sided alternative $$H_a:\mu \ne 100.$$ I want to test at the 5% level of significance.
t.test(x, mu = 100)
One Sample t-test
data: x
t = -1.798, df = 29, p-value = 0.08259
alternative hypothesis:
true mean is not equal to 100
95 percent confidence interval:
90.49593 100.61132
sample estimates:
mean of x
95.55363
The P-value is $$0.08259 > 0.05 = 5\%,$$ so I do not reject at the 5% level. Even though I happened to get a sample that is a bit strange, its sample mean 95.55 is not enough different from $$\mu_0 = 100$$ to reject the null hypothesis.
The P-value is the probability under the density curve of Student's t distribution with 29 degrees of freedom of getting a T statistic farther from $$0$$ than the observed $$-1.798.$$ In the figure below that amounts to the sum of the areas in the two tails outside the vertical red lines.
However, if I decide (perhaps after seeing the small value of $$\bar X)$$ to do a left-sided test of $$H_0:\mu=100$$ against $$H_a:\mu < 100,$$ also at the 5% level. Then I get P-value $$0.0413 < 0.05 = 5\%,$$ so I reject the null hypothesis. In the figure, this is the area in the left tail (only) to the left of the solid red line. (The P-value of the one-tailed test is half of the P-value of the two-sided test.)
t.test(x, mu=100, alte="less")
One Sample t-test
data: x
t = -1.798, df = 29, p-value = 0.0413
alternative hypothesis:
true mean is less than 100
95 percent confidence interval:
-Inf 99.75543
sample estimates:
mean of x
95.55363
Even though both the two-sided and the one-sided test were at the same 5% level, we failed to reject $$H_0$$ for the two-sided test and rejected for the one-tailed test. This is because the criteria for rejection are different for the two tests. So, in a practical application, it is important to decide from the start (preferably before data are available) whether you need to do a two-sided or a one-sided test.
Notes: (1) In case you're familiar with R and want the R code for the figure, it is shown below.
hdr = "Density of T(29)"
curve(dt(x, 20), -3.5, 3.5, lwd=2,
ylab="PDF", xlab="t", main=hdr)
abline(h=0, col="green2")
abline(v=0, col="green2")
abline(v=-1.798, col="red")
abline(v=1.798, col="red", lty="dotted")
In case you want to see how P-values can be obtained in R, here they are:
1 - diff(pt(c(-1.798, 1.798), 29))
[1] 0.08259619 # 2-sided P-value
pt(-1.798, 29)
[1] 0.04129809 | 2022-08-18T09:24:28 | {
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https://math.stackexchange.com/questions/477/cardinality-of-set-of-real-continuous-functions/479 | # Cardinality of set of real continuous functions
I believe that the set of all $$\mathbb{R\to R}$$ continuous functions is $$\mathfrak c$$, the cardinality of the continuum. However, I read in the book "Metric spaces" by Ó Searcóid that set of all $$[0, 1]\to\mathbb{R}$$ continuous functions is greater than $$\mathfrak c$$:
"It is demonstrated in many textbooks that $$\mathbb{Q}$$ is countable, that $$\mathbb{R}$$ is uncountable, that every non-degenerate interval is uncountable, that the collection of continuous functions defined on $$[0,1]$$ is of a greater cardinality than $$\mathbb{R}$$, and that there are sets of greater and greater cardinality."
I understand that (via composition with the continuous function $$\tan$$ or $$\arctan$$) these sets of continuous functions have the same cardinality. Therefore, which claim is correct, and how do I prove this?
• The result that you’ve found here is correct: there are $\mathfrak c=|\Bbb R|$ continuous real-valued functions on $[0,1]$. I find it hard to believe that Ó Searcóid made such an egregious error; could you quote exactly what he says? Apr 21, 2013 at 0:05
• This is from page 268 (first edition): "It is demonstrated in many textbooks that $\mathbb Q$ is countable, that $\mathbb R$ is uncountable, that every non-degenerate interval is uncountable, that the collection of continuous functions defined on $[0 , 1]$ is of a greater cardinality than $\mathbb R$, and that there are sets of greater and greater cardinality." Apr 21, 2013 at 0:29
• (Brian's comment and mine refer to a different version of the question, merged with this one as duplicate. The question was prompted by a claim in "Metric spaces", by Mícheál Ó Searcóid, where it is claimed that there are more continuous functions on $[0,1]$ than real numbers.) Apr 21, 2013 at 1:43
• Sorry to bring up an old question like that but I want to ask this. When we say the set of all $\mathbb{R\to R}$ continuous functions does that contain partial functions as well? for example does it contain a continuous function from $[0,1] \to \mathbb{R}$ ? Jul 8, 2015 at 22:11
• This question was asked on the first day on Math.SE, but was recently closed for not giving enough context etc. I have edited the question to improve it, basing the context on a question which was merged with it. (I've also used the above comments, to make the question clearer.) Jun 17, 2020 at 16:12
The cardinality is at least that of the continuum because every real number corresponds to a constant function. The cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $$\mathbb R^N$$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.
The Schroeder-Bernstein theorem now implies the cardinality is precisely that of the continuum.
Note that then the set of sequences of reals is also of the same cardinality as the reals. This is because if we have a sequence of binary representations $$.a_1a_2..., .b_1b_2..., .c_1c_2...$$, we can splice them together via $$.a_1 b_1 a_2 c_1 b_2 a_3...$$ so that a sequence of reals can be encoded by one real number.
• +1 Nice. Since the rational points are dense, this determines the function. - This is the trickiest claim in the argument, enough to count as a lacuna. It might make a good further question "Can there be two distinct, continuous functions that are equal at all rationals?" Jul 22, 2010 at 12:31
• @Kaestur: it works for countably many reals, which I think is all that was intended. Jul 22, 2010 at 12:32
• By popular demand, math.stackexchange.com/questions/505/… Jul 22, 2010 at 16:53
• @CharlesStewart That question doesn't make sense since $\mathbb{R}$ is a Hausdorff space (cf Stephen Willard's General Topology) Apr 28, 2019 at 17:10
• @Akerbeltz - The question Charles Stewart created does make sense, as you can tell from the (as of now) 52 upvotes and multiple good answers. What you mean is that the answer is "no" because $\mathbb{R}$ is Hausdorff. As none of the current answers make the connection to Hausdorffness explicit, perhaps you can add another good answer! Jun 23, 2020 at 16:16
Suppose $f:\mathbb R\to\mathbb R$ is a continuous function. Let $x\in\mathbb R$. Then there is a sequence of rational numbers $(q_n)_{n=1}^\infty$ that converges to $x$. Continuity of $f$ means that $$\lim_{n\to\infty}f(q_n) = f(\lim_{n\to\infty}q_n)=f(x).$$ This means that the values of $f$ at rational numbers already determine $f$. In other words, the mapping $\Phi:C(\mathbb R,\mathbb R)\to \mathbb R^{\mathbb Q}$, defined by $\Phi(f)=f|_{\mathbb Q}$, where $f|_{\mathbb Q}:\mathbb Q\to\mathbb R$ is the restriction of $f$ to $\mathbb Q$, is an injection. (Which implies that $|C(\mathbb R,\mathbb R)|<|\mathbb R^{\mathbb Q}|$). Here, $C(\mathbb R,\mathbb R)$ denotes the set of all continuous functions from $\mathbb R$ to $\mathbb R$, as usual.
Now, cardinal arithmetic tells us that $|\mathbb R^{\mathbb Q}| = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=|\mathbb R|$. (Namely, $(a^b)^c=a^{b\cdot c}$ holds for cardinal numbers.)
Let $x$ be any real number; there is a sequence $\langle q_n:n\in\Bbb N\rangle$ of rational numbers converging to $x$. If $f$ is continuous, then $f(x)=\lim_{n\to\infty}f(q_n)$, so $f(x)$ is completely determined by the values $f(q_n)$ for $n\in\Bbb N$ and hence by $f\upharpoonright\Bbb Q$.
For the cardinality part of the argument I’m going to follow the outline that you gave in the question; depending on what you know about cardinal arithmetic, there may be substantially shorter arguments. I’m also going to arrange the argument to use some techniques that are useful more generally, again perhaps at the expense of brevity.
I’m assuming that you know that $|\Bbb Q|=|\Bbb N|$ and hence that there is a bijection $\varphi:\Bbb Q\to\Bbb N$. This easily yields a bijection $\Phi:\Bbb R^{\Bbb N}\to\Bbb R^{\Bbb Q}$: if $f:\Bbb N\to\Bbb R$, then $$\Phi(f):\Bbb Q\to\Bbb R:q\mapsto f\big(\varphi(q)\big)\;,$$ i.e., $\Phi(f)=f\circ\varphi$. (I leave it to you to check that $\Phi$ is a bijection.)
Now define a map $$N:\Bbb R\to\wp(\Bbb N):x\mapsto\{\varphi(q):q\in\Bbb Q\text{ and }q\le x\}\;;$$
clearly $N$ is injective (one-to-one), and $N(x)$ is infinite for each $x\in\Bbb R$. Thus, we may write $$N(x)=\{n_x(k):k\in\Bbb N\}\;,$$ where $n_x(k)<n_x(k+1)$ for each $k\in\Bbb N$. This is nothing more complicated than listing $N(x)$ in increasing order, but it lets us define the sequence $\nu(x)=\langle n_x(k):k\in\Bbb N\rangle\in\Bbb N^{\Bbb N}$. We now have a map
$$\nu:\Bbb R\to\Bbb N^{\Bbb N}:x\mapsto\nu(x)=\langle n_x(k):k\in\Bbb N\rangle\;,$$
and it’s not hard to check that $\nu$ is injective. On the other hand, the map that takes a sequence $\langle n_k:k\in\Bbb N\rangle\in\Bbb N^{\Bbb N}$ to the real number whose continued fraction expansion is $$[n_0;n_1+1,n_2+1,n_3+1,\ldots]$$ is an injection from $\Bbb N^{\Bbb N}$ to $\Bbb R$ (in fact to $\Bbb R\setminus\Bbb Q$), so by the Cantor-Schröder-Bernstein theorem there is a bijection between $\Bbb R$ and $\Bbb N^{\Bbb N}$. (I write $n_k+1$ in the continued fraction expansion, because my $\Bbb N$ includes $0$.)
Clearly, then, there is a bijection between $\Bbb R^{\Bbb N}$ and $\left(\Bbb N^{\Bbb N}\right)^{\Bbb N}$. To finish off the argument along the lines that you sketched in your question, carry out the following steps.
• Find a bijection between $\left(\Bbb N^{\Bbb N}\right)^{\Bbb N}$ and $\Bbb N^{\Bbb N\times\Bbb N}$. (More generally, for any sets $A,B$, and $C$ there is a bijection between $\left(A^B\right)^C$ and $A^{B\times C}$; this fact is often useful and is well worth knowing.
• In the same way that I found a bijection between $\Bbb R^{\Bbb N}$ and $\Bbb R^{\Bbb Q}$, show that there is a bijection between $\Bbb N^{\Bbb N}$ and $\Bbb N^{\Bbb N\times\Bbb N}$.
• Conclude that there is a bijection between $\Bbb R^{\Bbb N}$ and $\Bbb N^{\Bbb N}$ and hence between $\Bbb R^{\Bbb N}$ and $\Bbb R$.
It is at least $c$, since all constant functions are continuous. Now consider the fact that $\mathbb{R}$ is separable.
On the one hand it is clear that the set of all the continuous functions from $$\mathbb{R}$$ to $$\mathbb{R}$$, which shall be denoted by $$\mathcal{C}^0$$, is such that:
$$|\mathbb{R}|\le|\mathcal{C}^0|$$
(because for each $$r\in \mathbb{R}$$, simply we consider the constant function $$f_r:\mathbb{R}\longrightarrow\mathbb{R}$$ defined by: for each $$x\in \mathbb{R},\;f_r(x)=r$$. Obviously, the assignation $$r\longmapsto f_r$$ is injective).
On the other hand, we know that $$\mathbb{R}$$ is a Hausdorff space, so if $$f,g\in\mathcal{C}^0$$ are two continuous functions such that they agree in the (dense) subset of the rational numbers, then $$f=g$$ (cf Stephen Willard, General Topology, 1970, Addison Wesley, page 89, 13.14).
This allows us to consider the function $$F:\mathcal{C}^0\longrightarrow ^\mathbb{Q}\mathbb{R}$$ defined by: for each $$f\in\mathcal{C}^0,\;F(f)=f|_\mathbb{Q}$$ (where $$^\mathbb{Q}\mathbb{R}$$ denotes the set of all the functions from $$\mathbb{Q}$$ to $$\mathbb{R}$$).
From the previous comment, it is clear that $$F$$ is then an injective function, therefore:
$$|\mathcal{C}^0|\le|^\mathbb{Q}\mathbb{R}|={\big(2^{\aleph_0}\big)}^{\aleph_0}=2^{\aleph_0\times\aleph_0}=2^{\aleph_0}=|\mathbb{R}|$$
From the Cantor-Bernstein theorem we conclude that $$|\mathcal{C}^0|=|\mathbb{R}|$$.
• Note that if $f\not=g$, then they will not agree in all of $\mathbb{Q}$ $-$ according to the quoted result from Willard's General Topology $-$. So $F(f)\not=F(g)$ Apr 28, 2019 at 17:39
I have a somewhat simple answer to this question. It is not as elaborate as the other but perhaps it will add some intuition.
let's look at the number of functions from $\mathbb R^n \to \mathbb R$.
For every element in $\mathbb R^n$ we need to choose a corresponding image in $\mathbb R$. There are $c$ elements in $\mathbb R$, and so if there are $\alpha$ elements in $\mathbb R^n$, there are $\alpha c$ functions (not continuous! just functions) from $\mathbb R^n \to \mathbb R$.
Let's find alpha: For the first element in an $n$ vector we have $c$ choices, for the second one $c$ choices, for the third one $c$ choices etc...overall $c^n=c$ choices.
So overall there are $c^{n+1}=c$ functions from $\mathbb R^n \to \mathbb R$.
since the continuous functions are a subset of this set, there are AT MOST $c$ continuous functions.
Now let's look at the function $f(x)=\xi ||x||_2$ where $\xi$ is some real number, $x$ is an $n$ vector, and $||x||_2$ is the euclidean norm of the vector $x$.
This function $f$ is continuous no matter which $\xi$ we pick. We have $c$ options to choose from, and so the set of continuous functions includes all the functions of the form $f(x)=\xi ||x||_2$ and so it is AT LEAST $c$.
Since it is at most $c$, and at least $c$, the conclusion is that it is exactly $c$.
• It seems that you claim that there are only $\mathfrak c$ functions from $\mathbb R$ to $\mathbb R$. This is not true; see here. The cardinality of the sets of all functions $\mathbb R\to\mathbb R$ is $\mathfrak c^{\mathfrak c} = 2^{\mathfrak c}$. Sep 3, 2014 at 11:27
• For every element in $\mathbb R^n$ we need to choose a corresponding image in $\mathbb R$. There are $c$ elements in $\mathbb R$, and so if there are $\alpha$ elements in $\mathbb R^n$, there are $\alpha c$ functions (not continuous! just functions) from $\mathbb R^n \to \mathbb R$. - Try this assuming $\mathbb{R}$ is a set with two elements. Mar 1, 2017 at 14:59 | 2022-05-21T13:54:03 | {
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https://math.stackexchange.com/questions/3678353/solving-int-01-x7001-x300-x3001-x700-dx | Solving $\int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx$
I am trying to solve the following integral:
$$\int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx$$
My intuition is that this integral is equal to zero but I am unsure as to which direction to take to prove this. I was thinking binomal expansion but I believe there must be a better way, possibly using summation notation instead.
• Well some ideas: always good to exploit symmetry. Also what happens if you replace $700$ and $300$ by very much smaller (different) numbers? [Does it matter that they are both even?] May 16 '20 at 21:20
• Set $x=-t$..... May 16 '20 at 21:23
• @MarkBennet : It doesn't matter if they are even or odd, nor whether they are integers. See my answer below. May 16 '20 at 21:23
• @MichaelHardy Quite. I was suggesting thoughts and questions which might possibly arise, and might be explored, rather than giving a direct answer. I put first the first thing to do (and if you work that out, I agree the other things are redundant). I put the "use small values" piece in, because that can help (eg be more obviously sketchable) and then the odd/even thing because if you don;'t know what's going on, you sometimes have to take a little care in building a small model of the problem. May 17 '20 at 5:12
\begin{align} \text{Let } u & = 1-x \\ \text{and consequently } x & = 1-u \\ du & = -dx \end{align} As $$x$$ goes from $$0$$ to $$1,$$ $$u$$ goes from $$1$$ to $$0.$$
This substitution shows that this integral is $$-1$$ times this integral. So it is $$0.$$
Appendix by the original poster:
Working with the second integral, his substitution shows that:
$$\int_0^1x^{300}(1-x)^{700}dx = \int_1^0(1-u)^{300}u^{700}(-du) = \int_0^1u^{700}(1-u)^{300}du$$
Thus \begin{align} & \int_0^1 [x^{700}(1-x)^{300} - x^{300}(1-x)^{700}] \, dx \\[8pt] = {} & \int_0^1 x^{700}(1-x)^{300} \, dx - \int_0^1(1-u)^{300}u^{700}\, du = 0 \end{align}
• Doesn't $u$ go from $1$ to $0$, not $1$ to $u$? May 16 '20 at 21:35
Let $$1/2 - x \to x^\prime$$ to see that the integral is $$0$$.
Convert the integrand to
$$(x^\prime /2)^{700} (x^\prime /2)^{300} - (x^\prime /2)^{300}(x^\prime /2)^{700}$$
and put in the right limits...
• Can you explain this solution further? I went with the answer but this seems like an interesting method May 16 '20 at 21:35
• @S.Miller : I think possibly he has in mind something about like this: \begin{align} & u = 2\left( \tfrac 1 2 - x \right) = 1 - 2x \\ {} \\ & x = \frac {1-u} 2 \quad \text{ and } \quad 1-x = \frac{1+u}2 \\ {} \\ & \text{As x goes from 0 to \tfrac 1 2,} \\ & \text{u goes from 1 to 0.} \\ {} \\ & x^{700} = \left( \frac {1-u} 2 \right)^{700} \\ {} \\ & (1-x)^{300} = \left( \frac{1+u} 2 \right)^{300} \end{align} $$\int_0^{1/2} x^{700} (1-x)^{300} \, dx = \frac 1 {2^{1000}} \int_1^0 (1-u)^{700} (1+u)^{300} \, \left( \frac{-du} 2 \right)$$ $$\text{and so on.}$$ May 16 '20 at 23:36
• Yep............ May 16 '20 at 23:37
• @S.Miller : Note that this $\quad\uparrow\quad$ is $\displaystyle \int_0^{1/2},$ not $\displaystyle \int_0^1.$ You'd also need to look at $\displaystyle \int_0^{1/2}$ of the other term, which has $700$ and $300$ interchanged and has a minus sign, and also at $\displaystyle \int_{1/2}^1$ of both functions. $\qquad$ May 16 '20 at 23:39
• @S.Miller : One thing that might suggest this method is that when $x=1/2$ then the value of the function being integrated is $0,$ so the polynomial must be divisible by $x - \tfrac 1 2. \qquad$ May 16 '20 at 23:44
Well, solving a more general case we have:
$$\mathcal{I}_\beta\left(\text{n},\text{k}\right):=\int_0^\beta\left(x^\text{n}\left(\beta-x\right)^\text{k}-x^\text{k}\left(\beta-x\right)^\text{n}\right)\space\text{d}x\tag1$$
Let $$\text{u}=\beta-x$$, so we get $$-\text{du}=\text{d}x$$, so:
$$\mathcal{I}_\beta\left(\text{n},\text{k}\right)=\int_\beta^0-\left(\left(\beta-\text{u}\right)^\text{n}\text{u}^\text{k}-\left(\beta-\text{u}\right)^\text{k}\text{u}^\text{n}\right)\space\text{du}=$$ $$\int_0^\beta\left(\text{u}^\text{k}\left(\beta-\text{u}\right)^\text{n}-\text{u}^\text{n}\left(\beta-\text{u}\right)^\text{k}\right)\space\text{du}=$$ $$\int_0^\beta\left(x^\text{k}\left(\beta-x\right)^\text{n}-x^\text{n}\left(\beta-x\right)^\text{k}\right)\space\text{d}x\tag2$$
So, we get:
$$\mathcal{I}_\beta\left(\text{n},\text{k}\right)+\mathcal{I}_\beta\left(\text{n},\text{k}\right)=0\tag3$$
First let $$f(x)=x^{700}(1-x)^{300}-x^{300}(1-x)^{700}$$ and see that \begin{aligned} f\left(\frac{1}{2}+x\right) &= \left(\frac{1}{2}+x\right)^{700}\left(\frac{1}{2}-x\right)^{300}-\left(\frac{1}{2}+x\right)^{300}\left(\frac{1}{2}-x\right)^{700} \\ f\left(\frac{1}{2}-x\right) &= \left(\frac{1}{2}-x\right)^{700}\left(\frac{1}{2}+x\right)^{300}-\left(\frac{1}{2}-x\right)^{300}\left(\frac{1}{2}+x\right)^{700} \end{aligned} Sum these to get $$f\left(\frac{1}{2}+x\right)+f\left(\frac{1}{2}-x\right)=0$$
$$f\left(\frac{1}{2}+x\right)=-f\left(\frac{1}{2}-x\right)$$ So $$f$$ is odd with respect to the point $$x_0=1/2$$ which means \begin{aligned} \int_0^1 f(x)\,dx&=\int_0^{1/2}f(x)\,dx+\int_{1/2}^1f(x)\,dx \\ &=-\mathcal{J}'+\mathcal{J'}\\ &=0. \end{aligned}
• The location of the very last sentence-ending punctuation in your answer was very strange. I fixed it. May 16 '20 at 23:07
• @MichaelHardy Thank you. May 16 '20 at 23:09
We know that the Beta Function is defined by $$\beta(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}\,\mathrm dx$$ Applying the King rule of integration $$\int_a^b f(x)\,\mathrm dx=\int_a^b f(a+b-x)\,\mathrm dx$$ we can write $$\beta(m,n)=\int_0^1(1-x)^{m-1}x^{n-1}\,\mathrm dx=\beta(n,m)$$
Our integral is of the form \begin{align}I&=\int_0^1x^{m-1}(1-x)^{n-1}\,\mathrm dx-\int_0^1x^{n-1}(1-x)^{m-1}\,\mathrm dx\\&=\beta(m,n)-\beta(n,m)\\&=\beta(m,n)-\beta(m,n)\\&=0\end{align} | 2021-12-01T16:30:18 | {
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https://www.physicsforums.com/threads/can-i-change-the-limits-of-this-double-integral.663070/ | # Can I change the limits of this double integral
1. Jan 8, 2013
### egroeg93
1. The problem statement, all variables and given/known data
R is the region bounded by y=x^2 and y=4. evaluate the double integral of f(x,y)=6x^2+2y over R
After drawing the region I was wondering if I could just work with the first quadrant and then double my solution, because both y=x^2 and y=4 are even functions so my question is does my solution work? If so would my very first line be correct? Oh and I'm not sure how to write what I'm integrating between so when I put ∫[a,b]f'(x)dx thats f(a)-f(b).
2. Relevant equations
3. The attempt at a solution
∫[4,0]∫[y^1/2,-y^1/2]6x^2+2y dxdy = 2*∫[4,0]∫[y^1/2,0] 6x^2+2y dxdy
following it through
= 2*∫[4,0] [[2(y^1/2)^3+2*y*y^1/2]-[0]] dy
= 2*∫[4,0] 4*y^3/2 dy
= 2*[[8/5*(4^5/2)]-[0]]
= 512/5
2. Jan 8, 2013
### lurflurf
That looks right at a glance. You can split it, but it does not help much.
write
$$\int_{-2}^2 \int_{x^2}^4 \! (6x^2+2y) \text{ dy dx}$$
or just
$$\int \limits_A \! (6x^2+2y) \text{ dA}$$
if you can't be bothered with the limits.
3. Jan 8, 2013
### egroeg93
Okay let me rephrase my question with the aid of proper symbols :)
$$\int_{-2}^2 \int_{x^2}^4 \! (6x^2+2y) \text{ dy dx}$$ =
$$2*{\int_{0}^4 \int_{0}^{y^{1/2}} \! (6x^2+2y) \text{ dx dy}}$$
Is this just coincidence or is this always true when you have a Region formed by two even functions as in the question?
4. Jan 8, 2013
### lurflurf
Yes that is true (with some conditions) it is called Fubini's theorem. It is not quite always true if you include improper integrals, for example this common example given in the above Wikipedia link.
$$\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{ dx dy}=-\int_0^1 \int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2} \text{ dy dx}$$
That is the type of thing considered in theoretical treatments of calculus, but it is good to be aware of. When the function either does not go to infinity or is absolutely integrable it is safe to interchange the integrals. Of course (6x^2+2y) is a very well behaved integrant. What you have done is express the region in two equivalent ways.
2<x<-2 and x^2<y<4
is the same as
0<y<4 and -sqrt(y)<x<sqrt(y)
and your symmetry argument that the integral over half the region equals half the integral over the whole region.
Last edited: Jan 8, 2013 | 2017-11-19T10:42:44 | {
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https://math.stackexchange.com/questions/linked/141196 | 68 questions linked to/from Highest power of a prime $p$ dividing $N!$
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### How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes? [duplicate]
Possible Duplicate: Highest power of a prime $p$ dividing $N!$ How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
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If we have some prime $p$ and a natural number $k$, is there a formula for the largest natural number $n_k$ such that $p^{n_k} | k!$. This came up while doing an unrelated homework problem, but it is ...
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I was just trying to work out the exponent for $7$ in the number $343!$. I think the right technique is $$\frac{343}{7}+\frac{343}{7^2}+\frac{343}{7^3}=57.$$ If this is right, can the technique be ...
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### What is the largest power of 2 that divides $200!/100!$. [duplicate]
What is the largest power of 2 that divides $200!/100!$. No use of calculator is allowed. I had proceeded in a brute force method which i know regret.. I would like to know your methods.
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### How many factors of $10$ in $100!$ [duplicate]
Possible Duplicate: Highest power of a prime $p$ dividing $N!$ How many factors of 10 are there in $100!$ (IIT Question)? Is it 26,25,24 or any other value Please tell how you have done it
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The formula for the exponent of the highest power of prime $p$ dividing $n!$ is $\sum \frac{n}{p^k}$, but the question is $n=1000!$ (really, it has the factorial) and $p=5$. When I use Wolfram Alpha ,...
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If $90! = (90)(89)(88)...(2)(1)$, then what is the exponent of the highest power of $2$ which will divide $90!$ ? How would I apply one of the easiest method from Here? I need help on applying the ...
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### Concecutive last zeroes in expansion of $100!$ [duplicate]
Possible Duplicate: Highest power of a prime $p$ dividing $N!$ In decimal form, the number $100!$ ends in how many consecutive zeroes. I am thinking of the factorization of $100!$ but I am stuck. ...
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### Determine the number of 0 digits at the end of 100! [duplicate]
I got this question, and I'm totally lost as to how I solve it! Any help is appreciated :) When 100! is written out in full, it equals 100! = 9332621...000000. Without using a calculator, determine ...
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### Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number. [duplicate]
Find the prime factor decomposition of $100!$ and determine how many zeros terminates the representation of that number. Actually, I know a way to solve this, but even if it is very large and ...
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### Dividing $61!$ by $3$ as we can. [duplicate]
Find maximum possible $n$ in the equation $61!=3^n\cdot m$. Some textbooks gives a solution to this question like this: \begin{align*} 61&=\textbf{20}\cdot 3+1 \\ 20&=\textbf{6}\cdot 3+2 \\ 6&...
112 views
### Why does $\sum\limits_{k=1}^\infty \lfloor m/(n^k)\rfloor$ give you the number of times that $n$ divides $m!$? [duplicate]
If $n$ is a prime less than $m$, with $n,m \in \mathbb N$, why does $$\sum_{k=1}^\infty \left\lfloor \frac{m}{n^k}\right\rfloor$$ give you the number of times that $n$ divides $m!$? Examples: $n=13$ ...
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### What is the logic/theorem/derivation behind finding the exponent of p in n! By [n/p] + [n/p^2] + [n/p^3] + …? [duplicate]
The exponent of prime number of 3 in 100! is 48. It means 100! is divisible by $3^48$ E_3(100!) = \left\lfloor\frac{100}3\right\rfloor + \left\lfloor\frac{100}{3^2}\right\rfloor + \left\lfloor\frac{...
### Prove the multiplicity property for $n!$ [duplicate]
I was given this hint in a different problem, Now use that a prime $p$ occurs in $n!$ with multiplicity exactly \$\lfloor n/p\rfloor + \lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor + \lfloor n/p^4\... | 2019-05-21T18:31:31 | {
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https://math.stackexchange.com/questions/3162558/curve-where-torsion-and-curvature-equal-arc-length | # Curve where torsion and curvature equal arc length
I study differential geometry independently in my free time as an undergraduate. I am using the book by Do Carmo.
I recently read the section and local theory of curves and learned about torsion and curvature.
My question is, does there exist a curve that has both torsion and curvature equal to arc length? I have tried deriving such a curve, but I’ve failed.
I speculate that it must be somewhat helical in nature.
Standard equation of helix is given by $$\alpha(s)=(a\cos(s/c),a\sin(s/c),b)$$. The curvature of such a curve is $$\kappa(s)=\frac{a}{a^2+b^2}$$ and torsion is $$\tau(s)=\frac{b}{a^2+b^2}$$.
Clearly $$a=\frac{1}{2}s^{-1}=b$$.
I’m not sure if this is the right approach to take. I feel as though the curve’s normal ought to trace out a curve on a sphere, but it doesn’t.
Any help is appreciated.
• For any nondegenerate curve $|{\bf N}(s)| = 1$, so ${\bf N}(s)$ traces a curve on the unit sphere. It's true for curves satisfying $\kappa(s) = \tau(s) = s$, however, that ${\bf N}(s)$ has image contained in a great circle on that sphere. See my answer for an explicit solution $\gamma(s)$. – Travis Willse Mar 26 '19 at 4:58
• The planar curve that has curvature proportional to the arc length is known as the clothoid. It resurfaces in the solution by Travis. – Yves Daoust Mar 26 '19 at 9:39
• For future reference, any curve with $\tau/\kappa$ constant is a generalized helix. – Ted Shifrin Mar 27 '19 at 0:18
In fact, given any functions $$\kappa, \tau : (a, b) \to \Bbb R$$ satisfying $$\kappa(s) > 0$$ for all $$s \in (a, b)$$,
1. there is a curve $$\gamma(s)$$ parameterized by arc length whose curvature is $$\kappa(s)$$ and whose torsion is $$\tau(s)$$, and
2. any two such curves $$\gamma_1, \gamma_2$$ are unique up to rigit motions of $$\Bbb R^3$$, that is, there is a rigid motion $$A$$ of $$\Bbb R^3$$ such that $$\gamma_2 = A \circ \gamma_1$$.
This appears in $$\S$$ 1.5 of do Carmo's text, where it's called the Fundamental Theorem of the Local Theory of Curves; see also the appendix to $$\S$$ 4. In Clelland's excellent From Frenet to Cartan: The Method of Moving Frames, this is Corollary 4.15, where it's presented as a motivating special case of more general result that applies far beyond Euclidean geometry.
Setting the curvature and torsion of a curve $$\gamma$$ to prescribed functions $$\kappa, \tau$$ results in a nonlinear, third-order system in three functions (the components of $$\gamma$$), so for general $$\kappa, \tau$$ one shouldn't expect to find explicit, closed-form solutions $$\gamma$$.
On the other hand, the conditions $$\kappa(s) = \tau(s) = s$$ are tractable enough to find an explicit solution. Substituting in the usual Frenet equations in matrix form gives \begin{align*} \pmatrix{{\bf T}'(s)&{\bf N}'(s)&{\bf B}'(s)} &= \pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} \pmatrix{\cdot&-\kappa(s)&\cdot\\\kappa(s)&\cdot&-\tau(s)\\\cdot&\tau(s)&\cdot} \\ &= \pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} \cdot s\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot} . \end{align*} Rearranging gives $$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)}^{-1} \frac{d}{ds}\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = s\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot},$$ and solving formally yields $$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = \pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)} \exp \left[\frac{1}{2} s^2\pmatrix{\cdot&-1&\cdot\\1&\cdot&-1\\\cdot&1&\cdot}\right] .$$ Since all solutions are the same up to rigid motions, we may as well take $$\pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)}$$ to be any (special orthogonal) matrix we like, and it turns out to be convenient to take (cf. J.M.'s comment) $$\pmatrix{{\bf T}(0)&{\bf N}(0)&{\bf B}(0)} = \pmatrix{ \frac{1}{\sqrt{2}}&\cdot&-\frac{1}{\sqrt{2}}\\ \cdot&1&\cdot\\ \frac{1}{\sqrt{2}}&\cdot&\frac{1}{\sqrt{2}}. }$$
We can also compute the matrix exponential explicitly, and putting this all together gives $$\pmatrix{{\bf T}(s)&{\bf N}(s)&{\bf B}(s)} = \pmatrix{ \frac{1}{\sqrt{2}} \cos \frac{1}{\sqrt{2}} s^2&\ast&\ast\\ \frac{1}{\sqrt{2}} \sin \frac{1}{\sqrt{2}} s^2&\ast&\ast\\ \frac{1}{\sqrt{2}} &\ast&\ast } .$$ For a curve parameterized by arc length, $${\bf T}(s) = \gamma'(s)$$, so we can recover an explicit formula for a solution $$\gamma(s)$$ by integrating $${\bf T}(s)$$. Taking the initial condition $$\gamma(0) = (0, 0, 0)$$ yields the solution $$\color{#df0000}{\boxed{\gamma(s) = \pmatrix{ \frac{1}{\sqrt{2}} \int_0^s \cos \frac{1}{\sqrt{2}} \tau^2 d\tau\\ \frac{1}{\sqrt{2}} \int_0^s \sin \frac{1}{\sqrt{2}} \tau^2 d\tau\\ \frac{1}{\sqrt{2}} s \\ }}}.$$ Optionally, we can rewrite $$\gamma$$ in terms of the Fresnel integrals, $$C(x) := \int_0^x \cos t^2 \,dt$$ and $$S(x) := \int_0^x \sin t^2 \,dt$$, as $$\gamma(t) = \pmatrix{\frac{1}{\sqrt[4]{2}} C\left(\frac{1}{\sqrt[4]{2}} s\right)\\\frac{1}{\sqrt[4]{2}} S\left(\frac{1}{\sqrt[4]{2}} s\right)\\\frac{1}{\sqrt{2}} s} .$$
A plot of our solution $$\gamma(s)$$, $$-12 \leq s \leq 12$$:
• For reference, here's the expression in terms of Fresnel integrals: $$\left(\frac{s}{2}+\frac{\sqrt{\pi}}{2 \sqrt[4]{2}}C\left(\frac{\sqrt[4]{2} s}{\sqrt{\pi }}\right)\quad\frac{\sqrt{\pi }}{2^{3/4}}S\left(\frac{\sqrt[4]{2} s}{\sqrt{\pi }}\right) \quad\frac{s}{2}-\frac{\sqrt{\pi }}{2 \sqrt[4]{2}}C\left(\frac{\sqrt[4]{2} s}{\sqrt{\pi }}\right)\right)^\top$$ tho this uses the convention where the Fresnel integrals have an extra $\pi/2$ factor. As for the Do Carmo reference... – J. M. is a poor mathematician Mar 26 '19 at 6:48
• Since I can't edit the last comment anymore: a clockwise rotation about the $y$-axis gives a simpler set of parametric equations: $$\left(\frac{\sqrt{\pi }}{2^{3/4}}C\left(\frac{\sqrt[4]{2} s}{\sqrt{\pi }}\right)\quad \frac{\sqrt{\pi }}{2^{3/4}}S\left(\frac{\sqrt[4]{2} s}{\sqrt{\pi }}\right)\quad \frac{s}{\sqrt{2}}\right)^\top$$ – J. M. is a poor mathematician Mar 26 '19 at 6:56
• Thanks for the helpful comments. I've added the reference to do Carmo and (I hope you don't mind) tweaked the choice of initial frame in my answer to yield your nicer parametric equation for $\gamma$. – Travis Willse Mar 26 '19 at 7:52
• Not at all! Choosing a different initial Frenet frame is entirely equivalent to rotating the curve, of course. :) – J. M. is a poor mathematician Mar 26 '19 at 7:55
• It's true that as $s \to \infty$ the curve is asymptotic to a straight line. For the convenient frame used in my answer, this line is $x = y = 2^{-7 / 4} \sqrt\pi$. Whether this means the curve becomes "more straight" is a matter of definition. Indeed, the angle between $\bf T$ and this asymptote is a constant $\frac{\pi}{4}$---this is possible because as $s$ increases the curve becomes more tightly wound, owing to the quadratic polynomial inside the arguments of $\cos$ and $\sin$. – Travis Willse Mar 26 '19 at 8:55 | 2020-03-29T10:34:45 | {
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https://math.stackexchange.com/questions/1717040/transformation-of-positive-semi-definite-matrices | # Transformation of positive semi-definite matrices
Let $a,b,c,d,e$ be positive reals such that the following matrix is positive semi-definite: $$\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\ a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a \\ \end{pmatrix}$$
Does it follow that also the following matrix is positive semi-definite? $$\begin{pmatrix} e & d & c \\ d & c & b \\ c & b & a \\ \end{pmatrix}$$
[By iterated subtraction of rows and columns, the two matrices have the same determinant; but do these operations preserve also this stronger property?]
Let
$$A = \begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a + 2b+c \\a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a\end{pmatrix}, \quad B = \begin{pmatrix} e & d & c \\ d & c & b \\ c & b & a \end{pmatrix}.$$
Then $A=UBU^T$ and $B=VAV^T$, where
$$U = \begin{pmatrix} 1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}, \quad V = U^{-1} = \begin{pmatrix} 1 & -2 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 1\end{pmatrix}$$
Therefore $A$ and $B$ represent the same quadratic form, but in different bases, hence $A$ is positive semi-definite if and only if $B$ is positive semi-definite.
Added to answer: $x^TBx = (V^Tx)^TA(V^Tx) \geq 0$, hence $B$ is positive semi-definite.
• The result generalizes to higher order matrices of those forms -- the rows of $U$ are the rows of the Pascal triangle. – Catalin Zara Mar 30 '16 at 1:07
Edit: I make a rectification of an erroneous extension to semi-definite matrices (following remarks of @Paolo Leonetti).
I will consider only the particular case of positive definite matrices according to Sylvester criterion (see https://en.wikipedia.org/wiki/Positive-definite_matrix). This criteria is applied by checking the positivity of determinants of "russian dolls matrices" beginning
• by the North-West corner $M(1,1), M(1:2,1:2)$ and $M(1:3,1:3)=M$, or
• by the South-East corner $M(3,3), M(2:3,2:3)$ and $M(1:3,1:3)=M$.
It is this latter criteria that we will consider.
$$A=\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\ a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a \\ \end{pmatrix} \ \ \text{and} \ \ B=\begin{pmatrix} e & d & c \\ d & c & b \\ c & b & a \\ \end{pmatrix}$$
Let $M_{23}=M(2:3,2:3)$ (we keep in a matrix $M$ the elements which are in column 2 and 3, i.e., we suppress line 1 and column 1).
Thus it suffices to check the equivalence:
$$det(A)>0 \ \& \ det(A_{23})> 0 \ \& \ a > 0 \ \ \Longleftrightarrow \ \ det(B)\geq0 \ \& \ det(B_{23})\geq 0 \ \& \ a \geq 0 \ \ \ (1)$$
But $det(A)=det(B)$ (you say it in your text), and one can verify that:
$$det(A_{23})=det(B_{23})=ac-b^2$$
Thus (1) is established.
Remark: coefficients $1,2,1$, then $1,3,3,1$, then $1,4,6,4,1$ that appear in matrix $A$ come from Pascal triangle. This remark could open the way to a new theorem and new proof, generalizable to larger matrices of this form.
• Where did you take such version of Sylvester criterion about positive semi-definite matrices? I believe too the answer is affirmative, but shouldn't we check all principal minors of $A$ and $B$? .. – Paolo Leonetti Mar 28 '16 at 21:12
• No, only the leading principal minors are to be checked, as explicitly said in the article of wikipedia I have cited. Here are their terms "The kth leading principal minor of a matrix M is the determinant of its upper-left k by k sub-matrix. It turns out that a matrix is positive definite if and only if all these determinants are positive". – Jean Marie Mar 28 '16 at 21:27
• Indeed, here we are not talking about positive definite matrices. – Paolo Leonetti Mar 28 '16 at 21:32
• What is the nature of my misunderstanding ? Do you mean positive semi-definiteness versus $definiteness$ (without the "semi") ? or is it the positivity of coefficients you are meaning ? – Jean Marie Mar 28 '16 at 21:41
• A matrix $M$ is positive semi-definite if $x^TMx$ is nonnegative for each vector $x$. On the other hand, positive definiteness require strictly positiveness. The version of Sylvester criterion which you are talking about is about the latter case. Of course, there are variants of such criterion also for PSD matrices, but probably your condition is not sufficient. [I am just aware of a version requiring to check all principal minors, which is not quoted in Wikipedia] – Paolo Leonetti Mar 28 '16 at 21:47 | 2019-09-18T07:54:42 | {
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https://math.stackexchange.com/questions/674730/if-we-specify-some-linear-map-why-is-it-tv-j-w-j | # If we specify some linear map, why is it $Tv_j = w_j$?
If we define a linear map $T: V \to W$ such that $(v_1,...,v_n)$ is a basis of $V$ and we have that:
$$T(a_1v_1 + ...+a_nv_n) = a_1w_1+....+a_nw_n$$ where $a_1,...,a_n \in \mathbb{F}$. Why is it the case that we must have $Tv_j = w_j$ for $j=1,...,n$?
Thanks!
Put $a_i=0$ for $i\neq j$ and $a_j=1$. Then the formula you give for $T$ gives $T(v_j)=w_j$.
• Yes that's exactly how you should think about $T$. – Brian Fitzpatrick Feb 14 '14 at 0:58 | 2019-11-16T21:13:23 | {
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https://math.stackexchange.com/questions/625806/trigonometry-identity-problem/625828 | # Trigonometry Identity Problem
Prove that:
$$\frac{\tan A + \sec A - 1}{\tan A - \sec A + 1} = \frac{1 + \sin A}{ \cos A}$$
I found this difficult for some reason. I tried subsituting tan A for sinA / cos A and sec A as 1/cos A and then simplifying but it didn't work.
• Sorry, I had a tough time with the MathJAX – user2130295 Jan 3 '14 at 12:19
• You are doing well with the MathJAX now. Putting a backslash before the functions will render them in the proper font. So \cos gives $\cos$ instead of cos giving $cos$ – Ross Millikan Jan 3 '14 at 12:20
• Try to simplify the left hand side. – mathlove Jan 3 '14 at 12:20
Setting $\displaystyle1=\sec^2A-\tan^2A$ in the numerator,
So, the numerator becomes $$\tan A+\sec A-1$$ $$=\tan A+\sec A-(\sec^2A-\tan^2A)$$ $$=(\sec A+\tan A)\{1-(\sec A-\tan A)\}$$ $$=\frac{(1+\sin A)}{\cos A}(1-\sec A+\tan A)$$
You may also simplify the expression by using the definition of the tangent and the secant function:
\begin{align} &\dfrac{\tan A +\sec A-1}{\tan A - \sec A +1} \\=&\dfrac{\dfrac{\sin A}{\cos A} +\dfrac1{\cos A}-1}{\dfrac{\sin A}{\cos A} - \dfrac1{\cos A} +1} \\=&\dfrac{\sin A -\cos A+1}{\sin A+\cos A-1} \\=&\dfrac{(\sin A -\cos A+1)\cdot(\sin A+\cos A+1)}{(\sin A+\cos A-1)\cdot(\sin A+\cos A+1)}=\dfrac{(\sin A +1)^2-\cos^2 A}{(\sin A+\cos A)^2-1} \\=&\dfrac{\sin^2 A+(1-\cos^2 A) +2\sin A}{2\sin A\cos A}\end{align}
The rest is fairly straightforward.
Multiplying the numerator & the denominator by $\cos A$
$$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$$
$$\cos A(\sin A+1-\cos A)=\cos A\sin A+\cos A-\cos^2A$$ $$=\cos A(1+\sin A)-(1-\sin^2A)$$ $$\implies\cos A(\sin A+1-\cos A)=(1+\sin A)(\cos A-1+\sin A)$$
$$\implies\frac{\sin A+1-\cos A}{\sin A-1+\cos A}=?$$
We can start with $(1+\sin A)(\sin A-1+\cos A)$ as well
Find some similar results here
• @user2130295, here is another method – lab bhattacharjee Jan 3 '14 at 19:13 | 2019-12-16T02:50:16 | {
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https://math.stackexchange.com/questions/2856250/suppose-x-is-an-integer-such-that-3x-equiv-15-pmod64-find-remainder-whe | Suppose $x$ is an integer such that $3x \equiv 15 \pmod{64}$. Find remainder when $q$ is divided by $64$.
Suppose $x$ is an integer such that $3x \equiv 15 \pmod{64}$. If $x$ has remainder $2$ and quotient $q$ when divided by $23$, determine the remainder when $q$ is divided by $64$.
I tried a couple things. By division algorithm, know $x = 23q + 2$. So $3(23q + 2) \equiv 15 \pmod{64}$. Not sure how to go from there.
Another thing I tried is $3x = 64q + 15$. If we let $q$ be zero, then $x$ is obviously $5$. This also doesn't wind up being that helpful, and I think $x$ can have other values asides from $5$.
• So $69q+6\equiv 5q+6\equiv 15\mod 64$. So $5q\equiv 9\mod 64$. Add or subtract 64 from 9 to get a multiple of 5. I'd do $5q\equiv 9-64 =-55 \mod 64$. So $q\equiv -11\equiv 53 \mod 64$ – fleablood Jul 19 '18 at 6:20
4 Answers
Use as many variables as you like, if you are not comfortable with the modulus notation. We can then work with them as ordinary equations, and see how to derive any conclusion.
For example, $3x \equiv 15 \mod 64$ means that $3x-15 = 64k$ for some integer $k$. Now, $3$ divides the left hand side, therefore the right as well. Using a lemma about primes, $3 | 64$ or $3 | k$. The former is not true, so $3 | k$ is true. Let $k = 3m$, then cancelling $3$ we get $x - 5 = 64m$.
Write $x = 23q + 2$. It follows that $23q = x - 2 = x-5+3 = 64m+3$.
So, $23q -3 = 64m$. What we will do now, is multiply both sides by a very special number, $39$. $$23q - 3 = 64m \implies 897 q - 117 = 64 \times 39m \implies q - 117 = 64 \times 39m - 64 \times 14q\\ \implies q = 64(39m - 14q) + 117 \implies q = 64(39m - 14q + 1) + 53$$
Therefore, $q$ leaves a remainder of $53$ when divided by $64$.
Why did I multiply by $39$? What I wanted to do, actually, is to show that $q$ is a multiple of $64$ pus some remainder. The only way to isolate a single $q$ from the given equation, rather than $23q$, was so that I could remove exactly one $q$, and the remainder would be a multiple of $64$ (namely, $64 \times 14 = 896$) which I could send to the quotient side. The smallest number with which I could do this was $39$, since $23 \times 39 = 64 \times 14 + 1$.
$39$ is said to be the inverse of $23$ modulo $64$ for this reason.
• Is there an easy way to calculate the inverse of any modulo expression? – SolidSnackDrive Jul 19 '18 at 19:23
• I would also like to know if we can still find the inverse even in the case where $ax \equiv r \text{ mod} p$, such that gcd$(a,p) \neq 1$ – john fowles Jul 19 '18 at 21:08
• @SolidSnackDrive It can be done using the reverse Euclidean algorithm. Essentially, you perform the Euclidean algorithm used to find the gcd of two numbers, and then a "reverse substitution" gives you the desired inverse. Taking a small example, if you want the inverse of $4$ modulo $7$, then we do : $7 = 1 \times 4 + 3$, and $4 = 1 \times 3 + 1$. Now, $1 = 4 - 3 = 4 - (7 - 4) = 2 \times 4 - 7$, so $2 \times 4 = 1 \times 7 + 1$ ,and the inverse of $4$ is $2$. However, read my next comment. – Teresa Lisbon Jul 21 '18 at 3:14
• @johnfowles If $\gcd(a,p) \neq 1$, then solutions to the equation you have given exist if and only if $r$ is a multiple of the gcd also. For example, there are no solutions to $2x \equiv 3 \mod 4$, since $\gcd(2,4) = 2$ but $3$ is not a multiple of $2$. IF $r$ is a multiple, then write $n = \gcd(a,p)$, so $r = kn$. Now, $ax \equiv kn \mod p$, so $ax - kn = lp$ for some $l$. Now, if $a = bn$ and $p = qn$, then we can cancel $n$ to get $bx - k = lq$, or $bx \equiv k \mod q$, where $\gcd(b,q) = 1$. Back to square one : solve for $x$! This $x$ will also satisfy the earlier equation. – Teresa Lisbon Jul 21 '18 at 3:19
Continuing from where you stopped: \begin{align} 3(23q + 2) &\equiv 15 \pmod{64} \Rightarrow \\ 69q+6 &\equiv 15 \pmod{64} \Rightarrow \\ 69q &\equiv 9 \pmod{64} \Rightarrow \\ 64q+5q &\equiv 9 \pmod{64} \Rightarrow \\ 5q &\equiv 9 \pmod{64} \Rightarrow \\ 5q\cdot 13 &\equiv 9\cdot 13 \pmod{64} \Rightarrow \\ 65q &\equiv 117 \pmod{64} \Rightarrow \\ 64q+q &\equiv 64+53 \pmod{64} \Rightarrow \\ q &\equiv 53 \pmod{64}. \end{align}
$3x\equiv 15\pmod{64}$ means
$64\mid (3x-15)$
$64\mid 3(x-5)$
Since $64$ and $3$ have no common factors, it follows that $64$ must divide the remaining factor $x-5$ : $$x\equiv 5\pmod {64}$$
Directly plugin the given info $x=23q+2$ : $$23q+2\equiv 5 \pmod{64}$$
Subtract $2$ both sides and see if you can try the rest.
By division algorithm, know $x = 23q + 2$. So $3(23q + 2) \equiv 15 (mod 64)$.
Hint: then $\,15 = 69 q + 6 \equiv 5q + 6 \pmod{64} \implies 5q \equiv 9 \implies 65q \equiv 117 \implies \ldots\,$ | 2021-01-28T03:29:15 | {
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https://www.physicsforums.com/threads/combinatorics-problem.890580/ | # Homework Help: Combinatorics problem
1. Oct 24, 2016
### Mr Davis 97
1. The problem statement, all variables and given/known data
In how many ways can we pick a group of 3 different numbers from the group $1,2,3,...,500$ such that one number is the average of the other two? (The order in which we pick the numbers does not matter.)
2. Relevant equations
3. The attempt at a solution
I start by noting that in order to get a number in the group $1,2,3,...,500$ from the average of two other numbers from the group, those two numbers must be odd, since their sum must be even. There are 250 odd numbers from 1 to 500. Since the numbers have to be different, we have $250 \cdot 249$ ways to find two odd numbers from the list. However we are given that order does not matter, so we must divide by 2 to get $\displaystyle \frac{250 \cdot 249}{2} = 31125$
However, the correct answer is apparently 62250, which is my answer times 2. My question is, since we are given that order does not matter, don't we have to divide by 2? Since x + y is not different than y + x? Where does my logic go wrong when I think that we should divide by 2?
2. Oct 24, 2016
### Fightfish
What is the sum of two even numbers?
3. Oct 24, 2016
### Mr Davis 97
Ohhh... Don't know how I missed that. So the 31125 is correct for the odd numbers, but then we must add this to 31125, which is the number of even number pairs, which gets us to 62250. | 2018-05-24T02:59:05 | {
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https://yutsumura.com/if-two-matrices-have-the-same-rank-are-they-row-equivalent/ | # If Two Matrices Have the Same Rank, Are They Row-Equivalent?
## Problem 644
If $A, B$ have the same rank, can we conclude that they are row-equivalent?
If so, then prove it. If not, then provide a counterexample.
## Solution.
Having the same rank does not mean they are row-equivalent.
For a simple counterexample, consider $A = \begin{bmatrix} 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 \end{bmatrix}$.
Both of these matrices have rank 1, but are not row-equivalent because they are already in reduced row echelon form.
## Another solution.
The problem doesn’t specify the sizes of matrices $A$, $B$.
Note that if the sizes of $A$ and $B$ are distinct, then they can never be row-equivalent.
Keeping this in mind, let us consider the following two matrices.
$A=\begin{bmatrix} 1 \\ 0 \end{bmatrix} \text{ and } B=\begin{bmatrix} 1 & 0 \end{bmatrix}.$ Then both matrices are in reduced row echelon form and have rank $1$.
As noted above, they are not row-equivalent because the sizes are distinct.
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##### Find a Row-Equivalent Matrix which is in Reduced Row Echelon Form and Determine the Rank
For each of the following matrices, find a row-equivalent matrix which is in reduced row echelon form. Then determine the...
Close | 2018-07-18T14:39:58 | {
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https://math.stackexchange.com/questions/1269342/to-find-eigenvalues/1269362 | # To Find Eigenvalues
Find the eigenvalues of the $6\times 6$ matrix
$$\left[\begin{matrix} 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ \end{matrix}\right]$$
The options are $1, -1, i, -i$
It is a real symmetric matrix and the eigenvalues of a real symmetric matrix are real. Hence $i$ and $-i$ can't be its eigenvalues. Then what else we can say?
Is there any easy way to find it?
• You know what, sum of eigenvalues is equal to the trace(matrix), also product of eigenvalues is equal to det(matrix). – Ajay May 6 '15 at 5:58
• To add to what RattusRattus said: The trace is quite obviously 0. As Tani has already found, i and -i can't be eigenvalues. That leaves only 1 and -1, the sum of which must equal 0. Thus the eigenvalues are 1, 1, 1, -1, -1, and -1. – Yay295 May 6 '15 at 8:06
Consider the following permutation, $$\sigma=\left(\begin{matrix}1 & 2 & 3 & 4 & 5 &6\\4&5&6&1&2&3\end{matrix}\right)$$where , the matrix $A$ is defined to be the one whose $i$-th column is the $\sigma(i)$-th column of the Identity matrix. Then order of the permutation $\sigma$ is $2$. So, $A^2=I$. So, $x^2-1$ is an anihilating polynomial of $A$. As, $A\not =\pm I$ so, $x^2-1$ is the minimal polynomial of $A$. So, $\pm 1$ are the eigen values of $A$.
• different and simple idea. – Tani May 6 '15 at 13:30
• excellent idea +1 – Learnmore May 9 '15 at 3:28
• If you wouldn't mind, could you explain how you deduce that $x^2-1$ is an annihilating polynomial? I haven't had advanced linear yet, but I understand all of your argument except that part. – Alex Mathers May 9 '15 at 3:59
• @mathers101 To say that $p(x)$ is annihilating polynomial just means that $p(A)=0$. Here, we have $A^2-I=0$. – Potato May 9 '15 at 4:22
• Did you not understand what moderators have told you about bumping your posts with insignificant edits? – Jyrki Lahtonen Jun 7 '15 at 16:11
Note that , $$A=\left[\begin{matrix}O&I\\I&O\end{matrix}\right].$$where, $I_{3\times 3}$ and $O_{3\times 3}$ are Identity matrix and Zero matrix respectively.
Now, $AA^T=A^2=I_{6\times 6}\implies A$ is orthogonal. So, eigen values are either $1$ or $-1$. Again, $det(A)=-1$.
If all eigen values are $1$ then $det(A)=1$ , contradiction. Again if all eigen values are $-1$ then $det(A)=1$ ,again a contradiction.
Hence , $1$ and $-1$ both are eigen values of $A$.
The matrix has the following effect on the standard basis vectors:
It interchanges $e_1$ with $e_4$, $e_2$ with $e_5$, and $e_3$ with $e_6$. So the three linearly independent (check it!) vectors $e_1+e_4, e_2+e_5, e_3+e_6$ are fixed and so they are eigenvectors with eigenvalue $+1$.
Also check that $e_1-e_4, e_2-e_5, e_3-e_6$ are sent to their negatives, hence are eigenvectors of eigenvalue $-1$.
As these 6 vectors form a basis, $+1$ and $-1$ are the only eigenvalues both of multiplicity $3$.
Let's call the matrix $A$, and write $A$ as
$A = \begin{bmatrix} 0_3 & I_3 \\ I_3 & 0_3 \end{bmatrix}, \tag{1}$
where $0_3$ and $I_3$ are the $3 \times 3$ zero and identity matrices, respectively. It is easy to see from block matrix multiplication that
$A^2 = I_6, \tag{2}$
which in turn implies that any eigenvalues $\lambda$ satisfy the equation
$\lambda^2 = 1 \tag{3}$
since
$Av = \lambda v \tag{4}$
for some $v \ne 0$; thus
$v = I_6v = A^2 v = \lambda A v = \lambda^2 v \tag{5}$
or
$(\lambda^2 - 1)v = 0, \tag{6}$
forcing (3) since $v$ doesn't vanish. Thus the only eigenvalues $A$ may have are $\lambda = \pm 1$. In fact, both possibilities occur: for any column three-vector $x \ne 0$, the six-vector
$w = \begin{pmatrix} x \\ x \end{pmatrix} \tag{7}$
satisfies
$Aw = w, \tag{8}$
whereas
$y = \begin{pmatrix} x \\ -x \end{pmatrix} \tag{9}$
solves
$Ay = -y; \tag{10}$
thus both possibilities $\lambda = \pm 1$ occur; the eigenvalues of $A$ are precisely $\pm 1$.
Well, that seems to me like a pretty easy way to do it; we didn't have to evaluate any $6 \times 6$ determinants or do a lot of arithmetic.
Finally, the above easily generalizes to show that the eigenvalues of the $2n \times 2n$ matrix
$\begin{bmatrix} 0_n & I_n \\ I_n & 0_n \end{bmatrix} \tag{11}$
are also exactly $1, -1$.
Here as you say it's a real symmetric matrix so all the eigen values are real. Now we know that the sum of eigen values is equal to the trace of the matrix. Here trace of the matrix is equal to $0$, so if we take only $1$ or $-1$ as the eigen value, the trace becomes non-zero, so both $1$ and $-1$ are the eigen values.
We can also change basis to get $$I_3 \otimes C_2 = \left(\begin{array}{ccc}C_2&0&0\\0&C_2&0\\0&0&C_2\end{array}\right)$$ where $C_2 = \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$ is a representation for the generating element of the cyclic 2-group. For this small matrix it is easy to find value 1,-1 and vectors $\left(\begin{array}{c}1\\1\end{array}\right)$ and $\left(\begin{array}{r}1\\-1\end{array}\right)$. Please note that the eigenvectors are different from my other answer above, but this is OK (why?)
• It is because there may be many matrix representations of a group but only one which is irreducible. The FFT one with circulant matrices above is not irreducible, but this one is. – mathreadler May 6 '15 at 7:19
• This is the easiest method to get all eigenvalues and their multiplicities at once. Moreover it would work (with different size cycle matrices but not with the tensor product) for any permutation matrix (once you know that the eigenvalues of a $k$-cycle matrix are precisely the $k$-th roots of unity). Moreover, of all the properties that one immediately reads off of that matrix, being a permutation matrix is one for the most obvious ones. – Marc van Leeuwen May 7 '15 at 10:03
The matrix is circulant. Therefore $SDS^{-1}$ is it's eigendecomposition, where $S$ is the fast fourier transform of $I_6$ and $D$ is the diagonal matrix with elements from the FFT of the first row, which happens to be $diag([1,-1,1,-1,1,-1])$. So the eigenvalues are $-1$ and $1$.
Because sum of elements of each row is equal to one so one is an eigen value of the matrix. Again if we put $x=-1$ in $|A-xI|$ we get two row are equal so $-1$ is also an eigen value of $A$.
• yes its right.. – neelkanth May 6 '15 at 7:44
• its simplest way to check eigen values... – neelkanth May 6 '15 at 7:45
• How you say that sum of each row is $1$ implies $1$ is an eigen value ? Please explain.. – Empty May 6 '15 at 8:16
• Yes corresponding vector will be (1,1,...1) now you can check easily... – user237521 May 6 '15 at 8:18
• Actually if the sum of entries is the same constant for all rows (think of magic squares) that constant will be an eigenvalue with $(1,1,\ldots1)^T$ as eigenvector. – P Vanchinathan May 8 '15 at 7:29
The given matrix is a non-zero orthogonal symmetric matrix with trace zero, so its eigenvalues are $1$ and $-1$. | 2019-06-17T19:03:56 | {
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https://ocw.mit.edu/courses/mathematics/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/video-lectures/lecture-9-four-ways-to-solve-least-squares-problems/ | Lecture 9: Four Ways to Solve Least Squares Problems
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Description
In this lecture, Professor Strang details the four ways to solve least-squares problems. Solving least-squares problems comes in to play in the many applications that rely on data fitting.
Summary
1. Solve $$A^{\mathtt{T}} Ax = A^{\mathtt{T}}b$$ to minimize $$\Vert Ax - b \Vert^2$$
2. Gram-Schmidt $$A = QR$$ leads to $$x = R^{-1} Q^{\mathtt{T}}b$$.
3. The pseudoinverse directly multiplies $$b$$ to give $$x$$.
4. The best $$x$$ is the limit of $$(A^{\mathtt{T}}A + \delta I)^{-1} A^{\mathtt{T}}b$$ as $$\delta \rightarrow 0$$.
Related section in textbook: II.2
Instructor: Prof. Gilbert Strang
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PROFESSOR: Well, OK. So first important things about the course, plans for the course. And then today I'm going to move to the next section of the notes, section 2, or part 2, I should say. And actually I'll skip for the moments section 2-1 and go to section 2-2, and all of chapter 2 will come to you probably today or latest tomorrow. So that's where we're going next.
I'm following the notes pretty carefully, except I'm going to skip the section on tensors until I learn more basically. Yeah. Yeah. I could say a little about tensors, but this flows naturally using the SVD. So it's just a terribly important problem, least squares. And of course, I know that you've seen one or two ways to do least squares. And really the whole subject comes together.
Here I want to say something, before I send out a plan for looking ahead for the course as a whole. So there's no final exam. And I don't really see how to examine you, how to give tests. I could, of course, create our tests about the linear algebra part. But I don't think it's-- it's not sort of the style of this course to expect you quickly to create a proof for something in class.
So I think, and especially looking at what we're headed for, and moving quite steadily in that direction, is all the problems that this linear algebra is is aimed at, right up to and including conjugate gradient descent and deep learning, the overwhelmingly important and lively, active research area. I couldn't do better than to keep the course going in that direction.
So I think what I would ask you to do is late in sort of April, May, the regular homeworks I'll discontinue at a certain point. And then instead, I'll be asking and encouraging a project-- I don't know if that's the right word to be using-- in which you use what we've done. And I'll send out a message on Stellar listing five or six areas and only-- I mean, one of them is the machine learning, deep learning part. But they're all the other parts, things we are learning how to do. How to find sparse solutions, for example, or something about the pseudo inverse. All kinds of things.
So that's my goal, is to give you something to do which uses the material that you've learned. And look, I'm not expecting a thesis. But it's a good chance. So it will be more than just, drag in some code for deep learning and some data matrix and do it. But we'll talk more as the time comes.
So I just thought I'd say, before sending out the announcement, I would say it's coming about what as a larger scale than single one week homeworks would be here before. Any thoughts about that? I haven't given you details. So let me do that with a message, and then ask again. But I'm open to-- I hope you've understood-- I think you have-- that if you make suggestions, either directly to my email or on Piazza or whatever, they get paid attention to.
OK. Shall I just go forward with least squares? So what's the least squares problem, and what are these four ways, each bringing-- so let me speak about the pseudo inverse first. OK, the pseudo inverse of a matrix. All right. Good.
So we have a matrix A, m by n. And the pseudo inverse I'm going to call A plus. And it naturally is going to be n by m. I'm going to multiply those together. And I'm going to get as near to the identity as I can. That's the idea, of course, of the pseudo inverse, The word pseudo is in there, so no one's deceived. It's not an actual inverse.
Oh, if the matrix is square and has an inverse, of course. Then if A inverse exists, which requires-- everybody remembers it requires the matrix to be square, because I mean inverse on both sides. And it requires rank n, full rank. Then the inverse will exist. You can check it. MATLAB would check it by computing the pivots in elimination and finding n pivots.
So if A inverse exists, which means A times A inverse, and A inverse times A, both give I, then A plus is A inverse, of course. The pseudo inverse is the inverse when there is one. But I'm thinking about cases where either the matrix is rectangular, or it has zero eigenvalues. It could be square, but it has a null space, other than just the 0 vector. In other words, the columns are dependent.
What can we do then about inverting it? We can't literally invert it. If A has a null space, then when I multiply by a vector x in that null space, Ax will be 0. And when I multiply by A inverse, still 0. That can't change the 0. So if there is an x in the null space, then this can't happen. So we just do the best we can. And that's what this pseudo inverse is.
And so let me draw a picture of the picture you know of the row space and the null space. OK, and it's there, you see. There is a null space. And over here I have the column space and the null space of A transpose. OK. So this is the row space, of course. That's the column space of A transpose, and there is the column space of A. OK.
So which part of that picture is invertible, and which part of the picture is hopeless? The top part is invertible. This is the r-dimensional row space, r-dimensional column space. A takes a vector in here, zaps it into every-- you always end up in the column space. Here I take a vector in the row space-- say, x-- and it gets mapped to Ax.
And between those two spaces, A is entirely invertible. You get separate vectors here, go to separate vectors in the column space, and the inverse just brings it back. So we know what the pseudo inverse should do. It will take A will go that way, and A plus, the pseudo inverse will be just-- on the top half of the picture, it'll give us A plus. We'll take Ax back to x in the top half.
Now, what about here? That's where we have trouble, when we don't have-- that's what spoils our inverse. If there is a null space vector, then it goes where? When you multiply by A, this guy in the null space goes to 0. Usually along a straighter line than I've drawn. But it goes there. It gets to 0.
So you can't raise it from the dead, so to speak. You can't recover it when there's no A inverse. So we have to think, what shall A inverse do to this space here, where nobody's hitting it? So this would be the null space of A transpose. Because A-- sorry-- yeah, what should the pseudo inverse do? I said what should the inverse do? The inverse is helpless.
But we have to define A plus. I've said what it should do on that guy, on the column space. It should take everything in the column space back where it came from. But what should it do on this orthogonal space, where-- yeah, just tell me, what do you think? If I have some vector here-- let's call it V r plus 1. That would be like-- so here I have a nice basis for the column space. I would use V's for the ones that come up in the SVD. They're orthogonal, and they come from orthogonal U's.
So the top half is great. What shall I do with this stuff? I'm going to send that back by A plus. And what am I going to do with it? Send it to-- nowhere else could it go. 0 is the right answer. All this stuff goes back to 0.
I'm looking for a linear operator, a matrix. And I have to think, once I've decided what to do with all those and what to do with all these, then I know what to do with any combination. So I've got it. I've got it. So the idea will be, this is true for x in the row space. For x in the row space, if x is in the row space, Ax is in the column space, and A inverse just brings it back as it should.
And in the case of an invertible matrix A, what happens to my picture? What is this picture looking like if A is actually a 6 by 6 invertible matrix? In that case, what's in my picture and what is not in my picture? All this null space stuff isn't there. And null space is just a 0 vector. But all that I don't have to worry about. But in general, I do have to say.
So the point is that A plus on the-- what am I calling this? It's the null space of A transpose, or whatever on V r plus 1 to Vn, all those vectors, the guys that are not orthogonal to the column space. Then we have to say, what does A plus do to them? And the answer is, it takes them all to 0.
So there is a picture using what I call the big picture of linear algebra, the four spaces. You see what A plus should do. Now, I need a little formula for it. I've got the plan for what it should be, and it's sort of the natural thing. So A plus A is, you could say it's a projection matrix. It's not the identity matrix because if x is in the null space, A plus A will take it to 0. So it's a projection.
A plus A is the identity on the top half, and 0 on the bottom half. That's really what the matrix is. And now, I want a simple formula for it. And I guess my message here is, that if we're looking for a nice expression, start with the SVD. Because the SVD works for any matrix. And it writes it as an orthogonal matrix times a diagonal matrix times an orthogonal matrix.
And now I want to invert it. Well, suppose A had an inverse. What would that be? This is if invertible, what would be the SVD of A inverse? What would be the singular value decomposition, if this is good? So when is this going to be good? What would I have to know about that matrix sigma, that diagonal matrix in the middle, if this is truly an invertible matrix?
Well, no. What's its name? Those are not eigenvalues. Well, they're eigenvalues of A transpose A. But they're singular values. Singular value, that's fine. So that's the singular value matrix. And what would be the situation if A had an inverse? There would be no 0's. All the singular values would be sitting there, sigma 1 to sigma n.
What would be the shape of this sigma matrix? If I have an inverse, then it's got to be square n by n. So what's the shape of the sigma guy? Also square, n by n. So the invertible case would be-- and I'm going to erase this in a minute-- the invertbile case would be when sigma is just that. That would be the invertible case.
So let's see. Can you finish this formula? What would be the SVD of A inverse? So I'm given the SVD of A. I'm given the U and the sigma is cool and the V transpose. What's the inverse of that? Yeah, I'm just really asking what's the inverse of that product of three matrices.
What comes first here? V. The inverse of V transpose is V. That's because V is a orthogonal matrix. The inverse of sigma, just 1 over it, is just the sigma inverse. It's obvious what that means. And the inverse of U would go here. And that is U transpose. Great. OK.
So this is if invertible. If invertible, we know what the SVD of A inverse is. It just takes the V's back to the U's, or the U's back to the V's, whichever. OK. OK. Now we've got to do it, if we're going to allow-- if we're going to get beyond this limit, this situation, allow the matrix sigma to be rectangular. Then let me just show you the idea here.
So now I'm going to say, now sigma, in general, it's rectangular. It's got r non 0's on the diagonal, but then it quits. So it's got a bunch of 0's that make it not invertible. But let's do our best and pseudo invert it. OK. So now help me get started on a formula for using-- I want to write this A plus, which I described up there, in terms of the subspaces. Now I'm going to describe A plus in terms of U, sigma, and V, the SVD guys.
OK. So what shall I start with here? Well, let me give a hint. That was a great start. My V is still an orthogonal matrix. V transpose is still an orthogonal matrix. I'll invert it. At the end, the U was no problem. All the problems are in sigma. And sigma, remember, sigma-- so it's rectangular. Maybe I'll make it wide, two wide. And maybe I'll only give it two non-zeros, and then all the rest. So the rank of my matrix A is 2, but the m and n are bigger than 2. It's just got two independent columns, and then it's just sort of totally singular.
OK. So my question is, what am I going to put there? And I've described it one way, but now I'm going to describe it another way. Well, let me just say, what I'll put there is the pseudo inverse of sigma. I can't put sigma inverse using that symbol, because there is no such thing. With this, I can't invert it. So that's the best I can do.
So I'm almost done, but to finish, I have to tell you, what is this thing? So sigma plus. I'm now going to tell you sigma plus. And then that's what should sit there in the middle. So if sigma is this diagonal matrix which quits after two sigmas, what should sigma plus be? Well, first of all, it should be rectangular the other way. If this was m by n column, n columns and m rows, now I want to have n rows and m columns.
And yeah, here's the question. What's the best inverse you could come up with for that sigma? I mean, if somebody independent of 18.065, if somebody asks you, do your best to invert that matrix, I think we'd all agree it is, yeah. One over the sigma 1 would come there. And 1 over sigma 2, the non zeros.
And then? Zeros. Just the way up there, when we didn't know what to do, when there was nothing good to do. Zero was the right answer. So this is all zeros. Of course, it's rectangular the other way. But do you see that if I multiply sigma plus times sigma, if I multiply the pseudo inverse times the matrix, what do I get if I multiply that by that? What does that multiplication produce? Can you describe the-- well, or when you tell me what it looks like, I'll write it down.
So what is sigma plus times sigma? If sigma is a diagonal, sigma plus is a diagonal, and they both quit after two guys. What do I have? One? Because sigma 1 times 1 over sigma 1 is a 1. And the other next guy is a 1. And the rest are all zeros. That's right. That's the best I could do. The rank was only two, so I couldn't get anywhere. So that tells you what sigma plus is.
OK. So I described the pseudo inverse then with a picture of spaces, and then with a formula of matrices. And now I want to use it in least squares. So now I'm going to say what is the least squares problem. And the first way to solve it will be to involve-- A plus will give the solution.
OK. So what is the least squares problem? Let me put it here. OK, the least squares problem is simply, you have an equation, Ax equals b. But A is not invertible. So you can't solve it. Of course, for which-- yeah, you could solve it for certain b's. If b is in the column space of A, then just by the meaning of column space, this has a solution. The vectors in the column space are the guys that you can get.
But the vectors in the orthogonal space you cannot get. All the rest of the vectors you cannot get. So suppose this is like so, but always A is m by n rank r. And then we get A inverse when m equals n equals r. That's the invertible case. OK.
What do we do with a system of equations when we can't solve it? This is probably the main application in 18.06. So you've seen this problem before. What do we do if Ax equal b has no solution? So typically, b would be a vector of measurements, like we're tracking a satellite, and we get some measurements. But often we get too many measurements. And of course, there's a little noise in them. And a little noise means that we can't solve the equations.
That may be the case everybody knows is, where this equation is like expressing a straight line going through the data points. So the famous example of least squares is fit a straight line to the b's, to b1, b2. We've got m measurements. We've got m measurements. The physics or the mechanics of the problem is pretty well linear. But of course, there's noise.
And a straight line only has two degrees of freedom. So we're going to have only two columns in our matrix. A will be only two columns, with many rows. Highly rectangular. So fit a straight line. Let me call that line Cx plus D. Say this is the x direction. This is the b's direction. And we've got a whole bunch of data points. And they're not on a line.
Or they are on the line. Suppose those did lie on a line. What would that tell me about Ax equal b? I haven't said everything I need to, but maybe the insight is what I'm after here. If my points are right on the line-- so there is a straight line through them-- the unknowns here-- so let me-- so Ax-- the unknowns here are C and D. And the right hand side is all my measurements. OK.
Suppose-- without my drawing a picture-- suppose these points are on the line. Here's the different x's, the measurement times. Here is the different measurements. But if they're on a line, what does that tell me about my linear system, Ax equal b? It has a solution. Being on a line means everything's perfect. There is a solution.
But will there usually be a solution? Certainly not. If I have only two parameters, two unknowns, two columns here, the rank is going to be two. And here I'm trying to hit any noisy set of measurements. So of course, in general the picture will look like that. And I'm going to look for the best C and D. So I'll call it Cx plus D. Yeah, right. Sorry. That's my line. So those are my equations.
Sorry, I often write it C plus dx. Do you mind if I put the constant term first in the highly difficult equation here for a straight line? So let me tell you what I'm-- so these are the points where you have a measurement-- x1, x2, up to xn. And these are the actual measurements, b1 up to bm, let's say .
And then my equations are-- I just want to set up a matrix here. I just want to set up the matrix. So I want C to get multiplied by ones every time. And I want D to get multiplied by these x's-- x1, x2, x3, to xm, the measurement places. And those are the measurements. Anyway.
And my problem is, this has no solution. So what do I do when there's no solution? Well, I'll do what Gauss did. He was a good mathematician, so I'll follow his advice. And I won't do it all semester, as you know. But Gauss's advice was, minimize-- I'll blame it on Gauss-- the distance between Ax and b squared, the L2 norm squared, which is just Ax minus b transpose Ax minus b. It's a quadratic. And minimizing it gives me a system of linear equations.
So in the end, they will have a solution. So that's the whole point of least squares. We have an unsolvable problem, not no solution. We follow Gauss's advice to get the best we can. And that does produce an answer.
So this is-- if I multiply this out, it's x transpose, A transpose, Ax. That comes from the squared term. And then I have probably these-- actually, probably I'll get two of those, and then a constant term that has derivative 0 so it doesn't enter. So this is what I'm minimizing. This is the loss function.
And it leads to-- let's just jump to the key here. What equation do I get when I look for-- what equation is solved by the best x, the best x? The best x solves the famous-- this is regression in statistics, linear regression. It's one of the main computations in statistics, not of course just for straight line fits, but for any system Ax equal b.
That will lead to-- this minimum will lead to a system of equations that I'm going to put a box around, because it's so fundamental. And are you willing to tell me what that equation is? Yes, thanks.
AUDIENCE: A transpose A.
PROFESSOR: A transpose A is going to come from there-- you see it-- times the best x equals A transpose b. That gives the minimum. Let me forego checking that. You see that the quadratic term has the matrix in it. So it's derivative. Maybe the derivative of this is 2 A transpose Ax, and then the 2 cancels that 2. And this could also be written as x transpose A transpose b. So it's x transpose against A transpose b. That's linear. So when I take the derivative, it's that constant.
That's pretty fast. 18.06 would patiently derive that. But here, let me give you the picture that goes with it, the geometry. So we have the problem. No solution. We have Gauss's best answer. Minimize the 2 norm of the error. We have the conclusion, the matrix that we get in. And now I want to draw a picture that goes with it. OK.
So here is a picture. I want to have a column space of A there in that picture. Of course, the 0 vector's in the column space of A. So this is all possible vectors Ax. Right? You're never forgetting that the column space is all the Ax's.
Now, I've got to put b in the picture. So where does this vector b-- so I'm trying to solve Ax equal b, but failing. So if I draw b in this picture, how do I draw b? Where do I put it? Shall I put it in the column space? No. The whole point is, it's not in the column space. It's not an Ax. It's out there somewhere, b. OK. And then what's the geometry that goes with least squares and the normal equations and Gauss's suggestion to minimize the error? Where will Ax be, the best Ax that I can do?
So what Gauss has produced is an A here. You can't find an x. He'll do as best he can. And we're calling that guy x hat. And this is the algebra to find x hat. And now, where is the picture here? Where is this vector Ax hat, which is the best Ax we can get?
So it has to be in the column space, because it's A times something. And where is it in the column space? It's the projection. That's Ax hat. And here is the error, which you couldn't do anything about, b minus Ax hat. Yeah. So it's the projection, right.
So all this is justifying the-- so we're in the second approach to least squares, solve the normal equations. Solve the normal equations. That would be the second approach to least squares. And most examples, if they're not very big or very difficult, you just create the matrix A transpose A, and you call MATLAB and solve that linear system. You create the matrix, you create the right hand side, and you solve it.
So that's the ordinary run of the mill least squares problem. Just do it. So that's method two, just do it. What's method three? For the same-- we're talking about the same problem here, but now I'm thinking it may be a little more difficult. This matrix A transpose A might be nearly singular.
Gauss is assuming that-- yeah, when did this work? When did this work? And it will continue to work in the next three-- this works, this is good, if assuming A has independent columns. Yeah, better just make clear. I'm claiming that when A has-- so what's the reasoning?
If A has independent columns-- but maybe not enough columns, like here-- it's only got two columns. It's obviously not going to be able to match any right hand side. But it's got independent columns. When A has independent columns, then what can I say about this matrix? It's invertible. Gauss's plan works. If A has independent columns, then this would be a linear algebra step. Then this will be invertible. You see the importance of that step.
If A has independent columns, that means it has no null space. Only x equals 0 is in the null space. Two independent columns, but only two. So not enough to solve systems, but independent. Then you're OK. This matrix is invertible. You can do what Gauss tells you.
But we're prepared now-- we have to think, OK. So what do I really want to do? I want to connect this Gauss's solution to the pseudo inverse. Because I'm claiming they both give the same result. The pseudo inverse will apply. But we have something-- A is not invertible. Just keep remembering this matrix. It's not invertible. But it has got independent columns.
What am I saying there? Just going back to the picture. If A is a matrix with independent columns, what space disappears in this picture? The null space goes away. So the picture is simpler. But it's still the null space of A transpose. This is still pretty big, because I only had two columns and a whole lot of rows. And that's going to be reflected here.
So what am I trying to say? I'm trying to say that this answer is the same as the pseudo inverse answer. We could possibly even check that point. Let me write it down first. I claim that the answer A plus b is the same as the answer coming from here, A transpose A, inverse A transpose b, when I guess the null space is 0, the rank is all of n, whatever you like to say.
I believe that method one, this two within one quick formula-- so you remember that this was V sigma plus U transpose, right? That's what A transpose was. That this should agree with this. I believe those are the same when the null space isn't in the picture.
So the fact that the null space is just a 0 vector means that this inverse does exist. So this inverse exists. But A A transpose is not invertible. Right? No inverse. Because A A transpose would be coming-- all this is the null space of A transpose. So A transpose is not invertible.
But A transpose A is invertible. How would you check that? You see what I'm-- it's taken pretty much the whole hour to get a picture of the geometry of the pseudo inverse. So this is the pseudo inverse. And this is-- that matrix there, it's really doing its best to be the inverse. In fact, everybody here is just doing their best to be the inverse.
Now, how well is this-- how close is that to being the inverse? Can I just ask you about that, and then I'll make this connection, and then we're out of time. How close is that to being the inverse of A? Suppose I multiply that by A. What do I get?
So just notice. If I multiply that by A, what do I get? I get, yeah? I get I. Terrific. But don't be deceived to thinking that this is the inverse of A. It worked on the left side, but it's not going to be good on the right hand side. So if I multiply A by this guy in that direction, I'll get as close to the identity as I can come, but I won't get the identity that way.
So this is just a little box to say-- so what's the point I'm making? I'm claiming that this is the pseudo inverse. Whatever. Whatever these spaces. The rank could be tiny, just one. This works when the rank is n. I needed independent columns.
So when the rank is n-- so this is rank equal n. That Gauss worked. Then I can get a-- then it's a one-sided inverse, but it's not a two-sided inverse. I can't do it. Look, my matrix there. I could find a one-sided inverse to get the 2 by 2 identity. But I could never multiply that by some matrix and get the n by n identity out of those two pathetic columns. OK.
Maybe you feel like just checking this. Just takes patience. What do I mean by checking it? I mean stick in the pseudo SVD. Just put it in the SVD and cancel like crazy. And I think that'll pop out. Do you believe me? Because it's going to be a little painful. 3 U sigma V transpose, all transposed, and then something there and something there. I've got nine matrices multiplying away. But it's going to-- all sorts of things will produce the identity. And in the end, that's what I'll have.
So this is a one-sided true inverse, where the SVD-- this fit formula is prepared to have neither side invertible. It's still-- we know what sigma plus means. Anyway. So under the assumption of independent columns, Gauss works and gives the same answer as the pseudo inverse.
OK. Three minutes. That's hardly time, but this being MIT, I feel I should use it. Oh my god.
Number three. So what's number three about? Number three has the same requirement as number two, the same requirement of no null space. But it says, if I could get orthogonal columns first, then this problem would be easy. So everybody knows that Gram-Schmidt is a way-- boring way-- to get from these two columns to get two orthogonal columns.
Actually, the whole idea of Gram-Schmidt is already there for 2 by 2. So I have two minutes, and we can do it. Let's do Gram-Schmidt on these two columns-- I don't want to use U and V-- column y and z. OK.
Suppose I want to orthogonalize those guys. What's the Gram-Schmidt idea? I take y. It's perfectly good. No problem with y. There is the y vector, the all 1's. Then this guy is not orthogonal probably to that. It'll go off in this direction, with an angle that's not 90 degrees.
So what do I do? I want to get orthogonal vectors. I'm OK with this first guy, but the second guy isn't orthogonal to the first. So what do I do? How do I-- in this picture, how do I come up with a vector orthogonal to y?
Project. I take this z, and I take its projection. So z has a little piece-- that z vector has a big piece already in the direction of y, which I don't want, and a piece orthogonal to it. That's my other piece. That's my other piece. So here's y. And here's the-- that is z minus projection, let me just say. Whatever. Yeah.
I don't know if I even drew that picture right. Probably I didn't. Anyway. Whatever. The Gram-Schmidt idea is just orthogonalize in the natural way. I'll come back to that at the beginning of next time and say a word about the fourth way. So this least squares is not deep learning. It's what people did a century ago and continue to do for good reason. OK. And I'll send out that announcement about the class, and you know the homework, and you know the new due date is Friday. Good. Thank you. | 2021-10-16T17:37:24 | {
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http://scottpelland.com/kc90d/d58b9e-bipartite-graph-gfg | Bipartite graph: a graph G = (V, E) where the vertex set can be partitioned into two non-empty sets V₁ and V₂, such that every edge connects a vertex of V₁ to a vertex of V₂. A Bipartite Graph is one whose vertices can be divided into disjoint and independent sets, say U and V, such that every edge has one vertex in U and the other in V. The algorithm to determine whether a graph is bipartite or not uses the concept of graph colouring and BFS and finds it in O(V+E) time complexity on using an adjacency list and O(V^2) on using adjacency matrix. Bipartite Graphs. It can be used to model a relationship between two different sets of points. Bipartite Graphs Mathematics Computer Engineering MCA Bipartite Graph - If the vertex-set of a graph G can be split into two disjoint sets, V 1 and V 2 , in such a way that each edge in the graph joins a vertex in V 1 to a vertex in V 2 , and there are no edges in G that connect two vertices in V 1 or two vertices in V 2 , then the graph G is called a bipartite graph. diagrams graphs. Bipartite Graphs ¶ Bipartite graphs (bi-two, partite-partition) are special cases of graphs where there are two sets of nodes as its name suggests. As discussed by Burgos et al. Using Net Flow to Solve Bipartite Matching To Recap: 1 Given bipartite graph G = (A [B;E), direct the edges from A to B. We start by introducing some basic graph terminology. 14:34. Bipartite graphs. Active 28 days ago. 4.1 Interdomain message passing through bipartite graph convolution. Details. Characterization of Bipartite Graphs. Then, if graph is bipartite, all vertices colored with 1 are in one group and with color 2 is in another respectively. That is, it is a bipartite graph (V 1, V 2, E) such that for every two vertices v 1 ∈ V 1 and v 2 ∈ V 2, v 1 v 2 is an edge in E. I can create a graph and display it like this. For example, The vertex set of can be partitioned into two disjoint and independent sets and ; All the edges from the edge set have one endpoint vertex from the set and another endpoint vertex from the set ; Let’s try to simplify it further. Also, König's talks about general case of r-paritite so if what you're saying is true, then the theorem is just a special case of general case. Every bipartite graph (with at least one edge) has a partial matching, so we can look for the largest partial matching in a graph. Nideesh Terapalli 3,662 views. I am solving Bipartite graph problem on Coursera. Bipartite graphs have a type vertex attribute in igraph, this is boolean and FALSE for the vertices of the first kind and TRUE for vertices of the second kind.. bipartite_projection_size calculates the number of vertices and edges in the two projections of the bipartite graphs, without calculating the projections themselves. A bipartite graph has two sets of vertices, for example A and B, with the possibility that when an edge is drawn, the connection should be able to connect between any vertex in A to any vertex in B. A bipartite graph is possible if the graph coloring is possible using two colors such that vertices in a set are colored with the same color. According to Wikipedia,. The node from one set can only connect to nodes from another set. This generates a dictionary of numeric positions that is passed to the pos argument of the drawing function. Lecture notes on bipartite matching February 5, 2017 5 Exercises Exercise 1-2. At the end of the proof we will have found an algorithm that runs in polynomial time. Given a graph, determine if given graph is bipartite graph using DFS. Bipartite graphs and matchings of graphs show up often in applications such as computer science, computer programming, finance, and business science. 2 Add new vertices s and t. 3 Add an edge from s to every vertex in A. 4 Add an edge from every vertex in B to t. 5 Make all the capacities 1. Bipartite graphs have both of these properties, however there are classes of non-bipartite graphs that have these properties. Enumerate all maximum matchings in a bipartite graph in Python Contains functions to enumerate all perfect and maximum matchings in bipartited graph. How does one display a bipartite graph in the python networkX package, with the nodes from one class in a column on the left and those from the other class on the right? I only care about whether all the subsets of the above set in the claim have a matching. Definition. Now in graph , we’ve two partitioned vertex sets and . If the graph does not contain any odd cycle (the number of vertices in the graph … Let’s consider a graph .The graph is a bipartite graph if:. A graph Gis bipartite if the vertex-set of Gcan be partitioned into two sets Aand B such that if uand vare in the same set, uand vare non-adjacent. Ask Question Asked 9 years, 9 months ago. $\endgroup$ – martin tassy Feb 6 '16 at 22:27 1. Theorem 5.6.5. This problem is often called maximum weighted bipartite matching, or the assignment problem.The Hungarian algorithm solves the assignment problem and it was one of the beginnings of combinatorial optimization algorithms. Since the graph is multipartite and given the provided data format, I would first create a bipartite graph, then add the additional edges. Before moving to the nitty-gritty details of graph matching, let’s see what are bipartite graphs. $\begingroup$ @Mike I'm not asking about a maximum matching, I'm asking about the overall matching. Note that although the resulting graph returns TRUE for is_bipartite() the type argument is specified as numeric instead of logical and may not work properly with other bipartite … Lecture notes on bipartite matching Matching problems are among the fundamental problems in combinatorial optimization. [ 14 ] and Kontou et al. For example, see the following graph. 5. Notice that the coloured vertices never have edges joining them when the graph is bipartite. By this we mean a set of edges for which no vertex belongs to more than one edge (but possibly belongs to none). It is obviously that there is no edge between two vertices from the same group. 1. Usually chordal graph is about chords, it is natural to think the same for chordal bipartite. Complete Bipartite Graphs. Then, if you can find a maximum perfect matching in this transformed graph, that matching is minimal in your original graph. A simple graph is bipartite if and only if it does not contain any odd cycles as a subgraph (i.e. 1 Bipartite graphs One interesting class of graphs rather akin to trees and acyclic graphs is the bipartite graph: De nition 1. An edge cover of a graph G= (V;E) is a subset of Rof Esuch that every vertex of V is incident to at least one edge in R. Let Gbe a bipartite graph with no isolated vertex. the linear program from Equation (2) nds the maximum cardinality of an independent set. It is not possible to color a cycle graph … Implemented following the algorithms in the paper "Algorithms for Enumerating All Perfect, Maximum and Maximal Matchings in Bipartite Graphs" by Takeaki Uno, using numpy and networkx modules of python. Where B is the full bipartite graph (represented as a regular networkx graph), and B_first_partition_nodes are the nodes you wish to place in the first partition. it does not contain any $$C_n$$ for $$n$$ odd). Bipartite Graph | Leetcode 785 | Graph | Breadth First Search - Duration: 14:34. in the textbook of Diestel, he mentiond König's theorem in page 30, and he mentiond the question of this site in page 14. he didn't say at all any similiarities between the two. In this set of notes, we focus on the case when the underlying graph is bipartite. Note that it is possible to color a cycle graph with even cycle using two colors. A bipartite graph is a graph whose vertices can be divided into two disjoint and independent sets U and V such that every edge connects a vertex in U to one in V.. Theorem 1 For bipartite graphs, A= A, i.e. u i and v j denote the ith and jth node in U and V , respectively, where i = 1, 2, …, M and j = 1, 2, …, N . $\begingroup$ I don't agree with you. Viewed 16k times 8. As with trees, there is a nice characterization of bipartite graphs. How can I do it? 4-2 Lecture 4: Matching Algorithms for Bipartite Graphs Figure 4.1: A matching on a bipartite graph. I want it to be a directed graph and want to be able to label the vertices. nx.algorithms.matching.max_weight_matching has the parameter maxcardinality which, if set to True , means that it will only allow for complete matchings if such a matching exists. Bipartite Graphs and Matchings (Revised Thu May 22 10:59:19 PDT 2014) A graph G = (V;E) is called bipartite if its vertex set V can be partitioned into two disjoint subsets L and R such that all edges are between L and R. For example, the graph G 1 below on the left 1 6 2 3 4 7 5 G 1 1 3 2 4 5 G 2 Try to debug this program and try to understand and analyze. A bipartite graph BG (U, V, E) is a graph G (U ∪ V, E) where U and V denote two sets of the two domains of vertices (nodes). The edges used in the maximum network Actual problem statement is as follows: I am using BFS to find if the given graph is bipartite or not but the grader is showing "time exceeded". The rest of this section will be dedicated to the proof of this theorem. Show that the cardinality of the minimum edge cover R of Gis equal to jVjminus A bipartite graph that doesn't have a matching might still have a partial matching. $\endgroup$ – Fedor Petrov Feb 6 '16 at 22:26 $\begingroup$ I sincerely appreciate your answer, thank you very much. Maximum Cardinality Bipartite Matching (MCBM) Bipartite Matching is a set of edges $$M$$ such that for every edge $$e_1 \in M$$ with two endpoints $$u, v$$ there is no other edge $$e_2 \in M$$ with any of the endpoints $$u, v$$. Image by Author. A complete bipartite graph is a graph whose vertices can be partitioned into two subsets V 1 and V 2 such that no edge has both endpoints in the same subset, and every possible edge that could connect vertices in different subsets is part of the graph. A bipartite graph, also referred to as a “bigraph,” comprises a set of graph vertices decomposed into 2 disjoint sets such that no 2 graph vertices within the same set are adjacent. $\endgroup$ – Violetta Blejder Dec 8 at 1:22 A bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint sets U and V such that every edge connects a vertex in U to one in V. It is possible to test whether a graph is bipartite or not using DFS algorithm. The nodes from one set can not interconnect. 4. Here is an example of a bipartite graph (left), and an example of a graph that is not bipartite. P, as it is alternating and it starts and ends with a free vertex, must be odd length and must have one edge more in its subset of unmatched edges (PnM) than in its subset of matched edges (P \M). In particular, a graph has the strong Hall property if-and-only-if it is stable - its maximum matching size equals its maximum fractional matching size. I want to draw something similar to this in latex. In a weighted bipartite graph, the optimization problem is to find a maximum-weight matching; a dual problem is to find a minimum-weight matching. 6 Solve maximum network ow problem on this new graph G0. I've researched some solutions regarding the degree of one side of a bipartite graph related to the other, but it is a bit confusing. Passed to the nitty-gritty details of graph matching, let ’ s see what are bipartite graphs 4.1. 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https://math.stackexchange.com/questions/1364092/generalizing-the-fibonacci-sum-sum-n-0-infty-fracf-n10n-frac10 | # Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$
Given the Fibonacci, tribonacci, and tetranacci numbers,
$$F_n = 0,1,1,2,3,5,8\dots$$
$$T_n = 0, 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 0, 1, 1, 2, 4, 8, 15, 29, \dots$$
and so on, how do we show that,
$$\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$$
$$\sum_{n=0}^{\infty}\frac{T_n}{10^n} = \frac{100}{889}$$
$$\sum_{n=0}^{\infty}\frac{U_n}{10^n} = \frac{1000}{8889}$$
or, in general,
$$\sum_{n=0}^{\infty}\frac{S_n}{p^n} = \frac{(1-p)p^{k-1}}{(2-p)p^k-1}$$
where the above were just the cases $k=2,3,4$, and $p=10$?
P.S. Related post.
• Have you tried to find explicit formulas for $S_n$ like Binet's? – Michael Galuza Jul 17 '15 at 3:51
• @MichaelGaluza: Yes, actually. If you look at the general formula (eq.5) at mathworld.wolfram.com/Fibonaccin-StepNumber.html, I gave that to Eric back in 2005. But I just found that empirically. – Tito Piezas III Jul 17 '15 at 4:00
• I might be overlooking something, but I believe that something similar to Thomas Andrew's solution to your other question (generating functions) should give you the value, while convergence might be proven by showing $S_n/p^n < 1$ for sufficiently large $p$. (I haven't actually followed this path of reasoning out, but it "feels right." If I follow through it through, I'll leave a comment or answer explaining how it turned out.) – apnorton Jul 17 '15 at 4:04
• By $S_n$ do you mean $S_i = F_i$ for $i < n$ and $S_i = S_{i-1} + \ldots + S_{i -n}$? – Marcus M Jul 17 '15 at 4:05
• @MarcusM I believe the standard definition is $S_i = 0$ for $i \leq 0$, $S_1=1$, and $S_i = S_{i-1} + \cdots + S_{i-k}$. – apnorton Jul 17 '15 at 4:07
Outline: This follows the spirit of Thomas Andrew's solution to the OP's other question to find the value of the series. Convergence is then proven by showing $S_n/p^n<1$ for sufficiently large p.
Finding the Value with Generating Functions:
Define $S_n$ for a fixed $k$ as follows: $$S_n = \begin{cases} 0 & n \leq 0 \\ 1 & n = 1 \\ \sum_{j=1}^k S_{n-j} & n > 1 \end{cases}$$ Let $\mathcal{S}(z) = \sum_{n\geq 0}S_nz^n$. I assert without proof (I'll seek a source paper, rather than derive it myself) that: $$\mathcal{S}(z) = \frac{z}{1 - z - z^2 -\dots - z^k}$$
It follows: \begin{align*} \mathcal{S}(z) &= \frac{z}{1 - z - z^2 -\dots - z^k} \\ &= \frac{z}{2 - \sum_{j=0}^k z^j} \\ &= \frac{z}{2 - \frac{z^{k+1}-1}{z-1}} \\ &= \frac{z(z-1)}{2z -z^{k+1}-1} \\ &= \frac{z(z-1)}{(2-z^k)z-1} \end{align*}
The value we seek is $\mathcal{S}(1/p)$: \begin{align*} \mathcal{S}(1/p) &= \frac{(1/p)\left(\frac{1}{p}-1\right)}{\left(2-\frac{1}{p^k}\right)\frac{1}{p}-1} \\ &= \frac{\frac{1-p}{p^2}}{\left(\frac{2p^k-1}{p^{k+1}}\right)-1} \\ &= \frac{1-p}{\frac{p^2}{p^{k+1}}\left(2p^k-1 -p^{k+1}\right)} \\ &= \frac{(1-p)p^{k-1}}{(2-p)p^k -1} \end{align*}
This is as desired.
Proving Convergence:
However, this only provides the value if the series converges; that proof is entirely separate.
It is generally known that $S_n \in \mathcal O( r^n )$, where $r$ is the largest real root of the equation $2-\sum_{j=0}^k z^j = 0$. To justify this, consider the last identity in this Wiki article and consider its asymptotic behavior. (I don't like the source of that, but I'll find a better one later).
Thus, so long as $r < p$, the series $\mathcal{S}(1/p)$ is bounded above by a convergent geometric series $\sum_{n\geq 0} \left(\frac{r}{p}\right)^n$.
Fortunately, based on Lemma 2.7 of this arxiv paper, we know that the largest real root of $2-\sum_{j=0}^kz^j = 0$ is strictly bounded between $1$ and $2$. Thus, so long as $p \geq 2$, we have convergence.
• Your assertion on the generating function is very easy to prove, not worth hunting a reference for; just compute $S(z)-(zS(z)+z^2S(z)+\cdots+z^kS(z))$ term-by-term, using the generating relation for $n\gt k$ and the explicit definition for $n\leq k$. – Steven Stadnicki Jul 17 '15 at 19:07
The difference equations given by the suggested series are: \begin{align} F_{n+2} &= F_{n+1} + F_{n} \\ T_{n+3} &= T_{n+2} + T_{n+1} + T_{n} \\ \tag{1} U_{n+4} &= U_{n+3} + U_{n+2} + U_{n+1} + U_{n} \end{align} and so on. In general they take on the form \begin{align}\tag{2} \phi_{n+m} = \sum_{k=0}^{m-1} \phi_{n+m-k-1}, \end{align} where $\phi_{0}, \phi_{1}, \phi_{2}, \cdots$ are the initial values. By considering the generating function defined by \begin{align} f_{m}(t) = \sum_{n=0}^{\infty} \phi_{n+m} \, t^{n} \end{align} then it is readily found that \begin{align} f_{m}(t) &= \frac{1}{ 2 - \sum_{k=0}^{m} t^{k}} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \\ &= \frac{1 - t}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, t^{k} \right] \tag{3} \end{align} if $t \to 1/t$ then \begin{align} f_{m}(t) &= \frac{t^{m} (t - 1)}{1 - 2 t + t^{m+1}} \, \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, \frac{1}{t^{k}} \right] \end{align} When $t = 10$ this reduces to \begin{align}\tag{4} f_{m}\left(\frac{1}{10}\right) = \frac{9}{(10)^{m+1}- 2 \, (10)^{m} + 1} \cdot \sum_{k=0}^{m-1} \left[\left( \phi_{k} - \sum_{s=0}^{k-1} \phi_{s} \right) \, (10)^{m-k} \right] \end{align}
As an example let $m=3$, which corresponds to the Tribonacci series, to obtain \begin{align} f_{3}\left(\frac{1}{10}\right) &= \sum_{n=0}^{\infty} \frac{T_{n}}{(10)^{n}} = \frac{9 \, (10)^{3}}{10^{4} - 2 \cdot 10^{3} + 1} \cdot \left(\frac{1}{10}\right) = \frac{100}{889}. \end{align}
• Although this gives the form requested, it does not technically prove the series converges; that must be done separately. But, +1 even so. – apnorton Jul 17 '15 at 4:41
That's because $\sum_{n=0}^{\infty} F_nx^n =\frac1{1-x-x^2}$.
Putting $x = \frac1{10^k}$ gives $\sum_{n=0}^{\infty} \frac{F_n}{10^{kn}} =\frac1{1-10^{-k}-10^{-2k}} =\frac{10^{2k}}{10^{2k}-10^{k}-1}$.
For the others, the generating function is $\sum_{n=0}^{\infty} G_n x^n =\frac1{1-x-x^2-...-x^m}$ where $m=2$ for Fib, $m=3$ for Trib, and $m=4$ for Tetra.
For each of these, $\sum_{n=0}^{\infty} \frac{G_n}{ 10^n} =\frac1{1-\frac1{10}-\frac1{100}-...-\frac1{10^{m}}} =\frac{10^m}{10^m-10^{m-1}-...-1} =\frac{10^m}{8...(m-1 \ 8s)9}$. | 2019-10-20T07:24:44 | {
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https://math.stackexchange.com/questions/2930970/questions-on-a-self-made-theorem-about-polynomials | # Questions on a self-made theorem about polynomials
I recently came up with this theorem:
For any complex polynomial $$P$$ degree $$n$$:
$$\sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) = 0\quad \forall a,b \in\mathbb{C}$$
Basically, if $$P$$ is quadratic, $$P(a) - 3P(a+b) + 3P(a+2b) - P(a+3b) = 0$$ (inputs of $$P$$ are consecutive terms of any arithmetic sequence). This can be generalized to any other degrees.
• Has this been discovered? If yes, what's the formal name for this phenomenon?
• Is it significant/Are there important consequences of this being true?
• Can this be generalized to non-polynomials?
• I do not know if this have a name, but your theorem comes as a linear combination of the special case $P(x) = x^m$ where $0 \leq m \leq n$, and for the monomial case, I believe it is well known. I am still searching for a more suitable reference, but this page explains a bit: mathworld.wolfram.com/FiniteDifference.html – Hw Chu Sep 26 '18 at 0:49
• For $b=1$, this follows from Theorem 1 and Theorem 2 in my post math.stackexchange.com/questions/1379172/… . (My argument is more or less the same as Steven Stadnicki's.) To get the case of general $b$, you can either generalize my argument, or you can apply my argument to the polynomial $Q$ (also of degree $n$) defined by $Q\left(x\right) = P \left(a+xb\right)$. – darij grinberg Sep 26 '18 at 2:02
• My answer here has a reference to an article by Gould that has a very extensive list of references on this result, which goes back to Euler. – Chappers Sep 26 '18 at 15:03
• What specific field of math is this? I want to learn it. – clathratus Oct 2 '18 at 3:22
• @clathratus I'm not sure, tbh. I guess it's algebra, since it involves polynomials (something you learn in Algebra class at high school), and also because most techniques I used to prove this are algebraic manipulations (i.e., divide both sides by something, etc.). I hope someone else can give you a definite answer. – Felix Fourcolor Oct 2 '18 at 5:02
In brief: this is well-known, but definitely important.
It's easiest to write this in terms of the finite difference operator $$\Delta$$: $$\Delta P(x)=P(x+1)-P(x)$$. You use $$P(x+b)$$ instead of $$P(x+1)$$, but it's easy to see that these two things are equivalent; to keep things consistent with your notation, I'll write $$\Delta_b$$ for your operator.
The most important feature of the $$\Delta_b$$ operator is how it affects the degree of a polynomial:
Theorem: for any nonconstant polynomial $$P(x)$$, the degree of $$\Delta_b P(x)$$ is one less than the degree of $$P(x)$$.
Proof outline: Note that the degree of $$\Delta_b P(x)$$ is no greater than the degree of $$P(x)$$. Now, write $$P(x) = a_dx^d+Q(x)$$, where $$Q(x)$$ is a polynomial of degree $$d-1$$ or less. Then $$P(x+b) =a_d(x+b)^d+Q(x+b)$$, so $$\Delta_b P(x) = a_d\left((x+b)^d-x^d\right)+\Delta_b Q(x)$$; by the binomial theorem $$(x+b)^d=x^d+{d\choose 1}bx^{d-1}+\ldots$$, so $$(x+b)^d-x^d={d\choose 1}bx^{d-1}+\ldots$$ is a polynomial of degree at most $$d-1$$, and thus $$\Delta_bP(x)$$ is the sum of two polynomials of degree at most $$d-1$$ (namely, $$a_d\left((x+b)^d-x^d\right)$$ and $$\Delta_b Q(x)$$), so it's of degree at most $$d-1$$ itself.
(It's slightly more challenging to prove that the degree of $$\Delta_bP(x)$$ is exactly $$d-1$$ when $$b \neq 0$$, but this can also be shown.)
Why does this matter? Because it can be shown by induction that your sum is exactly the result of applying the $$\Delta_b$$ operator $$d+1$$ times, where $$d$$ is the degree of the polynomial; since each application of $$\Delta_b$$ reduces the degree by one, then $$(\Delta_b)^dP(x)$$ is a polynomial of degree zero — a constant — and thus $$(\Delta_b)^{d+1}P(x)$$ will be identically zero. This is exactly your identity.
Now, you may know that the derivative of a polynomial of degree $$d$$ is also a polynomial of degree $$d-1$$. It turns out that this isn't a coincidence; $$\Delta$$ is very similar to a derivative in many ways, with the Newton polynomials $${x\choose d}=\frac1{d!}x(x-1)(x-2)\cdots(x-d)$$ playing the role of the monomial $$x^d$$ with respect to the derivative. For more details, I suggest starting with Wikipedia's page on finite difference calculus.
In fact, we can also prove the converse (and this answers the question about generalizing to non-polynomials in the negative). I'll work in terms of $$\Delta$$, rather than $$\Delta_b$$, but again all the results generalize readily.
Note that $$\Delta^n P(x)$$ only depends on the values of $$P(x+i)$$ for $$i$$ an integer between $$0$$ and $$n$$; thus, a function can take arbitrary values for $$0\lt x\lt1$$ and still satisfy the identity; we can't say much about general points. However, it does constrain the values at integers:
Theorem: suppose that $$\Delta^{d+1}f(x)\equiv 0$$ identically. Then there exists a polynomial $$P(x)$$ of degree $$d$$ such that $$f(n)=P(n)$$ for all integers $$n$$.
The proof works by induction. For simplicity's sake, I'll consider all functions as being on $$\mathbb{Z}$$ now, and not consider non-integer values at all. Note first of all that if $$\Delta f(x)=g(x)$$, then $$f(n)=f(0)+\sum_{i=0}^{n-1}g(i)$$. (Proof by induction: the case $$n=1$$ is true by definition, since $$g(0)=\Delta f(0)=f(1)-f(0)$$ implies that $$f(1)=f(0)+g(0)$$. Now, assuming it's true for $$n=k$$, at $$n=k+1$$ we have $$f(k+1)=f(k)+g(k)$$ $$=f(0)+\sum_{i=0}^{k-1}g(i)+g(k)$$ $$=f(0)+\sum_{i=0}^kg(i)$$.) In particular, if $$\Delta f(x)\equiv 0$$ identically, then $$f(n)=f(0)$$ for all integers $$n$$; $$f()$$ is constant on $$\mathbb{Z}$$.
This gives us the base case for our induction; to induct we just need to show that if $$\Delta f(x)$$ is a polynomial of degree $$d$$, then $$f(x)$$ is polynomial of degree $$d+1$$. But suppose for concreteness that $$\Delta f(x)=P(x)=\sum_{i=0}^da_ix^i$$. Then $$f(n)=f(0)+\sum_{k=0}^{n-1}P(k)$$ $$=f(0)+\sum_{k=0}^{n-1}\left(\sum_{i=0}^da_ik^i\right)$$ $$=f(0)+\sum_{i=0}^da_i\left(\sum_{k=0}^{n-1}k^i\right)$$. Now, for each $$i$$ the sum $$\sum_{k=0}^{n-1}k^i$$ in parentheses in this last expression is known to be a polynomial of degree $$i+1$$ (see e.g. https://en.wikipedia.org/wiki/Faulhaber%27s_formula ), so the whole expression is a polynomial of degree $$d+1$$, as was to be proved.
• "this is well-known, but definitely important." Did you mean this particular theorem, or the idea of finite difference in general? – Felix Fourcolor Sep 26 '18 at 4:47
• @FelixFourcolor Sort of both - the idea of finite differences in the broad, but also particularly this consequence that the difference operator behaves like a derivative especially with respect to polynomials (and that therefore the n+1st difference is zero, which is your theorem). It's used, for instance, in building up interpolating polynomials. – Steven Stadnicki Sep 26 '18 at 5:20
• Would you happen to know why I got those down-votes on my answer below with no explanation added? It seems to me that I have a perfectly valid formal power series proof. Maybe you can advise here. – Marko Riedel Sep 26 '18 at 15:04
• @MarkoRiedel I'm afraid I don't; I wasn't one of the downvoters and it looks broadly sensible to me. It might just be a matter of a gap in levels between the question and the proof, but I wouldn't expect that to garner quite so many downvotes in and of itself. – Steven Stadnicki Sep 26 '18 at 15:39
• @FelixFourcolor I'll try to amend my answer to address that; the short version is that your conditions impose no restrictions on the function's values from e.g. $0$ to $b$, but force the function to be 'polynomial at multiples of $b$'. I'll add the proof of that (along with a clearer definition) in my answer. – Steven Stadnicki Oct 2 '18 at 21:25
What you wrote is the finite difference operator of order $$n+1$$, acting on $$P$$, $$\sum\limits_{k=0}^{n+1}(-1)^k\binom{n+1}{k}P(a+kb) =\Delta^{n+1} P(a+kb).$$
Notice that by linearity it suffices for the property to hold for all monomials $$k^m, m\le n$$ and it is easily explained by the fact that the first order difference of a polynomial is a polynomial of degree one less.
$$(k+1)^m-k^m=k^m+mk^{m-1}+\cdots-k^m.$$
Illustration ($$n=3$$):
$$\Delta^4 k^m=((4^m-3^m)-(3^m-2^m))-((3^m-2^m)-(2^m-1^m)) \\-((3^m-2^m)-(2^m-1^m))-((2^4-1^m)-(1^4-0^m)) \\=4^m-4\cdot3^m+6\cdot2^m-4\cdot1^m+0^m.$$ and $$\begin{matrix} 1&&1&&1&&1&&1 \\&0&&0&&0&&0 \\&&0&&0&&0 \\&&&0&&0 \\&&&&0 \end{matrix}$$
$$\begin{matrix} 0&&1&&2&&3&&4 \\&1&&1&&1&&1 \\&&0&&0&&0 \\&&&0&&0 \\&&&&0 \end{matrix}$$
$$\begin{matrix} 0&&1&&4&&9&&16 \\&1&&3&&5&&7 \\&&2&&2&&2 \\&&&0&&0 \\&&&&0 \end{matrix}$$
$$\begin{matrix} 0&&1&&8&&27&&64 \\&1&&7&&19&&37 \\&&6&&12&&18 \\&&&6&&6 \\&&&&0 \end{matrix}$$
Final remark:
On can show that $$\Delta_{n+1}k^{n+1}=(-1)^nn!$$, so that for a polynomial of degree $$n+1$$ the sum is
$$(-1)^nn!p_{n+1}b^{n+1},$$ independently of $$a$$.
• You don't need to include "illustrations", as I was the one who discovered this phenomenon myself, I must have known it. Instead, you could talk more about your final remark and maybe a sketch of proof for it. For now, your answer brings very limited new information to the table. That's the reason for my downvote. – Felix Fourcolor Oct 4 '18 at 6:20
• @FelixFourcolor: posting such a naive question made me think you needed simple explanations. Sorry to have disturbed you. – Yves Daoust Oct 4 '18 at 7:11
This result is known and was demonstrated by a student named Ruiz.
Here's the reference :
Sebastián Martín Ruiz, An Algebraic Identity Leading to Wilson's Theorem, The Mathematical Gazette, 80 (489) 579-582 (Nov. 1996).
You accesss to it at JSTOR : http://www.jstor.org/stable/3618534
This is not a new result. What you wrote is the formula for the $$n$$-th order forward difference and can be found even on Wikipedia.
I will show a representation formula which is valid for $$C^n(\mathbb R)$$ functions, not just polynomials. The same can be found also here.
The key point is that finite differences and derivatives commute: $$D\Delta_h=\Delta_hD$$.
For $$f\in C^1(\mathbb R)$$ you can compute $$\frac1h\Delta_h[f](x) = \frac{f(x+h)-f(x)}h = \frac1h \int_0^h D[f](x+x_1) \,dx_1$$
For $$f\in C^n(\mathbb R)$$, iterating the above formula, you get $$\begin{split} \frac1{h^n}\Delta_h^n[f](x) &= \frac{\frac1{h^{n-1}}\Delta_h^{n-1}[f](x+h) - \frac1{h^{n-1}}\Delta_h^{n-1}[f](x)} {h} \\ &= \frac1h \int_0^h \frac1{h^{n-1}}D\Delta_h^{n-1}[f](x+x_1) \,dx_1 \\ &= \frac1{h^2} \int_0^h \int_0^h \frac1{h^{n-1}} D^2\Delta_h^{n-2}[f](x+x_1+x_2) \,dx_1dx_2 \\ &= \,\cdots \\ &= \frac1{h^n} \int_0^h\dotsi\int_0^h D^n[f](x+x_1+\dotsb+x_n) \,dx_1\dotsm dx_n . \end{split}$$
The last integral is a weighted average of $$D^n[f]$$ over the segment $$[x,x+nh]$$. More precisely, let $$X_1,\dotsc,X_n\sim\mathrm{Uniform}(0,h)$$ be i.i.d. and $$S=X_1+\dotsb+X_n$$. Then the previous expression can be viewed as $$\frac1{h^n}\Delta_h^n[f](x) = \mathbb{E}\left[D^n[f](S)\right]$$
The mean value theorem for integrals applies and tells us that there exists $$x^*\in(x,x+nh)$$ such that $$\frac1{h^n}\Delta_h^n[f](x) = D^n[f](x^*).$$
Regarding your proposition, it follows trivially from the fact that if $$P$$ is a polynomial of degree $$n$$, then $$D^{n+1}[P]=0$$ identically, hence also $$\Delta_h^{n+1}[P]=0$$.
I can recommend reading at least Finite difference, Indefinite sum and Newton polynomial to learn about related material. | 2019-06-20T21:31:41 | {
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http://math.stackexchange.com/questions/653331/probability-of-getting-more-number-of-heads | # Probability of getting more number of heads
Mercedes and Edmond are playing a game.They toss 25 coins and count the number of heads and tails.If the number of heads is more,mercedes wins,while if the number of tails is more,edmond wins.What is the probability that mercedes wins,given all coins are unbiased ?
I faced this question today in an aptitude exam.
My approach was like this !
Mercedes will win if number of heads is greater.The possibilities are 13-12,14-11,.....24-1,25-0...for mercedes and edmond correspondingly.
Sample space will 25 because there can be 25 possible splits in heads and tails.
Is the probability 13/25 ?
Can somebody help ? I think i'm 90% wrong ?
Thanks.
-
You said the coins are biased? ... Can your answer involve a parameter $p$? – snarski Jan 27 '14 at 14:41
Very sorry for the inconvenience.All are unbiased. – vaidy_mit Jan 27 '14 at 14:42
Note that there are not $25,$ but $26$ possible outcomes, ranging from $0$ heads to $25$ heads. Since the coins are all unbiased, and Mercedes will win in exactly $13$ of the possible outcomes, and because the outcomes' probabilities are symmetrically distributed between the two players, then....
To be more rigorous, let's suppose that $H$ represents the number of heads in a given round. Then (for integers $0\le n\le 25$) there are $\binom{25}n$ ways to choose the $n$ coins that will be heads, and each coin has a probability of $0.5$ of being heads (tails), so the probability that $H=n$ is $$\Bbb P(H=n)=\binom{25}n\cdot0.5^n\cdot0.5^{25-n}=\binom{25}n\cdot0.5^{25}.$$ Mercedes will win if $13\le n\le 25$ and Edmond will win if $0\le n\le 12$ (i.e.: $13\le 25-n\le 25$). Since $\binom{25}{25-n}=\binom{25}n,$ then the conclusion readily follows.
Added: The important thing to keep in mind, here, is the symmetric distribution of winning outcomes between the two players. It is not so simple as simply taking $13$ out of $26$ distinguishable outcomes to get the probability that Mercedes wins, even though that gives the correct answer (in this case).
I mention distinguishable outcomes because there are (for example) $25$ different ways to get exactly one coin landing heads, but there may not be any way to tell such outcomes apart. If each coin was distinguishable from the rest (e.g.: had a different date, denomination, color, shape), then our different outcomes would all be equally likely, and we would only have to figure out the total number of such outcomes, and the number of those which gave Mercedes the win--$2^{25}$ and $2^{24},$ respectively, as it turns out. Unfortunately, we don't know how (or if) we may distinguish between such outcomes, and even if we did, it would be a ludicrous task to try to actually list them all (though counting them is not terribly difficult).
We need also keep in mind that the outcomes in this situation are not of equal probability! For example, while there are $25$ different (but indistinguishable) ways to flip exactly $1$ head, there is only $1$ way to flip all tails, so it is more likely that $1$ head will be flipped than no heads at all. This is ultimately why we can't simply take $13/26$ to get the answer. If the game had been such that Mercedes would win if there were $0,1,2,3,4,5,6,20,21,22,23,24,$ or $25$ heads, then despite the fact that she would win in exactly $13$ of the $26$ possible distinguishable outcomes, the outcomes' probabilities are not symmetrically distributed among the players. In fact, Edmond has more than a 99% chance of winning in such a scenario!
For a simpler example, suppose they were just flipping $3$ identical coins. Note that we have $1$ way for no heads to be flipped, $3$ ways for exactly one head to be flipped, $3$ ways for exactly two heads to be flipped, and $1$ way for three heads to be flipped. (Why?) This gives us $8$ total ways to flip the coins, even though there are only $4$ distinguishable outcomes. In particular, $$\Bbb P(H=0)=\frac18,\Bbb P(H=1)=\frac38,\Bbb P(H=2)=\frac38,\Bbb P(H=3)=\frac18.$$ Now, let's consider two possible games.
Game 1. If there are more heads than tails, Mercedes wins; otherwise, Edmond wins:
$$\Bbb P(\text{Mercedes wins})=\Bbb P(H=2)+\Bbb P(H=3)=\frac38+\frac18=\frac12.$$
Game 2. If there are only heads or only tails, Mercedes wins; otherwise, Edmond wins:
$$\Bbb P(\text{Mercedes wins})=\Bbb P(H=0)+\Bbb P(H=3)=\frac18+\frac18=\frac14.$$
In both cases, Mercedes won in two of the four distinguishable outcomes, but in the second case, she was given the two least likely outcomes, so was at a disadvantage from the start.
-
So,1/2(13/26) is the answer ? – vaidy_mit Jan 27 '14 at 14:45
That's correct! – Cameron Buie Jan 27 '14 at 14:49
Although $\frac12=\frac{13}{26}$ is true, $\frac{13}{26}$ is a rather misleading way to put the answer, since the $26$ possible splits are far from being equally likely. It is just a basic symmetry observation that tells that $\frac12$ is the only possible correct answer. – Marc van Leeuwen Jan 27 '14 at 14:58
@Marc: My thanks. I should have drawn more attention to the fact that the outcomes were not of uniform probability in the first place. – Cameron Buie Jan 27 '14 at 15:03
The comparison of 13/26 possible outcomes and having a $< 1%$ chance of winning is a great one for building understanding, +1. – snarski Jan 27 '14 at 15:15
Since there is no chance of a tie, the answer -- $P(\text{M wins}) = 1/2$ -- can be found by symmetry: both $M$ and $E$ have an equal chance of winning.
\begin{gather*} 1 = P(\text{M wins}) + P(\text{M doesn't win}) \\ = P(\text{M wins}) + P(\text{E wins}) \\ = P(\text{M wins}) + P(\text{M wins}) \\ = 2P(\text{M wins}). \end{gather*}
Solve for $P(\text{M wins}).$
-
WoW ! Cool answer :D Thanks :D – vaidy_mit Jan 27 '14 at 14:48 | 2016-05-04T08:31:15 | {
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https://math.stackexchange.com/questions/2579511/if-f-is-continuous-and-lim-limits-x-to-inftyfx1-fx-0-does-this-mea | # If $f$ is continuous and $\lim\limits_{x\to \infty}f(x+1)-f(x)=0$, does this mean that $\lim\limits_{x\to \infty}\frac{f(x)}{x}=0$? [duplicate]
Suppose that the function $f$ is continuous in $\mathbb{R}$ and $$\lim_{x\to \infty}\left(f(x+1)-f(x)\right)=0$$ then does this mean that $\lim_{x\to \infty}\left(\frac{f(x)}{x}\right)=0$ ?
## marked as duplicate by Paramanand Singh calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Dec 25 '17 at 16:57
Yes, that is true. By hypothesis, for any $\epsilon>0$, there is some $t>0$ such that for all $x>t$, we have $|f(x+1)-f(x)|<\epsilon$. Let $M=\max\limits_{x\in[t,t+1]}|f(x)|$, which exists since $f$ is continuous.
For $x=t+h$ with $n+1>h\geq n$, we have that $$|f(x)-f(t+h-n)|\leq |f(t+h)-f(t+h-1)|+\cdots +|f(t+h-n+1)-f(t+h-n)|< n\epsilon$$ and $$|f(t+h-n)|\leq M$$ hence $$|f(x)|\leq|f(x)-f(t+h-n)|+|f(t+h-n)|< n\epsilon+M\leq (x-t)\epsilon+M$$ so that for any $x=t+h$ with $h>0$, $$\left|\frac{f(x)}{x}\right|=\left|\frac{f(t+h)}{x}\right|<\left|\frac{(x-t)\epsilon+M}{x}\right|=\left|\epsilon+\frac{M-t\epsilon}{x}\right|\leq \epsilon+\left|\frac{M-t\epsilon}{x}\right|$$ Thus, for $x>\max\{t,|\frac{M-t\epsilon}{\epsilon}|\}$, we have that $\left|\frac{f(x)}{x}\right|\leq 2\epsilon$. This suffices to show that $\lim_{x\to\infty}\left|\frac{f(x)}{x}\right|=0$.
• Ah... In the third line from the bottom, middle equality: that cleared the whole darn thing up! Also, the choice of epsilon...which isn't fixed until the last step. Very nice indeed. +1 – DonAntonio Dec 25 '17 at 10:31
• How we can prove it using sequences ? – S.H.W Dec 25 '17 at 11:52
• A very nice proof Zev! As far as I understand, continuity is too much here, we can only demand the function to be bound! – dmtri Dec 25 '17 at 16:07
• ....to be bounded in any interval [t,t+1]. – dmtri Dec 25 '17 at 16:30
Fix $\varepsilon > 0$. There exists $K\in\mathbb{Z}$ such that $$|f(x+1)-f(x)| < \varepsilon \qquad \forall x \geq K.$$ Let $M := \max_{x\in [K, K+1]} |f(x)|$. Let $x > K$ and $n\in\mathbb{Z}$ such that $n \leq x < n+1$. Then $$|f(x)| \leq |f(x-(n-K)| + \sum_{j=1}^{n-K} |f(x+1-j) - f(x-j)| \leq M + (n-K)\varepsilon \leq M + (x-K)\varepsilon,$$ so that $$\limsup_{x\to +\infty} \frac{|f(x)|}{x} \leq \varepsilon.$$
• (1) In the 4th line, how is that first inequality obtained? (2) Assuming this same line is correct, you got $\;|f(x)|\le M+(x-K)\epsilon\;$ How from here you get the last inequality with the lim sup? – DonAntonio Dec 25 '17 at 10:20
• (1) $|f(x)| \leq |f(x-(n-K))| + |f(x) - f(x-(n-k)| \leq \ldots$. (2) Divide the inequality by $x$ (that you can assume positive) and take the limsup. – Rigel Dec 25 '17 at 10:23
• I tried to: getting $$\frac{|f(x)|}x\le \frac Mx+\frac{x-K}x\epsilon ....\text{still unclear}$$ Only clear that the second summand on the right tends to $\;\epsilon\;$ when $\;x\to\infty\;$ . – DonAntonio Dec 25 '17 at 10:28
• See my comment under the answer by Zev... – DonAntonio Dec 25 '17 at 10:32
• I don't understand your point. If $h(x) \leq g(x)$, then $\limsup h(x) \leq \limsup g(x)$. In this case, if $g(x)$ is the r.h.s. of the inequality, then $\limsup g(x) = \lim g(x) = \varepsilon$. – Rigel Dec 25 '17 at 10:40
Subtle question. By Cesàro-Stolz we have $$\lim_{n\to +\infty}\frac{f(n)}{n} = \lim_{n\to +\infty}\frac{f(n+1)-f(n)}{(n+1)-n} = 0$$ and the same holds if we consider $\lim_{n\to +\infty}\frac{f(n+\theta)}{n+\theta}$ with $\theta\in(0,1)$. On the other hand continuity plus $\lim_{x\to +\infty}f(x+1)-f(x)$ ensure that $f$ is uniformly continuous over $\mathbb{R}^+$, since they ensure that $f$ is continuous and approximately $1$-periodic. In particular, for any $\theta\in(0,1)$,
$$\left|\frac{f(n+\theta)}{n+\theta}-\frac{f(n)}{n}\right|\leq\frac{n|f(n+\theta)-f(n)|+\theta|f(n)|}{n^2}=O\left(\frac{1}{n}\right)$$ and $$\lim_{x\to +\infty}\frac{f(x)}{x}=0$$ holds.
• (1) How "the same holds...", as CS theorem is for sequences? I suppose $\;\theta\;$ is fixed here, so I understand how the extension of CS holds (please do correct me if I'm wrong). (2) Also, why can you deduce unif. continuity? As far as I understand for UC we should evaluate the differences $\;|f(x)-f(y)|\;$ , and here the values of the varialbes are $\;l1\;$ unit away: x,\,x+1\;$...I guess that "approx. 1-periodic" thingy is unclear to me. (3) The last equality with the big "O" is incomprehensible to me, too. – DonAntonio Dec 25 '17 at 10:26 • @DonAntonio: is it clear that a continuous and$1$-periodic function is uniformly continuous? Well, here it is almost the same. We have$\lim_{n\to +\infty}\frac{f(n+\theta)}{n+\theta}=0$for any$\theta\$, we just have to "stick these limits together" by exploiting UC. – Jack D'Aurizio Dec 25 '17 at 10:39 | 2019-05-23T18:49:06 | {
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https://www.projectrhea.org/rhea/index.php/Invertibility_of_a_system_ECE301S11 | Topic: System Invertibility
## Question
The input x(t) and the output y(t) of a system are related by the equation
$y(t)=x(t+2)$
Is the system invertible (yes/no)? If you answered "yes", find the inverse of this system. If you answered "no", give a mathematical proof that the system is not invertible.
Yes, this system is invertible. The inverse is $y(t)=x(t-2)$
Proof:
$x(t) \to \Bigg[ system 1 \Bigg] \to y(t) = x(t+2) \to \Bigg[ inverse \Bigg] \to z(t) = y(t-2) = x((t-2)+2) = x(t)$
--Cmcmican 17:08, 24 January 2011 (UTC)
• Good job! For some reason, this is a problem that a lot of students get stuck on. -pm
• Why does z(t)=y(t-2)?
Instructor's answer: This is by the definition of the second system: it time delays its input. So if the input were x(t), then the output would be x(t-2). In this case, the input is called y(t), so the output is then y(t-2). You may want to look at this video for more clarification. -pm
• My question was poorly articulated. I should have asked rather how the inverse of the function was found. Once I have the inverse, I understand how to cascade; I did not understand how the inverse of y(t)=x(t+2) is y(t)=x(t-2). However, after reviewing it again, I see that since y(t)=x(t+2) then y(t-2)=x(t). Therefore the inverse of the signal is y(t)=x(t-2).
Instructor's comment: I see. But you know, in general, inverting an invertible system can be quite challenging. There is no method that works all the time. If you are lucky and figure out how to isolate x(t) in terms of y (e.g., y(t), y(t+1), t y(t), stuff like that), like you cleverly did above, then you are good to go. But sometimes isolating x(t) is hard. In such cases, thinking logically about what that the system does might tell you the answer. Like in this particular case: the system is a time delay, and once you realize that, it's a no brainer to invert it without doing any math. In other cases (think cryptography here...) finding the answer might even be NP hard! -pm
Write it here. | 2022-08-13T22:06:36 | {
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https://math.stackexchange.com/questions/424698/is-lnx-uniformly-continuous | # Is $\ln(x)$ uniformly continuous?
Let $x\in[1,\infty)$. Is $\ln x$ uniformly continuous? I took this function to be continuous and wrote the following proof which I'm not entirely sure of.
Let $\varepsilon>0$, $x,y\in[1, ∞)$ and $x>y$. Then, $\ln x< x$ and $\ln y< y$ and this follows that $0<|\ln x-\ln y|<|x-y|$ since $x> y$. Choose $δ=ϵ$. Now suppose $|x-y|< δ$. Then, $|\ln x-\ln y|<|x-y|<\varepsilon$
It would be much appreciated if someone could validate my proof
• Dear Rajinda, As Jonas Meyer notes in his answer below, your argument is not correct. Just because $\ln x < x$and $\ln y < y$ you can't conclude anything about $\ln x - \ln y$ vs. $x - y$; for that, you have to know something about the disance between $\ln x$ and $x$ and the distance between $\ln y$ and $y$. (Just to give an example with numbers, $3 < 10$ and $1 < 9$, but it is not true that $3 - 1$ is less than $10 - 9$.) Regards, – Matt E Jun 19 '13 at 19:16
• A rookie mistake I suppose. However $lnx$ increases at lower rate than $x$ increases. Hence doesn't lnx<xand lny<y imply the inequality based on rate on increase i mentioned? In your example the lower side of the inequality decreases by a higher proportion than the higher side – Heisenberg Jun 19 '13 at 19:27
• Dear Rajinda, Yes, your discussion of rates of increase is exactly the point, and this is what you have to incorporate into your proof! My suggestion is that, if you really want to get your head round this, you try writing out a more careful proof on your own, trying to incorporate this idea of rate of increase in a careful way. (Maybe you will just end up coming back to one of the arguments given below, but maybe you will find a different argument!) Best wishes, – Matt E Jun 19 '13 at 19:33
• Thanks for helping me clear things up. Realized the flaw in the proof. – Heisenberg Jun 19 '13 at 19:35
• Qualitatively, the natural log function does not seem uniformly continous, because as we approach zero, it rapidly becomes large and negative. There is no upper bound on how big of a vertical delta corresponds to any given horizontal delta. – Kaz Jun 19 '13 at 20:38
You can prove something more general:
PROP Suppose $f:[a,\infty)\to\Bbb R$ has bounded derivative. Then $f$ is uniformly continuous on its domain.
P Pick $x,y\in[a,\infty)$ arbitrarily. By the mean value theorem, we can write $$|f(x)-f(y)|=|f'(\xi)||x-y|$$
Let $M=\sup\limits_{x\in[a,\infty)}|f'(x)|$. Then $$|f(x)-f(y)|\leq M|x-y|$$
Thus, for any $\epsilon$ we may take $\delta=\frac{\epsilon}{2M}$. Note that in your case $M=1$. I only divide by $2$ to turn $\leq$ into $<$.
ADD This means, for example, that $\log x$ (over $[a,\infty)$, $a>0$), $\sin x$, $\cos x$, $x$, and similar functions are all uniformly continuous. Note, for example, that $\sin(x^2)$ is not uniformly continuous. Note that we actually prove $f$ is $1$-Lipschitz with constant $M$, so this might be of interest.
• A very useful answer! Thanks alot. – Heisenberg Jun 19 '13 at 18:44
An easier argument is to note that the derivative of $\ln x$ is bounded by 1 on the interval $[1,\infty)$. Therefore $\ln x$ is Lipschitz and in particular uniformly continuous.
• OK but I was hoping to validate my proof as well. Is it correct? – Heisenberg Jun 19 '13 at 18:14
• I am not sure how you deduce that $|\ln x-\ln y|<|x-y|$. I don't think it is enough to write that $\ln x < x$. – Mikhail Katz Jun 19 '13 at 18:19
• If $x∈[1,∞)$ then $e^x> x$ so $ln x < x$. Why is it wrong ? – Heisenberg Jun 19 '13 at 18:27
• $x<a$ and $y<b$ does not imply $|x-y|<|a-b|$, for example let $x=1$ $y=5$ and $a=5$ and $b=6$ $|1-5|$ is not less than $|5-6|$ – user10444 Jun 19 '13 at 18:41
Alternatively, you can prove a function is uniformly continuous based off the following idea:
$f$ is uniformly continuous if and only if for any sequence $\{a_n\},\{b_n\}$
$$\lim\left(a_n-b_n\right)=0 \Rightarrow \lim\left(f\circ a_n-f\circ b_n\right)=0.$$
Let $\{a_n\}, \{b_n\}$ satisfy our hypothesis ($\lim\left(a_n-b_n\right)=0$), then we have $\lim a_n = \lim b_n$ and so
$$\lim\left(f\circ a_n-f\circ b_n\right) =\lim\left(\ln(a_n)-\ln(b_n)\right)=\lim\ln\left(\frac{a_n}{b_n}\right) = \ln(1) = 0.$$
• What you used, is it a theorem ? Haven't come across it – Heisenberg Jun 19 '13 at 18:40
• It is, Uniform continuity via sequences. It is worthwhile to prove this result, I was given a homework assignment to prove this and it was very useful for many uniform continuity proofs. – Kenny Hegeland Jun 19 '13 at 18:46
• True, at times using sequences does make life much easier. – Heisenberg Jun 19 '13 at 18:49
• As a tip, there are a lot of things which can be proven to be not uniformly continuous from this theorem. Many times you can take $a_n=n+\frac{1}{n}, b_n = n$ and contradict this characterization. – Kenny Hegeland Jun 19 '13 at 18:53
• I think your proof is not correct, since for any two sequences ${a_n},{b_n}$ such that $\lim(a_n-b_n)=0$ we do not know if $\lim{a_n}$ and $\lim{b_n}$ exist. And even if the limits $\lim{a_n}$ and $\lim{b_n}$ exist, we do not have $\lim \frac{a_n}{b_n}=1$ (unless the limits are not equal to zero): $a_n=\frac{1}{n}$ and $b_n=\frac{2}{n}$, we have $\lim(a_n-b_n)=0$ but $\lim \frac{a_n}{b_n}\neq 0$. In fact this can be used to prove that $x\mapsto ln(x)$ is not uniformly continuous on $(0,\infty)$ – user50618 Dec 21 '14 at 22:01
No, your proof has a problem. If $f(x)<x$ for all $x\in[1,\infty)$, it does not follow that $|f(x)-f(y)|<|x-y|$ for all $x$ and $y$.
You have $x>y$, and using the fact that $\ln$ is increasing, $|\ln x -\ln y|=\ln x - \ln y$. But how do you conclude that this is less than $x-y=|x-y|$? We know that $\ln x<x$, which gives $\ln x - \ln y <x-\ln y$. But $\ln y<y$ applied to the last expression gives $x-\ln y>x-y$, which doesn't help. Replacing $\ln x$ with $x$ makes the expression bigger, while replacing $\ln y$ with $y$ makes the expression smaller. To ensure that the net result is bigger, you need to know that $x-\ln x > y- \ln y$. But this is just a rearrangement of the inequality that you want to prove.
In summary: The conclusion that $|\ln x -\ln y|\leq |x-y|$ for all $x,y\geq 1$ is true, but more is needed to show it. Some methods to complete the proof are given in the other answers.
• So basically it lacks detail but I haven't stated anything incorrect? – Heisenberg Jun 19 '13 at 19:09
• @Rajinda: It is incorrect to say that from $\ln x<x$ and $\ln y<y$ it follows that $|\ln x-\ln y|<|x-y|$. The inequality is correct, but the stated justification is not. – Jonas Meyer Jun 19 '13 at 19:10
• @Rajinda: I don't know what you mean by "putting forward the idea". You can state the inequality, and that puts forward the idea. But in your proof, I guess you want to give valid justification for why the inequality is true. Their are several methods of doing so. For example, the mean value theorem can be used as in Peter's answer. Here is another: If $x>y\geq 1$, then $\ln x-\ln y=\ln(x/y)=\ln(1+((x/y)-1))\leq (x/y)-1\leq x-y$. Here I used the inequality $\ln(1+t)\leq t$ for all $t\geq 0$, which can be proved using positivity of the derivative of $t-\ln(1+t)$ for all $t>0$. – Jonas Meyer Jun 19 '13 at 19:23
• Sorry, I meant how can I justify in another way. Thanks for the help! – Heisenberg Jun 19 '13 at 19:32
Assume $$x>y>1$$, Then by triangle inequality and the fact that $$y>1$$:
$$\frac{x}{y} < \frac{|x-y|}{|y|} + 1 < |x-y|+1$$
Let $$\epsilon > 0$$, choose $$\delta = e^{\epsilon}-1$$, then we have
$$|\ln(x)- \ln(y)| = \ln(\frac{x}{y}) < \ln(|x-y|+1) < \ln(e^{\epsilon} - 1+1) = \epsilon$$ | 2021-05-14T01:55:02 | {
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https://math.stackexchange.com/questions/10664/calculating-probability-with-a-piecewise-density-function | Calculating probability with a piecewise density function
I tried solving this, and I am pretty sure I am integrating this correctly, however, my solution manual shows -1 in the equation when doing this and I do not know why. The answer in the solution manual is correct.
Problem: Find the corresponding distribution function and use it to determine the probability that a random variable having the distribution function will take on a value between 0.4 and 1.6.
f(x) = x for 0 < x < 1
2-x for 1 <= x < 2
0 elsewhere
so for F(0.4 < x < 1.6) I did after integrating:
2(1.6) - [(1.6)^2 / 2] - [(0.4)^2 / 2] = 1.84
however the correct answer is 0.84. The solution manual has a -1 in their equation, but I do not know how they got it.
• Note that in this simple case, you can just calculate areas of rectangles and triangled and get the result. – Raphael Nov 17 '10 at 15:35
Since $f(t)$ is defined piecewise, you have to be careful with the integral. This is the source of your mistake.
If $0 < x < 1$, $$F(x) = \int_0^x f(t) dt = \int_0^x t dt = \left.\frac{1}{2}t^2\right|_0^x = \frac{1}{2}x^2,$$ which you have.
However, if $1 \leq x < 2$, then you have to break the integral that yields $F(x)$ up into pieces. This is $$F(x) = \int_0^x f(t) dt = \int_0^1 t dt + \int_1^x (2-t) dt = \left.\frac{1}{2}t^2\right|_0^1 + \left[2t - \frac{1}{2}t^2\right]_1^x$$ $$= \frac{1}{2} + 2x - \frac{1}{2}x^2 - 2 + \frac{1}{2} = 2x - \frac{1}{2}x^2 - 1.$$ Here's where the $-1$ comes in.
This is a common mistake. If it makes you feel any better, my probability students trip up over this all the time. :)
• Wow, thank you very much for the well explained solution. – Raptrex Nov 17 '10 at 6:43
First, probabilities are from $0$ to $1$, so you're certainly wrong.
Second, you calculated the cdf incorrectly for the interval $1 \leq x < 2$. It should be $$F(x) = \int_0^1 t \, \mathrm{d}t + \int_1^x 2-t \, \mathrm{d}t.$$ You forgot the first part, and integrated the second part from $0$ instead of from $1$. Since $$\int_0^1 t \, \mathrm{d}t = 1/2$$ whereas $$\int_0^1 2-t \, \mathrm{d}t = 3/2,$$ you counter $1/2$ too little and $3/2$ too much, for a total gain of $+1$. | 2019-05-25T23:55:38 | {
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https://math.stackexchange.com/questions/2312614/how-to-solve-3037-mod-77-without-calculator/2312622 | # How to solve $30^{37} \mod 77$ without calculator?
I tried doing something like this: $$(30^2)^{15}(30^7)\mod 77$$ but it is not effective, maybe someone knows some tips and tricks to solve this ?
• Have you heard about the Chinese remainder theorem? It says that because $77 = 7\cdot 11$, where $7$ and $11$ are coprime, you can focus on calculations modulo $7$ and modulo $11$ separately, and put it together to modulo $77$ afterwards. – Arthur Jun 6 '17 at 20:29
• Hint: $30=2\times 3 \times 5$ and $2$, $3$ and $5$ are coprime to $77$ and $\phi(77)=60$ – Amin235 Jun 6 '17 at 20:31
• – Ethan Bolker Jun 6 '17 at 20:36
By Fermat Little Theorem $$30^{6} \equiv 1 \pmod{7} \\ 30^{10} \equiv 1 \pmod{11}$$
Therefore $$30^{37}=(30^6)^6 \cdot 30 \equiv 30\equiv 2 \pmod{7} \\ 30^{37} \equiv 30^7\equiv (-3)^7 \equiv -3^4 \cdot 3^3 \equiv 7 \cdot 5 \equiv 2\pmod{11}$$
This shows that $7,11 |30^{37}-2$ and hence $77 | 30^{37}-2$. Thus $$30^{37}\equiv 2 \pmod{77}$$ unless I made a small mistake in the computations.
• I got the same answer (but you beat me by 19 seconds). – TonyK Jun 6 '17 at 20:43
• @TonyK I know the feeling, happened to me few times in the past too :) – N. S. Jun 6 '17 at 20:43
• @N.S thanks for advice. – Walter White Jun 6 '17 at 20:45
Use repeated squaring, for example $30^{32}=((((30^2)^2)^2)^2)^2$ and take mod 77 at each step.
when confronted with $f(x) \equiv y \pmod {77}$
consider
$f(x) \equiv y \pmod {7}$ and $f(x) \equiv y \pmod {11}$
Fermat's little theorem says that for all prime $p$ and if $p$ does not divide $a$
$a^{p-1} \equiv 1 \pmod p$
$30^{37} \equiv 2 \pmod 7\\ 30^{37} = (-3)^7\equiv 2 \pmod {11}$
$30^{37}\equiv 2 \pmod{77}$
Do the calculation separately modulo $7$ and modulo $11$.
Modulo $7$:
$30^{37}\equiv 2^{37} \equiv 2$
where using Fermat's Little Theorem $2^{36}=(2^6)^6\equiv 1$.
Modulo $11$:
$30^{37}\equiv 8^{37} \equiv 2^{3×37}=2^{111}\equiv 2$
where using Fermat's Little Theorem $2^{110}=(2^{10})^{11}\equiv 1$.
So $30^{37}\equiv 2 \bmod 7$ and $\bmod 11$, thus the Chinese Remainder Theorem gives
$30^{37}\equiv 2 \bmod 77$ | 2019-12-09T10:30:00 | {
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https://math.stackexchange.com/questions/3959099/how-many-eigenvalues-does-an-n-times-n-matrix-have-and-how-does-this-relate-t | # How many eigenvalues does an $n\times n$ matrix have, and how does this relate to the algebraic multiplicity of the dominant eigenvalue?
I'm having to reevaluate my understanding of eigenvalues and how many eigenvalues an $$n\times n$$ matrix possesses. Previously, I had thought that such a matrix $$A$$ possessed $$d\leq n$$ complex eigenvalues, and that this number $$d$$ was determined by the number of distinct roots of the matrix's characteristic polynomial. Furthermore, the sum of the complex eigenvalues' algebraic multiplicities equals $$n$$, since the characteristic polynomial of $$A$$ necessarily has degree $$n$$ and therefore has $$n$$ complex roots (where a root of multiplicity $$m$$ is counted $$m$$ times). However, I came across a problem involving dominant eigenvalues that prompted me to question this interpretation:
If $$A$$ has a dominant eigenvalue $$\lambda_1$$, prove that the eigenspace $$E_{\lambda_1}$$ is one-dimensional.
Solution: If $$\lambda$$ is dominant, then $$|\lambda|>|\gamma|$$ for all other eigenvalues $$\gamma$$. But this means that the algebraic multiplicity of $$\lambda$$ is $$1$$, since it appears only once in the list of eigenvalues listed with multiplicity, so its geometric multiplicity is $$1$$ and thus its eigenspace is one-dimensional.
So if I'm reading this explanation correctly, a list of all of the eigenvalues of $$A$$ should include $$i$$ instances of an eigenvalue with algebraic multiplicity $$i$$. In other words, every $$n \times n$$ matrix has exactly $$n$$ complex eigenvalues, and there is a distinction between the number of eigenvalues that a matrix possesses and the number of distinct eigenvalues that a matrix possesses. This subtle distinction seemed arbitrary until I considered the solution to this problem, which seems to require that all eigenvalues be treated as separate entities, even if they possess the same scalar values. For example, an eigenvalue $$\lambda =2$$ with algebraic multiplicity $$2$$ should actually be thought of as two eigenvalues $$\lambda _1 = \lambda _2 = 2$$. With this understanding, it is clear that neither $$\lambda _1$$ nor $$\lambda _2$$ can be a dominant eigenvalue, since it is not true that $$|\lambda _1|>|\lambda _2|$$, nor is it true that $$|\lambda _2|>|\lambda _1|$$. In fact, it is impossible for any eigenvalue with algebraic multiplicity greater than $$1$$ to be dominant. From here, I am comfortable with the fact that a dominant eigenvalue (if it exists) must also have geometric multiplicity $$1$$, since the geometric multiplicity of an eigenvalue is always less than or equal to the corresponding algebraic multiplicity.
Is this the correct way to interpret the preceding proof/eigenvalues in general? Hopefully I've articulated my thought process clearly, and thank you for taking the time to help!
## 1 Answer
I don't think a convention is well-established: in some contexts, I see "different eigenvalues" refer to a set of distinct values with associated algebraic multiplicities, while in other contexts, I see "different eigenvalues" refer to the set of $$n$$ eigenvalues, possibly with repetitions due to multiplicity. Typically one can either discern which convention is being used, or the author should take care to clarify what is meant.
In your case, I think you just have to read the definition of "dominant eigenvalue" carefully. Based on the problem writing "dominant eigenvalue $$\lambda_1$$," I suspect the definition is written as
if $$\lambda_1, \ldots, \lambda_n$$ are the eigenvalues of $$A$$, then $$\lambda_1$$ is considered dominant if $$|\lambda_1| > |\lambda_i|$$ for all $$i \ne 1$$
or something like that, which is unambiguous compared to
$$|\lambda| > |\gamma|$$ for all other eigenvalues $$\gamma$$
which is very ambiguous for the reasons you raise.
Now that we know that the context of your question is the power method, then my above guess on what "dominant eigenvalue" means is incorrect.
Let $$\lambda_1, \ldots, \lambda_m$$ be the distinct eigenvalues of $$A$$ with multiplicities $$n_1, \ldots, n_m$$. If $$|\lambda_1| > |\lambda_i|$$ for all $$i \ne 1$$, then $$\lambda_1$$ is said to be the dominant eigenvalue. The power method will converge to something in the eigenspace corresponding to $$\lambda_1$$. To ensure that it does not converge to zero, the initial vector must not be orthogonal to the eigenspace.
• Thank you, that clears up some of the ambiguity. The book I'm reading (Poole's Linear Algebra) can't seem to make up its mind on which definition it espouses; it gives another example of a $3\times 3$ matrix with eigenvalues $17$, $17$, and $-1$, and boldly states that $17$ is the dominant eigenvalue of the matrix . . . despite this eigenvalue having an algebraic (and geometric) multiplicity of 2. So I don't know. What seems to be the important characteristic of a dominant eigenvalue is that the power method converges to a vector in its eigenspace, regardless of its algebraic multiplicity(?) – RyanC Dec 23 '20 at 22:11
• @RyanC See my edit. – angryavian Dec 23 '20 at 22:59 | 2021-05-14T11:27:15 | {
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https://www.itl.nist.gov/div898/handbook/prc/section2/prc262.htm | 7. Product and Process Comparisons
7.2. Comparisons based on data from one process
7.2.6. What intervals contain a fixed percentage of the population values?
## Percentiles
Definitions of order statistics and ranks For a series of measurements $$Y_1, \, \ldots, \, Y_N$$, denote the data ordered in increasing order of magnitude by $$Y_{[1]}, \, \ldots, \, Y_{[N]}$$. These ordered data are called order statistics. If $$Y_{[j]}$$ is the order statistic that corresponds to the measurement $$Y_i$$, then the rank for $$Y_i$$ is $$j$$; i.e., $$Y_{[j]} \sim Y_i \,\, \Longrightarrow \,\, r_i = j \, .$$
Definition of percentiles Order statistics provide a way of estimating proportions of the data that should fall above and below a given value, called a percentile. The $$p$$th percentile is a value, $$Y_{(p)}$$, such that at most $$(100 p)$$ % of the measurements are less than this value and at most $$100(1-p)$$ % are greater. The 50th percentile is called the median.
Percentiles split a set of ordered data into hundredths. (Deciles split ordered data into tenths). For example, 70 % of the data should fall below the 70th percentile.
Given n points, the percentile corresponding to the i-th point is
$$\frac{i}{n+1}$$
More typically we start with a desired percentile value and this percentile of interest may not correspond to a specific data point. In this case, interpolation between points is required. There is not a standard univerally accepted way to perform this interpolation. After describing our default method, several alternative methods are given. All of the methods discussed here are used in practice.
Estimation of percentiles Percentiles can be estimated from $$N$$ measurements as follows: for the $$p$$th percentile, set $$p(N+1)$$ equal to $$k + d$$ for $$k$$ an integer, and $$d$$, a fraction greater than or equal to 0 and less than 1.
1. For $$0 \lt k \lt N, \,\,\,\,\, Y_{(p)} = Y_{[k]} + d \left( Y_{[k+1]} - Y{[k]} \right)$$
2. For $$k = 0, \,\,\,\,\, Y_{(p)} = Y_{[1]}$$
Note that any p ≤ 1/(N+1) will simply be set to the minimum value.
3. For $$k ≥ N, \,\,\,\,\, Y_{(p)} = Y{[N]}$$
Note that any pN/(N+1) will simply be set to the maximum value.
Example and interpretation For the purpose of illustration, twelve measurements from a gage study are shown below. The measurements are resistivities of silicon wafers measured in ohm.cm.
i Measurements Order stats Ranks
1 95.1772 95.0610 9
2 95.1567 95.0925 6
3 95.1937 95.1065 10
4 95.1959 95.1195 11
5 95.1442 95.1442 5
6 95.0610 95.1567 1
7 95.1591 95.1591 7
8 95.1195 95.1682 4
9 95.1065 95.1772 3
10 95.0925 95.1937 2
11 95.1990 95.1959 12
12 95.1682 95.1990 8
To find the 90th percentile, $$p(N+1)$$ = 0.9(13) = 11.7; $$k$$ = 11, and $$d$$ = 0.7. From condition (1) above, $$Y_{(90)}$$ is estimated to be 95.1981 ohm.cm. This percentile, although it is an estimate from a small sample of resistivities measurements, gives an indication of the percentile for a population of resistivity measurements.
Note that there are other ways of calculating percentiles in common use Hyndman and Fan (1996) in an American Statistician article evaluated nine different methods (we will refer to these as R1 through R9) for computing percentiles relative to six desirable properties. Their goal was to advocate a "standard" definition for percentiles that would be implemented in statistical software. Although this has not in fact happened, the article does provide a useful summary and evaluation of various methods for computing percentiles. Most statistical and spreadsheet software use one of the methods described in Hyndman and Fan.
The method described above corresponds to method R6 of Hyndman and Fan. This is the default method used by Dataplot.
The method advocated by Hyndman and Fan is R8. For the R8 method, set $$p(N+(1/3) + (1/3))$$ and proceed as above. Note that any p ≤ (2/3)/(N+(1/3)) will be set to the minimum value and any p ≥ (N-(1/3))/(N+(1/3)) will be set to the maximum value. Both R and Dataplot can optionally use this method. For the example given above, R8 gives 95.1972 (compared to 95.1981) for the 90-th percentile.
Some software packages set $$1 + p(N-1)$$ equal to $$k + d$$ and then proceed as above. This is method R7 of Hyndman and Fan. This is the method used by Excel and is the default method for R (the R quantile function can optionally use any of the nine methods discussed in Hyndman and Fan). For the example given above, R7 gives 95.1957.
The R6, R7, and R8 methods give fairly similar, but not exactly the same (particularly for small samples), results. For most purposes, any of these three methods should be acceptable.
Another method of calculating percentiles (given in some elementary textbooks) starts by calculating $$p N$$. If that is not an integer, round up to the next highest integer $$k$$ and use $$Y_{[k]}$$ as the percentile estimate. If $$p N$$ is an integer $$k$$, use $$0.5 \left( Y_{[k]} + Y_{[k+1]} \right)$$. One of R6, R7, or R8 would typically be preferred to this method.
Definition of Tolerance Interval An interval covering population percentiles can be interpreted as "covering a proportion $$p$$ of the population with a level of confidence, say, 90 %." This is known as a tolerance interval. | 2020-12-03T23:32:15 | {
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https://mathoverflow.net/questions/334244/stationary-distribution-of-a-markov-process-defined-on-the-space-of-permutations | Stationary distribution of a Markov process defined on the space of permutations
Let $$S$$ be the set of $$n!$$ permutations of the first $$n$$ integers. Let $$p\in(0,1)$$. Consider the Markov Process defined on the elements of $$S$$.
1. Let $$x\in S$$. Choose $$1\le i uniformly at random.
2. If $$x_i < x_{i+1}$$, swap $$x_i$$ and $$x_{i+1}$$ with probability $$p$$, otherwise do nothing. If $$x_i > x_{i+1}$$, swap $$x_i$$ and $$x_{i+1}$$ with probability $$1-p$$, otherwise do nothing.
This process is ergodic, because there is path between any two states with non-zero probability. It has a stationary distribution. I conjecture that the stationary distribution of $$p(x)$$ depends only on $$p$$ and on the number of mis-rankings of $$x$$, defined as $$\sum_{i\le j} 1\{x_i < x_j\}$$. But am not able to prove it. I also wonder whether this simple model has been studied somewhere, maybe in Statistical Mechanics. Any literature reference is appreciated.
• Is this well-defined? Suppose $p=1/3$ and you pick $2$ and $3$. We should swap $2$ and $3$ with probability $1/3$ because $2<3$. Then again, we should swap $3$ and $2$ with probability $2/3$ because $3>2$... Jun 18 '19 at 4:38
• Of course there are $n(n-1)/2$ pairs... if it is swap $i,j$ with probability $p$, then you have a random walk on $S_n$ driven by $\nu\in M_p(G)$ given by $\nu((i\qquad j))=\frac{2p}{n(n-1)}$ on transpositions and $\nu(e)=1-p$. Random walks on groups have as stationary distributions, due to the fact that the stochastic matrix is doubly stochastic. For $p=1-1/n$ this is the random transposition shuffle. See Diaconis and Shahshahani 1981. Jun 18 '19 at 7:25
• *"... have as stationary distributions,..., the uniform distribution". Jun 18 '19 at 8:13
• @JP McCarthy: we don't swap $2$ and $3$, but instead $x_2$ and $x_3$. It seems well-defined to me. But there is a typo: there are $n(n-1)/2$ choices for $i$ and $j$, not $n(n+1)/2$. Jun 18 '19 at 10:05
• Ah I missed $i<j$ which means the pair $\{i,j\}$ is an ordered pair $(i,j)$. Take the permutation $\{(1,3),(2,1),(3,2)\}$. I take it the notation means that $x_1=3$, $x_2=1$, and $x_3=2$? Jun 18 '19 at 10:30
Your conjecture is correct. In fact, provided $$0 < p < 1$$, the Markov process is recurrent and reversible with unique stationary distribution proportional to $$\pi_\sigma = \Bigl(\frac{p}{1-p}\Bigr)^{\ell(\sigma)},$$ where $$\ell(\sigma)$$ is the Coxeter length of $$\sigma$$. (This is the number of 'misrankings' in your question.)
Proof. Suppose that $$0 < p < 1$$. Let $$p_{\sigma\tau}$$ be the probability of a step from $$\sigma \in S$$ to $$\tau\in S$$. We solve the detailed balance equations $$\pi_\sigma p_{\sigma\tau} = \pi_\tau p_{\tau\sigma}$$. Suppose that $$\tau = \sigma_1 \ldots \sigma_{i+1} \sigma_i \ldots \sigma_n$$ where $$1 \le i < \le n$$. Then we step from $$\sigma$$ to $$\tau$$ with probability either $$p/n$$ or $$(1-p)/n$$. Explicitly,
\begin{align*} n \pi_\sigma p_{\sigma\tau} &= \Bigl(\frac{p}{1-p}\Bigr)^{\ell(\sigma)} \begin{cases} p & \text{if \sigma_i < \sigma_j} \\ 1-p & \text{if \sigma_i > \sigma_j} \end{cases}\\ &= \Bigl(\frac{p}{1-p}\Bigr)^{\ell(\sigma)} \begin{cases} \frac{p}{1-p} (1-p) &\text{if \sigma_i < \sigma_j} \\ \frac{1-p}{p} p & \text{if \sigma_i > \sigma_j} \end{cases} \\ &= \Bigl(\frac{p}{1-p}\Bigr)^{\ell(\tau)} \begin{cases} 1-p & \text{if \tau_i > \tau_j} \\ p & \text{if \tau_i < \tau_j} \end{cases} \\ &= n\pi_\tau p_{\tau\sigma}. \end{align*} If $$\tau$$ is not of this form and $$\tau \not= \sigma$$ then $$p_{\sigma\tau} = p_{\tau\sigma} = 0$$. Hence the detailed balance equations hold. You observed in your question that there is a single communicating class of states. The walk is aperiodic because there is a positive chance of staying put at each step. Hence the invariant distribution is unique. $$\quad \Box$$
For completeness, suppose that $$p=0$$ or $$p=1$$ and that the walk starts at $$\sigma \in S$$. It is clear that if $$p=0$$ then after $$\ell(\sigma)$$ steps the walk reaches the identity permutation; if $$p = 1$$ then after $$\binom{n}{2} - \ell(\sigma)$$ steps the walk reaches the order reversing permutation $$1 \mapsto n$$, $$2 \mapsto n-1$$, $$\ldots$$, $$n\mapsto 1$$ of maximum Coxeter length. The only randomness arises from the order in which inversions are removed/added.
Remark In another version of the problem, we step according to a general transposition $$(i,j)$$ chosen uniformly at random. In this case the process is not reversible. When $$n \le 3$$ the invariant distribution depends only on the Coxeter length: for example if $$n=3$$ then, ordering permutations $$123,213,132,312,231,321$$, the transition matrix is
$$\frac{1}{3} \left( \begin{matrix} 3(1-p) & p & p & 0 & 0 & p \\ 1-p & 2-p & 0 & p & p & 0 \\ 1-p & 0 & 2-p & p & p & 0 \\ 0 & 1-p & 1-p & 1+p & 0 & p \\ 0 & 1-p & 1-p & 0 & 1+p & p \\ 1-p & 0 & 0 & 1-p & 1-p & 3p \end{matrix} \right).$$
A computer algebra calculation shows that the invariant distribution is proportional to $$(\alpha,\beta,\beta,\gamma,\gamma,\delta)$$ where $$\alpha = (1-p)(6-11p+7p^2)$$, $$\beta = (1-p)p(8-7p)$$, $$\gamma = (1-p)p(1+7p)$$ and $$\delta = p(2-3p+7p^2)$$. For $$n=4$$ the invariant distribution is more complicated. For example, if $$p=3/4$$ then the invariant probabilities for $$2134$$ and $$1324$$ are $$5325/485760$$ and $$8749/485760$$, respectively.
• Thank you Mark! Jun 18 '19 at 19:43
• I have two questions: 1. What is in the interpretation of $n\choose 2$ in the balance equation? 2. It seems that the third equality requires that $\ell(\tau)=\ell(\sigma)\pm 1$. This is always the case when $i$ and $j$ are adjacent, but not in general. Jun 21 '19 at 5:16
• I'm sorry: my answer is wrong because I wrongly assumed the Coxeter length goes up by 1 on every swap. As you noticed, it would be correct for the similarly defined random walk where $i$ and $j$ are required to be adjacent. Please could you unaccept my answer so that I can delete it. Jun 21 '19 at 10:11
• Mark, I think I am going to restate the formulation of my problem by restricting the assumption so that your answer is correct (just change the i, j to i, i+1). It's a special case, and it's still interesting. It's now interpretable as a noisy bubble-sort algorithm. Also, it is useful because I think now that my original conjecture was wrong! Jun 22 '19 at 1:33
• Okay! In that case I will edit the answer above so I can delete my second answer. Jun 22 '19 at 10:45
This stationary distribution is known as the Mallows measure, see e.g. the references in http://www.sc.ehu.es/ccwbayes/members/ekhine/tutorial_ranking/data/slides.pdf
For the Markov chain connection, see e.g. | 2021-09-25T18:18:22 | {
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https://math.stackexchange.com/questions/978851/is-the-plane-minus-the-integer-lattice-homeomorphic-to-the-plane-minus-the-integ | # Is the plane minus the integer lattice homeomorphic to the plane minus the integers?
The question, more rigorously posed, is:
Is $\Bbb R^2-\Bbb Z^2$ homeomorphic to $\Bbb R^2-\Bbb Z\times\{0\}$?
This question has been bugging me in the back of my head for a while now. Sometimes, I think it's clear that they are, and then I think it's clear that they aren't. Nothing from Point-set topology is helping me solve this, at least as far as I see. And the only help from algebraic topology that I'm familiar with is the fundamental group; however, these two spaces seem to have the same fundamental group.
Any help in alleviating my torment is appreciated.
• You can probably construct a proof for "yes" by enumerating $\mathbb{Z}^2$ with a spiral, and using this to construct the homeomorphism on successive open squares. – Slade Oct 17 '14 at 22:39
• Or use classification of open surfaces: Both have genus zero and homeomorphic spaces of ends. – Moishe Kohan Oct 17 '14 at 22:42
• @Slade that was one approach that made me think 'yes, they are'. However, how do you strech the spiral onto the $x$-axis? In order to do so would require the spiral to be 'two-sided'; that is, two concentric spirals that meet at the origin. But how do you stretch that onto the $x$-axis without creating some tear between two lattice points that are close together? – Robert Wolfe Oct 17 '14 at 22:42
• @Bryan You don't need to stretch it onto the $x$-axis, just some neighborhood of the $x$-axis. It should probably correspond to the integers $0,1,-1,2,-2,\cdots$. This is messy, of course. – Slade Oct 17 '14 at 22:53
• What is the fundamental group? – Seth Oct 17 '14 at 23:08
1. Open connected surfaces are classified by a theorem of Brown and Messer "The classification of two-dimensional manifolds", Transactions of AMS, 1979. The first invariant is the genus (and genera of ends): In your case genus equals $0$ since both surfaces are obtained by removing compact subsets $E_1, E_2$ from $S^2$. The next invariant is orientability (and orientability of ends): Both surfaces are oriented. The last and the most interesting invariant is the "set of ends" of the surface, which is a certain compact Hausdorff topological space. In the case of surfaces $S^2\setminus E$, the set of ends is $E$ with the subspace topology, provided that $E$ is compact and totally disconnected. In the examples you have, both sets $E_1, E_2$ are 1-point compactifications of countably infinite discrete spaces, hence $E_1$ is homeomorphic to $E_2$. In particular, $S^2 \setminus E_1$ is homeomorphic to $S^2\setminus E_2$.
2. For the fundamental group: It is a corollary of one of Whitehead's theorems that each open connected surface is homotopy equivalent to a bouquet of circles, hence, has free fundamental group. To compute rank of the group, it suffices to look at the 1st homology group, which, in your case has countably infinite rank (say, by the Alexander duality). Hence, $\pi_1(R^2\setminus Z^2)\cong F_{\aleph_0}$, where $\aleph_0$ is the cardinality of the set of natural numbers.
It's easy enough to explicitly construct a homeomorphism for this. All we require is that there is some homeomorphism $f:\mathbb{R^2\rightarrow R^2}$ such that the set $$S=\{x:\exists y\in\mathbb{R}[f(x,y)\in\mathbb{Z^2}]\}$$ has no accumulation points and has that the associated $y$ to each $x$ is unique - that is, we imagine that, for a given $x$, the image of $f(x,y)$ where $y\in\mathbb{R}$ is a curve in $\mathbb{R}^2$. As we vary $x$, this curve sweeps through the space, and, at a series of values $x\in S$, it encounters, on this curve, a single integer point.
We can create such a function by choosing some $g(x)$ such that the set $\{\frac{1}{g(n)}\:n\in \mathbb{N}\}$ contains no pair of elements whose ratio is rational and has no accumulation point. Something like $g(n)=e^{-4n^2-n}$ would suffice, since the exponent is never equal for distinct integers and $e^{x}$ is never rational for rational, non-zero $x$. Then, we define $f(x,y)=xg(y)$ - representing sweeping the plane by curves which are simply scalings of the curve $(y,g(y))$. It is easy enough to see that the set $S$ of $x$ such that the curve $f(x,y)$ passes through a point on the integer lattice has no accumulation points.
Thus, what is remaining is to find a map $p:\mathbb{Z}\rightarrow S$ which preserves order. This clearly exists, since $S$ has no accumulation points, meaning each element has a "next" element, and enumerating elements like so would reach every element. We can extend this (for instance as a piecewise linear function) to a continuous bijection $p':\mathbb{R}\rightarrow \mathbb{R}$ such that $p'|_{\mathbb{Z}}=p$ - that is, it interpolates all the pairs of coordinates $(n,p(n))$. We can also create a continuous function $s$, by similar means, such that, for integer $n$, the function $s(n)$ is the solution for $y$ to $f(p(n),y)\in \mathbb{Z}^2$.
From here, we just note that the following function is a homeomorphism from $\mathbb{R^2 - Z\times \{0\}}$ to $\mathbb{R^2-Z^2}$: $$(x,y)\mapsto f(p'(x),y+s(x))$$ since the image $p'[\mathbb{R}-\mathbb{Z}]=\mathbb{R}-S$ and the image $\mathbb{R}+s(x)=\mathbb{R}$ as well, and the image $f[\mathbb{(R}-S)\mathbb{\times R}]=\mathbb{R}^2-\mathbb{Z}^2$ and all functions involved are continuous.
This generalizes to show that the plane minus any countably large set of points with no accumulation point is homeomorphic.
Alright, I came up with my own proof.
A couple of lemmas:
Lemma 1: Let $X$ and $Y$ be topological spaces. And let $\mathcal{B}$ be a locally-finite collection of closed sets such that $\bigcup\mathcal{B}=X$. Suppose, in addition, that there is a continuous function $f_B:B\rightarrow Y$ for each $B\in\mathcal{B}$ such that $f_B(x)=f_C(x)$ for all $B\in\mathcal{B}$, all $C\in\mathcal{B}$ such that $C\cap B\neq\varnothing$, and all $x\in B\cap C$. Then there is a unique continuous function $f:X\rightarrow Y$ such that $f|_B=f_B$ for all $B\in\mathcal{B}$. (This is an infinite pasting lemma)
Lemma 2: Let $v\in B_1(0)\subseteq\Bbb R^2$ (this is the open ball). Then there is a homeomorphism $h:\Bbb R^2\rightarrow \Bbb R^2$ such that $h(x)=x$ for all $x\in\Bbb R^2-B_1(0)$ and such that $h(0)=v$.
Lemma 3: Let $U\subseteq \Bbb R^2$ be a connected open set. For any two $u$ and $v\in U$ there is a finite number of points $\{x_1, x_2, \ldots, x_n\}\subseteq U$ and a finite number of strictly positive real numbers $\{\epsilon_1, \epsilon_2, \ldots, \epsilon_n\}$ such that $u=x_1$, and $v=x_n$, and $B_{\epsilon_{i}}(x_i)\cap B_{\epsilon_{i+1}}(x_{i+1})\neq\varnothing$ and $B_{\epsilon_i}(x_i)\subseteq U$ for all $i$.
Lemma 4: Let $U\subseteq \Bbb R^2$ be a connected, open subset. Then $U$ is strongly $n$-homogeneous for all $n\geq 1$.
I leave these nice lemmas for the interested reader. They are fun.
Now choose a series of radii $\{r_n\}_{n=0}^\infty$ such that $B_{r_n}(0)$ contains the square of inradius $n$ (for $n=0$ choose $r_0=1/2$) and such that $\overline{B_{r_n}}(0)$ is disjoint from the square of inradius $n+1$. Finally, consider the collection $\{\overline{B_{r_0}}(0), A_1, A_2, A_3, \ldots\}$ where we have $A_i=\overline{B_{r_i}}(0)-B_{r_{i-1}}(0)$. This collection consists of a closed ball containing the point $(0,0)$ and a sequence of concentric closed annuli which contain one of the squares in question in their interior.
Now we use the above lemmas to create a homeomorphism on each of the annuli (leave the first closed ball alone) to bring the lattice points they contain onto the $x$-axis while still fixing the boundaries of each annulus. We then stitch these together using the first lemma to create a global homeomorphism which pulls the integer lattice onto a discrete countable subset of the $x$-axis, which is clearly homeomorphic to the $\Bbb R^2-\Bbb Z\times\{0\}$. | 2019-05-22T11:28:16 | {
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https://math.stackexchange.com/questions/1012103/let-a-b-in-mathbbr-show-that-a4b48-ge-8ab/1012117#1012117 | # Let a,b $\in$ $\mathbb{R}.$ Show that $a^4+b^4+8\ge 8ab.$
Let $$a,b \in \mathbb{R}.$$ Show that $$a^4+b^4+8\ge 8ab.$$
The question is from the inequalities section of An Excursion in mathematics by Bhaskaraycharya Pratisthanan. My heuristics include using the AM-GM inequality. I am unable to design the problem to proceed further.
You're on right path just a small trick using AM-GM Write $$a^4+b^4+8=a^4+b^4+4+4$$ And then apply AM-GM $$\frac{a^4+b^4+4+4}{4}\ge \sqrt[4]{16a^4b^4}$$
$$\frac{a^4+b^4+4+4}{4}\ge 2ab$$
$${a^4+b^4+8}\ge 8ab$$
Or You can just SOS it $$(a^2-b^2)^2\ge0\implies a^4+b^4-2a^2b^2\ge0$$
$$2(ab-2)^2\ge0\implies2a^2b^2+8-8ab\ge0$$ Add above two inequalities to get $$a^4+b^4+8\ge8ab$$
• You may also add to your answer how the inequality can be obtained from the Rearrangement Inequality. I think it will help the OP more.
– user170039
Nov 8 '14 at 16:41
• @user170039 how can we obtain it from rearrangement inequality ?
– r9m
Nov 9 '14 at 21:43
• @r9m: The AM-GM Inequality can be proved from Rearrangement Inequality.
– user170039
Nov 10 '14 at 3:05
• @user170039 thanks for clearing it ! :)
– r9m
Nov 10 '14 at 4:24
We have $$\frac{a^4+b^4}{2}\geq a^2b^2$$ by using AM-GM inequality, therefore we will obtain $$a^4+b^4+8-8ab\geq 2a^2b^2+8-8ab=2(ab-2)^2\geq 0$$
• Do you ever look at your answers and think "that could look a lot better if I had bothered to make it pretty". Nov 17 '14 at 21:34
• i try to find the answer as fast as i can Sarah Nov 17 '14 at 21:35
• I think you would be doing this community a great favour by later revising your answer and making it high quality. Nov 17 '14 at 21:36
• Proper grammar. Good spacing between formulas. Look at other answers as an example. Nov 17 '14 at 21:39
• I do not know if you are aware @Dr.SonnhardGraubner but several users have been complaining about the quality of your posts. Posting answers is fine but the ultimate point of SE is to provide high quality questions and answers. Your contribution is welcome as long as you are making an effort to participate. You will find that you will not have as much negative feedback if your answers are well written and formatted as Sarah suggested. Nov 17 '14 at 21:43
Another way to look at it is to consider the following. $$(a^2-b^2)^2\geq0$$ Therefore
$$a^4+b^4\geq2a^2b^2$$
Then from there you proceed the same way as Dr.Sonnhard did.. | 2021-12-05T01:02:40 | {
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https://proofwiki.org/wiki/Inverse_of_Group_Product | # Inverse of Group Product
## Theorem
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a, b \in G$, with inverses $a^{-1}, b^{-1}$.
Then:
$\paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
### General Result
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $a_1, a_2, \ldots, a_n \in G$, with inverses $a_1^{-1}, a_2^{-1}, \ldots, a_n^{-1}$.
Then:
$\paren {a_1 \circ a_2 \circ \cdots \circ a_n}^{-1} = a_n^{-1} \circ \cdots \circ a_2^{-1} \circ a_1^{-1}$
## Proof 1
$\displaystyle \paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} }$ $=$ $\displaystyle \paren {\paren {a \circ b} \circ b^{-1} } \circ a^{-1}$ Group Axiom $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle \paren {a \circ \paren {b \circ b^{-1} } } \circ a^{-1}$ Group Axiom $\text G 1$: Associativity $\displaystyle$ $=$ $\displaystyle \paren {a \circ e} \circ a^{-1}$ Group Axiom $\text G 3$: Existence of Inverse Element $\displaystyle$ $=$ $\displaystyle a \circ a^{-1}$ Group Axiom $\text G 2$: Existence of Identity Element $\displaystyle$ $=$ $\displaystyle e$ Group Axiom $\text G 3$: Existence of Inverse Element
The result follows from Group Product Identity therefore Inverses:
$\paren {a \circ b} \circ \paren {b^{-1} \circ a^{-1} } = e \implies \paren {a \circ b}^{-1} = b^{-1} \circ a^{-1}$
$\blacksquare$
## Proof 2
We have that a group is a monoid, all of whose elements are invertible.
The result follows from Inverse of Product in Monoid.
$\blacksquare$
## Proof 3
$\displaystyle \paren {a \circ b} \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle e$ Definition of Inverse Element $\displaystyle \leadsto \ \$ $\displaystyle a \circ \paren {b \circ \paren {a \circ b}^{-1} }$ $=$ $\displaystyle e$ Group Axiom $G1$: Associativity $\displaystyle \leadsto \ \$ $\displaystyle b \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle a^{-1}$ Group Product Identity therefore Inverses $\displaystyle \leadsto \ \$ $\displaystyle b^{-1} \circ b \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle b^{-1} \circ a^{-1}$ $\displaystyle \leadsto \ \$ $\displaystyle e \circ \paren {a \circ b}^{-1}$ $=$ $\displaystyle b^{-1} \circ a^{-1}$ Definition of Inverse Element $\displaystyle \leadsto \ \$ $\displaystyle \paren {a \circ b}^{-1}$ $=$ $\displaystyle b^{-1} \circ a^{-1}$ Definition of Identity Element
$\blacksquare$ | 2020-01-19T15:32:18 | {
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https://math.stackexchange.com/questions/2046830/uniform-and-absolute-convergence-of-improper-integral | # Uniform and absolute convergence of improper integral
What is an example of an improper integral , $\int_a^\infty f(u,v)du$, that converges uniformly for $v$ is some subset $S$, but where $\int_a^\infty|f(u,v)|du$ converges pointwise but NOT uniformly on $S$?
When Weierstrass’s Test shows that Riemann improper integral with a parameter $v$, $\int_a^\infty f(u,v)du,$ is uniformly convergent it also shows that $\int_a^\infty |f(u,v)|du,$ is uniformly convergent.
I can also find examples where the integral is uniformly convergent but not absolutely convergent, like $\int_a^\infty \sin(vu)/u du$ where $v \in [c,\infty)$.
Or is it always true that a uniformly and absolutely convergent improper integral must also be uniformly convergent with the absolute value of the integrand taken?
No -- uniform and absolute convergence does not imply uniform-absolute convergence. For a counterexample, consider the improper integral
$$\int_0^\infty \frac{v}{u^2 + v^2} \sin u \, du,$$
where $v \in [0, \infty)$ .
It is straightforward to show the integral is absolutely convergent since for $v > 0$,
$$\int_0^ \infty\left|\frac{v}{u^2 + v^2} \sin u \right| \, du \leqslant \int_0^\infty \frac{v}{u^2 + v^2} \, du = \int_0^\infty \frac{1/v}{1 + (u/v)^2} \,du = \pi/2.$$
and uniformly convergent by the Dirichlet test.
However, we do not have uniform-absolute convergence. For all $n \in \mathbb{N}$ we have, $$\int_{n\pi}^{2n\pi} \frac{v}{u^2 + v^2} |\sin u| \, du \geqslant \frac{v}{(2n\pi)^2 + v^2}\int_{n\pi}^{2n\pi} |\sin u| \, du \\ = \frac{v}{(2n\pi)^2 + v^2}\sum_{k=0}^{n-1}\int_{n\pi + k\pi}^{n\pi + k\pi + \pi} |\sin u| \, du \\= \frac{2nv}{4\pi^2n^2 + v^2},$$
Choosing a sequence of points $v_n = n \in [0, \infty)$, we see that for all $n$
$$\int_{n \pi}^{2n\pi} \frac{n}{u^2 + n^2} |\sin u| \, du \geqslant \ \frac{2n^2}{4\pi^2n^2 + n^2} = \frac{2}{4\pi^2+1} > 0.$$
Thus, the convergence fails to be uniform for all $v \in [0,\infty).$
• Thank you. I understand the Dirichlet test. How is it carried out for showing the first integral is unif. conv? – scobaco Dec 6 '16 at 22:38
• @scobaco: You're welcome. To use Dirichlet you show (1) that the integral of $\sin u$ over$[0,x]$ is uniformly bounded for all $x \geqslant 0$ -- which is obviously true -- and (2) that $v/(u^2 + v^2)$ is monotonically decreasing and uniformly convergent to $0$ as $u \to \infty$ for all $v \in [0,\infty)$. The monotone convergence to $0$ is clear. That it converges uniformly is shown with the estimate $v/(u^2 + v^2) \leqslant v/(2uv) = 1/2u$. This follows from $u^2 + v^2 \geqslant 2uv.$ – RRL Dec 6 '16 at 23:12 | 2021-01-28T09:52:46 | {
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http://math.stackexchange.com/questions/297578/solve-for-x-frac12x2-2x-5-0 | # Solve for $x$, $-\frac{1}{2}x^2 + 2x + 5 = 0$
I'm having trouble solving this equation for $x$:
$$-\frac{1}{2}x^2 + 2x + 5 = 0$$
What's the steps to take to solve it?
Thanks.
-
The equation in the question.. – Cypras Feb 7 '13 at 23:02
Not sure how to format properly? It should be -1/2 – Cypras Feb 7 '13 at 23:04
@Cypras: the equation is not written properly and it is not clear what you meant as the LaTex is a bit hosed. Regards – Amzoti Feb 7 '13 at 23:04
Are you familiar with the quadratic formula? – Gerry Myerson Feb 7 '13 at 23:05
Multiply by -2 and use the ABC formula – Slugger Feb 7 '13 at 23:05
$$x^2-4x-10=0$$ $$x^2-2\cdot x\cdot2+2^2-2^2-10=0$$ $$x^2-2\cdot x\cdot2+2^2=4+10$$ $$(x-2)^2=14$$ $$x-2=\pm \sqrt{14}$$ $$x=2\pm \sqrt{14}$$
-
+1 simple but leading approach. – B. S. Feb 8 '13 at 17:39
@Thank you very much Babak. – Adi Dani Feb 8 '13 at 18:30
If you multiply by $-2$ you get $x^2-4x-10=0$, then you can use the quadratic formula to get $x=\frac 12 (4 \pm \sqrt {16+40})=2 \pm \sqrt {14}$
-
How did you get 14? – Cypras Feb 7 '13 at 23:11
sqrt(16 + 40) = sqrt(56) = sqrt(2 * 2 * 14) = 2 * sqrt(14) – oks Feb 7 '13 at 23:14
@oks: I am not sure OP will see your comment. You can preface it with <at>Cypras as Rustyn Yazdanpour did to make sure s/he does. You can only notify one user with the at sign per comment. – Ross Millikan Feb 7 '13 at 23:18
@Rustyn That makes no sense to me. Square root of 56 equals 7.4something? How do you change it like that? – Cypras Feb 7 '13 at 23:20
@Cypras: Rustyn Yazdanpour is doing arithmetic under the square root sign, then using the fact that $\sqrt{ab}=\sqrt a \cdot \sqrt b$. The $40$ in the next to last should be $14$. – Ross Millikan Feb 7 '13 at 23:26
If you multiply $$-\frac{1}{2}x^2 + 2x + 5 = 0\tag{1}$$
by $-2$ you get $$x^2-4x-10=0$$ Using the quadratic formula: $$ax^2 + bx + c = 0 \iff x = \frac{1}{2a}\left(-b \pm \sqrt{b^2 - 4ac}\right),$$ where in this case, $a = 1,\;b= -4,\; c = -10$ we have $$x=\frac 12 (4 \pm \sqrt {16+40})=2 \pm \sqrt {14}\tag{2}$$
Walking through the simplification of $(2)$
$$x = \frac12 (4 \pm \sqrt{16 + 40}) \;=\; \frac12(4 \pm \sqrt{56})\; = \;\frac12 (4 \pm \sqrt{4\cdot 14}) \;=\; \frac12 (4 \pm \sqrt{4}\sqrt{14})\;$$ $$= \;\frac12 (4 \pm 2\sqrt{14})\;=\; \frac 12 \cdot 4 \pm \frac12 \cdot 2\sqrt{14}\;\; = \;\;2 \pm \sqrt{14}$$
-
We, here, denote the discriminate by $\Delta=b^2-4ac$. + – B. S. Feb 9 '13 at 5:34 | 2014-07-28T20:30:31 | {
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