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https://math.stackexchange.com/questions/2796049/proper-containment-of-finite-index-subgroups-of-infinite-groups-of-same-index | # Proper Containment of Finite Index Subgroups of Infinite Groups of Same Index.
Just trying to get my head around the proper way to understand the index of subgroups in infinite groups. Is the following claim true? If so, how would one prove it, and if not, can you help me come up with a counter example? And are there tighter restrictions (such as normality?) that make the claim true in this case?
Suppose that $G$ is a group (not necessarily finite), with subgroups $H$, and $K$ with $H \subseteq K \subseteq G$ and $\left| G: K \right| = \ell = \left| G: H \right|$ for some $\ell \in \mathbb{N}_{ > 1}$. Then $H = K$.
Edit (to add some context, and disclaimer, some of what I say below may be inaccurate since I'm just studying this)
I am currently studying quotients of free groups, and their finite index subgroups, and in particular those that are kernels of surjective groups homomorphisms onto another group.
That is if $G$ is a quotient of a free group on $n$ generators, by $m$ relations, so that $G \cong \left< x_{1}, \dots , x_{n} \mid r_{1}, \dots , r_{m} \right>$, and $\phi : G \rightarrow S$ is a surjective group homomorphism onto a finite group $S$ of size $s$ say. Then for $K = \operatorname{ker}(\phi)$, we see that $K$ is an index $s$ normal subgroup of $G$.
Then, $G$ can be realised as the fundamental group of a finite cell complex with $1$ $0$-cell, (labelled $b$) $n$ $1$-cells (labelled $x_{1}$ through $x_{n}$), and $m$ $2$-cells (attached along $r_{1}$ through $r_{m}$ respectively). This can then be triangulated to realise $G$ as the fundamental group of a finite path-connected simplicial complex $M$ say. Then, since $K$ is normal in $G$, it corresponds uniquely to a regular based covering map $p : (\tilde{M},\tilde{b}) \rightarrow (M, b)$ such that $p_{*}(\pi_{1}(\tilde{M})) = K$ (where $p_{*}$ denotes the group homomorphism induced by $p$).
Then, since $M$ is a finite path-connected simplicial complex, $\tilde{M}$ is too, and additionally $p$ is a simplicial map. Then, we can calculate a finite generating set for $p_{*}(\pi_{1}(\tilde{M}))$ by calculating the maximal tree $T$ in $\tilde{M}$ and then there is a generator for each edge in $\tilde{M}$ that is not in $T$. Then the relations come from the loops transversed in $\tilde{M}$ corresponding to each $r_{i}$ starting at each vertex in $p^{-1}(b)$. Then we know that the degree of $p$ corresponds to the index of $K$ in $G$, so $\tilde{M}$ has $s$ vertices. Then, since $p$ is regular and $M$ has $2n$ edges coming out of its singular vertex, $\tilde{M}$ has $2ns$ edges counting each edge twice, and so $\tilde{M}$ has $ns$ edges. Then $T$ has $s-1$ edges since it is a maximal tree of a graph with $s$ vertices. So we conclude that $K$ has a presentation with exactly $ns - (s-1) = (n-1)s + 1$ generators and $ms$ relations (we have $m$ relations for each point in $p^{-1}(b)$).
The motivation for this questions comes from the desire to be able to verify my solution. Since it is relatively easy to check if all of my generators are in the kernel, I can show containment easily. Then if I calculate indexes and show they are the same, would I be able to conclude that my solution is correct? | 2021-05-15T11:36:00 | {
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http://math.stackexchange.com/questions/110201/how-do-i-eliminate-useless-subsets | # How do I eliminate useless subsets?
BACKGROUND / MOTIVATION:
The other day, I was given a finite covering $\mathcal{B} = \{I_1, \ldots, I_n\}$ of $[0, 1]$ by open intervals, and I wanted to refine it by eliminating useless intervals, that is, those $I_k$ which are contained in the union of the others: $$I_k \subseteq I_1 \cup \dots \cup \widehat{I_k} \cup \dots \cup I_n~~(\text{I_k is ommited.)}$$ Since my covering was finite, it was easy to establish an algorithm for doing this: check if $I_1$ is useless; if it is, remove it from the list and proceed to $I_2$, considering the new list now, and so on.
However, I couldn't help but notice that this process depends on the given order. For example, if $I_1 = [0, 1]$, $I_2 = [0, 3/4)$ and $I_3 = (1/4, 1]$, I would eliminate just $I_1$ and I'd be done. If, on the other hand, $I_3$ and $I_1$ switched places, my algorithm would end up eliminating $I_2$ and $I_3$ instead.
I then started wondering whether there is a more direct, non-algorithmic argument to show that such a refinement always exists. A naive guess would be to define $$\mathcal{B}^{\prime} = \bigg\{ I_k \in \mathcal{B} : I_k \not\subseteq \bigcup_{j \neq k} I_j \bigg\},$$ but this may be empty. In the finite case, one could settle for the given algorithm, but the situation becomes troublesome when dealing with arbitrary coverings. This has lead me to the following question.
QUESTION: Suppose you have a set $X$ and an arbitrary family of subsets $\mathcal{A} = \{A_i\}_{i \in I}$ of $X$ such that their union is $X$. How can you prove that there exists a refinement $\mathcal{A}^{\prime} = \{A_j\}_{j \in J}$, $J \subseteq I$, such that no member of $\mathcal{A}^{\prime}$ is useless (contained in some union of other members of $\mathcal{A}^{\prime}$) but $\mathcal{A}^{\prime}$ still covers $X$?
Thanks.
-
There is an exercise in a homework set from April 2010 in Ron Freiwald's topology course that generalizes Brian M. Scott's counterexample: It says that every noncompact topological space has an open cover with no irreducible subcover. – Jonas Meyer Feb 17 '12 at 1:06
You can’t, because it isn’t always true. Let $X=\mathbb{N}$, and for $n\in\mathbb{N}$ let $A_n=\{0,1,\dots,n\}$. Let $\mathscr{A}=\{A_n:n\in\mathbb{N}\}$. Clearly $\mathscr{A}$ is a cover of $\mathbb{N}$. However, if $I$ is any subset of $\mathbb{N}$ such that $\mathscr{A}'=\{A_i:i\in I\}$ covers $\mathbb{N}$, then $I$ is infinite, and $\{A_i:i\in I\setminus\{\min I\}\}$ is a proper subset of $\mathscr{A}'$ that still covers $\mathbb{N}$. Thus, $\mathscr{A}$ has no irreducible subcover.
Added: It’s a useful theorem in general topology that if the cover $\mathscr{A}$ is point-finite, i.e., if $$\operatorname{st}(x,\mathscr{A})\triangleq\{A\in\mathscr{A}:x\in A\}$$ is finite for every $x\in X$, then it has an irreducible subcover. This can be proved using Zorn’s lemma. Let $\mathfrak{R}=\{\mathscr{R}\subseteq\mathscr{A}:\bigcup\mathscr{R}=X\}$, and consider the partial order $\langle\mathfrak{R},\supseteq\rangle$. It’s easy to see that a maximal element in this partial order is an irreducible subcover of $\mathscr{A}$, so we need only show that every chain in $\langle\mathfrak{R},\supseteq\rangle$ has a $\supseteq$-maximal element. Let $\mathfrak{C}$ be a chain in $\langle\mathfrak{R},\supseteq\rangle$, and let $\mathscr{C}=\bigcap\mathfrak{C}$. Suppose that $\mathscr{C}$ does not cover $X$; then there is some $x\in X\setminus\bigcup\mathscr{C}$. Suppose that $x\in A\in\mathscr{A}$; then $A\notin\mathscr{C}$, so there is some $\mathscr{C}_A\in\mathfrak{C}$ such that $A\notin\mathscr{C}_A$. But $\operatorname{st}(x,\mathscr{A})$ is finite, so we can enumerate it as $\{A_1,\dots,A_n\}$ for some $n$, and without loss of generality $$\mathscr{C}_{A_1}\supseteq\dots\supseteq\mathscr{C}_{A_n}\;.$$ But then $\mathscr{C}_{A_n}\cap\operatorname{st}(x,\mathscr{A})=\varnothing$, so $x\notin\bigcup\mathscr{C}_{A_n}$, contradicting the assumption that $\mathscr{C}_{A_n}\in\mathfrak{R}$. Thus, $\mathscr{C}$ does cover $X$, every chain in $\langle\mathfrak{R},\supseteq\rangle$ has a $\supseteq$-maximal element, and Zorn’s lemma applies to yield an irreducible subcover of $\mathscr{A}$. | 2016-02-08T17:46:49 | {
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https://yamazo.work/tdzz/20663 | # Rectangular equation to polar equation
Formula Used: R = sqrt(x * x + y * y) , angle=atan(y/x) Where, Rectangle coordinates: x and y - horizontal and vertical distances from the origin. Polar coordinates(r,q): r - the distance from the origin to the point. Q - the angle measured from the positive x axis to the point. T - angle (in degrees). Related Calculator Precalculus : Convert Rectangular Equations To Polar Form Study concepts, example questions & explanations for Precalculus. CREATE AN ACCOUNT Create Tests & Flashcards. Home Embed All Precalculus Resources . 12 Diagnostic Tests 380 Practice Tests Question of the Day Flashcards Learn by Concept. Example Questions The equation of the circle can be transformed into rectangular coordinates using the coordinate transformation formulas in Equation 7.8. Example 7.14 gives some more examples of functions for transforming from polar to rectangular coordinates. Together, the four equations for r, x, and y allow you to change (x, y) coordinates into polar . Coordinates and back again anytime. For example, to change the polar coordinate . To a rectangular coordinate, follow these steps: Find the x value. Use the unit circle to get . Which means that . … Write each rectangular equation in polar form. Y=-3. 🎉 The Study-to-Win Winning Ticket number has been announced! Go to your Tickets dashboard to see if you won! 🎉
푟 = ඥ푥 ଶ + 푦 ଶ ⎯⎯⎯⎯⎯⎯⎯ θ = 푡푎푛 ିଵ ൫ ೣ ⎯⎯ ൯ 푥 = 푟푐표푠휃 푦 = 푟푠푖푛휃 Example #1: Convert 푥 ଶ + 푦 ଶ = 4 to polar form. Because 푟 = ඥ푥 ଶ + 푦 ଶ ⎯⎯⎯⎯⎯⎯⎯ we know that 푟 ଶ = 푥 ଶ + 푦 ଶ .
Question: Find the polar equation(s) to the rectangular equation {eq}x^2 + y^2 = 10x +3y. {/eq} Rectangular and Polar Form: The given equation is the general equation of a circle, so let's compare So I've got three rectangular equations. I want to convert them to polar form, to see what they look like in polar. Now these are pretty simple ones like y equals 5 is a horizontal line. Remember how we convert back and forth between polar and rectangular. We use these conversion equations. For example, for y equals 5, I could use this equation
To convert from rectangular coordinates to polar coordinates, use one or more of the formulas: cos θ = x r, sin θ = y r, tan θ = y x. \displaystyle \cos \theta =\frac … Solution for Convert the rectangular equation (x - 1)2 + y2 = 1 to a polar equation that expresses r in terms of θ. Square both sides and substitute r 2 = x 2 + y 2, y = r sin. . θ (hence y 2 = r 2 sin 2. . θ ), x 2 = a 2 θ 2 − r 2 sin 2. . θ (hence x 2 = a 2 θ 2 − a 2 θ 2 sin 2. . θ) we get. $\begingroup$ You may multiply your polar equation through by $\ r \$ to get $\ r^2 = r - r \sin \theta \ ,$ and then write $\ x^2 + y^2 = r - y \ .$ Then solve for $\ r \$ in this equation, square both sides again, and replace the resulting $\ r^2 \$ with $\ x^2 + y^2 \$ and … Two Examples: Change from Rectangular to Polar Coordinates and Sketch; Three Examples: Change from Polar Coordinates to Cartesian Coordinates; Examples #1-6: Express each Equation in Polar Form; Examples #7-10: Express each Equation in Rectangular Form; Graphing Polar Equations. 1 hr 14 min 15 Examples. Introduction to Video: Graphing Polar Convert the polar equation to rectangular form. 1.) r sec(theta) = 3 2.) r = 4 cos(theta) - 4 sin(theta) convert from rectangular equation to polar form. 1.) x^2 + (y-1)^2 = 1 2.) (x-1)^2 + (y+4)^2 = 17 For polar to rectangular you need to do these things. I = iota. 5 ∗ a n g ( 40) = 5 ∗ c o s ( 40) + 5 ∗ i. S i n ( 40) You will have to do like this. In case of terms with imaginary terms. ( a + i b) ∗ ( c + i d) = ( a c − b d) + i ( a d + b c) You. Continue Reading. Virtual calculator has some cons.
If you have the polar equation $r=4,$ it means that regardless of the angle, the distance from the origin is always 4. It doesn't matter if the angle is $\pi/2, 33\pi/37,$ or $0$, it will always be 4. This means th... To change a rectangular equation to a polar equation just replace x with r cos θ and y with r sin θ. Use completing the square to obtain standard circle equation Therefore polar equation is converted to rectangular form and is the graph of a circle of radius 1 centered at (0,-1). ) convert the rectangular equation to polar form: x^2+y^2=a^216. Convert Differential Equation Into Polar Coordinates. Convert Differential Equation Into Polar Coordinates The graph of this polar equation is a circle. ESolutions Manual - Powered by Cognero The graph has a rectangular equation y = −4 and a Page 24 polar equation r = −4 csc . 9-3 Polar and Rectangular Forms of Equations The graph has a rectangular equation y = − a polar equation = . X and Write rectangular and polar equations for each graph. 63. Polar equations give us a different mathematical perspective on graphing. In rectangular coordinates, we use two axes which meet at the origin and are perpendicular to one another. In polar coordinates, we start with a fixed point, O, called the pole or origin and then we construct an initial ray called the polar axis. View 11.3 Converting Equations to Rectangular.Pdf from 8TH GRADE 101 at Acellus Academy. 11.3 Converting Equations: Polar to Rectangular Date: April 2, 2020 Essential Question: How do you convert a Firstly, remember the rules for converting from rectangular to polar: x = rcos (theta) y = rsin (theta) r = sq. Rt. (x^2 + y^2) theta = tan^-1 (y/x) To convert a rectangular equation into polar form, remove the numerators. Then, start changing rectangular values into polar form as per the rules above. Keep solving until you isolate the variable r. Parametric Equations and Polar Coordinates. Eliminate the Parameter, Set up the parametric equation for to solve the equation for . Rewrite the equation as . Subtract from both sides of the equation. Replace in the equation for to get the equation in terms of . Simplify . Tap for more steps... R = 6 sec θ Simplify. Example 2: Convert the following polar equation to rectangular equations. A) r = 5 b) θ = π / 6. Step 1: Square both sides of r = 5 and substitute for r2. R 2 = x 2 + y 2. R = 5. R 2 = 5 2 = 25 Square. X 2 + y 2 = 25 Sub. Step 2: Determine the value of t a n θ and equate this to y x . Textbook solution for Precalculus: Mathematics for Calculus (Standalone… 7th Edition James Stewart Chapter 8.1 Problem 45E. We have step-by-step solutions for your textbooks written by Bartleby experts! | 2020-11-30T01:03:29 | {
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https://math.stackexchange.com/questions/2948005/multiplying-by-10s-place-voodoo-why-is-30-times-50-15-times-100 | # Multiplying by $10$'s place voodoo: Why is $30\times 50= 15\times 100$?
This is a very trivial question, but I can't seem to reason out why $$30\times 50= 15\times 100$$
As a kid, I never really thought about why it works, but now I can't figure it out and the idea is really troubling me. I understand that we can break up the problem like this: $$3\times 10\times 5\times 10$$ but at this point I feel like I've lost the intuitive aspect of the problem. Can someone plz help and provide some intuition?
• +1 for the title – clathratus Oct 9 '18 at 3:58
• Also, is it ok to ask such questions? I sometimes feel like I am going insane (or being stupid) when I think of these problems, but I feel like I accepted these facts as a kid rather than pondering them too deeply. Ever since I took Calculus, I can't seem to accept any fact without a formal proof. – Dude156 Oct 9 '18 at 4:10
$$\begin{array}{|c|c|c|c|c|} \hline 1 & 2 & 3 & 4 & 5 \\ \hline 6 & 7 & 8 & 9 & 10 \\ \hline11 & 12 & 13 & 14 & 15 \\ \hline \end{array}$$
You own this piece of land, you have $$15$$ squares in total, the size of the square is $$10$$ m by $$10$$m. Each square is $$100m^2$$. What is the total area? $$15 \times 100$$
One dimension of the land is $$3\times 10$$. The other dimension if $$5 \times 10$$.
If you have land of $$a \times c$$ number of rectangles land of size $$b \times d$$ each, size of each rectangle is $$b \times d$$. Total area would be $$(a \times c)(b \times d)$$.
One dimension of your land is $$a \times b$$ and the other dimension would be $$c \times d.$$ Hence total area is $$(a \times b)(c \times d)$$.
• Now I say, Eureka! – Dude156 Oct 9 '18 at 3:51
• Also, is it ok to ask such questions? I sometimes feel like I am going insane (or being stupid) when I think of these problems, but I feel like I accepted these facts as a kid rather than pondering them too deeply. Ever since I took Calculus, I can't seem to accept any fact without a formal proof. – Dude156 Oct 9 '18 at 4:16
• The question is fine. mine is not a formal proof though. I was just trying to give an intuition. to come up with a proof, we need to know which set are we working on and what is a multiplication. Check the commutative and associative property. – Siong Thye Goh Oct 9 '18 at 4:27
• @Dude156 I think that it is a sign of you thinking about things more deeply. Mathematicians aren't supposed to assume anything, and there's nothing wrong with questioning the basics that we've always just assumed. – clathratus Oct 9 '18 at 18:01
$$30\cdot50=(3\cdot10)(5\cdot10)=(3\cdot5)(10\cdot10)=15\cdot100$$ It's all just the property that $$(ab)(cd)=(ac)(bd)$$
• Can you please provide a proof (or a link to the proof)? – Dude156 Oct 9 '18 at 3:57
• $$(ab)(cd)=abcd=acbd=(ac)(bd)$$ I think it's just like reliant on an extension of the associative property for multiplication – clathratus Oct 9 '18 at 3:58
If you're looking for an intuitive explanation, maybe consider it like a change of units. $$30\cdot50$$ is, say, the area of a rectangle 30 millimeters by 50 millimeters in square millimeters, but you can switch back and forth between units. So Instead you think of it as a rectangle 3 centimeters by 5 centimeters, and so the area is just 15 square centimeters, which, converting back to millimeters, is $$15 \text{cm}^2 \cdot \left(\frac{10\text{mm}}{1 \text{cm}}\right)^2=1500\text{mm}$$
Intuitively why your particular problem works is because you are doubling one of the numbers and halving the other with no net change in the product. This always works for any two numbers, say $$8$$ and $$13$$, whereby $$8\times 13 = 4\times 26 = 104$$ | 2019-05-22T19:47:08 | {
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http://math.stackexchange.com/questions/267544/probability-of-a-12-symbol-string-containing-3-even-digits | Probability of a 12 symbol string containing 3 even digits
From Carol Ash's, "Probability Tutoring Book", pg. 19, prob. 1-3.4.
If a 12 symbol string is formed from the 10 digits and 26 letters, repetition not allowed, what is the prob that it contains 3 even digits?
I assumed string implies order matters, so I counted of $P(36,12)=\frac{36!}{(36-12)!}$ total possible strings.
Then for the numerator I counted:
1. Pick 3 spots for the 3 even digits
2. Fill the first, second and third even digit slots
3. From the remaining 31 non-even number characters pick something for each of the remaining 9 slots.
So I got $\frac{\binom{12}{3}\times P(5,3)\times P(31, 9)}{P(36,12)}$ where $P(n,k)=\frac{n!}{(n-k)!}$.
However Ash says:
For the total, pick a committee of 12 symbols from the 36. For the fav, pick a subcommittee of 3 evens and a subcommittee of 9 others. $$\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}$$
My question is:
1. Did I misunderstand the question and order doesn't matter?
2. Order does matter and I'm just wrong
3. These actually work out the same ? ( if so why? )
-
They are the same but IMHO for such a solution the text should be "If a 12 symbol string is formed from the 10 digits and 26 letters, repetition not allowed, what is the prob that it contains exactly 3 even digits?". Otherwise I would assume an 'at least' is implied.Anyway I am not a native English speaker and the book surely knows better than me :) – AndreasT Dec 30 '12 at 10:20
@AndreasT It is somewhat ambiguous. But it's just a matter of getting used to the particular author's writing style. As long as they're consistent it's not too much of a problem. – Robert S. Barnes Dec 30 '12 at 10:43
They’re the same. Since $$P(n,k)=\binom{n}kk!\;,$$ we have
\begin{align*} \frac{\binom{12}{3}P(5,3)P(31, 9)}{P(36,12)}&=\frac{\binom{12}3\binom533!\binom{31}99!}{\binom{36}{12}12!}\\\\ &=\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}\cdot\frac{\binom{12}33!9!}{12!}\\\\ &=\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}\cdot\frac{12!}{12!}\\\\ &=\frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}\;. \end{align*}
You’re quite right that order matters, but its effect on numerator and denominator is identical, so in the end it has no effect. Look at it this way: Ash’s computation has the number of acceptable sets of symbols in the numerator and the number of possible sets of symbols in the denominator. Each set of symbols can be permuted in $12!$ ways, so there are $$\binom53\binom{31}912!$$ acceptable strings of symbols and $$\binom{36}{12}12!$$ possible strings of symbols. That’s the ratio that you calculated. The numbers of acceptable and possible objects have changed, but the ratio itself has not.
-
Here is one way of looking at it:
The total number of strings is $\frac{36!}{24!}$.
First pick 3 of 5 even digits. Order doesn't matter yet, so there are $\binom{5}{3}$ ways. Pick the remaining 9 digits from the remaining 31 non-even digits, order still doesn't matter, so there are $\binom{31}{9}$ ways.
Now place the 3 even digits. There are $(12)(11)(10) = \frac{12!}{9!}$ ways of doing this, and there are $9!$ ways of placing the remaining 9.
So the probability is $\frac{\binom{5}{3}\binom{31}{9}\frac{12!}{9!} 9!}{\frac{36!}{24!}} = \frac{\binom{5}{3}\binom{31}{9}}{\frac{36!}{12!24!}}= \frac{\binom{5}{3}\binom{31}{9}}{\binom{36}{12}}$.
One way of looking at it is that when you select, order does not matter, but when you finally place an object, order matters.
In your Step 1, you are selecting, so order does not matter. In Step 2, you are placing these objects, so order matters. In Step 3, you are combining selecting and placing in one step (select 9 out of 31 objects and place them into 9 slots). The combining cancels relevant quantities (and possibly confuses?).
I separated the selection from the placing.
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I really like the separation of selecting from placing. It makes the overall process much clearer. – Robert S. Barnes Dec 30 '12 at 10:39 | 2016-06-29T16:51:41 | {
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http://mathhelpforum.com/algebra/3037-sa-volume-half-sphere-print.html | # SA and volume of half a sphere?
• May 20th 2006, 12:46 AM
xanith_47
SA and volume of half a sphere?
Hello,
I need help with volume and surface areas of half a sphere (hemispheres)
I know that v= 4/3x pi x r cubed
SA= 4x pi x r squared
However, when I do this for half a sphere, i time the answer by half and when i refer to the answers at the back of my textbook, its different. Im not sure if the sphere is open or closed, since the book doesnt state that, and are there different formulas for finding out SA and V of half open and closed spheres?
Any help is greatly appreciated.
• May 20th 2006, 02:56 AM
ticbol
Quote:
Originally Posted by xanith_47
Hello,
I need help with volume and surface areas of half a sphere (hemispheres)
I know that v= 4/3x pi x r cubed
SA= 4x pi x r squared
However, when I do this for half a sphere, i time the answer by half and when i refer to the answers at the back of my textbook, its different. Im not sure if the sphere is open or closed, since the book doesnt state that, and are there different formulas for finding out SA and V of half open and closed spheres?
Any help is greatly appreciated.
You mentioned those for a sphere, so half of those should be for a hemisphere.
If your answers and those at the back of the book are not the same, it could be that you misunderstood the problems so you could have used wrong formulas.
Can you post a question, with your answer and that answer at the back of the book? Let us investigate.
• May 20th 2006, 05:02 AM
xanith_47
Calculate the surface area of the following sphere to the nearest m squared:
A hemisphere with diameter of 17m
The textbook doesnt mention whether its open or closed..if that makes a difference..
The answer at the back is 681m squared.... and i got 453.96m squard= 454m squard. I used 2 x pi x 8.5 squard = formula for the SA of half a sphere.
Can you explain to me how this works?
• May 20th 2006, 05:32 AM
earboth
Quote:
Originally Posted by xanith_47
Calculate the surface area of the following sphere to the nearest m squared:
A hemisphere with diameter of 17m
The textbook doesnt mention whether its open or closed..if that makes a difference..
The answer at the back is 681m squared.... and i got 453.96m squard= 454m squard. I used 2 x pi x 8.5 squard = formula for the SA of half a sphere.
Can you explain to me how this works?
Hello,
your result is OK - and the result in the book is OK too.
The book defines the surface of a hemisphere as the half of a complete sphere plus the area of the covering circle .
So you get: $2*\pi*(8.5m)^2 + \pi*(8.5m)^2 = 3*pi*(8.5)^2\approx 680.94 m^2$
Greetings
EB
• May 20th 2006, 06:09 AM
xanith_47
Umm.. so would the "the area of the covering circle" be the area INSIDE the circle? How come then, when the hemisphere is closed, we dont count that area which closes the circle, but we count the area inside it when its open. So your saying that the extra area inside the circle= pi x r squard= area of a normal circle?
• May 20th 2006, 06:48 AM
earboth
Quote:
Originally Posted by xanith_47
Umm.. so would the "the area of the covering circle" be the area INSIDE the circle? How come then, when the hemisphere is closed, we dont count that area which closes the circle, but we count the area inside it when its open. So your saying that the extra area inside the circle= pi x r squard= area of a normal circle?
Hello,
I've attached a diagram to show you where you find this additional area.
Greetings
EB
• May 20th 2006, 03:22 PM
xanith_47
Ok thanks for the help.. so if there was no covering circle (hence its open), then we dont calculate the area inside the circle..just the exterior area? Thanks
• May 20th 2006, 03:50 PM
ticbol
Quote:
Originally Posted by xanith_47
Ok thanks for the help.. so if there was no covering circle (hence its open), then we dont calculate the area inside the circle..just the exterior area? Thanks
My half-cent.....
Supposed to be, a sphere is solid. It is not hollow--unless it is specified to be hollow.
Like a cube. You won't think of a cube as hollow if all is said is it is a cube.
So the hemisphere in the problem is supposed to be solid. Nothing was mentioned that it was hollow or solid.
Hence its surface area is the sum of the areas of the curved surface of the dome and of the flat surface of the base.
The SA of the dome is (1/2)[4pi(r^2)]
The base is a great circle of the sphere where the hemisphere came from (a great circle of a sphere has the center of the sphere as its center also). Its SA is pi(r^2). | 2017-10-22T19:49:12 | {
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http://www.maplesoft.com/support/help/Maple/view.aspx?path=Student/NumericalAnalysis/TaylorPolynomial | Student[NumericalAnalysis] - Maple Programming Help
Home : Support : Online Help : Education : Student Package : Numerical Analysis : Computation : Student/NumericalAnalysis/TaylorPolynomial
Student[NumericalAnalysis]
TaylorPolynomial
compute a Taylor polynomial approximation
Calling Sequence TaylorPolynomial(f, x, opts) TaylorPolynomial(f, x = x0, opts)
Parameters
f - algebraic; expression in the variable x x - name; the name of the independent variable in f x0 - realcons; the point about which the series is expanded opts - (optional) equation(s) of the form keyword = value where keyword is one of digits, errorboundvar, order; the options for computing the Taylor polynomial
Options
• digits = posint
A positive integer; the environment variable Digits will be set to this integer during the execution of this procedure. By default, digits = 10.
• errorboundvar = name, algebraic
The name to assign to the independent variable in the error term. An error term for the taylor approximation is returned if the errorboundvar option is specified. Additionally, if the errorboundvar option and the extrapolate option are specified, the error bound is given for the extrapolated point as well.
• extrapolate = realcons, list(realcons)
A point or list of points to be extrapolated. If this option is specified, the approximate value(s) of the taylor polynomial at the specified point(s), the exact value(s) of f at the specified point(s), and the error bound(s) are returned after the taylor polynomial.
• order = posint, posint..posint, list(posint)
The order of the Taylor polynomial. If the order is specified as a range or list, the Taylor polynomial for each order in the range or list will be computed. By default, order = 6.
Description
• The TaylorPolynomial command computes the taylor series expansion of f about x0 and converts it into a polynomial.
• If x0 is not specified, then the Taylor expansion is calculated about the point x=0.
• The possible return values of the TaylorPolynomial command are:
– P
P is the n-th order Taylor polynomial of f calculated about x = x0
– [P, R]
P is the n-th order Taylor polynomial of f calculated about x = x0 and R is the remainder term, such that P+R=f. This output is returned when the errorboundvar option is specified and the extrapolate option is not specified.
– [P, R, B]
P is the n-th order Taylor polynomial of f calculated about x = x0, R is the remainder term, such that P+R=f and B is a list containing the extrapolated point(s), the value(s) of P at the extrapolated point(s), the value(s) of f at the extrapolated point(s) and the associated error bound(s). This output is returned when both the errorboundvar and the extrapolate options are specified.
– If order is specified as a list or range the return will be an expression sequence of the output described above, with an expression sequence member for each order.
Examples
> $\mathrm{with}\left(\mathrm{Student}[\mathrm{NumericalAnalysis}]\right):$
> $\mathrm{TaylorPolynomial}\left(\mathrm{sin}\left(x\right),x\right)$
${x}{-}\frac{{1}}{{6}}{}{{x}}^{{3}}{+}\frac{{1}}{{120}}{}{{x}}^{{5}}$ (1)
> $\mathrm{TaylorPolynomial}\left(\mathrm{sin}\left(x\right),x,\mathrm{errorboundvar}='\mathrm{ξ}'\right)$
$\left[{x}{-}\frac{{1}}{{6}}{}{{x}}^{{3}}{+}\frac{{1}}{{120}}{}{{x}}^{{5}}{,}{-}\frac{{1}}{{5040}}{}{\mathrm{cos}}{}\left({\mathrm{ξ}}\right){}{{x}}^{{7}}\right]$ (2)
> $\mathrm{TaylorPolynomial}\left(\mathrm{sin}\left(x\right),x,\mathrm{errorboundvar}='\mathrm{ξ}',\mathrm{extrapolate}=1.3\right)$
$\left[{x}{-}\frac{{1}}{{6}}{}{{x}}^{{3}}{+}\frac{{1}}{{120}}{}{{x}}^{{5}}{,}{-}\frac{{1}}{{5040}}{}{\mathrm{cos}}{}\left({\mathrm{ξ}}\right){}{{x}}^{{7}}{,}\left[{1.3}{,}{0.9647744166}{,}{0.9635581854}{,}{0.001245010258}\right]\right]$ (3)
> $\mathrm{TaylorPolynomial}\left({ⅇ}^{x},x,\mathrm{order}=3\right)$
${1}{+}{x}{+}\frac{{1}}{{2}}{}{{x}}^{{2}}{+}\frac{{1}}{{6}}{}{{x}}^{{3}}$ (4)
> $\mathrm{TaylorPolynomial}\left(\mathrm{ln}\left(x\right),x=2,\mathrm{order}=3,\mathrm{errorboundvar}='\mathrm{ξ}'\right)$
$\left[{\mathrm{ln}}{}\left({2}\right){+}\frac{{1}}{{2}}{}{x}{-}{1}{-}\frac{{1}}{{8}}{}{\left({x}{-}{2}\right)}^{{2}}{+}\frac{{1}}{{24}}{}{\left({x}{-}{2}\right)}^{{3}}{,}{-}\frac{{1}}{{4}}{}\frac{{\left({x}{-}{2}\right)}^{{4}}}{{{\mathrm{ξ}}}^{{4}}}\right]$ (5) | 2017-07-27T02:41:43 | {
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http://mymathforum.com/algebra/49268-optimal-number-shares-print.html | My Math Forum (http://mymathforum.com/math-forums.php)
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- - optimal number of shares (http://mymathforum.com/algebra/49268-optimal-number-shares.html)
get42n8 January 3rd, 2015 11:23 AM
optimal number of shares
here is a problem:
I have \$2000 I want to buy 2 different shares YYY (share price 109) and ZZZ (share price 14) what is the optimal number of YYY and ZZZ shares to buy to spend the most of \$2000 and have the remainder of that as close to zero as possible?
109y+14z=2000
what would be another equation in my Systems of equations? I cannot think of it.
Timios January 3rd, 2015 12:10 PM
I am not sure that there is another equation.
My thinking is that one should maximize the 14 dollar shares. To do that, I would buy just one 109 dollar share. This would leave 1891 to spend on the 14 dollar shares.
1891÷14=135.0717...
If you buy 135 of these shares you will spend 1890 dollars.
This amount when added to the 109 dollars sums to 1999 dollars.
That's pretty darn close. But is there's an algebraic way to do it, I would like to become aware of it.
And if there is another combination that comes closer to 2000 dollars, I would be surprised.
mathman January 3rd, 2015 12:11 PM
The only constraint you have is that x and y be non-negative integers and the left side cannot exceed the right side. By brute force try y between 0 and 18 to see what gets you the closest.
soroban January 3rd, 2015 05:08 PM
Hello, get42n8!
Quote:
I have \$2000. I want to buy 2 different shares YYY (\$109 each) and ZZZ (\$14 each). What is the optimal number of YYY and ZZZ shares to buy to spend the most of \$2000 and have the remainder of that as close to zero as possible? $109y+14z\:=\:2000$
There is no second equation.
This is a Diophantine equation, requiring integer solutions.
This can be solved by Modular Arithmetic,
but here is an algebraic solution.
Solve for $z\!:\;\; z \;=\;\dfrac{2000-109y}{14} \quad\Rightarrow\quad z\;=\;142 - 8y + \dfrac{12+3y}{14}$
Since $z$ is an integer, $3(4+y)$ must be a multiple of 14.
The first time this happens is: $\:y = 10$
Hence: $\:z \:= \: \dfrac{2000-109(10)}{14} \:=\:\dfrac{910}{14} \:=\:65$
Therefore: $\:\boxed{y \,=\,10,\;z\,=\,65}$
Check: $\:109(10) + 14(65) \:=\:2000$
Timios January 3rd, 2015 08:35 PM
Good job, soroban!
$z \;=\;\dfrac{2000-109y}{14} \quad\Rightarrow\quad z\;=\;142 - 8y + \dfrac{12+3y}{14}$
buiduchuy January 3rd, 2015 11:50 PM
Quote:
Good job, soroban! Please help me to understand why $$z=\frac{2000−109y}{14}⇒z=142−8y+\frac{12+3 y}{14}$$
Because $$z = \frac{2000-109y}{14} = \frac{2000}{14} - \frac{109y}{14} \\ =142 + \frac{12}{14} - (8y - \frac{3y}{14}) \\ = 142 - 8y + \frac{12+3y}{14}$$
talip January 5th, 2015 11:21 AM
This problem can be modeled as integer programming model as follows:
Min x
Subject to
109y+14z+x=2000
x,y,z >=0 and integer
In operations research area, there are some methods which can be ussed to solve these models. Also some softwares like Lindo solver can also be used.
All times are GMT -8. The time now is 03:19 AM. | 2018-04-26T11:19:18 | {
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https://plainmath.net/1639/parametric-information-explaining-explanation-parametric-counterclockwise | Ask question
# That parametric equations contain more information than just the shape of the curve. Write a short paragraph explaining this statement. Use the following example and your answers to parts (a) and (b) below in your explanation. The position of a particle is given by the parametric equations x = sin (t) and y = cos (t) where t represents time. We know that the shape of the path of the particle is a circle. a) How long does it take the particle to go once around the circle? Find parametric equations if the particle moves twice as fast around the circle. b) Does the particle travel clockwise or counterclockwise around the circle? Find parametric equations if the particle moves in the opposite direction around the circle.
Question
That parametric equations contain more information than just the shape of the curve. Write a short paragraph explaining this statement. Use the following example and your answers to parts (a) and (b) below in your explanation. The position of a particle is given by the parametric equations $$x =\ \sin\ (t)\ and\ y =\ \cos\ (t)$$ where t represents time. We know that the shape of the path of the particle is a circle. a) How long does it take the particle to go once around the circle? Find parametric equations if the particle moves twice as fast around the circle. b) Does the particle travel clockwise or counterclockwise around the circle? Find parametric equations if the particle moves in the opposite direction around the circle.
## Answers (1)
2020-10-28
Step 1 (a) Note that, the position of the particle is given by the parametric equations $$x =\ \sin t,\ and\ y =\ \cos\ t$$. The parametric equations contain more than just shape of the curve. They also represent the direction of curve as traveling. If a position of a particle is determined by the equation $$x =\ \sin\ t,\ y =\ \cos\ t,$$ this set of equations denotes which direction the particle is traveling based on different times t. For example, at $$t = 0,\ \text{the particle is at the point}\ (0,\ 1)\ \text{but at time}\ t=\ \frac{\pi}{2}\ \text{the particle has moved to the point}\ (1,\ 0)$$ in a clockwise direction As the period of the parametric equations is $$2\ \pi$$, to find for the particle to travel a full rotation around the circle. It will take the time $$t = 2\ \pi$$ to traverse the circle in a clockwise direction. To travel the circle twice as fast simply double the coefficient inside each trigonometric function and the parametric equations are $$x =\ \sin\ 2t,\ y =\ \cos\ 2t.$$ Thus, the time that will be taken by the particle to go once around the circle is $$t = 2\ \pi\ \text{and the parametric equations, the particle moves twice as fast around the circle are}\ x =\ \sin\ 2t,\ y =\ \cos\ 2t.$$ Step 2 (b) Note that, the particle travels clockwise. For example, at $$t = 0,\ \text{the particle is at the point}\ (0,\ 1),\ \text{but at the time}\ t =\ \frac{\pi}{2}\ \text{the particle has moved to the point}\ (1,\ 0)$$ in a clockwise direction. The parametric equations when the particle travels in the opposite direction, the parametric equations will be exchanged. That are, $$x =\ \cos\ t,\ y =\ \sin\ t.$$ Thus, the particle travels clockwise and if the particle travels in opposite direction around the circle, the parametric equations are $$x =\ \cos\ t,\ y =\ \sin\ t.$$
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... | 2021-06-22T14:40:20 | {
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https://stats.stackexchange.com/questions/359768/probability-of-a-difference-between-two-sampling-means-of-two-populations | # probability of a difference between two sampling means of two populations
If I have two normally distributed populations $\mu_1 = 7, \sigma_1 = 0.5$ and $\mu_2 = 6.6, \sigma_2 = 0.5$ and I sample each of the populations say with 10 samples each. If I want to work out the probability that $(\overline X_1 - \overline X_2) > 0.6$, would I use the following formula?
$$Z = \frac{(\overline X_1 - \overline X_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} = \frac{0.6-0.4}{\sqrt{\frac{0.5^2}{10}+\frac{0.5^2}{10}}}= \frac{0.2}{0.2236}=0.8945$$
Therefore, $$P((\overline X_1 - \overline X_2)>0.6)=P(Z>0.8945) = 1 - P(Z<0.8945) = 0.1855$$ That's my theoretical answer. However, I believe the distribution of (\overline X_1 - \overline X_2) should be normal with $\mu_{\overline X_1 - \overline X_2} = \mu_1-\mu_2 = 0.4$ and the standard deviation of this distribution should be: $$\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}=\sqrt{\frac{0.5^2}{10}+\frac{0.5^2}{10}} = 0.2236$$
My question is, that if I run the code to build the sampling distributions of the samples, and replicate the sampling for each population 1000 times, and then run histograms, as below. Can I establish this probability from the histograms instead of the theoretical way above? i.e. the probability that the difference between the sampling means $\overline X_1 - \overline X_2 > 0.6$
simulate.diff.norm = function(n, R, mu, sigma){
x_bar = rep(NA, R)
for(i in 1:R){
data = rnorm(n, mean=mu, sd=sigma)
x_bar[i] = mean(data)
}
return(x_bar)
}
SampleDist1 = simulate.diff.norm(n=9, R=10000, mu=7, sigma=0.5
SampleDist2 = simulate.diff.norm(n=9, R=10000, mu=6.6, sigma=0.5
hist(SampleDist1, main = paste("Histogram of Pop A Sampling Mean"))
plot.new()
hist(SampleDist2, main = paste("Histogram of Pop B Sampling Mean"))
• The very first formula I think holds if both the samples are drawn independently. – StubbornAtom Jul 30 '18 at 13:35
• You cannot "establish" any theoretical result from a simulation: all you can do is verify that the simulation is consistent with some theoretical assertion. – whuber Jul 30 '18 at 14:07
I ended up changing that code to simulate 10000 experiments of the two sample distributions together and built another distribution consisting of the difference between the means of each of the nine-sample experiments, for each of the 1000 loops. This resulted in three vectors to feed into the histogram, the sampling distribution for population A, the sampling distribution for population B, and the sampling distribution for the difference between the means of each ith sample of the populations i:10000
Then I could compare the theoretical result with the result of the simulation
simulate.diff.norm = function(n, R, mu1, mu2, sigma1, sigma2){
SchoolA_x_bar = rep(NA, R)
SchoolB_x_bar = rep(NA, R)
X_bar1_minus_X_bar2 = rep(NA, R)
for(i in 1:R){
SchoolA = rnorm(n, mean=mu1, sd=sigma1)
SchoolB = rnorm(n, mean=mu2, sd=sigma2)
SchoolA_x_bar[i] = mean(SchoolA)
SchoolB_x_bar[i] = mean(SchoolB)
X_bar1_minus_X_bar2[i] = SchoolA_x_bar[i] - SchoolB_x_bar[i]
}
return(list(SA=SchoolA_x_bar, SB=SchoolB_x_bar, X1X2=X_bar1_minus_X_bar2))
}
# Theoretical model
mydata = rnorm(10000, mean=0.4, sd=0.2236)
theoretical.diff = sum(mydata>0.6)/10000
theoretical.diff
#start the simulation
Simulated_SA_Minus_SB_Sampling_Mean = simulate.diff.norm(n=10, R=10000, mu1=7, mu2=6.6, sigma1=0.5, sigma2=0.5)
hist(mydata, main = bquote(paste("Theoretical ",E(bar(XA)-bar(XB)))), xlab="Mean difference between sample School A and sample School B")
## now plot the histograms of the simulation
plot.new()
hist(Simulated_SA_Minus_SB_Sampling_Mean[["SA"]], main = "Simulated Sampling Mean for School A")
hist(Simulated_SA_Minus_SB_Sampling_Mean[["SB"]], main = "Simulated Sampling Mean for School B")
hist(Simulated_SA_Minus_SB_Sampling_Mean[["X1X2"]], main = bquote(paste("Simulated Mean Difference ",E(bar(XA)-bar(XB)))))
simulated.diff = sum(Simulated_SA_Minus_SB_Sampling_Mean[["X1X2"]]>0.6)/10000
simulated.diff
with the following output:
> # using a sample size of 9 run 10000 simulations of sampling of each of the two populations.
> # find the mean of each sample and find the difference between these two means for each i:10000 sampling experiments
> simulate.diff.norm = function(n, R, mu1, mu2, sigma1, sigma2){
+ SchoolA_x_bar = rep(NA, R)
+ SchoolB_x_bar = rep(NA, R)
+ X_bar1_minus_X_bar2 = rep(NA, R)
+ for(i in 1:R){
+ SchoolA = rnorm(n, mean=mu1, sd=sigma1)
+ SchoolB = rnorm(n, mean=mu2, sd=sigma2)
+ SchoolA_x_bar[i] = mean(SchoolA)
+ SchoolB_x_bar[i] = mean(SchoolB)
+ X_bar1_minus_X_bar2[i] = SchoolA_x_bar[i] - SchoolB_x_bar[i]
+ }
+ return(list(SA=SchoolA_x_bar, SB=SchoolB_x_bar, X1X2=X_bar1_minus_X_bar2))
+ }
>
> # Theoretical model
> mydata = rnorm(10000, mean=0.4, sd=0.2236)
> theoretical.diff = sum(mydata>0.6)/10000
> theoretical.diff
[1] 0.1866
> #start the simulation
> Simulated_SA_Minus_SB_Sampling_Mean = simulate.diff.norm(n=10, R=10000, mu1=7, mu2=6.6, sigma1=0.5, sigma2=0.5)
> hist(mydata, main = bquote(paste("Theoretical ",E(bar(XA)-bar(XB)))), xlab="Mean difference between sample School A and sample School B")
> ## now plot the histograms of the simulation
> plot.new()
> hist(Simulated_SA_Minus_SB_Sampling_Mean[["SA"]], main = "Simulated Sampling Mean for School A")
> hist(Simulated_SA_Minus_SB_Sampling_Mean[["SB"]], main = "Simulated Sampling Mean for School B")
> hist(Simulated_SA_Minus_SB_Sampling_Mean[["X1X2"]], main = bquote(paste("Simulated Mean Difference ",E(bar(XA)-bar(XB)))))
> simulated.diff = sum(Simulated_SA_Minus_SB_Sampling_Mean[["X1X2"]]>0.6)/10000
> simulated.diff
[1] 0.1882
Very similar results throughout. Thanks. | 2019-06-17T23:38:16 | {
"domain": "stackexchange.com",
"url": "https://stats.stackexchange.com/questions/359768/probability-of-a-difference-between-two-sampling-means-of-two-populations",
"openwebmath_score": 0.6135420203208923,
"openwebmath_perplexity": 7062.78679767971,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9702399060540358,
"lm_q2_score": 0.8633916011860785,
"lm_q1q2_score": 0.8376969860226243
} |
https://kokecacao.me/page/Course/F22/15-210/Lecture_008_-_Randomized_Quicksort.md | # Lecture 008 - Randomized Quicksort
## Randomized Algorithm
Monte Carlo algorithms: might be wrong Las Vegas algorithms: might be fast
### Quick Select
Order Statistics: find the $k$th minimum item in a sequence.
Quick Select:
1. given an array of $n$ elements
2. find the $k$th element
3. put everything less than the $k$th element on the left
4. put everything greater than the $k$th element on the right
Complexity:
• Average Complexity: $O(n)$ work with high probability, $O(\log^2 n)$ span with high probability
• Worst Complexity $O(n^2)$
• Worst When:
• pick the smallest element as pivot, asked for biggest element
• pick the biggest element as pivot, asked for smallest element
• Anti-adversarial: random chosen a pivot
Algorithm:
1. pick the first element in array as a pivot
2. sort the array against the pivot, we get pivot index $p$
3. if the pivot is at $k$th position ($p = k$), return
4. otherwise, find the $k$th element in left array if $k$ is less than pivot index $p$
5. otherwise, find the $k-(p+1)$th element in the right array if $k$ is greater than pivot index $p$
Pseudo Code // TODO
#### Expected Complexity of Quick Select
We want to bound the expected input length at each level:
• Let $Y_d = \text{input length at level d}$
• Let $Z_d = \text{pivot index at level d}$
\begin{align*} E[Y_{d+1}] =& \sum_y \sum_z Pr\{Y_d=y \cap Z_d = z\} \cdot f(y, z)\\ =& \sum_y \sum_z Pr\{Y_d=y \cap Z_d = z\} \cdot \max(0, z, y-z-1)\tag{either left half, right half, or get lucky}\\ =& \sum_y \sum_z Pr\{Y_d=y\} \cdot Pr\{Z_d = z | Y_d = y\} \cdot \max(0, z, y-z-1)\\ =& \sum_y \sum_z Pr\{Y_d=y\} \cdot \frac{1}{y} \cdot \max(0, z, y-z-1)\\ =& \sum_y Pr\{Y_d=y\} \cdot \frac{1}{y} \cdot (\sum_z\max(0, z, y-z-1))\\ \leq& \sum_y Pr\{Y_d=y\} \cdot \frac{1}{y} \cdot (2\sum_{z=y/2}^{y-1}z)\\ =& \sum_y Pr\{Y_d=y\} \cdot \frac{1}{y} \cdot (2\cdot \frac{1}{2}(y/2+y-1)(y-1-y/2+1))\\ \leq& \sum_y Pr\{Y_d=y\} \cdot \frac{1}{y} \cdot \frac{3}{4}y^2\\ =& \frac{3}{4} \sum_y yPr\{Y_d=y\}\\ =& \frac{3}{4} E[Y_d]\\ E[Y_d] =& n \cdot (\frac{3}{4})^d \end{align*}
Therefore, the expected work is:
\begin{align*} E[W] =& \sum_d E[W_d]\\ =& \sum_d E[Y_d]\\ =& \sum_d n \cdot (\frac{3}{4})^d\\ =& n\sum_d (\frac{3}{4})^d\\ \in& O(n) \tag{by Geometric Series}\\ \end{align*}
The expected span is: assuming we have $O(\log n)$ levels with high probability, then
\begin{align*} E[S] =& \#\text{levels} \cdot E[S_d]\\ =& \#\text{levels} \cdot \log n \tag{by filter}\\ =& \log n \cdot \log n \tag{w.h.p}\\ =& \log^2 n \tag{w.h.p}\\ \end{align*}
### Quick Sort
Pseudo Code // TODO
The probability of comparing indices $i, j$ (assuming $j > i$) is:
E[X_{ij}] = Pr\{i \text{ compaired to } j\} = \frac{2}{j - i + 1}
This is because
• We only compare $i, j$ if one of them is a pivot (nominator)
• We know eventually $i, j$ will end up with different partition
• If we choose a pivot that is less than $i$ or greater than $j$, we never make progress
• If we choose a pivot between $i, j$, then we make progress (denominator)
The overall number of comparison is
\begin{align*} E[W] =& \sum_{i < j} \frac{2}{j - i + 1}\\ =& 2\sum_{i = 0}^{n - 1} \sum_{j = i+1}^{n - 1} \frac{1}{j - i + 1}\\ =& 2\sum_{i = 0}^{n - 1} \sum_{k = 1}^{n - i - 1} \frac{1}{k + 1} \tag{let $k = j - 1$}\\ =& 2\sum_{i = 0}^{n - 1} (H_{n - i} - 1)\\ \leq& 2\sum_{i = 0}^{n - 1} (1 + \log(n - i) - 1)\\ =& 2\sum_{i = 0}^{n - 1} \log(n - i)\\ \in& O(n \log n)\\ \end{align*}
For span analysis, we use high probability bound. This is because the max of two high probability bound is usually the same high probability bound.
### Binary Search Tree
(* 15-210 Fall 2022 *)
(* Parametric implementation of binary search trees *)
(* INCOMPLETE AND UNTESTED *)
(* Live-coded in Lecture 11, Wed Oct 5, 2022 *)
(* Frank Pfenning + students *)
signature KEY =
sig
type t
val compare : t * t -> order
end
structure K :> KEY =
struct
type t = int
val compare = Int.compare
end
signature ParmBST =
sig
type T (* abstract *)
datatype E = Leaf | Node of T * K.t * T
val size : T -> int
val expose : T -> E (* exposes structure, not internal info *)
val joinMid : E -> T (* rebalance *)
end
structure P :> ParmBST =
struct
datatype T = TLeaf | TNode of T * K.t * int * T
datatype E = Leaf | Node of T * K.t * T
fun size TLeaf = 0
| size (TNode (L, k, s, R)) = s
fun expose T = case T
of TLeaf => Leaf
| TNode (L, k, s, R) => Node (L, k, R)
fun joinMid E = case E
of Leaf => TLeaf
| Node (L, k, R) => TNode (L, k, size L + size R + 1, R)
end
signature BST =
sig
type T (* abstract *)
val empty : T
val find : T -> K.t -> bool
val insert : T -> K.t -> T
val delete : T -> K.t -> T
val union : T * T -> T
val intersection : T * T -> T
(* more... *)
end
functor Simple (structure P : ParmBST) :> BST =
struct
type T = P.T
val empty = P.joinMid (P.Leaf)
fun split T k = case P.expose T of
P.Leaf => (empty, false, empty)
| P.Node (L, k', R) => case K.compare (k, k') of
LESS => let val (LL, b, LR) = split L k (* LL < k < LR *)
in (LL, b, P.joinMid(P.Node(LR, k', R))) end
| EQUAL => (L, true, R)
| GREATER => let val (RL, b, RR) = split R k
in (P.joinMid(P.Node(L, k', RL)), b, RR) end
fun insert T k =
let val (L, _, R) = split T k
in P.joinMid(P.Node(L, k, R)) end
fun find T k = case P.expose T of
P.Leaf => false
| P.Node(L, k', R) => case K.compare (k, k') of
LESS => find L k
| EQUAL => true
| GREATER => find R k
end
### Expected Span
Consider a game in which we draw some number of tasks at random such that a task has length $n$ with probability $1/n$ and has length $1$ otherwise. The expected length of a task is therefore bounded by $2$. Imagine now drawing $n$ tasks and waiting for all them to complete, assuming that each task can proceed in parallel independently of other tasks. Prove that the expected completion time is not constant.
\begin{align*} &\left(\frac{n-1}{n}\right)^{n}\cdot1+\left(1-\left(\frac{n-1}{n}\right)^{n}\right)n\\ \to& 1/e + (1 - 1/e)n\tag{take the limit}\\ \in& O(n)\\ \end{align*}
Repeat the same exercise with slightly different probabilities: a randomly chosen task has length $n$ with probability $1/n^3$ and $1$ otherwise. Prove now that the expected completion time is bounded by a constant.
\begin{align*} &\left(\frac{n^{3}-1}{n^{3}}\right)^{n}\cdot1+\left(1-\left(\frac{n^{3}-1}{n^{3}}\right)^{n}\right)n\\ \to& 1+(1-1)n\tag{take the limit}\\ <&2\\ \in& O(1) \end{align*}
Table of Content | 2023-03-21T23:52:00 | {
"domain": "kokecacao.me",
"url": "https://kokecacao.me/page/Course/F22/15-210/Lecture_008_-_Randomized_Quicksort.md",
"openwebmath_score": 0.9949945211410522,
"openwebmath_perplexity": 11047.485590825418,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES\n\n",
"lm_q1_score": 0.9702399034724604,
"lm_q2_score": 0.8633915976709976,
"lm_q1q2_score": 0.837696980383242
} |
https://www.cdslab.org/paramonte/notes/examples/c/mvn/ | ## The mathematical objective function
Suppose we want to sample random vectors
$\mathbf{x}=[x_1,x_2,x_3,x_4] ~~,$
from a 4-dimensional Multivariate Normal (MVN) Probability Density Function (PDF),
$\mathrm{PDF}_\mathrm{MVN} \equiv \pi ( \mathbf{x} ~|~ \mu, \Sigma) = (2\pi)^{-\frac{k}{2}} \sqrt{|\Sigma|} \exp\bigg( -\frac{1}{2} (\mathbf{x}-\mu)^\mathrm{T}\Sigma^{-1}(\mathbf{x}-\mu) \bigg) ~~,$
that has a mean vector,
$\mu = [0,0,0,0]$
with a covariance matrix,
$\Sigma = \begin{bmatrix} 1.0 & 0.5 & 0.5 & 0.5 \\ 0.5 & 1.0 & 0.5 & 0.5 \\ 0.5 & 0.5 & 1.0 & 0.5 \\ 0.5 & 0.5 & 0.5 & 1.0 \end{bmatrix}$
This will serve as our mathematical objective function. Clearly, all of the four dimensions of this MVN are correlated with each other with a Pearson’s correlation coefficient of $0.5$.
### Working with the natural logarithm of the objective function
In general, since probabilities are often tiny numbers, we’d want to always work with their natural logarithms in the world of computers to avoid unwanted overflow or underflow of numbers, which can crash our simulations. To highlight the importance of taking the natural logarithms of all input objective functions to the ParaMonte routines, we always name them like getLogFunc().
## Implementing the objective function in C
Here is a minimalistic implementation of the 4-D multivariate normal distribution objective function in C, which can also be downloaded as logfunc.c.
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//
// Description:
// + Return the natural logarithm of an ndim-dimensional Multivariate Normal (MVN)
// probability density function (PDF) with the Mean and Covariance Matrix as defined below.
// Reference: https://en.wikipedia.org/wiki/Multivariate_normal_distribution
// Input:
// + ndim: The number of dimensions of the domain of the objective function.
// + point: The input 64-bit real-valued vector of length ndim,
// at which the natural logarithm of objective function is computed.
// Output:
// + logFunc: A 64-bit real scalar number representing the natural logarithm of the objective function.
// Author:
// + Computational Data Science Lab, Monday 9:03 AM, May 16 2016, ICES, UT Austin
// Visit:
// + https://www.cdslab.org/paramonte
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#include "logfunc.h"
double getLogFunc (
int32_t ndim,
double Point[]
)
{
// multivariate normal (MVN) distribution specifications:
const double LOG_INVERSE_SQRT_TWO_PI = log(0.398942280401432); // log(1/sqrt(2*Pi))
const double MEAN[NDIM] = {0., 0., 0., 0.}; // mean vector of the MVN
const double COVMAT[NDIM][NDIM] = {
{1.0,0.5,0.5,0.5},
{0.5,1.0,0.5,0.5},
{0.5,0.5,1.0,0.5},
{0.5,0.5,0.5,1.0}
}; // covariance matrix of the MVN
const double INVCOVMAT[NDIM][NDIM] = {
{+1.6, -0.4, -0.4, -0.4},
{-0.4, +1.6, -0.4, -0.4},
{-0.4, -0.4, +1.6, -0.4},
{-0.4, -0.4, -0.4, +1.6}
}; // inverse covariance matrix of the MVN
const double LOG_SQRT_DET_INV_COV = 0.581575404902840; // logarithm of square root of the determinant of the inverse covariance matrix
// subtract mean vector from the input point
int i;
double NormedPoint[NDIM];
for(i = 0; i < NDIM; i++){
NormedPoint[i] = Point[i] - MEAN[i];
}
// compute the log probability density function of the MVN
double exponentTerm = 0.;
for(i = 0; i < NDIM; i++){
double MatMulResult[NDIM];
MatMulResult[i] = 0.;
int j;
for(j = 0; j < NDIM; j++){
MatMulResult[i] += INVCOVMAT[i][j] * NormedPoint[j];
}
exponentTerm += NormedPoint[i] * MatMulResult[i];
}
double logFunc = NDIM * LOG_INVERSE_SQRT_TWO_PI + LOG_SQRT_DET_INV_COV - 0.5*exponentTerm;
return logFunc;
}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
Here is the contents of the corresponding C header file, which can also be downloaded as logfunc.h.
#ifndef PM_LOG_FUNC
#define PM_LOG_FUNC
#include <math.h>
#include <stdint.h>
#define NDIM 4 // The number of dimensions of the domain of the objective function.
double getLogFunc (
int32_t , // ndim
double [] // Point
);
#endif
## Calling the ParaMonte samplers
Suppose we want to randomly sample the above objective function via the Delayed-Rejection Adaptive Metropolis-Hastings Markov Chain Monte Carlo sampler of ParaMonte (the ParaDRAM sampler).
The following C main code can be used to call the ParaMonte library routine runParaDRAM(),
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//
// Description:
// + Run the Monte Carlo sampler of the ParaMonte library given the input log-target density function getLogFunc().
// Output:
// + The simulation output files.
// Author:
// + Computational Data Science Lab, Monday 9:03 AM, May 16 2016, ICES, UT Austin
// Visit:
// + https://www.cdslab.org/paramonte
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#include <stdio.h>
#include <stdint.h>
#include <string.h>
#include "logfunc.h" // getLogFunc, NDIM
#include "paramonte.h" // runParaDRAM
int main(int argc, char *argv[])
{
char inputFile[] = "./paramonte.in";
int32_t inputFileLen = strlen(inputFile);
int32_t ndim = NDIM;
// C rules for argument passing apply here
, &getLogFunc
, inputFile
, inputFileLen
);
return 0;
}
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
### Calling the ParaMonte samplers: The signature header file
Here is the contents of the C paramonte.h signature header file for the ParaMonte library routines,
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////
////
//// ParaMonte: plain powerful parallel Monte Carlo library.
////
//// Copyright (C) 2012-present, The Computational Data Science Lab
////
//// This file is part of the ParaMonte library.
////
//// Permission is hereby granted, free of charge, to any person obtaining a
//// copy of this software and associated documentation files (the "Software"),
//// to deal in the Software without restriction, including without limitation
//// the rights to use, copy, modify, merge, publish, distribute, sublicense,
//// and/or sell copies of the Software, and to permit persons to whom the
//// Software is furnished to do so, subject to the following conditions:
////
//// The above copyright notice and this permission notice shall be
//// included in all copies or substantial portions of the Software.
////
//// THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND,
//// EXPRESS OR IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF
//// MERCHANTABILITY, FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT.
//// IN NO EVENT SHALL THE AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM,
//// DAMAGES OR OTHER LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR
//// OTHERWISE, ARISING FROM, OUT OF OR IN CONNECTION WITH THE SOFTWARE
//// OR THE USE OR OTHER DEALINGS IN THE SOFTWARE.
////
//// ACKNOWLEDGMENT
////
//// ParaMonte is an honor-ware and its currency is acknowledgment and citations.
//// As per the ParaMonte library license agreement terms, if you use any parts of
//// this library for any purposes, kindly acknowledge the use of ParaMonte in your
//// work (education/research/industry/development/...) by citing the ParaMonte
//// library as described on this page:
////
//// https://github.com/cdslaborg/paramonte/blob/main/ACKNOWLEDGMENT.md
////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//
// Description:
// - The C header file containing the prototypes of the ParaMonte library routines to be called from C/C++.
// Prototypes:
// Author:
// - Computational Data Science Lab, Monday 9:03 AM, May 16 2016, ICES, UT Austin
//
////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#ifndef ParaMonte
#define ParaMonte
#include <stdint.h>
#if defined DLL_ENABLED
extern __declspec(dllimport)
#endif
// ndim: dimension of the domain of the LogFunc
int32_t ,
// getLogFunc(ndim, Point(ndim)): procedure pointer to the LogFunc
double (*) (
int32_t ,
double []
),
// inputFile: ParaDRAM input file, containing the path the file containing a list of
// all optional input variables and values
char [],
// inputFileLen: the length of the inputFile char vector: int32_t inputFileLen = strlen(inputFile);
int32_t
);
#endif
### Calling the ParaMonte samplers: The input file
We will store the simulation specifications in a separate external input file whose path "./paramonte.in" is given by the variable inputFile in the main code above. See this page for a detailed description of the structure and convention rules of the input files. Also, see this page for detailed descriptions of the simulation specifications of the ParaDRAM sampler.
Here is the contents of this paramonte.in input file,
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
!%
!% Description:
!% + Run the Monte Carlo sampler of the ParaMonte library given the input log-target density function getLogFunc().
!% Output:
!% + The simulation output files.
!% Author:
!% + Computational Data Science Lab, Monday 9:03 AM, May 16 2016, ICES, UT Austin
!% Visit:
!% + https://www.cdslab.org/paramonte
!%
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
!
! USAGE:
!
! -- Comments must begin with an exclamation mark (!).
! -- Comments can appear anywhere on an empty line or, after a value assignment.
! -- All variable assignments are optional and can be commented out. In such cases, appropriate default values will be assigned.
! -- Use ParaDRAM namelist (group) name to group a set of ParaDRAM simulation specification variables.
! -- The order of the input variables in the namelist groups is irrelevant and unimportant.
! -- Variables can be defined multiple times, but only the last definition will be considered as input.
! -- All variable names are case insensitive. However, for clarity, this software follows the camelCase code-writing practice.
! -- String values must be enclosed with either single or double quotation marks.
! -- Logical values are case-insensitive and can be either .true., true, or t for a TRUE value, and .false., false, or f for a FALSE value.
! -- All vectors and arrays in the input file begin with index 1. This is following the convention of
! the majority of science-oriented programming languages: Fortran, Julia, Mathematica, MATLAB, and R.
!
! For comprehensive guidelines on the input file organization and rules, visit:
!
!
! To see detailed descriptions of each of variables, visit:
!
!
! Base specifications
description = "
This\n
is a\n
multi-line\n
description." ! strings must be enclosed with "" or '' and can be continued on multiple lines.
! No comments within strings are allowed.
!outputColumnWidth = 25 ! this is an example of a variable that is commented out and
! therefore, its value won't be read by the sampler routine.
! To pass it to the routine, simply remove the ! mark at
! the beginning of the line.
outputRealPrecision = 17
!outputDelimiter = ","
outputFileName = "./out/" ! the last forward-slash character indicates that this
! is the folder where the output files will have to stored.
! However, since no output filename prefix has been specified,
! the output filenames will be assigned a randomly-generated prefix.
!sampleSize = 111
randomSeed = 2136275,
chainFileFormat = "compact"
variableNameList = "variable1" ! Notice the missing fourth variable name here.
, "variable2" ! Any missing variable name will be automatically assigned an appropriate name.
, "variable3"
domainLowerLimitVec = 4*-1.e300 ! repetition pattern rules apply again here. 4 dimensions => 4-element vector of values.
domainUpperLimitVec = 4*1.e300 ! repetition pattern rules apply again here. 4 dimensions => 4-element vector of values.
parallelizationModel = "single chain" ! "singleChain" would also work. Similarly, "multichain", "multi chain", or "multiChain".
!targetAcceptanceRate = 0.23e0
progressReportPeriod = 1000
maxNumDomainCheckToWarn = 100
maxNumDomainCheckToStop = 1000
restartFileFormat = "binary"
overwriteRequested = true ! FALSE, false, .false., .f., and f would be also all valid logical values representing False
silentModeRequested = false ! FALSE, false, .false., .f., and f would be also all valid logical values representing False
!mpiFinalizeRequested = true ! TRUE, true, .true., .t., and t would be also all valid logical values representing True
! MCMC specifications
chainSize = 30000
startPointVec = 4*1.e0 ! four values of 1.e0 are specified here by the repetition pattern symbol *
!sampleRefinementCount = 10
sampleRefinementMethod = "BatchMeans"
randomStartPointDomainLowerLimitVec = 4*-100.e0 ! repetition pattern rules apply again here. 4 dimensions => 4-element vector of values.
randomStartPointDomainUpperLimitVec = 4*100.0 ! repetition pattern rules apply again here. 4 dimensions => 4-element vector of values.
randomStartPointRequested = false
! DRAM specifications
scaleFactor = "2*0.5*Gelman" ! The asterisk here means multiplication since it is enclosed within quotation marks.
proposalModel = "normal" ! or "uniform" as you wish.
!delayedRejectionCount = 5
!delayedRejectionScaleFactorVec = 5*1.
!proposalStartStdVec = 4*1.0 ! repetition pattern rules apply again here. 4 dimensions => 4-element vector of values.
!proposalStartCorMat = 1,0,0,0, ! 2-dimensional correlation-matrix definition, although it is commented out and won't be read.
! 0,1,0,0,
! 0,0,1,0,
! 0,0,0,1,
!proposalStartCovMat = 100,0,0,0,
! 0,100,0,0,
! 0,0,100,0,
! 0,0,0,100,
/
## Compiling and linking to generate the executable
Now, follow the instructions on this page on how to build your C example for serial and parallel simulations on the Operating System (OS) of you choice: Windows, Linux, or macOS / Darwin.
## Running the ParaMonte simulations
Once you have built your application, running the simulation is as simple as calling the generated executable on the command-prompt for serial applications, or invoking the MPI/Coarray launcher on the command-prompt for MPI/Coarray parallel simulations. Follow the instructions and tips on this page on how to run your C example on either Windows, Linux, or macOS / Darwin Operating System (OS), on either a single processor or in parallel, on multiple processors or clusters of nodes of processors.
### Running the ParaMonte simulations on a single processor
If the executable main.exe has been built for serial simulations (using one processor), simply call the executable on the command line. The calling method is identical on all three Operating Systems: Windows, Linux, and Darwin (macOS). Here is an example call on Windows OS,
D:\example>main.exe
************************************************************************************************************************************
************************************************************************************************************************************
**** ****
**** ****
**** ParaMonte ****
**** Version 1.0.0 ****
**** May 23 2018 ****
**** ****
**** Department of Physics ****
**** Computational & Data Science Lab ****
**** Data Science Program, College of Science ****
**** The University of Texas at Arlington ****
**** ****
**** originally developed at ****
**** ****
**** Multiscale Modeling Group ****
**** Center for Computational Oncology (CCO) ****
**** Oden Institute for Computational Engineering and Sciences ****
**** Department of Aerospace Engineering and Engineering Mechanics ****
**** Department of Neurology, Dell-Seton Medical School ****
**** Department of Biomedical Engineering ****
**** The University of Texas at Austin ****
**** ****
**** ****
**** Amir Shahmoradi ****
**** ****
**** ****
**** cdslab.org/pm ****
**** ****
**** https://www.cdslab.org/paramonte/ ****
**** ****
**** ****
************************************************************************************************************************************
************************************************************************************************************************************
************************************************************************************************************************************
**** ****
**** Setting up ParaDRAM environment ****
**** ****
************************************************************************************************************************************
ParaDRAM - NOTE: Variable outputFileName detected among the input variables to ParaDRAM:
ParaDRAM - NOTE: ./out/
ParaDRAM - NOTE: Absolute path to the current working directory:
ParaDRAM - NOTE: D:\example
ParaDRAM - NOTE: Generating the requested directory for ParaDRAM output files:
ParaDRAM - NOTE: .\out\
ParaDRAM - NOTE: No user-input filename prefix for ParaDRAM output files detected.
ParaDRAM - NOTE: Generating appropriate filenames for ParaDRAM output files from the current date and time...
ParaDRAM - NOTE: ParaDRAM output files will be prefixed with:
ParaDRAM - NOTE: Searching for previous runs of ParaDRAM...
ParaDRAM - NOTE: No pre-existing ParaDRAM run detected.
ParaDRAM - NOTE: Starting a fresh ParaDRAM run...
ParaDRAM - NOTE: Generating the output report file:
ParaDRAM - NOTE: Please see the output report and progress files for further realtime simulation details.
Accepted/Total Func. Call Dynamic/Overall Acc. Rate Elapsed/Remained Time [s]
========================= ========================= =========================
30000 / 130490 0.179 / 0.2291 2.2880 / 0.000
ParaDRAM - NOTE: Computing the Markov chain's statistical properties...
ParaDRAM - NOTE: Computing the final decorrelated sample size...
ParaDRAM - NOTE: Generating the output sample file:
ParaDRAM - NOTE: Computing the output sample's statistical properties...
ParaDRAM - NOTE: Mission Accomplished.
### Running the ParaMonte simulations in parallel on multiple processors
If the executable main.exe has been built for MPI-enabled parallel simulations (on multiple processors), you will need to invoke the MPI launcher to start the simulation.
• On Windows
For parallel simulations on a single node (of multiple processors), for example on a single computer, the command typically looks like the following,
mpiexec -localonly -n 3 main.exe
where the flag -localonly ensures the simulation is run only on a single node. This option avoids the invocation of the Hydra service which would require prior registration. The flag -n 3 assigns three MPI tasks to three physical processors for the simulation. To run your simulation on a cluster of nodes, follow the guidelines and instructions provided by the Intel MPI library development team.
• On Linux / Darwin (Mac)
The MPI launcher command for parallel simulations on Linux and Darwin are identical and look very similar to the Windows command (except for the flag -localonly which is not needed),
mpiexec -n 3 main.exe
where the flag -n 3 assigns three MPI tasks to three physical processors for the simulation.
For example, the following command will launch the simulation on 3 processors on a Windows OS,
D:\example>mpiexec -localonly -n 3 main.exe
************************************************************************************************************************************
************************************************************************************************************************************
**** ****
**** ****
**** ParaMonte ****
**** Version 1.0.0 ****
**** May 23 2018 ****
**** ****
**** Department of Physics ****
**** Computational & Data Science Lab ****
**** Data Science Program, College of Science ****
**** The University of Texas at Arlington ****
**** ****
**** originally developed at ****
**** ****
**** Multiscale Modeling Group ****
**** Center for Computational Oncology (CCO) ****
**** Oden Institute for Computational Engineering and Sciences ****
**** Department of Aerospace Engineering and Engineering Mechanics ****
**** Department of Neurology, Dell-Seton Medical School ****
**** Department of Biomedical Engineering ****
**** The University of Texas at Austin ****
**** ****
**** ****
**** Amir Shahmoradi ****
**** ****
**** ****
**** cdslab.org/pm ****
**** ****
**** https://www.cdslab.org/paramonte/ ****
**** ****
**** ****
************************************************************************************************************************************
************************************************************************************************************************************
************************************************************************************************************************************
**** ****
**** Setting up ParaDRAM environment ****
**** ****
************************************************************************************************************************************
ParaDRAM - NOTE: Variable outputFileName detected among the input variables to ParaDRAM:
ParaDRAM - NOTE: ./out/
ParaDRAM - NOTE: Absolute path to the current working directory:
ParaDRAM - NOTE: D:\example
ParaDRAM - NOTE: Generating the requested directory for ParaDRAM output files:
ParaDRAM - NOTE: .\out\
ParaDRAM - NOTE: No user-input filename prefix for ParaDRAM output files detected.
ParaDRAM - NOTE: Generating appropriate filenames for ParaDRAM output files from the current date and time...
ParaDRAM - NOTE: ParaDRAM output files will be prefixed with:
ParaDRAM - NOTE: Searching for previous runs of ParaDRAM...
ParaDRAM - NOTE: No pre-existing ParaDRAM run detected.
ParaDRAM - NOTE: Starting a fresh ParaDRAM run...
ParaDRAM - NOTE: Generating the output report file:
ParaDRAM - NOTE: Please see the output report and progress files for further realtime simulation details.
Accepted/Total Func. Call Dynamic/Overall Acc. Rate Elapsed/Remained Time [s]
========================= ========================= =========================
30000 / 130260 0.200 / 0.2298 1.3140 / 0.000
ParaDRAM - NOTE: The time cost of calling the user-provided objective function must be at least 52.5143 times more
ParaDRAM - NOTE: (that is, ~24.6651 microseconds) to see any performance benefits from singlechain parallelization
ParaDRAM - NOTE: model for this simulation. Consider simulating in the serial mode in the future (if used within the
ParaDRAM - NOTE: same computing environment and with the same configuration as used here).
ParaDRAM - NOTE: Computing the Markov chain's statistical properties...
ParaDRAM - NOTE: Computing the final decorrelated sample size...
ParaDRAM - NOTE: Generating the output sample file:
ParaDRAM - NOTE: Computing the output sample's statistical properties...
ParaDRAM - NOTE: Mission Accomplished.
You may have noticed the following note in the above simulation output,
ParaDRAM - NOTE: The time cost of calling the user-provided objective function must be at least 52.5143 times more
ParaDRAM - NOTE: (that is, ~24.6651 microseconds) to see any performance benefits from singlechain parallelization
ParaDRAM - NOTE: model for this simulation. Consider simulating in the serial mode in the future (if used within the
ParaDRAM - NOTE: same computing environment and with the same configuration as used here).
This indicates that the simulation in parallel does not lead to any efficiency gain. Given the computer specifications on which the above simulation is performed (a decently powerful computer), this may not be surprising. Based on the CPU power and inter-processor communication costs, our example simulation in the above seems to be a good candidate for serial simulations instead.
### Final notes
Mission accomplished. We have now used the adaptive Markov chain sampler routine of the ParaMonte library to generate random numbers from our objective function of interest. This simulation will generate five files in a folder named ./out/ in the current directory, each of which contains some specific information about the simulation. The purpose of each file is indicated by the suffix at the end of its name: _chain, _sample, _report, _progress, _restart. If multichain parallelism is requested via the input simulation specification variable parallelizationModel = 'multiChain', then there will be five simulation output files generated for each processor separately and independently of each other. This is as if we have simulated several independent but simultaneous ParaDRAM simulations of our objective function. More details about the contents of each of the ParaDRAM simulation output files can be found on this page.
## Post-processing the ParaMonte simulation results
The post-processing of the output files generated by the ParaMonte routines can be nicely and quickly done via either the ParaMonte Python library or the ParaMonte MATLAB library.
If you have any questions about the topics discussed on this page, feel free to ask in the comment section below, or raise an issue on the GitHub page of the library, or reach out to the ParaMonte library authors. | 2021-05-17T00:59:02 | {
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https://math.stackexchange.com/questions/588314/matrix-raised-to-14th-power/588320 | Matrix raised to 14th power
Calculate $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)^{14}$
Whould I do it one by one, and then find a pattern? I sense $6^{14}$ on the diagonal, and zeroes in the "lower triangle", but the "upper triangle" I'm not sure. Was thinking $14 \cdot 6^{13}$ but that's not correct.
• HINT Try multiplying the matrix out and finding a pattern by the 3rd or 4th power. – Don Larynx Dec 1 '13 at 17:11
• Closely related: for a $3 \times 3$ matrix A ,value of $A^{50}$ is – hardmath Dec 1 '13 at 17:23
• 14 is small enough that you can even do it by hand; doing only 5 matrix multiplications; namely: $A^2=A.A,A^4=A^2.A^2,A^6=A^2.A^4,A^8=A^4.A^4,A^{14}=A^6.A^8$! – Ali Dec 1 '13 at 19:04
Write the matrix as $6I+N$ where $$N=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$ and expand $(6I+N)^{14}$ using the binomial formula (which is valid here because $I$ and $N$ commute). Note that since $N^3=0$, you don't need to compute most of the coefficients.
• You beat me to it and also remembered to mention that the two matrices commute. – Carsten S Dec 1 '13 at 17:15
• What's happened to your avatar, Harald? I remember a bee was on it!? :) – mrs Dec 1 '13 at 17:21
• @B.S. I still see a bee. – Git Gud Dec 1 '13 at 17:21
• @B.S. It still is. No, it's a fly actually. If you don't see it, it's a problem with the gravatar site. Most likely temporary. – Harald Hanche-Olsen Dec 1 '13 at 17:21
Write it as $\left(\begin{matrix} 6&1&0\\0&6&1\\0&0&6\end{matrix}\right)=\left(\begin{matrix} 6&0&0\\0&6&0\\0&0&6\end{matrix}\right)+\left(\begin{matrix} 0&1&0\\0&0&1\\0&0&0\end{matrix}\right)$, use the binomial formula and see what you can say about powers of these two matrices.
$$A^n = \begin{bmatrix} 6^n & n \cdot 6^{n-1} & \dbinom{n}2 6^{n-2}\\ 0 & 6^n & n \cdot 6^{n-1}\\ 0 & 0 & 6^n\end{bmatrix}$$ Prove this by induction.
• Does that mean you didn't induce it, but just found the pattern? – Thomas Ahle Dec 5 '13 at 15:46
• @ThomasAhle Yes. – user17762 Dec 5 '13 at 15:48
I didn't notice that this was suggested by Harald Hanche-Olsen until just now. Consider this an expansion on his answer.
Since the identity matrix commutes with any matrix, we can use the binomial theorem with $$\left(6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}\right)^n$$ while noting that $$\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix} \quad\text{and}\quad \begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}^3=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}$$ To get $$\begin{bmatrix}6&1&0\\0&6&1\\0&0&6\end{bmatrix}^n =6^n\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}+6^{n-1}\binom{n}{1}\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix}+6^{n-2}\binom{n}{2}\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix}$$
Denote the element in the top center (now 1) in $A^n$ by $a_n$. From the matrice multiple and the diagonal values we can see that $$a_{n+1}=6^na_n+6^na_n=2*6^na_n$$ from this we can make the general nth element $$a_n=2*6^{n-1}*2*6^{n-2}*...*2*1=2^n*6^{(n-1)+(n-2)+...+1}=2^n*6^{\frac{n(n-1)}{2}}$$ Putting n=14 gives: $$a_{14}=2^{14}*6^{91}$$ | 2019-04-23T22:31:55 | {
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https://math.stackexchange.com/questions/2354280/intuitive-reason-why-sqrtnn-to-1-as-n-to-infty/2354290 | # Intuitive reason why $\sqrt[n]n\to 1$ as $n\to\infty$?
We are aware of the limit $$\lim_{n\to\infty}\sqrt[n]n = 1;$$ is there any geometric or otherwise intuitive reason to see why this limit holds?
Edit: I am adding some context, since this question was previously put on-hold, and I think one of the main reasons was that it was poorly motivated. From theorem 8.1 of Baby Rudin, suppose the series $$\sum_{n=0}^\infty c_nx^n$$ converges for $|x|<R$, and define $$f(x) = \sum_{n=0}^\infty c_nx^n \qquad (|x|<R). \tag{1}$$ Among other conclusions, the function $f$ is differentiable in $(-R,R)$, and $$f'(x) = \sum_{n=0}^\infty nc_n x^{n-1} \qquad (|x|<R). \tag{2}$$ Rudin uses the fact that $\sqrt[n]n\to 1$ as $n\to\infty$ to justify that the series in $(1)$ and the series in $(2)$ have the same radius of convergence. I recognized the limit, but it is just such a nice combination of $n$ and the $n$th-root, that I thought there ought to be some nice intuitive way to understand it, hence this question.
• taking the root shrinks the magnitude at a faster rate than it grows. Under the root, n is growing at a linear rate. – CogitoErgoCogitoSum Jul 11 '17 at 1:25
• For those who are voting to close, pray tell why? – Alex Ortiz Jul 11 '17 at 1:51
• Just to inform if you have close rights you can see the reason for close votes. Currently there are two reasons: 1) off-topic and 2) unclear what you are asking. But i think your question has good intent. – Paramanand Singh Jul 11 '17 at 5:08
• Note that this is equivalent to $\lim_{n\to0}n^n=1$. – Akiva Weinberger Jul 12 '17 at 18:08
$$n^{1/n} = \exp\left(\frac{\log(n)}n\right)$$ $\log(n)$ grows, but very slowly, slower than $n$ or any positive power of $n$. So $\log(n)/n \to 0$ as $n \to \infty$, and $n^{1/n} \to \exp(0) = 1$.
• Does this mean that $\lim_{n\to\infty}\sqrt[\log n] n=e$? – abiessu Jul 11 '17 at 1:08
• @abiessu $$n^\frac{1}{\log{n}} = \exp\left(\frac{\log n}{\log n} \right) = \exp(1)$$ so yes. – Trevor Gunn Jul 11 '17 at 3:23
• I've quite literally never noticed this simple fact before. (Although for me, it almost feels as though it does more to explain the fact that $\frac{\log(n)}{n} \to 0$ as $n \to \infty$ as a consequence of $n^\frac{1}{n} \to 1$.) Anyway, (+1) for this. – Chris Jul 14 '17 at 22:19
If you know the Arithmetic-Geometric Mean Inequality, then you can see that
$$1\le\sqrt[n]n\le{\sqrt n+\sqrt n+1+\cdots+1\over n}={2\sqrt n+(n-2)\over n}=1+{2\over\sqrt n}-{2\over n}$$
and the Squeeze Theorem gives the limit $\sqrt[n]n\to1$.
Here the issue is that $n \rightarrow \infty$, but for any fixed $x > 0$ we have $x^{1/n} \rightarrow 1$. So to consider $n^{1/n}$ you have to ask which "wins": the $n$ at the base or the $1/n$ in the exponent. To think about this, it might help to compare to, say, $(2^n)^{1/n}$. This tends to (and is equal to) $2$. The exponent of $1/n$ has the power to take a huge number like $2^n$ and reduce it to a constant. Since $n$ is much, much smaller than $2^n$, you might expect then that the power of $1/n$ would "win" in the end, giving a result of $1$. This is not a proof, but it gives the right intuition.
If "intuitive reason" contains arguments which are natural, then the following may be of interest.
A recurring argument about convergence of sequences are through subsequences. For example, to see that $\sqrt[n]{a}$ must converge to $1$ if it converges at all (which it does, since it is decreasing) notice that $\sqrt[2n]{a}$ must converge to the same limit $L$, since it is a subsequence. But $(\sqrt[2n]{a})^2=\sqrt[n]{a}$, and therefore $L^2=L \implies L=1$, since it obviously cannot be $0$.
This is just to exemplify a natural argument which commonly applies. Trying to apply this directly in the case $n^{1/n}$ (taking the subsequence $2n$) has problems since the $\sqrt[n]{2}$ will become $1$. Therefore, we must take a subsequence which is increasing faster. It is natural to consider $n^2$. We then have that, if $n^{1/n}$ converges, then $n^{2/n^2}=((n^{1/n})^{1/n})^2$ converges to the same limit. But since $n^{1/n}$ is assumed to be convergent, we have that for sufficiently large $n$: $$1<((n^{1/n})^{1/n})^2 < ((L+1)^{1/n})^2.$$ By the squeeze theorem, $n^{2/n^2}$ then converges to $1$, and hence so does $n^{1/n}$. In fact, this is a full proof, except that we are supposing that $n^{1/n}$ converges (but it does, since this sequence is eventually decreasing).
PS: This argument has sparkles of the philosophy of the Cauchy condensation test.
\begin{align} & \underbrace{2\times \cdots\cdots\cdots\cdots\times 2}_{\Large n \text{ factors}} \gg n. \\[15pt] \text{Therefore } & \frac n {\quad\underbrace{2\times \cdots\cdots\cdots\cdots\times 2}_{\Large n \text{ factors}}\quad} \approx 0 \text{ when } n\approx\infty. \\[15pt] \text{Let } m = 2^n. \text{Then } & \frac{\log_2 m} m \approx 0 \text{ when } m \approx\infty \\[10pt] & 2^{(\log_2 m)/m} \approx 1 \\[10pt] & m^{1/m} \approx 1. \end{align}
• I'm sorry, but I don't see how this suggests the limit in question. – Alex Ortiz Jul 11 '17 at 3:00
• @AOrtiz : I've added to the answer since you posted your comment. – Michael Hardy Jul 11 '17 at 3:07
One way to see this is: for example, $(1.01)^n$ grows faster than $n$, so for large enough $n$, $n$ will eventually be less than $(1.01)^n$. Once you have reached this point, you will have $1 < n^{1/n} < 1.01$. Now, just replace $1.01$ with $1 + \epsilon$ for any $\epsilon > 0$, and you will have a proof of the limit.
To reduce the first statement even further: let $a_n := \frac{n}{(1.01)^n}$. Then $\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = \lim_{n\to\infty} \frac{1 + 1/n}{1.01} = \frac{1}{1.01} < 1$. Therefore, suppose $N$ is such that $\frac{a_{n+1}}{a_n} < 0.995$ for $n \ge N$; then for $n > N$, $0 < a_n < a_N (0.995)^{n-N}$, so by the squeeze theorem, $\lim_{n\to \infty} a_n = 0$. So, the intuitive point here is: even though the base of the exponent is only marginally greater than 1, it eventually starts making $a_n$ decrease approximately like a geometric sequence with ratio $\frac{1}{1.01}$.
(Of course, the other answers expressing $n^{1/n}$ as $e^{\frac{1}{n} \ln n}$ will give a much better idea of how fast $n^{1/n}$ approaches 1: namely, $n^{1/n}$ is approximately equal to $1 + \frac{\ln n}{n}$ for large $n$.)
• I liked this reasoning. +1 And this can be made into a proof by contradiction. The sequence $n^{1/n}$ is decreasing and is never less than $1$. If it tends to $L > 1$ then we should have $n^{1/n} \geq L$ or $$n \geq L^{n} = (1 + (L - 1))^{n} \geq \frac{n(n - 1)}{2}(L - 1)^{2}$$ (last inequality via binomial theorem) and we get an obvious contradiction. – Paramanand Singh Jul 11 '17 at 5:16
The intuition is that the "$n$" in the index of the radical is dominant when compared to the "$n$" in the radicand.
So, for example, suppose $n=10^6$. One million is a large number. What happens if we take its millionth root [ie., $(10^6)^{10^{-6}}$]? Well, let's do this one step at a time.
If you take the square root of $10^6$, it is downsized considerably, and you get $1000$.
If you take its cube root, you get an even smaller number, namely $100$.
If you take its sixth root, you get $10$.
Well, we still have $999994$ "roots" to go. Hence, it's fairly reasonable to assume that after all of these "roots", we get a number quite close to $1$. And, as it so happens, as we take larger and larger $n$, we can make $n^{\frac 1n}$ arbitrarily close to $1$.
You can write the limit like this $\lim\limits_{n→∞}\sqrt[n]{n}=\lim\limits_{n→∞}e^{\frac{ln(n)}{n}}=e^{\lim\limits_{n→∞}\frac{ln(n)}{n}}$ the limit of $\frac{log(n)}{n}$ is zero, and $e^0=1$ .
The simplest proof I know of uses only Bernoulli's inequality (which is simpler than the AM-GM inequality):
$(1+n^{-1/2})^n \ge 1+n^{1/2} \gt n^{1/2}$ so, raising to the $2/n$ power, $n^{1/n} \lt (1+n^{-1/2})^2 =1+2n^{-1/2}+n^{-1} \le 1+3n^{-1/2}$.
$$\lim_{n \to \infty} n^{1/n} = L$$ $$\lim_{n \to \infty} \frac{\ln(n)}{n} = \ln(L)$$ $$0 = \ln(L)$$ $$L = 1$$
But alternatively,
$$\lim_{n \to \infty} n^{1/n} = \lim_{u \to 0^+} (\frac{1}{u})^u$$
Which, at zero is usually intuitively given as $\frac{1}{1}$.
• Okay, so the first one uses the continuity of the log function. I'm not familiar with the second identity. – Alex Ortiz Jul 11 '17 at 0:43
• Replace $n$ with $1/u$. As $n$ approaches $\infty$, $1/u$ approaches $0$ from the positive side. Therefore we switch the limit. – John Lou Jul 11 '17 at 0:47
• If you're talking about the last part, $0^0$ is more often than not given as $1$. – John Lou Jul 11 '17 at 0:48
Let's take the obvious $\displaystyle \sqrt[n]{c^n}=c$.
But we know the following facts :
• for $0\le a<1$ then $a^n\to 0$ and $n\gg a^n$
• for $b>1$ then $b^n\to+\infty$ and $n\ll b^n$
So in fact the squeeze $\mathbf{a^n\ll n\ll b^n}$ becomes when taking the $n-$root : $\quad\mathbf{a\le \sqrt[n]{n}\le b}$
For any $a<1$ and any $b>1$ which coerce it to be $1$.
As you can notice the gap between $a^n\ll ...\ll b^n$ is huge and many sequences can fit there, not only $n$, but $n^2, n^{100}$ or any polynomial as well, making $\sqrt[n]{\ }$ a powerful reducing tool, in the same way $\ln(x^k)=k\ln(x)$ flattens powers.
This in indeed what happens in the theoretical $\displaystyle \sqrt[n]{n^k}=e^ {\frac 1n\ln(n^k)}=e^{k\overbrace{\frac{\ln(n)}{n}}^{\to 0}}\to 1$
The question asks for an intuitive reason for the value of the limit $L$ if it exists. Look at a slightly more general limit.
Set $f(n):=\sqrt[n]{n^k}$ for some integer $k>0$ and assume $f(n) \to L$ as $n\to\infty$.
So $f(n)^n = n^k$ and we need $L^n \approx n^k$. Now $L<1$ is false since $L^n<1<n^k$ and is decreasing. So suppose $L>1$. Then we have $L^n \approx n^k$ as $n$ gets big.
But we know that exponential growth exceeds any polynomial function of $n$ such as $n^k$, so $L=1$.
• I am not following your first claim about if $L < 1$. Nor am I really following your second claim to be honest. Maybe you can surrender more details? – Alex Ortiz Jul 11 '17 at 1:00
As a child you couldn't resist calculating the sequence $\{1,2,4,8,16,\dots\}$ in your mind, fascinated by the pattern of growth.
Years later, you are staring at $$\tag 1 \lim_{n \to \infty} n^{1/n} = \,?$$ You know that for all $n \ge 0$, $n \le 2^n$, and as $n$ grows, you feel awkward even looking at the inequality—there is no comparison!
So, $(n)^{1/n} \le {(2^n)}^{1/n} = 2$, and if the sequence (1) converges it has to be between $1$ and $2$.
You now look at $$\tag 2 (n)^{1/n} \le {(s^n)}^{1/n} = s$$ with $1 \lt s \le 2$.
You realize that if (2) holds true for sufficiently large $n$, then you've squeezed the convergence of (1) further, between $1$ and $s$. You let $s = 1.5$ and $n = 4$ and now 'you know' that the sequence (1) converges to $1$.
Proof
Let $0 \lt p \le 1$ be fixed. If we can show that $$\tag 3 n \le (1 + p) ^ n$$ for all large $n$, then we proved that (1) converges to $1$.
For fun, we prove the case $p = 1$ again. The second term of the binomial expansion of the RHS of (3) is exactly equal to $n$, so not much to do there.
If $p \lt 1$, the second term (= $np$) will not work. So, hoping for the best, we examine the third term:
\begin{align*} n &\le \frac{1}{2} n(n-1) p^2 \text{ iff} \\ 2 &\le (n-1) p^2 \text{ iff} \\ n &\ge \frac{2}{p^2} + 1. \end{align*} The proof is complete.
• I really like this line of reasoning because it paints a picture of how one might determine this limit by experimentation. I also like how you shared the insight that we can bound the limit from above by progressively smaller limits. I hope you don't mind the formatting edits I made to your answer—I think it's a little prettier this way. – Alex Ortiz Jul 14 '17 at 3:59
• It flows better with your formats - thanks! – CopyPasteIt Jul 14 '17 at 11:49
I see it as $\sqrt[n]{n}$ is decreasing, because its derivitaive is $\frac{1}{n^2}\sqrt[n]{n}\left(1-\ln(n)\right)<0$. And $1$ is a lower bound. So there is a limit $L$, and question is whether $L$ is something greater than $1$. If the limit were $L>1$, then for large $n$ you have $$\sqrt[n]{n}= L+\varepsilon$$ $$n=(L+\varepsilon)^n>L^n$$ and we can't have a linear $n$ exceeding an exponential with base greater than $1$.
Change $n=2^m:$ $$\lim_\limits{n\to\infty} n^{\frac{1}{n}} =\lim_\limits{m\to\infty} \left({2^m}\right)^{\frac{1}{2^m}}=$$ $$2^{\lim_\limits{m\to\infty} \frac{m}{2^m}}=2^0=1.$$
• The OP asked for the intuition behind $\sqrt[n] n \to 1$, not for a formal proof. – Alex M. Aug 7 '17 at 9:46
• @AlexM, thank you for taking time to review my answer and comment. I thought there was an intuition in it, because when the same $n$ is both in base and exponent, the difference in their rates is not conspicuous, but when $n=2^m$, the difference in rates of increase is obvious. Don't you think? – farruhota Aug 7 '17 at 9:56 | 2021-01-27T02:27:37 | {
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http://math.stackexchange.com/questions/89525/binary-function-that-is-distributive-associative-commutative | # Binary function that is distributive, associative, commutative
Is there such an operation that is distributive over addition, and is not multiplication?
Also, please no operations that are defined piece-wise, or that are trivial.
It must apply to all integers, but for all reals for example, the result does not matter.
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Do you also want a unit element for your binary operation? – Zhen Lin Dec 8 '11 at 6:20
Pardon my ignorance, but what does that mean? – soandos Dec 8 '11 at 6:21
Addition doesn't distribute over addition... – Arturo Magidin Dec 8 '11 at 6:22
@soandos: A unit element is an element $u$ such that $a\odot u = a$ for all $a$. – Arturo Magidin Dec 8 '11 at 6:22
@ArturoMagidin, fixed, thanks. No need, but it would be interesting – soandos Dec 8 '11 at 6:23
The only such operations are those of the form $m\otimes n=mna$ for some fixed integer $a$. When $a=1$ this is ordinary multiplication, and when $a=0$ it’s trivial.
Suppose that $\otimes$ is such an operation. Then for any integer $n$ we have $$n\otimes 0=n\otimes(0+0)=(n\otimes 0)+(n\otimes 0)\,;$$ the only integer satisfying $x+x=x$ is $0$, so $n\otimes 0=0$ for every $n\in\mathbb{Z}$. Now let $a=1\otimes 1$. Then $$1\otimes 2=1\otimes(1+1)=(1\otimes 1)+(1\otimes 1)=a+a=2a\,,$$ and an easy induction shows that $1\otimes n=na$ for every positive integer $n$. We already know that $1\otimes 0=0$, so in fact $1\otimes n=na$ for every integer $n\ge 0$.
Now suppose that $n$ is a negative integer. Then $$0=1\otimes 0=1\otimes\big(n+(-n)\big)=(1\otimes n)+\big(1\otimes(-n)\big)=(1\otimes n)-na\,,\tag{1}$$ so $1\otimes n=na$, and we’ve now shown that $1\otimes n=na$ for every $n\in\mathbb{Z}$.
We can now repeat the argument to generalize the operand $1$ to any integer:
$$2\otimes n=(1+1)\otimes n=(1\otimes n)+(1\otimes n)=2an\,,$$ and another easy induction gives us $m\otimes n=mna$ for every non-negative integer $m$ and every integer $n$. Finally, the trick that I used in $(1)$ can be used again to show that $m\otimes n=mna$ for all $m,n\in\mathbb{Z}$.
Added: Because $\otimes$ is just scaled multiplication, it’s certainly both commutative and associative: $$m\otimes n=amn=anm=n\otimes m$$ and $$(k\otimes m)\otimes n=a(akm)n=ak(amn)=k\otimes(m\otimes n)$$ for any $k,m,n\in\mathbb{Z}$.
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How can I turn this into a function with conventional symbols? – soandos Dec 8 '11 at 6:40
@soandos: For each $a\in\mathbb{Z}$ you have a function $f_a$ defined by $f_a(m,n)=amn$. – Brian M. Scott Dec 8 '11 at 6:42
So thats just multiplication? – soandos Dec 8 '11 at 6:58
@soandos: It’s multiplication with an extra scaling factor. For instance, the operation $f_3$ gives you $f_3(2,5)=30$, $f_3(-4,7)=-84$, and so on. The operation $f_1$ is ordinary multiplication. – Brian M. Scott Dec 8 '11 at 7:09
@Brian It would be helpful to add a remark about associativity, which is completely absent above. – Bill Dubuque Dec 8 '11 at 15:19
The answer to your question is essentially no: every non-degenerate associative binary operation which distributes over $+$ is multiplication in some (possibly non-commutative) rng. Let us abstract the situation a little: so suppose we have an abelian group $A$, whose operation we write as $+$. As with the addition of numbers, we have a zero element, commutativity, associativity, and negative elements.
Suppose we have a binary operation $\odot$ which distributes over $+$: in terms of equations, we have
\begin{align} a \odot (b + c) & = a \odot b + a \odot c \newline (a + b) \odot c & = a \odot c + b \odot c \end{align}
The first equation tells us that $a \odot -$ is a group homomorphism $A \to A$, and the second equation tells us that the map $a \mapsto (a \odot -)$ is a group homomorphism $A \to \textrm{End}(A)$. But $\textrm{End}(A)$ is naturally a (non-commutative unital) ring, and if we have the non-degeneracy condition
$$a \odot b = 0 \text{ for all } b \text{ in } A \implies a = 0$$
then every element $a$ is mapped to a unique endomorphism $a \odot -$ in $\textrm{End}(A)$. This allows us to identify $A$ with a subset $A'$ of $\textrm{End}(A)$. The associativity condition
$$(a \odot b) \odot c = a \odot (b \odot c)$$
means we can identify $\odot$ with composition of endomorphisms, and this means that $A'$ is a subrng of $\textrm{End}(A)$. If we demand that $\odot$ have a right unit, then the non-degeneracy condition is automatically satisfied; on the other hand, if we demand that $\odot$ have a left unit and be non-degenerate, then $A'$ is even a subring of $\textrm{End}(A)$, and the left unit is also a right unit.
- | 2015-07-05T17:44:00 | {
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https://math.stackexchange.com/questions/1319611/whats-the-parametric-equation-for-the-plane-through-a-point-x-y-z-perpendicul | What's the parametric equation for the plane through a point (x,y,z) perpendicular to (a,b,c)?
Find the parametric vector and Cartesian equations for the following planes:
a. The plane thru point $(2,1,-2)$ perpendicular to vector $(-1,1,2)$.
b. The plane thru the three points $(2,2,-2)$, $(-1,1,2)$ and $(2,3,1)$.
• What did you try, which level are you, what are you suppose to know already? Please give information in order for us to help you UNDERSTAND, and not give you answers that you will just copy/paste and not gain anything by it. – Martigan Jun 10 '15 at 7:24
• I'm in Year 3 Calculus. I tried to plug the numbers into an equation given by my lecturer where x(t) = a + lamda (b-a) +mu (c-a) and tried to modify it since for (a) I was only given the point and vector. Am I using the wrong equation? I would like to know the equations you think need to be used. – user239904 Jun 10 '15 at 7:32
• Welcome to math.SE! As a heads up, this site is not a solution manual, instead it emphasizes collaboration and human interaction. :) – Alp Uzman Jun 10 '15 at 7:32
Hint:
Let $P=(x,y,z)$ a generic point on the plane and $P_0=(x_0,y_0,z_0)$ the point through it passes, than a vector parallel to the plane is $P-P_0=(x-x_0,y-y_0,z-z_0)^T$. This vector is orthogonal to a ''normal'' $\vec n=(a_n,a_n,a_n)^T$ if the dot product between them is null: $$(P-P_0) \cdot \vec n=0 \iff (x-x_0,y-y_0,z-z_0)(x_n,y_n,z_n)^T=0$$ Calculate the dot product and you have the equation of the plane.
For a plane through three given points you can simultaneously solve the three equation obtained by the general equation of a plane $ax+by+cz+d=0$ whan substitute the coordinate of the three points or $(x,y,z)$.
You can see here.
• Hello. Thank you for your help. Will this equation produce a vector output? My professor gave us the answer for (a): (2,1,-2) +s(-2,2,2) +t(3,3,0). First, all the equations I've tried to use do not give me vector outputs like the answer. I thought that because the vector in (a) was perpendicular to the point, it would equal n⃗ but I don't understand the steps from that point. – user239904 Jun 10 '15 at 7:39
Fiest be aware that you can define a plan by different means.
One of them is through parametric equations such as
$(1)$ $x=a_1+\lambda u_1 +\mu v_1$, $y=a_2+\lambda u_2 +\mu v_2$ and $y=a_3+\lambda u_3 +\mu v_3$ with $(a_1,a_2,a_3)$ a point of the plane and $(u_1,u_2,u_3)$, $(v_1,v_2,v_3)$ two non colinear vectors.
An other one is through a cartesian equation such as
$(2)$ $ax+by+cz+d=0$
For your question $(a)$, your professor gave you a parametric equation, which is different from the statement of your problem (give the cartesian equation).
In any case, let's find a parametric equation for the problem $(a)$.
Since you already have a point of the plane $(2,1,-2)$, you just need to find two non-colinear vectors of the plane. Each of them should be perpendicular to the normal vector, that is, the dot product should be $0$.
One of the vector could be very easily $(1,1,0)$, whom dot product with $(1,-1,2)$ is clearly $0$.
The second one could be $(0,2,1)$. Since it is not colinear with $(1,1,0)$, this works.
Hence an equation $(2,1,-2)+\lambda (1,1,0)+ \mu (0,2,1)$
This is different from the answer of your professor, but this is OK, there are an infinity of possible parametric equations for the same plan. Bear in mind that a method to find the last vector, instead of finding our one fitting one, could be to find a vector which is normal to both the vector normal to the plane and the one you already found, with the vector cross product. It would be in that case $(1,-1,2) \times (1,1,0)$. It is probably what your professor did.
In the second case, you can take the first point $A$ as your defining point, and the vectors $AB$ and $AC$ as vectors for the parametrization of the plane.
Or you can also (for cartesian equation) find out the three unknowns by resolving this set of equations:
$ax_1+by_1+cz_1+1=0$, $ax_2+by_2+cz_2+1=0$, $ax_3+by_3+cz_3+1=0$ | 2019-11-14T21:15:39 | {
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https://math.stackexchange.com/questions/1523282/proving-that-the-existence-of-strongly-inaccessible-cardinals-is-independent-fro | # Proving that the existence of strongly inaccessible cardinals is independent from ZFC?
ZFC can't prove that strongly inaccessible cardinals exist, or else it would prove that a model of itself exists and hence $Con(ZFC)$. So this leaves us with two options:
1. ZFC proves there are no strongly inaccessible cardinals
2. The existence of strongly inaccessible cardinals is independent from ZFC
I've heard, however, that you can't actually prove that it's #2. That is, you can't prove the independence of large cardinal axioms from ZFC. Why is this?
At first, I thought it was because proving independence would mean that ZFC proves the consistency of a larger theory that contains ZFC as a model. However, I don't think my reasoning was right. That wouldn't prove that a model exists - just that a model could exist. And isn't it proven that Con(ZFC) is independent from ZFC, and hence that ZFC + Con(ZFC) is consistent iff ZFC is?
So how do you show that you can't show the independence of large cardinal axioms?
• You don't necessarily have to talk about models. Working with weak arithmetical theories that suffice to develop an internal formalization of first-order logic are usually enough. Then you don't have models, but you can talk about a proof of contradiction or the lack thereof. – Asaf Karagila Nov 10 '15 at 23:21
Well. In order to establish "true independence" you need to show that $\sf ZFC$ is consistent with both the statement and its negation, at least assuming that $\sf ZFC$ is consistent.
However just assuming that $\sf ZFC$ is consistent is not enough to establish that $\sf ZFC$+large cardinal axioms are consistent, because from large cardinal axioms we can prove the consistency of $\sf ZFC$; thus a proof of consistency from $\sf ZFC$ of large cardinal axioms will ultimately prove the consistency of $\sf ZFC$, and by the second incompleteness theorem this is impossible.
So you can only prove that $\sf ZFC$ does not prove the existence of large cardinals, not that it does not refute them.
And indeed every now and then you can find people claiming to have proofs that inaccessible cardinals are inconsistent with $\sf ZFC$.
(The situation is similar with $\sf ZFC+\operatorname{Con}(ZFC)$, which is also strictly stronger than just $\sf ZFC$ in terms of consistency. But I don't think anyone who seriously considers $\sf ZFC$ a reasonable theory thinks it is inconsistent, so this assumption gets questioned far less often than inaccessible cardinals.)
Caveat lector:
Relative consistency results about $\sf ZFC$ and assumptions stronger than $\sf ZFC$ itself will depend a lot on the meta-theory. If you work in a theory which assumes $\sf\operatorname{Con}(ZFC)+\lnot\operatorname{Con}(ZFC+LC)$, then $\sf ZFC$ will refute large cardinals in that meta-theory. Because there will be a "natural number" encoding a proof of contradiction in a way recognizable by $\sf ZFC$.
The point is that when you want to talk about the ability to refute something stronger than $\sf ZFC$ you need to ask yourself what sort of meta-theory you have. If inaccessible cardinals are refutable from $\sf ZFC$, that's great (or actually the other thing, sucky); if not, then the question if they are refutable becomes meta-theory dependent.
• I think you're restating the same thing I said in different terms. For large cardinals to be independent, you'd have to prove that you can't prove their nonexistence. My understanding is that you can prove that you can't prove that you can't prove their nonexistence. Right? Why is that? (I think I worded that right...) – Mike Battaglia Nov 10 '15 at 23:26
• No, I'm saying that you need to show it is consistent they exist and that it is consistent they don't. The latter is easy; the former is impossible, because you can't start just from $\sf ZFC$ and prove the consistency of $\sf ZFC+LC$. So $\sf ZFC$ cannot do the former, therefore it is impossible to establish independence. Only unprovability. And who knows, maybe inaccessible cardinals are inconsistent with $\sf ZFC$. – Asaf Karagila Nov 10 '15 at 23:28
• OK, then in those terms, what I'm asking is: why can't ZFC prove the consistency of ZFC + LC? – Mike Battaglia Nov 10 '15 at 23:29
• Because from the consistency of ZFC+LC you can prove the consistency of ZFC. So if ZFC could do it, it would prove its own consistency and thus be inconsistent, subjected to the second incompleteness theorem. – Asaf Karagila Nov 10 '15 at 23:30
• Well. Yes. But there is a huge caveat here about your metatheory. I've already went to bed, so this will have to wait a couple of hours until morning finds me. Sorry for that. But in a naive nutshell, this is the argument, yes. – Asaf Karagila Nov 10 '15 at 23:51 | 2021-08-01T17:48:51 | {
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https://stats.stackexchange.com/questions/64938/if-linear-regression-is-related-to-pearsons-correlation-are-there-any-regressi | # If linear regression is related to Pearson's correlation, are there any regression techniques related to Kendall's and Spearman's correlations?
Maybe this question is naive, but:
If linear regression is closely related to Pearson's correlation coefficient, are there any regression techniques closely related to Kendall's and Spearman's correlation coefficients?
• As a simple example where you have one explanatory and a dependent variable: A linear regression of the ranks of $x$ and $y$ would yield Spearman's correlation coefficient as regression coefficient. And in this case, $x$ and $y$ are interchangeable in the regression. Jul 20, 2013 at 9:28
• Just a few thoughts. Kendall's $\tau$ and Spearman's $\rho$ are both correlation coefficients based on ranks. The sought after relationship between $x$ and $y$ would then need to involve their ranks. However, computing the ranks introduces dependence between the observations, which in turn imposes dependence between the error terms, eliminating linear regression. However, in a different setting, modeling the dependence structure between $x$ and $y$ with copulas would make a link with Kendall's $\tau$ and/or Spearman's $\rho$ possible, depending on the choice of copula. Jul 20, 2013 at 10:35
• @QuantIbex does that dependence necessarily imply $E[\varepsilon_i\varepsilon_j]\neq 0$? Jul 31, 2014 at 3:07
There's a very straightforward means by which to use almost any correlation measure to fit linear regressions, and which reproduces least squares when you use the Pearson correlation.
Consider that if the slope of a relationship is $\beta$, the correlation between $y-\beta x$ and $x$ should be expected to be $0$.
Indeed, if it were anything other than $0$, there'd be some uncaptured linear relationship - which is what the correlation measure would be picking up.
We might therefore estimate the slope by finding the slope, $\tilde{\beta}$ that makes the sample correlation between $y-\tilde{\beta} x$ and $x$ be $0$. In many cases -- e.g. when using rank-based measures -- the correlation will be a step-function of the value of the slope estimate, so there may be an interval where it's zero. In that case we normally define the sample estimate to be the center of the interval. Often the step function jumps from above zero to below zero at some point, and in that case the estimate is at the jump point.
This definition works, for example, with all manner of rank based and robust correlations. It can also be used to obtain an interval for the slope (in the usual manner - by finding the slopes that mark the border between just significant correlations and just insignificant correlations).
This only defines the slope, of course; once the slope is estimated, the intercept can be based on a suitable location estimate computed on the residuals $y-\tilde{\beta}x$. With the rank-based correlations the median is a common choice, but there are many other suitable choices.
Here's the correlation plotted against the slope for the car data in R:
The Pearson correlation crosses 0 at the least squares slope, 3.932
The Kendall correlation crosses 0 at the Theil-Sen slope, 3.667
The Spearman correlation crosses 0 giving a "Spearman-line" slope of 3.714
Those are the three slope estimates for our example. Now we need intercepts. For simplicity I'll just use the mean residual for the first intercept and the median for the other two (it doesn't matter very much in this case):
intercept
Pearson: -17.573 *
Kendall: -15.667
Spearman: -16.285
*(the small difference from least squares is due to rounding error in the slope estimate; no doubt there's similar rounding error in the other estimates)
The corresponding fitted lines (using the same color scheme as above) are:
Edit: By comparison, the quadrant-correlation slope is 3.333
Both the Kendall correlation and Spearman correlation slopes are substantially more robust to influential outliers than least squares. See here for a dramatic example in the case of the Kendall.
• (+1) Great explanation! Is there any reason why Kendall seems to be more preferred to Spearman in this context (at least judging from the fact that Kendall correlation corresponds to a slope estimator that has a name, Theil-Sen, whereas Spearman one does not)? Jul 31, 2014 at 10:45
• There are a number of reasons why this seems to be the case. First is that the Theil-Sen line has a simply-described estimator (median of the pairwise slopes), which the Spearman lacks; in small samples it's very suitable for hand calculation. The Kendall correlation approaches normality faster and is more mathematically tractable. See also here and here. Jul 31, 2014 at 11:35
The proportional odds (PO) model generalizes Wilcoxon and Kruskal-Wallis tests. Spearman's correlation when $X$ is binary is the Wilcoxon test statistic simply translated. So you could say that the PO model is a unifying method. Since the PO model can have as many intercepts as there are unique values of $Y$ (less one), it handles both ordinal and continuous $Y$.
The numerator of the score $\chi^2$ statistic in the PO model is exactly the Wilcoxon statistic.
The PO model is a special case of a more general family of cumulative probability (some call cumulative link) models including the probit, proportional hazards, and complementary log-log models. For a case study see Chapter 15 of my Handouts.
Aaron Han (1987 in econometrics) proposed the Maximum Rank Correlation estimator that fits regression models by maximizing tau. Dougherty and Thomas (2012 in the psychology literature) recently proposed a very similar algorithm. There is an abundance of work on the MRC illustrating its properties.
Aaron K. Han, Non-parametric analysis of a generalized regression model: The maximum rank correlation estimator, Journal of Econometrics, Volume 35, Issues 2–3, July 1987, Pages 303-316, ISSN 0304-4076, http://dx.doi.org/10.1016/0304-4076(87)90030-3. (http://www.sciencedirect.com/science/article/pii/0304407687900303)
Dougherty, M. R., & Thomas, R. P. (2012). Robust decision making in a nonlinear world. Psychological review, 119 (2), 321. Retrieved from http://damlab.umd.edu/pdf%20articles/DoughertyThomas2012Rev.pdf. | 2022-05-25T01:37:57 | {
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http://math.stackexchange.com/questions/736525/computing-lim-limits-n-to-infty11-n2n | # Computing $\lim\limits_{n\to\infty}(1+1/n^2)^n$
Why is $\lim\limits_{n\to\infty}(1+\frac{1}{n^2})^n = 1$?
Could someone elaborate on this? I know that $\lim\limits_{n\to\infty}(1+\frac{1}{n})^n = e$.
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Write out the binomial expansion. You'll see why. – Frank Apr 2 '14 at 12:54
Try the substitution $n = \frac{1}{h}$ with $h \to 0$. And then exponentiate the expression i.e. $f(x) = e^{\ln f(x)}$. – Mussé Redi Apr 2 '14 at 13:00
Estimate the terms by $e^{\frac 1n}$ which is greater than each term, and has limit $1$. It is clear that the individual terms are greater than $1$. Squeeze. – Mark Bennet Apr 2 '14 at 13:01
I know that $\lim_{n\to\infty}(1+\frac{1}{n})^n = e$
Indeed. A consequence of this statement that you know, is that there exists some finite $K\gt1$ such that, for every positive $n$, $$1\lt\left(1+\frac1n\right)^n\lt K,$$ (It happens that the optimal upper bound is $K=\mathrm e$, but, to solve your problem, one can forget such a refinement.) Hence, $$1\lt\left(1+\frac1{n^2}\right)^n=\left(\left(1+\frac1{n^2}\right)^{n^2}\right)^{1/n}\lt K^{1/n}.$$ Since $K^{1/n}\to1$ irrespectively of the value of $K$, this proves that $$\lim_{n\to\infty}\left(1+\frac1{n^2}\right)^n=1.$$
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+1 Well explained! – Sawarnik Apr 2 '14 at 16:06
Hint: prove that $$\log (1+u) \le u$$
then use the continuity of $\exp$.
details:
$$\left(1+\frac 1{n^2} \right)^n =\exp \left(n \log\left(1+\frac 1{n^2} \right)\right) \\ 0\le n \log\left(1+\frac 1{n^2}\right) \le \frac 1n \\ 1\le \left(1+\frac 1{n^2} \right)^n \le\exp\frac 1n\to 1$$
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How do justify $0\leq n\log\left(1 + 1/n^2\right)\leq 1/n$? – rookie Apr 2 '14 at 16:58
because $\log(1+u)\le u$, see the hint. – mookid Apr 2 '14 at 16:59
As you know that $$\lim_{n\to \infty} (1+\frac{1}{n})^n=e,$$ you also have $$\lim_{n\to \infty} (1+\frac{1}{n^2})^{n^2}=e,$$ which yields $$\lim_{n\to \infty} (1+\frac{1}{n^2})^{n}=\lim_{n\to \infty} \left((1+\frac{1}{n^2})^{n^2}\right)^{1/n}=\lim_{n\to \infty} e^{1/n}=1.$$ EDIT: The last two equalities should be seen together. It works because the interior of the parenthesis converges towards $e$ and you take the $n$-th root of some number closer and closer to $e$. When $n$ goes to infinity, this is the same as the value of the limit of $e^{1/n}$.
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I think this not rigorous. What is the justification for $$\lim_{n\to \infty} \left((1+\frac{1}{n^2})^{n^2}\right)^{1/n}=\lim_{n\to \infty} e^{1/n}$$ – Amr Apr 2 '14 at 13:10
I agree that it is a bit "unclassical". I added an edit. – Jérémy Blanc Apr 2 '14 at 13:25
A bit shaky. Look at Did's answer for a more precise proof... – mookid Apr 2 '14 at 14:37
There are some good answers here, but let me share a method that is more computational (in that it doesn't require noticing certain inequalities and using the squeeze theorem), and may help you with other similar limits.
First, one convenient way of taking exponents out of a limit expression is to take a logarithm:
$$\lim_{n \rightarrow \infty} f(n)^{g(n)} = \exp\left[\lim_{n \rightarrow \infty} g(n)\ln(f(n))\right]$$
Provided that $f(n) > 0$ (and either limit exists). In your case, this becomes:
$$\lim_{n \rightarrow \infty} (1 + \frac{1}{n^2})^{n} = \exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right]$$
To evaluate the limit inside the brackets on the right, you can use l'Hospital's rule. You need to change the indeterminate form to $0/0$ and replace $n$ with a 'continuous variable' $x$:
$$\lim_{n \rightarrow \infty} n \ln(1 + \frac{1}{n^2}) = \lim_{x \rightarrow \infty} \frac{\ln(1 + \frac{1}{x^2})}{\frac{1}{x}} = \lim_{x \rightarrow \infty} \frac{(-\frac{2}{x^3})/(1 + \frac{1}{x^2})}{-\frac{1}{x^2}} = \lim_{x \rightarrow \infty} \frac{-\frac{2}{x^3}}{-\frac{1}{x^2}(1 + \frac{1}{x^2})}$$
To simplify this last fraction, multiply the numerator and denominator by $-x^4$:
$$= \lim_{x \rightarrow \infty} \frac{2x}{x^2 + 1} = 0$$
$$\exp\left[\lim_{n \rightarrow \infty} n\ln(1 + \frac{1}{n^2})\right] = \exp[0] = 1$$
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In the first two identities there should be noted that $\exp$ is a continuous function and thus we can pass the limit to the exponent. – Mussé Redi Apr 2 '14 at 18:13
It's really not necessary to know anything about the subtler limit, $e=\lim_{n\to\infty}\left(1+{1\over n}\right)^n$. A binomial expansion and some crude estimates are sufficient here. Note first that
$${n\choose k}\left({1\over n}\right)^k={n\over n}\cdot{n-1\over n}\cdot\cdots\cdot{1\over n}\le1$$
and therefore
$$\left(1+{1\over n^2}\right)^n=\sum_{k=0}^n{n\choose k}\left({1\over n^2}\right)^k=\sum_{k=0}^n{n\choose k}\left({1\over n}\right)^k{1\over n^k}\le\sum_{k=0}^n{1\over n^k}\le1+{1\over n}+{1\over n^2}(n-2)$$
The Squeeze Theorem takes over from here.
- | 2016-02-06T18:53:32 | {
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https://mathematica.stackexchange.com/questions/104014/fitting-data-in-log-log-scale | # Fitting data in Log-Log scale
I want to fit my data from a txt file to a curve in log-log scale. What I tried to do was this:
data = Import["Path", "Table"][[All, {2, 1}]];
FindFit[Log[data], Log[1 - Gamma[A, B/x]/Gamma[A]], {A, B}, x]
When I plot this, this is not the scale I get from ListLogLogPlot. How do I fit data on log-log scale to a curve in log-log scale? The curve I want to fit is $f(x)=1-\Gamma[a,b/x]/\Gamma[a]$ (Data and everything is imported correctly - I just want to change the scale on which I'm fitting)
• Are you aware that fitting the curve in log-space will generally yield a different fit? Also, assumptions about distribution of errors (e.g, that they are normally distributed) will often be violated. Jan 13, 2016 at 23:01
• Yes, I'm aware of that. That's why I want to do it. (My data has a lot of points near the origin of axis and I'm mostly interested in the points far away). Jan 13, 2016 at 23:03
data = Table[{x, 1 - Gamma[1, 2/x]/Gamma[1] + Random[]/10}, {x, 1, 10, .1}];
res = FindFit[data, 1 - Gamma[A, B/x]/Gamma[A], {A, B}, x];
(* {A -> 0.913848, B -> 2.06033} *)
Show[
ListLogLogPlot[data],
LogLogPlot[1 - Gamma[A, B/x]/Gamma[A] /. res // Evaluate, {x, 1, 10}],
Frame -> True
]
You can fit in log-log space, but then don't forget to scale x in your gamma expression as Exp[x], because you scaled it as Log[x] in the input:
res2 = FindFit[Log[data], Log[1 - Gamma[A, B/Exp[x]]/Gamma[A]], {A, B}, x]
(* {A -> 0.923818, B -> 2.09093} *)
Show[
ListLogLogPlot[data],
LogLogPlot[1 - Gamma[A, B/x]/Gamma[A] /. res // Evaluate, {x, 1, 10}],
LogLogPlot[1 - Gamma[A, B/x]/Gamma[A] /. res2 // Evaluate, {x, 1, 10},
PlotStyle -> {Red, Dashed}],
Frame -> True
]
The two fits are very similar, but if you inspect the fitted parameters you'll see they differ slightly. As I said in the comments above, fitting in the log domain is not the same as in the linear domain.
• I don't think that's what I want to do. Doesn't your code fit my curve to a "linear" data and just shows log-log plot of it? (Meant the first one) Jan 13, 2016 at 23:20
• @Caims What about the second one? I wasn't too sure what you actually wanted. Jan 13, 2016 at 23:30
• It is perfect, Thank you so much! That's what I wanted. Jan 13, 2016 at 23:33 | 2022-05-16T16:23:59 | {
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http://math.stackexchange.com/questions/14231/4-mathbbz-isomorphic-to-5-mathbbz | # $(4\mathbb{Z}, +)$ isomorphic to $(5\mathbb{Z}, +)$
The question asks me to determine if $4\mathbb{Z}$ and $5\mathbb{Z}$ (with standard addition) are isomorphic and if so to give the isomorphism.
My attempts: What I am having difficulty with is showing a mapping that preserves the operation. i.e., $\phi(a+b) = \phi(a) + \phi(b)$.
What I have so far is:
$\phi(a+b) = \phi(5a/4) + \phi(5b/4)$.
$\phi(a+b) = 5/4(a+b) = 5/4a + 5/4b = \phi(a) + \phi(b)$.
Therefore, my conclusion is $4\mathbb{Z}$ and $5\mathbb{Z}$ are isomorphic and the isomorphism is $\phi(n) = \frac{5}{4}n$.
Can someone either confirm this or point me in the right direction? I really appreciate everyone willingness to help one another! Thank you!
-
Seems fine to me. Since $n$ is divisible by 4, $\frac{5}{4}n\in 5\mathbb{Z}$. You just need to show injectivity and surjectivity. – Joe Johnson 126 Dec 14 '10 at 0:11
Also, there is only one infinite cyclic group. Each group you listed is an infinite cyclic group. – Sean Tilson Dec 14 '10 at 0:41
@Sean Tilson: "There is only one infinite cyclic group" up to isomorphism. – Arturo Magidin Dec 14 '10 at 2:53
@ Arturo:but of course... (and that isomorphism need not be unique!) (good catch though) @user: you should take my mistake to mean that cook people don't really worry about two things being different when they happen to be isomorphic. Unless they are super cool, then they worry a lot about the choices of isomorphisms. – Sean Tilson Dec 14 '10 at 5:16
Bill's argument is of course the simplest, but to discuss your effort...
Two comments: your argument only shows that $\phi$ is a homomorphism; you haven't shown it is an isomorphism (one-to-one and onto). This is a major thing: you cannot conclude it is an isomorphism just form knowing it is a homomorphism. You need to show it is invertible, or show it is one-to-one and show it is onto.
Second comment is more of a nit-pick: you should not say that $\phi$ is the isomorphism; there may be more than one isomorphism (in fact, there is; you can also take $\psi(4n) = -5n$). You should say (once you establish it, of course), that $\phi$ is an isomorphism.
-
To add to Arturo's answer: both groups are cyclic. Any homomorphism from a cyclic group is uniquely determined by what it does to a generator. Now, to be onto a cyclic group, you need to hit a generator of that group. A little bit of thought will tell you that what I just said implies that $\pm\phi$ are the only two isomorphisms there can be. – Alex B. Dec 14 '10 at 4:12
HINT $\rm\ \ \ \mathbb Z\ \cong n\ \mathbb Z\$ via $\rm\ x\to\ n\ x\$ for $\rm\: n\ne 0\:.\$ Thus $\rm\ n\ \mathbb Z\ \cong\ \mathbb Z\ \cong\: n'\ \mathbb Z\$ by transitivity of $\rm\: \cong\:$ | 2014-07-13T04:57:01 | {
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https://math.meta.stackexchange.com/questions/26841/labeling-simultaneous-equations/26842 | # Labeling Simultaneous Equations?
I'm aware that you can label a single line with \tag:
$$2x+3y=5\tag{1}$$
But can you do this with simultaneous equations? For example, each of the following in:
$$\begin{cases} 2x+3y=5 \\ x-y=19 \end{cases}$$
One option for you: You can use the "align" environment. The symbol, in the first case, that I'll align is =. To do that, for every occurrence of $=$, precede it with & separate lines with \\ or \\ \\ as desired, but at the end of each line to be, prior to \\, you can use \tag{*} to label each and every line.
For example, given the following code:
\begin{align} 2x+3y &= 5 \tag{1} \\ x-y &=19\tag{2} \end{align}
We get \begin{align} 2x+3y &= 5 \tag{1} \\ x-y&=19\tag{2} \end{align}
In order to left align the equations, we move the & to the beginning of each equation, as follows:
\begin{align} & 2x+3y = 5 \tag{1} \\ & x-y =19\tag{2} \end{align}
\begin{align} & 2x+3y = 5 \tag{1} \\ & x-y =19\tag{2} \end{align}
There are so many neat things you can do with formatting. I've learned many of those neat things on this site. Look around at other posts; e.g., search systems of equations, and look at the answers; when you see something you'd like to replicate, hover over it, right-click, click "Show Math As: and then click "Tex Commands"
• Thank you for taking the time and answering. I would still like to have the left bracket symbolic of simultaneous equations however... is there a way to do that? – AmateurMathPirate Aug 21 '17 at 22:08
• Try this: \begin{cases} 2x+3y=5 & (1) \\ x-y=19 & (2)\end{cases} to get \begin{cases} 2x+3y=5 & (1) \\ x-y=19 & (2)\end{cases} – amWhy Aug 21 '17 at 22:14
• That's an improvisation. You can use & to describe or label the the equation. – amWhy Aug 21 '17 at 22:15
• Yep that's the only solution I've come up with myself. It gets a little hazy if the two equations don't have exactly the same line length, but it is what it is.... – AmateurMathPirate Aug 21 '17 at 22:16
• Note, though, that using & (1) and &(2) the labels align. You can get more space by ending each equation with & & (1), etc. – amWhy Aug 21 '17 at 22:20
• Indeed that seems to be the best solution. Nothing so nice as using the tag though. – AmateurMathPirate Aug 21 '17 at 22:22
• :O Never knew you could use multiple ampersands like that. – Simply Beautiful Art Aug 21 '17 at 23:42
\begin{align} A + B & = C \\ D + E & = F \\ G + H & = I \\ J + K & = L \end{align} \tag 1
Here is the code for what appears above.
\begin{align} A + B & = C \\ D + E & = F \\ G + H & = I \\ J + K & = L \end{align} \tag 1
If you put \tag 1 before rather than after \begin{align} then the number is on the right after the first line.
$$\begin{cases} 2x+3y=5 \\ x-y=19 \end{cases} \tag 1$$
• I think the asker is looking for a way to number each equation in the set of simultaneous equations. So given your example, we have both equations identified as (1) ? – amWhy Aug 23 '17 at 16:37
• @amWhy : The poster does use the word "each" and I was aware that that was probably the immediate concern. But $(1)$ several people had already dealt with that and $(2)$ that's not substantially different from what is done with just one line: you merely put a tag on each line separately, and $(3)$ it is sometimes worth knowing about what I point out in my posting. – Michael Hardy Aug 23 '17 at 20:08
• I agree that what you say is worth knowing. Please understand, I did not down-vote your answer. – amWhy Aug 23 '17 at 20:11 | 2020-10-22T18:32:42 | {
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https://www.ntminfo.org/4dmapx5g/find-transpose-of-a-matrix-02237e | # find transpose of a matrix
The program below then computes the transpose of the matrix and prints it on A transpose of a matrix is a new matrix in which the rows of the original are the columns now and vice versa. Calculate the transpose of the matrix. That is, if $$P$$ =$$[p_{ij}]_{m×n}$$ and $$Q$$ =$$[q_{ij}]_{r×s}$$ are two matrices such that$$P$$ = $$Q$$, then: Let us now go back to our original matrices A and B. So, we can observe that $$(P+Q)'$$ = $$P’+Q'$$. Transpose of a matrix A is defined as - A T ij = A ji; Where 1 ≤ i ≤ m and 1 ≤ j ≤ n. Logic to find transpose of a matrix. To calculate the transpose of a matrix, simply interchange the rows and columns of the matrix i.e. Join our newsletter for the latest updates. Transpose of a matrix is given by interchanging of rows and columns. How to calculate the transpose of a Matrix? But before starting the program, let's first understand, how to find the transpose of any matrix. That’s because their order is not the same. If a matrix is multiplied by a constant and its transpose is taken, then the matrix obtained is equal to transpose of original matrix multiplied by that constant. © Parewa Labs Pvt. Then we are going to convert rows into columns and columns into rows (also called Transpose of a Matrix in C). We can clearly observe from here that (AB)’≠A’B’. For Square Matrix : The below program finds transpose of A[][] and stores the result in B[][], … Ltd. All rights reserved. Some properties of transpose of a matrix are given below: If we take transpose of transpose matrix, the matrix obtained is equal to the original matrix. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. the screen. Free matrix transpose calculator - calculate matrix transpose step-by-step This website uses cookies to ensure you get the best experience. Recommended: Please solve it on “ PRACTICE ” first, before moving on to the solution. M <-matrix(1:6, nrow = 2) JAVA program to find transpose of a matrix. C++ Programming Server Side Programming. I already defined A. Let’s say you have original matrix something like - x = [ … Your email address will not be published. Now, there is an important observation. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. Python Basics Video Course now on Youtube! rows and columns. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. write the elements of the rows as columns and write the elements of a column as rows. The transpose of a matrix can be defined as an operator which can switch the rows and column indices of a matrix i.e. Then $$N’ = \begin{bmatrix} 22 &85 & 7 \\ -21 & 31 & -12 \\ -99 & -2\sqrt{3} & 57 \end{bmatrix}$$, Now, $$(N’)'$$ = $$\begin{bmatrix} 22 & -21 & -99 \\ 85 & 31 & -2\sqrt{3} \\ 7 & -12 & 57 \end{bmatrix}$$. Transpose of the matrix B1 is obtained as B2 by inserting… Read More » Transpose of a matrix is the process of swapping the rows to columns. Transpose of an addition of two matrices A and B obtained will be exactly equal to the sum of transpose of individual matrix A and B. and $$Q$$ = $$\begin{bmatrix} 1 & -29 & -8 \\ 2 & 0 & 3 \\ 17 & 15 & 4 \end{bmatrix}$$, $$P + Q$$ = $$\begin{bmatrix} 2+1 & -3-29 & 8-8 \\ 21+2 & 6+0 & -6+3 \\ 4+17 & -33+15 & 19+4 \end{bmatrix}$$= $$\begin{bmatrix} 3 & -32 & 0 \\ 23 & 6 & -3 \\ 21 & -18 & 23 \end{bmatrix}$$, $$(P+Q)'$$ = $$\begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & -18 \\ 0 & -3 & 23 \end{bmatrix}$$, $$P’+Q'$$ = $$\begin{bmatrix} 2 & 21 & 4 \\ -3 & 6 & -33 \\ 8 & -6 & 19 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 17 \\ -29 & 0 & 15 \\ -8 & 3 & 4 \end{bmatrix}$$ = $$\begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & -18 \\ 0 & -3 & 23 \end{bmatrix}$$ = $$(P+Q)'$$. To understand this example, you should have the knowledge of the following C++ programming topics: To learn other concepts related to matrices, download BYJU’S-The Learning App and discover the fun in learning. The following statement generalizes transpose of a matrix: If $$A$$ = $$[a_{ij}]_{m×n}$$, then $$A'$$ =$$[a_{ij}]_{n×m}$$. The transpose of a matrix is defined as a matrix formed my interchanging all rows with their corresponding column and vice versa of previous matrix. A new matrix is obtained the following way: each [i, j] element of the new matrix gets the value of the [j, i] element of the original one. One thing to notice here, if elements of A and B are listed, they are the same in number and each element which is there in A is there in B too. Transpose of a matrix is obtained by changing rows to columns and columns to rows. write the elements of the rows as columns and write the elements of a column as rows. What basically happens, is that any element of A, i.e. A matrix P is said to be equal to matrix Q if their orders are the same and each corresponding element of P is equal to that of Q. To find the transpose of a matrix, we will swap a row with corresponding columns, like first row will become first column of transpose matrix and vice versa. Hence, for a matrix A. Transpose of a Matrix can be performed in two ways: Finding the transpose by using the t() function. The transpose of a matrix can be defined as an operator which can switch the rows and column indices of a matrix i.e. For Square Matrix : The below program finds transpose of A[][] and stores the result in B[][], we can change N for different dimension. Submitted by IncludeHelp, on May 08, 2020 . The following is a C program to find the transpose of a matrix: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 2… The transpose of matrix A is represented by $$A'$$ or $$A^T$$. Enter a matrix. the orders of the two matrices must be same. Initialize a 2D array to work as matrix. For example if you transpose a 'n' x 'm' size matrix you'll get a … So let's say I have the matrix. The addition property of transpose is that the sum of two transpose matrices will be equal to the sum of the transpose of individual matrices. A transpose of a matrix is simply a flipped version of the original matrix. Add Two Matrices Using Multi-dimensional Arrays, Multiply two Matrices by Passing Matrix to a Function, Multiply Two Matrices Using Multi-dimensional Arrays. The transpose of a matrix is a new matrix whose rows are the columns of the original. HOW TO FIND THE TRANSPOSE OF A MATRIX Transpose of a matrix : The matrix which is obtained by interchanging the elements in rows and columns of the given matrix A is called transpose of A and is denoted by A T (read as A transpose). In Python, we can implement a matrix as a nested list (list inside a list). So, Your email address will not be published. $$B = \begin{bmatrix} 2 & -9 & 3\\ 13 & 11 & 17 \end{bmatrix}_{2 \times 3}$$. This program can also be used for a non square matrix. For the transposed matrix, we change the order of transposed to 3x2, i.e. r*c). $$a_{ij}$$ gets converted to $$a_{ji}$$ if transpose of A is taken. Find the transpose of that matrix. C++ Program to Find Transpose of a Matrix. So, taking transpose again, it gets converted to $$a_{ij}$$, which was the original matrix $$A$$. Find Largest Number Using Dynamic Memory Allocation, C Program Swap Numbers in Cyclic Order Using Call by Reference. In this program, the user is asked to enter the number of rows Let's do B now. The multiplication property of transpose is that the transpose of a product of two matrices will be equal to the product of the transpose of individual matrices in reverse order. Transpose is a new matrix formed by interchanging each the rows and columns with each other, we can see the geometrical meaning of this transformation as it will rotate orthogonality of the original matrix. Let's say I defined A. I'll try to color code it as best as I can. There can be many matrices which have exactly the same elements as A has. Consider the following example-Problem approach. Given a matrix, we have to find its transpose matrix. The number of rows in matrix A is greater than the number of columns, such a matrix is called a Vertical matrix. Input elements in matrix A from user. it flips a matrix over its diagonal. There are many types of matrices. Transpose. Before answering this, we should know how to decide the equality of the matrices. r and columns c. Their values should be less than 10 in Here is a matrix and its transpose: The superscript "T" means "transpose". Declare another matrix of same size as of A, to store transpose of matrix say B. Then, the user is asked to enter the elements of the matrix (of order r*c). The answer is no. C++ Program to Find Transpose of a Matrix C++ Program to Find Transpose of a Matrix This program takes a matrix of order r*c from the user and computes the transpose of the matrix. Let’s understand it by an example what if looks like after the transpose. So, let's start with the 2 by 2 case. Though they have the same set of elements, are they equal? But actually taking the transpose of an actual matrix, with actual numbers, shouldn't be too difficult. Thus, there are a total of 6 elements. That is, $$(kA)'$$ = $$kA'$$, where k is a constant, $$\begin{bmatrix} 2k & 11k \\ 8k & -15k \\ 9k &-13k \end{bmatrix}_{2×3}$$, $$kP'$$= $$k \begin{bmatrix} 2 & 11 \\ 8 & -15 \\ 9 & -13 \end{bmatrix}_{2×3}$$ = $$\begin{bmatrix} 2k & 11k \\ 8k & -15k \\ 9k &-13k \end{bmatrix}_{2×3}$$ = $$(kP)'$$, Transpose of the product of two matrices is equal to the product of transpose of the two matrices in reverse order. We can transpose a matrix by switching its rows with its columns. In this C++ tutorial, we will see how to find the transpose of a matrix, before going through the program, lets understand what is the transpose of By using this website, you agree to our Cookie Policy. If order of A is m x n then order of A T is n x m. link brightness_4 code # R program for Transpose of a Matrix # create a matrix with 2 rows # using matrix() method . Here, we are going to implement a Kotlin program to find the transpose matrix of a given matrix. For example, consider the following 3 X 2 matrix: 1 2 3 4 5 6 Transpose of the matrix: 1 3 5 2 4 6 When we transpose a matrix, its order changes, but for a square matrix, it remains the same. To calculate the transpose of a matrix, simply interchange the rows and columns of the matrix i.e. C Program to Find Transpose of a Matrix - In this article, you will learn and get code on finding the transpose of given matrix by user at run-time using a C program. this program. The transpose of a matrix is a new matrix that is obtained by exchanging the rows and columns. edit close. This JAVA program is to find transpose of a matrix. The horizontal array is known as rows and the vertical array are known as Columns. Solution- Given a matrix of the order 4×3. $$M^T = \begin{bmatrix} 2 & 13 & 3 & 4 \\ -9 & 11 & 6 & 13\\ 3 & -17 & 15 & 1 \end{bmatrix}$$. Those were properties of matrix transpose which are used to prove several theorems related to matrices. it flips a matrix over its diagonal. For example, for a 2 x 2 matrix, the transpose of a matrix{1,2,3,4} will be equal to transpose{1,3,2,4}. Transpose of a matrix is obtained by interchanging rows and columns. So, is A = B? Here you will get C program to find transpose of a sparse matrix. filter_none. The number of columns in matrix B is greater than the number of rows. Transpose of a matrix can be calculated by switching the rows with columns. Store values in it. To understand this example, you should have the knowledge of the following C programming topics: The transpose of a matrix is a new matrix that is obtained by exchanging the That is, $$A×B$$ = $$\begin{bmatrix} 44 & 18 \\ 5 & 4 \end{bmatrix} \Rightarrow (AB)’ = \begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix}$$, $$B’A'$$ = $$\begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix}$$, = $$\begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix}$$ = $$(AB)'$$, $$A’B'$$ = $$\begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 40 & 9 \\ 26 & 8 \end{bmatrix}$$. Required fields are marked *, $$N = \begin{bmatrix} 22 & -21 & -99 \\ 85 & 31 & -2\sqrt{3} \\ 7 & -12 & 57 \end{bmatrix}$$, $$N’ = \begin{bmatrix} 22 &85 & 7 \\ -21 & 31 & -12 \\ -99 & -2\sqrt{3} & 57 \end{bmatrix}$$, $$\begin{bmatrix} 22 & -21 & -99 \\ 85 & 31 & -2\sqrt{3} \\ 7 & -12 & 57 \end{bmatrix}$$, $$\begin{bmatrix} 2 & -3 & 8 \\ 21 & 6 & -6 \\ 4 & -33 & 19 \end{bmatrix}$$, $$\begin{bmatrix} 1 & -29 & -8 \\ 2 & 0 & 3 \\ 17 & 15 & 4 \end{bmatrix}$$, $$\begin{bmatrix} 2+1 & -3-29 & 8-8 \\ 21+2 & 6+0 & -6+3 \\ 4+17 & -33+15 & 19+4 \end{bmatrix}$$, $$\begin{bmatrix} 3 & -32 & 0 \\ 23 & 6 & -3 \\ 21 & -18 & 23 \end{bmatrix}$$, $$\begin{bmatrix} 3 & 23 & 21 \\ -32 & 6 & -18 \\ 0 & -3 & 23 \end{bmatrix}$$, $$\begin{bmatrix} 2 & 21 & 4 \\ -3 & 6 & -33 \\ 8 & -6 & 19 \end{bmatrix} + \begin{bmatrix} 1 & 2 & 17 \\ -29 & 0 & 15 \\ -8 & 3 & 4 \end{bmatrix}$$, $$\begin{bmatrix} 2 & 8 & 9 \\ 11 & -15 & -13 \end{bmatrix}_{2×3}$$, $$k \begin{bmatrix} 2 & 11 \\ 8 & -15 \\ 9 & -13 \end{bmatrix}_{2×3}$$, $$\begin{bmatrix} 9 & 8 \\ 2 & -3 \end{bmatrix}$$, $$\begin{bmatrix} 4 & 2 \\ 1 & 0 \end{bmatrix}$$, $$\begin{bmatrix} 44 & 18 \\ 5 & 4 \end{bmatrix} \Rightarrow (AB)’ = \begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix}$$, $$\begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} \begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix}$$, $$\begin{bmatrix} 44 & 5 \\ 18 & 4 \end{bmatrix}$$, $$\begin{bmatrix} 9 & 2 \\ 8 & -3 \end{bmatrix} \begin{bmatrix} 4 & 1 \\ 2 & 0 \end{bmatrix} = \begin{bmatrix} 40 & 9 \\ 26 & 8 \end{bmatrix}$$. How to Transpose a Matrix: 11 Steps (with Pictures) - wikiHow Transpose of a matrix is obtained by changing rows to columns and columns to rows. 1 2 1 3 —-> transpose In this program, the user is asked to enter the number of rows r and columns c. Their values should be less than 10 in this program. Dimension also changes to the opposite. Transpose of a matrix is obtained by changing rows to columns and columns to rows. Transpose of a Matrix Description Calculate the transpose of a matrix. Definition. To obtain it, we interchange rows and columns of the matrix. A matrix is a rectangular array of numbers that is arranged in the form of rows and columns. For 2x3 matrix, Matrix a11 a12 a13 a21 a22 a23 Transposed Matrix a11 a21 a12 a22 a13 a23 Example: Program to Find Transpose of a Matrix Transpose a matrix means we’re turning its columns into its rows. A matrix is a rectangular array of numbers or functions arranged in a fixed number of rows and columns. Thus Transpose of a Matrix is defined as “A Matrix which is formed by turning all the rows of a given matrix into columns and vice-versa.”, Example- Find the transpose of the given matrix, $$M = \begin{bmatrix} 2 & -9 & 3 \\ 13 & 11 & -17 \\ 3 & 6 & 15 \\ 4 & 13 & 1 \end{bmatrix}$$. The transpose of matrix A is written A T. The i th row, j th column element of matrix A is the j th row, i th column element of A T. In another way, we can say that element in the i, j position gets put in the j, i position. Okay, But what is transpose! Transpose of a matrix: Transpose of a matrix can be found by interchanging rows with the column that is, rows of the original matrix will become columns of the new matrix. play_arrow. For example X = [[1, 2], [4, 5], [3, 6]] would represent a 3x2 matrix. Program to find the transpose of a given matrix Explanation. Transpose of a Matrix in C Programming example This transpose of a matrix in C program allows the user to enter the number of rows and columns of a Two Dimensional Array. To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. So, we have transpose = int[column][row] The transpose of the matrix is calculated by simply swapping columns to rows: transpose[j][i] = matrix[i][j] Here's the equivalent Java code: Java Program to Find transpose of a matrix Below is the step by step descriptive logic to find transpose of a matrix. 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The above matrix A is of order 3 × 2. (This makes the columns of the new matrix the rows of the original). Watch Now. In this program, we need to find the transpose of the given matrix and print the resulting matrix. Commands Used LinearAlgebra[Transpose] See Also LinearAlgebra , Matrix … Here, the number of rows and columns in A is equal to number of columns and rows in B respectively. We can treat each element as a row of the matrix. row = 3 and column = 2. Such a matrix is called a Horizontal matrix. The first row can be selected as X[0].And, the element in the first-row first column can be selected as X[0][0].. Transpose of a matrix is the interchanging of rows and columns. Thus, the matrix B is known as the Transpose of the matrix A. So. The algorithm of matrix transpose is pretty simple. $$A = \begin{bmatrix} 2 & 13\\ -9 & 11\\ 3 & 17 \end{bmatrix}_{3 \times 2}$$. To transpose matrix in C++ Programming language, you have to first ask to the user to enter the matrix and replace row by column and column by row to transpose that matrix, then display the transpose of the matrix on the screen. Let us consider a matrix to understand more about them. Transpose of a matrix in C language: This C program prints transpose of a matrix. Then, the user is asked to enter the elements of the matrix (of order | 2021-03-02T10:22:02 | {
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https://math.stackexchange.com/questions/1242759/when-does-the-probability-of-dealing-2-cards-that-ive-already-dealt-in-a-previo | # When does the probability of dealing 2 cards that I've already dealt in a previous shuffle reach 50%
I play poker and create poker software. So questions about card shuffling and dealing interest me. I'm trying to understand the likelihood of receiving the same hand twice over a number of shuffles.
Here's how my hand is created: Take a freshly shuffled deck of cards. Deal the first two cards (more specifically, the first two cards whose faces I see.) Consider order to not be important. There are (52 x 51) / 2 possibilities, that is 1,326 possibilities. That suggests to me that it is reasonably likely that in any online poker session (which can involve playing hundreds of hands), I'll receive the same two cards twice.
How many hands would I need to play for the probability to see a hand I've already had to be at least 50%?
(Perhaps to be answered in the comments: how can I educate myself to answer such questions in the future? Is there a good book or online resource?)
This is the birthday problem.
Instead of determining the probability of (at least) a duplicate deal, because that "at least" raises difficulties in computation, let us instead determine the probability $q_k$ that no duplicate arises in $k$ deals.
$q_1 = 1 = 1326/1326$, obviously, because there can be no duplicate in a single deal. In two deals, a duplicate occurs if the second deal duplicates the first. As you've determined there are $(52)(51)/2 = 1326$ deals, so that probability that a duplicate does not occur is $q_2 = q_1(1325/1326) = (1326/1326)(1325/1326)$.
For the third deal not to be a duplicate, it has to avoid both the first two deals, which happens with probability $1324/1326$, so $q_3 = q_2(1324/1326) = (1326/1326)(1325/1326)(1324/1326)$. And for the fourth deal not to be a duplicate, it has to avoid all of the first three deals, so $q_4 = q_3(1323/1326) = (1326/1326)(1325/1326)(1324/1326)(1323/1326)$. It should be clear by now that
$$q_k = \frac{1326}{1326} \cdot \frac{1325}{1326} \cdot \frac{1324}{1326} \cdot \cdots \cdot \frac{1326-k+2}{1326} \cdot \frac{1326-k+1}{1326} = \frac{1326!}{(1326-k)!1326^k}$$
and is terminated by $q_{1327} = 0$, since by the pigeonhole principle, a duplicate is certain to occur by the $1327$th deal. (By convention, we can write $(-1)! = \Gamma(0) = +\infty$, so in some sense, the expression above is applicable even to $k = 1327$.)
The sequence $q_k$ drops much faster than one generally thinks (if one hasn't encountered the birthday problem previously), and drops to below $1/2$ (that is, attains a probability greater than $1/2$ that a duplicate has occurred) at about $k \doteq 1/2 + \sqrt{2N\ln 2}$, where $N$ is the number of choices ($365$ in the case of the birthday problem, and $1326$ here). That expression yields a critical point of about $k \doteq 1/2 + \sqrt{(2)(1326)(0.69315)} \doteq 43.375$; hence, the $43$ provided by David in his answer.
• Thank you. I have encountered the birthday problem, but didn't work out myself that this is the same problem. – Steve McLeod Apr 24 '15 at 5:52
Looks like about $43$ hands before the prob of a repeat of the first $2$ cards reaches at least $50$% but it depends what you mean by previous. For example, if you draw cards A and B initially, are you then only looking for the next occurrence of A and B and want to know when it is $50$% or greater chance of getting A and B as the first 2 cards or do you mean if you get A,B on the first hand, then $2$ different cards, C and D on the 2nd hand, then you are now looking for either an A,B pair or a C,D pair...?
• I mean, getting any hand that has previously appeared. So yes, A,B or C,D or E,F or... – Steve McLeod Apr 20 '15 at 3:05 | 2019-09-17T19:34:05 | {
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https://la.mathworks.com/help/symbolic/functionalderivative.html | Documentation
# functionalDerivative
Functional derivative
## Syntax
``D = functionalDerivative(f,y)``
## Description
example
````D = functionalDerivative(f,y)` returns the Functional Derivative of the functional $F=\int f\left(x,y\left(x\right),y\text{'}\left(x\right)...\right)dx$ with respect to the function y = y(x), where x represents one or more independent variables. If `y` is a vector of symbolic functions, `functionalDerivative` returns a vector of functional derivatives with respect to the functions in `y`, where all functions in `y` must depend on the same independent variables.```
## Examples
### Find Functional Derivative
Find the functional derivative of the function given by $f\left(y\right)=y\left(x\right)\mathrm{sin}\left(y\left(x\right)\right)$ with respect to the function `y`.
```syms y(x) f = y*sin(y); D = functionalDerivative(f,y)```
```D(x) = sin(y(x)) + cos(y(x))*y(x)```
### Find Functional Derivative of Vector of Functionals
Find the functional derivative of the function given by $H\left(u,v\right)={u}^{2}\frac{dv}{dx}+v\frac{{d}^{2}u}{d{x}^{2}}$ with respect to the functions `u` and `v`.
```syms u(x) v(x) H = u^2*diff(v,x)+v*diff(u,x,x); D = functionalDerivative(H,[u v])```
```D(x) = 2*u(x)*diff(v(x), x) + diff(v(x), x, x) diff(u(x), x, x) - 2*u(x)*diff(u(x), x)```
`functionalDerivative` returns a vector of symbolic functions containing the functional derivatives of `H` with respect to `u` and `v`, respectively.
### Find Euler-Lagrange Equation for Spring
First find the Lagrangian for a spring with mass `m` and spring constant `k`, and then derive the Euler-Lagrange equation. The Lagrangian is the difference of kinetic energy `T` and potential energy `V` which are functions of the displacement `x(t)`.
```syms m k x(t) T = sym(1)/2*m*diff(x,t)^2; V = sym(1)/2*k*x^2; L = T - V```
```L(t) = (m*diff(x(t), t)^2)/2 - (k*x(t)^2)/2```
Find the Euler-Lagrange equation by finding the functional derivative of `L` with respect to `x`, and equate it to `0`.
`eqn = functionalDerivative(L,x) == 0`
```eqn(t) = - m*diff(x(t), t, t) - k*x(t) == 0```
`diff(x(t), t, t)` is the acceleration. The equation `eqn` represents the expected differential equation that describes spring motion.
Solve `eqn` using `dsolve`. Obtain the expected form of the solution by assuming mass `m` and spring constant `k` are positive.
```assume(m,'positive') assume(k,'positive') xSol = dsolve(eqn,x(0) == 0)```
```xSol = -C3*sin((k^(1/2)*t)/m^(1/2))```
Clear assumptions for further calculations.
`assume([k m],'clear')`
### Find Differential Equation for Brachistochrone Problem
The Brachistochrone problem is to find the quickest path of descent under gravity. The time for a body to move along a curve `y(x)` under gravity is given by
`$f=\sqrt{\frac{1+y{\text{'}}^{2}}{2gy}},$`
where g is the acceleration due to gravity.
Find the quickest path by minimizing `f` with respect to the path `y`. The condition for a minimum is
`$\frac{\delta f}{\delta y}=0.$`
Compute this condition to obtain the differential equation that describes the Brachistochrone problem. Use `simplify` to simplify the solution to its expected form.
```syms g y(x) assume(g,'positive') f = sqrt((1+diff(y)^2)/(2*g*y)); eqn = functionalDerivative(f,y) == 0; eqn = simplify(eqn)```
```eqn(x) = diff(y(x), x)^2 + 2*y(x)*diff(y(x), x, x) == -1```
This equation is the standard differential equation for the Brachistochrone problem.
### Find Minimal Surface in 3-D Space
If the function u(x,y) describes a surface in 3-D space, then the surface area is found by the functional
`$F\left(u\right)=\iint f\left(x,y,u,{u}_{x},{u}_{y}\right)dxdy=\iint \sqrt{1+{u}_{x}^{2}+{u}_{y}^{2}}dxdy,$`
where ux and uy are the partial derivatives of u with respect to x and y.
Find the equation that describes the minimal surface for a 3-D surface described by the function `u(x,y)` by finding the functional derivative of `f` with respect to `u`.
```syms u(x,y) f = sqrt(1 + diff(u,x)^2 + diff(u,y)^2); D = functionalDerivative(f,u)```
```D(x, y) = -(diff(u(x, y), y)^2*diff(u(x, y), x, x)... + diff(u(x, y), x)^2*diff(u(x, y), y, y)... - 2*diff(u(x, y), x)*diff(u(x, y), y)*diff(u(x, y), x, y)... + diff(u(x, y), x, x)... + diff(u(x, y), y, y))/(diff(u(x, y), x)^2... + diff(u(x, y), y)^2 + 1)^(3/2) ```
The solutions to this equation `D` describe minimal surfaces in 3-D space such as soap bubbles.
## Input Arguments
collapse all
Expression to find functional derivative of, specified as a symbolic variable, function, or expression. The argument `f` represents the density of the functional.
Differentiation function, specified as a symbolic function or a vector, matrix, or multidimensional array of symbolic functions. The argument `y` can be a function of one or more independent variables. If `y` is a vector of symbolic functions, `functionalDerivative` returns a vector of functional derivatives with respect to the functions in `y`, where all functions in `y` must depend on the same independent variables.
## Output Arguments
collapse all
Functional derivative, returned as a symbolic function or a vector of symbolic functions. If input `y` is a vector, then `D` is a vector.
collapse all
### Functional Derivative
Consider functionals
`$F\left(y\right)=\underset{\Omega }{\int }f\left(x,y\left(x\right),y\text{'}\left(x\right),y\text{'}\text{'}\left(x\right),...\right)dx,$`
where Ω is a region in x-space.
For a small change in the value of y, δy, the change in the functional F is
The expression $\frac{\delta f\left(x\right)}{\delta y}$ is the functional derivative of f with respect to y. | 2020-01-22T23:48:10 | {
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https://www.physicsforums.com/threads/limit-of-function-with-two-variables.851119/ | # Limit of function with two variables
1. Jan 6, 2016
### Austin
I have a question for determining the limit of a function with two variables. My textbook says that the limit (x,y)->(0,0) of 4xy^2/(x^2+y^2)=0. This is true if we evaluate the limit if it approaches along the x-axis (y=0) or the y-axis (x=0) or any line on the plane y=kx. I am wondering if this is sufficient to prove the limit=0 if we only need approach with lines.
If for example we let y=x^(1/3) then the limit does not equal zero.
I am just starting multivariable calculus so the idea of multivariable limits is new to me, so I am not sure if the direction we choose has to be a straight line or if it can be along any path like y=x^(1/3)
THanks
2. Jan 6, 2016
### Khashishi
Most functions you deal with at your level will be analytic functions. https://en.wikipedia.org/wiki/Analytic_function
Analytic functions are continuous and give the same value for the limit no matter what direction you approach the limit from. Moreover, the derivatives exist and don't depend on the direction. But there are also discontinuous functions which can have different limits depending on direction. An example is the step function.
Oops I should have read your post more carefully before responding. You claim that the limit is not 0 along the curve y=x^(1/3) but can you show what limit you get?
Last edited: Jan 6, 2016
3. Jan 7, 2016
### Samy_A
$\displaystyle \lim_{(x,y) \rightarrow 0} {f(x,y)}=0$ means that the limit should be 0 no matter how (x,y) approaches (0,0). You are right about that.
However, the textbook is right that the the given function $f(x,y)=\frac{4xy²}{x²+y²}$ approaches 0 as (x,y) approaches (0,0).
You can deduce that from the following inequality:
$|f(x,y)|=|\frac{4xy²}{x²+y²}|\leq4\frac{|x|y²}{y²}=4|x|$
Last edited: Jan 7, 2016
4. Jan 7, 2016
### Silicon Waffle
Concerning the inequality as a starting point, you can also go with something like this,
$\forall x,y\in\mathbb R$, $(x-y)^2≥0$ always holds true.
That means
$x^2+y^2≥2xy$
Or
$\frac{1}{x^2+y^2}\leq\frac{1}{2xy}$
Given that $\forall x,y\neq 0$
Now multiply both sides with $4xy^2$ to obtain
$\frac{4xy^2}{x^2+y^2}\leq\frac{4xy^2}{2xy}=2y$
So you can find the limit of your problem easier.
5. Jan 7, 2016
### Samy_A
Better use absolute values, because $\frac{1}{x^2+y^2}\leq\frac{1}{2xy}$ won't be true if one of x or y are negative.
6. Jan 7, 2016
### Silicon Waffle
Yes, that is correct. Thanks Samy_A.
7. Jan 7, 2016
### Staff: Mentor
I once had a thesis (computer science) in hand in which exactly this has been used to prove a central theorem of it.
8. Jan 7, 2016
### Samy_A
Ouch. Was it fixable, or did they have to add the condition that x and y are >0 to that central theorem?
9. Jan 7, 2016
### Staff: Mentor
I don't remember the outcome. I mentored a student whose task it was to elaborate the thesis. He found the error and it wasn't easily fixable. It was simply wrong. As we revealed it to the professor he simply replied: "So find another proof. The result is correct."
However, that was a bit beyond our scope. Examinations of algorithms in special computational classes tend to be very specific. Plus the author came from another place on this globe and had a different way to describe stuff.
10. Jan 8, 2016
### HallsofIvy
Staff Emeritus
Since $r= \sqrt{x^2+ y^2}$f measures the distance from (0, 0) to (x, y) no matter what the angle is, limits in two variables can often be done by converting to polar coordinates. Here, $\frac{4xy^2}{x^2+ y^2}= \frac{4(r cos(\theta))(r^2 sin^2(\theta))}{r^2}= \frac{r^3}{r^2}cos(\theta)sin^2(\theta)= 4 cos(\theta)sin^2(\theta)$. As r goes to 0, that goes to 0 for any $\theta$. | 2017-12-14T19:28:35 | {
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https://math.stackexchange.com/questions/2524878/help-me-understand-criteria-for-triangle-abc-sim-triangle-abc | # Help me understand criteria for $\triangle ABC \sim \triangle A'B'C'$
I have a hard time properly understanding the definitions of similarity between two triangles (symbolically denoted as $\triangle ABC \sim \triangle A'B'C'$). My professor defined similarity like this:
$$\text{If } a': a = b' : b = c': c \text{ then } \triangle ABC \sim \triangle A'B'C'.$$
This deduction is obtained via the study of homotetia. For a certain triangle $\triangle ABC$ in Euclidean plane, we can choose a point $O$ and connect it to points $A$, $B$ and $C$ with three rays originating in $O$ – we now have $OA$, $OB$ and $OC$. We then choose a point on one of those rays – thus, we get either $OA'$, $OB'$ or $OC'$. The quotient of the length of these line segments will help us to deduce our coefficient of "stretch", $k$. All in all, we can say that $|OA|:|OA'| = 1 : k$. It follows that $k = \frac{|OA'|}{|OA|} = \frac{|OB'|}{|OB|} = \frac{|OC'|}{|OC|}$, which in simpler notation reaffirms the claim $a': a = b' : b = c': c$, and also proves that all of these quotients are equal to the coefficient $k$ ($a': a = b' : b = c': c = k$). This deduction from my professor was very clear and I quickly grasped it. But then I found the following definition in my textbook:
\begin{align*} &\triangle ABC \sim \triangle A'B'C' \text{ if } a : b : c = a': b ': c'. \\ &\triangle ABC \sim \triangle A'B'C' \text{ if } (\alpha=\alpha')\wedge (\beta=\beta') \wedge (\gamma = \gamma').\\ \end{align*}
I don't have a problem when it comes to the second line of criteria (I understand the correlation between similarity and angles completely). I do however have a big problem with the first line. The textbook dropped it from nowhere, no explanation, no proof, nothing, unlike my professor, who led me through the process of understanding his definition of similarity. What I'd like to do is prove that $a': a = b' : b = c': c$ follows from $a : b : c = a': b ': c'$ and/or vice versa.
So, if possible, I'd prefer if you wrote down proof of the claim $a': a = b' : b = c': c \Longleftrightarrow a : b : c = a': b ': c'$. If not, I'll have to settle with a proof of $a': a = b' : b = c': c \Longrightarrow a : b : c = a': b ': c'$ or $a : b : c = a': b ': c' \Longrightarrow a': a = b' : b = c': c$.
The thing that really confuses me though is the inapplicability of the definition from the textbook in exercises. Here's an example of my claim.
TRANSLATION: In a square $ABCD$ we fold one of its sides in such a way so that point $B$ "maps" (please, ignore the clumsy terminology) to the midpoint of the line segment $CD$. Show that triangles $HGC$, $EPF$ and $EHD$ are similar and that the lenght of their sides are in relationship $3 : 4 : 5$.
The exercise is not particularly hard. We prove the similarity via proving that the triangles have two (and therefore three) angles of the same size. We get the $x$ relatively quickly by applying the Pythagorean theorem ($x=\frac{5}{8}$). With this value discovered, we can get $a = \frac{3}{8}$, $b = \frac{4}{8}$ and $c = \frac{5}{8}$. After this, we're practically finished. I applied the method my professor suggested and got the following values for the other two triangles:
\begin{align*} &\triangle EHD; \quad a' = \frac{3}{6} \quad b' = \frac{4}{6} \quad c' = \frac{5}{6} \\ &\triangle EPF; \quad a'' = \frac{3}{27} \quad b'' = \frac{4}{27} \quad c'' = \frac{5}{27}. \end{align*}
If we accept the definition and the premise of similarity as written in the textbook, then we need to show that the quotient $3 : 4 : 5$ is true for only one of the three similar triangles and we automatically know that the other two also have the quotient $3 : 4 : 5$. At least that is the implication of the dubious definition. But I wanted to make sure. This is my stab at it:
\begin{align*} a : b : c &= \frac{3}{8} : \frac{4}{8} : \frac{5}{8} = \frac{6}{5}, \\ a' : b' : c' &= \frac{3}{6} : \frac{4}{6} : \frac{5}{6} = \frac{9}{10}. \end{align*}
Proposition states: If two triangles are similar, then $a : b : c = a' : b' : c'$. But
$$\frac{6}{5} \neq \frac{9}{10} \Longrightarrow a : b : c \neq a' : b' : c'.$$
The same goes for other combinations of triangles (e.g. $a'' : b'' : c'' = a' : b' : c'$); no equality, only inequalities.
My second wish: please tell me why I get this result. It seems like the definition of similarity in my textbook is false.
There are two possibilities: either the proposition in my textbook is false or my understanding of the operand "$:$" (and consequently my calculations) is false. I am certain that the latter one is to blame for my confusion. It appears that I have a deep underlying issues of improper understanding of these definitions, and in the end, improper understanding of the concept of similarity as a whole.
Please help me. Explain everything in great detail, so I may finally truly understand. I am currently in high school and intend on studying pure mathematics in college. I want to get to the bottom of things, understand, not just pick up some pattern that I was taught and spew it on the test sheet. Without this sense of true understanding, all of my efforts are meaningless.
• Yes, you're definitely misunderstanding something. Did you go to school in a country where $:$ is a division symbol? – Henning Makholm Nov 17 '17 at 16:15
• Yes he is, they use it in the picture, see? – Aqua Nov 17 '17 at 16:17
• @HenningMakholm Yes. For example, $a : b$ is $\frac{a}{b}$. This shouldn't be a problem in the context of my textbook, as it cites the "additional" criteria (e.g. $a' : a = b' : b = c' : c$) also with "$:$" right alongside this "problematic" definition, and it works (my professor's definition, that is)! So I am truly confused... – Gregor Perčič Nov 17 '17 at 16:20
Your first question: Show that $a:a'=b:b'\implies a:b=a':b'$.
This is Euclid's proposition V.16. Your actual request is for a version of this involving $a,b$ and $c$, but it can be handled two at a time. We assume, with Euclid, that all quantities being put into ratios are strictly greater than $0$. With proportions, we have that the product of the means equals the product of the extremes, thus:
$$a:a'=b:b' \Leftrightarrow ab'=a'b \Leftrightarrow a:b=a':b'.$$
In more conventional notation,
$$\frac{a}{a'}=\frac{b}{b'} \Leftrightarrow ab'=a'b \Leftrightarrow \frac{a}{b}=\frac{a'}{b'},$$
where we can simply think of the moves as multiplication and division.
Now, that answer will only extend to the three-ratio case if we clear up what's going on with the notation $a:b:c$. When we write $a:b:c=d:e:f$, this is shorthand for the two proportions $a:b=d:e$ and $b:c=e:f$, which together imply that $a:c=d:f$. It does not represent the equation $(a/b)/c=(d/e)/f$. I think this is why that notation for ratios is not very common anymore.
Thus, when you calculate a numeric value for the expression $\frac38:\frac48:\frac58$, it is not meaningful. That expression is not intended to simplify to a single number.
Answering your initial question, with this notational clarification in mind, we can proceed as follows:
\begin{align} a:a'=b:b'=c:c' &\implies a:a'=b:b' \text{ and } b:b'=c:c' \text { (separating the compound equality)}\\ &\implies a:b=a':b' \text{ and } b:c=b':c' \text{ (alternating proportions by V.16)}\\ &\implies a:b:c=a':b':c' \text{ (definition of triplicate proportion)} \end{align}
• Thank you for your answer! Two things though: 1. I still do not understand how we can construct $a : b : c$ with "handling two at a time". 2. Could you please Show me how to get $a : b : c = d : e : f$ from $a : b = d : e$ and $b : c = e : f$. – Gregor Perčič Nov 17 '17 at 16:33
• That's a definition. $a:b:c=d:e:f$ literally just means $a:b=d:e$ and $b:c=e:f$. – G Tony Jacobs Nov 17 '17 at 16:34
• Wow. Could we represent this definition as well-ordered triples in classical number theory (as @HenningMakholm mentioned)? – Gregor Perčič Nov 17 '17 at 16:36
• @GTonyJacobs: Not complextly exactly, since your proposed unfolding would imply that $1:0:1=1:0:2$. – Henning Makholm Nov 17 '17 at 16:37
• @GregorPerčič: Except for the question of whether some of the quantities cam be zero (or negative), the concept Tony is explaining is the same one I am explaining, just from a slightly different focus. A minor nitpick though: you're talking about ordered triples here, not "well-ordered" which has a different (and completely unrelated), meaning. – Henning Makholm Nov 17 '17 at 16:45
You're misunderstanding the notation $$x:y:z = p:q:r$$ You've been taught that $:$ means division, so you think that this is an equality asserting that the results of the divisions $\frac{x/y}z$ and $\frac{p/q}r$ are the same. But that is not what it means!
Something like $x:y:z = p:q:r$ is a (nowadays not extremely common) notation for the claim that the triples $(x,y,z)$ and $(p,q,r)$ are "in proportion" to each other -- or less fancily, that the relation between $x$, $y$ and $z$ is the same as the relation between $p$, $q$ and $r$, in a particular sense.
There are various ways to make precise what this means; one that fits a modern number-centric understanding well is
By $x:y:z=p:q:r$ we mean that there is some number $k\ne 0$ such that $p=k\cdot x$ and $q=k\cdot y$ and $r=k\cdot z$.
It should be clear how this generalizes to different number of operands such as $x:y:z:w=p:q:r:s$, or indeed just $x:y=p:q$.
In particular if $x:y=p:q$ means that there is $k\ne 0$ such that $p=kx$ and $q=ky$, it is a simple matter of algebra to see that this is the case exactly when $\frac xy = \frac pq$ -- where the fractions are now truly dvisions -- except when $y$ and/or $q$ is zero. So in that case this usage is mostly compatible with writing $:$ for division. But this connection does not generalize neatly to more than two elements on each site.
The above defines $x:y:z=p:q:r$ as a single, monolithic relation between six quantities, notated with the compound symbol "$::=::$". Some people prefer to write it as $x:y:z\sim p:q:r$ instead, to emphasize that it shouldn't be read as an equality between two things notated $x:y:z$ and $p:q:r$.
But if we really want it to be an actual equality between $x:y:z$ and $p:q:r$, we could achieve it by defining
The notation $x:y:z$ denotes the set $\{(kx,ky,kz)\mid k\ne 0\}$.
And we can then prove that these sets are identical exactly if the monolithic definition of $x:y:s=p:q:r$ above is satisfied.
(But I don't think this definition of $x:y:z$ as a set is common. I have seen it a few times in the context of projective geometry, but not in elementary geometry or arithmetic).
• I think it was a common notation a long time ago, when everyone still learned geometry from Euclid's Elements. – G Tony Jacobs Nov 17 '17 at 16:46 | 2019-10-17T00:02:30 | {
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https://mathoverflow.net/questions/321751/can-all-countable-cat0-cube-complexes-be-isometrically-embedded-in-l1-mat | # Can all countable $CAT(0)$ cube complexes be isometrically embedded in $l^1(\mathbb{N},\mathbb{R})$?
In this paper (theorem 2), Chepoi & Hagen say
There exists an infinite $$CAT(0)$$ cube complex $$X$$ with constant maximum degree which cannot be isometrically embedded into a Cartesian product of a finite number of trees, i.e., the chromatic number of its crossing graph is infinite.
so that you can't hope to isometrically embed it in a $$\mathbb{R}^n$$ with $$l^1$$ metric. But what about $$l^1(\mathbb{N})$$:
Question 1: Under what conditions does a $$CAT(0)$$ cube complex, with the "polyhedral complex" metric induced by the $$l^1$$ metric on cubes, embed in $$l^1(\mathbb N)$$ ?
Question 2: Same question but only about embedding the $$0$$ skeleton of said complex.
My reason for the question is the following: It is known that the $$0$$-skeleton of a $$CAT(0)$$ cube complex is a median space, and intuitively it makes sense since it looks a lot like a nice subset of $$\mathbb{Z}^n$$ for some $$n$$ big enough, which is itself median. I'm therefore wondering if you could prove median-ness of this $$0$$-skeleton by embedding your complex in $$l^1(\mathbb{N})$$, and then arguing that the embedding is actually submedian, or whatever allows you to argue median-ness back for the skeleton.
EDIT: added the obvious "countable" condition. Then an obvious embedding would be to enumerate the vertices and send them to the "basis" of $$l^1$$ … I wonder if/where that would break.
• @YCor: I made the question explicit. I hope this is precise enough now. Jan 26, 2019 at 14:35
– YCor
Jan 26, 2019 at 15:58
All CAT(0) cube complexes $$C$$ with $$\ell^1$$-metric embed isometrically into $$\ell^1$$. If the set of vertices is countable, one can choose $$\ell^1$$ of a countable set.
Indeed, say that a subset $$B$$ of the vertex set $$V_C$$ of $$C$$ is (totally) convex if it contains vertices of all geodesic paths between any two elements of $$B$$. Call it $$B$$ "biconvex" if both $$B$$ and its complement in $$V_C$$ are convex. (It's usually called "halfspace" but I don't think it's a good choice of terminology.) Let $$\mathcal{B}$$ be the set of biconvex subset. For every oriented edge $$(x,y)$$, there exists a unique $$B=B_{x,y}\in\mathcal{B}$$ such that $$y\in B$$ and $$x\notin B$$ (namely $$B_{x,y}=\{z\in V_C:zy\le zx\}$$). It is known that all biconvex subsets have this form. This shows that the cardinal of $$\mathcal{B}$$ is bounded above by the cardinal of $$V_C$$ (if infinite).
An isometric embedding $$f$$ of $$C$$ into $$\ell^1(\mathcal{B})$$ consists in the following: for $$x\in C$$, let $$\mathcal{B}(x)$$ be the set of biconvex subsets containing $$x$$. Fix a vertex $$x_0$$; map $$x\in V_C$$ to $$f(x)=1_{\mathcal{B}(x)}-1_{\mathcal{B}(x_0)}$$. Here we endow $$\mathcal{B}$$ with the atomic measure for which singletons have measure $$1/2$$, so that $$f$$ is an isometric embedding: indeed for all $$x,y\in V_C$$, each of $$\mathcal{B}(x)\smallsetminus \mathcal{B}(y)$$ and $$\mathcal{B}(y)\smallsetminus \mathcal{B}(x)$$ have exactly $$xy$$ elements. It is not hard to check that $$f$$ has a canonical affine extension to cells.
• Mmh, that's a nice argument. I assume for vertices $x,y$, $xy$ means the distance from $x$ to $y$ ? Also, is that a standard kind of argument ? Jan 26, 2019 at 15:12
• Yes it's indeed standard. One reference is my survey arxiv.org/abs/1302.5982 : essentially Corollary 7C5 is the same. I'm not sure of an earliest reference: this was rather often done with unnecessary extra-assumptions such as finiteness assumptions on the complex, or equivariance with respect to a group action with some properties, etc.
– YCor
Jan 26, 2019 at 15:56
From the point of view of graph theory, it is well known that median graphs (i.e. one-skeleta of CAT(0) cube complexes) embeds isometrically in Hamming cubes. An approach I like uses restricted quotients, which have other nice applications.
Restricted quotients. Given a CAT(0) cube complex $$X$$ and a collection of hyperplanes $$\mathcal{J}$$, define the pseudo-metric $$d_\mathcal{J} : X^{(0)} \times X^{(0)} \to \mathbb{N}$$ as $$(x,y) \mapsto \text{number of hyperplanes in \mathcal{J} separating x,y.}$$ The restricted quotient $$X_\mathcal{J}$$ is the cube complex obtained by cubulating the wallspace $$(X, \mathcal{J})$$. Such a cube complex can be defined in many different ways, but the key point to keep in mind is that its vertex-set is the quotient of $$X^{(0)}$$ by the equivalence relation: $$x \sim_\mathcal{J} y$$ if $$d_\mathcal{J}(x,y)=0$$.
For instance, start from the dual graph $$\Gamma$$ of $$X \backslash \bigcup_{J\in \mathcal{J}} J$$, i.e. the graph whose vertices are the connected components of $$X \backslash \bigcup_{J\in \mathcal{J}} J$$ and whose edges link two components if they are separated by a unique hyperplane. Then $$X_\Gamma$$ coincides with the cube complex obtained from $$\Gamma$$ by filling in all the subgraphs isomorphic to one-skeleta of $$k$$-cubes with $$k$$-cubes for every $$k \geq 2$$.
There is clearly a surjective map $$\pi_\mathcal{J} : X \twoheadrightarrow X_\mathcal{J}$$ (obtained by sending each vertex of $$X$$ to the vertex of $$X_\mathcal{J}$$ corresponding to the component that contains $$x$$). Moreover, $$d_{X_\mathcal{J}}(\pi_\mathcal{J}(x),\pi_\mathcal{J}(y)) = d_\mathcal{J}(x,y) \text{ for all vertices x,y \in X.}$$ (Here, my cube complexes are endowed with $$\ell^1$$-metrics.)
Constructing embeddings. Because the distance in $$X$$ between any two vertices coincides with the number of hyperplanes that separate them, it follows that, for every partition $$\{ \mathcal{J}_i, \ i \in I\}$$ of the hyperplanes of $$X$$, then $$\bigoplus\limits_{i \in I} \pi_{\mathcal{J}_i} : X \to \bigoplus\limits_{i \in I} X_{\mathcal{J}_i}$$ is an isometric embedding of $$X$$ into a product of CAT(0) cube complexes (namely, the restricted quotients).
Application 1: If our partition is $$\{ \{J\}, \text{ J hyperplane}\}$$, then each restricted quotient is a single edge and one obtains an isometric embedding of $$X$$ into an infinite-dimensional cube $$[0,1]^I$$.
Application 2: Let $$\Delta X$$ denote the crossing graph of $$X$$, i.e. the graph whose vertices are the hyperplanes of $$X$$ and whose edges link two hyperplanes if they are transverse. Let $$\chi$$ denote the chromatic graph of $$\Delta X$$. By considering the partition of the hyperplanes of $$X$$ induced by a coloring of $$\Delta X$$ with $$\chi$$ colors, one obtains an isometric embedding of $$X$$ into a product of $$\chi$$ simplicial trees. | 2022-07-02T17:24:30 | {
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https://math.stackexchange.com/questions/3793142/number-of-ways-to-color-sectors-of-circle | # Number of ways to color sectors of circle.
A circular tabletop is divided into four congruent sectors by two diameters that are perpendicular to each other. Each sector is to be painted with one of four colors. How many distinct ways can the table be painted? (A color may be used on more than one sector, but paintings that are the same after a rotation are not considered distinct.)
I got the answer 58 through bashy casework, but I'm not sure if that answer is even correct. Is there a clever-er or quicker way to do it?
Thanks!
• Burnside's lemma seems to be the way to go here. Are reflections counted as distinct, or not? Edit: in context I suppose reflecting the table wouldn't make sense. Aug 16 '20 at 18:34
• What is burnside's lemma? Aug 16 '20 at 18:50
I'll put a full answer, assuming only rotations.
Burnside's lemma states that the number of ways to color an object with a group of symmetries $$G$$ is the average number of colorings fixed by each of the symmetries in $$G$$. That is, if $$X^g$$ is the number of colorings fixed by a group element $$g \in G$$, the total number of colorings under these symmetries is $$\frac{1}{|G|} \sum_{g\in G} X^g.$$
The group of rotations $$G = \langle r \rangle$$ on the table is a cyclic group of order $$4$$, consisting of the identity $$e$$ (which doesn't move anything), the $$90^{\circ}$$ rotation clockwise $$r$$, the $$180^{\circ}$$ rotation $$r^2$$, and the $$90^{\circ}$$ rotation counterclockwise $$r^3$$.
Let us represent each coloring with the colors $$c_1c_2c_3c_4$$.
The number of colorings fixed under $$e$$, the identity, is simply every possible one. There are four possibilities for each of $$c_1, c_2, c_3, c_4$$, giving us $$4^4 = 256$$.
Now, $$r$$ acts on the coloring $$c_1c_2c_3c_4$$ by turning it into $$c_2c_3c_4c_1$$. This coloring is fixed if $$c_1 = c_2 = c_3 = c_4$$. The colorings fixed by the rotation $$r$$ are thus precisely the colorings for which there is only one color, and there are thus clearly $$4$$ of these. By symmetry, this is the number of colorings fixed by $$r^3$$ as well.
Finally, $$r^2$$ turns $$c_1c_2c_3c_4$$ into $$c_3c_4c_1c_2$$. These two colorings are the same if $$c_1 = c_3, c_2 = c_4$$. Thus, the colorings fixed by $$r^2$$ are precisely the colorings with at most two alternating colors. These colorings are determined by $$c_1$$ and $$c_2$$, and so there are exactly $$4 \cdot 4 = 16$$ of these.
Thus, the number of colorings required is $$\frac{1}{4}(256 + 4 + 4 + 16) = 70$$.
• Could you direct me to a proof of Burnside's Lemma? Aug 16 '20 at 18:57
• Here's the usual Wikipedia article, but the language is quite group-theoretic. en.wikipedia.org/wiki/Burnside%27s_lemma Aug 16 '20 at 19:01
• 1/ I like to think of it, combinatorially, as a double-counting correction of sorts. This is a bit more obvious in the following example: suppose you have $n$ people to be seated at a circular table. Naively, you would count $n!$ ways to arrange them. Of course, if we're treating rotationally equivalent arrangements as the same, there's overcounting of sorts. To avoid this, fix one of these $n$ people as an anchor, seat them at any of $n$ possible seats. This tells us that the $n!$ is an overcount by a factor of $n$, and so there are $(n - 1)!$ seatings. Aug 16 '20 at 19:12
• 2/2 Now, do the calculation above, using Burnside's lemma with the group of rotations by $0, 1, 2, \dots, n - 1$ seats. Aug 16 '20 at 19:12
First, we have $$4$$ colourings with only one colour.
Then, we have $${4\choose 2}=6$$ colourings with two colours and a repeating pattern (e.g. black/white/black/white).
Out of $$4^4=256$$ colourings of the table distinguishing the orientation, the first $$4$$ colourings correspond to $$4$$ colourings. The other $$6$$ colourings correspond to $$6\times 2=12$$ colourings.
The remaining $$256-4-12=240$$ colourings are all asymmetrical, and so groups of $$4$$ of them correspond to a single colouring disregarding the orientation. Thus, the number of those colourings is $$240/4=60$$.
Putting it all together, we end up with $$60+4+6=70$$ colourings. | 2021-10-25T04:39:24 | {
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https://mathhelpboards.com/threads/the-set-of-all-integers-z-2.5739/ | # [SOLVED](The set of all integers) Z^2
#### katye333
##### New member
Hello all,
I'm having a lot of trouble when it comes to set notation.
For instance, what does (the set of all integers) $$\displaystyle Z^2$$ mean?
What values are contained in this set?
Sorry if I didn't use the MATH tags right.
#### Fantini
MHB Math Helper
Hello Katye! I assume you mean $\mathbb{Z}^2$. This is shorthand notation for $\mathbb{Z} \times \mathbb{Z}$, which is a cartesian product.
When we write $\mathbb{Z} \times \mathbb{Z}$ we mean a set with elements of the form $(a,b)$, where $a$ (or the first component) and $b$ belong to $\mathbb{Z}$. In general, whenever we have $A \times B$, with $A$ and $B$ sets, we write in set notation as follows:
$$A \times B = \{ (a,b) \in A \times B : a \in A, b \in B \}.$$
This is saying that the set $A \times B$ constructed from the sets $A$ and $B$ have elements denoted $(a,b)$ where the first component belongs to $A$ and the second belongs to $B$. Likely if you have more than three components: you'll always read each component as belonging to the corresponding set in order.
For example, try to tell in the following cases how are the elements in each set:
$$\mathbb{R} \times \mathbb{Z},$$ $$\mathbb{R} \times \mathbb{Q},$$ $$\mathbb{Z} \times \mathbb{Q} \times \mathbb{Z}.$$
Cheers!
#### katye333
##### New member
Hello Katye! I assume you mean $\mathbb{Z}^2$. This is shorthand notation for $\mathbb{Z} \times \mathbb{Z}$, which is a cartesian product.
When we write $\mathbb{Z} \times \mathbb{Z}$ we mean a set with elements of the form $(a,b)$, where $a$ (or the first component) and $b$ belong to $\mathbb{Z}$. In general, whenever we have $A \times B$, with $A$ and $B$ sets, we write in set notation as follows:
$$A \times B = \{ (a,b) \in A \times B : a \in A, b \in B \}.$$
This is saying that the set $A \times B$ constructed from the sets $A$ and $B$ have elements denoted $(a,b)$ where the first component belongs to $A$ and the second belongs to $B$. Likely if you have more than three components: you'll always read each component as belonging to the corresponding set in order.
For example, try to tell in the following cases how are the elements in each set:
$$\mathbb{R} \times \mathbb{Z},$$ $$\mathbb{R} \times \mathbb{Q},$$ $$\mathbb{Z} \times \mathbb{Q} \times \mathbb{Z}.$$
Cheers!
Thank you for responding!
So, for the first one of yours:
A would be $\mathbb{R}$ and B would be in $\mathbb{Z}$?
Now if I have a function defined as
Cheers! | 2021-01-16T05:35:48 | {
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https://math.stackexchange.com/questions/1649196/proof-verification-if-x-is-a-nonnegative-real-number-then-big-sqrtx-b | # Proof Verification: If $x$ is a nonnegative real number, then $\big[\sqrt{[x]}\big] = \big[\sqrt{x}\big]$
Let $x$ be a nonnegative real number and denote $[x]$ as the greatest integer less than or equal to $x$. We will attempt to prove that $\big[\sqrt{x}\big] = \big[\sqrt{[x]}\big]$.
First suppose that $x$ is a perfect square. Then the equation trivially holds.
Assuming that $x$ is not a perfect square, we have $\sqrt{x} \not\in \mathbb{Z}$.
If $[x] \leq x$, then it clear that $\sqrt{[x]} \leq \sqrt{x}$ where $\sqrt{[x]}$ may or may not be an integer.
Further, it is true that there are no integers in the interval $\big(\sqrt{[x]},\sqrt{x}\big)$; because if such an integer $q$ existed, we would deduce that
$$[x] < q^2 < x$$
which is a blatant contradiction since $q^2 \in \mathbb{Z}$.
Because there are no integers in $\big(\sqrt{[x]},\sqrt{x}\big)$, it follows that
$$\big[\sqrt{x}\big] \leq \sqrt{[x]} \leq \sqrt{x} \tag{1}$$
But since $\big[\sqrt{x}\big]$ is the closest integer to $\sqrt{x}$ that satisfies $(1)$, it follows that it must also be the closest integer to $\sqrt{[x]}$, that is to say, it is the greatest integer less than or equal to $\sqrt{[x]}$.
Therefore, $\big[\sqrt{x}\big] = \big[\sqrt{[x]}\big]$.
Are there any problems with the logic of the above proof? Thank you.
• What is $[x]$? – YoTengoUnLCD Feb 10 '16 at 15:29
• My apologies. It is the floor of $x$. – Jeremiah Dunivin Feb 10 '16 at 15:29
• If $[x]\leq x$ then you can only conclude that $\sqrt{[x]}\leq \sqrt{x}$. Don't confuse $\leq$ with $<$. – Thomas Andrews Feb 10 '16 at 15:38
• You're right. But even if I change the inequality to $\leq$, the proof still seems to hold. – Jeremiah Dunivin Feb 10 '16 at 15:45
(An alternate approach, which is really very close to your proof, but perhaps clearer.)
Let $n=\lfloor \sqrt{x}\rfloor$, so $$n\leq \sqrt{x}<n+1\implies \\n^2\leq x<(n+1)^2\implies\\n^2\leq \lfloor x\rfloor <(n+1)^2\implies \\ n\leq \sqrt{\lfloor x\rfloor}<n+1$$
The key trick is that $n^2\leq x$ means $n^2\leq \lfloor x\rfloor$, because $n^2$ is an integer.
To prove $\lfloor f(x)\rfloor = \lfloor g(x)\rfloor$, we merely need to show that:
$$\lfloor f(x)\rfloor \leq g(x)\text{ and }\\ \lfloor g(x)\rfloor \leq f(x)$$
(Worth figuring out why this is sufficient.) Often, one of these will be "obvious." For example, as in your case, you have $f(x)\leq g(x)$ for all $x$, so $\lfloor f(x)\rfloor \leq g(x)$.)
So the only real part you needed above was that $$n=\lfloor\sqrt{x}\rfloor\leq \sqrt{\lfloor x\rfloor}$$
• Nevermind, I see how $n$ is the greatest integer. I would hate to use your answer for my homework, so I was wondering if there are any other flaws besides the one that you mentioned in your comment above. – Jeremiah Dunivin Feb 10 '16 at 16:13
$x$ lies between two perfect squares
$$n^2\le x<(n+1)^2$$ and as $n^2$ is an integer
$$n^2\le\lfloor x\rfloor\le x<(n+1)^2.$$
Then taking the square root
$$n\le\sqrt{\lfloor x\rfloor}\le\sqrt x<n+1$$
which is another way to say
$$\lfloor \sqrt{\lfloor x\rfloor}\rfloor=\lfloor \sqrt x\rfloor.$$ | 2019-09-19T14:57:29 | {
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http://math.stackexchange.com/questions/291768/gcdm-n-1-and-gcd-mn-a-1-implies-a-cong-1-pmod-mn | # $\gcd(m,n) = 1$ and $\gcd (mn,a)=1$ implies $a \cong 1 \pmod{ mn}$
I have $m$ and $n$ which are relatively prime to one another and $a$ is relatively prime to $mn$ and after alot of tinkering with my problem i came to this equality: $a \cong 1 \pmod m \cong 1 \pmod n$ why is it safe to say that $a \cong 1 \pmod {mn}$?..
-
fyi: having a good title makes things more easily searchable in the future. Doing this lets us minimize duplicate efforts. why not alter the title to something more appropriate such as $\gcd(m,n) = 1$ and $\gcd (mn,a)=1$ implies $a \cong 1 (mod mn)$ – MSEoris Jan 31 '13 at 23:26
To add to what @MSEoris said, your title is a false statement, but the body of your post is true. – Rick Decker Feb 1 '13 at 1:13
A useful reference, with a wider scope, is en.wikipedia.org/wiki/Chinese_remainder_theorem – Andreas Caranti Feb 1 '13 at 7:30
It looks as if you are asking the following. Suppose that $m$ and $n$ are relatively prime. Show that if $a\equiv 1\pmod{m}$ and $a\equiv 1\pmod{n}$, then $a\equiv 1\pmod{mn}$.
So we know that $m$ divides $a-1$, and that $n$ divides $a-1$. We want to show that $mn$ divides $a-1$.
Let $a-1=mk$. Since $n$ divides $a-1$, it follows that $n$ divides $mk$. But $m$ and $n$ are relatively prime, and therefore $n$ divides $k$. so $k=nl$ for some $l$, and therefore $a-1=mnl$.
Remark: $1.$ There are many ways to show that if $m$ and $n$ are relatively prime, and $n$ divides $mk$, then $n$ divides $k$.
One of them is to use Bezout's Theorem: If $m$ and $n$ are relatively prime, there exist integers $x$ and $y$ such that $mx+ny=1$.
Multiply through by $k$. We get $mkx+nky=k$. By assumption, $n$ divides $mk$, so $n$ divides $mkx$. Clearly, $n$ divides $nky$. So $n$ divides $mkx+nky$, that is, $n$ divides $k$.
$2.$ Note that there was nothing special about $1$. Let $m$ and $n$ be relatively prime. If $a\equiv c\pmod{m}$ and $a\equiv c\pmod{n}$, then $a\equiv c\pmod{mn}$.
-
so easy! thanks I'm just really tired..keep on the good work! – Rachel Bernoulli Jan 31 '13 at 23:31
$\gcd(m,n)=1$ implies there is an $x$ and $y$ so that $$mx+ny=1\tag{1}$$ $a\equiv1\pmod{m}$ and $a\equiv1\pmod{n}$ imply there is a $j$ and $k$ so that $$a-1=jm=kn\tag{2}$$ multiplying $(1)$ by $j$ and using $(2)$ $$nkx+jny=j\tag{3}$$ Plugging $(3)$ back into $(2)$ yields $$a-1=jm=n(kx+jy)m\tag{4}$$ which implies that $a\equiv1\pmod{mn}$.
Idea of the Preceding Argument
The idea of $(3)$ is to show that since $\gcd(m,n)=1$, by $(1)$, and $jm$ is a multiple of $n$, by $(2)$, we have that $j$ is a multiple of $n$. Thus, $a-1=jm$ is a multiple of $mn$, by $(4)$.
-
+1, since I'm a Darboux fan. – Rick Decker Feb 1 '13 at 1:10
@Rick Probably you mean Bezout, not Darboux. In any case, you can see clearly the relation between the Bezout and gcd proofs in my answer. – Math Gems Feb 1 '13 at 1:52
@MathGems. Gaah! Of course you're right. Bezout $\ne$ Darboux. (slinks away in embarassment) – Rick Decker Feb 1 '13 at 1:54
@RickDecker: they both have 'b' and 'oo' sounds :-) – robjohn Feb 1 '13 at 2:49
@Rob Darn Bezoukas! – Math Gems Feb 1 '13 at 3:14
Hint $\rm\,\ \ m,n\,|\,a\!-\!1\!\iff\! lcm(m,n)\,|\,a\!-\!1\ \$ [proof below]
And, further, recall that $\rm\,\ lcm(m,n) = \dfrac{mn}{gcd(m,n)}\ [= mn\ \ if\ \ gcd(m,n)=1]$
Here's a proof of $\rm\ m,n\,|\,b\:\Rightarrow\:mn\,|\,bd,\ \ d = gcd(m,n) [= mx+ny\$ by Bezout]
$\rm\begin{eqnarray} \rm\quad m,n\,|\,b\:\Rightarrow\:mn\,|\,bm,bn &\Rightarrow&\,\rm mn\ |\ \,bmx\, +\, bny \ =\ b(mx\!+\!ny)\, = bd\quad [Bezout\ form] \\ \rm\quad\ \ m,n\,|\,b\:\Rightarrow\:mn\,|\,bm,bn &\Rightarrow&\,\rm mn\ |\ \gcd(bm, bn)\, =\: b \gcd(m,n) = bd\quad [GCD\ form] \\ \end{eqnarray}$
- | 2015-04-19T14:46:50 | {
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https://www.physicsforums.com/threads/how-do-i-evaluate-this-log-function.629812/ | # Homework Help: How do I evaluate this log function?
1. Aug 19, 2012
### feihong47
1. The problem statement, all variables and given/known data
2. Relevant equations
3. The attempt at a solution
I assume I can't use a calculator obviously.. so I'm quite stuck. The answer is 5, but I have no idea how to get that.
(log23)(log34)(log45) ... (log3132)
2. Aug 19, 2012
### Mute
Use the fact that
$$\log_b a = \frac{\log_c a}{\log_c b}$$
for any positive, real numbers a, b and c (with c > 1).
3. Aug 19, 2012
### feihong47
That was a fast response, and a fact that I didn't know. How do I create equations on this forum BTW?
4. Aug 20, 2012
### DeIdeal
You use LaTeX, that is, the [ itex ][ /itex ] -tags (remove the spaces) . There's a guide in the forums somewhere if you're not familiar with it, I'll edit a link to it to this post if I find it. You can also open the "LaTeX Reference" by clicking the Σ in the toolbar.
If you don't have to write anything complicated, you can just use the quick symbols and x2 and x2 buttons in the toolbar. LaTeX looks a lot nicer & it's easier to read, though.
EDIT: Here's the LaTeX guide.
5. Aug 20, 2012
### SammyS
Staff Emeritus
You could also approach this as follows:
Let $\displaystyle y=(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\,\dots\,( \log_{31}32)$
Then, $\displaystyle 2^y=2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$
By laws of exponents and the definition of a logarithm,
$2^{(\,(\log_{2}3)\,(\log_{3}4)\,(\log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=\left(2^{(\log_{2}3)}\right)^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=3^{(\,(\log_{3}4)\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=\left(3^{(\log_{3}4)}\right)^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
$=4^{(\,( \log_{4}5)\, \dots\,( \log_{31}32)\,)}$
etc.
$=\left(31^{(\log_{31}32)}\right)$
$=32$
6. Aug 20, 2012
### HallsofIvy
So, just in case others misunderstand, $2^y= 32= 2^5$ and therefore, y= 5 as the original poster said.
7. Aug 22, 2012
### feihong47
Very nice approach. Thanks! | 2018-09-23T21:50:17 | {
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https://mathoverflow.net/questions/255252/two-statistics-on-the-permutation-group/255260 | # Two statistics on the permutation group
Let $\mathfrak{S}_n$ be the permutation group on an $n$-element set. For each fixed $k\in\mathbb{N}$, consider the two sets $$A_n(k)=\{\sigma\in\mathfrak{S}_n\vert\,\, \text{\exists i,\,\, 1\leq i\leq n\, such that \,\sigma(i)-i=k}\}$$ and $$B_n(k)=\{\sigma\in\mathfrak{S}_n\vert\,\, \text{\exists i,\,\, 1\leq i\leq n\, such that \,\sigma(i+1)-\sigma(i)=k}\}.$$
QUESTION. I believe the following is true, for each $n$ and $k$: $$\# A_n(k)=\# B_n(k).$$ Is there a combinatorial proof of this? If it is known, then can you provide references?
• Isn't $A_n(0)$ the set of non-derangements (i.e., those permutations having a fixed point), while $B_n(0) = \emptyset$? – Christian Stump Nov 21 '16 at 17:59
• Okay, it appears to hold for $k>0$ (tested for $n \leq 8, 1 \leq k \leq n$. – Christian Stump Nov 21 '16 at 18:05
• usual suspects here :-) - yes, apparently $\mathbb N=\{1,2,3,\dots\}$. – Martin Rubey Nov 21 '16 at 18:05
• Since this generalises the equidistribution of descents and excedences, you might get lucky with one of the bijections showing this. – Martin Rubey Nov 21 '16 at 18:09
• Indeed, I just checked that - experimentally -- for each $k$ the distribution of the number of $k$-excedences equals the distribution of $k$-descents. – Martin Rubey Nov 21 '16 at 18:14
A simple variant of the "transformation fondamentale" of Rényi and of Foata-Schützenberger does the trick. Write a permutation $\sigma$ in disjoint cycle form, with the smallest element of each cycle first, and the cycles arranged in decreasing order of the smallest element, e.g., $(7,8)(5,6,9)(3)(1,4,2)$. Erase the parentheses to get another permutation $\hat{\sigma}$, written as a word, e.g., $785693142$. This gives a bijection $\mathfrak{S}_n\to\mathfrak{S}_n$ with the property that $\sigma(i)-i=k>0$ if and only if $\hat{\sigma}(j+1)-\hat{\sigma}(j)=k$, where $\hat{\sigma}(j)=i$.
• I think that this is precisely what Find Stat found, reproduced in the other answer. – Martin Rubey Nov 21 '16 at 19:40
• @Stanley: I like this cute and crisp proof. – T. Amdeberhan Nov 21 '16 at 20:04
You can use sage and www.findstat.org to find a candidate for a bijection as follows. First define the statistics you are interested in:
def A_num(s, k):
return len([1 for i,e in enumerate(s,1) if e-i==k])
def B_num(s, k):
return len([1 for e,f in zip(s, s[1:]) if f-e==k])
sage: findstat("Permutations", lambda s: A_num(s, 2), depth=3)
0: (St000534: The number of 2-rises of a permutation., [Mp00066: inverse, Mp00087: inverse first fundamental transformation, Mp00064: reverse], 200)
sage: findstat("Permutations", lambda s: B_num(s, 2), depth=3)
0: (St000534: The number of 2-rises of a permutation., [], 200)
sage: findstat("Permutations", lambda s: A_num(s, 1), depth=3)
0: (St000237: The number of indices $i$ such that $\pi_i=i+1$., [], 200)
1: (St000214: The number of adjacencies (or small descents) of a permutation., [Mp00066: inverse, Mp00087: inverse first fundamental transformation], 200)
2: (St000441: The number of successions (or small ascents) of a permutation., [Mp00066: inverse, Mp00087: inverse first fundamental transformation, Mp00064: reverse], 200)
sage: findstat("Permutations", lambda s: B_num(s, 1), depth=3)
0: (St000441: The number of successions (or small ascents) of a permutation., [], 200)
1: (St000214: The number of adjacencies (or small descents) of a permutation., [Mp00064: reverse], 200)
2: (St000237: The number of indices $i$ such that $\pi_i=i+1$., [Mp00064: reverse, Mp00086: first fundamental transformation, Mp00066: inverse], 200)
So, this suggests that using the composition of the maps http://www.findstat.org/MapsDatabase/Mp00066, http://www.findstat.org/MapsDatabase/Mp00087 and http://www.findstat.org/MapsDatabase/Mp00064 might be a good idea. No guarantee, of course.
• I think this is an answer -- I also tested this, and this combination of maps does yield the desired bijection for $n \leq 8$ and all $k$. – Christian Stump Nov 21 '16 at 18:44
• I guess I should now create a new mathoverflow account Find Stat. Is Shalosh Ekhad a mathoverflow user? – Martin Rubey Nov 21 '16 at 18:51
• To see this bijection in action, you can go to www.findstat.org/St000650 and then click on "search for values" to find that the third hit is exactly this map. – Christian Stump Nov 21 '16 at 18:57
• @MartinRubey: this is a good tool. Thank you! – T. Amdeberhan Nov 21 '16 at 20:03
• @MartinRubey: Regarding Ekhad, you may contact Doron Zeilberger. – T. Amdeberhan Nov 22 '16 at 0:02 | 2020-05-29T16:40:35 | {
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http://mathhelpforum.com/math-topics/8009-compound-interest-help-please.html | can anyone help me with a formula for this?
What interest rate is required to earn $250 interest on a$999 investment over 3 years compounding daily
Thank You
2. Originally Posted by xxpink__saltxx
can anyone help me with a formula for this?
What interest rate is required to earn $250 interest on a$999 investment over 3 years compounding daily
Thank You
$\displaystyle 999=250\times i^x$ where $\displaystyle i$ is interest (plus 1) and $\displaystyle x$ is the number of time periods.
So how many days are in 3 years, let's assume there's no leap-year. The answer is 1095 days
So: $\displaystyle 999=250\times i^{1095}$
Thus: $\displaystyle \frac{999}{250}=i^{1095}$
Then: $\displaystyle \sqrt[1095]{\frac{999}{250}}=i\approx 1.00126591$
So interest is: $\displaystyle \approx .126591\%$
3. Hello, xxpink__saltxx!
What interest rate is required to earn $250 interest on a$999 investment
over 3 years compounding daily?
The compound interest formula is: .$\displaystyle A \;= \;P(1 + i)^n$
where: $\displaystyle A$ = final amount, $\displaystyle P$ = principal, $\displaystyle i$ = periodic interest rate, $\displaystyle n$ = number of periods.
We are given: .$\displaystyle A = 999,\;P = 250$
Since the interest is compounded daily, the interest rate is: $\displaystyle \frac{I}{365}$
. . and the number of periods is: $\displaystyle 3 \times 365 \,=\,1095$
So we have: .$\displaystyle 999 \;=\;250\left(1 + \frac{I}{365}\right)^{1095}$
. . and we must solve for $\displaystyle I$, the annual interest rate.
We have: .$\displaystyle \left(1 + \frac{I}{365}\right)^{1095} \:=\:\frac{999}{250} \:=\:3.996$
Raise both sides to the $\displaystyle \frac{1}{1095}$ power:
. . $\displaystyle \left[\left(1 + \frac{I}{365}\right)^{1095}\right]^{\frac{1}{1095}} \;=\;(3.996)^{\frac{1}{1095}} \quad\Rightarrow\quad 1 + \frac{I}{365}\:=\:1.001265909$
. . Then: .$\displaystyle \frac{I}{365}\:=\:0.001265909\quad\Rightarrow\quad I \:=\;0.462056836$
Therefore, the annual interest rate is about $\displaystyle 46.2\%.$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Check
$\displaystyle A \;= \;\$250\left(1 + \frac{0.462}{365}\right)^{1095} \:=\:\$998.8298956 \:\approx\:\$999$. . . Yes! 4. Originally Posted by Soroban Therefore, the annual interest rate is about$\displaystyle 46.2\%.\$
Yes, but the question asks for daily interest rate
5. Soroban was right. The solutions posted were correct.
Plus, the question does not ask for the "daily" interest rate. It only asks for the "interest rate", and in reality, all banks state annual interest rates. | 2018-04-19T18:01:02 | {
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https://collegephysicsanswers.com/openstax-solutions/calculate-pressure-due-ocean-bottom-marianas-trench-near-philippines-given-its-0 | Change the chapter
Question
Calculate the pressure due to the ocean at the bottom of the Marianas Trench near the Philippines, given its depth is 11.0 km and assuming the density of sea water is constant all the way down. (b) Calculate the percent decrease in volume of sea water due to such a pressure, assuming its bulk modulus is the same as water and is constant. (c) What would be the percent increase in its density? Is the assumption of constant density valid? Will the actual pressure be greater or smaller than that calculated under this assumption?
1. $1.09 \times 10^{3}\textrm{ atm}$
2. $5.0\%$ decrease in volume
3. $5.3\%$ increase in density. The actual pressure will be greater than calculated in part (a) since, with greater density, there will be greater mass, and thus weight, in the vertical column of water. This greater weight will cause greater pressure.
Solution Video
# OpenStax College Physics for AP® Courses Solution, Chapter 11, Problem 82 (Problems & Exercises) (4:51)
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Video Transcript
This is College Physics Answers with Shaun Dychko. We are going to calculate the pressure at the bottom of the Mariana trench which is 11.0 kilometers below sea level and we convert that into meters. The density of seawater we look up in our table [11.2] and it is 1.025 times 10 to the 3 kilograms per cubic meter and the pressure or the gauge pressure and that is to say the amount of pressure in excess of the atmospheric pressure is the density of the fluid times acceleration due to gravity times the height of the fluid column and so that's 1.025 times 10 to the 3 kilograms per cubic meter times 9.80 newtons per kilogram times 1.10 times 10 to the 4 meters—height— and that is this many pascals which we convert into atmospheres by multiplying by 1 atm for every 1.013 times 10 to the 5 pascals and that is 1.09 times 10 to the 3 atmospheres. So that's a 1090 atm of pressure at the bottom of the Mariana trench—that's huge! And it was okay to just ignore atmospheric pressure and just talk about the gauge pressure and I guess the question is sort of asking just for the gauge pressure because it says "what's the pressure due to the seawater?" and so suggesting that just the gauge pressure is enough. We could add 1 atmosphere to that but you wouldn't even see it because it's not one of the significant digits in the ones place here. Okay! Part (b) says calculate the percent decrease in volume of the seawater due to this pressure. So we have to look back to chapter 5 and we say that the change in volume divided by the original volume in other words, the fractional change in volume is 1 over the Bulk modulus of the material multiplied by the force per area applied on the material but force per area is pressure so we are gonna write P in place of F over A and so we look up the Bulk modulus for seawater and it is 2.2 times 10 to the 9 newtons per square meter and so we have 1.10495 times 10 to the 8 pascals which we reuse in this figure here because we want mks units—meters, kilograms and seconds type units here— and divide it by the Bulk modulus and that is 0.05023 which is 5.0 percent and that's a decrease in volume. There's no negative sign in this formula but it's just understood that this is a decrease in volume. And so the next question in part (c) is what is the percent change in the density because since the volume is changing and the masses can stay the same... the masses being squished into a smaller volume now so its density is going to increase and so by how much is the question? So we have this final density which will be some mass divided by the original volume plus this change minus the original density which is mass divided by original volume all divided by the original density and then we can divide this term by this term to make 1 and this term divided by this is the same as this multiplied by its reciprocal so times by V naught over m and the m's cancel and we are left with V naught over V naught plus ΔV. Okay... and all this is minus 1. And then divide top and bottom by 1 over V naught and that's convenient because we end up with this expression where we have this term which we have already calculated. So we have 1 over 1 plus the fractional change in volume and I have introduced the negative sign there because our understanding is that this is a decrease so we can just put it there and this all works out to 0.05289. So this is an increase in density, 5.3 percent. And now because we have noticed that the density is going to increase by a significant amount— 5.3 percent is significant— the actual pressure for part (a) is going to be greater than what we have calculated because our calculation makes the assumption that the density is constant over this entire vertical column of seawater— this 11.0 kilometer high column of seawater— but in our work here in parts (b) and (c), we can see that the density is not constant and so the actual pressure will be greater because with greater density, there's gonna be greater mass stuffed into the given volume and therefore the weight of this vertical column will be greater and therefore its pressure will be greater. | 2021-05-12T02:24:20 | {
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https://math.stackexchange.com/questions/1425859/evaporation-of-spherical-droplet-of-liquid | # Evaporation of spherical droplet of liquid
Suppose a spherical droplet of liquid evaporates at a rate proportional to its surface area.
$${dV\over dt} = -kA$$
where $V=volume (mm^3)$, $t=time(min)$, $k=$ evaporation rate $(mm/min)$, and $A=$ surface area $(mm^2)$. Use Euler's method to compute the volume of the droplet from t=0 to 10 min using a step size of 0.25 min. Assume k = 0.1, and that the initial droplet has a radius of 3mm. Assess the validity of your results by determining the radius of your final computed volume and verifying that it is consistent with the evaporation rate.
I think I have a handle of the Euler's method aspect, so now I'm trying to determine the radius of the liquid mathematically and compare the two.
I tried converting the equation into terms of radius and time and integrating:
$${d {4\over 3} \pi r^3 \over dt} = -k4\pi r^2$$ Simplifying... $${d {1\over 3} r \over dt} = -k$$
$$\int d r = \int -3k dt$$
$$r = -3kt + C$$ Using r(t=0) to solve for C... $$3 = -3k(0) + C$$ $$r = -3kt + 3$$
but the result isn't right. I don't think I can convert the volume derivative to radius and integrate like that, but I don't know how else to go about it.
I would appreciate, in addition to help on the question, if anyone could point me to similar examples. Thanks!
What you did to solve the ODE by substituting the formula for $V$ and $A$ was correct.
The error you made is when you dropped the $r^2$ term while $r^3$ was still in differntial form. So you should have differentiated $r^3$ first then you can factorise. Anyway here's the correct approach:
$$\frac{dV}{dt}=-kA\\ \frac{d\frac{4\pi}{3}r^3}{dt}=-k4\pi r^2\\ \frac{4\pi}{3}\frac{d}{dt}r^3=-4\pi k r^2\\ \frac{1}{3}\times 3r^2\frac{dr}{dt}=-kr^2\\ r^2\frac{dr}{dt}=-kr^2\\ \frac{dr}{dt}=-k\\ r(t)=-kt+C$$
And since $r(t=0)=3$ then:
$$C=3$$
So:
$$r(t)=-kt+3$$
• Thanks a lot! I completely forgot that was something you could do. – SBoots Sep 7 '15 at 21:28
• @SBoots Glad I helped. Always remember to differentiate first until you get $f(r)dr$ then you can deal with that $f(r)$. – Oussama Boussif Sep 7 '15 at 21:30 | 2020-11-30T21:19:45 | {
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https://ece.uwaterloo.ca/~dwharder/NumericalAnalysis/Appendices/quadratic/ | # Classic Definition
a x2 + b x + c = 0
we may find the roots by calculating
Figure 1. Classic form of the quadratic formula.
For example, the roots of x2 + 8x + 15 are given by
or -3 and -5.
# Alternate Form
Considering simply the positive case, let us rationalize the numerator (see below):
This gives us the formula:
If we consider also the negative case, we get an alternate formula for the two roots, namely:
Figure 2. Alternate form of the quadratic formula.
Thus, the roots of the above quadratic polynomial are:
which gives us the two roots -5 and -3.
# Subtractive Cancelation
Consider the quadratic equation x2 + 23500 x + 0.0000823, we may use Matlab and the classic form (Figure 1) to calculate the two roots:
```>> format long
>> a = 1; b = 23500; c = 0.0000823;
>> (-b-sqrt(b^2 - 4*a*c))/2/a
ans = -23499.9999999965
>> (-b+sqrt(b^2 - 4*a*c))/2/a
ans = -3.50337359122932e-09
```
If, however, we use the alternate form (Figure 2) to calculate the roots, we get:
```>> -2*c/(b - sqrt(b^2 - 4*a*c))
ans = -23491.6425145288
>> -2*c/(b + sqrt(b^2 - 4*a*c))
ans = -3.50212765957499e-09
```
These results are tabulated in Table 1 together with the correct answers to the given number of decimal places:
Table 1. Comparison of results with correct values.
Larger RootSmaller Root
Classic Formula -23499.9999999965-3.50337359122932e-09
Alternate Formula-23491.6425145288-3.50212765957499e-09
Notice that the classic form gives an excellent approximation of the larger root, while the alternate form gives an excellent approximation of the the smaller root.
This occurs because 4ac is very small, so b2 - 4acb2 and therefore we are calculating b - |b| and b + |b|. This leads to subtractive cancelation, a loss of precision due to the subtraction of similar numbers. For this example, we calculated:
```>> b - sqrt(b^2 - 4*a*c)
ans = 0 01111100011 1110000110000000000000000000000000000000000000000000
>> b + sqrt(b^2 - 4*a*c)
ans = 0 10000001110 0110111100101111111111111111111111111111110000111101
```
If we had been able to maintain full precision, the two floating point numbers would have been given by the second number in each case, and thus the red highlights those bits in the mantissa which match:
``` 0 01111100011 1110000110000000000000000000000000000000000000000000
0 01111100011 1110000101010100001010011010010100110100011101001100
0 10000001110 0110111100101111111111111111111111111111110000111101
0 10000001110 0110111100101111111111111111111111111111110000111101
```
The first answer is correct to only 8 bits, whereas the second is correct to all 52 bits.
Therefore we see that subtractive cancelation has occurred, giving us a poor approximation when we used b - sqrt(b^2 - 4*a*c).
Thus we should use a different formula when finding either the larger or smaller roots:
## Larger Root
We should use the classic formula (Figure 1) if we are searching for the larger root (choosing the appropriate sign to ensure that -b and sqrt(b^2 - 4*a*c) are either both positive or both negative.
## Smaller Root
We should use the alternate formula (Figure 2) if we are searching for the smaller root (choosing the appropriate sign to ensure that b and sqrt(b^2 - 4*a*c) are either both positive or both negative.
# Rationalizing an expression
Recall that:
so that we lose the radical (square root) of the expression. | 2017-04-29T07:30:37 | {
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https://crypto.stackexchange.com/tags/secret-sharing/hot | Tag Info
Hot answers tagged secret-sharing
20
In Shamir's scheme is a secret sharing scheme, that is, someone that has fewer shares than is required get no information about the secret. For example, if we have a system where we require 3 shares to reveal the shared secret, then someone with 2 shares cannot be able to reconstruct it. This is true if we make the shared secret the zero-th coefficient; ...
12
It's simply not secure. Sure, it "works", in the sense that you can generate shares and reconstruct the secret from a sufficient number of them, but the essential security property of Shamir's secret sharing — namely, that knowing less than the required threshold number of shares reveals no information about the secret — does not hold. Since it'...
12
Here is an active attack on the privacy of out-of-the-box SSS. For this attack, we'll assume that the attacker (without a valid share) is allowed to participate (with $T-1$ friends with honest key shares), jointly use the protocol to recover a 'shared secret' (which might not be the real shared secret); we'll assume that this shared secret recovery process ...
10
As you note, Shamir's threshold secret sharing is perfectly secure (or information theoretically secure), yet does leak some information about the size of the secret (same thing with one-time pad). If you are worried about leaking some information about the size of the secret, then padding could be used to lower the information leakage (instead of knowing ...
10
Shamir's (m,n) secret sharing scheme has a secret $s_0$ which is represented as an element of a finite field $\mathbb F_q$ of $q$ elements. There are also $m-1$ other "randomly chosen" elements $s_1, s_2, \ldots, s_{m-1}$ that the designer uses. The scheme creates a polynomial $$S(x) = s_0 + s_1x + \cdots + s_{m-1}x^{m-1}$$ and evaluates $S(x)$ at $n$ ...
10
The process is pretty simple. As you say, each party multiplies their two shares. They then use Shamir secret sharing to share the resulting value with the other parties. Once they have received a "subshare" from each other party, each party simply runs Lagrangian interpolation on the subshares they received (plus their own subshare). The result is a share ...
10
No, the Runge phenomenon is known not to affect Shamir's scheme. Remember, the point of Shamir's scheme is not actually to form an approximation over an interval; instead, it's to encode a secret in a randomly chosen polynomial, and then divide up clues to that polynomial so that, with enough clues (shares), someone can reconstruct the entire polynomial (...
10
In RSA, assuming knowledge of the public key but not the private key, analyzing any number of triplets of matching message, encrypted message, and signature $(m,M,sg)$, does not help (as far as we know) towards recovering the private key $s$ (nor an equivalent). That's regardless of the sensible padding or RSA variant used (as long as neither the padding nor ...
10
let's say have used a 64-bit secret key to encrypt a file and then we split the key into 2 pieces of 32-bit That right there is your misunderstanding. Shamir secret sharing does not split a secret into pieces that are smaller than the original secret. Instead, it splits the secret into pieces that are the same size as the original secret. A simple ...
8
Shamir's secret-sharing scheme has $n$ shares of a secret. The shares are of the form $(x_0,f(x_0)), (x_1,f(x_1)), \ldots , (x_{n-1},f(x_{n-1}))$ where the $x_i$ are $n$ distinct nonzero elements of a finite field $\mathbb F$, and $f(x)$ is a polynomial of degree $k-1$ with coefficients in $\mathbb F$. One coefficient, say $f_0$, of $f(x)$ is the secret ...
8
In the scenario you describe, any of the non-cheating participants can contact each of the others and arrange to swap shares and reconstruct the secret. (Equivalently, all the participants can agree to publish their shares, at which point any of them can pair their share with each of the others.) If there's only one cheater, the participant who does this ...
8
There is no reason in Shamir's scheme for the finite field $\mathbb F$ to have a prime number $p$ of elements; the field can have $p^m$ elements for suitable prime $p$ and integer $m \geq 1$. So, using $F_{2^8}$, the field with $2^8$ elements is perfectly all right. However, choosing $m = 1$ has the advantage that calculations in $\mathbb F_p$ can be done ...
8
The point is that the dealer generating the update needn't know what the shared secret is. If we had a dealer that remembered what the shared secret was (or we asked enough people to contribute their shares so that the dealer could reconstruct it), then yes, the dealer could generate new shares. However, this would require is a dealer that did know the ...
8
Actually, you can do Shamir Secret Sharing over any finite field $GF(p^k)$, for any prime $p$ and any integer $k$. If $k=1$, you have the $GF(p)$ field you mentioned; however it works on extension fields as well. We often pick $p=2$ and $k$ a multiple of 8; this makes everything nice even number of bytes (at the cost of doing our calculations in $GF(2^k)$). ...
8
This cannot be achieved information-theoretically. This is typically the task that requires multiparty computation protocol to be achieved. In particular, the common method for what you want is called "secure multiplication protocol", and is in general constructed from an additively homomorphic encryption scheme (over the ring $R$), or from oblivious ...
8
With Shamir's Secret Sharing you can add as many shares as you want, as long as the threshold and the secret is unchanged. You can see that neither the generation of the polynomial that produces the shares nor the reconstruction of the secret by polynomial interpolation require the parameter $N$ (number of shares), but only the threshold $k$. Of course ...
8
Yes. It is possible to construct secret sharing schemes for general (monotone) access structures. You can read about the first construction in the paper called Secret Sharing Scheme Realizing General Access Structure. I also suggest reading this survey by Amos Beimel.
8
The proposed sharing scheme does allow reconstruction of key $K$ from any two of the three shares $K_A$, $K_B$, $K_C=(X_A,X_B)$, because: $K=K_A+K_B$ from $K_A$ and $K_B$ $K=K_A\oplus X_A$ from $K_A$ and $K_C$ $K=K_B\oplus X_B$ from $K_B$ and $K_C$ Problem is, the scheme leaks information about key $K$: always less than 1 bit worth to participants $A$ and $... 8 The maximum number of shares in Shamir's secret sharing is limited by the size of the underlying finite field. In particular, the maximum is one less than the number of elements in the field, since each share must be associated with a distinct element of the field, and one of the elements (usually the zero element) must be reserved for the secret being ... 7 It has to do with which modulus you use. You did all your arithmetic modulo 11. However, when using Feldman's VSS, you gotta use two different moduli (using each one in the appropriate spot). In your example, you shouldn't do all arithmetic modulo 11. Instead, you should be doing some arithmetic modulo 11, and some arithmetic modulo 5 (the order of$g$... 7 Shamir's secret sharing scheme provides only confidentiality against shareholders who want to try to learn the secret. It does not prevent denial-of-service attacks (or attacks on integrity), where a malicious shareholder submits a bogus share to try to cause the reconstruction of the shares to fail. If you want security against that sort of attack, don't ... 7 Shamir's secret sharing works in any finite field. A field is a mathematical structure that follows the usual laws of addition and multiplication. A finite field is a field with a finite number of elements, unlike for example the real numbers, which have an infinite number of elements. Fields exist for all prime powers pk where p is a prime and k a positive ... 7 Full disclosure: In 2007 I founded an association aiming at voting transparency. I'm proud that my efforts may have had some role, however small, in the fact that the number of French cities using electronic voting machines for political elections, then growing, has been declining since then. The book defining the protocol of the question is made freely ... 7 Let us first consider the problem without involving Shamir secret-sharing at all. Suppose that$n = 140$and that the secret$\sigma$is a 140-byte Twitter message. The space is thus restricted considerably, from all possible$256$byte values to the printable characters permitted to be used in Twitter messages, and the distribution in this restricted space ... 7 Yes, preprocessing Beaver triples in an offline phase leads to a faster online phase. The online phase of an AND gate requires just two openings plus local computations. But there are other advantages as well. Define a "linear representation"$[x]$to be any way of representing/distributing a value$x$among parties such that the following properties hold: ... 7 Your understanding is correct. The SPDZ protocol can be used for any number of two or more parties. In fact, this is one of the strengths of the SPDZ protocol. Namely, many recent secure computation protocols such as the various versions of the Yao protocol or the TinyOT protocol are limited to two parties. So it may sometimes be overemphasized that SPDZ ... 7 Here's one more way in which a dishonest participant can mess with Shamir's secret sharing: Let's briefly review how secret reconstruction in Shamir's$(k,n)$secret sharing works. Given the$x$-coordinates of$k$participants$(x_1, x_2, \dots, x_k)$, one way to reconstruct the secret is to compute the Lagrange basis polynomials:$$\ell_j(x) = \prod_{1 \... 7 Is there an algorithm that can realistically generate keys for value of k between 1,000 and 1,000,000? How about Shamir's Secret Sharing method? For$k = 1000000$, generating a share would take a million field operations (multiplications and addition); this can be done in a few milliseconds per core (depending somewhat on the field you pick; there are CPU ... 7 Each person might be allotted more than one share of the secret. Let$G$,$C$and$D$denote the number of shares allotted to a General, a Colonel, and a Desk Clerk respectively, and let$T\$ denote the Threshold of the secret sharing scheme. Then, we have that \begin{align} T &\leq 5G,\\ T &\leq 4C + 3D,\\ T &\leq 3G + 3D. \end{align} Can you ...
Only top voted, non community-wiki answers of a minimum length are eligible | 2020-02-22T04:03:58 | {
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https://math.stackexchange.com/questions/2536219/determine-if-sum-n-1-infty-fracn4nn6n-converges | Determine if $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ converges
I am trying to determine if the sum $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ is convergent or divergent.
In attempting to solve this problem I began by seeing if I could do the divergence test - in this case it is not helpful. The terms $4^n,6^n$ look like geometric series type terms but the $n$ in the numerator and denominator are troublesome. I first tried factoring $n$ out of the expression:
$$\sum_{n=1}^\infty \frac{n+4^n}{n+6^n} = \sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$$
Next I tried comparing to the series $$\sum_{n=1}^\infty \frac{\frac{4^n}{n}}{\frac{6^n}{n}} = \sum_{n=1}^\infty \frac{{4^n}}{{6^n}} = \sum_{n=1}^\infty{\left(\frac{2}{3}\right)}^n$$
This is a convergent geometric series as $r = \frac{2}{3}\lt1$
now I reason that the series $\sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$ is neither smaller not greater than the series $\sum_{n=1}^\infty{(\frac{2}{3})}^n$ Am I able to conclude anything in this case by a direct comparison test or can I only conclude that the first series is convergent if the second is greater than the first?
Thanks
Even though you can't do any useful direct comparison with $\left(\frac{2}{3}\right)^n$ because it gives lower bound which converges, you can find upper bound that converges.
In general we would like either lower bound that diverge, or an upper bound that converges. And how to find upper bound for a fraction? Usually by finding upper bound for nominator and lower bound for denominator.
For example $n+4^n < 4^n+ 4^n = 2\cdot 4^n$ and $n+6^n>6^n$, so
$$\frac{4^n+n}{6^n+n} < \frac{ 2\cdot 4^n}{6^n} = 2 \left(\frac{2}{3}\right)^n.$$
• Thanks @Sil, i like this answer. That's a really neat trick i haven't seen before! – Blargian Nov 25 '17 at 7:17
By the direct comparison test, you cannot conclude anything, as
$$\frac{4^n+n}{6^n+n} > \frac{4^n}{6^n}.$$
However, you can use the limit comparison test, which will allow you to deduce the result.
• You could also use direct comparison by noting that $$\frac{n+4^n}{n+6^n} \le \frac{4^n + 4^n}{6^n} = 2 \left( \frac 23 \right)^n$$ – User8128 Nov 25 '17 at 7:12
• I thought of that however i wasn't sure what i should take as my $b_n$ term? If i factor out the $n$ on the top and the bottom would i take $b_n$ as $\frac{1}{1+\frac{6^n}{n}}$? – Blargian Nov 25 '17 at 7:12
• @Blargian Since the things you're summing are $\frac{4^n+n}{6^n+n}$ and $\frac{4^n}{6^n}$, simply pick one of them to be $a_n$ and the other to be $b_n$. – Carl Schildkraut Nov 25 '17 at 7:18
• @Carl Schildkraut I take $b_n = \frac{4^n}{6^n}$ then $\lim_{n \to \infty}\frac{a_n}{b_n} = \lim_{n \to \infty}\frac{4^n+n}{6^n+n} \frac{6^n}{4^n}$ and i simplify this to $\frac{1+\frac{n}{4^n}}{1+\frac{n}{6^n}}$ I am unsure of how to proceed with finding this limit. I know that if the result is $\gt 1$ then the series will converge. – Blargian Nov 25 '17 at 7:32
• With the limit comparison test, if the limit is any positive real number, then the series either both converge or both diverge. In this case, you’ll get a limit of $1$, which means the test series converges, just like the known series. You get $1$ by noting that both $\frac{n}{4^n}$ and $\frac{n}{6^n}$ go to $0$ as $n\to\infty$. – G Tony Jacobs Nov 25 '17 at 7:59
One way to use direct comparison is this:
$$\frac{4^n+n}{6^n+n}\le \frac{5^n}{6^n}=\left(\frac56\right)^n.$$ | 2019-08-24T21:53:56 | {
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https://math.stackexchange.com/questions/3864711/maximizing-the-value-of-int-01-fxf-1x-mathrm-dx | # Maximizing the value of $\int_0^1 f(x)f^{-1}(x)\ \mathrm dx$
I am trying to find the maximum size of the integral $$\int_0^1 f(x)f^{-1}(x)\ \mathrm dx$$ for differentiable, increasing $$f$$ with $$f(0)=0$$ and $$f(1)=1$$. I made up this exercise for myself and thought it would be easy, but I can't do it.
I feel the answer should be $$\frac 1 3$$ intuitively, which comes from $$f(x)=x$$. So far I've tried integration by parts but then I don't know what to do.
Edit: here is the integration by parts I tried, though I think it doesn't lead anywhere: $$\int^1_0 f(x)f^{-1}(x)\ \mathrm dx=\int_0^1f^{-1}(x)\ \mathrm dx-\int_0^1f'(x)\left(\int_0^x f^{-1}(t)\ \mathrm dt\right)\ \mathrm dx\text.$$ I thought this could help because $$f'(x)>0$$ since $$f$$ is increasing and the other factor in this integral is also positive by default.
• Interesting question, nicely presented, which I do not know the answer to. The only reason that I did not upvote is because you haven't shown any work. Please edit your query to show your work. For example, if you have tried integration by parts, show it. In fact, show anything that you have tried. – user2661923 Oct 13 at 23:51
• Just an observation: for $f(x)=x^{1/a}, \; a \in \mathbb{R}^+$, where $f^{-1}(x)=x^{a}$, the integral evaluates to $\frac{1}{a+1/a+1}$ which is maximized at $a=1$ or $f(x)=x$. – Ty. Oct 14 at 1:31
• With the substitution $x = f(t)$, it becomes $$\int_{0}^{1}f\left(f\left(t\right)\right)f'\left(t\right)t\ \mathrm{d}t$$ but I don't know how much that helps. – Varun Vejalla Oct 14 at 2:42
The answer is indeed 1/3, which can be proved using the Fenchel-Young inequality for Legendre transforms.
Define $$F(t):=\int_0^t f(x)dx$$ so that $$F$$ is convex on $$[0,1]$$. The Legendre transform of $$F$$ is given by $$G(t)=\sup_{u\in [0,1]} (tu-F(u))=\int_0^t f^{-1}(x)dx$$ for $$t \in [0,1]$$.
Young's inequality (also called Fenchel's inequality) says that $$ab \leq F(a)+G(b)$$ for any $$a,b \in [0,1]$$.
Consequently we see that $$f(x)f^{-1}(x)\leq F(f(x))+G(f^{-1}(x))$$. Now notice from Fubini that $$\int_0^1F(f(x))dx = \int_0^1\int_0^{f(x)}f(u)dudx$$$$= \int_0^1 \int_0^1 f(u)1_{\{u and symmetrically we obtain that $$\int_0^1G(f^{-1}(x))dx = \int_0^1 f^{-1}(u)(1-f(u))du.$$ Now integrating the identity $$f(x)f^{-1}(x)\leq F(f(x))+G(f^{-1}(x))$$ from $$0$$ to $$1$$, we get that $$\int_0^1 f(x)f^{-1}(x)dx \leq \int_0^1 f(u)(1-f^{-1}(u))du+\int_0^1 f^{-1}(u)(1-f(u))du.$$ Since $$\int_0^1f(u)du+\int_0^1 f^{-1}(u)du=1$$, the previous expression reduces to $$3\int_0^1 f(x)f^{-1}(x)dx\leq 1.$$
• In case anyone is wondering why $\sup_{u\in [0,1]} (tu-F(u))=\int_0^t f^{-1}(x)dx$, this is because the supremum must be attained wherever the derivative of the thing being maximized equals zero, which is at $F'(u)=t$, i.e., $u=f^{-1}(t)$. Consequently the sup equals $tf^{-1}(t)-F(f^{-1}(t))$, which (using integration by parts or just drawing a picture) conicides with $\int_0^t f^{-1}(x)dx$. – shalop Oct 14 at 4:09
Prove $$\int_0^1 f(x)f^{-1}(x)dx \leq \frac{1}{3}$$.
First we notice that $$f(x) \leq x$$ leads to $$x \leq f^{-1}(x)$$, and $$f(x) \geq x$$ leads to $$x \geq f^{-1}(x)$$, so $$[f(x)-x][f^{-1}(x)-x] \leq 0$$.
Integrate it.Then we get $$\int_0^1 f(x)f^{-1}(x)dx+\int_0^1 x^2 dx \leq \int_0^1 x(f(x)+f^{-1}(x))dx=\int_0^1 xf(x) dx+\int_0^1 xf^{-1}(x)dx$$.
Let $$y=f^{-1}(x)$$,then the second integral in the right is$$\int_0^1 yf(y)df(y)=\frac{1}{2}\int_0^1 ydf^2(y)=\frac{1}{2}[1-\int_0^1 f^2(y)dy]$$.
So$$\int_0^1 f(x)f^{-1}(x)dx+\int_0^1 x^2 dx \leq \frac{1}{2}[1+\int_0^1 f(x)(2x-f(x))dx] \leq \frac{1}{2}[1+\int_0^1 x^2 dx]$$.
done.
"=" iff $$2x-f(x)=f(x)$$,i.e. $$f(x)=x$$
• Really nice solution! – Bennett Gardiner Oct 14 at 5:22
• It took me an embarrassingly long time to figure this out, but the reason why $f(x)(2x-f(x))\leq x^2$ is that $2xy-y^2 \leq x^2$ where $y=f(x)$. Nice. – shalop Oct 14 at 6:44
• And if anyone else was stumped by $\int_0^1 y\,df^2(y) = 1-\int_0^1f^2(y)\,dy$, it follows from en.wikipedia.org/wiki/Integral_of_inverse_functions. – Milten Oct 14 at 9:28
I believe you can prove it using the calculus of variations. Here is a sketch of how it would go.
Consider $$h(x) = f(x) + \delta g(x)$$ where $$g(x)$$ is a function in some class of nicely-behaved functions and $$\delta$$ is a small number. Then $$h^{-1}(x) \approx f^{-1}(x) - \delta g(x)$$. Now consider
$$u(\delta) := \int_0^1 h(x)h^{-1}(x) = \int_0^1 (f(x) + \delta g(x)) (f^{-1}(x) - \delta g(x))$$
Then
$$u'(\delta)|_{\delta = 0} = \int_0^1 (g(x)f^{-1}(x) - g(x) f(x)) = \int_0^1 g(x)(f(x) - f^{-1}(x))dx$$
Since this must be zero for all $$g$$, you get $$f^{-1}(x) = f(x)$$. But since $$f$$ is increasing, you get
$$x \le f(x) \le f(f(x)) = x$$
and therefore $$f(x) = x$$, so your answer is correct.
• Slight subtlety: It seems like you proved that if a maximizer exists then it must be the identity function. However we don’t know a priori that a maximizer does in fact exist. Something like Arzela Ascoli should work in this regard. – shalop Oct 14 at 4:39
• I am having trouble seeing why $h^{-1}(x)$ is close to $f^{-1}(x)-\delta g(x)$. When I take the derivative of $h^{-1}(x)$ with respect to $\delta$ I get $g(f^{-1}(x))/f’(f^{-1}(x))$ which doesn’t seem compatible. – 83964 Oct 14 at 5:20
• @83964 I'm having trouble seeing it too. Maybe I was too careless there! Luckily people have already come up with nicer solutions. – Flounderer Oct 14 at 5:36
We can make rigorous the argument given by Flounderer using the comment given by Varun Vejalla. Also we can prove that the identity function is indeed a maximizer computing the second variation which turns out to be strongly concave.
Using the change of variable $$x=f(t)$$ we have $$\int_0^1 f(x) f^{-1}(x)\,dx=\int_0^1 f(f(t))f'(t) t \,dt.$$ We compute the Euler-Lagrange equation: let $$h(t)=f(t)+\delta g(t)$$ where $$g\in S:=\{G \in C^1[0,1] : g(0)=g(1)=0\}$$ and $$\delta$$ real number. We $$u[g](\delta)=\int_0^1 t (f(t)+\delta g(t))(f(f(t)+\delta g(t))+\delta g(f(t)+\delta g(t)))\,dt$$ Now the Euler-Lagrange equations is $$0=\frac{d}{d\delta}u[g](0)=\int_0^1 t \{ g'(t) f(f(t))+ f'(t) f'(f(t)) g(t)+f'(t) g(f(t)) \}\,dt=\int_0^1 t [f(f(t))g(t)]' \,dt+ \int_0^1 t f'(t) g(f(t)) \,dt.$$ Integrate by parts the first integral and substitute $$f(t)=x$$ to obtain $$-\int_0^1 f(f(t)) g(t)+\int_0^1 g(x) f^{-1}(x) dx=\int_0^1 g(t) \lvert f^{-1}(t)-f(f(t))\rvert\, dt \quad \forall g\in S.$$ An application of the fundamental lemma gives $$f^{-1}(t)=f(f(t)),\quad \text{that is} \quad t=f(f(f(t))).$$ By symmetry we can suppose $$t\leq f(t)$$ and since $$f$$ increases we obtain $$t\leq f(t)\leq f(f(t))\leq f(f(f(t))=t$$. Hence $$f(t)=t$$ is the unique critical point for the functional.
We can prove that $$f(t)=t$$ is indeed a maximum since $$u''[g](0) \leq - \alpha \int_0^1 g^2(t)\,dt$$ for some $$\alpha>0$$.
If we compute the second variation in $$\delta=0$$ in Flounderer's argument we obtain $$-2 \int_0^1 g^2(t)\, dt$$, namely the strongly concave condition with $$\alpha =2$$. We can make the argument rigorous using the approach above
• I am still confused why this argument says that $f(x)=x$ should give the maximum. It might just be a local maximum, right? When finding the maximum of a function on an interval we also have to check the boundary points. Is there an analogue to that here? – 83964 16 hours ago | 2020-10-24T23:24:57 | {
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https://cs.stackexchange.com/questions/56672/what-is-the-language-generated-by-a-given-grammar | # What is the language generated by a given grammar
Given the grammar
$s \to aSb \mid bSb \mid a \mid b$;
what is the language generated by the grammar over the alphabet $\{a,b\}$?
When I was solving this question I was a bit confused about the language generated by this grammar. Would be set of all palindromes? Or would the language generated by the above grammar be that of all odd length palindromes?
Is it possible that a palindrome generated by above grammar be of odd length only as there is no rule for $S \to \varepsilon$?
The language of this grammar is all strings of the form $wb^n$ where $w\in \{a,b\}^*$, and $|w| = n+1$. If I abuse the notation a bit, it is $\Sigma^{n+1}b^n$.
If perchance you mean $S \rightarrow aSa \ | \ bSb \ | \ a\ | \ b$, then the language contains all palindromic strings of odd length.
S→aSa | bSb | a | b
This grammar accepts all palindromic strings of odd length For Ex: aaa is generated by S->aSa and S->a and for string abbba : S->aSa , S->bSb and S->b aSa abSba abbba and hence we get abbba whose palindrome would be abbba itself.
S->aSa S->bSb S->a S->b solution now put any value in "S" in the L,H,S it will give a string of odd palindrome e.g aaa,aba,bbb,bab, abSba=ababa or abbba etc
• What does this add over existing answers? – Raphael Feb 12 '19 at 11:45
The strings accepted by language are {a, b, aaa, bbb, aba, bab, ..}.Therefore more specific answer is " All of these strings are odd length palindromes." Yes it is odd length because S doesn't go to epsilon. | 2020-02-27T00:15:59 | {
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http://mathhelpforum.com/algebra/18637-factoring-questions-help.html | # Math Help - Factoring Questions - Help!
1. ## Factoring Questions - Help!
We were recently given a worksheet with some problems on it and I'm struggling with a few of the problems. Can anyone lead me in the direction or show me how to do some of these factoring problems?
Factor Fully.
1) 28s^2 - 8st - 20t^2
2) y^2 - (r - n)^2
3) 10x^2 + 38x + 20
4) y^5 - y^4 + y^3 - y^2 + y - 1
I'd greatly appreciate any help given for any of these problems. Thank you!
2. Originally Posted by Jeavus
1) 28s^2 - 8st - 20t^2
2) y^2 - (r - n)^2
3) 10x^2 + 38x + 20
4) y^5 - y^4 + y^3 - y^2 + y - 1
1) $28s^2-8st-20t^2=4(7s^2-2st-5t^2)$, it remains to factorice $7s^2-2st-5t^2=7s^2-(7st-5st)-5t^2$, can you take it from there?
2) Use $a^2-b^2=(a+b)(a-b)$
3) Similar as 1)
4) Factorice by grouping
3. Originally Posted by Jeavus
We were recently given a worksheet with some problems on it and I'm struggling with a few of the problems. Can anyone lead me in the direction or show me how to do some of these factoring problems?
Factor Fully.
1) 28s^2 - 8st - 20t^2
$28s^2 - 8st - 20t^2$
First rule, factor any constant out of the whole thing that you can:
$= 4(7s^2 - 2st - 5t^2)$
Now, since 7 and 5 are prime all we need to do is arrange the 7 and 5 in some way to multiply out correctly. Trial and error gives me:
$= 4(7s + 5t)(s - t)$
Originally Posted by Jeavus
2) y^2 - (r - n)^2
Remember $a^2 - b^2 = (a + b)(a - b)$?
Let a = y, b = r - n.
$y^2 - (r - n)^2 = (y + (r - n))(y - (r - n)) = (y + r - n)(y - r + n)$
Originally Posted by Jeavus
3) 10x^2 + 38x + 20
$10x^2 + 38x + 20$
$= 2(5x^2 + 19x + 10)$
This can't be factored any further than this.
Originally Posted by Jeavus
4) y^5 - y^4 + y^3 - y^2 + y - 1
This is a fun one, isn't it? Factor by grouping:
$y^5 - y^4 + y^3 - y^2 + y - 1$
$= (y^5 - y^4 + y^3) + (- y^2 + y - 1)$
$= y^3(y^2 - y + 1) - (y^2 - y + 1)$
Notice that both terms in parenthesis are the same, so we can factor again:
$= (y^3 - 1)(y^2 - y + 1)$
Now, the second factor can't be factored any further, but the first term can:
$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$
So
$y^5 - y^4 + y^3 - y^2 + y - 1 = (y^3 - 1)(y^2 - y + 1) = (y - 1)(y^2 + y + 1)(y^2 - y + 1)$
-Dan
4. 1) $28s^{2} - 8st - 20t^{2}$
The trick here is to factor out 4 first.
Then we have $4(7s^{2}-2st-5t^{2})$
Now, look at the middle term in the parentheses.
Rewrite -2st as 5st-7st:
$4(7s^{2}+5st-7st-5t^{2})$
$4((7s^{2}+5st)-(7st+5t^{2}))$
$4(s(7s+5t)-t(7s+5t))$
$4(s-t)(7s+5t)$
5. Thanks so much for all the help! That all made perfect sense.
I have a few more questions that I am struggling with if you'd be so kind as to direct me in the correct path:
1) 3v^2 - 11v - 10
Personally I don't think this one can be factored, as nothing multiplies to (-30) and adds to (-11).
2) h^3 + h^2 + h + 1
3) 9(x + 2y + z)^2 - 16(x - 2y + z)^2
All help will be greatly appreciated.
6. They all have the same idea, try it by your own!
7. I'm completely stumped with the first one. =/
I've tried various grouping methods with the second.
And I'm assuming there is some kind of trinomial factoring but again, I'm lost as to finding the light.
Any hints? :|
8. Here's where I'm stuck for the 2nd one:
h^3 + h^2 + h + 1
= h(h^2 + h + 1) + 1
Now I need two numbers that multiply to 1, but also add to 1 - which doesn't make sense to me. =/
EDIT AGAIN:
Solved this one!
Anyone have any ideas about the other two?
9. Originally Posted by Jeavus
Thanks so much for all the help! That all made perfect sense.
I have a few more questions that I am struggling with if you'd be so kind as to direct me in the correct path:
1) 3v^2 - 11v - 10
Are you sure that you have the signs right? Could it be:
$v^2-11v+10$?
RonL
10. Yes, I have the signs right.
Un-factorable, right?
11. It has real nasty roots, but you could factorice it finding its roots.
12. Krizalid/anyone else, do you have any hints towards the third one?
9(x + 2y + z)^2 - 16(x - 2y + z)^2
13. Originally Posted by Jeavus
9(x + 2y + z)^2 - 16(x - 2y + z)^2
$9\left( {x + 2y + z} \right)^2 - 16\left( {x - 2y + z} \right)^2 = \left[ {3\left( {x + 2y + z} \right)} \right]^2 - \left[ {4\left( {x - 2y + z} \right)} \right]^2$
Does that make sense?
Now apply the identity $a^2-b^2=(a+b)(a-b)$
14. 9(x + 2y + z)^2 - 16(x - 2y + z)^2
= [3(x + 2y + z)]^2 - [4(x - 2y + z)]^2
a^2 - b^2
a = [3(x + 2y + z)]
b = [4(x - 2y + z)]
(a + b)(a - b)
= ([3(x + 2y + z)] + [4(x - 2y + z)])([3(x + 2y + z)] - [4(x - 2y + z)])
Where do I go from this stage?
15. Originally Posted by Jeavus
Yes, I have the signs right.
Un-factorable, right?
No real factors
RonL
Page 1 of 2 12 Last | 2015-05-28T09:19:29 | {
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https://math.stackexchange.com/questions/946712/proving-the-summation-formula-using-induction-sum-k-1n-frac1kk1 | # Proving the summation formula using induction: $\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
I am trying to prove the summation formula using induction:
$$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$
So far I have...
Base case:
• Let n=1 and test
$\frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
$\frac{1}{1(1+1)} = 1-\frac{1}{1+1}$
$\frac{1}{2} = \frac{1}{2}$
• True for n=1
Induction Hypothesis:
• Assume the statement is true for the n-th case
$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$
Inductive Step:
• Prove, using the Inductive Hypothesis as a premise, that
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} = 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)}{(n+1)(n+2)}+\frac{-2-n}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)} = \frac{(n+1)(n+2)-2-n+1)}{(n+1)(n+2)} = \frac{(n+1)(n+2)-n-1}{(n+1)(n+2)} = \frac{n^2+2n+1}{(n+1)(n+2)} = \frac{(n+1)(n+1)}{(n+1)(n+2)} = \frac{n+1}{n+2}$$
To prove $$1-\frac{1}{n+2} = \frac{n+1}{n+2}$$ Multiply both sides by $n+2$ to get an equivalent expression. $$(1-\frac{1}{n+2}) * (n+2) = (\frac{n+1}{n+2}) * (n+2)$$ $$n+1=n+2−1$$
Does this all make sense? How can this be improved upon?
• Uhm, probably you wanted to say "assume that it is true for $n$", and move to the $n+1$ case. – Avitus Sep 26 '14 at 8:32
• This is a telescoping series. – Lucian Sep 26 '14 at 9:23
• @Avitus Thank you. I just worry that I am unclear with my writing! – mar10 Sep 26 '14 at 9:55
• Ah I think I understand what you mean. I will try to correct that to make my notation a bit more professional or formal. – mar10 Sep 26 '14 at 10:04
• Great! The user @Ant has a nice introduction to the notation of proofs by induction. You can take it as inspiration for your amendments. – Avitus Sep 26 '14 at 10:07
What do you know about induction proof?
You assume that statement is valid for $P(n)$, and show that is then valid for $P(n+1)$. (basically, you prove $P(n) \implies P(n+1)$.
Then, if the statement is valid for $P(0)$, is valid for $P(1)$, then is valid for $P(2)$ and so on.
This way you proved your statement for every $n \in \mathbb N$.
Assume $$\sum_{k=1}^n \frac{1}{k(k+1)} = 1-\frac{1}{n+1}$$
Your goal is to show that
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = 1-\frac{1}{n+2}$$
which should not be too difficult given the previous assumption.
Ask if you have any troubles!
EDIT
How do you manipulate that expression? The goal is to make the premises appear! So just do
$$\sum_{k=1}^{n+1} \frac{1}{k(k+1)} = \sum_{k=1}^{n} \frac{1}{k(k+1)} + \frac{1}{(n+1)(n+2)} =$$
thanks to the inductive step
$$= 1-\frac1{n+1} + \frac{1}{(n+1)(n+2)}$$
You just have to prove that this equals $\displaystyle 1- \frac{1}{n+2}$ and you are done
EDIT 2
How do you prove that $$\frac{n+1}{n+2} = 1 - \frac{1}{n+2}$$?
You can multiply both sides by $n+2$ to get an equivalent expression.
$$n+1 = n+2 - 1$$ which is true, and so $\displaystyle \frac{n+1}{n+2} = 1 - \frac{1}{n+2}$ is also true
• Well, I understand the statement is valid for Premise(n) and that I must show it will be valid for p(n+1) meaning we prove for every n the statement must be true. The trouble I am having is manipulating one side of the equation (in the inductive step I have created) to show n always with +1 next to it; As in I thought I could manipulate it so whenever you see any n in the equation it will be (n+1), then from there I can simplify to get my answer. I am unsure how to proceed. The concept behind it makes sense, I just think I have trouble getting the algebraic steps in between? – mar10 Sep 26 '14 at 8:47
• @mar10 edited. Is it better now? – Ant Sep 26 '14 at 8:52
• Ah I think I see what you mean, let me edit my first post! I must sleep shorty, if I cannot figure this out, is there a chance you will be online tomorrow? It is very late where I am. – mar10 Sep 26 '14 at 8:54
• @mar10 I probably will. – Ant Sep 26 '14 at 8:55
• @mar10 is it correct. To show that the two numbers are the same just multiply both of them by $n+2$ and simplify – Ant Sep 26 '14 at 9:34
True for $\color{brown}{n=1}$: $$\color{brown}{\sum_{k=1}^1\frac1{k(k+1)}}=\frac1{1\cdot2}=\color{brown}{1-\frac1{1+1}}.$$ If true for $\color{blue}{n-1}$, then true for $\color{green}n$: $$\color{green}{\sum_{k=1}^n\frac1{k(k+1)}}=\color{blue}{\sum_{k=1}^{n-1}\frac1{k(k+1)}}+\frac1{n(n+1)}=\color{blue}{1-\frac1n}+\frac1{n(n+1)}=\color{green}{1-\frac1{n+1}}.$$
• Your method is much shorter than mine. Is this prefered and complete? I feel as though I need to put more steps in to be more specific and show I really understand it. – mar10 Sep 26 '14 at 9:53
• The principle of our computations is the same: add a term to the sum. This is what really matters, the rest is "routine" work. I think you are putting a little too many intermediate steps. Also note that I used $n-1\to n$ rather than $n\to n+1$, as some terms are shorter. But the appropriate level of details depends on the situation. Isn't it nice with explanatory colors? – Yves Daoust Sep 26 '14 at 10:06
Why are you trying to prove it by induction when you can do it in a much more elegant way? Use the fact that $\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}$ and the result follows directly, as all other terms except $1$ and $\frac{1}{n+1}$ cancels out.
• To better my understanding with induction is why I am using induction in this case! The much more elegant way you suggest is great though. – mar10 Sep 26 '14 at 8:49
$$\sum_{k=1}^{m+1} \frac1{k(k+1)} = \sum_{k=1}^m\frac1{k(k+1)} +\frac1{(m+1)(m+2)}$$
$$= 1-\frac1{m+1}+\frac1{(m+1)(m+2)}=1-\frac{m+2-1}{(m+2)(m+1)}=\cdots$$
• What are you trying to say with this? Sorry for my ignorance. – mar10 Sep 26 '14 at 8:48
• @mar10, By inductive hypothesis, $$\sum_{k=1}^m\frac1{k(k+1)}= 1-\frac1{m+1}$$ – lab bhattacharjee Sep 26 '14 at 8:51
Assume for $n-1$ then you will be very clear. I am giving the final step only... $$\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n-1}\frac{1}{k(k+1)}+\frac 1 {n(n+1)}=1-\frac 1 n+\frac 1 {n(n+1)}=1+\frac {-n-1+1}{n(n+1)}\\=1-\frac {1}{n+1}$$
• This looks like a whole solution! How is this only the final step? – mar10 Sep 26 '14 at 9:29 | 2021-04-20T12:51:45 | {
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# Eight dogs are in a pen when a sled owner comes to choose four dogs to
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Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
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28 Sep 2017, 05:06
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The total number of ways of choosing 4 dogs from 8 is $$8C4 = \frac{8*7*6*5}{4*3*2} = 70$$
Since there are 4 dogs in order to place in 4 spots, and each pairing of 2 dogs is considered
a different team, there are a total of $$\frac{4!}{2!*2!} = 3*2 = 6$$ ways
Hence, the different sled teams the owner can have are $$70*6 = 420$$(Option D)
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28 Sep 2017, 05:15
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
8c2*6c2\2
210(C)
not sure whether it is right or not.
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28 Sep 2017, 07:05
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Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
First we need 4 dogs in total out of eight = 8C4
Now we need to just pick two dogs for the first row so that remaining two dogs will be automatically be selected for other row = 4C2
Total ways = 8C4 * 4C2 = 70*6 = 420
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
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03 Oct 2017, 09:28
1
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.
1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
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07 Oct 2017, 05:52
JeffTargetTestPrep wrote:
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
There are 8C4 = 8!/[4!(8-4)!] = (8 x 7 x 6 x 5)/4! = (48 x 7 x 5)/24 = 2 x 7 x 5 = 70 ways to choose 4 dogs from a total of 8 dogs. Once 4 dogs are chosen, let’s see how many pairings of two rows of 2 dogs are possible. Let’s say the 4 dogs are A, B, C, and D.
1) Row 1: A-B, Row 2: C-D
2) Row 1: A-C, Row 2: B-D
3) Row 1: A-D, Row 2: B-C
4) Row 1: B-C, Row 2: A-D
5) Row 1: B-D, Row 2: A-C
6) Row 1: C-D, Row 2: A-B
Thus, for each group of 4 dogs chosen, there could be 6 teams. Therefore, the total number of teams that can be formed is 70 x 6 = 420.
Why have we done 4,5 and 6 selection?What i can discern from the question is different pairings form different teams. So if AB is paired once whether in row 1 or 2, its pairing is done.
Please explain what is wrong in my interpretation of the question?
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Posts: 5927
Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink]
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07 Oct 2017, 06:13
1
Bunuel wrote:
Eight dogs are in a pen when a sled owner comes to choose four dogs to form a sled team. If the dogs are to be placed in two rows of two dogs each and different pairings of dogs are considered different teams, how many different sled teams can the owner form?
A. 24
B. 70
C. 210
D. 420
E. 1,680
Hi..
let's do the Q step by step and understand what is the TRICK in it
STEP I;-
make teams of 4
choosing 4 out of 8 is 8C4=$$\frac{8!}{4!4!}$$
TIME saver- dont simplify as more steps are involved
STEP II :-
Make different pairs in 4
choosing 2 out of 4 = 4C2 = $$\frac{4!}{2!2!}$$
The TRICKY part - when we choose two out of 4, the other 2 are already there as SECOND pair. BUT 4C2 counts them as 2 separate ways
so divide by 2!
so $$\frac{4!}{2!2!*2}$$
Also it is also valid for the second set of 4..
answer = $$\frac{8!}{4!4!}*\frac{4!}{2!2!*2}*2=\frac{8*7*6*5}{4}=7*6*5*2=420$$
D
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Re: Eight dogs are in a pen when a sled owner comes to choose four dogs to [#permalink] 07 Oct 2017, 06:13
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# If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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Updated on: 20 Mar 2017, 05:06
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If $$2^{2x+5} - 65.2^{x+1} = - 8$$, then which of the following statement is definitely correct?
A. x can only be a positive integer
B. x can only be a negative integer
C. x can be either a positive integer or a negative integer
D. No value of x exists which satisfies the equation
E. x is not an integer
Thanks,
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Originally posted by EgmatQuantExpert on 20 Mar 2017, 04:30.
Last edited by Bunuel on 20 Mar 2017, 05:06, edited 1 time in total.
Renamed the topic and edited the question.
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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20 Oct 2017, 23:56
3
2
Mahmud6 wrote:
Hi Mahmud6
$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$
$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$
Let $$2^{(x+1)}=a$$, so we have
$$8a^2-65a+8=0$$ or
$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$
so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)
or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)
Option C
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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20 Mar 2017, 04:31
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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20 Oct 2017, 23:09
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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21 Oct 2017, 00:54
niks18 wrote:
Mahmud6 wrote:
Hi Mahmud6
$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$
$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$
Let $$2^{(x+1)}=a$$, so we have
$$8a^2-65a+8=0$$ or
$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$
so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)
or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)
Option C
Hi,
Thank you.
Paul
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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21 Oct 2017, 00:57
Paulli1982 wrote:
niks18 wrote:
Mahmud6 wrote:
Hi Mahmud6
$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$
$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$
Let $$2^{(x+1)}=a$$, so we have
$$8a^2-65a+8=0$$ or
$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$
so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)
or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)
Option C
Hi,
Thank you.
Paul
Hi Paulli1982
Dot in algebra mean multiplication so it is $$65*2^{x+1}$$ and not a decimal number
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Re: If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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21 Oct 2017, 06:47
niks18 wrote:
Mahmud6 wrote:
Hi Mahmud6
$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$
$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$
Let $$2^{(x+1)}=a$$, so we have
$$8a^2-65a+8=0$$ or
$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$
so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)
or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)
Option C
Hi,
May I please know what is wrong with the below approach:
2^x * 32 - 65*2*2^x = -8
2^x = (2/7)^2
Please let me know where I am wrong in this.
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If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement [#permalink]
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21 Oct 2017, 06:52
rahul16singh28 wrote:
niks18 wrote:
Mahmud6 wrote:
Hi Mahmud6
$$2^{(2x+5)} - 65.2^{(x+1)}=-8$$
$$2^{(2x+2)}.2^3-65.2^{(x+1)}+8=0 => 2^{2(x+1)}.2^3-65.2^{(x+1)}+8=0$$
Let $$2^{(x+1)}=a$$, so we have
$$8a^2-65a+8=0$$ or
$$(8a-1)(a-8)=0 => a=\frac{1}{8}$$ or $$a=8$$
so $$2^{(x+1)}=\frac{1}{8}=2^{-3} => x=-4$$ (negative)
or $$2^{(x+1)}=8=2^3 =>x=2$$ (positive)
Option C
Hi,
May I please know what is wrong with the below approach:
2^x* 32 - 65*2*2^x = -8
2^x = (2/7)^2
Please let me know where I am wrong in this.
hi rahul16singh28
the highlighted portion is not correct. it is $$2^{2x}$$ and not $$2^x$$
If 2^(2x+5) - 65.2^(x+1) = - 8, then which of the following statement &nbs [#permalink] 21 Oct 2017, 06:52
Display posts from previous: Sort by | 2018-12-13T15:39:15 | {
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https://math.stackexchange.com/questions/1232408/how-can-i-factorize-this-quadratic-expression | # How can I factorize this quadratic expression
Going by the exercises of a book I have been factorizing quadratic equations the following way, let's say I have:
$${x^2 - 7x + 12 = 0}$$
I know that
$${a \times b = 12 \\ \text{ and } \\ a + b = -7}$$
So it's easy enough to figure out that -3 and -4 will do.
The result is obviously:
$${(x - 3)(x - 4) = 0}$$
I a trying to apply the same logic to the following:
$${4x^2 - 4x - 15 = 0}$$
so
$${a \times b = -15 \\ \text{ and } \\ a + b = -4}$$
There is no obvious answer that I can think of, I guess I could solve the equation system and get an a and a and a b but this makes no sense in this context since it would basically only add complexity to the original context.
Am I missing something obvious here?
How can I factorize this? Should modify the original equation some way so that it's possible?
• hint : let $(2x-a)(2x-b)=4x^2-4x-15$ now find a,b – Khosrotash Apr 13 '15 at 4:37
• @daryakhosrotash ok, I'm trying... thanks – Trufa Apr 13 '15 at 4:39
• $$(2x-a)(2x-b)=4x^2-2x(a+b)+ab\\so\\-2(a+b)=-4\\ab=-15$$ – Khosrotash Apr 13 '15 at 4:40
• @daryakhosrotash ohh, I think this is making sense now.... – Trufa Apr 13 '15 at 4:44
• @daryakhosrotash: that works this time, but it could be $(4x-a)(x-b)$ that works the next time. You can change that to $(2x-\frac a2)(2x-2b)$, but if $a$ is odd the division doesn't come out even. – Ross Millikan Apr 13 '15 at 4:44
In the case of a non-one coefficient in front of the $x^2$ term, like $ax^2 + bx + c$, you have to find two numbers that add to $b$ and that multiply to $ac$ instead of to $c$. So for $4x^2 - 4x - 15$ we need two numbers that multiply to $4\times(-15) = -60$ and add to $-4$; $6$ and $-10$ work. Then: $$4x^2 - 4x - 15 = 4x^2 + 6x - 10x - 15 = 2x(2x+3) - 5(2x+3) = (2x-5)(2x+3).$$
• Ok, didn't know that, but 60 has a lot of factors and it would take (at least) me a while to find them, they wouldn't come intuitively. – Trufa Apr 13 '15 at 4:46
• Since $4$ is small compared to $60$, your pair of factors is not very far apart. You could start looking around the "middle": $\sqrt{60}$, which is between $7$ and $8$ (because $60$ is between $49$ and $64$). Going down from $8$ we test for divisors of $60$ until we get to $6$, which works. – Unit Apr 13 '15 at 18:15
Your rule for finding $a$ and $b$ depends on the fact that the coefficient of the $x^2$ term is $1$. In that case you are expanding $(x-a)(x-b)=x^2-(a+b)x+ab$. If the leading coefficient is not $1$, you can divide it out. In your example, that gives $4(x^2-x-\frac {15}4)$ and now you can use your technique.
We can always use the quadratic formula to find the roots of such an equation. Recall that $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$ Hence, \begin{align*} x&=\frac{-(-4)\pm\sqrt{4^2-4\cdot4\cdot(-15)}}{2\cdot 4}\\ &=\frac{4\pm\sqrt{16+16\cdot 15}}{8}\\ &=\frac{4\pm16}{8}\\ &=\frac{1}{2}\pm2. \end{align*} Therefore, the roots of the equation are $$x=\frac{5}{2} \ \text{or} \ x=\frac{-3}{2}.$$ From this, we can see that $$x-\frac{5}{2}=0 \ \Longrightarrow \ 2x-5=0$$ and $$x+\frac{3}{2}=0 \ \Longrightarrow \ 2x+3=0.$$ This means that we can rewrite the quadratic equation in the form $$4x^2-4x-15=(2x-5)(2x+3).$$
• Thanks, I am aware of this, but this section of the book in particular tries to teach factorization as a method for solving. Thank you very much anyway! – Trufa Apr 13 '15 at 4:56
• You're welcome. As a high school student I was always really bad at factorizing quadratics by sort of guessing at the factors. I always used the quadratic equation in the end. – user230944 Apr 13 '15 at 5:03
Hint $\,\ 4x^2\!-4x-15\, =\, (2x)^2\!-2(2x)-15\, = \color{#c00}{X^2}-2X-15,\$ for $\, X = 2x$
Remark $\$ The idea generalizes to a way to reduce polynomial factorization to the case where the leading coefficient $\color{#c00}{=\bf 1}$ (i.e. is $\rm\color{#c00}{monic})$, see the AC method. It is very handy for such problems.
Use what is known as the "$ac$ method". Multiply $a$ by $c$ to get $-60 (4*-15)$
Thus you find two numbers $(p,q)$ whose product is $-60$ and whose sum is $-4$. In this case, a very quick check of the factors of $60$ will show that $p=6, q=-10$.
Now, if $(ax+b)(cx+d)=4x^2-4x-15, ab=6, cd=-10, ac=4, bd=-15$, based on what we just worked out. From this, we can deduce $a=2, b=3,c=2,d=-5$, thus we get $(2x+3)(2x-5)$ and solve for $x=-\frac{3}{2}, x=\frac{5}{2}$ | 2019-07-19T13:12:04 | {
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http://math.stackexchange.com/questions/17908/why-is-it-that-if-a-is-an-n-by-m-matrix-and-both-ba-and-ab-are-indentity-matric | # Why is it that if A is an n by m matrix, and both BA and AB are indentity matrices, then A is square and can't be rectangular?
Some background: I was in class today, and the professor was proving that given a matrix $A$ that is $n$ by $m$ and a $B$ such that $AB=I$ where $B$ is obviously $m$ by $n$ and $I$ is $n$ by $n$, if it is also the case that $BA=I$, then $A$ must be a square matrix (that is, $m=n$). We didn't manage to get through however, because we couldn't find a reason why $A$ actually had to be square in that case -- it seemed to me that it would just be $BA=I$ where $I$ is $m$ by $m$. So in both cases, we would have an Identity matrix, but of different dimensions (sounds strange, but as far as I recall, as long as the product yields an Identity, $B$ counts as an inverse, even if dimensions vary depending on whether you multiply it from left or right). I was thinking something along the lines of overdetermined systems -- if $A$ is invertible (and not square) at least from the right, it has to have more equations than variables, but that did not bring me anywhere. Thanks!
-
@Arturo: whoops. I slightly misread the question. (I was under the impression that the question was asymmetric in m and n.) – Qiaochu Yuan Jan 17 '11 at 23:16
## 3 Answers
Assuming your matrices have real coefficients, you can use the Rank-Nullity Theorem, which is essentially what you are talking about. All you need to know is that a homogeneous system of linear equations with more unknowns than equations always has nontrivial solutions.
Suppose that $A$ is $n\times m$, $B$ is $m\times n$, and $AB=I_n$, $BA=I_m$.
Consider the system of equation $A\mathbf{x}=\mathbf{0}$, where $\mathbf{x}$ is the column vector with $m$ unknowns. A solution $\mathbf{s}$ must be equal to $\mathbf{0}$, because if $A\mathbf{s}=\mathbf{0}$, then $$\mathbf{0} = B\mathbf{0} = B(A\mathbf{s}) = (BA)\mathbf{s} = I_m\mathbf{s}=\mathbf{s}.$$ So the only solution to $A\mathbf{x}=\mathbf{0}$ is the trivial solution, which tells you that you must have at most as many unknowns as equations. Since $A\mathbf{x}=\mathbf{0}$ has $n$ equations and $m$ unknowns, you know $n\geq m$.
Now, repeat the argument with the system $B\mathbf{x}=\mathbf{0}$; multiplying by $A$ on the left you conclude that the system has at least as many equations as unknowns, because the trivial solution is the only solution. So $m\geq n$.
Putting the two together you get $n=m$.
(Notice that you do not need to assume that it is the same matrix $B$ that works on both sides at the outset. If you have some $m\times n$ matrices $B$ and $C$ such that $CA=I_m$ and $AB=I_n$, then $C=CI_n = C(AB) = (CA)B = I_mB = B$; or you can just do the argument above using $C$ in the first part instead of $B$ to conclude $m=n$ first, then conclude that $C=B=A^{-1}$).
(The argument actually breaks down if your matrices have coefficients in more general rings; in particular, if you allow the entries to come from noncommutative rings, then you can have nonsquare matrices $A$ and $B$ such that $AB$ and $BA$ are different-sized identities.)
-
That makes sense -- simple yet powerful. Thanks a lot! – InterestedGuest Jan 17 '11 at 23:17
@InterestedQuest: Bill's answer is by far simpler; it also highlights exactly how commutativity enters into the picture, which mine does not. You might want to switch your "accepted" answer! – Arturo Magidin Jan 18 '11 at 2:33
Arturo, I do not know what Bill means by trace -- I have just started studying linear algebra though. – InterestedGuest Jan 18 '11 at 2:36
@InterestedQuest: The trace of a square matrix is the sum of the entries along the main diagonal. It is an easy exercise that if $A$ is $m\times n$ and $B$ is $n\times m$, then the traces of $AB$ and of $BA$ are the same (find a formula for the $ij$-th entry of $AB$ in terms of the entries of $A$ and of $B$). And the trace of the $k\times k$ identity is easily seen to be $k$. – Arturo Magidin Jan 18 '11 at 2:50
HINT $\ \$ $\rm\ n\: =\:$ trace $\rm\: AB\ =\:$ trace $\rm\: BA\: = m$
-
Almost "Proof without a word"!! – Tapu Feb 21 '12 at 11:43
Here is another nice exercise which also results in the desired conclusion: Let $A$ be an $n\times m$ matrix and $B$ be an $m \times n$ matrix. If $m < n$ then $\det(AB) = 0$ (note that the argument is pretty much the same as Arturo's answer. All you need is that the image of a linear map has at most the same dimension as the domain)
- | 2014-10-25T17:52:38 | {
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https://physics.stackexchange.com/questions/543416/confusion-regarding-the-direction-of-static-friction-and-point-of-contact | # Confusion regarding the direction of static friction and point of contact
I'm doing a practice question for my Physics 1 class, and after solving this I believe the answer key is incorrect. However, I'd like to confirm this and make sure I'm not incorrectly deciding upon the direction of static friction in my free body diagram; it's not entirely clear to me:
A block is pushed up against a vertical wall by a force of 20N. The force is at an angle of 40 degrees above the horizontal. The coefficient of static friction between the block and the wall is 0.50. Find the maximum mass, in kg, that the force can prevent from sliding down.
Here is my free body diagram:
My equations:
$$20\sin(40)+\mu_sN=mg \tag{1}$$ $$N=20\cos(40) \tag{2}$$ $$m=\frac{20\sin(40)+(0.50)(20\cos(40))}{g} \tag{3}$$ $$m\approx2.1kg \tag{4}$$
My answer is m is approximately equal to $$2.1$$ kg, assuming the direction of static friction is upward. The answer key says that m is approximately equal to $$0.52$$ kg. I noticed that this would be the case if the direction of static friction were downward.
From a little bit of research, it's my understanding that the direction of static friction would be dependent upon whether the vertical component of the applied force $$(F\sin(40))$$ was greater or less than the weight of the object ($$mg$$). If $$F \sin (40)$$ is less $$mg$$, then the direction of static friction should be upward, and if greater, downward. Adding to the confusion is that the weight of the object, in this case, is an unknown. The problem is asking for the maximum mass.
So my question is, is there anyone that can provide more clarification on the concepts and this particular problem? What is the direction of static friction in this particular problem, and why?
• You already have it very clear. Friction opposes the movement which would result if the friction force were removed. – Charles Francis Apr 13 '20 at 7:33
## 1 Answer
Friction is always opposite to the resultant force as you pointed out. The modulus of the static force will be equal to the resultant force if this is less than $$\mu N$$.
So in your exercise if you want to prevent it from sliding down, you can assume than $$mg > F_y$$, where $$F_y = 20sin(40)$$. This would be the case if $$m>\frac{F_y}{g} = 1.31$$. And your answer would be correct.
However, if $$F_y > mg$$ (if there were no friction it would slide upwards) You can find the minimum mass that is necessary in order to "stop" $$F$$. So if you assume this:
$$F_y - m_{min}g - \mu N= 0$$ $$m_{min} = \frac{F_y-\mu N}{g}$$ $$m_{min} = 0.53$$
Which pretty much fit the answer key you have been given. In conclusion, to prevent it from sliding the mass should be between $$0.53$$ and $$2.1$$, but the maximum mass to prevent from sliding down is the one you calculated. | 2021-01-15T17:38:53 | {
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https://math.stackexchange.com/questions/200141/inequality-1-frac1kk-leq-3 | # Inequality $(1+\frac1k)^k \leq 3$
How can I elegantly show that: $(1 + \frac{1}{k})^k \leq 3$ For instance I could use the fact that this is an increasing function and then take $\lim_{ k\to \infty}$ and say that it equals $e$ and therefore is always less than $3$
1. Is this sufficient?
2. What is a better wording than "increasing function"
• A bounded monotonic sequence $\{x_n\}$ is convergent. – mrs Sep 21 '12 at 6:16
• 1.- You only have to show that the function/sequence is monotonic increasing ($a_{n+1}>a_{n} \forall n \in \mathbb{N}$). And you can use the value of the limit if known to compare it to the given value. – 3d0 Jun 22 '15 at 14:37
• The algebraic expression $\bigg(1 + \dfrac 1k\bigg)^k$ is never equal to $3$, but is always less than $3$. $$\lim_{k\to\infty}\bigg(1+\frac 1k\bigg)^k = e = 2.7182818284590452353602874713526\ldots < 3$$ – Mr Pie Feb 18 '18 at 4:21
Using binomial theorem, $(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$
Now, consider ${k\choose r}\frac{1}{k^r}=\frac{k!}{r!(k-r)!}\frac{1}{k^r}=\frac{(1)(1-\frac{1}{k})(1-\frac{2}{k})...(1-\frac{r-1}{k})}{r!}\lt \frac{1}{r!}$
Thus, $$(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$$ $$\lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}$$ $$\lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}+\frac{1}{(k+1)!}+\cdots =e$$
Hence $$(1+\frac{1}{k})^k\lt e\lt 3$$
We will use the facts that $\ln(1+x) < x$ and $(1 + \frac{1}{k})^k = {\rm e}^{k\,\ln( 1 + \frac{1}{k} )} \,.$
$$\ln\left( 1 + \frac{1}{k} \right) < \frac{1}{k} \Rightarrow k\,\ln\left( 1 + \frac{1}{k} \right) < 1 \Rightarrow {\rm e}^{k\,\ln( 1 + \frac{1}{k} )} < {\rm e} < 3$$
$$\Rightarrow {\left( 1 + \frac{1}{k}\right)}^{\frac{1}{k}} < 3 \,.$$ | 2019-05-23T15:06:53 | {
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https://www.physicsforums.com/threads/conversion-of-temperature-scales.703606/ | Homework Help: Conversion of Temperature Scales
1. Jul 30, 2013
whoareyou
Can someone explain the last paragraph of this slide? It doesn't make sense to me.
2. Jul 30, 2013
Staff: Mentor
First, since it is addressing rates, the constants are no longer needed (does that make sense?).
Second, each of those fractions is equal to 1. Just like 1km/1000m = 1. You can use fractions like these to convert from one unit to another, or one rate to another involving the units.
So if you have a rate like 0.2mV per degree Celcius, you could convert that into a rate involving the Rankine temperature scale if that were of some help for some reason. Which one of those fractions would you use for such a conversion?
3. Jul 30, 2013
whoareyou
I get how to do the converions, I just can't quite understand why when you want to convert to another temperature scale, you don't use the first set of equations listed. I mean 0°C != 1.8(32°F).
4. Jul 30, 2013
Staff: Mentor
That's the rate issue that I mention in the first part of my post. If you differentiate an equation, what happens to the constants? The paragraph is discussing when you are dealing with rates that involve temperatures.
My example of 0.2mV per degree C is the temperature coefficient of voltage in a diode junction...
5. Jul 30, 2013
whoareyou
Constants become 0 when differentiated. But ... I don't get it lol :(
Btw, your example would be 0.2mV/°C x 1°C/1.8°R = 1mV/9°R.
6. Jul 30, 2013
lurflurf
So if we have a temperature difference we do not care about the values only the differences thus for differences
°R= °F
K= °C
Thus we can use ratios like for absolute temperature
7. Jul 30, 2013
Staff: Mentor
I've added in delta symbols to show that we are talking about changes...
8. Jul 31, 2013
whoareyou
Ok I see how the math works out with temperature changes. But in this example:
It's not a temperature change is it (ie. ΔK to Δ°R)?
9. Jul 31, 2013
lurflurf
K and °R are absolute temperature scales so conversion can be done by ratios. Absolute temperature can always be regarded as a change, that is absolute temperature=Absolute temperature-temperature absolute zero. °F and °C are not absolute so conversion requires also addition and subtraction. Temperature conversion in general can be effected by two reference temperatures say A and B
$$T^\prime =T_A ^\prime + \frac{T_B ^\prime-T_A ^\prime}{T_B -T_A }(T-T_A)$$
primes denote measurement in the new system
example
$$T_{human} (\,^{\circ}\mathrm{F}) =T_{water freeze} (\,^{\circ}\mathrm{F}) + \frac{T_{water boil} \,^{\circ}\mathrm{F}-T_{water freeze}\,^{\circ}\mathrm{F} }{T_{water boil}\,^{\circ}\mathrm{C} -T_{water freeze} \,^{\circ}\mathrm{C}}(T_{human}\,^{\circ}\mathrm{C}-T_{water freeze}\,^{\circ}\mathrm{C}) \\ =32\,^{\circ}\mathrm{F}+\frac{212\,^{\circ}\mathrm{F}-32\,^{\circ}\mathrm{F}}{100\,^{\circ}\mathrm{C}-0\,^{\circ}\mathrm{C}}(37\,^{\circ}\mathrm{C}-0\,^{\circ}\mathrm{C})=98.6\,^{\circ}\mathrm{F}$$
(that is just to illustrate conversion it is a common but questionable human temperature)
common reference temperatures include
absolute zero
freezing point water
triple point water
human body temperature
boiling point water
when we work in absolute temperature we have absolute zero=0
so
$$T^\prime =T_A ^\prime + \frac{T_B ^\prime-T_A ^\prime}{T_B -T_A }(T-T_A)$$
becomes
$$T^\prime =0 + \frac{T_B ^\prime-0}{T_B -0 }(T-0)=\frac{T_B ^\prime}{T_B }T$$
Last edited: Jul 31, 2013
10. Jul 31, 2013
whoareyou
I'm kinda slow ... lol. How does this relate to my example (the example above)?
11. Jul 31, 2013
lurflurf
Your example pertains to temperature conversion. That is how temperature conversion works. In particular that absolute temperatures can be converted by using ratios. If the temperature is not absolute only differences can be converted by ratios. Units can be converted by ratios when 0 is the same point for all units.
12. Aug 1, 2013
whoareyou
So let me see if I understand, conversion between absolute temperatures can be done with a ratio because both scales start at the same point (ie. 0) and conversion between non-absolute scales can only be done with temperature differences since those scales aren't aligned.
So then in that same example slide that I posted previously, if you were to convert °R to °C, the final answer would have to be Δ°C and not °C?
13. Aug 3, 2013
lurflurf
That is pretty much it. To define a temperature scale we fix a zero and a size of unit. This is different than most units where we fix the same zero. Note that °C and K have units the same size and °R and °F do as well. I do not think Δ°C is standard notation, but it makes clear we are talking about the size of the unit and not its actual value. It might be best to only use absolute temperature units in R, the bigger problem is not R but the fact the equations R is used in would need to be changed. For example the ideal gas law pv=nRT would need to be written pv=nR(T-T(absolute zero))
the temperature unit in R would them be the size of the temperature used. In this case we in fact have a temperature difference. Absolute temperatures can always be regarded as differences as T=T-T(absolute vero)=T-0 | 2018-05-20T10:28:41 | {
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http://bootmath.com/prove-or-find-a-counterexample-for-all-real-numbers-x-and-y-it-holds-that-x-y-is-irrational-if-and-only-if-both-x-and-y-are-irrational.html | Prove or find a counterexample: For all real numbers x and y it holds that x + y is irrational if, and only if, both x and y are irrational.
The following is a Homework Question that I’ve been working on and I would like some feedback on my answer: Prove or find a counterexample: For all real numbers $x$ and $y$ it holds that $x + y$ is irrational if, and only if, both $x$ and $y$ are irrational.
So I’ve got (or at least like to think I’ve got) a counter example.
Let $x = 1$ and $y = \sqrt{2}$.
Thus $x$ is rational and $y$ is irrational. Adding $x$ and $y$ gives us $1 + \sqrt{2}$ which cannot be simplified any further (right?) and is an irrational number.
Thereby proving that $x + y$ can be irrational without both x and y being irrational.
Is this okay?
Any feedback is greatly appreciated thank you!
Solutions Collecting From Web of "Prove or find a counterexample: For all real numbers x and y it holds that x + y is irrational if, and only if, both x and y are irrational."
Hint: $\sqrt{2}-\sqrt{2}=0$ is rational.
Your counterexample works fine. I think a better version would be $$(\sqrt{2}-1)+(1)=\sqrt{2}$$ is clearly irrational, whereas $1$ is rational. Otherwise you would have to justify why the sum of an irrational and a rational number is irrational.
Note that an “if and only if” statement is really two statements:
• if $x$ and $y$ are both irrational then $x+y$ is irrational, and
• if $x+y$ is irrational then $x$ and $y$ are both irrational.
To prove the “iff” statement false you need an example where one of these is false: you could use the example given by @mathse or @LAcarguy.
It’s not necessary to prove that both parts are false, but if you want to, you could simply use both those examples: @mathse’s example shows that the second part of the statement is false, while @LAcarguy’s example does it for the first part.
It is not true. Example is: $x = 1-\sqrt{2}$ and $y = \sqrt{2}$, then both $x$ and $y$ are irrationals but $x + y = 1$ a rational number.
Here’s a hint, although the question has already been answered. Try turning the statement around, and seeing if you can find a rational number (or even a whole number…) and an irrational number whose difference is irrational.
Another counterexample is $x=0$ and $y=$ something irrational. | 2018-07-19T23:14:32 | {
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http://www.hhdy87.com/ore2/pythagorean-identities-proof-problems.html | # Pythagorean identities proof problems
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The Pythagorean Theorem is one of the most important theorems in geometry. The upcoming discussion covers the fundamental trigonometric identities and their proofs. Type your expression into the box to the right. . The Pythagorean Theorem says that, in a right triangle, the square of a (a 2) plus the square of b (b 2) is equal to the square of c (c 2): Pythagorean Theorem calculator, formula, work with steps, step by step calculation, real world and practice problems to learn how to find any unknown side length of a right triangle. 2These identities are so named because angles formed using the unit circle also describe a right tri-angle with hypotenuse 1 and sides of length x and y: These identities are an How to use the Pythagorean identity The Formulas following from the basic trigonometric identity . Press play! 0% 0:00. cos ' Y -sin . This worksheet explains how to solve problems and simplify expressions using Pythagorean Identities. D. Sum and difference formulas are useful in verifying identities. Verify Trigonometric Identities. The identities + = and + = are also called Pythagorean trigonometric identities. The hypotenuse opposes the right angle. Verify: P = J 8 T L a m q 6 ë 5 ? q g l 6 ë 5. I would recommend that before you read my method, you should have a look at User-10603123151986444072's hexagonal method if you are facing difficulties with those basic formulae. A similar problem presents itself when testing a proposed identity with angle Learn how to simplify and prove quotient identities and reciprocal identities in trigonometry. Pythagorean identities are identities in trigonometry that are extensions of the Pythagorean theorem. 2: Verifying Trigonometric Identities One of the skills required in more advanced mathematics, especially calculus, is the ability to use identities to write expressions in alternative forms. Practice: Use the Pythagorean identity · Pythagorean Given the sine (or cosine) of an angle, find its cosine (or sine) using the Pythagorean identity. Pythagorean Identities. Write tan in terms of sin . csc. SWBAT prove and apply the Pythagorean Identity. Trigonometric Identities (Revision : 1. 1. Use Pythagorean Identities to find missing trigonometric values. 2These identities are so named because angles formed using the unit circle also describe a right tri-angle with hypotenuse 1 and sides of length x and y: These identities are an The proof of the last identity is left to the reader. Draw a right triangle and then read through the problems again to determine the length of the legs and the hypotenuse. This video will give you an introduction to the Pythagorean Identities like sin^2(x) + cos^2(x) = 1. But where does the Pythagorean identity even come from? We want to see a proof. Simplifying and Proving Trigonometric Identities: An Introduction For this unit, we're going to be simplifying and proving trigonometric identities. Show all work. An identity is an equation that is true for all values of xfor which the expressions in the equation are de ned. tan 2 θ + 1 = sec 2 θ. Syllabus Objective: 3. Prove each of the following identities. Suggestions Learn well the formulas given above (or at least, know how to find them quickly). Since we will make use of the basic trigonometric identities, a list of these Trigonometric Identities is available in this site. Ptolemy's theorem implies the theorem of Pythagoras. OTHER TOPICS Profit and loss shortcuts. By rearranging it we get 1−cos2x=sin2x Lesson 1: Reciprocal, Quotient, and Pythagorean Identities. See more ideas about Precalculus, Trigonometry and Teaching math. w j XM\afdeet bwHiItthz pIZn\fgiCnuidt_e^ mPSrceUcwaplic[uylnues\. The Pythagorean identities are deduced from the Pythagorean Theorem. TRIGONOMETRY Proving Trigonometric Identities 2. How to Do Algebra Problems, aged problems helper, short math poems mathematics algebra, composition of a function word problems, is there an easy way to learn algebra. doc For cases where you cannot find a clean solution (as in the case where sin θ = 1), you may need to use a calculator. 8. Solve 2 2sin ( ) 3cos(t t ) for all solutions t 0 2 In addition to the Pythagorean identity, it is often necessary to rewrite the tangent, secant, Prep up with a thorough knowledge of the identities from the fundamental trigonometric identities chart. 4 name: 2. See and . 1) sin y+ sin y • cot² y = csc y sin y+ sin y(1+cot² y) = csc y 1) Pythagorean Identity sin y(csc ² y) = csc y sin y(1/sin ² y) = csc y 2)Simplify csc ² y to 1/sin ² y Improve your math knowledge with free questions in "Pythagorean Theorem" and thousands of other math skills. Remember our steps for how to use this theorem. That will definitely come in handy: imagine trying to find an angle's cosine without any of the triangle's measurements except for sine. Yes, this problem can also be solved without the use of a Pythagorean Identity. 1 + cot2(x) = csc2(x) tan2(x) + 1 = sec2(x) Pythagorean Identities Circle Identities Trig identities relating sine with cosine , tangent with secant , and cotangent with cosecant . References to complexity and mode refer to the overall difficulty of the problems as they appear in the main program. The focus is on Pythagorean Identites, Even vs Odd Properties, Cofunction Properties and Reciprocal Identities. We have already Proofs Using The "Other Two" Pythagorean Identities: tan 2 X+1 = sec 2 X & 1+cot 2 X = csc 2 X These two identities show up all the time in trig proofs, but they're really easy to get mixed up (wait, does tan go with sec or csc?). 6. Using the theorem of Pythagoras, we can write the sine and cosine functions in . Pythagorean Theorem Practice Problems Worksheets This Pythagorean Theorem Problems Worksheet will produce problems for practicing solving the lengths of right triangles. These involve squares of the basic trig functions and are know as the Pythagorean Identities. Yesterday, I presented a solution using a Pythagorean identity, but I was unable to be certain if the final answer was a positive or negative without drawing a picture. The Pythagorean Theorem allows you to work out the length of the third side of a right triangle when the other two are known. RHS = cscxcosx tanx+cotx = 1 sinx cosx sinx cosx + cosx sinx = 1 sinx cosx 1 sin2 Proof of the tangent and cotangent identities. Example 5. secx - tanx SInX - - secx 3. Though they are not the same, and the differences are what can cause you problems. Sample Problems. The high-school students can get an in-depth knowledge of identities like quotient, reciprocal, cofunction and Pythagorean. This theorem is basically used for the right-angled triangle and by which we can derive base, perpendicular and hypotenuse formula. · COS X. How To Solve Pythagorean Theorem proofs and geometry problems. – Identify the known values and substitute these into Pythagoras’ theorem. 2 Presentation: 2. (If it is not a Right Angled Triangle go to the Triangle Identities page. Use the Pythagorean Theorem to solve problems 4. Each of these identities is true for all values of u for which both sides of the identity are defined. 49 Math 111: Summary of Trigonometric Identities Reciprocal Identities sin = 1 csc cos = 1 sec tan = 1 cot csc = 1 sin sec = 1 cos cot = 1 tan Quotient Identities tan = sin cos cot = cos sin Pythagorean Identities 1 = sin2 +cos2 sec2 = tan2 +1 csc2 = 1+cot2 Even/OddIdentities sin( ) =sin csc( csc cos( ) = cos sec( ) = sec tan( ) =tan cot( cot Using only those, you can prove the Pythagorean Theorem. Sine and Cosecant Functions (Special Property) Sine Identity; Cosine and Secant Functions (Special Property) Cosine Identity; Tangent and Cotangent Functions (Special Property) Tangent Identity; Cofunction Identities We can express these identities as fractions that contain 1 in the numerator. Solution of exercise 2. 1-5. For this particular activity, I will focus on some trigonometric identities that can be derived using the Pythagorean Theorem. •Angle of elevation- the angle created by a horizontal line and an upward line of sight to an object above the observer. Demonstrate a proof of the Pythagorean Theorem 3. But there are a lot of them and some are hard to remember. Generally we use three types of Pythagorean identities(as per Gujarat Secondary Education Board), they are: – Previous section Negative Angle Identities Next section Additional Trigonometric Identities Take a Study Break Literary Characters Summed Up in Quotes from The Office Sep 19, 2019 expressions using double-and half-angle formulas. Your expression may contain sin, cos, tan, sec, etc. Knowing that cos α = ¼ , and that 270º <α <360°, calculate the remaining trigonometric ratios of angle α. The Pythagorean identity can handle that and so much more. In a scenario where a certain section of a wall needs to be painted, the Pythagorean Theorem can be used to calculate the length of the ladder needed if the height of the wall and the distance of the base of the ladder from the wall are known. An example of a trigonometric identity is The Pythagorean identity follows by squaring both definitions above, and adding; the left-hand side of the MCR3U. Great photo of free problems examples Very nice work, photo of problems examples online You may want to see this photo of examples online problem Online problem verifying photos taken in 2015 Great problem verifying exercises image here, check it out Hello students, in this blog we are going to discuss Pythagorean identities. Example. The Pythagorean Theorem can be used in any real life scenario that involves a right triangle having two sides with known lengths. Why you should learn it GOAL 2 GOAL 1 What you should Chapter 4/5 Part 2 Outline Unit Goal: By the end of this unit, you will be able to solve trig equations and prove trig identities. Example 2: Pythagorean Theorem Word Problems A 15 foot ladder is leaning against a wall. • Provide copies of Extension: Proving the Pythagorean Theorem using Similar Triangles as necessary. and . sin 2 θ + cos 2 θ = 1. 6. We will again run into the Pythagorean identity, sin2 x+cos2 x = 1. Power-Reducing/Half Angle Formulas. • Each small group of students will need a large sheet of paper, copies of the Sample Methods to Discuss, and the Comparing Methods of Proof sheet. Most scientific calculators include inverse trig functions; just remember to appropriately set your calculator for radian or degree mode, depending on which angle system you're using. problem icon Can you find a way to prove the trig identities using a diagram? problem (1). Pythagoras . Click here to visit our frequently asked questions about HTML5 video. 4. proof by induction (2 I will post a number of problems that I have been asked to Verifying trigonometric identities (VTI) involves many domains of mathematics, such as the use of algebraic skills to manipulate expressions and the equality concept embedded in verification and identities. You can also derive the equations using the "parent" equation, sin 2 (θ) + cos 2 (θ) = 1. 8along with the Quotient and Reciprocal Identities in Theorem10. (try to relate with the sin formulae) sin (-x)=-sinx Sin(x+y)=sin Pythagorean Theorem worksheets contain skills on right triangles, missing leg or hypotenuse, Pythagorean triple, word problems, printable charts and more. Pr 10 Use trig identities to change to. Trigonometric Identities Formula. Even-Odd Identities. In the chart below, please focus on memorizing the following categories of trigonometric identities: 1) Reciprocal Identities 2) Quotient Identities 3) Pythagorean Identities 4) Even/Odd Identities 5) Double-Angle Formulas While the other identities and formulas in the chart are good to know, they will not be essential to your success in our The proof of the last identity is left to the reader. Types of angles Using and verifying identities Fundamental trigonometric functions Pythagorean identities Simplifying expressions Periodicity of trigonometric identities Co-function identities Difference identities Sum identities Double angle identities Proof of double angle sine identities Proof of double angle cosine identities Negative angle formulas Actually, the Pythagorean identity of cosecant and cot functions is proved mathematically in trigonometry by the geometrical method. The half‐angle identities for the sine and cosine are derived from two of the cosine identities described earlier. A summary of Negative Angle Identities in 's Trigonometric Identities. But before solving trigonometric identities we will first see derivatives of trigonometric functions. cos 2 θ + sin 2 θ = 1. Pythagorean Theorem Practice Problems. 4 Review Old 5. Simplify the expression. There are several methods to prove the Pythagorean Theorem. The Pythagorean identities are derived from the basic Pythagoras' Theorem. The cofunction identities apply to complementary angles and pairs of reciprocal functions. By the Pythagorean Theorem, any triangle with sides proportional to a 3x4x5 triangle is a right triangle: 3^2 + 4^2 = 9 + 16 = 25 = 5^2. Cymath is an online math equation solver and mobile app. 3: Pythagorean Identities. Right Triangle. Also included in this video is a brief proof why it works, and how they are all connected. Step 2: Use the Pythagorean Theorem (a 2 + b 2 = c 2) to write an equation to be solved. It is similar to the proof provided by Pythagoras. The proof of each of those follows from the definitions of the trigonometric functions, Topic 15. Work on the most complex side and simplify it so that it has the same form as the simplest side. 3, we saw the utility of the Pythagorean Identities in Theorem10. k. State the reciprocal identities for csc , sec , and cot . Many people ask why Pythagorean Theorem is important. The Introduce the Pythagorean Identity mathematically and geometrically. \sin^2 \theta + \cos^2 \theta = 1. Name Math 4 review problems March 18, 2016 page 1 Review: trigonometric identities On Tuesday (3/22) we will have a test covering sections 5. Derivation of Pythagorean Identities. The Procedure for Simplifying Trigonometric Expressions: When you are simplifying, at each step you can: 1. back to top. and squares are made on each of the three sides, This page demonstrates the concept of Trigonometric Identities. Though you'll use many of the same techniques, they are not the same, and the differences are what can cause you problems. The Unit Circle shows us that. ) Each side of a right triangle has a name: Trigonometry (trig) identities. Download as PDF file [Trigonometry] [Differential Equations] the Pythagorean theorem (Lecture 2), proof by contradiction (Lecture 16), limits (Lecture 18) and proof by induction (Lecture 23). If one leg of a right triangle has length 1, then the tangent of the angle adjacent to that leg is the length of the other leg, and the secant of the angle is the length of the hypotenuse. Trigonometric Identities Practice Worksheet Use the quotient and reciprocal identities . Let us know the standard identities of binomials and trinomials equations. C. For example, cos 2 u1sin2 u51 is true for all real numbers and 1 1 tan2 u5sec2 u is true for all real numbers except u5 when n is an integer. The Pythagorean Theorem and its many proofs . State the ratio identities for tan and cot . When asked to prove an identity, if you see a negative of a variable inside a trig function, you automatically use an even/odd identity. If an equation is valid only for certain replacement values of the variable, then it is called a conditional equation . Proof of the Pythagorean trig identity · Using the Pythagorean trig 2 e. So, let us study how to derive the proof of the Pythagorean identity for the co-secant and co-tangent functions. Now we can recognize the Pythagorean identity cos2(x)+sin2(x)=1, Free trigonometric identity calculator - verify trigonometric identities step-by-step. Double Angle Formulas. – Solve for the unknown value. Using these identities, we can solve various mathematical problems. By definition: Given the sine (or cosine) of an angle, find its cosine (or sine) using the Pythagorean identity. Tuesday, 2/27 . Problems 20, 25 and 35 review other fundamental identities and help students develop using algebra skills such as subtracting fraction. We can prove this identity using the Pythagorean theorem in the unit circle with x²+y²=1. This identity is important because it sets an expression involving trig functions equal to 1, and this simplification is very helpful for solving equations. The two identities labeled a') -- "a-prime" -- are simply different versions of a). Up till now we have finished almost all trigonometric identities, now its time to find out how to solve trigonometric identities. Maria walked 3 km west and 4 km south. opposite sin hypotenuse q= hypotenuse csc opposite q= adjacent cos hypotenuse q= hypotenuse sec adjacent q= opposite tan adjacent q= adjacent cot opposite q= Unit circle definition For this definition q is any Pythagorean identities, and the negative angle identities. REVIEW Quotient Identities Reciprocal Identities Pythagorean Identities 3. Practice Problem 1: A sail on a sailboat is in the shape of a right triangle. 1 sin 1 1 cos cos 1 sin sin 802 C HAPTER 1 1 Trigonometric Identities and Equations PROBLEM The proofs for the Pythagorean identities using secant and cosecant are very similar to the one for sine and cosine. The proof is very geometric and uses the distance formula (a. As we proceed through the course, more rules will become available to you. the Pythagorean Theorem in disguise). tan2 x sin' x = tan' x - sin' x Trigonometric identities are equalities involving trigonometric functions. ) Target Task A task that represents the peak thinking of the lesson - mastery will indicate whether or not objective was achieved List of all algebraic identities in algebra with proofs in algebraic and geometrical methods and examples to understand use of them as formulas. We will again run into the Pythagorean identity, sinº x + cos2 x = 1. We can 27 Jul 2015 Curriculum Map: Trig Identities. Perfect for acing essays, tests, and quizzes, as well as for writing lesson plans. 1) Use a cofunction identity 2) Use a pythagorean identity 3) Verify. Pr 12 Use trig identities to change to. Pythagoras' Theorem. Your textbook contains a proof of subtraction identity for cosine in section 7. And Opposite is opposite the angle. Practice Problem 2: Given a triangle \Delta ABC, as it is shown in the picture below. Use these fundemental formulas of trigonometry to help solve problems by re-writing expressions in another equivalent form. Free practice questions for Precalculus - Prove Trigonometric Identities. Proof of the Pythagorean identities. Learn basic trig formulas and simple steps to solve trig identities. Warning: Techniques used to solve equations, such as adding the same term to each side, and multiplying each side by the same term. 70. Similarly, an equation which involves trigonometric ratios of an angle represents a trigonometric identity. Some problems are different but other are the same. Time, speed and distance shortcuts. Percent of a number word problems. Some are easy and some can be quite challenging, but in every case the identity itself frames your work with a beginning and an ending. In these problems, it is best to look at what is being asked for. The Pythagorean configuration is known under many names, the Bride's Chair; being probably the most popular. Conceptual Use of the Pythagorean Theorem by Ancient Greeks to Estimate the Distance From the Earth to the Sun Significance The wisp in my glass on a clear winter’s night Is home for a billion wee glimmers of light, Each crystal itself one faraway dream With faraway worlds surrounding its gleam. All trigonometric derivations and values are based on the Pythagorean Identities. The proofs of the power-reducing formulas for the other five functions are similar. Students will understand why the Pythagorean Theorem works and how to prove it using various manipulatives Curriculum Expectations Trig identitys are very important u need to have them on your finger tips. Problem Set. You first replace all trig Prove the Pythagorean Identity sin^2(θ) + cos^2(θ) = 1 use all the time in trigonometry, and if you know them well, you can use them to solve tons of problems. To derive the other Pythagorean Identities, divide the entire equation by Note that the equations in bold are the trig identities used when simplifying. Geometrically, these are identities involving certain functions of one or more angles. Return to Precalculus Home Page. They are distinct from triangle identities, which are identities potentially involving angles but also involving side lengths or other lengths of a triangle. Trig Identities. Section Subject Learning Goals Use the Pythagorean theorem to calculate the value of X. The sign of the two preceding functions depends on the quadrant in which the resulting angle is located. The Trigonometric Identities are equations that are true for Right Angled Triangles. Trig Cheat Sheet Definition of the Trig Functions Right triangle definition For this definition we assume that 0 2 p <<q or 0°<q<°90. When working out problems on trig identities, you may need to do a proof. How high up the wall is the ladder? Practice Problems: Pythagorean Theorem Word Problems 7) If the length of a rectangular television screen is 20 inches and its height is 15 inches, what is the length of Here is an example of a more or less general approach to proving trig identities, which won’t always give the shortest or most elegant proof, but at least it’s a backup: Proving Trigonometric Identities. Identities for the following problems. \cos^2\theta+\sin^2\theta=1. The converse of the Pythagorean theorem and special triangles. In math, an "identity" is an equation that is always true, every single time. problems. It is also sometimes called Pythagorean Theorem. 2 e tan2 e -1 = tan. Try changing them to a Pythagorean identity and see whether anything interesting happens. Pythagorean Identity in Trigonometry One of the most important trigonometric identities is the Pythagorean Identity, which is closely In this first section, we will work with the fundamental identities: the Pythagorean identities, the even-odd identities, the reciprocal identities, and the quotient identities. The Pythagorean identities can be used to simplify problems by transforming it is possible to derive a number of relationships between the angles. ). Carpenters and builders use this idea all the time. If you have ever wondered why the Pythagorean identity, sin2θ + cos2θ = 1, is so important, and where it came from, then read on. hu, matus@utia. Trigonometry is useful when setting up problems involving right triangles. You will also learn how these identities can help you solve more complicated trig problems. cos2 x = cscxcosx tanx+cotx Solution: We will start with the right-hand side. 4 name: This will relieve stress in memorizing all the trigonometric identities and focus more on applying the identity to problems. Once you do that, applying these identities in different problems will become a piece of the pie. Using the Pythagorean identity, sin 2 α+cos 2 α=1, two additional cosine identities can be derived. An example of a trigonometric identity is. 3, and part of 5. 4) 1 Trigonometric Identities you must remember The “big three” trigonometric identities are sin2 t+cos2 t = 1 (1) sin(A+B) = sinAcosB +cosAsinB (2) cos(A+B) = cosAcosB −sinAsinB (3) Using these we can derive many other identities. In Pythagorean Theorem, c is the triangle’s longest side while b and a make up the other two sides. sinθ = cosθ tanθ. Read More 3D Pythagoras and Trigonometry worksheets. Builders measure the diagonal and make sure it is 50 ft. Apply the Pythagorean Identity in the context of solving algebra problems. Besides the statement of the Pythagorean theorem, Bride's chair has many interesting Directions: Verify the given problems. Study concepts, example questions & explanations for Precalculus. is equal to 1. While they may seem hard to memorize and understand, they’re actually quite the opposite, as you only need to practice them in a few questions to get the gist of them. Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree. Basic Trigonometric Identities. Co-Function Identities. Study your basic trig identities before you start this game. Pythagorean Theorem In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. find the values of the remaining trigonometric functions, using a Pythagorean Identity. Domain and range of trigonometric functions Domain and range of inverse trigonometric functions. As well as giving a geometric Solution of exercise 1. Use algebra: 2 22 csc 1 csc 1 csc 1 cos cos Proving Trigonometric Identities 1. A sample problem is solved, and two practice problems 1 Jun 2018 Having discussed trigonometric identities on Monday, let's make this Trig Week, by 1 with \sec^2\theta – \tan^2\theta (one of the Pythagorean identities). Your triangle may have a different shape, but it has to be a right triangle. Free trigonometric identity calculator - verify trigonometric identities step-by-step Proving Trigonometric Identities (page 3 of 3) When you were back in algebra, you rationalized complex and radical denominators by multiplying by the conjugate; that is, by the same values, but with the opposite sign in the middle. • use addition and subtraction formulas to solve problems of the Pythagorean Theorem through one possible proof of addition and subtraction identities 10. sin2 x + cos2 x = 1. 4 Trigonometric Identities In Section10. Pre-Algebra giving you a hard time? Shmoop's free Basic Geometry Guide has all the exercises, quizzes, and practice problems you've been craving. To get the most benefit from these problems, work them without first looking at the solutions. U u JMfa odNeC lw 7i6tHhe gI EnqfziInsi rt 8eC cP Or Te L- yA Dllg 0eVbhrMaT. Therefore, the reciprocal trig identities are as defined follows: Reciprocal Identities – Proof. You can then just read off the Pythagorean identities. D K ^AolplE krHiugHhdtRsB ErxeqsQecrsv^etd_. This graph is a great tool to use int he classroom because it uses only the six basic Trig Identities and creates many different formulas that they students will use multiple times through out the life time of math. The three Pythagorean identities are After you change all trig terms in the expression to The Pythagorean identity tells us that no matter what the value of θ is, sin²θ+cos²θ is equal to 1. 31 Jul 2019 You will be expected to be able to prove a trigonometric identity such as Learn the Pythagorean identities below and then try the examples that follow. Simplifying Problems Step-by-Step Lesson- These problems require you to combine trigonometric identities that many people find difficult. Volume. Trigonometric identities Problems on trigonometric identities Trigonometry heights and distances. I like to give students problems that include old topics along with the new material. 4 about trigonometric identities. These are called Pythagorean identities, because, as we will see in their proof, they are the trigonometric version of the Pythagorean theorem. In these problems, you are commonly asked to "prove" that one side of an equation is equal to the other side of an equation, and you will need to simplify expressions using quotient and reciprocal identities to do so. Chapter 6 Trigonometric Identities Sec 6. Formula for the Pythagorean Identities sin^2 \theta ; + cos^2 \theta = 1 tan^2 \theta + 1 = sec^2 \theta 1 + cot ^2 \theta = csc^2 \theta 5. 3 Problem Solving with the Pythagorean Theorem and Trigonometry 5. Instead, a group of shapes will be given, with just enough information to eventually calculate a particular side, area, or angle. Pythagorean Theorem - Sample Math Practice Problems The math problems below can be generated by MathScore. pythagorean id proof#1. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. Pythagorean Trigonometric Identity (1) Trig ID Movie (II) Trig ID Movie (III) Even/Odd Identities. Includes full solutions and score reporting. tan x sin x + . These identities will be used as our starting point for proving more identities. A trigonometric identity is an equation that involves trigonometric functions and is true for every single value substituted for the variable (assuming both sides are "defined" for that value) You will find that trigonometric identities are especially useful for simplifying trigonometric expressions. Recall that these identities work both ways. Pr 11 Use trig identities to change to. These are the only rules that are available to you. The domain . MENSURATION. Using Pythagorean Theorem in Complex Problems Many GMAT questions will not simply give a right triangle with one missing side to calculate. Equations such as (x 2)(x+ 2) = x2 4 or x2 1 x 1 = x+ 1 are referred to as identities. The Pythagorean Theorem calculator, formula, example calculation (work with steps), real world problems and practice problems would be very useful for grade school students (K-12 education) in classifying triangles, especially in studying right triangles. The latter serves as a foundation of Trigonometry, the branch of mathematics that deals with relationships between the sides and angles of a triangle. Sum-Difference Formulas. 3. Pythagorean Identities - Independent Practice Worksheet Complete all the problems. Directions: Utilize your knowledge of Pythagorean Identities to solve the following problems. Go to Solutions. In the main program, all problems are automatically graded and the difficulty adapts dynamically based on performance. Could anyone please help? If you can, can you also tell me where I can find the information if there is a link to this sort of info? Also, what's the difference between Pythagorean Identities and Use identities to find the value of each expression. 1) Change to sines and cosines 2) Find common denominators 3) Add fractions Double Bonus: The Pythagorean Identities. The fundamental identity states that for any angle θ, \theta, θ, cos 2 θ + sin 2 θ = 1. from MathTV(You-tube) Pythagorean ID's: Compnd. We can use the eight basic identities to write other equations that There are three trigonometry identities based on of the Pythagorean theorem. 1 Reciprocal, Quotient, and Pythagorean Identities and Proving Trigonometric Identities A Trigonometric Identity is a trigonometric equation that is true for all permissible values of the variable in the expressions on both sides of the equation. Angle Formulae <Introduction> <Unit Circle> <Radians> <Angle sum and difference formulas> <Trigonometry in Triangles> <Graphs> <Problems> Identities are equations that are always true. Lecture Notes Trigonometric Identities 1 page 4 6. Pr 9 Use trig identities to change to. Proof of Pythagorean Identities : Lets drow an unit circle as showing in picture and draw an angle θ since it is a unit circle so line CP = 1, let draw the perpendicual lines to x and y axis as PN and PM. After watching this video lesson, you will be able to spot the Pythagorean identities when you see them. Ratio and proportion shortcuts Pythagorean identity for secant and tangent functions and proof to learn how to derive Pythagorean identity in terms of secant and tangent functions. “Doing a proof” means that you need to check your work, using a different formula. Content Descriptors: Prove and Apply trigonometric identities. Lastly we apply the Pythagorean identity: sin2x+cos2x=1 ! (one of the most useful identities for these types of problems). Let us prove the three Pythagoras trigonometric identities, which are commonly used. The Pythagorean Theorem takes place in a right triangle. a² + b² = c² . Additionally, while verification is proof-making, students engage in problem solving in order to complete the task. Now I notice a Pythagorean identity in the numerator, allowing me to simplify:. Find the hypotenuse of a triangle with a base of 11 cm and height of 9 cm. Moreover, the trigonometric identities also help when working out limits, derivatives and integrals of trig functions. Round your answer to the nearest hundredth. The last types of questions you may be asked that deal with quotient and reciprocal identities may be "proof" questions. The Pythagorean Identities are true for every x. Pr 13 Use trig identities to change to. Trigonometric identities are equalities involving trigonometric functions. 5 Jun 2019 To prove that an equation is an identity, we need to apply known The following questions are meant to guide our study of the material in this section. Eventually, you come up with Trigonometric Identities Worksheet Introduction to Identities 1. Try it. ) Each side of a right triangle has a name: Adjacent is always next to the angle. His argument Statement [3] is the Pythagorean identity. Reciprocal and Quotient Identities basic trigonometric identities. Quotient Identities. complete list of identities to keep around when you're working on trig. Next, we move to algebraic identities. We will not prove the subtraction identity for cosine during lecture, but you may add it to the list of rules that you can use when proving identities. Could anyone please help? If you can, can you also tell me where I can find the information if there is a link to this sort of info? In this Pythagorean identities worksheet, 11th graders solve 10 different problems that include applying various Pythagorean identities in each. 9. ➢ Using Basic Trig Identities We can derive a couple of other Pythagorean-like identities from the first. 1 General. Starting with sin 2 ( x) + cos 2 ( x) = 1, and using your knowledge of the quotient and reciprocal identities, derive an equivalent identity in terms of tan( x) and sec( x ). Precalculus Help » Trigonometric Functions » Proving Trig Identities » Prove Trigonometric Identities Notice that a Pythagorean Identity is present. How can the Pythagorean Theorem be proved using a mean proportional in a 2-column format? What was the original proof that Pythagoras himself used to prove his theorem? If #hatQ# is a Right-angle in #DeltaPQR and PR-PQ=9 , PR-QR=18# then what is the perimeter of the triangle? Best Answer: In order to prove the identity you need to start with a right triangle whose legs are equal to 1 and the hypotenuse is equal to sqr(2). The foot of the ladder is 7 feet from the wall. Solution: From the pythagorean identities we have: sin 2 t + cos 2 t = 1 sin 2 t + (3/5) 2 = 1 sin 2 t = 1 - 9/25 = 16/25 sint = 4/5. Pythagorean Theorem Word Problems- Independent Practice Worksheet 1. tan 2 θ + 1 = sec 2 Pythagorean Theorem By Joy Clubine, Alannah McGregor & Jisoo Seo Teaching Objectives For students to discover and explore the Pythagorean Theorem through a variety of activities. Video Clip : Trigonometrical Identities . Brief history: Pythagoras lived in the 500’s BC, and was one of the first Greek mathematical thinkers. There are typically two types of problems you’ll have with trig identities: working on one side of an equation to “prove” it equals the other side, and also solving trig problems by substituting identities to make the problem solvable. Generalized minimizers of convex integral functionals and Pythagorean identities Imre Csisz ar1 and Franti sek Matu s2 csiszar@renyi. Times table shortcuts. The longest side in a right triangle is called hypotenuse, and the other two sides are called legs. Trig identities are trigonometry equations that are always true, and they’re often used to solve trigonometry and geometry problems and understand various mathematical properties. Sub in those negative angle identities to get the cosine difference identity: cos(α – β) = cos(α)cos(β) + sin(α)sin(β) Now let's take our hard-earned sum and difference identities, and use them to solve problems. }$$ Homework Problem 77 offers a proof of the Pythagorean identity. In a 30°-60°-90° triangle the length of the hypotenuse is always twice the length of the shorter leg and the length of the longer leg is always √3 times the length of the shorter leg. Even if we commit the other useful The Pythagorean Theorem along with the sum and difference formulas can be used to find multiple sums and differences of angles. 2These identities are so named because angles formed using the unit circle also describe a right tri-angle with hypotenuse 1 and sides of length x and y: These identities are an Bhaskara's First Proof Bhaskara's proof is also a dissection proof. identities that it knows about to simplify your expression. For Pre-Algebra giving you a hard time? Shmoop's free Basic Geometry Guide has all the exercises, quizzes, and practice problems you've been craving. Rewrite the terms inside the second parenthesis by using the quotient identities 5. Section 7. He used the following diagrams in proving the Pythagorean Theorem. If you looking for either a or b, make sure that the value you enter for a or b is not bigger or equal to c. All these trig identities can be derived from first principles. Specific Outcome 6: Prove trigonometric identities, using: reciprocal identities, quotient identities, However, we can use them in a number of problems. Chord AB is a side of the given square from the Circle and Square problem and is bisected by symmetry into two segments, each of length a. 1) Find the exact ALL answers should be EXACT values! Trigonometric Identities Questions And Answers Pdf Read/Download fundamental trigonometric identities worksheet answers 1 trigonometric identities pdf kuta software using trigonometric identities unit 05 answer key 2010. If you aren’t going to be given all of the Pythagorean Identities in your Trigonometry class, you don’t have to worry about memorizing all of them. (1- cos2x) (cosec x) 2. Solving word problems in trigonometry. 0 A shoutout is a way of letting people know of a game you want them to play. How to derive and use the Reciprocal, Quotient, and Pythagorean Identities, Regents When simplifying problems that have reciprocal trig functions, start by to solve problems. Print this page as a handy quick reference guide. that there isn't just one correct method for solving a math problem. Derive the Pythagorean Identity geometrically. A^2 = B^2 + C^2 + 2*B*C*Cos(a), the Pythagorean theorem is just a special case of this. Sample Problem. COS X. My first math droodle has also related to the Pythagorean theorem. From these facts, the primary Pythagorean identity can be shown. add 0, or 3. Basic Trigonometric Identities Trigonometric identities are equations involving the trigonometric functions that are true for every value of the variables involved. Since this point is in quadrant IV, sint is negative, so we get: __ Proof: To find the power-reducing formula for the sine, we start with the cosine double angle formula and replace the cosine squared term using the Pythagorean identity. Pay attention and look for trig functions being squared. The proof of the last identity is left to the reader. k Worksheet by Kuta Software LLC Let’s look at a “proof” of ! causes big problems, so you are better off using Law Microsoft Word - The Logic of Proving Trig Identities. Pythagorean theorem was proven by an acient Greek named Pythagoras and says that for a right triangle with legs A and B, and hypothenuse C See this lesson on Pythagorean Theorem, animated proof See How to generate triples of sizes that are natural See In Depth Wikipedia article on Pythagorean theorem Precalculus Notes: Unit 5 – Trigonometric Identities Page 2 of 23 Precalculus – Graphical, Numerical, Algebraic: Pearson Chapter 4 cot 1 cos sin22 cos cos sin sin sin sin cos sin b) tan csc sin tan cos 1 csc sin sin tan csc 1 cos sin 1 sec cos Ex2: Simplify the expression 2 csc 1 csc 1 cos xx x. If csc = 9 7 tan à L 7 8, find the values of the remaining trigonometric functions, using a (Suggested problems) Sec 5. The magic hexagon can help us remember that, too, by going clockwise around any of these three triangles: And we have: sin2(x) + cos2(x) = 1. Area and perimeter. In reference to the right triangle shown and from the functions of a right triangle: a/c = sin θ Introduction: In this lesson, three trigonometric identities will be derived and applied. This page will try to simplify a trigonometric expression. By using the ratio identities, the Pythagorean Identity sin cos 1,22xx and a little algebra you can derive the other two Pythagorean Identities: 1 tan sec 22 and 1 cot csc . How do you prove the three Pythagorean identities? Apparently each proof is very short. Trigonometric Identities You might like to read about Trigonometry first! Right Triangle. Practice with Pythagorean Identities. Hard Examples. It is named after Pythagoras, a mathematician in ancient Oct 29, 2018- Explore cindyjeanalbert's board "trig identities" on Pinterest. 1 + cos x = esc x + cot x sinx. How to verify trigonometric identities? Several examples with detailed solutions are presented. Concept 1: Pythagorean Identities, Sum/Difference & The Basic 8 Trig Identities (worksheet). Divide both sides by cos 2 (θ) to get the identity 1 + tan 2 (θ) = sec 2 (θ). In addition, the solutions of many types of applied problems require the use of trigonometric identities and the ability to manipulate these identities in Now, while converting everything to sine and cosine is a solid strategy ---it's certainly what I try first--- it's worth noting that the exercise is pretty straightforward when you're familiar with other Pythagorean identities, namely $$\sec^2\theta \equiv \tan^2\theta + 1 \qquad\qquad \csc^2\theta \equiv \cot^2\theta + 1$$ ©B w2m0C1f6k mKQuZtear mS[olfdtbwraLrweX `LvLaCi. Proving the Sine Addition and Subtraction Identities Proving the Cosine Addition and Subtraction Identities A Distance Formula Proof for the Pythagorean Theorem - How to use the Pythagorean Theorem, Converse of the Pythagorean Theorem, Worksheets, Proofs of the Pythagorean Theorem using Similar Triangles, Algebra, Rearrangement, examples, worksheets and step by step solutions, How to use the Pythagorean Theorem to solve real-world problems Fundamental Identities The equation x 2 + 2 x = x ( x + 2), for example, is an identity because it is valid for all replacement values of x . — Problem Set (It is not necessary to write a formal proof; just show the steps of the identity. The length of the longest side of the sail is $220$ centimeters, and the length of the other side of the sail is $5$ meters. The principle can be used on a rectangle of any size, of course. Trig Identity / Pythagorean Theorem confusion? you should have no problems understanding where the next two identities originated from. And locked in the realm of each tiny sphere Proving Trigonometric Identities. 2. Trigonometric Identities The Pythagorean Theorem, sin2 x+cos2 x = 1, has other forms, like (1) sin2 x cos2 x cos2 x cos2 x 1 cos2 x) tan2 x+1 = sec2 x and (2) sin2 x sin2 x cos2 x sin2 x 1 sin2 x) 1+cot2 x = csc2 x. apply a known rule, 2. Let r be the radius of circle C. Included is a list of essential identities, examples, and tips on proving identities. Pythagorean Theorem calculator. of analytical reasoning that is needed to prove trigonometric identities is essential for the study of calculus and other higher topics in mathematics. Consider a right angle ∆ABC which is right-angled at B as shown in the given figure. We already know the basics of algebra, we know why we use algebra and what are the general terms one needs to know, in order to solve an algebraic equation. Join thousands of learners improving their maths marks online with Siyavula Practice. Pythagorean Theorem Algebra Proof What is the Pythagorean Theorem? You can learn all about the Pythagorean Theorem, but here is a quick summary:. Here’s a third solution that also use a Pythagorean trig identity but avoids this difficulty. set about proving the identity using algebra and a previous identity. Using the trig identities for substituting a Pythagorean looking expression. State the three Pythagorean identities. some solved examples using Pythagorean identities. When you click the button, this page will try to apply 25 different trig. Identities involving trig functions are listed below. sec8sin8 tan8+ cot8 sin' 8 5 . PreCalculus Session - Trig Identities Card Sort etc In the morning precalculus sessions at #TMC13 we were asked to work with group members to develop a lesson, activity, assessment, etc to fit a topic that we found difficult to teach. (1 – sin (a)) (1 + sin (a)) 4. Problems 33 and 34 have students us the co-function properties. Use double- and half-angle formulas to solve real-life problems, such as finding the mach number for an airplane in Ex. 4 review. Starting with sin2(x) + cos2(x) = 1, and using your knowledge of the quotient and reciprocal identities, derive an equivalent identity in terms of tan(x) and sec(x). Note that the triangle below is only a representation of a triangle. ©K 12 p0W1y29 yK qu BtaE ZSMoyf0t swNaxr 0eF 2L 7LiCR. ♦ Half-Angle Theorem Derivation of Basic Identities; Derivation of Cosine Law; Derivation of Pythagorean Identities; Derivation of Pythagorean Theorem; Derivation of Sine Law; Derivation of Sum and Difference of Two Angles; Derivation of the Double Angle Formulas; Derivation of the Half Angle Formulas; Formulas in Solid Geometry Practical problems • When using Pythagoras’ theorem to solve practical problems, draw a right-angled triangle to represent the problem. Unlike a proof without words, a droodle may suggest a statement, not just a proof. Let’s start by working on the left side of the equation…. Also you should know the Pythagorean Theorem Definition. Learn exactly what happened in this chapter, scene, or section of Trigonometric Identities and what it means. So, replace sin . Questions* to toss out to the class might be: How do we account for angles in different quadrants? This enables us to solve equations and also to prove other identities. cas. The Pythagorean identities pop up frequently in trig proofs. -----= cos8 cot8. The Pythagorean Identities 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. Proving the Pythagorean Theorem (revisited). Practice Problems. Free trigonometric identities - list trigonometric identities by request step-by-step A. 5 – The student will solve application problems involving triangles 23 Feb 2013 Problem: Prove that the equation below is an identity: 2 sin2? – sin4? This is because sin2?=(1 – cos2?) from the Pythagorean identity 19 Sep 2012 OK, here's what an identity proof is: it's a list of expressions that are exactly the same as one side of the identity. Not only did these identities help us compute the values of the circular functions for angles, they were also useful in simplifying expressions involving the circular The purpose of the Pythagorean Theorem in real life is the same as it is in math class—to assist in solving right triangles. A short equation, Pythagorean Theorem can be written in the following manner: a²+b²=c². 1) If sin , find cos ( 2) If tan ( ) , find cot ( Engaging students with trigonometric identities may seem daunting, but I believe the key to success for this unit lies within allowing students to make the discovery of the identities themselves. pythagorean identities proof problems
ci, t6hnmv, plx6, zlltolsvb7, bfda, hgnn131, fzpqf, mnt, znmu, hgju38r, 3kth, | 2019-11-17T03:11:55 | {
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https://math.stackexchange.com/questions/4043519/theres-8-black-balls-and-7-white-balls-3-of-the-balls-are-drawn-at-random-pro | # There's 8 black balls and 7 white balls. 3 of the balls are drawn at random. Probability of drawing 2 of one color and 1 of the other color?
A bin has 8 black balls and 7 white balls. 3 of the balls are drawn at random. What is the probability of drawing 2 of one color and 1 of the other color?
Here's what I tried:
Case 1: 2 black balls and 1 white ball
8/15 * 7/14 * 7/13 = 392/2730 = 28/195
Case 2: 2 white balls, 1 black ball
7/15 * 6/14 * 8/13 = 336/2730 = 24/195
28/195 + 24/195 = 52/195 = 4/15
My teacher said it was wrong and I don't get why. She didn't give me the solutions either and just told me the answer was 4/5. Can someone tell me what I'm doing wrong? Thanks!
• You need to multiply by $3$ as for example, in the first case, the white ball can be either the first, second or third. – Math Lover Feb 28 at 18:50
Case $$1$$: $$2$$ black balls and $$1$$ white ball. The correct probability will be
$$P(2B,1W) = \displaystyle \frac{ 3 \cdot8 \cdot 7 \cdot 7}{15 \cdot 14 \cdot 13} = \frac{28}{65}$$.
$$\big[\text{It is basically } \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{8}{15} \cdot \frac{7}{14} \cdot \frac{7}{13} + \frac{7}{15} \cdot \frac{8}{14} \cdot \frac{7}{13}\big]$$
I would suggest think as selecting $$2$$ black balls out of $$8$$ and $$1$$ white ball out of $$7$$ vs. selecting any $$3$$ balls out of $$15$$ which can be written as -
$$P(2B,1W) = \displaystyle {8 \choose 2}{7 \choose 1} / {15 \choose 3}$$
Case $$2$$: $$2$$ white balls, $$1$$ black ball
$$P(2W,1B) = \displaystyle {8 \choose 1}{7 \choose 2} / {15 \choose 3} = \frac{24}{65}$$.
Adding them, we get probability as $$\displaystyle \frac{4}{5}$$.
What you have written corresponds to the case when first two balls of one color and then the ball of the other color are drawn. Therefore your calculations miss the factor 3 accounting for possible permutations of the balls.
As stated in the comments to your question, you have ascertained the probability of drawing two black balls and a white ball in that order and the possibility of drawing two white balls and a black ball in that order. You have to multiply by $$3$$ to account for all possible orders of drawing the balls you want.
Here's another approach. The only other possibility is to draw three balls of the same color. The probability of drawing three black balls is $$\frac{8}{15} \cdot \frac {7}{14} \cdot \frac{6}{13}=\frac{336}{15 \cdot 14 \cdot 13}$$ and the probability of drawing three white balls is $$\frac{7}{15} \cdot \frac{6}{14} \cdot \frac{5}{13}=\frac{210}{15 \cdot 14 \cdot 13}$$, so the probability of one of these two occurrences is $$\frac{446}{15 \cdot 14 \cdot 13}=\frac 15$$, and the probability that doesn't happen (i.e., you end up with two balls of one color and one ball of another color) must therefore be $$1-\frac 15=\frac 45$$.
The answer follows from the story of the Hypergeometric distribution! A random variable that counts the number of successes (in this case, let's call a success drawing a white ball) from a fixed number of draws (3, in our case) without replacement is said to have a Hypergeometric distribution. (If the scenario were instead with replacement, we'd have a Binomially distributed random variable on our hands.)
Let's let $$X$$ be this particular random variable. We know that at a minimum $$X = 0$$, and at a maximum, $$X = 3$$, but we're interested in the probability that $$X = 2$$ (2 white balls, 1 black ball) or $$X = 1$$ (1 white ball, 2 black balls). Since these cases are disjoint, we can simply add the probabilities of these two cases: $$P(X=1) + P(X=2)$$.
There are $$\binom{15}{3}$$ ways to pick 3 balls. To get exactly 2 white balls (and therefore, 1 black ball), we need to choose 2 white balls from the 7 possible white balls, and 1 black ball from the 8 possible black balls. The same logic can be used to evaluate the other case:
\begin{align} P(X=1) &= \frac{\binom{7}{1}\binom{8}{2}}{\binom{15}{3}}\\ P(X=2) &= \frac{\binom{7}{2}\binom{8}{1}}{\binom{15}{3}} \end{align}
Adding the probabilities together, we arrive at:
$$P(X=1 \cup X=2) = \frac{4}{5}$$ | 2021-04-11T04:13:06 | {
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https://calculus.subwiki.org/wiki/Derivative_of_differentiable_function_on_interval_satisfies_intermediate_value_property | # Derivative of differentiable function satisfies intermediate value property
## Name
The result is sometimes called Darboux's theorem, and is attributed to the mathematician Darboux. However, there is another more famous theorem named after Darboux.
## Statement
### Endpoint one-sided version
Suppose $f$ is a function defined on an interval $[a,b]$. Suppose:
• $f'$ (the derivative of $f$) exists on the open interval $(a,b)$.
• The right hand derivative $f'_+(a)$ exists
• The left hand derivative $f'_-(b)$ exists
Suppose further that $f'_+(a) \ne f'_-(b)$ and $w$ is a real number strictly between these two numbers. Then, there exists $c \in (a,b)$ such that $f'(c) = w$.
### Two-sided version
Suppose $f$ is a differentiable function on an open interval $(p,q)$. Then, the derivative $f'$ satisfies the intermediate value property on $(p,q)$: for $a < b$, both in $(p,q)$, and any value $w$ is strictly between $f'(a)$ and $f'(b)$, there exists $c \in (a,b)$ such that $f'(c) = w$.
## Proof
### Proof idea
The proof idea is to find a difference quotient that takes the desired value intermediate between $f'(a)$ and $f'(b)$, then use Fact (3).
### Proof details for one-sided endpoint version using the mean value theorem
The two-sided version follows from the one-sided endpoint version, so we only prove the latter.
Given: $f$ is a function defined on an interval $[a,b]$. Suppose:
• $f'$ (the derivative of $f$) exists on the open interval $(a,b)$.
• The right hand derivative $f'_+(a)$ exists
• The left hand derivative $f'_-(b)$ exists
Further, $f'_+(a) \ne f'_-(b)$ and $w$ is a real number strictly between these two numbers.
To prove: There exists $c \in (a,b)$ such that $f'(c) = w$.
Proof: Consider the function $g$ defined on the interval $[0,1]$ as follows. By $\Delta f$ we denote the difference quotient. Note that we use $f'_+$ for the right hand derivative and $f'_-$ for the left hand derivative:
$g(t) := \left\lbrace\begin{array}{rrl} f'_+(a), & & t = 0 \\ (\Delta f)(2tb + a(1 - 2t),a) = & \frac{f(2tb + a(1 - 2t)) - f(a)}{2t(b - a)}, & 0 < t < 1/2 \\ (\Delta f)(b,a) = & \frac{f(b) - f(a)}{b - a}, & t = 1/2\\ (\Delta f)(b,(2 - 2t)a + b(2t - 1)) = & \frac{f(b) - f((2 - 2t)a + b(2t - 1))}{(2 - 2t)(b - a)}, & 1/2 < t < 1\\ f'_-(b), & & t = 1 \\\end{array}\right.$
Note that for $0 < t < 1$, $g(t)$ is a difference quotient between two points in $[a,b]$, at least one of which is one of the endpoints $a,b$.
Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $g$ is continuous on $(0,1/2)$ and $(1/2,1)$. Fact (1) $f$ is differentiable on $(a,b)$ and one-sided differentiable at the endpoints. Follows directly from continuity of $f$ and the nature of the expressions.
2 $g$ is right continuous at $t = 0$ definition of derivative as a limit of a difference quotient [SHOW MORE]
3 $g$ is continuous at $t = 1/2$ Fact (1) $f$ is differentiable We take the limit by plugging $t = 1/2$ in the left side and right side definition and check. We again use the continuity of $f$ by Fact (1).
4 $g$ is left continuous at $t = 1$ definition of derivative as a limit of a difference quotient [SHOW MORE]
5 $g$ is continuous on $[0,1]$. Steps (1)-(4) step-combination direct.
6 There exists $t \in (0,1)$ such that $g(t) = w$. In particular $w$ is a difference quotient between two points, both of them in $[a,b]$. Fact (2) $w$ is between $f'(a)$ and $f'(b)$. Step (5) By definition, $f'_+(a) = g(0)$ and $f'_-(b) = g(1)$. Step (5) and Fact (1) tell us that since $w$ is between these, there exists $t \in (0,1)$ such that $g(t) = w$.
7 There exists $c \in (a,b)$ such that $f'(c) = w$. Fact (3) $f$ is differentiable on $(a,b)$ Step (6) Step-fact combination direct.
### Proof details for direct proof of one-sided version
There is a direct proof that does not involve any appeal to the mean value theorem. This proof is shorter, but relies on the extreme value theorem. Note that the previous proof that relies on the mean value theorem indirectly relies on the extreme value theorem, whereas the proof below makes a direct appeal to the extreme value theorem.
Fill this in later | 2021-02-27T21:58:27 | {
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http://neda.psdeg-psoe.org/htlyz6/4c1425-shortest-distance-between-two-parallel-lines | This command calculates the 2D distance between entities. Classes. Shortest distance between two parallel lines. 2 mins read. So it's a fairly simple "distance between point and line" calculation (if the distances are all the same, then the lines are parallel). If you know the lines are parallel, you can solve the problem using the formula for the distance between a point and a line: form a vector from a point on the first line to a point on the second line and cross it with the normalized direction vector of one of the lines. The given two parallel lines = + s and = + t are denoted by L 1 and L 2 respectively. Before we proceed towards the shortest distance between two lines, we first try to find out the distance formula for two points. Find a line parallel to two planes and intersecting two lines. Short answer: yes, for both 2d and 3d. So sqr(m^2+1) times height of parallelogram = abs(b1-b2) and finally, the shortest distance between the two lines = … 0. shortest distance between 2 vectors vectors question Edxecel AS core pure paper show 10 more Why is the shortest distance the perpendicular distance for parallel lines? How to Find Find shortest distance between two lines and their Equation. – Michelle Aug 21 '13 at 19:55 You must make note that the shortest distance between parallel lines is actually the length of the perpendicular between them or joining the two lines. If two lines intersect at a point, then the shortest distance between is 0. The shortest distance between two skew lines lies along the line which is perpendicular to both the lines. 0. Calculates the shortest distance between two lines in space. As we know that the vector equation of a line is of form r = a + mb where a is the position vector through which line is passing, b is a vector parallel to line and m is a constant. Consider two parallel lines and .Pick some point on .Now pick a point to vary along .Say is a point on such that is perpendicular to both lines. Dear friends, Situation: There's 2 roads next to eachother. The are not parallel, and have curves in them. The distance between the intersection points A´ 1 and A´ 2 is at the same time the distance between given lines, thus: Distance between two skew lines Through one of a given skew lines lay a plane parallel to another line and calculate the distance between any point of that line and the plane. Two lines will either be parallel or skew. Alternatively we can find the distance between two parallel lines as follows: Considers two parallel lines $\begin{gathered} ax + by + c = 0 \\ ax + by + {c_1} = 0 \\ \end{gathered}$ Now the distance between two parallel lines can be found with the following formula: Keywords: Math, shortest distance between two lines. Am I right in thinking, that the shortest distance between two parallel lines, say L 1 = r 1 + λt and L 2 = r 2 + μt, is always going to be: Nearly, it's . Skip to primary navigation; ... and the shortest distance between a point and a line is the length of a perpendicular line from that point to the line. It should be pretty simple to see why intuitively. A line parallel to Vector (p,q,r) through Point (a,b,c) is expressed with x − a p = y − b q = z − c r x − a p = y − b q = z − c r If two equations of line. Shortest Distance between a Pair of Skew Lines. Now: i need to give the distance between them roads along the full road. Line passing through the point B(a2,b2,c2) parallel to the vector V2(p2,q2,r2) Point B (,,) Vector V2 (,,) Shortest distance between two lines(d) 1. Otherwise, draw a diagram and consider Pythagoras' Theorem. The shortest distance between the two parallel lines can be determined using the length of the perpendicular segment between the lines. We will look at both, Vector and Cartesian equations in this topic. Solution of I. Distance between two parallel line in 3D . But area of the parallelogram is also base times height where the height of the parallelogram is the shortest distance between the parallel lines. The distance between two parallel planes is understood to be the shortest distance between their surfaces. This lesson lets you understand the meaning of skew lines and how the shortest distance between them can be calculated. Say the perpendicular distance between the two lines is , and the distance varies since our point B varies, call this distance . Think about that; if the planes are not parallel, they must intersect, eventually. Consider two parallel lines whose equations in vector form are given by. If they intersect, then at that line of intersection, they have no distance -- 0 distance -- between them. The distance between two lines in $$\mathbb R^3$$ is equal to the distance between parallel planes that contain these lines.. To find that distance first find the normal vector of those planes - it is the cross product of directional vectors of the given lines. Ex 11.2, 14 Find the shortest distance between the lines ⃗ = ( ̂ + 2 ̂ + ̂) + ( ̂ − ̂ + ̂) and ⃗ = (2 ̂ − ̂ − ̂) + (2 ̂ + ̂ + 2 ̂) Shortest distance between the lines with vector equations ⃗ = (1) ⃗ + (1) ⃗and ⃗ = (2) ⃗ + (2) ⃗ is AS Further Maths question vector help Further vectors help please! The shortest distance between the lines is the distance which is perpendicular to both the lines given as compared to any other lines that joins these two skew lines. The shortest distance between two intersecting lines is zero. Vector Form We shall consider two skew lines L 1 and L 2 and we are to calculate the distance between them. The blue lines in the following illustration show the minimum distance found. Proof that if two lines are parallel, then all the points on one line are an equal distance ("Equidistant") from the other line. I have two line segments: X1,Y1,Z1 - X2,Y2,Z2 And X3,Y3,Z3 - X4,Y4,Z4. I am trying to find the shortest distance between the two segments. Finding plane equation given two lines. Shortest Distance Between Skew Lines with Basic Geometry. If so, the answer is simply the shortest of the distance between point A and line segment CD, B and CD, C and AB or D and AB. It does not matter which perpendicular line you are choosing, as long as two points are on the line. Elevations are not considered in the calculations. If you're looking for a proof, the math SE is probably a better place. Note that the distance between two intersecting lines is zero. 6. If the selected entities cross or are collinear, the distance is displayed as zero The shortest distance between two skew lines is the length of the shortest line segment that joins a point on one line to a point on the other line. Formula of Distance If there are two points say A(x 1 , y 1 ) and B(x 2 , y 2 ), then the distance between these two points is given by √[(x 1 -x 2 ) 2 + (y 1 -y 2 ) 2 ]. The shortest distance between skew lines is equal to the length of the perpendicular between the two lines. If two lines are parallel, then the shortest distance between will be given by the length of the perpendicular drawn from a point on one line form another line. In three-dimensional geometry, skew lines are two lines that do not intersect and are not parallel.A simple example of a pair of skew lines is the pair of lines through opposite edges of a regular tetrahedron.Two lines that both lie in the same plane must either cross each other or be parallel, so skew lines can exist only in three or more dimensions. In the following section, we shall move on to explore how the distance between parallel lines can be measured. Now comparing these equations with standard form , and write , and vectors ,we get. Hot Network Questions The line segment is perpendicular to both the lines. The distance between two skew lines is naturally the shortest distance between the lines, i.e., the length of a perpendicular to both lines. This concept teaches students how to find the distance between parallel lines using the distance formula. The line1 is passing though point A (a 1 ,b 1 ,c 1 ) and parallel to vector V 1 and The line2 is passing though point B(a 2 ,b 2 ,c 2 ) and parallel to vector V 2 . I have been looking for a solution for hours, but all of them seem to work with lines rather than line segments. Shortest Distance between two lines in the 3D plane. In the same way, the shortest distance between two skew lines is defined as the length of the line segment perpendicular to both the skew lines. Experiment to confirm for 2D: draw two parallel lines on a sheet of paper, draw a line between them perpendicular to both, and then try to draw a shorter line between them that isn't perpendicular to both. Is this in 2 dimensions? Problem 1. Online space geometric calculator to find the shortest distance between given two lines in space, each passing through a point and parallel to a vector. 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https://math.stackexchange.com/questions/2550852/nested-integral-volume-of-a-right-simplex | # Nested integral: volume of a right simplex
Consider the integral $$I_n \equiv \int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots \int_{t_n = t_{n-1}}^T dt_n \, .$$ I suspect the result should be $I_n = T^n/n!$ but would like to prove it. Doing the $n^\text{th}$ integral gives \begin{align} I_n &= \int_{t_1=0}^T dt_1 \cdots \int_{t_{n-1}=t_{n-2}}^T dt_{n-1} \, (T-t_{n-1}) \\ &= T I_{n-1} - \int_{t_1=0}^T dt_1 \dots \int_{t_{n-1}=t_{n-2}}^T dt_{n-1} \, t_{n-1} \, . \end{align} Is there some way to evaluate the second term or to come up with a useful recurrence relation or induction step?
I think induction is your best bet here, and it doesn't even really require any $I_n$ notation. We can simply evaluate the integral on it's own. Starting with your first step, which you did correctly, we begin to integrate. $$\int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots \int_{t_n = t_{n-1}}^T dt_n = \int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots\int_{t_{n-1}=t_{n-2}}^T (T-t_{n-1})dt_{n-1}$$ Integrating by power rule (and chain rule for the negative), we get $$\int_{t_1=0}^T dt_1 \int_{t_2 = t_1}^T dt_2 \cdots\int_{t_{n-2}=t_{n-3}}^T -\frac{(T-t_{n-1})^{2}}{2!} \Bigg|_{t_{n-2}}^{T} dt_{n-2} \\ =\cdots\int_{t_{n-2}=t_{n-3}}^T -\frac{\left(T-T\right)^2}{2!} -\left(-\frac{\left(T-t_{n-2}\right)^2}{2!}\right) dt_{n-2} \\ = \cdots\int_{t_{n-2}=t_{n-3}}^T \frac{\left(T-t_{n-2}\right)^2}{2!}dt_{n-2}$$ And finally, we see a pattern, so let us prove the induction step $$\int_{t_{k}=t_{k-1}}^T\frac{(T-t_k)^a}{a!} dt_k = -\frac{(T-t_k)^{a+1}}{(a+1)!}\Bigg|_{t_{k-1}}^T = \frac{(T-t_{k-1})^{a+1}}{(a+1)!}$$ Now that we have that, we integrate along until we get to the final integral with $t_1$ $$\int_{t_1=0}^T \frac{(T-t_1)^{n-1}}{(n-1)!}dt_1 = -\frac{(T-t_1)^n}{n!} \Bigg|_0^T = \frac{T^n}{n!}$$ And there we have the answer: the volume of a regular right simplex!
• I don't understand where $-(T-T)^2/2!$ comes from. I submitted an edit to remove it, but apparently it was rejected. – DanielSank Dec 6 '17 at 19:18
• It comes from the fact that we are evaluating a definite integral. I will add a line above to show this. – Isaac Browne Dec 6 '17 at 19:52
• Oh I see. I would have integrated the $T$ and $-t_{n-1}$ separately, but your way is better. – DanielSank Dec 6 '17 at 20:02
Here's an alternative, albeit perhaps a less inspiring approach. The way to read off that the integral $$I_n = \int_0^T dt_1 \int_{t_1}^T dt_2 \cdots \int_{t_{n-1}}^T dt_n$$ is measuring volume of the regular right $n$-simplex of height $T$ is to see that the domain of the integral is $V = \{(t_1, t_2, \cdots, t_n) \in \Bbb R^{n+1} : t_{k-1} \leq t_k \leq T; t_0 = 0 \}$. This is the convex subset of $\Bbb R^{n+1}$ defined by the set of linear inequalities $0 \leq t_1 \leq \cdots \leq t_n \leq T$; one can easily see that this is nothing other than a simplex (inductive argument: fix a $t_1 = t_{01}$ and observe that at that slice it's just the same convex subset in one dimension lower. It's clearly a simplex for $n = 1$.)
Here's a picture for $n = 2$ (stolen from Shifrin's "Multivariable Mathematics"):
where $t_1 = x$ and $t_2 = y$ are the variables in the convention of the image. Here we have changed the slices of the volume from $(t_1 = c)$'s for various $c \in [0, T]$ to $(t_2 = c)$'s for various $c \in [0, T]$. What this accomplishes is the following; we change the limits of our integral to $0 \leq t_2 \leq T$ and $0 \leq t_2 \leq t_1$, and accordingly switches the order of the integrals:
$$I_2 = \int_0^T dt_1 \int_{t_1}^T dt_2 = \int_0^T dt_2 \int_0^{t_2} dt_1$$
We can similarly generalize. Change the limits (and order) of the integral $I_n$ from the inequalities $0 \leq t_1 \leq T, t_1 \leq t_2 \leq T, \cdots, t_{n-1} \leq t_n \leq T$ to $0 \leq t_n \leq T, 0 \leq t_{n-1} \leq t_n, \cdots, 0 \leq t_1 \leq t_2$. Hence, we obtain the equality $$I_n = \int_0^T dt_n \int_0^{t_n} dt_{n-1} \cdots \int_0^{t_2} dt_1$$
Notice that this is secretly nothing but Fubini; we have changed the order of integrals. Now, why is this useful? Just notice that we can start evaluating the integral from the innermost iterated integral
\begin{align} I_n = \int_0^T dt_n \int_0^{t_n} dt_{n-1} \cdots \int_0^{t_3} dt_2 \int_0^{t_2} dt_1 & = \int_0^T dt_n \int_0^{t_n} dt_{n-1} \cdots \int_0^{t_{k+1}} dt_k \; t_k^{k-1}/(k-1)! \\ &= \int_0^T dt_n \; t_n^{n-1}/(n-1)! \\ &= T^n/n!\end{align}
So your integral is indeed the volume of the regular right-angled $n$-simplex of height $T$ which is indeed equal to $T^n/n!$. | 2021-04-21T10:38:15 | {
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https://math.stackexchange.com/questions/2107373/how-is-lim-limits-n-to-infty-left-frac-lnn1-lnn-left1-frac | How is $\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)}{\ln(n)}\left(1 + \frac{1}{n}\right) \right)^n = e$?
Tried this question here How to calculate $\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)^{n+1}}{\ln n^n} \right)^n$? and was curious about the result. The answer according to Wolfram Alpha is $e$, so I wanted to try it.
$\lim\limits_{n \to \infty} \left( \frac{\ln((n+1)^{n+1})}{\ln (n^n)} \right)^n$
$\lim\limits_{n \to \infty} \left( \frac{(n+1)\ln(n+1)}{n\ln (n)} \right)^n$
$\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)}{\ln(n)}\left(1 + \frac{1}{n}\right) \right)^n$
This is similar to the typical definition $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$ but it has the extra log factors.
How come these two happen to be equivalent? Is it valid to apply L'Hospital's Rule to the logs even though they're inside the $()^n$? Or can it be applied to just part of the function and not the other half? What's the correct way to handle this extra log multiplier?
For instance:
$\lim\limits_{n \to \infty}\frac{\ln(n+1)}{\ln(n)} = \lim\limits_{n \to \infty}\frac{\frac{d}{dn}\ln(n+1)}{\frac{d}{dn}\ln(n)} = \lim\limits_{n \to \infty}\frac{n}{1+n} = \lim\limits_{n \to \infty}\frac{1}{1/n+1} = 1$
but I don't think we can necessarily analyze this "separately" from the main result; I think they must be taken together somehow. I also considered squeeze theorem but couldn't think of another function approaching $e$ from the other side.
• The log fraction goes to 1 as n goes to infinity, no? – pie314271 Jan 21 '17 at 13:58
• Yes, but I want a more formal proof; in practice if I get in the habit of trying to eyeball things, I get it wrong – Jay Smith Jan 21 '17 at 13:58
• The intuition says that $\ln(n+1)/\ln(n)\approx 1$ as $n$ gets large, but of course this does not constitute a rigorous proof. – Eff Jan 21 '17 at 13:59
• @pie314271 (and Eff too): that is an invalid argument. The $(1+\frac{1}{n})$ part tends to $1$ too, but still the limit is not $1$. – TonyK Jan 21 '17 at 14:00
• I don't think I showed that – Jay Smith Jan 21 '17 at 14:06
When $n\to\infty$, we get $$\frac{\ln(n+1)}{\ln n}= \frac{\ln n+\ln\left(1+\frac{1}{n}\right)}{\ln n} = 1+ \frac{\ln\left(1+\frac{1}{n}\right)}{\ln n} = 1 + \frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right) \tag{1}$$ (using that $\ln(1+x)=x+o(x)$ when $x\to0$) so that \begin{align} \frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right) &= \left(1 + \frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\right)\left(1+\frac{1}{n}\right) = 1+\frac{1}{n}+\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\\ &= 1+\frac{1}{n}+o\left(\frac{1}{n}\right) \tag{2} \end{align} and from (2) and the same Taylor expansion of $\ln(1+x)$ at $0$ we get \begin{align} \left(\frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right)\right)^{n} &= e^{n\ln \left(\frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right)\right)} = e^{n\ln \left(1+\frac{1}{n}+o\left(\frac{1}{n}\right)\right)} = e^{n\left(\frac{1}{n}+o\left(\frac{1}{n}\right)\right)} = e^{1+o\left(1\right)} \\&\xrightarrow[n\to\infty]{} e^1 = e \end{align} as claimed.
• Also how do you go from $1+\frac{1}{n}+\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)$ to $1+\frac{1}{n}+o\left(\frac{1}{n}\right)$? What does $o$ mean here exactly? – Jay Smith Jan 21 '17 at 14:19
• @JaySmith I am heavily biased on using Taylor expansions whenever possible (as it's a systematic way to tackle limits, and works almost every time). With that in mind, you want to make "standard" quantities appear, with something tending to $0$: the thing that is mainly the issue here is the ratio of logarithms, so the first thing to do is massage it. $$\ln(n+1) = \ln n + \ln\left(1+\frac{1}{n}\right)$$ is a good thing to do to achieve our goals, because of that. After this step, it's pretty much on autopilot. – Clement C. Jan 21 '17 at 14:20
• The second (and last) standard and very useful step is to write the quantity of interest $f(n)^{g(n)}$ in its exponential form $\exp(g(n)\ln f(n))$: by continuity of $\exp$, one then only has to care about the exponent $g(n)\ln f(n)$, and this does almost always make things much clearer. – Clement C. Jan 21 '17 at 14:21
• @JaySmith $o(\cdot)$ ("little o") is the Landau notation. One writes $f(n)=o(g(n))$ (at some specified point, here when $n\to \infty$) if, basically, $\frac{f(n))}{g(n)}\xrightarrow[n\to\infty]{} 0$. In our case, we have $\frac{\frac{1}{n\ln n}+o\left(\frac{1}{n\ln n}\right)}{\frac{1}{n}} = \frac{1}{\ln n} + o\left(\frac{1}{\ln n}\right) \xrightarrow[n\to\infty]{} 0$. – Clement C. Jan 21 '17 at 14:23
Use my comment in the question mentioned to use that if $a_n\to a$ then $$\left(1+\frac{a_n}{n}\right)^n \to e^a$$ in this case $$a_n=n\frac{\ln(n+1)-\ln n}{\ln n}=\frac{1}{\ln n}\ln \left(1+\frac{1}{n}\right)^n\to 0$$ and thus $$\left(\frac{\ln (n+1)}{\ln n}\right)^n\to 1$$ | 2019-06-17T04:50:27 | {
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http://mathhelpforum.com/discrete-math/154199-proof-induction.html | # Math Help - Proof by induction.
1. ## Proof by induction.
I have a question:
Prove that $n^{5}-n$ is divisible by 10 for all $n\ge2$.
Since my current chapter is on induction, I figure it has something to do with induction, I don't really know. But all the induction I've learned in this chapter looks something like
$2,6,12...n(n+1) = \frac{n(n+1)(n+2)}{3}$ etc..
Any help?
2. Originally Posted by spycrab
I have a question:
Prove that $n^{5}-n$ is divisible by 10 for all $n\ge2$.
Since my current chapter is on induction, I figure it has something to do with induction, I don't really know. But all the induction I've learned in this chapter looks something like
$2,6,12...n(n+1) = \frac{n(n+1)(n+2)}{3}$ etc..
Any help?
follow the procedure for proof by induction.
1. show the statement is true for n = 2
2. assuming the statement is true for n, show that the statement is true for n+1
3. Yes, I know that, but what I'm trying to say is how do you show that it is divisible by 10
4. someone may offer a shorter solution, but here's the way I did it ...
$2^5 - 2 = 30 = 3 \cdot 10$ ... true for $n = 2$
$(n+1)^5 - (n+1) =$
$n^5 + 5n^4 + 10n^3 + 10n^2 + 5n + 1 - n - 1 =$
$n^5 - n + 5n^4 + 10n^3 + 10n^2 + 5n =$
$(n^5 - n) + 10(n^3 + n^2) + 5n(n^3 + 1)$
since the statement is true for $n$, then $n^5 - n$ can be written in the form $10k$, where $k$ is an integral value ...
$10k + 10(n^3 + n^2) + 5n(n^3 + 1)$
the first term is a multiple of 10, as is the second term.
for the 3rd term, $5n(n^3 + 1)$ ...
if $n$ is even, then $5n$ is a multiple of 10.
if $n$ is odd, then $n^3 + 1$ is even, making the 3rd term also a multiple of 10.
5. thank you!
6. Yes, skeeter's method is exactly how I did it also.
I would word it slightly differently, but that is just my way!
P(k)
$k^5-k$ is divisible by 10 ?
P(k+1)
$k^5-k$ being divisible by 10 causes $(k+1)^5-(k+1)$ to also be divisible by 10 ?
If we can show that $(k+1)^5-(k+1)$ must be divisible by 10 if $k^5-k$ is,
then we've shown that if P(k) is true for k=2, this causes P(k) to be true for k=3,
which causes P(k) to be true for k=4....
a never-ending chain of cause and effect.
Then P(k) being true for k=2, causes P(k) to be true for all k above 2
in the natural number system.
We never know that the original statement is true until after we have
validated P(k+1) and checked the formula for k=2.
K and k+1 are used instead in the notation instead of n and n+1 merely to focus in on
$(n+1)^5-(n+1) = n^5 - n + 5n(n+1)(n^2+n+1)$
and
$10| n^5-n$ and $10|5n(n+1)$.
8. We can also get to the result without induction.
$N = n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)=n(n-1)(n+1)(n^2+1)$
The RHS contains the product of three consecutive integers, so one of them must be even, so $N$ is divisible by 2.
If one of these three is divisible by 5 also then we are done, so assume that this isn't the case, in which case either $n + 1$ is 1 less than a multiple of 5 or $n - 1$ is 1 more than a multiple of 5.
Taking the first of these possibilities,
let $n+1 = 5p-1,$ say, then $n=5p-2,$
so $n^2+1=(5p-2)^2+1=25p^2-20p+5$ which is divisible by 5, and in which case so is $N$ and so $N$ is divisible by 10. Similarly for the other possibility. | 2014-03-07T07:34:59 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=k441o32850lkospornhudrbhs7&topic=281.0;wap2 | MAT244-2013S > Easter and Semester End Challenge
Easter challenge
(1/2) > >>
Victor Ivrii:
Draw phase portraits :
\begin{gather}
\left\{\begin{aligned}
&x'=-\sin(y),\\
&y'= \sin (x);
\end{aligned}\right. \tag{a}\\
\left\{\begin{aligned}
&x'=-\sin(y),\\
&y'= \,2\sin (x);
\end{aligned}\right. \tag{b}
\end{gather}
Explain the difference between portraits and its reason
Hareem Naveed:
Attached are the two phase portraits.
In terms of difference between the two; they are level curves of the following functions:
$$H_{a}(x,y) = \cos(y)+\cos(x) \\ H_{b}(x,y) = \cos(y) + 2\cos(x)$$
Level curves are also attached.
From the level curves, in a, the centres are hemmed in by 2 defined separatrices, not so in b where there is only one.
How could I formalize these statements? Intuitively, I can "see" the answer.
Victor Ivrii:
--- Quote from: Hareem Naveed on March 29, 2013, 11:38:46 AM ---Attached are the two phase portraits.
In terms of difference between the two; they are level curves of the following functions:
$$H_{a}(x,y) = \cos(y)+\cos(x) \\ H_{b}(x,y) = \cos(y) + 2\cos(x)$$
Level curves are also attached.
From the level curves, in a, the centres are hemmed in by 2 defined separatrices, not so in b where there is only one.
How could I formalize these statements? Intuitively, I can "see" the answer.
--- End quote ---
Your solution is definitely correct but I want a bit more observation about dynamics, not only about centers but about all trajectories. Start from the very informal description as I am interested more in an understanding than the formal expression
Alexander Jankowski:
This is a nice problem. I can give a little bit of input: in system (b), the rate of change $y'(x)$ is twice that of the same rate in system (a). Perhaps we can treat (b) as a vertical expansion of (a), which is what the contour maps suggest. We can also note that each system has the same critical points $(2 \pi n,2 \pi n)$, where $n$ is an integer.
By inspection, we see that the separatrices in system (a) are lines with slopes $\pi$ and $-\pi$. In fact, by looking more closely at the contour maps, the separatrices appear to have the equations $$y = ± \pi x + n \pi,$$ where $n$ is an integer. In system (b), the separatrices are sinusoidal functions that oscillate about an equilibrium line that is parallel to the $y$-axis. Finally, because there are infinitely many critical points in each system, there are also infinitely many separatrices.
Victor Ivrii:
--- Quote from: Alexander Jankowski on March 29, 2013, 06:11:46 PM ---This is a nice problem. I can give a little bit of input: in system (b), the rate of change $y'(x)$ is twice that of the same rate in system (a). Perhaps we can treat (b) as a vertical expansion of (a), which is what the contour maps suggest. We can also note that each system has the same critical points $(2 \pi n,2 \pi n)$, where $n$ is an integer.
By inspection, we see that the separatrices in system (a) are lines with slopes $\pi$ and $-\pi$. In fact, by looking more closely at the contour maps, the separatrices appear to have the equations $$y = ± \pi x + n \pi,$$ where $n$ is an integer. In system (b), the separatrices are sinusoidal functions that oscillate about an equilibrium line that is parallel to the $y$-axis. Finally, because there are infinitely many critical points in each system, there are also infinitely many separatrices.
--- End quote ---
I know that this is a nice problem, I just invented it :D You observe interesting things (correctly) albeit separatrices are only sinusoid-looking but not sinusoid lines.
But what about other trajectories which are not separatrices?
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http://mathhelpforum.com/statistics/124781-problem-about-probability.html | So you have 3 phone companies: Company A has a 41% share on the market, B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.
Now they say that 35% of A and B and 30% of C company customers are unsatisfied with the service.
What is probability of a satisfied customer be connected to company B?
Can anyone help me solve this, thank you!
2. Originally Posted by Wright
B has 38% ....
Now they say that 35% of B are unsatisfied with the service.
What is probability of a satisfied customer be connected to company B?
I would suggest 35% of 38% will 'stay' connected.
$\displaystyle \frac{35}{100}\times\frac{38}{100}= \dots$
3. Originally Posted by Wright
So you have 3 phone companies: Company A has a 41% share on the market, B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.
Now they say that 35% of A and B and 30% of C company customers are unsatisfied with the service.
What is probability of a satisfied customer be connected to company B?
I interpreted the question to mean: What proportion of all satisfied customers are connected to company B? Since 65% of A and B customers, and 70% of C customers, are satisfied, the answer should then be $\displaystyle \frac{38\times0.65}{41\times0.65 + 38\times0.65 + 21\times0.7}$.
4. Hello, Wright!
I agree with Opalg . . . this is Conditional Probability.
You have 3 phone companies: Company A has a 41% share on the market,
B has 38% and C has 21%, summing up gives 100%, so they are really the only ones.
Now they say that: 35% of A and B and 30% of C company customers are unsatisfied with the service.
What is probability of a satisfied customer be connected to company B?
We have: .$\displaystyle \begin{array}{ccc} P(A) &=& 0.41 \\ P(B) &=& 0.38 \\ P(C) &=& 0.21 \end{array}$
Let: . $\displaystyle \begin{array}{ccc}S &=& \text{satisfied} \\ U &=& \text{unsatisfied} \end{array}$
Then: .$\displaystyle \begin{array}{ccccccc} P(A_U) &=& 0.35 & \Rightarrow & P(A_S) &=& 0.65 \\ P(B_U) &=& 0.35 & \Rightarrow & P(B_S) &=& 0.65 \\ P(C_U) &=& 0.30 & \Rightarrow & P(C_S) &=& 0.70 \end{array}$
Bayes' Theorem: .$\displaystyle P(B\,|\,S) \;=\;\frac{P(B\,\wedge\,S)}{P(S)}\;\;{\color{blue}[1]}$
Numerator: .$\displaystyle P(B\,\wedge S) \:=\:(0.38)(0.65) \:=\:0.247\;\;{\color{blue}[2]}$
Denominator:
. . $\displaystyle \begin{array}{ccccccc}P(A \wedge S) &=& (0.41)(0.65) &=& 0.2665 \\ P(B \wedge S) &=& (0.38)(0.65) &=& 0.2470 \\ P(C \wedge S) &=& (0.21)(0.30) &=& 0.1470 \\ \hline \text{Total:}& & P(S) &= & 0.6605 & {\color{blue}[3]} \end{array}$
Substitute [2] and [3] into [1]: .$\displaystyle P(B\,|\,S) \;=\;\frac{0.247}{0.6605} \;\approx\;37.4\%$
5. Thanks for the kind answers everyone. | 2018-05-25T20:54:23 | {
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https://math.stackexchange.com/questions/4532364/conjecture-about-product-of-lengths-in-a-circle | # Conjecture about product of lengths in a circle
In a circle of radius $$r>1$$, point $$P$$ is a distance $$\sqrt{r^2-1}$$ from the centre of the circle. From $$P$$, draw $$n$$ line segments to the circle, such that the angles between neighboring line segments are equal. (There is more than one way to do this, because all the line segments can be rotated together; any of these ways will do.)
Here is an example with $$r=\sqrt2$$ and $$n=12$$.
Call the lengths of the line segments $$l_1, l_2, l_3, ..., l_n$$.
I am trying to prove (or disprove) the following conjecture:
$$\lim\limits_{n\to\infty}\prod\limits_{k=1}^n l_k=1$$
My attempt
If $$n$$ is even, then consider an arbitrary chord $$AB$$ through $$P$$, and diameter $$CD$$ through $$P$$. Using the intersecting chords theorem, we have
$$(AP)(PB)=(CP)(PD)=(r-\sqrt{r^2-1})(r+\sqrt{r^2-1})=1$$
$$\therefore\prod\limits_{k=1}^n l_k=1$$
$$\therefore \lim\limits_{n\to\infty}\prod\limits_{k=1}^n l_k=1$$
If $$n$$ is odd, then we can easily verify that $$\prod\limits_{k=1}^n l_k$$ does not always equal $$1$$. For example, let $$r=\sqrt2$$ and $$n=3$$, and let one of the line segments go through the centre of the circle. We can easily calculate that $$\prod\limits_{k=1}^n l_k=\frac{1}{2}(3+3\sqrt2-\sqrt5-\sqrt{10})\approx 0.922$$.
But since the product with even $$n$$ always equals $$1$$ (with or without taking the limit), it seems plausible that the product with odd $$n$$ approaches $$1$$ as $$n\to\infty$$. But I have not found a convincing way to prove this.
(This question was inspired by this related question and @person's answer.)
• When $n$ is odd you can extend the chords and this procedure gives $2n$-chords case. And the product of the added chords must be close to the product of $n$ chords we started. Sep 15 at 22:17
• @Bob Dobbs The problem is that you an multiply an ever-growing number of multiples close to starting ones. E.g. $e^{1/n}$ tends to $1$, but a product of $n$ copies of $e^{1/n}$ does not. Sep 15 at 23:25
• You should first try to prove that the answer is independent of $r$, then prove that it is independent of the ways of rotating the lines so that equiangularity is ensured. Sep 16 at 9:35
• For odd $n$ product can be far from 1 for big $r$, for example at $r=100$, $n=9$ product can be from $1/a$ to $a$, where $a>16$. But I believe $a$ value is approaching 1 with $n$ increase. Sep 16 at 10:05
We can calculate the line segment lengths as $$x_k=l_{k+1}=\sqrt{1+(r^2-1)\cos^2(\frac{2\pi k}{n})}-\sqrt{r^2-1}\cos(\frac{2\pi k}{n})$$ where $$k=0,1,2,...,n-1$$.
For $$n=4m+1$$, I wrote the product in the form: $$\prod_{k=0}^{n-1}x_k=x_{\frac{n-1}{4}}\prod_{k=0}^{\frac{n-5}{4}}x_{k}x_{k+\frac{n+1}{2}}\prod_{k=\frac{n+3}{4}}^{\frac{n-1}{2}}x_{k}x_{k+\frac{n-1}{2}}.$$ Notice that $$x_{\frac{n-1}{4}}\rightarrow 1$$ as $$n\rightarrow\infty$$. Now, it is enough to show that $$x_{k+\frac{n\pm 1}{2}}=x_{k+\frac{n}{2}}+O(\frac{1}{n^2}),\tag{1}$$ since then $$\prod_{k=0}^{n-1}x_k=(1+O(\frac{1}{n^2}))^{\frac{n-1}{2}}$$ and when $$n\rightarrow\infty$$, it tends to $$1$$.
We can show that $$\cos(\frac{2\pi}{n}(k+\frac{n\pm 1}{2}))=-\cos(\frac{2\pi k}{n})+O(\frac{1}{n^2})$$ and since also $$\sqrt{1+\frac{1}{n^2}}=1+O(\frac{1}{n^2})$$, $$(1)$$ holds.
$$n=4m+3$$ case is similar.
• I think there is no need to separate the two odd cases. Let $x_0$ be the length of the line segment that is closest to being perpendicular to the line through $P$ and the centre of the circle. So $\lim\limits_{n\to\infty}x_0=1$, and for any odd $n$, $\prod\limits_{k=1}^n x_k = x_0\prod\limits_{k=1}^{\frac{n-1}{2}} x_k x_{\left(k+\frac{n-1}{2}\right)}$ . If you simplify your answer like this, then in your expression for $x_k$, you would need to change cos to sin. (The big-O thing would still work out.) | 2022-09-30T08:55:56 | {
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https://math.stackexchange.com/questions/237871/inverse-matrixs-eigenvalue/237873 | # Inverse matrix's eigenvalue?
It's from the book "linear algebra and its application" by gilbert strang, page 260.
$(I-A)^{-1}$=$I+A+A^{2}+A^{3}$+...
Nonnegative matrix A has the largest eigenvalue $\lambda_{1}$<1.
Then, the book says, $(I-A)^{-1}$ has the same eigenvector, with eigenvalue $\frac{1}{1-\lambda_{1}}$.
Why? Is there any other formulas between inverse matrix and eigenvalue that I don't know?
A matrix $A$ has an eigenvalue $\lambda$ if and only if $A^{-1}$ has eigenvalue $\lambda^{-1}$. To see this, note that $$A\mathbf{v} = \lambda\mathbf{v} \implies A^{-1}A\mathbf{v} = \lambda A^{-1}\mathbf{v}\implies A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}$$
If your matrix $A$ has eigenvalue $\lambda$, then $I-A$ has eigenvalue $1 - \lambda$ and therefore $(I-A)^{-1}$ has eigenvalue $\frac{1}{1-\lambda}$.
• OMG! That's brilliant! Thanks a lot. Nov 15, 2012 at 10:51
• Let $\mathbf{v}$ be an eigenvector of $A$ under $\lambda$. Then $$(I-A)\mathbf{v} = I\mathbf{v} - A\mathbf{v} = \mathbf{v} - \lambda\mathbf{v} = (1-\lambda)\mathbf{v}$$
– EuYu
Nov 15, 2012 at 10:59
• Why I can't accept your answer? The box keeps saying me to wait in 6 minutes. Nov 15, 2012 at 11:06
• @EuYu Hope I can revive this post for a quick question. What happened when going from the second implication to the third in your answer. To be more clear, how did u go from $A^{-1}A\mathbf{v} = \lambda A^{-1}\mathbf{v}$ to $A^{-1}\mathbf{v} = \frac{1}{\lambda}\mathbf{v}$? Is it because we know that $A^{-1}\mathbf{v}=\lambda^{-1}\mathbf{v}$? Thanks in advance. Nov 27, 2018 at 7:32
• @S.Crim We have $\lambda A^{-1}\mathbf{v} = A^{-1}A\mathbf{v} = \mathbf{v}$. We know that $\lambda \neq 0$ since $A$ is invertible, so we can divide through by $\lambda$ to get the desired result.
– EuYu
Nov 27, 2018 at 7:51
If you are looking at a single eigenvector $v$ only, with eigenvalue $\lambda$, then $A$ just acts as the scalar $\lambda$, and any reasonable expression in $A$ acts on $v$ as the same expression in $\lambda$. This works for expressions $I-A$ (really $1-A$, so it acts as $1-\lambda$), its inverse $(I-A)^{-1}$, in fact for any rational function of $A$ (if well defined; this is where you need $\lambda_1<1$) and even for $\exp A$. | 2022-05-29T00:28:24 | {
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https://math.stackexchange.com/questions/2916502/can-we-assume-that-any-two-events-are-within-the-same-sample-space | # Can we assume that “any two events” are within the same sample space?
I had a question regarding the properties of probability.
If we were to say that we have "any two events $A$ and $B$," is it okay to assume that they both belong to the same sample space $S$ (i.e. $A \subseteq S$ and $B \subseteq S$)?
I'll give a specific example exercise that prompted me to ask this question:
$\\$
"Show that for any events $A$ and $B$, $\$ $P(A) + P(B) - 1 \le P(A \cap B)$."
What I did is move the 1 over, so now we have
$$P(A) + P(B) \le 1 + P(A \cap B)$$
which is true if we assume that $P(A) + P(B) \le P(S) = 1$.
$\\$
I hope my question makes sense. Thank you for the feedback!
• The question certainly intends for you to assume that $A$ and $B$ are subsets of the same sample space, but perhaps this (or something to this effect) should be stated explicitly in the question. I'd be interested in hearing suggestions for a more precise and yet non-cumbersome way of phrasing the example question. – littleO Sep 14 '18 at 7:40
• I may misunderstood you, but do you claim that assuming $A, B\subseteq S$ is equivalent with $P(A) +P(B) \leq P(S)$? – Shashi Sep 14 '18 at 7:46
• For your example exercise you only need to apply inclusion-exclusion. Practically I do not see people will explicitly stating this as it is obvious(?). If there is a probability measure $P$ such that $P(A), P(B)$ are well-defined, there is also a corresponding sample space $\Omega$ and sigma-algebra $\mathcal{F}$ to form a probability space, and by definition $A, B \subseteq \Omega$ and $A, B \in \mathcal{F}$. – BGM Sep 14 '18 at 7:52
• $P(A)+P(B)\le1+P(A\cap B)$ is true, but $P(A)+P(B)\le1$ is not necessarily true. For example, if $A=B=S,$ then $P(A)+P(B)=2.$ – bof Sep 14 '18 at 8:21
• @Sean You are very confused. Probabilities can't exceed one. The sum of two probabilities can be as big as 2. Let's look at a concrete example. The random experiment is rolling a fair $6$-sided die. The sample space $S$ consists of the $6$ outcomes $1,2,3,4,5,6.$ Let $A$ be the event $X\le3$ and $B$ the event $X\ne2.$ Then $P(A)=\frac12$, $P(B)=\frac56$,and $P(A)+P(B)=\frac12+\frac56=\frac43\gt1.$ On the other hand, $P(A\cup B)=1.$ – bof Sep 14 '18 at 10:55
Recall the definition. A probability space is a triple: $\left(S, \mathcal F, \mathbb P\right)$ where $\mathcal F$ is a set of "events" (Subsets of $S$) and $\mathbb P$ is a function from $\mathcal F$ to $\left[ 0,1 \right]$.
So the expression $$\mathbb P(A) + \mathbb P(B) - 1 \leq \mathbb P(A\cap B)$$ only makes sense if $A, B \in \mathcal F$ The function $\mathbb P$ carries its domain around with it, meaning that it can only ever refer to one sample space.
In addition the term event is ambiguous if there are two sample spaces involved. So the use of the term event usually means that there is only one sample space under consideration. Which means that "Any two events $A$ and $B$" can be taken to imply that $A$ and $B$ are events in the same sample space. | 2019-04-25T22:44:17 | {
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https://www.physicsforums.com/threads/sequence-convergence-proof.388851/ | # Homework Help: Sequence Convergence Proof
1. Mar 22, 2010
1. The problem statement, all variables and given/known data
Let $$X=(x_n)$$ be a sequence of strictly positive numbers such that $$\lim(x_{n+1}/x_n)<1$$. Show for some $$0<r<1$$, and for some $$C>0$$, $$0<x_n<Cr^n$$
2. Relevant equations
3. The attempt at a solution
Let $$\lim(x_{n+1}/x_n)=x<1$$
By definition of the limit, $$\lim(x_{n+1}/x_n)=x \Rightarrow \forall \epsilon>0$$ there exists $$\: K(\epsilon)$$ such that $$. \: \forall n>K(\epsilon)$$
$$|\frac{x_{n+1}}{x_n}-x|<\epsilon$$
Since i can pick any epsilon, let epsilon be such that $$\epsilon + x = r <1$$. Also, I know that since this is a positive sequence, $$\frac{x_{n+1}}{x_n}>0$$. Therefore, for large enough $$n$$,
$$0<\frac{x_{n+1}}{x_n}<r<1.$$
From here I am not sure where to go, any hints would be much appreciated! I cannot find out what this tells me about $x_n$
2. Mar 22, 2010
### jbunniii
For large enough $n$ you also have the following:
$$\frac{x_{n+k}}{x_n} = \frac{x_{n+1}}{x_n} \frac{x_{n+2}}{x_{n+1}} \cdots \frac{x_{n+k}}{x_{n+k-1}} < r^k$$
3. Mar 22, 2010
Can I let $$x_n=C$$ and therefore say the sequence $$x_{n+k}<Cr^k$$? If so then n would fixed and k would be the index, correct? Thanks for the speedy response!
4. Mar 22, 2010
### jbunniii
You're not quite there yet.
First note that you need
$$x_{n} < C r^n$$
not
$$x_{n+k} < Cr^k$$
Also, it needs to be true for EVERY $n$, not just sufficiently large $n$.
First let's introduce some notation. It is easier to discuss if we give this "sufficiently large" $n$ a name, say $N$. For all $n \geq N$,
$$0 < \frac{x_{n+1}}{x_n} < r < 1$$
which yields
$$x_{N+k} < r^k x_N$$
for all $$k \geq 0$$.
I can get the index and exponent to agree by writing, equivalently,
$$x_{N+k} < r^{N+k} (x_N r^{-N})$$
Then if I set $$C_1 = x_N r^{-N}$$ I get
$$x_{N+k} < C_1 r^{N+k}$$
which looks pretty promising. However, this $C_1$ may not be large enough to work for ALL $n$, i.e. it may not be true that
$$x_n < C_1 r^n$$
for all $n < N$.
But note that there are only finitely many $x_n$ with $n < N$. Can you use that fact to find a $C$ that does work for all $n$?
Last edited: Mar 22, 2010
5. Mar 22, 2010
So, for every $$j<N$$, Let $$M_j$$ be such that $$x_j<M_jr^j$$ (there exists such an M by the archimedean property). Now, if I take $$C=\sup\{M_j,C_1| j\in\mathbb{N},j<N\}$$, that should do the job since there are only finitely many M's, correct? | 2018-12-19T09:47:25 | {
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https://math.stackexchange.com/questions/3750248/splitting-a-continuous-monotonically-increasing-function-fx-as-hxhx-ep/3750435 | # Splitting a continuous monotonically-increasing function $f(x)$ as $h(x)+h(x+\epsilon) = f(x)$
Given a continuous monotonically-increasing function $$f: [0,1]\to \mathbb{R}$$ and a parameter $$\epsilon>0$$, does there exist a continuous monotonically-increasing function $$h$$ such that, for all $$x\in[0,1]$$:
$$h(x)+h(x+\epsilon) = f(x)?$$
If $$\epsilon=0$$ then $$h(x)=f(x)/2$$. But when $$\epsilon>0$$, the function $$f$$ should be split into two parts with a "phase difference" of $$\epsilon$$. It seems easy, but I could not find the formula for this $$h$$.
• well, just a trivial thought, for linear functions, something like $h(x):=f(x-\frac{\epsilon}{2})$ does the job – alphaomega Jul 8 '20 at 20:23
• When $f$ is a generic polynomial (even not increasing), there is unique polynomial solution $h$ of the same degree: drive.google.com/file/d/1aWaRJ-m9nRMccxbhGT9cx5pLhg2k_Nee/… – enzotib Jul 8 '20 at 20:24
• Is $h(x)$ supposed to be independent of $\epsilon$? – herb steinberg Jul 8 '20 at 20:37
• @herbsteinberg $h$ may depend on $\epsilon$. – Erel Segal-Halevi Jul 8 '20 at 20:41
No, this is not generally true. For any $$\epsilon < 1/2$$, we can construct a strictly increasing differentiable function $$f$$ such that no monotonically-increasing function $$h$$ satisfies your property.
Outline of construction: let $$f$$ be flat on the intervals $$[0, \epsilon +\delta]$$ and $$[\epsilon +2\delta, 1]$$ but steep in between.
Fix any $$\epsilon<1/2$$ and define $$\delta>0$$ such that $$\delta < \min\{1/2 - \epsilon, \epsilon/2\}$$. Construct $$f$$ to be linear for $$x \leq \epsilon+\delta$$ with slope $$\gamma>0$$:
1. $$f(x)=c + \gamma x$$ for $$x \leq \epsilon+\delta$$.
Lemma 1: $$(c - \gamma)/2 \leq h(x) \leq (c + \gamma)/2$$ for $$x \leq \epsilon+\delta$$.
Proof: First observe that $$h(x) \leq f(x)/2$$ for all $$x \in [0,1]$$, otherwise $$h(x) + h(x+\epsilon)>f(x)$$ by monotonicity, which gives the upper bound for $$x \leq \epsilon+\delta$$. The lower bound follows by substituting this upper bound for $$h(\epsilon)$$ in the expression: $$h(0) + h(\epsilon) = c$$.
Lemma 2: $$h(x) \leq c/2 + \gamma$$ for $$x \in [\epsilon+\delta, 2\epsilon+\delta]$$.
Proof: This follows by substituting the lower bound from Lemma 1 for $$h(x-\epsilon)$$ in the expression: $$h(x-\epsilon) + h(x) = c + \gamma(x-\epsilon)$$.
--
Let $$f$$ be linear for $$x \geq \epsilon+2\delta$$ with slope $$\gamma$$:
1. $$f(x) = d + \gamma x$$ for $$x \geq \epsilon+2\delta$$.
Lemma 3: $$(d - \gamma)/2 \leq h(x) \leq (d + \gamma)/2$$ for $$x \in [\epsilon+2\delta, 1]$$.
Proof: Same as in Lemma 1.
--
Notice that both Lemmas 2 and 3 apply to the point $$x = \epsilon + 2\delta$$.
1. Choose $$c$$, $$d$$, and $$\gamma$$ such that:
$$c/2 + \gamma < (d - \gamma)/2$$
$$\Longleftrightarrow \gamma < (d-c)/3$$
This gives the contradiction that: $$h(\epsilon+2\delta) \leq c/2 + \gamma < (d - \gamma)/2 \leq h(\epsilon+2\delta)$$
--
Finally, it doesn't matter what $$f$$ is for $$x \in (\epsilon+\delta, \epsilon + 2\delta)$$; any valid (smooth strictly increasing) construction here would work.
Conjecture: There exists such an $$h$$ for all $$f$$ satisfying a bound on the ratio of derivatives: $$f'(x)/f'(y) \leq M(\epsilon)$$ for all $$x,y \in [0,1]$$. Basically, the slope cannot fluctuate too much.
(This trivially holds in the linear case where $$M=1$$, but a higher/the highest bound would be interesting.)
• Interesting. But what if $f$ must be strictly monotonically-increasing? – Erel Segal-Halevi Jul 9 '20 at 0:48
• @ErelSegal-Halevi I've updated the proof now to include that constraint. – Sherwin Lott Jul 9 '20 at 3:21
• Great answer, and very interesting conjecture too! – Erel Segal-Halevi Jul 9 '20 at 7:25 | 2021-05-16T22:08:38 | {
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https://math.stackexchange.com/questions/977790/how-do-you-add-two-fractions | # How do you add two fractions?
I have a fraction I am trying to solve. I know the answer already, as Wolfram says it is $\frac{143}{300}$.
The fraction is: $$\frac{5}{12} + \frac{3}{50} = \space ?$$ Please explain why and how your method works.
Null has given you a good way. Here's a way without worrying about the LCM: $${a\over b}+{c\over d}={ad+bc\over bd}$$ In the example, $${5\over12}+{3\over50}={(5)(50)+(12)(3)\over(12)(50)}={286\over600}={143\over300}$$
The price of not worrying about the LCM is that you get an answer, $286/600$, that isn't in lowest terms, so you have the extra step at the end of reducing the fraction.
• I've tried that method, it doesn't always work? – kinesis Oct 17 '14 at 6:30
• It does always work [in fact it is usually taken as the definition of addition], but sometimes it doesn't give you the answer in simplest form (lowest terms). – Eric Stucky Oct 17 '14 at 6:45
• Then again, no method is guaranteed to give the answer in lowest terms. E.g., $(1/2)+(1/2)=2/2=1$. – Gerry Myerson Oct 17 '14 at 10:57
The fractions need to have the same denominator in order to add them together. The denominator must be the least common multiple of the two denominators, or a multiple of it. The least common multiple of 12 and 50 is 300.
First multiply $5/12$ by $25/25$. Since $25/25 = 1$ the product is the same as the original fraction. You simply multiply the numerators together and the denominators together:
$$\frac{5}{12}\times\frac{25}{25} = \frac{125}{300}$$
Then multiply $3/50$ by $6/6$. Again, the product is the same as the original fraction:
$$\frac{3}{50}\times\frac{6}{6} = \frac{18}{300}$$
Now both fractions have the same denominator and you can simply add the numerators:
$$\frac{125}{300}+\frac{18}{300} = \frac{143}{300}$$
• What's the significance of the 25? Where do you pull this from? Why did you divide 50 by 2? – kinesis Oct 17 '14 at 6:10
• @kinesis You need the denominators to be equal to do the addition. The multiples of 50 (the other denominator) are 100, 150, 200, etc. The first multiple of 12 (the first denominator) that is equal to a multiple of 50 is 300, which requires multiplying by 25. Ultimately, I'm multiplying the denominators until they equal the least common multiple, which in this case is 300. – Null Oct 17 '14 at 6:13
• Great. When would you not use LCM to get equal denominators? Isn't it ineffective to have to write down all the multiples on a test? – kinesis Oct 17 '14 at 6:18
• @kinesis Generally you use the LCM but you can use any multiple of it (maybe you come up with a multiple of it first, or maybe it's easier to work with). For example, I could have simply multiplied 12 by 100 to get 1200, which would require multiplication of 50 by 24 to get 1200 for it as well. It doesn't matter what you use as long as you use the same denominator for the two fractions. – Null Oct 17 '14 at 6:21
• Isn't it ineffective to have to write down all the multiples while doing a test? I see.. so any multiple ... – kinesis Oct 17 '14 at 6:21
Here is another equivalent idea, put $$x=\frac{5}{12}+\frac{3}{50}$$ Multiply by 12 $$12x=5+\frac{36}{50}=5+\frac{18}{25}$$ Multiply by 25 $$300x=125+18$$ Divide by 300 $$x=\frac{143}{300}$$
If $a=b$ then for any function $f(a)=f(b)$. Suppose $\displaystyle x=\frac{5}{12}+\frac{3}{50}$. Then $\displaystyle (12\cdot 50)\cdot x=(12\cdot 50)\cdot\left(\frac{5}{12}+\frac{3}{50}\right)$, so $\displaystyle 600x=\frac{12\cdot 50\cdot 5}{12}+\frac{12\cdot 50\cdot 3 }{50}=$ $=50\cdot 5+12\cdot 3=286$, why $\displaystyle x=\frac{286}{600}=\frac{143}{300}$. | 2019-06-24T23:53:42 | {
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https://crypto.stackexchange.com/questions/5889/calculating-rsa-private-exponent-when-given-public-exponent-and-the-modulus-fact/5894 | # Calculating RSA private exponent when given public exponent and the modulus factors using extended euclid
When given $p = 5, q = 11, N = 55$ and $e = 17$, I'm trying to compute the RSA private key $d$.
I can calculate $\varphi(N) = 40$, but my lecturer then says to use the extended Euclidean algorithm to compute $d$. That's where I get stuck.
Here's my work so far:
First I use the Euclid algorithm to calculate:
40 = 2(17) + 6
17 = 2(6) + 5
6 = 1(5) + 1 = gcd
So I know the GCD is 1. Applying the 'extended' section of the algorithm:
6 = 40-2(17)
5 = 17-2(6)
1 = 6-1(5)
1 = 6-1(17-2(6))
3(6) = 1 (17)
3(40 - 2(17)) - 1(17)
3(40) - 3(17)
I know the answer is $33$, but I have no idea how to get there using the extended Euclidean algorithm. I can't figure out why I'm getting $3(40) - 3(17)$ when I know the answer should contain $33$.
• – CodesInChaos Jan 3 '13 at 13:30
• Unfortunately it has to be computed by hand... I've edited the question to included more of my work. – DougalMaguire Jan 3 '13 at 13:32
• This YouTube video shows this in a really easy way to understand. – Vaibhav Shah Aug 30 '16 at 7:29
• This YouTube video explains in a very good way how to calculate public and private keys for RSA: youtube.com/watch?v=kYasb426Yjk. It also explains the extended euclidean algorithm. – Aedvald Tseh Jan 2 at 10:05
The extended Euclidean algorithm is essentially the Euclidean algorithm (for GCD's) ran backwards.
Your goal is to find $d$ such that $ed \equiv 1 \pmod{\varphi{(n)}}$.
Recall the EED calculates $x$ and $y$ such that $ax + by = \gcd{(a, b)}$. Now let $a = e$, $b = \varphi{(n)}$, and thus $\gcd{(e, \varphi{(n)})} = 1$ by definition (they need to be coprime for the inverse to exist). Then you have:
$$ex + \varphi{(n)} y = 1$$
Take this modulo $\varphi{(n)}$, and you get:
$$ex \equiv 1 \pmod{\varphi{(n)}}$$
And it's easy to see that in this case, $x = d$. The value of $y$ does not actually matter, since it will get eliminated modulo $\varphi{(n)}$ regardless of its value. The EED will give you that value, but you can safely discard it.
Now, we have $e = 17$ and $\varphi{(n)} = 40$. Write our main equation:
$$17x + 40y = 1$$
We need to solve this for $x$. So apply the ordinary Euclidean algorithm:
$$40 = 2 \times 17 + 6$$
$$17 = 2 \times 6 + 5$$
$$6 = 1 \times 5 + 1$$
Write that last one as:
$$6 - 1 \times 5 = 1$$
Now substitute the second equation into $5$:
$$6 - 1 \times (17 - 2 \times 6) = 1$$
Now substitute the first equation into $6$:
$$(40 - 2 \times 17) - 1 \times (17 - 2 \times (40 - 2 \times 17)) = 1$$
Note this is a linear combination of $17$ and $40$, after simplifying you get:
$$(-7) \times 17 + 3 \times 40 = 1$$
We conclude $d = -7$, which is in fact $33$ modulo $40$ (since $-7 + 40 = 33$).
As you can see, the basic idea is to use the successive remainders of the GCD calculation to substitute the initial integers back into the final equation (the one which equals $1$) which gives the desired linear combination.
As for your error, it seems you just made a calculation error here:
3(40 - 2(17)) - 1(17)
which incorrectly became:
3(40) - 3(17)
It seems you forgot the factor of 3 for the left 17, the correct result would be:
3(40 - 2(17)) - 1(17) = 3 * 40 - 3 * 2 * 17 - 1 * 17 = 3 * 40 + (-7) * 17
Which is the -7 expected.
• The error is in the line before already, unless that equal sign should be a minus sign and he left out the $1=$ on the left side. Otherwise, that line reads as $17 = 18$. – tylo May 6 '15 at 14:19
The method in the other answer is didactic, but requires backtracking earlier calculations, and thus having kept these or use of recursion, which is undesirable in constrained environments as often used for crypto.
Another commonly taught method is the full extended Euclidean algorithm, which finds Bézout coefficients without recursion. However that requires keeping track of 6 quantities beyond inputs, when for the modular inverse we can do with 4. Plus, the usual description manipulates negative integers; requires a final correction of sign; and makes use of simultaneous double assignment or variable swap, which are not directly available in some computer languages.
Here is a step-by-step method to compute $$e^{-1}\bmod m$$ (and test if that's defined) for non-negative integer $$e$$ and positive integer $$m$$ . It uses the half-extended Euclidean algorithm, modified to deal only with non-negative quantities (always at most the largest input) and simple assignments.
1. $$a\gets e$$, $$b\gets m$$, $$x\gets0$$ and $$y\gets1$$.
Note: $$ax+by=m$$ will keep holding, with $$m$$ the modulus.
2. if $$a=1$$, then output "the desired inverse of $$e$$ modulo $$m$$ is $$y$$" and stop.
3. If $$a=0$$, then output "the desired inverse of $$e$$ modulo $$m$$ does not exist" and stop ($$b$$ is the GCD).
4. $$q\gets\lfloor b/a\rfloor$$ .
5. $$b\gets b-aq$$ and $$x\gets x+qy$$ .
6. if $$b=1$$, then output "the desired inverse of $$e$$ modulo $$m$$ is $$m-x$$" and stop.
7. If $$b=0$$, then output "the desired inverse of $$e$$ modulo $$m$$ does not exist" and stop ($$a$$ is the GCD).
8. $$q\gets\lfloor a/b\rfloor$$ .
9. $$a\gets a-bq$$ and $$y\gets y+qx$$ .
10. Continue at 2.
The question asks to apply that with $$a=e=17$$ and $$m=\varphi(N)=40$$.
• At step 1: $$a=17$$ , $$b=40$$ , $$x=0$$ , $$y=1$$ .
• At steps 4/5: $$q=2$$ , $$b=6$$ , $$x=2$$, ( $$a=17$$ and $$y=1$$ unchanged)
• At steps 8/9: $$q=2$$ , $$a=5$$ , $$y=5$$, ( $$b=6$$ and $$x=2$$ unchanged)
• At steps 4/5: $$q=1$$ , $$b=1$$ , $$x=7$$, ( $$a=5$$ and $$y=5$$ unchanged)
• At step 6, we output "the desired inverse of $$17$$ modulo $$40$$ is" $$m-x=33$$.
Thus $$d=33$$ is one (out of several) possible private exponents.
A useful way to understand the extended Euclidean algorithm is in terms of linear algebra. (This is somewhat redundant to fgrieu's answer, but I decided to post this anyway, since I started writing this before fgrieu expanded their answer. Hopefully the slightly different perspective may still be useful.)
Let's say we're trying to find the inverse of $e$ modulo $\varphi$, i.e. a number $d$ such that $$de \equiv 1 \pmod \varphi.$$ In other words, given $e$ and $\varphi$, we wish to find an integer solution $(d, k)$ to the linear equation $$de + k\varphi = 1.$$ Of course, we know that this equation is only solvable if $\gcd(e,\varphi) = 1$. More generally, if that's not the case, the best we can hope for is a solution to the generalized equation $$de + k\varphi = r,$$ where $r = \gcd(e,\varphi)$ is the smallest positive integer for which such a solution exists.
As it happens, we already have several trivial solutions to this equation, including \begin{aligned}d_0 &= 0,& k_0 &= 1,& r_0 &= \varphi,& \text{and} \\ d_1 &= 1,& k_1 &= 0,& r_1 &= e.\end{aligned}
However, as noted above, we're specifically interested in solutions that minimize $r$, which these trivial solutions usually don't. However, we hopefully remember from high school algebra that subtracting both sides of a valid equation from the respective two sides of another valid equation yields yet another valid equation: if $x = y$ and $p = q$, then $x - p = y - q$.
Thus, assuming that $r_0 = \varphi > r_1 = e > 0$, we can obtain another solution with an even smaller (but still non-negative) $r$ by repeatedly subtracting both sides of the equation $d_1e + k_1\varphi = r_1$ from the corresponding sides of $d_0e + k_0\varphi = r_0$ until the resulting solution \begin{aligned}d_2 &= d_0-a_1d_1,& k_2 &= k_0-a_1k_1,& r_2 &= r_0-a_1r_1,\end{aligned} where $a_1$ is the number of times we've subtracted the smaller solution from the larger one, satisfies $r_2 < r_1$. In fact, we can even directly calculate the multiplier $a_1 = \left\lfloor\frac{r_0}{r_1}\right\rfloor$ (where $\lfloor x \rfloor$ denotes $x$ rounded down, i.e. the largest integer no greater than $x$) without having to do any actual repeated subtraction.
Now we have a new solution $d_2e + k_2\varphi = r_2$, but the new $r_2$ might still not be minimal. However, it is smaller than $r_1$, so we can repeat the same subtraction trick again to obtain yet another new solution \begin{aligned}d_3 &= d_1-a_2d_2,& k_3 &= k_1-a_2k_2,& r_3 &= r_1-a_2r_2,\end{aligned} where $a_2 = \left\lfloor\frac{r_1}{r_2}\right\rfloor$, and so on.
More generally, given the two trivial initial solutions, we can keep constructing new solutions with smaller and smaller $r$ using the recurrence \begin{aligned}d_{i+1} &= d_{i-1}-a_i d_i,& k_{i+1} &= k_{i-1}-a_i k_i,& r_{i+1} &= r_{i-1}-a_i r_i,\end{aligned} where $a_i = \left\lfloor\frac{r_{i-1}}{r_i}\right\rfloor$.
We can keep repeating this process until, eventually, we find that $r_i$ evenly divides $r_{i-1}$ (which implies that $r_{i+1}$ would be zero, which we don't want). At that point, if $r_i = 1$, then the corresponding coefficient $d_i$ (reduced modulo $\varphi$) is the modular inverse of $e$ that we wanted.
Otherwise, it's not hard to show that $r_i$ in fact evenly divides all $r_j$ for $0 \le j < i$, including $r_0 = \varphi$ and $r_1 = e$, and is thus a nontrivial common divisor (in fact, the greatest common divisor) of $e$ and $\varphi$. In particular, this implies that $e$ is not invertible modulo $\varphi$.
Ps. As fgrieu notes in their answer, it's not actually necessary to keep track of the $k_i$ coefficients if we're only interested in $d$ (and $r$). Thus, an implementation of this algorithm only needs to store $r_i$, $d_i$, $r_{i-1}$, $d_{i-1}$ and the temporary value $a_i$. (Some other temporary values may also be needed in practice, although it should be noted that $r_{i+1}$ and $d_{i+1}$ do not need to be stored separately in general, since they can immediately overwrite $r_{i-1}$ and $d_{i-1}$.)
Here's a simple implementation in Python (which, conveniently, has arbitrary-precision integers built in):
def modinv(e, phi):
d_old = 0; r_old = phi
d_new = 1; r_new = e
while r_new > 0:
a = r_old // r_new
(d_old, d_new) = (d_new, d_old - a * d_new)
(r_old, r_new) = (r_new, r_old - a * r_new)
return d_old % phi if r_old == 1 else None
Try it online! | 2021-07-30T09:57:11 | {
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https://math.stackexchange.com/questions/2768930/fa-0-implies-x-a-mid-fx-polynomial-zeros-imply-factors | # $f(a) = 0 \implies (x-a) \mid f(x)$? (polynomial zeros imply factors)
Hopefully this is a foolish question with a simple answer.
I've seen it said that, for polynomial $f(x)$
$$f(a)=0 \iff (x-a) \mid f(x)$$
I can see the implication from factor to zero: $(x-a) \mid f(x) \implies f(a)=0$, because anything multiplied by zero is zero, and if $x=a$, then $(x-a) = 0$.
But I'm not sure of the converse: $f(a) = 0 \implies (x-a) \mid f(x)$.
If $f(a)=0$, is the only way it can be zero is if it includes a factor $(x-a)$? How can we be sure there isn't some other circumstance that could make it zero? I guess I'm looking for a proof.
Hopefully it is a foolish question and simple answer!
You can divide $f(x)$ by $(x-a)$ to get $$f(x) = q(x)(x-a) + r(x)$$ where $\deg(r) < \deg(x-a)=1$. Therefore $\deg(r)= 0$ so $r$ is a constant and we can drop the $x$. Now evaluate at $a$: $$f(a) = q(a)(a-a) + r$$
The LHS is zero by assumption, the RHS is equal to $r$. Thus $r=0$ and $(x-a)$ divides $f(x)$ without remainder.
$\,$
Edit: It seems to me like you're confused about why polynomial division always works, so here's the proof to clear things up.
Theorem. For any $f(x), g(x) \in \Bbb Q[x]$, $g(x) \neq 0$, there exist $q(x), r(x) \in \Bbb Q[x]$ such that $$f(x) = q(x)g(x) + r(x)$$ and where $\deg(r) < \deg(g)$.
Proof: First, choose any $q(x), r(x)$ such that the above equation is satisfied; i.e. $q(x) = 0, r(x) = f(x).$ If $\deg(r) < \deg(g)$, we are done.
Therefore assume $\deg(r) \geq \deg(g)$. Let $r(x) = r_nx^n + r_{n-1}x^{n-1} + \dots + r_0$, $g(x) = g_mx^m + g_{m-1}x^{m-1} + \dots + g_0$. Now let $r'(x) = r(x) - x^{n-m}r_ng_m^{-1}g(x)$, let $q'(x) = q(x) + x^{n-m}r_ng_m^{-1}$. Then $\deg r' < \deg r$ (just look at the coefficient of $x_n$) and again we have $$f(x) = q'(x)g(x) + r'(x).$$
If $\deg(r') < \deg(g)$ now, we can stop. If not, repeat the above with $q'$ and $r'$. Because $\deg(r) < \infty$, this procedure must stop after finitely many iterations, so eventually we will have a remainder of degree less than $\deg(g)$.
This concludes the proof.
• Thanks! I can follow this, but don't "get" it. BTW I see that $r(x)$ being a constant is crucial, so that $r(x)=0$ for all x, not just for $x=a$. – hyperpallium May 9 at 8:16
• This proof means that there could be some weird interaction between a polynomial's terms that lead to $f(a)=0$... and if that happens, the terms can be rewritten as having a factor of $(x-a)$. e.g. for $f(x)=x^2-2x$, there's a weird interaction for $x=2$ where the terms cancel $f(2)=2^2-2\cdot 2=0$, which this proof says means $f(x)$ can be written with a factor of $(x-2)$. Which it can be: $x^2-2x = x(x-2)$. This is astonishing to me, partly because it's non-constructive, just showing it must be true, not how. But, using polynomial long division, we can construct the other factor $q(x)$. – hyperpallium May 9 at 8:16
• This proof relies on it always being possible to rewrite $f(x)$ as $q(x)(x-a)+r(x)$, Polynomial Remainder Theorem (PRT). I don't fully "get" polynomial long division, and perhaps getting that will enable me to "get" PRT and the proof given here. I do get the algorithm of polynomial long division, and why it always works - but not why it means $f(x)$ can be written as a term $q(x)b(x)$ with the divisor $b(x)$ as a factor, and a remainder term $r(x)$. It makes sense to multiply the division form by divisor, but not as polynomials (also, divide by zero ($b(x)=(x-a), b(a)$) is undefined). – hyperpallium May 9 at 8:37
• I've added a proof for the existence of a polynomial division algorithm. The resulting $q(x), r(x)$ are unique too. Tell me if this helps. – Lukas Kofler May 9 at 10:24
• It might take less than $\deg(f) - \deg(g) + 1$ steps as we might cancel out not only the highest power but, by chance, the next highest ones as well. It seems like you understood everything else perfectly well though. – Lukas Kofler May 11 at 8:48
Here is an explicit way of showing it. A general polynomial can be written on the form $$f(x) = a_0 + a_1x + a_2 x^2 + \ldots + a_n x^n$$ From this we have that $f(x) - f(a)$ is given by $$f(x) - f(a) = a_1(x-a) + a_2 (x^2-a^2) + \ldots + a_n (x^n - a^n)\tag{1}$$ Now we can try to factor each of the terms above: \begin{align}x^2-a^2 &= (x-a)(x+a)\\x^3-a^3 &= (x-a)(x^2+ax+a^2)\\x^4-a^4 &= (x-a)(x^3+ax^2+a^2x + a^3)\end{align} and in general $x^k - a^k = (x-a)(x^{k-1} + ax^{k-2} + \ldots + a^{k-2}x + a^{k-1})$. This shows that every term on the right hand side of $(1)$ is divisible by $(x-a)$ so in general $(x-a) \mid f(x) - f(a)$. A special case of this is that if $f(a) = 0$ then $(x-a) \mid f(x)$.
• Thanks! In getting the formula $x^k-a^k$, I can guess that when you multiply the series by $x$ and take away the series multiplied by $a$, all the terms in the middle must cancel out, leaving those at the ends - but I can't see how the details actually do that, nor work out how to see it. It seems it should be easy, using $\Sigma$ sums with indices, but I'm not making much progress. – hyperpallium May 9 at 9:10
• @hyperpallium That's exactly how it works. Making a proof for the general case is not that hard: we compute $(x-a)\sum_{k=0}^n x^{n-k}a^{k} = \sum_{k=0}^n x^{n-k+1}a^{k} - \sum_{k=0}^n x^{n-k}a^{k+1}$ and by changing the summation index $k+1 \to K$ in the last sum it becomes $\sum_{K=1}^{n+1} x^{n-K+1}a^{K}$ and you can see that the two sums exactly cancel excepts for $k=0$ in the first term and $K=n+1$ in the last term which leaves us with $x^{n+1} - a^{n+1}$. – Winther May 9 at 9:22
• Thanks! My attempt was similar.... I think I can see it from the difference, $\Sigma - \Sigma$ I'll need to sit down and go through this step by step myself - the complexity is a too much for me, and any little misunderstanding trips me up. The $n$ here is $k-1$ from the answer; the $k$ here is an index. The change of index "$k+1 \to K$" I haven't seen before in concept or notation. – hyperpallium May 12 at 5:47
• Oh! I see, "the last sum" is (of course) the second $\Sigma$, I thought it meant the $\Sigma + -\Sigma$ (since subtraction is an addition of a negated operand... a little knowledge is a dangerous thing). So, the summed amount(?) is identical to the first $\Sigma$, only the indices differ, making it as you say, very clear which ones cancel out. Nice! I end up learning more about the methods than the result - which is what I really need! – hyperpallium May 12 at 5:48 | 2018-10-19T15:55:51 | {
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https://math.stackexchange.com/questions/969781/a-beautiful-game-of-gold-and-silver-coins/969837 | # A beautiful game of gold and silver coins
A stack of silver coins is on the table.
For each step we can either remove a silver coin and write the number of gold coins on a piece of paper, or we can add a gold coin and write the number of silver coins on another piece of paper.
We stop when only gold coins are left.
Prove that the sums of numbers on these two papers are equal.
Tried playing the game and the problem seems right each time, but I can't proceed from there. Is it done by induction?
• "Write number of golden coins." Number of golden coins from where? Stack contains only silver coins, right? – taninamdar Oct 12 '14 at 7:24
• @taninamdar starts at 0 obviously – Flowers Oct 12 '14 at 7:33
• -1: The game is badly posed... Please improve: which "first" paper? where are the golden coins? – Jorge Leitão Oct 12 '14 at 22:04
The state of the game can be desribed by $$(g,s,G,S),$$ where $g$ is the number of golden coins on the table, $s$ is the number of silver coins on the table, $G$ is the sum of the numbers in the first paper, and $S$ is the sum of the numbers in the second paper. The initial state is $(0,n,0,0)$, and we want to show that if the state of the game is $(g,0,G,S)$, then $G=S$.
If we are at $(g_i,s_i,G_i,S_i)$ and add a golden coin, the state changes to $$(g_{i+1},s_{i+1},G_{i+1},S_{i+1}) = (g_i+1,s_i,G_i,S_i+s_i),$$ and if we remove a silver coin, the state changes to $$(g_{i+1},s_{i+1},G_{i+1},S_{i+1}) = (g_i,s_i-1,G_i+g_i,S_i).$$
One plan to solve the problem is to find an invariant, for example, a function from $(g,s,G,S)$ to integers, such that these transformations do not change the value of that function. Looking at the equations for a while suggests something with $gs$ because that's how we would get changes of size $g$ and $s$. A bit more looking gives us $$f(g,s,G,S) = gs+G-S.$$ Once we have found the above formula, it is easy to verify that a step does not affect the value of $gs+G-S$.
Thus if we start from $f(0,n,0,0)=0$ and end with $f(g,0,G,S) = G-S$, we can see that $G=S$.
• beautiful. will try to use this on more game theory probs. – Jackie Poehler Oct 12 '14 at 13:04
• @JackiePoehler is this even game theory? – Soham Chowdhury Oct 13 '14 at 11:19
• this is simply outstanding, the power of well-organized thought – hunter Oct 13 '14 at 12:04
When you add a gold coin, you write $n$ for the number of silver coins left.
Every time you remove one of those $n$ silver coins, that gold coin gets counted once as part of the number of gold coins - a total of $n$ times, since all the silver coins are eventually removed.
Say you start with $n$ silver coins. As long as you remove only silver coins, you keep writing $0$s. These moves don’t affect the sum on either paper, so you might as well ignore them and assume that the first move is to add a gold coin, writing $n$ on the second paper. Each time you remove a silver coin before you add another gold coin, you’ll write another $1$ on the second paper, so if you remove $s_1$ coins, you’ll add $s_1$ to the total on the first paper.
Then you add a second gold coin, which adds $n-s_1$ to the total on the second paper. Say you then remove $s_2$ silver coins (where $s_2$ could be $0$); that adds $2s_2$ to the total on the first paper and leaves you with $n-(s_1+s_2)$ silver coins.
When you then add a third gold coin, you write $n-(s_1+s_2)$ on the second paper, and if you then remove another $s_3$ silver coins, you’ll write $3$ on the first paper $s_3$ times.
In general, when you’ve added the $k$-th gold coin, you’ll write $n-\sum_{i=1}^{k-1}s_i$ on the second paper, and when you then remove another $s_k$ silver coins, you’ll write $k$ on the first paper $s_k$ times. At this point the total on the first paper will be $$\sum_{i=1}^kis_i\;,$$ and the total on the second paper will be
$$\sum_{j=0}^{k-1}\left(n-\sum_{i=1}^js_i\right)=kn-\sum_{j=1}^{k-1}\sum_{i=1}^js_i=kn-\sum_{i=1}^{k-1}\sum_{j=i}^{k-1}s_i=kn-\sum_{i=1}^{k-1}(k-i)s_i\;.$$
(Note that $\sum_{i=1}^0s_i=0$.)
Suppose that at the end there are $m$ gold coins. Write out expressions for the totals on the two papers in terms of $m$ and $n$; the expressions will involve summations. Then show that the expressions are equal; it’s a fairly simple algebraic manipulation at that point. If you get completely stuck, mouse over the spoiler-protected block below.
If there are $m$ gold coins at the end, the total on the first paper will be $$\sum_{i=1}^mis_i\;,$$ and the total on the second will be $$mn-\sum_{i=1}^{m-1}(m-i)s_i=mn-\sum_{i=1}^m(m-i)s_i\;.$$ You’ll need to realize that $\sum_{i=1}^ms_i$ can be simplified enormously.
Let the number of gold coins and silver coins be x and y. Imagine the xy-grid and let our starting state and ending state be represented by (0, y) and (x, 0) respectively.
With each step, (x, y) becomes either (x+1, y) or (x, y-1), corresponding to when a gold coin is added or a silver coin is removed. Consider the path traced by the point moving from (0, y) to (x, 0), and also consider the area bounded by this curve, the positive x-axis, and the positive y-axis.
The area mentioned above may be evaluated in two ways, either as sum of areas of rectangles with width 1 and largest possible height, or sum of areas of rectangles with height 1 and largest possible width. Each piece of paper lists the areas of the individual rectangles, and since the total area is the same, the sum on each paper is the same and both are equal to the value of area under the path traced.
Let's start with just a single silver coin. We can now add $k$ gold coins, until we remove our $1$ silver coin and the game ends.
For each of those gold coins, we wrote $1$ on our piece of paper, so that's a total of $k$ 1's. When we finally removed the silver coin, we wrote $k$ on the other piece of paper, once. It's easy to see that both sums are $k$.
Let's move on to $2$ silver coins. We now add $m$ gold coins, then we remove the first silver coin, then we add another $k$ gold coins and remove the second silver coin.
On one piece of paper we have $m$ 2's and $k$ 1's, on the other we have $m$ (from when we removed the first silver coin) and $m + k$ (from when we removed the second silver coin). As both sum to $2m + k$, again, the sums are equal.
If we define $g_s$ as the number of gold coins added between between removal of silver coins $s$ and $s-1$, we can see that $s$ is listed $g_s$ times, while $g_s$ occurs $s$ times as a summand of a number in the other list, for all $g, s \in \mathbb Z_{\ge 0}$
Let $s_k$ be the number of silver coins remaining when the $k$th gold coin is added. The numbers written down when gold coins are added therefore add up to $$s_1+s_2+s_3+s_4+\cdots\ .$$ Each time a silver coin is removed, the number of gold coins is written down. That is, $0$ is written down $s_1$ times; $1$ is written down $s_1-s_2$ times; $2$ is written down $s_2-s_3$ times; $3$ is written dow $s_3-s_4$ times; and so on; the total is $$0s_1+1(s_1-s_2)+2(s_2-s_3)+3(s_3-s_4)+\cdots$$ which is the same as above.
(There would be a problem if the series were "really" infinite, but the values of $s_k$ will eventually be zero.) | 2019-05-25T17:47:49 | {
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https://math.stackexchange.com/questions/2803102/what-are-hyperbolic-trig-functions-functions-of | # What are Hyperbolic Trig Functions Functions of?
Circular trig functions take in an angle and spit out a ratio. What do hyperbolic functions take in (I know it's a number, but what geometrically does it represent)?
I've seen images that suggest they're a function of area, and others describe $(\cosh(t),\sinh(t))$ as a parametric, which makes me think that t is arc length covered as you move along the unit hyperbola, just like how with a circular parametric trig function t can be interpreted as arc length. It seems like it should be the function of an angle, but it isn't.
• In relativity, the parameter $\zeta$ in the matrix $\begin{bmatrix} \cosh \zeta & \sinh\zeta \\ \sinh\zeta & \cosh \zeta\end{bmatrix}$ is called rapidity. See Wikipedia (notice that my $\zeta$ is that page's $w$). HTH – Giuseppe Negro May 31 '18 at 14:15
• It doesn't seem to be the arc length, but it is the area between the hyperbola, the ray from the origin to the point, and the $x$-axis. Which is also true for the usual trigonometric functions on the circle. See Wikipedia's discussion of hyperbolic angles and circular angles for more. – user856 May 31 '18 at 14:35
• From physics: the shape of en.wikipedia.org/wiki/Catenary is described by $\cosh (\bullet)$. – Oleg567 May 31 '18 at 14:41
• @Rahul I was about to post that as an answer, but since you commented it first I will leave it for you – gen-z ready to perish May 31 '18 at 15:12
• I wonder what you really want to know. Why do you expect the input to $\sinh$ and $\cosh$ to have special meaning? I assume because they are named similarily to $\sin$ and $\cos$ which seem to have special meaning for their input. But the names are more for historical reasons and that they emerge from similar considerations. $\sin$ and $\cos$ more or less coincidentally work well with angles. One can try to define "hyperbolic angles" or find geometric interpretations for the inputs to $\sinh$/$\cosh$, but it is more artificial and more driven by the similar names than necessity/geometry. – M. Winter May 31 '18 at 15:17
The connecting factor amongst the hyperbolic and circular functions and the unit circle and hyperbola is not the angle subtended by the curve but rather the area bounded by it. Here are two images I pulled from the internet:
Consider the point $\mathrm H(\cosh w, \sinh w)$ on the unit hyperbola and the point $\mathrm C(\cos z, \sin z)$ on the unit circle, both plotted on the $xy$-plane. Then
• $w$ equals twice the area bounded by $\rm OH$, the hyperbola, and the $x$-axis,
and since the area of the unit circle is $\pi(1)^2=\pi$, then
• $z$ equals twice the area bounded by $\rm OC$, the circle, and the $x$-axis.
So to summarize, the relationship amongst the argument $t$ of any of the aforementioned functions and the area $a$ bounded by their characteristic locus is $t=2a$.
You mention “arc length covered as you move along the unit hyperbola.” Let’s consider the integral formula for arc length
$$L = \int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$
and compare the arc length $L(\mathrm H(T))$ traced along the unit hyperbola from point $(1,0)$ to $(\cosh T, \sinh T)$ and the arc length $L(\mathrm C(T))$ traced along the unit circle from point $(1,0)$ to $(\cos T, \sin T)$:
\begin{align} \mathrm H &: \begin{cases} x=\cosh t \\ y=\sinh t\end{cases} \\[2ex] &: \begin{cases} {dx}/{dt} = \sinh t \\ {dy}/{dt} = \cosh t\end{cases} \\[4ex] \mathrm C &: \begin{cases} x=\cos t \\ y=\sin t\end{cases} \\[2ex] &: \begin{cases} {dx}/{dt} = -\sin t \\ {dy}/{dt} = \cos t\end{cases} \end{align}
So
$$L(\mathrm H(T)) = \int_0^T \sqrt{ \cosh^2 t + \sinh^2 t } \, dt$$ $$L(\mathrm C(T)) = \int_0^T \sqrt{ \cos^2 t + \sin^2 t } \, dt$$
Now invoke the identities $\sin^2t+\cos^2t = 1$ but $\sinh^2t+\cosh^2t = \exp2t-\sinh(t)\cosh(t)$ and you will see that
$$L(\mathrm C(T)) = \int_0^T \sqrt{ 1 } \, dt = t\bigr|_0^T = T$$
\begin{align} L(\mathrm H(T)) &= \int_0^T \sqrt{ e^{2x} - {e^x-e^{-x}\over2} \cdot {e^x+e^{-x}\over2} } \, dt \\[2ex] &= \int_0^T \sqrt{ e^{2x} - {e^{2x}-1-1-e^{-2x}\over4} } \, dt \\[2ex] &= \int_0^T \sqrt{ {4e^{2x} - e^{2x}+2+e^{-2x}\over4} } \, dt \quad \cdots \end{align}
Does that help you see why the nice relationship seen between $t$ and $L(\mathrm C(T))$ does not emerge with $t$ and $L(\mathrm H(T))$? | 2020-05-26T23:41:46 | {
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http://openstudy.com/updates/55de0faae4b03aeb8dc28489 | ## mathmath333 one year ago Counting question
1. mathmath333
\large \color{black}{\begin{align} & \normalsize \text{How many differenrt signals can be made by hoisting }\hspace{.33em}\\~\\ & \normalsize \text{5 different coloured flags one above the other ,when }\hspace{.33em}\\~\\ & \normalsize \text{any number of them may be hoisted at a time }\hspace{.33em}\\~\\ & a.)\ 2^{5} \hspace{.33em}\\~\\ & b.)\ ^{5}P_{5} \hspace{.33em}\\~\\ & c.)\ 325 \hspace{.33em}\\~\\ & d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}
2. anonymous
is hoisting 0 flags considered a signal or no?
3. Michele_Laino
I think that the requested number is given by the subsequent computation: $\Large \left( {\begin{array}{*{20}{c}} 5 \\ 1 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 2 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 3 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 5 \\ 5 \end{array}} \right) = {\left( {1 + 1} \right)^5}$
4. anonymous
@Michele_Laino you forgot 5C0
5. ganeshie8
i think it should be $\sum\limits_{k=1}^5k!\binom{5}{k}$ as we can permute the different flags too...
6. Michele_Laino
yes! @pgpilot326
7. anonymous
"any number" so since 0 is a number I guess that means 0 flags hoisted is a signal.
8. anonymous
so like @Michele_Laino said with the addition of 5 C 0 to make 2^5.
9. Michele_Laino
I think that you are right @ganeshie8 :)
10. anonymous
the number for @ganeshie8 's expression is just 2^5 - 1 (doesn't include the 5 C 0 term).
11. ganeshie8
:) i still feel that they should have explicitly mentioned whether permutations are considered or not..
12. Michele_Laino
I think that the permutations are admitted, since the order of the flags is not specified!
13. phi
it make sense permutations are allowed e.g. black over red means something different from red over black
14. mathmath333
why u not considered $$^{5}P_{0}$$
15. mathmath333
by the way answer in book is 325
16. anonymous
actually, order does matter since they are stacked. thus it should $\sum_{k=0}^{5}\left(\begin{matrix}5 \\ k\end{matrix}\right)\cdot k!= \sum_{k=0}^{5}\frac{ 5! }{ \left( 5-k \right)!\cdot \cancel{k!} }\cdot \cancel{k!} =\sum_{k=0}^{5} {_5}P_{k}$
17. mathmath333
Does at least one flag is necessary for a signal .
18. anonymous
=5+20+60+240 so 5P0 doesn't count as a signal. So has to have at least 1 flag
19. anonymous
yes but red over green is another signal, different from green over red
20. anonymous
order is implied in the question... "How many differenrt signals can be made by hoisting 5 different coloured flags $$\underline{\text{one above the other}}$$"
21. ganeshie8
|dw:1440617925160:dw| | 2016-10-24T23:53:56 | {
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https://mathhelpboards.com/threads/determine-the-angle-pqr.8318/ | # Determine the angle PQR
#### anemone
##### MHB POTW Director
Staff member
In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.
#### Plato
##### Well-known member
MHB Math Helper
In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.
From the given you know $$m\left( {\angle PDR} \right)\;\& \;m\left( {\angle DPR} \right)~.$$
You also know $$m\left( {\angle PQD} \right)+m\left( {\angle QPD} \right)=135^o$$.
Can you finish?
#### anemone
##### MHB POTW Director
Staff member
From the given you know $$m\left( {\angle PDR} \right)\;\& \;m\left( {\angle DPR} \right)~.$$
You also know $$m\left( {\angle PQD} \right)+m\left( {\angle QPD} \right)=135^o$$.
Can you finish?
Of course! But only because this is a challenge problem and I am not seeking for help for this problem, Plato!
#### mente oscura
##### Well-known member
In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.
Hello.
1º)$$\angle DPR=180º-(180º-45º)-30º=15º$$
2º) Draw the height from $$Q \ to \ \overline{PR}$$, getting the point $$O$$
$$\sin 30º=\dfrac{1}{2} \ then \ \overline{QO}=\overline{QD}=\overline{DR}$$
3º) $$\angle PDO=60º-45º=15º$$
4º) For 1º) $$\overline{OP}=\overline{OD}=\overline{QO} \rightarrow{} \angle PQO=\angle QPO=45º$$
5º) For 4º): $$\angle QPD=45º-15º=30º$$
Regards.
#### anemone
##### MHB POTW Director
Staff member
Hello.
1º)$$\angle DPR=180º-(180º-45º)-30º=15º$$
2º) Draw the height from $$Q \ to \ \overline{PR}$$, getting the point $$O$$
$$\sin 30º=\dfrac{1}{2} \ then \ \overline{QO}=\overline{QD}=\overline{DR}$$
3º) $$\angle PDO=60º-45º=15º$$
4º) For 1º) $$\overline{OP}=\overline{OD}=\overline{QO} \rightarrow{} \angle PQO=\angle QPO=45º$$
5º) For 4º): $$\angle QPD=45º-15º=30º$$
Regards.
WOW! That's a brilliant way to draw a line $QO$ such that it is perpendicular to $PR$ and everything immediately becomes obvious.
Bravo, mente oscura! | 2022-05-24T17:51:05 | {
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https://math.stackexchange.com/questions/2523322/prove-that-there-exist-c-in-0-1-such-that-fc2 | # Prove that there exist $c \in (0,1)$ such that $|f'(c)|>2$
Let $f$ be continuous on $[0,1]$ and differentiable on $(0,1)$, with $f(0) = f(1) = 0$ and $f(k)=1$ for some $k\in (0,1)$. Prove that there exist $c \in (0,1)$ such that $|f'(c)|>2$
I tried the taylor expansion of $f$ at $x=k$,ie $f(x) = f(k) + f'(\xi)(x-k) = 1+f'(\xi)(x-k)$.
$f(0) = 1+ f'(\xi_0)(-k) = 1+ f'(\xi_1)(1-k) = f(1) = 0$
But i am not sure how to show the desired inequality. Any hints?
• You might use the mean value theorem. – Brian Borchers Nov 16 '17 at 17:04
• I would try MVT – Rodrigo Dias Nov 16 '17 at 17:05
• hmm im not sure what function i should use MVT on, its probably not on f and the intervals [0,k] and [k,1], right? – Timothy Nov 16 '17 at 17:10
• Do you mean $|f'(c)| \geq 2$? – Gibbs Nov 16 '17 at 17:17
• @Gibbs I don't think so, perhaps you have some counterexamples in mind? – Timothy Nov 16 '17 at 17:19
Use the mean value theorem:
$$f'(c_0)=\frac{f(k)-f(0)}{k-0}=\frac{1}{k}\\f'(c_1)=\frac{f(1)-f(k)}{1-k}=\frac{-1}{1-k}$$ Now, there are 3 cases: $\begin{cases}k\in(0,\frac12)\\k=\frac12\\k\in(\frac12,1)\end{cases}$
In case 1, $f'(c_0)>\frac1{\frac12}=2$
In case 3, $f'(c_1)<-\frac1{1-\frac12}=-2\implies |f'(c_1)|>2$
In case 2, both $f'(c_0),|f'(c_1)|$ are equal to 2, if every point between $0$ to $k$ and between $k$ to $1$ has slope of $2$ in one side and $-2$ in the other then $f'(k)$ is undefined, so I know that it can't be, but I also know that the average is $2$ in one side and $-2$ in the other, I know that because the average value of a function is: $$\frac1{b-a}\int_a^b g(x) dx$$ put in this $b=\frac12,a=0$ or $b=1, a=\frac12$, and $g(x)=f'(x)$ and you will get $2$ or $-2$
So there exist at leas one point between $0$ and $k$ that has slope is greater than $2$ and at least one point between $k$ and $1$ that has slope that is less than $-2$
• I can see that "the average is or −2−2" is pretty intuitive but is there a more precise way of phrasing it? – Timothy Nov 16 '17 at 17:36
• @Timothy I added it to the answer – Holo Nov 16 '17 at 17:51
• We can handle the second case without using integrals. See my answer. – Paramanand Singh Nov 16 '17 at 18:41
• @ParamanandSingh i understand this but i still think that using average like this is easier, the integral may look like it require some work but all it really says is that the average slope is the slope of the line that pass through the endpoints. computing it is really easy – Holo Nov 16 '17 at 18:53
• There is nothing wrong with the integral approach but it requires the derivative $f'$ to be integrable. – Paramanand Singh Nov 16 '17 at 19:41
It is easy to dispose of the cases $k<1/2$ and $k>1/2$ using the equations $$f(0)+kf'(c_{1})=f(k)=1=f(k)=f(1)+(k-1)f'(c_{2})$$ The case when $k=1/2$ requires a bit more analysis. If $f'(1/2)>0$ then there is a point $k'>1/2$ such that $f(k') >f(1/2)=1$ and then $$1<f(k')=f(1)+(k'-1)f'(c_{3})$$ so that $|f'(c_{3})|>2$. Similarly we can deal with the case $f'(1/2)<0$.
Let's consider the case when $f'(1/2)=0$ and $f(1/2)=1$ is the maximum value of $f$ in $[0,1]$. If $f'(c) \leq 2$ for all $c\in(0,1/2)$ then we can see that $$f(x)=f(0)+xf'(c)\leq 2x,f(x)=f(1/2)+(x-1/2)f'(d)\geq 1+2(x-1/2)=2x$$ so that $f(x) =2x$ for all $x\in[0,1/2]$ and this contradicts $f'(1/2)=0$. Thus we must have some point $c\in(0,1/2)$ for which $f'(c) >2$.
• Sorry, I don't see how $f(x) \ge 2x$ since $f'(d) /le 2$. Can you elaborate on it please? – Timothy Nov 17 '17 at 7:07
• @Timothy: $f'(d) \leq 2$ and multiply this with negative number (actually non-positive number) $(x-1/2)$ then the inequality is reversed. – Paramanand Singh Nov 17 '17 at 7:15 | 2019-06-25T11:31:53 | {
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https://math.stackexchange.com/questions/321610/convergence-of-sequence-of-averages-the-other-way-around | # convergence of sequence of averages the other way around
In a vector normed space, if $\{x_n\} \longrightarrow x$ then $z_n = \dfrac{x_1 + \cdots+x_n}{n} \longrightarrow x$
Is it true the other way arround too? meaning: if $z_n = \dfrac{x_1 + \cdots+x_n}{n} \longrightarrow x$ then $\{x_n\} \longrightarrow x$?
My intuition says that it is true because $z_n = \dfrac{x_1 + \cdots+x_{n-1}}{n} + \frac{x_n}{n}\longrightarrow x$
Thanks :)
• If $x_n=(-1)^n$ then $x_n$ does not converge yet $z_n \to 1/2$. The topic is methods of summation, I think "Cesaro sum"... – coffeemath Mar 5 '13 at 17:10
• @coffeemath I think you mean $x_n = (1 + (-1)^n)/2$. – Antonio Vargas Mar 5 '13 at 17:50
• @Antonio Yes, I have the wrong limit. If $x_n=(-1)^n$ as I have it, then $z_n \to 0$. And with your adjustment $x_n=(1+(-1)^n)/2$ the limit is $z_n \to 1/2$. Any limit like this shows the "other way around" idea doesn't work. – coffeemath Mar 5 '13 at 18:02
If the average of the first $n$ terms of a sequence tends to a limit, does the sequence itself tend to a limit?
The answer is no in general, as is discussed in the comments. The simplest counterexamples are the sequences which oscillate between two different values $\alpha$ and $\beta$; we would expect that the average of the first $n$ terms of such a sequence will tend to the average of $\alpha$ and $\beta$. As a concrete example let's define
$$x_n = \frac{1+(-1)^n}{2},$$
so that $x_n$ alternates between $0$ (when $n$ is odd) and $1$ (when $n$ is even). We then have
$$n \, z_n = x_1 + x_2 + \cdots + x_n = \begin{cases} \frac{n}{2} & \text{if } n \text{ is even}, \\ \frac{n-1}{2} & \text{if } n \text{ is odd}, \end{cases}$$
from which we can deduce that
$$\frac{1}{2} - \frac{1}{2n} \leq z_n \leq \frac{1}{2}$$
and thus
$$\lim_{n \to \infty} z_n = \frac{1}{2}.$$
There are, however, many cases where one can deduce the convergence of the original sequence from the convergence of this average.
For example, if $(x_n)$ is a positive, monotonic sequence, then the convergence of $(z_n)$ implies the convergence of $(x_n)$.
A slightly more difficult (and more useful) example is discussed in this thread.
Results of this general shape are called Tauberian theorems. A nice reference is the book Divergent Series by G. H. Hardy. | 2019-09-23T01:15:47 | {
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# Math: Absolute value (Modulus)
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Re: Math: Absolute value (Modulus) [#permalink]
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23 Aug 2018, 01:37
How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.
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Re: Math: Absolute value (Modulus) [#permalink]
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23 Aug 2018, 02:12
1
1
onyx12102 wrote:
How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.
______________
I'd say 1 or 2.
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Re: Math: Absolute value (Modulus) [#permalink]
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23 Aug 2018, 04:02
1
onyx12102 wrote:
How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.
Not more than a couple. The problem is that often GMAT likes to combine topics. So, you might have a geometry question which will make you use some basic absolute value concept such as area bounded by the graph of |x| + |y| = 4. So a basic understanding of all topics is a good idea even if you plan to ignore some topics.
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Re: Math: Absolute value (Modulus) [#permalink]
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01 Sep 2018, 14:16
Quote:
2. Solve new equations:
a) x−1=4x−1=4 --> x=5
b) −x+1=4−x+1=4 --> x=-3
3. Check conditions for each solution:
a) x=5x=5 has to satisfy initial condition x−1>=0x−1>=0. 5−1=4>05−1=4>0. It satisfies. Otherwise, we would have to reject x=5.
b) x=−3x=−3 has to satisfy initial condition x−1<0x−1<0. −3−1=−4<0−3−1=−4<0. It satisfies. Otherwise, we would have to reject x=-3.
So...just to clarify...the answer can either be 5 or -3?
Just need the clarification on what this is saying
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Re: Math: Absolute value (Modulus) [#permalink]
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03 Sep 2018, 20:55
In the following example of 3-steps approach for complex problems
Example #1
Q.: |x+3|−|4−x|=|8+x||x+3|−|4−x|=|8+x|. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) x<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not less than -8) Can any one please explain me how x<-8 because whenever I am doing it I am getting x>=-8 for both |8+x| and -|8+x|
b) −8≤x<−3−8≤x<−3. −(x+3)−(4−x)=(8+x)−(x+3)−(4−x)=(8+x) --> x=−15x=−15. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) --> x=9x=9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not more than 4)
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Re: Math: Absolute value (Modulus) [#permalink]
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22 Sep 2018, 02:50
IN Example #1
Q.: |x+3|−|4−x|=|8+x||x+3|−|4−x|=|8+x|. How many solutions does the equation have?
C condition has typo error
c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) --> x=9x=9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
its not -15 but 9
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Re: Math: Absolute value (Modulus) [#permalink]
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29 Sep 2018, 05:26
chetan2u Hi, sorry I keep bugging you but your explanations are easiest to understand
Can you explain to me the "trick" mentioned in the post? I don't get it :/ What are we doing after finding the mid-point?
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29 Sep 2018, 09:02
hibobotamuss wrote:
chetan2u Hi, sorry I keep bugging you but your explanations are easiest to understand
Can you explain to me the "trick" mentioned in the post? I don't get it :/ What are we doing after finding the mid-point?
No problems. Pl keep asking questions and I would reply to each whenever I get time
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Re: Math: Absolute value (Modulus) [#permalink]
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29 Sep 2018, 09:24
"Problem: 1<x<9. What inequality represents this condition?
A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5
Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D."
This is the "trick" mentioned in the post, don't know what it means chetan2u
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Re: Math: Absolute value (Modulus) [#permalink]
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20 Nov 2018, 12:54
gettinit wrote:
Let’s consider following examples,
Example #1 I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem?
Q.: $$|x+3| - |4-x| = |8+x|$$. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:
a) $$x < -8$$. $$-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)
b) $$-8 \leq x < -3$$. $$-(x+3)-(4-x) =+ (8+x)$$ --> $$x = -15$$. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)
c) $$-3 \leq x < 4$$. $$+ (x+3)-(4-x) =+ (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)
d) $$x \geq 4$$. $$+(x+3) + (4-x) = + (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)
I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!
I didn't understand how the sign for each expression was determined.
I'd appreciate any help.
Thanks!!
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Re: Math: Absolute value (Modulus) [#permalink]
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26 Nov 2018, 15:07
Bunuel wrote:
When are we supposed to use the 3-step method? I mean, on which kind of problems?
Re: Math: Absolute value (Modulus) &nbs [#permalink] 26 Nov 2018, 15:07
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Display posts from previous: Sort by | 2019-01-19T01:19:37 | {
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https://math.stackexchange.com/questions/2183248/are-set-notations-emptyset-and-emptyset-different | # Are set notations $\emptyset$ and $\{\emptyset\}$ Different?
Question : Let $A_0=\emptyset$ (the empty set).For each $i=1,2,3,...,$ define the set -
$A_i=A_{i-1}$ $\cup$ {$A_{i-1}$}. The set $A_3$ is :
• $\emptyset$
• $\{\emptyset\}$
• $\{\emptyset, \{\emptyset\}\}$
• $\{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}\}$
My Work : I Googled this, but didn't get anything useful.
• Yes they are different. $\{ \emptyset \}$ is a singleton set and $\emptyset$ is null set. – Error 404 Mar 12 '17 at 13:28
• Set $\{\varnothing\}$ has an element: $\varnothing$. Set $\varnothing$ has no elements (another notation for this set is $\{\}$). So the sets are different. – drhab Mar 12 '17 at 13:29
• I am curious, how have you googled that? As for your question, think about sets as boxes. Is an empty box different from a box containing an empty box? – user251257 Mar 12 '17 at 13:30
• One of infinitely many: math.stackexchange.com/questions/1951267/…; another would be math.stackexchange.com/questions/1845389/… – Asaf Karagila Mar 12 '17 at 13:38
• Yes, the set $A_n$ has $n$ elements; it's the standard representative of the ordinal number $n$ – Henno Brandsma Mar 12 '17 at 14:13
$\emptyset \subset \emptyset$ but $\emptyset \in \{\emptyset\}$ (and also $\emptyset \subset \{\emptyset \}$ since $\emptyset$ is included in every set.)
Now $$\{\emptyset \}\in \{\emptyset ,\{\emptyset \}\}\quad \text{and}\quad \{\emptyset \}\subset \{\emptyset ,\{\emptyset \}\}.$$
• That last example is amusing. A set is an element as well as a subset of another set. +1 – Paramanand Singh May 21 '17 at 17:19
Yes, they're different.
$\emptyset$ denotes a set with no elements (null set) whereas $\{\emptyset\}$ denotes a set with 1 element, namely, the empty set being the element of this set.
Therefore, $A_i \neq A_j$ $\forall i \neq j$
Also, observe that, $|A_i| = i$ $\forall i \in \mathbb{Z}$ | 2019-12-15T05:38:53 | {
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https://math.stackexchange.com/questions/2936269/how-can-you-simplify-sqrt9-6-sqrt2/2936299#2936299 | # How can you simplify $\sqrt{9-6\sqrt{2}}$?
How do you simplify: $$\sqrt{9-6\sqrt{2}}$$
A classmate of mine changed it to $$\sqrt{9-6\sqrt{2}}=\sqrt{a^2-2ab+b^2}$$ but I'm not sure how that helps or why it helps.
This questions probably too easy to be on the Math Stack Exchange but I'm not sure where else to post it.
• See How to simplify a square root for several applicable approaches.
– dxiv
Sep 30 '18 at 5:06
• Find $a,b$ such that $\sqrt{9-6\sqrt 2}=\sqrt{a+b-2\sqrt{ab}}$ since $\sqrt{a+b-2\sqrt{ab}}=\sqrt{(\sqrt a-\sqrt b)^2}$. Sep 30 '18 at 5:07
Try to use the formula your classmate gave. In this situation, $$9-6\sqrt2={\sqrt3}^2-2{\sqrt{3\times6}}+{\sqrt6}^2\Rightarrow(1)$$ That is because $$6{\sqrt2}=2{\sqrt{3\times6}}$$ Expression (1) now looks similar to $$a^2-2ab+b^2$$ where $$a=\sqrt3$$ and $$b=\sqrt6$$ Using this we can conclude that $$9-6\sqrt2={\sqrt3}^2-2{\sqrt{3\times6}}+{\sqrt6}^2=(\sqrt3-\sqrt6)^2$$ We can subsitute in the original expression $$\sqrt{9-6\sqrt2}=\sqrt{(\sqrt3-\sqrt6)^2}=-(\sqrt3-\sqrt6)=\sqrt6-\sqrt3$$ The simplest form will be $$\sqrt6-\sqrt3$$
Let's forget about $$9-6\sqrt{2}$$ for a second and just think about the expression your classmate thinks is useful:$$a^2-2ab+b^2.$$
And let's keep in mind our goal here. We're looking for something which is a perfect square (since we want it to play well inside a $$\sqrt{\quad}$$...
Well, this should remind us of $$a^2\color{red}{+}2ab+b^2=(a+b)^2.$$ But that "$$-$$" on the $$2ab$$ term is throwing me off! Is there any way to fix it?
This is where we get something for free from just doing a small change of variable: if we let $$c=-b$$, we get $$a^2-2ab+b^2=a^2+2ac+c^2.$$ That right hand side is of course just $$(a+c)^2$$, or better yet $$(a-b)^2$$. So we now know: $$\color{green}{\sqrt{a^2-2ab+b^2}=a-b}$$ (or rather, fine, $$\vert a-b\vert$$. FINE.).
That's why what your friend wants to do is reasonable. So nowthe question is: how do we do it?
Ultimately this can just feel like trial-and-error at first, but my instinct here is to say that "$$-6\sqrt{2}$$" looks a lot like "$$-2ab$$." Because they both have a minus sign. And $$6$$ is even. This doesn't work immediately, but when we factor out a $$3$$ things get much cleaner ...
\begin{align} 9 - 6\sqrt2 &= 3 (3-2\sqrt2) \\ &= 3((\sqrt2)^2 - 2(1)\sqrt{2} +1^2) \\ &= 3(\sqrt2-1)^2 \end{align} Hence, $$\sqrt{9-6\sqrt2} = \sqrt{3}(\sqrt2 - 1)$$
Your class mate is being.... clever.
If $\sqrt {9-6\sqrt 2}=a-b$ then $9-6\sqrt 2=a^2-2ab+c^3$
Let $2ab=6\sqrt 2$ and $a^2+b^2=9$.
Can we do that?
If we let $b^2=k$ and $a^2=9-k$ then $ab=\sqrt {k (9-k)}=3\sqrt 2=\sqrt {18}$. Solving $k (9-k)=18$ for $k$ (if it isn't visiblely obvious that we can do it in our heads, it is just a quadratic that we can solve by quadratic formula) and, for covenience, choicing the smaller solution (because we want $a>b$), we get $k =3$ is a good solution..
So $a=\sqrt 6$ and $b=\sqrt 3$.
I.e. in other words
$\sqrt {9-6\sqrt 2}=$
$\sqrt {6-2\sqrt 6\sqrt 3 +3}=$
$\sqrt {(\sqrt 6- \sqrt 3)^2}=$
$|\sqrt 6 - \sqrt 3|=$
$\sqrt 6 -\sqrt 3$.
The reason for doing that is that $$\sqrt{a^2-2ab+b^2} = \sqrt{(a-b)^2} = a-b$$. Now try to put your radical in the form your classmate suggested! | 2021-12-02T01:29:54 | {
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https://math.stackexchange.com/questions/2834273/solving-tan4-theta-dfrac-cos-theta-sin-theta-cos-theta-sin | # Solving $\tan{4\theta} = \dfrac{\cos{\theta} - \sin{\theta}}{\cos{\theta} + \sin{\theta}}$
Solve: $\tan{4\theta} = \dfrac{\cos{\theta} - \sin{\theta}}{\cos{\theta} + \sin{\theta}}$ for acute angle $\theta$
I need help solving that problem. I have tried to do both side, but no result yet.
What I have done. $\dfrac{\sin{4\theta}}{\cos{4\theta}} = \dfrac{\cos{2\theta}}{1 + \sin{2\theta}}$
I got that by multiply the RHS with $\dfrac{\cos{\theta} + \sin{\theta}}{\cos{\theta} + \sin{\theta}}$
• Don't post a picture. Write up your question. And if you must post a picture, orient it so that I don't get a crick in my neck when I read it. – Doug M Jun 28 '18 at 1:20
• You only need $1$ picture also rotate it $90$ degrees anticlockwise. – Chris2018 Jun 28 '18 at 1:21
• How are we supposed to read this? – Ryan Jun 28 '18 at 1:21
• I read it by pressing the downward pointing arrow on the upper left ofvthe question. -1. – Oscar Lanzi Jun 28 '18 at 1:23
• I am sorry. I use mobile phone, its hard to write equation. I will edit my question later – Muhamad Abdul Rosid Jun 28 '18 at 1:32
We need to solve $$\tan4\theta=\frac{\frac{1}{\sqrt2}\cos\theta-\frac{1}{\sqrt2}\sin\theta}{\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta}$$ or $$\tan4\theta=\frac{\sin(45^{\circ}-\theta)}{\cos(45^{\circ}-\theta)}$$ or $$\tan4\theta=\tan(45^{\circ}-\theta)$$ or $$4\theta=45^{\circ}-\theta+180^{\circ},k$$ where $k$ is an integer number, or $$\theta=9^{\circ}+36^{\circ}k,$$ which gives the answer: $$\left\{9^{\circ}, 45 ^{\circ}, 81^{\circ}\right\}$$ By the way, your idea is also works.
Indeed, we need to solve that $$\frac{\sin4\theta}{\cos4\theta}=\frac{\cos2\theta}{1+\sin2\theta}$$ or $$\frac{2\sin2\theta\cos2\theta}{\cos4\theta}=\frac{\cos2\theta}{1+\sin2\theta}.$$ Now, $\cos2\theta=0$ gives $$2\theta=90^{\circ}+180^{\circ}k,$$ where $k$ is an integer number, which gives $\theta=45^{\circ}$.
Also, $$\frac{2\sin2\theta}{\cos4\theta}=\frac{1}{1+\sin2\theta}$$ or $$4\sin^22\theta+2\sin\theta-1=0$$ or $$\sin2\theta=\frac{-1+\sqrt5}{4}$$ or $$\sin2\theta=\sin18^{\circ},$$ which gives $$2\theta=18^{\circ}$$ or $$2\theta=162^{\circ}$$ and we get the same answer.
• Thank you. I prefer the first one. Honestly, I get the answer myself with different steps. – Muhamad Abdul Rosid Jun 28 '18 at 5:41
• You are welcome! – Michael Rozenberg Jun 28 '18 at 5:42
• I think the problem is solved now. – Muhamad Abdul Rosid Jun 28 '18 at 5:44
(Remark: This answer is much like Michael Rozenberg's, which I didn't see until posting. It differs enough, I think, to be worth keeping.)
Since $\sin(\pi/4)=\cos(\pi/4)$, we have
$${\cos\theta-\sin\theta\over\cos\theta+\sin\theta}={\sin(\pi/4)\cos\theta-\cos(\pi/4)\sin\theta\over\cos(\pi/4)\cos\theta+\sin(\pi/4)\sin\theta}={\sin(\pi/4-\theta)\over\cos(\pi/4-\theta)}=\tan(\pi/4-\theta)$$
Writing $\phi=\pi/4-\theta$, we get
$$\tan\phi=\tan(\pi/4-\theta)={\cos\theta-\sin\theta\over\cos\theta+\sin\theta}=\tan(4\theta)=\tan(\pi-4\phi)=-\tan4\phi$$
It's clear that $\phi=0$, which corresponds to $\theta=\pi/4$, is a solution, and it's also easy to see that we get a solution from $\phi=\pi-4\phi$, i.e., $\phi=\pi/5$, which corresponds to $\theta=\pi/4-\pi/5=\pi/20$. To round this out, the symmetry of $\tan\phi=-\tan4\phi$ for $\phi$ and $-\phi$ tells us $\phi=-\pi/5$ is also a solution, corresponding to $\theta=\pi/4+\pi/5=9\pi/20$.
It remains to see if there are any other solutions in the interval $-\pi/4\le\phi\le\pi/4$, which corresponds to acute angles $0\le\theta\le\pi/2$. But the curves $y=\tan\phi$ and $y=-\tan4\phi$ are straightforward to sketch: the first is stricly increasing, and the latter is strictly decreasing (because of the minus sign) in each interval where it's defined, so it's easy to see the two curves intersect exactly three times. Thus $\phi=0$ and $\phi=\pm\pi/5$ are the only three solutions corresponding to acute angles $\theta$, which are
$$\theta=\pi/20,\pi/4,\text{and }9\pi/20$$ | 2019-05-25T05:38:51 | {
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https://math.stackexchange.com/questions/2442042/what-is-a-subsequence-in-calculus/2442044 | What is a subsequence in calculus?
For example, if I have the sequence $(1,2,3,4,5,6,7,8,\ldots)$ i.e $x(n) = n$ for all natural numbers, then is the subsequence $(1,1,1,1,1...)$ valid? Or can I only take one element from the sequence once? Would the subsequence $(1,2,1,2,1,2...)$ be a valid subsequence?
Thanks.
In mathematics, a subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements
Formally, a subsequence of the sequence $(a_n)_{n \in \mathbb{N}}$ is any sequence of the form $(a_{n_k})_{k \in \mathbb{N}}$ where $(n_k)_{k \in \mathbb{N}}$ is a strictly increasing sequence of positive integers.
Hence, your two examples are not valid, as $1$ appears exactly once in the original sequence.
However if the original sequence is $(1,2,3,4,5,6,7,8,1,2,3,4,5,6,7,8,\ldots)$ that is the $8$ numbers are periodic, then yes, it is a valid subsequence.
• Thank you. You mentioned 'without changing the order of the remaining elements'. What does that mean? Does it mean (5,1,3,7,9....) is not a valid subsequence, since we made the 5 appear before the 1 and 3? Also, yes, I did intend it to be an increasing sequence of positive integers.
– John
Sep 23 '17 at 17:51
• yup, in the original sequence $5$ appears after $1$, hence the ordered need to be kept that way. a valid subsequence is $(1,3,5,7,9,\ldots)$. Sep 23 '17 at 17:52
• +1 Note: most of the time "subsequence" means "infinite subsequence" so you have to keep infinitely many of the elements. Sep 23 '17 at 17:57
• @Saad I think that another way to look at it (which is what this answer is saying in the less formal initial definition) is that if you imagine writing out your entire sequence (which you can't really do in practice but you can imagine what that would look like if you could) and then crossed out some (possibly an infinite number) of terms, what you would be left with is a subsequence. A pointed out by Ethan Bolker, you can't leave just a finite number of terms, but you could still cross out an infinite number of them (every other term for example). Sep 24 '17 at 4:27
A sequence of real numbers is a map $x:\Bbb N\to\Bbb R$; we typically write it as $n\mapsto x_n$ (rather than $n\mapsto x(n)$).
A subsequence of this sequence is a map $y=x\circ\phi$ where $\phi:\Bbb N\to \Bbb N$ is a strictly increasing map. Then $y_n=x_{\phi(n)}$. Neither of your examples is a subsequence of your given sequence which has $x_n=n$.
• Thanks. Could you explain why the order of terms is preserved if ϕ is a strictly increasing map? i.e If I have the harmonic series, (1+1/2+1/3... ) and then I take the map ϕ(n) = 2^n, why does that guarantee that the order of terms is preserved? I'm just trying to get some intuition for why we need a strictly increasing map.
– John
Sep 23 '17 at 18:09
• @Saad You are now confusing sequences with series. Sep 23 '17 at 18:12
• Sorry, I meant the sequence (1,1+1/2,1+1/2+1/3,... ), and then forming the subsequence (1+1/2,1+1/2+1/3+1/4,1+1/2+1/3+1/4+1/5+1/6+1/7+1/8,..)
– John
Sep 24 '17 at 1:33
This may be easier to visualize: Write your sequence (any sequence) $$7,8,5,6,4,4,5,7,7,7,7,8,7,3,2,6,1,2,6,2,8,3,3,2,3,\ldots$$ Then select some elements $$7,\underline{8},5,6,\underline{4},4,5,\underline{7},7,7,7,8,\underline{7},3,\underline{2},6,\underline{1},\underline{2},6,\underline{2},8,3,\underline{3},2,3,\ldots$$ and erase the rest $$8,4,7,7,2,1,2,2,3,\ldots$$ That is a subsequence...
Consider a sequence $$a_1,a_2,a_3,a_4,a_5,a_6,a_7,a_8, \dots$$
Intuitively, we can create a subsequence by selecting an infinite amount of random terms of this sequence, and throwing what's left away. However, it is important that if we select $a_i$ and $a_j$ with $i <j$, then also $i < j$ in the subsequence. In other words: we keep the order of the elements in the original sequence.
For example, the sequence:
$$a_1,a_3,a_4,a_7, \dots$$ would be a subsequence of the sequence written earlier.
In this case, we can then find a strictly increasing function $k: \mathbb{N} \to \mathbb{N}$ such that $$1 \mapsto 1$$ $$2 \mapsto 3$$ $$3 \mapsto 4$$ $$4 \mapsto 7$$ $$\dots$$
and write $b_n = a_{k_{n}}$ to denote the subsequence .
The strictly increasing part is important to preserve the order of the elements as in the original sequence.
This leads us to the following definition:
A sequence $(y_n)_n$ is subsequence of a function $(x_n)_n$ iff there exists a strictly increasing function $k: \mathbb{N} \to \mathbb{N}$ such that for all $n \in \mathbb{N}$, we have $y_n = x_{k(n)}$
No.
A subsequence is a sequence taken from the original, where terms are selected in the order they appear.
For example, let $x_n = \frac{1}{n}$. Let's take a subsequence $x_{n_j}$ where we pick every other term, i.e.
$$x_{n_j} = \left(1, \frac{1}{3}, \frac{1}{5}, \frac{1}{7}, \ldots \right)$$
Notice how these terms in the subsequence are taken in the order they appear.
you should not ask whether you can take just one element from the sequence once or not as $(1,1,1,1.....)$ is surely is not a subsequence of $(1,2,3,4,5.....)$ but $(1,1,1,1,1,1.....)$ is subsequence of $(-1,1,-1,1,-1,1.....)$. Hence you should consider your subsequence as if your original sequence is say $\{x(n)\}_{n=1}^\infty$ then your subsequence will be anything of the form $\{x(n_i)\}_{i=1}^\infty$ for some $n_1<n_2<n_3.....$. | 2021-09-27T13:29:06 | {
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https://math.stackexchange.com/questions/2445992/in-how-many-ways-can-we-choose-3-objects-from-28-objects-on-circle-such-that-the | # In how many ways can we choose 3 objects from 28 objects on circle such that they are neither adjacent nor diametrically opposite?
Suppose 28 objects are placed along a circle at equal distances. In how many ways can 3 objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite? I have done this problem in the following manner: There are 28 ways to choose the first object.For the second object, there are two cases:
Case 1:
It is the next-to-next object from the first object. This can be done in two ways. For each of these two ways, we can choose the third object in 21 ways excluding the 7 as per restrictions. (1 for position of first object,1 for second object,3 for their adjacent positions and 2 for their opposite positions). Therefore, for this case, there are 28*2*21 ways to choose the object.
Case 2:
Second object occupies any other position except for the next to next positions of first object. There are 22 ways to choose the second object then. Now for each of these ways we can choose the third object in 18 ways. So the number of ways is 28*18*22.
Now the answer is 28*(18*22+2*21) ways which is way larger than the given answer:2268. The given solution uses complementary method.Though I understood the solution,I am not able to figure out the flaw in my approach but I understand that something is seriously wrong.
Kindly help me figure out the mistake.
Suppose $28$ objects are placed along a circle at equal distances. In how many ways can three objects be chosen from among them so that no two of the three chosen objects are adjacent nor diametrically opposite?
There are $\binom{28}{3}$ ways to select three of the $28$ objects. From these, we must exclude those cases in which two or more objects are adjacent or diametrically opposite.
There are $28$ ways to select a pair of adjacent objects since there are $28$ possible starting points as we move clockwise around the circle. For each such pair, there are $26$ ways to choose the third object.
The only way to have two pairs of adjacent objects is to select three consecutive objects, which can be done in $28$ ways since, again, there are $28$ possible starting points as we move clockwise around the circle.
By the Inclusion-Exclusion Principle, there are $$\binom{28}{3} - 28 \cdot 26 + 28$$ ways to select three objects so that no two of them are adjacent.
In considering the case of one pair of adjacent objects, we have already excluded those cases in which the third object is diametrically opposite one of the objects in the adjacent pair. We still need to remove those cases in which two objects are diametrically opposite each other but no two of the objects are adjacent.
There are $14$ pairs of diametrically opposite objects. For each such pair, there are $22$ ways of selecting a third object that is not adjacent to either of these objects. Hence, there are $14 \cdot 22$ pairs of diametrically opposite objects which we have not previously excluded.
Hence, the number of permissible selections is $$\binom{28}{3} - 28 \cdot 26 + 28 - 14 \cdot 22 = 2268$$
I am not able to find the flaw in my approach.
You are counting the same arrangements multiple times.
Suppose we number the objects from $1$ to $28$ as we proceed clockwise around the circle.
Case 1: There are two subcases, each of which you have counted multiple times.
Type 1: There is exactly one pair of next-to-next objects.
You count each such selection twice. For example, you count the selection $1, 3, 7$:
• once when you select $1$, then select $3$ as the next-to-next object, and $7$ as the additional object
• once when you select $3$, then select $1$ as the next-to-next object, and $7$ as the additional object
Type 2: There are two pairs of next-to-next objects. You count these selections four times.
For example, you count the selection $1, 3, 5$:
• once when you select $1$, then select $3$ as the next-to-next object, and $5$ as the additional object
• once when you select $3$, then select $1$ as the next-to-next object, and $5$ as the additional object
• once when you select $3$, then select $5$ as the next-to-next object, and $1$ as the additional object
• once when you select $5$, then select $3$ as the next-to-next object, and $1$ as the additional object
There are $28$ cases of the second type, corresponding to the $28$ possible starting points as we move clockwise around the circle.
For the first type, there are $28$ pairs, corresponding to the $28$ possible starting points as we move clockwise around the circle. There are $19$ ways to place the additional object so that it satisfies the restrictions and does not fall into the second type.
We can separate your $21$ ways of selecting the third object into $19$ ways of the first type and $2$ of the second type (choosing the third object to be one that is next-to-next of one end of the pair). Hence, there are actually $$\frac{1}{2} \cdot 28 \cdot 2 \cdot 19 + \frac{1}{4} \cdot 28 \cdot 2 \cdot 2 = 28 \cdot 19 + 28 = 28 \cdot 20 = 560$$ such cases.
Case 2: You count each such selection six times, once for each permutation of the three objects you selected.
However, you have to be careful how you count this case. Notice, for instance, that the selections $1, 4, 19$ and $1, 7, 20$ both satisfy your criteria. However, $1$ and $4$ share excluded spaces, while $1$, $7$, and $20$ do not. Whenever two of the objects are separated by two or three other objects, they share excluded spaces, which you did not take into account.
Taking the circle to be numbered, but the $3$ objects to be identical, we can divide the inadmissible cases out of $\binom{28}3$ total placements into three disjoint types for ease of computation:
• all three together: $28$
• exactly two together: $28\cdot24$
• two diametrically opposite and the third apart: $14\cdot22$
Thus admissible placements $= \binom{28}3 - (28+28\cdot 24+ 14\cdot 22) = 2268$ | 2020-04-03T02:31:22 | {
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https://math.stackexchange.com/questions/3049047/diagonalizable-matricies-and-eigenvalues | # Diagonalizable matricies and eigenvalues
Let $$A$$,$$B$$,$$C$$ be three different real $$3 \times 3$$ matricies with the following properties:
• $$A$$ has the complex eigenvalue $$\lambda=3-5i$$
• $$B$$ has eigenvalues $$\lambda=0$$, $$\lambda=5$$, $$\lambda=-5$$
• $$C=M M^T$$ for some real $$3 \times 2$$ matrix $$M$$.
Which of the matrices are necessarily diagonalizable? In the case of complex eigenvectors, diagonalization is over $$\mathbb{C}$$.
$$(A)$$ Only $$B$$
$$(B)$$ Only $$A$$ and $$B$$
$$(C)$$ Only $$B$$ and $$C$$
$$(D)$$ All three of them
$$(E)$$ None of them
Okay so first of all, I am almost certain that $$(B)$$ is diagonalizable. There are $$3$$ distinct eigenvalues for a $$3 \times 3$$ matrix so it can definitely be diagonalized.
I know for $$(A)$$ that because we have a real matrix, that the complex numbers must come in conjugate pairs so there must be a 2nd complex eigen value and one more real value so that $$(A)$$ can diagonalized too.
I'm not sure about $$(C)$$ though. I know that $$(C)$$ is a $$3\times 3$$ matrix but I am not sure if that justifies in there being 3 distinct eigenvalues so thus I could conclude that all three are diagonalizable (Thus $$(D)$$ being the answer).
Is my reasoning relatively on the right track?
• Well, $\;A\;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real... – DonAntonio Dec 22 '18 at 1:30
• There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers. – Future Math person Dec 22 '18 at 1:42
Our OP Future Math person has correctly argued that $$A$$ and $$B$$ are diagonalizable, $$A$$ over $$\Bbb C$$ and $$B$$ over $$\Bbb R$$; in each case for the reason that the matrix has $$3$$ distinct eigenvalues, hence $$3$$ linearly independent eigenvectors.
$$C = MM^T? \tag 1$$
$$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; \tag 2$$
that is, $$C$$ is a real symmetric matrix; as such, it too may be diagonalized, whether or not the eigenvalues are distinct; that real symmetric matrices are diagonalizable is a well-known result.
• It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:\mathbb{R}^3 \to \mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel. – Dionel Jaime Dec 22 '18 at 4:54 | 2021-05-16T13:04:43 | {
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https://math.stackexchange.com/questions/4059310/what-does-point-wise-convergence-in-the-sup-norm-mean | # What does point-wise convergence in the sup norm mean?
I am starting out with functional analysis and learning about point-wise and uniform convergence. In particular, I need help with my proof that the space $$\mathcal{B} = \left\{ f:X\rightarrow \mathbb{R} \, \middle| \, f \, \text{bounded} \right\}$$ of bounded real functions on an arbitrary space $$X$$ is complete in the sup norm $$\lVert f \rVert_{\infty} := \sup_{x\in X} \lvert f\left(x\right) \rvert.$$
My book (Intro to Topology and Modern Analysis by Simmons) poses this proof as a problem before any discussion on convergence of function sequences, so I will not use terms like "point-wise" or "uniform" convergence in the proof itself.
### My Proof
Let $$\left(f_n\right)_{n\in \mathbb{N}} \subset \mathcal{B}$$ be a Cauchy sequence.
Letting $$\epsilon >0$$, there is an $$N\in \mathbb{N}$$ such that $$m,n\geq N \implies \lVert f_m - f_n\rVert_{\infty} < \epsilon.$$ Thus, by definition of the sup norm we have for every $$x \in X$$, $$m,n \geq N \implies \lvert f_m\left(x\right) -f_n\left(x\right) \rvert \leq \lVert f_m-f_n\rVert_{\infty} < \epsilon$$ This shows that, for fixed $$x$$, the real sequence $$\left(f_m\left(x\right)\right)_{m\in \mathbb{N}}$$ is Cauchy and converges since $$\mathbb{R}$$ is complete. Hence, we define $$f:X\rightarrow \mathbb{R}$$ by $$f\left(x\right) = \lim_{m\rightarrow \infty} f_m\left(x\right).$$
So far we have merely constructed a function $$f$$ to which our sequence $$\left(f_m\right)_{m\in \mathbb{N}}$$ may converge (see discussion after proof); it remains to show that it does converge as a bona fide sequence in the sup norm.
To this end, let $$\epsilon >0$$ and take $$N$$ in the Cauchy hypothesis so that for any $$x\in X$$and $$m,n> N$$ we have $$\lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}$$. Then for this same $$n>N$$ we have $$\lvert f\left(x\right) - f_n\left(x\right)\rvert = \lvert \lim_{m\rightarrow \infty} f_m\left(x\right) - f_n\left(x\right) \rvert = \lim_{m\rightarrow \infty} \lvert f_m\left(x\right) - f_n\left(x\right) \rvert < \frac{\epsilon}{2}$$ (pulling the limit out of the absolute value required a small side lemma). Thus, we have $$n>N \implies \lVert f-f_n\rVert_{\infty} = \sup_{x\in X} \lvert f\left(x\right) -f_n\left(x\right)\rvert \leq \frac{\epsilon}{2} < \epsilon$$ so that $$\left(f_n\right)_{n\in \mathbb{N}}$$ converges to our $$f$$ in the sup norm.
Proving the boundedness of the limit function $$f$$ is a whole other can of worms which is irrelevant to my confusion, so I will avoid typing it out here. Once the limit function is bounded, we know that the Cauchy sequence converges in $$\mathcal{B}$$ and so $$\mathcal{B}$$ is complete.
### My Confusion
I was happy with this proof until I learned more about point-wise versus uniform convergence. My proof shows that the sequence $$\left(f_n\right)_{n\in\mathbb{N}}$$ converges in the sup norm to the $$f$$ we constructed. In doing so, I see in retrospect that I also proved that this is uniform convergence. This uniform convergence seemed to come naturally out of the proof since we used the sup norm: we found an $$N\in \mathbb{N}$$ for which the $$\epsilon$$ property holds for all $$x\in X$$. However, I didn't set out to prove uniform convergence, only convergence in the norm we had defined on the function space, so it looks like this convergence in this norm might imply uniform convergence.
My question is: where does the notion of point-wise convergence enter in? The construction of my limit function $$f$$ is done in the usual "point-wise" manner, and so I assume that by construction I can say that "$$\left(f_n\right)_{n \in \mathbb{N}}$$ converges point-wise to $$f$$," but I am bothered by when I specialize to a point $$x\in X$$, my "point-wise convergence" ends up using only the standard norm on $$\mathbb{R}$$ and not the sup norm.
Hence, I can imagine we might have function sequences which converge "point-wise" but not uniformly; how are we justified in saying that they "converge," then, since they won't converge as sequences in $$\mathcal{B}$$, where the norm is the sup norm?
It seems to me that you are overthinking this problem. You proved that, for every $$\varepsilon>0$$, there is some $$N\in\Bbb >N$$ such that if $$n\geqslant N$$ and if $$x\in X$$, then$$\sup_{x\in X}\bigl|f(x)-f_n(x)\bigr|<\varepsilon;$$in other words, you proved that, for every $$\varepsilon>0$$, there is some $$N\in\Bbb >N$$ such that if $$n\geqslant N$$ and if $$x\in X$$, then $$\|f-f_n\|_\infty<\varepsilon$$, which is what you were supposed to proved.
Concerning your remark that “using only the standard norm on R and not the sup norm”, well… what's the sup norm on $$\Bbb R$$. On $$\Bbb R^n$$, you can define$$\|(x_1,\ldots,x_n\|_{\text{sup}}=\sup\{|x_1|,|x_2|,\ldots,|x_n|\}=\max\{|x_1|,|x_2|,\ldots,|x_n|\},$$but, when $$n=1$$, this is just the usual distance in $$\Bbb R$$. Besides, on $$\Bbb R^n$$ all norms are equivalent.
After focusing a bit more on the basic definitions, I think I've made sense of this on my own. I'll leave my notes here in this answer to help anyone else who hits upon this.
I will generalize the definitions of convergence that I was using to more general spaces. Let $$\left(X, d_X\right)$$ and $$\left(Y,d_Y\right)$$ be metric spaces and $$f$$ and $$\left\{f_n\right\}_{n\in\mathbb{N}}$$ be functions from $$X$$ to $$Y$$.
We say that "$$\left(f_n\right)_{n\in\mathbb{N}}$$ converges pointwise to $$f$$" if $$\forall \epsilon>0$$ and $$x\in X$$ we can find some $$N \in \mathbb{N}$$ so that $$n>N \implies d_{Y}\left(f_n\left(x\right), f\left(x\right) \right) < \epsilon.$$ Notice that we don't use the metric on $$X$$ anywhere in this definition, since we are concerned with a single point in the domain; the metric on $$Y$$ allows us to quantify the distance between the function values (in my original problem, this was simply the usual metric on $$\mathbb{R}$$ since my functions were real-valued).
We say that "$$\left(f_n\right)_{n\in\mathbb{N}}$$ converges uniformly to $$f$$" if $$\forall \epsilon >0$$ we can find some $$N\in \mathbb{N}$$ so that for all $$x\in X$$ we have $$n>N \implies d_Y\left(f_n\left(x\right), f\left(x\right) \right) < \epsilon.$$ Again we do not use the metric on $$X$$.
These two notions of convergence of function sequences are certainly not equivalent, though uniform convergence implies pointwise. My earlier proof proceeded by finding a pointwise limit of the sequence and showing that --- by the grace of its Cauchy-ness --- this convergence was also uniform. My confusion started when trying to consider these function sequences as sequences in a function space equipped with the uniform metric. My proof below shows that uniform convergence of a function sequence is equivalent to convergence of the sequence in a function space equipped with the uniform norm. Thus, I am happily convinced that these notions of convergence coincide and it makes sense to consider uniform convergence as true convergence of a sequence of elements (functions) in a (function) space.
## Proof
First let $$\left(f_n\right)_{n\in \mathbb{N}}$$ converge uniformly to $$f$$, all functions from metric space $$\left(X, d_X\right)$$ to $$\left(Y, d_Y\right)$$ belonging to some function space, call it $$\mathcal{C}$$. Define the uniform metric on $$\mathcal{C}$$ to be the usual $$d_{\mathcal{C}}\left(f,g\right) = \lVert f - g\rVert_{\infty} := \sup_{x\in X} d_Y\left(f\left(x\right), g\left(x\right)\right).$$
Let $$\epsilon >0$$ arbitrary. Then by uniform convergence there exists some $$N \in \mathbb{N}$$ so that $$\forall x\in X, n > N \implies d_Y\left(f_n\left(x\right), f\left(x\right) \right) < \frac{\epsilon}{2}.$$ Now let $$B = \left\{ d_Y\left(f_{n'}\left(x'\right), f\left(x'\right)\right) \, \middle| \, x' \in X, n' > N\right\}.$$ We have just shown that this set $$B\subset \mathbb{R}$$ is bounded, and as a subset of $$\mathbb{R}$$ it has a supremum $$\beta = \sup B.$$ But then $$\beta \leq \frac{\epsilon}{2} < \epsilon$$. Now, notice that since for a fixed n > N, $$\left\{ d_Y\left(f_n\left(x\right), f\left(x\right)\right)\, \middle| \, x\in X \right\} \subseteq B,$$ we thus have $$n>N \implies \lVert f_n - f\rVert_{\infty} = \sup_{x\in X} d_Y\left(f_n\left(x\right), f\left(x\right)\right) \leq \beta < \epsilon.$$ That is, $$\left(f_n\right)_{n\in \mathbb{N}}$$ converges to $$f$$ in the uniform norm on $$\mathcal{C}$$.
Conversely, let $$\left(f_n\right)_{n\in \mathbb{N}}$$ converge to $$f$$ in the uniform norm on $$\mathcal{C}$$. Then let $$\epsilon >0$$ and take $$N \in \mathbb{N}$$ so that $$n>N \implies \lVert f_n - f\rVert_{\infty} = \sup_{x\in X} d_Y\left( f_n\left(x\right), f\left(x\right)\right) < \epsilon.$$ But then for arbitrary $$x\in X$$ we have $$n>N \implies d_Y\left(f_n\left(x\right), f\left(x\right)\right) \leq \lVert f_n - f\rVert_{\infty} < \epsilon,$$ so that $$\left(f_n\right)_{n\in\mathbb{N}}$$ converges uniformly to $$f$$. $$\blacksquare$$ | 2022-05-18T10:55:49 | {
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https://math.stackexchange.com/questions/1641966/probability-of-second-card-being-an-ace?noredirect=1 | # Probability of second card being an ace
Consider choosing a card from a well-shuffled standard deck of 52 playing cards.
• Suppose that, after the first extraction, the card is not reinserted in the deck. What is the probability that the second card is an ace?
• Suppose that, after the first extraction, the card is reinserted in the deck. What is the probability that the second card is an ace?
For the first case, I've been thinking about $$\left(\frac{4*3}{52*51}\right) + \left(\frac{48*4}{52*51}\right)$$ Where I counted two examples: a)the ace was drawn already, and b) it wasn't. For the second part of the task, would it be $$\left(\frac{4*4}{52*52}\right) + \left(\frac{52*4}{52*52}\right)$$ I'm pretty sure that I'm assuming wrong, but I can't come up with anything else. Any help would be great!
• Where does it say in the problem statement that the first card drawn was an ace? – David K Feb 5 '16 at 16:13
• for both questions the answer is $4/52 = 1/13$ – Ant Feb 5 '16 at 16:16
• @DavidK well it doesn't, that's why I counted it for both cases, since we do not know what was the first card. Could've been an ace, could've not been. – Jake Feb 5 '16 at 16:17
• where do you have 52/52 from in the second case? – Scavenger23 Feb 5 '16 at 16:19
• @Pauline mainly because of a symmetry argument. There is nothing special about the "first" card. You can very well say "look at the 13th card" or whatever, it does not make any difference; it would be different if you look at the first card and you see that it's NOT an ace (or you see that is IS an ace) but if you do not have any information, then probability does not change and it's still 1/13 for whatever card in the deck – Ant Feb 5 '16 at 16:24
As Ant said in the comments, in both cases, the answer is $\frac{4}{52} = \frac{1}{13}$. Your first equation simplifies to this: \begin{align*} \left(\frac{4\cdot 3}{52 \cdot 51}\right) + \left(\frac{48 \cdot 4}{52 \cdot 51}\right) &= \left(\frac{4}{52}\right)\left( \frac{3}{51} + \frac{48}{51}\right) \\ &= \left(\frac{4}{52}\right)(1) \\ &= \frac{1}{13} \end{align*} But this is actually overcomplicating things; drawing the first card doesn't affect the odds at all, regardless of whether or not it's replaced. We don't know anything about what the first card drawn was, so it doesn't add any information, and thus doesn't affect the probability that the second card is an ace.
The first question can be interpreted as:
If you give the cards a number then what is the probability that number $2$ is given to an ace?
The second question can be interpreted as:
If you give the cards a number then what is the probability that number $1$ is given to an ace?
(Here I mean the numbers $\{1,2,\dots,52\}$)
In the second cases the numbering takes place after the (irrelevant) extraction and reinserting of a card.
In both cases the answer is $\frac4{52}=\frac1{13}$
From a probabilistic point of view, extracting a card and placing it on same table of your deck without looking at it is like not extracting it at all: you have the same information you had before. More precisely given a probability space $(\Omega,\mathcal F,\mathbb P)$ your $\sigma$-algebra $\mathcal F$ is not changed, so your probability space is not changed.
Mathematically if $E_1=\{$Ace at the first extraction$\}$ and $E_2=\{$Ace at the second extraction$\}$ then $$P(E_2)=P(E_2\cap E_1)+P(E_2\cap E_1^c)$$ because $E_1$ and $E_1^c$ are a partition of $\mathcal F$.
From the definition of conditional probability $P(E_2|E_1)=\frac{P(E_2\cap E_1)}{P(E_1)}$, so $P(E_2\cap E_1)=P(E_2|E_1)P(E_1)$ and $P(E_2\cap E_1^c)=P(E_2|E_1^c)P(E_1^c)$.
Now, know the first card was an ace, means that you have 51 cards in your deck now and only 3 aces in that, so $P(E_2|E_1)=\frac 3 {51}$ and obviously $P(E_1)=\frac 4 {52}$. Similarly, knowing your first card wasn't an ace means you have 4 aces in a deck of 51 cards , so $P(E_2|E_1^c)=\frac 4 {51}$ and $P(E_1^c)=1-P(E_1)=\frac {48} {52}$. So $$P(E_2)=P(E_2\cap E_1)+P(E_2\cap E_1^c)\\=\frac 3 {51}\cdot\frac 4 {52}+\frac 4 {51}\cdot\frac {48} {52}\\=\frac 4 {52}[\frac 3 {51}+\frac {48} {51}]\\=\frac 4 {52}=\frac 1 {13}$$
For the first part,
$$\left(\frac{4*3}{52*51}\right) + \left(\frac{48*4}{52*51}\right) = \frac{4}{52} \left(\frac{3}{51} + \frac{48}{51}\right) = \frac{1}{13} \left(\frac{51}{51}\right) = \frac{1}{13}$$
which is the correct answer. The answer has to be $1/13$, because the answer must be the same if you ask the question with the word "king" instead of "ace", or "seven" instead of "ace", etc. There are exactly $13$ different words you can have where you wrote "ace", each producing the same non-zero probability.
The answer for the second one has to be $1/13$ for the same reason. But in this case,
$$\left(\frac{52*4}{52*52}\right) = \frac{1}{13},$$
so the term $\frac{4*4}{52*52}$ should not be added. It looks like you considered the case where the first card was an ace as a $4/52$ probability, but when you considered the "other" case you used a $52/52$ probability, so basically you counted twice the case where the first card is an ace. | 2019-11-18T19:47:25 | {
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http://mathhelpforum.com/trigonometry/95853-trigonometric-general-solutions.html | # Math Help - Trigonometric General Solutions
1. ## Trigonometric General Solutions
I guess Trigonometry question go in the Trigonometry section. That makes sense.
I was having trouble with this question (See attachment) as I'm not sure how to change it into a sum or something similar to find the general solution. I’m not sure if I need to use identities or a product to sum formula or something similar. If someone could work me through it, it would be greatly appreciated.
2. Try double angle formula
3. I thought of using the double angle formula on sin4x but I'm not sure how to use this to find the general solutions. If you could show me what you mean it would help a lot.
EDIT : That's a good start to use double angle formula on sin 4x
5. Using the double angle formula sin4x-cosx=0,
=2sin2xcos2x-cosx=0
This is all I could manage to do. I'm not sure if I should use the double angle formula on cosx either?
6. Use double angle formula again on sin 2x first, then factorise the equation
7. songoku,
I'm not sure how to use the double angle formula a second time and then factorise it. It would be helpful if you could show me how to do that and then showhow to factorise it.
Thanks.
I guess Trigonometry question go in the Trigonometry section. That makes sense.
I was having trouble with this question (See attachment) as I'm not sure how to change it into a sum or something similar to find the general solution. I’m not sure if I need to use identities or a product to sum formula or something similar. If someone could work me through it, it would be greatly appreciated.
A simple approach:
$\sin (4x) = \cos (x)$
$\Rightarrow \sin (4x) = \sin \left( \frac{\pi}{2} - x \right)$
using the complementary angle formula.
Case 1: $4x = \frac{\pi}{2} - x + 2n \pi$
Case 2: $4x = \pi - \left( \frac{\pi}{2} - x\right) + 2n \pi$
where n is an integer. Solve for x in each case.
9. wow, really nice work mr_fantastic
btw, I mean like this :
2sin2xcos2x-cosx = 0
4 sin x cos x cos 2x - cos x = 0
(cos x) (4 sin x cos 2x - 1 ) = 0
cos x = 0 or 4 sin x cos 2x - 1 = 0
Use the double angle formula again on cos 2x and state the equation in term of sin x
but mr_fantastic has given a really elegant way | 2014-03-08T19:14:23 | {
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https://ai.stackexchange.com/questions/12612/which-function-haty-y2-or-y-haty2-should-i-use-to-compute-th/12620#12620 | # Which function $(\hat{y} - y)^2$ or $(y - \hat{y})^2$ should I use to compute the gradient?
The MSE can be defined as $$(\hat{y} - y)^2$$, which should be equal to $$(y - \hat{y})^2$$, but I think their derivative is different, so I am confused of what derivative will I use for computing my gradient. Can someone explain for me what term to use?
The derivative of $$\mathcal{L_1}(y, x) = (\hat{y} - y)^2 = (f(x) - y)^2$$ with respect to $$\hat{y}$$, where $$f$$ is the model and $$\hat{y} = f(x)$$ is the output of the model, is
\begin{align} \frac{d}{d \hat{y}} \mathcal{L_1} &= \frac{d}{d \hat{y}} (\hat{y} - y)^2 \\ &= 2(\hat{y} - y) \frac{d}{d \hat{y}} (\hat{y} - y) \\ &= 2(\hat{y} - y) (1) \\ &= 2(\hat{y} - y) \end{align}
The derivative of $$\mathcal{L_2}(y, x) = (y - \hat{y})^2 = (y - f(x))^2$$ w.r.t $$\hat{y}$$ is
\begin{align} \frac{d}{d \hat{y}} \mathcal{L_2} &= \frac{d}{d \hat{y}} (y - \hat{y})^2 \\ &= 2(y -\hat{y}) \frac{d}{d \hat{y}} (y -\hat{y}) \\ &= 2(y - \hat{y})(-1)\\ &= -2(y - \hat{y})\\ &= 2(\hat{y} - y) \end{align}
So, the derivatives of $$\mathcal{L_1}$$ and $$\mathcal{L_2}$$ are the same.
The MSE can be defined as $$(\hat{y} - y)^2$$, which should be equivalent to $$(y - \hat{y})^2$$
They are not just "equivalent". It is actually the exact same function, with two different ways to write it.
$$(\hat{y} - y)^2 = (\hat{y} - y)(\hat{y} - y) = \hat{y}^2 -2\hat{y}y + y^2$$
$$(y - \hat{y})^2 = (y -\hat{y})(y - \hat{y}) = y^2 -2y\hat{y} + \hat{y}^2$$
These are exactly the same function. Not just "equivalent" or "equivalent everywhere", but actually the same function. It is therefore no surprise that any derivative is also the same - including the partial derivative with respect to $$\hat{y}$$ which is what you typically use to drive gradient descent.
The two ways of writing the function is because it is a square and thus has two factorisations. When you write it as a square you can choose which form to use for the inner term.
Which function [form] should I use to compute the gradient?
You can use either form, it does not matter. They represent the same function and have the same gradient.
The derivative is the same as far as I understand it.
If $$y$$ is constant and $$\hat{y}$$ is the variable the result will be:
$$((\hat{y} - y)^2)' = 2(\hat{y} - y)$$
and for the other formula:
$$((y - \hat{y})^2)' = -2(y - \hat{y})$$
which is the same.
• I can see that the second equation can only be derived if you have taken the correct route for partial derivative (I commented earlier that it looked wrong - but actually I was wrong to say that). Usually tutorials don't consider that $y$ is a "constant", but that this is a partial derivative, where we only care about the gradient w.r.t. $\hat{y}$. The result is much the same, but either way it may help to show a step of expansion May 31 '19 at 16:09 | 2021-10-27T06:42:41 | {
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https://math.stackexchange.com/questions/569901/linear-basis-for-a-quotient-ring | # Linear basis for a quotient ring
Question:
Let $k$ be a real number, and let $A$ denote the ring $\mathbb{R}[x]/(x^2+k)$. Find an $\mathbb{R}$-linear basis for $A$ and describe the multiplication law in terms of this basis.
I am not quite sure about the meaning of $\mathbb{R}$-linear basis. For example, if $k=1$, then $\mathbb{R}[x]/(x^2+k)$ is isomorphic to $\mathbb{C}$. So in this case, the $\mathbb{R}$-linear basis is simply a real scalar or just 1. Right?
In addition, what would be the general form of the basis for any $k$, please?
Also what does it mean by "describing the multiplication law"?
Extension Question:
As a variation to the above question, what would change if we change $\mathbb{R}$ to $\mathbb{Z}$, please? To be more specific, what is the basis for $\mathbb{Z}[x]/(x^2-2)$ this time, please? Thank you!
• Nothing is changed: $\Bbb Z[x]/(x^2-2)$ is a free $\Bbb Z$-module of rank 2 and a basis is also $\{1,x\}$. (The division with remainder of polynomials in $\Bbb Z[x]$ by $x^2-2$ still holds.) – user89712 Nov 19 '13 at 8:38
• I understand that $\Bbb R[x]/(x^2-1)$ is isomorphic to $\mathbb{R}\times \mathbb{R}$. Then what is $\Bbb Z[x]/(x^2-2)$ isomorphic to? – LaTeXFan Nov 19 '13 at 9:30
• To $\Bbb Z\times\Bbb Z$ (as $\Bbb Z$-modules, or as abelian groups if you like, but not as rings!). – user89712 Nov 19 '13 at 9:32
If $A=\Bbb R[x]/(x^2+k)$, then let's look at the elements of this factor ring: take $f(x)\in\Bbb R[x]$ and consider $f(x)\pmod{(x^2+k)}$. In $\Bbb R[x]$ we have $f(x)=(x^2+k)g(x)+r(x)$ with $\deg r\le 1$, so $r(x)=ax+b$ and $f(x)\pmod{(x^2+k)}=r(x)\pmod{(x^2+k)}$. This shows us that the residue classes of polynomials modulo $(x^2+k)$ are in fact residue classes of polynomials of degree at most one. In conclusion, $A=\{ax+b:a,b\in\Bbb R\}$ and then $\{1,x\}$ is an $\Bbb R$-basis of $A$. (Actually, in this context, $x$ is in fact the residue class of $x$ modulo $(x^2+k)$. If you like, you can denote it by other letter.)
Now let's describe the multiplication law in terms of basis: $$(ax+b)(cx+d)=(ad+bc)x+bd-kac.$$ (Here I have used that $x^2=-k$ in $A$.)
• Thank you for your answer. It is very helpful. – LaTeXFan Nov 18 '13 at 23:12
You have to remember polynomial division. If $f(x),g(x)\in\mathbb{R}[x]$, with $g\ne0$, there exist unique polynomials $q(x)$ and $r(x)$ such that $f(x)=g(x)q(x)+r(x)$ and the degree of $r$ is strictly less than the degree of $g$.
In your case, for any $f(x)\in\mathbb{R}[x]$ there are $q(x)$ and $a,b\in\mathbb{R}$ such that $$f(x)=(x^2+k)q(x)+ax+b$$ because $x^2+k$ has degree $2$, so the remainder $r(x)=ax+b$ for some real $a$ and $b$.
Can you see the dimension of $A$ as a vector space over $\mathbb{R}$, now? Do you have any restriction about what $a$ and $b$ can be? Can you deduce that any element of $A$ can be represented as the equivalence class of a linear polynomial? How do you multiply those linear polynomials?
(Hint: the dimension is $2$.)
• Thank you for your answer. It is quite clear. – LaTeXFan Nov 18 '13 at 23:10
By modding out $\mathbb{R}[x]$ by $(x^2+k)$, all you are doing is "extending" $\mathbb{R}$ by adjoining the roots of the polynomial $x^2+k=0$. In particular, as you pointed out, letting $k=1$ gives you $\mathbb{R}[x]/(x^2+1)\cong \mathbb{R}[i]\cong \mathbb{C}$. In this case, a $\mathbb{R}$-basis would be $\{1,i\}$, or if you were to express the basis as elements of $\mathbb{R}[x]/(x^2+1)$, you could say $\{1,x\}$ with multiplication law $x^2=-1$.
I hope this helps!
• Thank you for your help. It makes sense to me now. – LaTeXFan Nov 17 '13 at 6:51
• I'm glad I could help. – userCaltech Nov 17 '13 at 7:13 | 2019-07-23T00:28:40 | {
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https://math.stackexchange.com/questions/3279393/is-the-reasoning-neg-p-vee-neg-q-p-therefore-neg-q-valid | # Is the reasoning $\neg p\vee\neg q;\;p\therefore\neg q$ valid?
Is it valid to say that: $$\begin{array}{l}\neg p\vee\neg q\\p\\\hline\therefore\neg q\end{array}$$ knowing that?: $$\begin{array}{l}p\vee q\\\neg p\\\hline\therefore q\end{array}$$ (the last reasoning is called "Disjunctive syllogism").
I think yes, because:
$$\begin{array}{lll} (1)&\neg p\vee\neg q&\text{Premise}\\ (2)&p&\text{Premise}\\ (3)&p\to\neg q&\text{Conditional equivalence in (1)}\\ (4)&\neg q&\text{Modus Ponens (2)-(3)}\\ \end{array}$$
Therefore the first reasoning is valid.
Is my deduction correct?
• Everything is correct! – azif00 Jul 1 '19 at 3:27
• The "disjunctive syllogism" is a special case of Resolution, as used (at least theoretically, implementations may differ) in Prolog – David Tonhofer Jul 1 '19 at 16:34
Yes, this is valid. As you mention, it's called the Disjunctive Syllogism (or Modus Tollendo Ponens) and is one of the fundamental building blocks (inference rules) of propositional calculus.
The "standard" proof goes something like this, using conjunction introduction and De Morgan:
• Given $$\neg p \vee \neg q$$
• Given $$p$$
• Hypothetically suppose $$q$$:
• Conjunction introduction gives $$p \wedge q$$
• De Morgan gives $$\neg (\neg p \vee \neg q)$$
• Let $$X$$ stand for $$\neg p \vee \neg q$$
• We now have $$X \wedge \neg X$$: a contradiction
• Therefore $$q \rightarrow \bot$$
• Therefore, by reductio ad absurdum, $$\neg q$$
You can also get this without De Morgan if you have to:
• Given $$\neg p \vee \neg q$$
• Given $$p$$
• Hypothetically suppose $$\neg p$$:
• Now we have $$p \wedge \neg p$$, which is a contradiction
• So $$\neg p \rightarrow \bot$$, and $$\bot \rightarrow \neg q$$ (principle of explosion)
• Therefore $$\neg p \rightarrow \neg q$$
• $$\neg q \rightarrow \neg q$$, by conditional introduction
• So $$\neg q$$ by disjunction elimination
• Thank you!! Another question could be "But we did not use the disjunctive syllogism to prove the validity of this reasoning", and we could answer "But that reasoning was not necessary to prove this one", right? – manooooh Jul 1 '19 at 3:31
• @manooooh Correct! I added a short proof. – Draconis Jul 1 '19 at 3:34
• @manooooh There's a rule called "ex falso sequitur quodlibet" or the "principle of explosion", which says basically that $\bot \rightarrow X$ for any $X$. I'll clarify. – Draconis Jul 1 '19 at 3:48
• Alternatively, use disjunction elimination, negation elimination, and negation introduction.$$\def\fitch#1#2{\quad\begin{array}{|l} #1\\\hline #2\end{array}}\small\fitch{1.~\lnot p\lor\lnot q\hspace{5ex}\text{Premise}\\2.~p\hspace{11.5ex}\text{Premise}}{\fitch{3.~q\hspace{8ex}\text{Assume}}{\fitch{4.~\lnot p\hspace{3ex}\text{Assume}}{5.~\bot\hspace{4ex}\text{Negation Elim.},2,4}\\\fitch{6.~\lnot q\hspace{3ex}\text{Assume}}{7.~\bot\hspace{4ex}\text{Negation Elim.},3,6}\\8.~\bot\hspace{7.5ex}\text{Disjunction Elim.},1,4{-}5,6{-}7}\\9.~\lnot q\hspace{10ex}\text{Negation Intro.},3{-}8}$$ – Graham Kemp Jul 1 '19 at 6:46
• @GrahamKemp I didn't know one could write comments like that. How? – David Tonhofer Jul 1 '19 at 16:29 | 2020-10-31T16:34:20 | {
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https://math.stackexchange.com/questions/483238/prove-that-roots-are-real | # Prove that roots are real
I am stuck with this equation, I need to prove the roots are real when $a, b, c \in R$
The equation is $$(a+b-c)x^2+ 2(a+b)x + (a+b+c) = 0$$
If someone could tell me the right way to go about this, so I can attempt it.
Thank you
EDIT: I have made an error in the question. I have now corrected it.
• Do you know the relationship between the discriminant of a polynomial and it's roots? – Tom Oldfield Sep 3 '13 at 19:19
• You must have misread the question. The roots are only real if $|a+b+c| \le |a+b|$. – TonyK Sep 3 '13 at 19:29
• I had an error in the question, it is now corrected – user Sep 3 '13 at 19:33
We look at the discriminant of the the polynomial, which for a quadratic $ax^2 +bx +c$ is $b^2 -4ac$, plugging the values in for our polynomial gives $$\Delta = 4(a+b)^2-4(a+b-c)(a+b+c)\\ = 4[(a+b)^2 - (a+b)^2+c^2]\\ = 4c^2$$
Since the square of a real number is positive, we know that the roots must be real, by looking at the quadratic forumla and seeing that the solutions are $$\frac{-b\pm\sqrt\Delta}{2a}$$
And the square root of a positive real number is real. We used the discriminant because it makes computation so much easier, than if we were doing everything that we did in the first step underneath the radical, and it would be rather ugly. Inspection shows that if $\Delta > 0$, there are two distinct real roots, if $\Delta < 0$, there are two complex roots, which are conjugate, and if $\Delta = 0$ then you have a real double root.
• Sorry, typo in question. – user Sep 3 '13 at 19:35
• All fixed with the new problem! – guest196883 Sep 3 '13 at 19:41
• we know that the roots must be positive - probably a typo, they must be real. – Džuris Sep 3 '13 at 19:45
• Yes yes, thanks. It was a typo. – guest196883 Sep 3 '13 at 19:46
• All answers were really good but this one just that bit clearer. Thanks – user Sep 3 '13 at 21:01
Set $a+b-c=A$ and $a+b+c=B$; then $2(a+b)=A+B$ and your equation is $$Ax^2+(A+B)x+B=0$$ that can be written $$Ax^2+Ax+Bx+B=0$$ or $$Ax(x+1)+B(x+1)=0$$ and finally $$(Ax+B)(x+1)=0.$$ If $A=0$, then the equation is $B(x+1)=0$. If also $B=0$, the equation is an identity; otherwise it has the only solution $-1$. If $A\ne0$, the solutions are $-1$ and $-B/A$.
Note that $a$ and $b$ only appear in the expression $a+b$. So let us simplify our lives and denote $a+b =:d$, so we need to show that the following has real roots:
$$(d-c)x^2 + 2dx + (c+d) = 0.$$
There is a useful object called the discriminant. For an equation $Ax^2 + Bx + C$ it is given as: $\Delta = B^2 - 4 AC$. The roots of a polynomial are real iff $\Delta \geq 0$. For our polynomial, we have:
$$\Delta = 4d^2 - 4(d+c)(d-c) = 4 c^2 \geq 0.$$
Because $\Delta \geq 0$, the roots are real, and you are done.
By the way, the roots of $Ax^2 + Bx + C$ are $\frac{-B \pm \Delta}{2A}$, which is very often very useful.
• Typo in the question, Sorry – user Sep 3 '13 at 19:35
$$ax^2 +bx^2 + cx^2 + 2ax+2bx + a+b+c = 0$$
or
$$x^2(a+b-c) + 2x(a+b) +a+b+c = 0$$
Let us now find the expression for the discriminant,
$$\Delta = 4(a+b)^2-4(a+b-c)(a+b+c)$$
If we prove that, $\Delta\geq 0$, we done.
$$4(a+b)^2-4(a+b+c)(a+b-c)=$$
$$=4a^2+8ab+4b^2-4(a^2+2ab+b^2-c^2)=$$
$$=4a^2+8ab+4b^2-4a^2-8ab-4b^2+4c^2=$$
$$4c^2\geq 0$$
Hence, indeed
$$\Delta\geq 0$$
and therefore the quadratic equation has real roots.
You could note that $x=-1$ is a root.
If this is not obvious at first, write (motivated by the fact that $a,b$ always appear as $a+b$)
$$0=(a+b-c)x^2+2(a+b)x+(a+b+c)=(a+b)(x+1)^2-c(x^2-1)$$$$=(x+1)\left((a+b)(x+1)-c(x-1)\right)=(x+1)\left((a+b-c)x+(a+b+c)\right)$$
Note that if $a+b=c$ the equation is linear (just one root), and that if $c=0, a+b\neq 0$ there is a double root at $x=-1$
Obviously going via the discriminant is a reliably effective way. With coefficients like these it is worth testing $x=\pm1$ just in case it saves a lot of work. | 2019-11-18T04:07:07 | {
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# What is the remainder when the positive integer n is divided by 5 ?
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What is the remainder when the positive integer n is divided by 5 ?
(1) When n is divided by 3, the quotient is 4 and the remainder is 1.
(2) When n is divided by 4, the remainder is 1.
DS07502.01
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Re: What is the remainder when the positive integer n is divided by 5 ? [#permalink]
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26 Apr 2019, 04:28
1
From S1:
n = 3*4+1
n = 13
Sufficient.
When 13 is divided by 5, the remainder is 3.
From S2:
n/4 = 1
n can be 1, 5, 9....
Insufficient.
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Re: What is the remainder when the positive integer n is divided by 5 ? [#permalink]
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27 Apr 2019, 07:02
Bunuel wrote:
What is the remainder when the positive integer n is divided by 5 ?
(1) When n is divided by 3, the quotient is 4 and the remainder is 1.
(2) When n is divided by 4, the remainder is 1.
DS07502.01
OG2020 NEW QUESTION
from given info
#1
When n is divided by 3, the quotient is 4 and the remainder is 1.
n= 13 , remainder divided by 5 ; 3
sufficient
#2
When n is divided by 4, the remainder is 1
n=5,15,25 , we get remainder 0
and for all other values remainder as 1 for n = 9, 21
insufficeint
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Re: What is the remainder when the positive integer n is divided by 5 ? [#permalink]
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27 Apr 2019, 07:09
Bunuel wrote:
What is the remainder when the positive integer n is divided by 5 ?
(1) When n is divided by 3, the quotient is 4 and the remainder is 1.
(2) When n is divided by 4, the remainder is 1.
DS07502.01
OG2020 NEW QUESTION
We know that $$n$$ is a positive integer. The original question: $$n \bmod 5=?$$
1) We know that $$n=3\cdot 4+1=13$$, so $$n \bmod 5=3$$. Thus, the answer to the original question is a unique value. $$\implies$$ Sufficient
2) We know that $$n=4q+1$$ and can test possible cases. If $$n=1$$, then $$n \bmod 5=1$$. However, if $$n=5$$, then $$n \bmod 5=0$$. Thus, we can't get a unique value to answer the original question. $$\implies$$ Insufficient
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What is the remainder when the positive integer n is divided by 5 ? [#permalink]
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30 Apr 2019, 08:10
Quote:
What is the remainder when the positive integer n is divided by 5 ?
(1) When n is divided by 3, the quotient is 4 and the remainder is 1.
(2) When n is divided by 4, the remainder is 1.
This question primarily hinges on our understanding of the relationship between dividend, divisor, quotient, and remainder. If we can either solve for n, or solve for the units digit of n, we will have sufficiency (as divisibility by 5 depends entirely on the units digit). Now we're ready to take a look at our statements!
Statement (1) tells us that "When n is divided by 3, the quotient is 4 and the remainder is 1." If we set this up mathematically using our understanding that the dividend = (divisor)*(quotient) + remainder, we can solve for "n," as n = 3*4 + 1, or 13. If we know "n," we certainly know the remainder when we divide "n" by 5 - in this case, 3. (Sufficient)
With Statement (2), we know that n/4 gives us a remainder of 1, so we know that n = "some multiple of 4, plus 1." However, we can quickly disprove sufficiency, as we could have 4+1 = 5, giving us a remainder of 0, or 8+1 = 9, giving us a remainder of 4. As soon as we can pick permissible values that give us different values for this value Data Sufficiency quesiton, we have disproven sufficiency. (Not Sufficient)
In this case, taking a moment to preempt what we need to have sufficiency, and recognizing that if we're given enough information to solve for "n," we have sufficiency, and that if we are able to quickly choose permissible values for "n" that give us different answers to our value DS question, we can efficiently disprove sufficiency, we are left with (A).
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Re: What is the remainder when the positive integer n is divided by 5 ? [#permalink]
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30 Apr 2019, 09:39
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Bunuel wrote:
What is the remainder when the positive integer n is divided by 5 ?
(1) When n is divided by 3, the quotient is 4 and the remainder is 1.
(2) When n is divided by 4, the remainder is 1.
Target question: What is the remainder when the positive integer n is divided by 5 ?
Statement 1: When n is divided by 3, the quotient is 4 and the remainder is 1.
There's a nice rule that says, "If N divided by D equals Q with remainder R, then N = DQ + R"
For example, since 17 divided by 5 equals 3 with remainder 2, then we can write 17 = (5)(3) + 2
Likewise, since 53 divided by 10 equals 5 with remainder 3, then we can write 53 = (10)(5) + 3
So, from statement 1, we can write: n = (3)(4) + 1 = 13
If n = 13, then we get a remainder of 3 when we divide 13 by 5
Since we can answer the target question with certainty, statement 1 is SUFFICIENT
Statement 2: When n is divided by 4, the remainder is 1.
We have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc.
For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
Some possible values of n are: 1, 5, 9, 13, 17, . . . etc.
Case a: If n = 1, then we get a remainder of 1 when we divide 1 by 5.
Case b: If n = 5, then we get a remainder of 0 when we divide 5 by 5.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT
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Re: What is the remainder when the positive integer n is divided by 5 ? [#permalink]
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06 May 2019, 20:15
Bunuel wrote:
What is the remainder when the positive integer n is divided by 5 ?
(1) When n is divided by 3, the quotient is 4 and the remainder is 1.
(2) When n is divided by 4, the remainder is 1.
DS07502.01
OG2020 NEW QUESTION
Statement One Alone:
When n is divided by 3, the quotient is 4 and the remainder is 1.
We see that n must be 3 x 4 + 1 = 13 and 13/5 = 2 remainder 3.
Statement one alone is sufficient to answer the question.
Statement Two Alone:
When n is divided by 4, the remainder is 1.
Thus, n can be values such as 1, 5, 9, 13…
1/5 = 0 remainder 1, however, 5/5 = 1 remainder 0.
Statement two alone is sufficient to answer the question.
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Re: What is the remainder when the positive integer n is divided by 5 ? [#permalink]
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16 May 2019, 14:38
Hi All,
We're told that N is a positive integers. We're asked for the remainder when N is divided by 5. This question can be solved with a mix of Arithmetic and TESTing VALUES.
(1) When N is divided by 3, the quotient is 4 and the remainder is 1.
Fact 1 gives us remarkably specific information....
N/3 = 4r1
This outcome can only occur when N = 13, since 13/3 = 4r1. No other value of N fits this information, so we have 13/5 = 2r3 and the answer to the question must be 3.
Fact 1 is SUFFICIENT
(2) When N is divided by 4, the remainder is 1.
Fact 2 isn't quite as 'restrictive' as Fact 1 is. There are lots of different values of N that will fit here:
IF...
N = 1, then 1/4 = 0r1 and the answer to the question is 1/5 = 0r1.... 1
N = 5, then 5/4 = 1r1 and the answer to the question is 5/5 = 1r0.... 0
Fact 2 is INSUFFICIENT
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Re: What is the remainder when the positive integer n is divided by 5 ? [#permalink] 16 May 2019, 14:38
Display posts from previous: Sort by | 2019-11-21T11:19:06 | {
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https://math.stackexchange.com/questions/974973/continuous-function-with-continuous-one-sided-derivative | # Continuous function with continuous one-sided derivative
Simple example of the absolute value function $x \mapsto |x|$ on $\mathbb{R}$ shows that it is possible for a continuous function to posses both the right-hand and the left-hand side derivatives and still not being differentiable on $\mathbb{R}$.
I was wondering if it is possible to assume something about one of the one-hand side derivatives to obtain differentiability.
The obvious came to my mind:
Is it true that if a continuous function $f \in C(\mathbb{R})$ has left-hand-side derivative $f_{-}^{'}$ that is continuous on $\mathbb{R}$, then the function $f$ is differentiable?
• The necessary and sufficient condition is that $f'_-=f'_+$. Can you deduce this identity from the continuity of $f'_-$? – Siminore Oct 15 '14 at 13:37
• @Siminore: Thanks for your comment. I need to prove that the right-hand-side derivative $f'_+$ exists and is equal to $f'_{-}$. Is that what you meant? If so, then I have trouble with that. – xen Oct 15 '14 at 13:43
• See this question. – Tony Piccolo Oct 15 '14 at 14:38
• @TonyPiccolo Many thanks! I somehow overlooked that question. – xen Oct 15 '14 at 18:32
Yes.
The keystone is:
Lemma. Let $f\colon [a,b]\to\mathbb R$ be continuous and assume that $f'_+(x)$ exists and is $>0$ for all $x\in [a,b)$. Then $f$ is strictly increasing.
Assume otherwise, i.e. $f(a)\ge f(b)$. We recursively define a map $g\colon \operatorname{Ord}\to [a,b)$ such that $g$ and $f\circ g$ are strictly inreasing. Since the class $\operatorname{Ord}$ of ordinals is a proper class and $g$ is injective, we arrive at a contradiction, thus showing the claim.
• Let $g(0)=a$.
• For a successor $\alpha=\beta+1$ assume we have already defined $g(\beta)$. For sufficently small positive $h$ we have that $g(\beta)<g(\beta)+h<b$ and $\frac{f(g(\beta)+h)-f(g(\beta))}{h}\approx f_+(g(\beta))>0$. Pick one such $h$ and let $g(\alpha)=g(\beta)+h$.
• If $\alpha$ is a limit ordinal, assume $g(\beta)$ is defined for all $\beta<\alpha$. Let $x=\sup_{\beta<\alpha} g(\beta)$. A priori only $x\le b$, but we need $x<b$. Because $f$ is continuous and $f\circ g$ is strictly increasing, we conclude that $f(x)=\sup_{\beta<\alpha} f(g(\beta))\ge f(g(1))>f(g(0))=f(a)=f(b)$. Therefore $x<b$ as desired and we can let $g(\alpha)=x$.
$\square$
Corollary 1. (something like a one-sided Rolle theorem) Let $f\colon [a,b]\to\mathbb R$ be continuous with $f(a)=f(b)$. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=0$ for some $x\in[a,b)$.
Proof. Assume otherwise. Then either $f_+(x)>0$ for all $x$ or $f_+(x)<0$ for all $x$. In the first case the lemma applies and gives us a contradiction to $f(a)=f(b)$; in the other case, we consider $-f$ instead of $f$. $\square$
Corollary 2. (something like a one-sided IVT) Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_+$ exists and is continuos in $[a,b)$. Then $f'_+(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in[a,b)$.
Proof. Apply the previous corollary to $f(x)-\frac{f(b)-f(a)}{b-a}x$. $\square$
By symmetry, we have
Corollary 3. Let $f\colon [a,b]\to\mathbb R$ be continuous. Assume $f_-$ exists and is continuos in $(a,b]$. Then $f'_-(x)=\frac{f(b)-f(a)}{b-a}$ for some $x\in(a,b]$. $\square$
Theorem. Let $f\in C(\mathbb R)$ be a function with $f'_-$ continuous on $\mathbb R$. Then $f\in C^1(\mathbb R)$.
Proof. Consider aribtrary $a\in \mathbb R$. Let $\epsilon>0$ be given. Then by continuity of $f'_-$, for some $\delta>0$ we have $|f'_-(x)-f'_-(a)|<\epsilon$ for all $x\in(a,a+\delta)$. Thus for $0<h<\delta$ we have $\left|\frac{f(a+h)-f(a)}{h}-f'_-(a)\right|<\epsilon$ by corollary 3. We conclude that $f'_+(a)=f'_-(a)$, i.e. $f$ is differentiable at $a$. $\square$
• Great answer! Thanks. – xen Oct 15 '14 at 18:35
• Just one remark: shouldn't it be $f(x) = \sup_{\beta < \alpha} f(g(\beta))$? – xen Oct 15 '14 at 18:44
• And one more question: what does $\mathrm{Ord}$ means here? The class of all ordinals or the class of countable ordinals? Moreover, since I only "know" that "proper class" means that this class is not a set, how the fact that $g$ is injective and $f \circ g$ is stricly increasing contradicts $f(a) \geq f(b)$? Since $g$ is not a bijection I can not see why $f$ must be increasing. – xen Oct 15 '14 at 19:30
• @xen Thanks for hinting at the typo ($f(\beta)$ isn't even defined). I mean the class of all ordinals. The countable ordinals are a set (e.g. a subset o fthe smallest uncountable ordinal). The contradiction is that by strictly increasing, the map $f$ gives a bijection between a proper class and a set (a subset of $[a,b]$); this is already a contradiction for the first ordinal of cardinality beyond the continuum. – Hagen von Eitzen Oct 16 '14 at 6:43 | 2019-10-18T21:44:57 | {
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https://www.johndcook.com/blog/2017/09/12/making-a-problem-easier-by-making-it-harder/?utm_source=feedburner&utm_medium=feed&utm_campaign=Feed%3A+TheEndeavour+%28The+Endeavour%29 | # Making a problem easier by making it harder
In the oral exam for my PhD, my advisor asked me a question about a differential equation. I don’t recall the question, but I remember the interaction that followed.
I was stuck, and my advisor countered by saying “Let me ask you a harder question.” I was still stuck, and so he said “Let me ask you an even harder question.” Then I got it.
By “harder” he meant “more general.” He started with a concrete problem, then made it progressively more abstract until I recognized it. His follow-up questions were logically harder but psychologically easier.
This incident came to mind when I ran across an example in Lawrence Evans’ control theory course notes. He uses the example to illustrate what he calls an example of mathematical wisdom:
It is sometimes easier to solve a problem by embedding it within a larger class of problems and then solving the larger class all at once.
The problem is to evaluate the integral of the sinc function:
He does so by introducing the more general problem of evaluating the function
which reduces to the sinc integral when α = 0.
We can find the derivative of I(α) by differentiating under the integral sign and integrating by parts twice.
Therefore
As α goes to infinity, I(α) goes to zero, and so C = π/2 and I(0) = π/2.
Incidentally, note that instead of computing an integral in order to solve a differential equation as one often does, we introduced a differential equation in order to compute an integral.
## 8 thoughts on “Making a problem easier by making it harder”
1. This well known phenomenon even has a name, “the inventor’s paradox.”
2. Thanks. I didn’t know that term, but it’s a handy one.
The idea of “liberating constraints” is similar: becoming more creative by giving yourself fewer options. Maybe they’re dual concepts. The inventor’s paradox weakens your hypotheses and liberating constraints strengthen your conclusion.
3. Jonathan
There was a beautiful example of this on 538.com’s The Riddler a while back.
https://fivethirtyeight.com/features/can-you-bake-the-optimal-cake/
Working out the math for the 3-layer cake is a horrible mess of algebra, but the riddle ends with two questions, ” What percentage of the cone’s volume does the cake fill?”, and “What if he had asked for an N-layer cake?”. If you concentrate on these you get a short, neat inductive formula with just a fraction of the algebraic pain.
I particularly liked the way those questions were stuck on at the end, as if they were mere afterthoughts, but they’re actually the keys to the elegant solution,
4. It is true that making a problem more general can be an effective problem solving technique. But there’s something else going on in some of these examples.
When my advisor asked me a more general question, he was also giving me information. He knew the answer, and was steering me toward it by focusing my attention. The same with the cake problem. When someone generalizes the problem for you, they’re giving you a hint.
5. I remember a trick like this for evaluating gamma function integrals too.
btw. Looks like you dropped a sign on the second line of the I'(\alpha) calculation.
6. Peeter, Thanks. I fixed the dropped sign.
7. Regular reader of your emails, John, but this one in particular was superb. | 2017-09-25T15:48:06 | {
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https://brilliant.org/discussions/thread/telescoping-series/ | # Telescoping Series
A telescoping series is a series where each term $u_k$ can be written as $u_k = t_{k} - t_{k+1}$ for some series $t_{k}$. The benefit of such a series, is that it allows us to easily add up the terms, because $u_1 + u_2 + u_3 + \ldots + u_n = (t_1 - t_2) + (t_2 - t_3) + (t_3 - t_4 ) + \ldots + (t_n - t_{n+1} ) = t_1 - t_ {n+1}.$
Observe that most of the $t_k$ terms cancel out with their counterparts in other brackets, and hence we are only left with $t_1 - t_{n+1}$. This is comparable to a collapsible telescope, in which the long spyglass is easily retracted into a small instrument that fits into your pocket.
As you work through Arron's Telescoping Series Investigation, you would realize that for the series $u_k = \frac{1}{k(k+1)}$ and terms $t_k = \frac{ 1}{k}$, we have $u_k = t_k - t_{k+1}$ since
$\frac{ 1} { k} - \frac{ 1} { k+1} = \frac{ k+1} { k(k+1) } - \frac{ k} { k(k+1) } = \frac{ 1}{ k (k+1) } .$
As such, we have our telescoping series. This allows us to conclude that
$\sum_{i=1}^n \frac{1}{i(i+1) } = \frac{1}{1} - \frac{ 1}{n+1}.$
In particular, since $\frac{1}{n+1}$ approaches 0 as $n$ gets large, we get that
$\sum_{i=1}^\infty \frac{1}{i(i+1) } = 1.$
###### Image credit: Wikipedia
Note by Calvin Lin
6 years, 1 month ago
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You may want to update the link to Arron's Set. Here it is: https://brilliant.org/profile/arron-udsft3/sets/telescoping-series-investigation/
Also, typo: "we have $u_k=t_k-t_{k+1}$"
- 6 years, 1 month ago
Thanks! Edited :)
Staff - 6 years, 1 month ago
This method of reduction is known as Partial Fraction. Often useful in integration also.
- 5 years, 11 months ago
This is not the method of partial fractions. Please read again to clear doubts.
- 5 years, 10 months ago
Partial-Fraction Decomposition is the full name.
- 5 years, 10 months ago
$(x-4),~(x -2),~x,~(x+2),~are~one~ degree,~binomiamls~in~A.P.~~with~ difference~d=2.\\ They ~are ~n=4~terms.~~ G(x)= \dfrac 1 {(x-4)(x -2)x(x+2)}.\\ If~a~series~starts~at~S=6,~and~ends~at~E=12,~~Find~~\displaystyle~\sum_{x=S}^ E G(x).\\ Apart ~from~Partial~Fraction~there~is~another~method~given~below.\\ Name~~~L~~~as~Last-term,~L=(x+2) ,~~~F~as~First-term,~F=(x-4).~~K=L-F=6.\\ P(x)=\dfrac F K*G(x),~~~~Q(x)=\dfrac L K*G(x). ~~\scriptsize~ both~ expressions ~have~ only~(n-1)=3~terms~because ~of~ cancellation.\\ \displaystyle Using~ the~ formula~\large \color{#20A900}{\sum_{x=S}^E G(x)=\frac 1 K*\Bigg(\sum_{x=S}^{S+d-1} Q(x) - \sum_{x=E-d+1}^{E} P(x)\Bigg ).}\\ \displaystyle Substituting~given~values,~ \sum_{x=6}^{12} G(x)=\frac 1 6*\Big\{\sum_{x=6}^7 Q(x) - \sum_{x=11}^{12} P(x) \Big \}.\\ P(x)=\dfrac{x-4}{(x-4)(x-2)x(x+2)}=\dfrac 1 {(x-2)x(x+2)}.~~~Q(x)=\dfrac{x+2}{(x-4)(x-2)x(x+2)}=\dfrac 1 {(x-4)(x-2)x}.\\ \displaystyle \therefore~\frac 1 6*\sum_{x=6}^7 Q(x) =\frac 1 6*\Big \{\dfrac 1 {2*4*6}+\dfrac 1 {3*5*7} \Big \}=\dfrac{51}{18*560}...(a)\\ \displaystyle \therefore~\frac 1 6*\sum_{x=11}^{12} P(x) =\frac 1 6*\Big \{\dfrac 1 {9*11*13}+\dfrac 1 {10*12*14} \Big \}=\dfrac{\frac{989}{429}}{18*560}...(b)\\ \displaystyle \color{#D61F06}{ \sum_{x=6}^{12} \dfrac 1 {(x-4)(x -2)x(x+2)}=\dfrac{51 -\frac{989}{429}}{18*560}}=\dfrac{2089}{432432}\\ \text{Below we see how this formula is developed.}\\ G(x)=\dfrac 1 {(x-4)(x -2)x(x+ 2)}=\frac 1 {(x+2) - (x-4)}*\dfrac{ \color{#EC7300}{(x+2)} - \color{#20A900}{(x-4)} } { {\color{#20A900}{(x-4)}}(x-2)x \color{#EC7300}{(x+2)}}\\ =\frac 1 {(x+2) - (x-4)}*\Big \{{\color{#EC7300}{\dfrac 1{ (x-4)(x-2)x} }}- {\color{#20A900}{\dfrac 1{ (x-2)x (x+2)}}}\Big \}\\ \scriptsize \qquad \qquad \qquad \qquad ~Let ~{\color{#EC7300}{Q(x)=\dfrac 1{ (x-4)(x-2)x} }}, ~~~{\color{#20A900}{P(x)=\dfrac 1{ (x-2)x (x+2)}}} \\ \therefore~G(x)=\frac 1 {(x+2) - (x-4)}*\Big \{Q(x) -P(x)\Big \} \\ \implies~\displaystyle~ \sum_{x=6}^{12} G(x)=\frac 1 {(x+2) - (x-4)}*\Big \{\sum_{x=6}^{12} Q(x) - \sum_{x=6}^{12} P(x).\Big \} \\ \qquad \qquad \qquad \scriptsize \color{#20A900}{But~~\displaystyle \sum_{n=6}^{12}Q(x)=\sum_{n=6}^7 Q(x)+ \sum_{n=6}^{10} P(x)...Proof~in~Notes ~below.}\\ = \displaystyle \frac 1 {(x+2) - (x-4)}*\Big \{\sum_{x=6}^7 Q(x) +\sum_{x=6}^{10} P(x)- \sum_{x=8}^{10} P(x) - \sum_{11}^{12} P(x).\Big \}\\ \displaystyle \frac 1 {(x+2) - (x-4)}*\Big \{\sum_{x=6}^7 Q(x) - \sum_{11}^{12} P(x).\Big \}\\$
TO COMPLETE SOON
$\color{#20A900}{Note-~~\displaystyle \sum_{n=6}^{12}Q(x) \\ \displaystyle = \sum_{n=6}^{7} Q(x) + \sum_{n=8}^{12} Q(x)\\ \displaystyle =\sum_{n=6}^{7} Q(x)+ \sum_{n=8}^{12} \dfrac 1 {(x-4)(x -2)x(x+2)}.\\ \displaystyle =\sum_{n=6}^{7} Q(x)+ \sum_{n=6}^{10} \dfrac 1 {(x -2)x(x+2)(x+4)}.\\ \displaystyle = \sum_{n=6}^{7} Q(x)+ \sum_{n=6}^{10} P(x)}.$
- 3 years ago
You're right to point out that there is a better way in certain cases like this. Partial fractions allow us to quickly reduce to a "known problem", instead of having to "Creatively determine the telescoping terms $t_k$".
Note: You're missing a factor of $\frac{1}{7}$.
Staff - 3 years ago
Thanks for your comment. I have added $\frac 1 7$.
I have changed problem so it is made more clear. Avoid two 4 in (x-4) and (x+4). Also have an upper limit so can show its working. I will add more explanation soon.
- 3 years ago | 2020-05-26T21:45:58 | {
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https://math.stackexchange.com/questions/3196618/inequality-w-r-t-determinant-of-a-non-negative-definite-matrix | # Inequality w.r.t determinant of a non-negative definite matrix.
I am reading a paper where the author mentioned the following property without proof. Neither can I prove it nor can I find the proof in various textbooks.
For any non-negative definite (i.e. positive semidefinite) matrix $$A$$, $$\text{det}(A) \leq \Pi_iA_{ii}$$
How to prove it?
First, there is the fact that for any $$n\times n$$ matrix $$P$$, if one writes $$P=[v_1, v_2, ..., v_n]$$, where $$v_k$$ is the $$k$$-th column of $$P$$, then $$|\text{det}P|\leq |v_1|\cdot|v_2|\cdot ...\cdot|v_n|;$$ here $$|v_k|$$ is the length of the vector $$v_k$$. This actually reaches the equal sign if $$v_1, ...,v_n$$ are mutually perpendicular; in that case $$[v_1/|v_1|, ..., v_n/|v_n|]$$ is an orthogonal matrix, thus has determinant $$\pm 1$$, on the other hand its determinant is also $$\text{det} P/(|v_1|\cdot ...\cdot |v_n|)$$. For the general case, we do a procedure similar to Gram-Schmidt: replace $$v_2$$ by $$v_2'$$, which is $$v_2$$ minus its projection to $$v_1$$; that will reduce $$|v_2|$$ but not changing $$\text{det}P$$. Replace $$v_3$$ by $$v_3'$$, which is $$v_3$$ minus its projection to the $$v_1, v_2$$ plane, ... and in the end get $$[v_1, v_2', v_3'...]$$ with orthogonal columns. Thus $$|\text{det} P|= |v_1|\cdot |v_2'|\cdot..\cdot|v_n'|\leq |v_1|\cdot...\cdot|v_n|$$.
Now $$A$$ is nonnegative, it is "well-known" that we can find orthogonal matrix $$Q$$ so that $$A=Q^T\Lambda Q$$, where $$\Lambda$$ is diagonal with $$\geq 0$$ entries. Write $$\sqrt \Lambda$$ be the diagonal matrix with $$(\sqrt\Lambda)^2=\Lambda$$, thus $$A=Q^T\sqrt\Lambda\sqrt\Lambda Q=(Q^T\sqrt\Lambda Q)(Q^T\sqrt\Lambda Q)$$. in conclusion, we can write $$A=P^TP$$, with $$P=Q^T\sqrt\Lambda Q$$.
Let $$P=[v_1, ...,v_n]$$. Then from $$A=P^TP$$ we see $$a_{11}=|v_1|^2$$, ... , $$a_{nn}=|v_n|^2$$. Thus $$\text{det}A= (\text{det} P)^2\leq (|v_1|\cdot...\cdot|v_n|)^2=|v_1|^2\cdot...\cdot|v_n|^2=a_{11}\cdot ...\cdot a_{nn}.$$
• Thanks, your proof is amazing. The first fact is also known as Hadamard's inequality. If you assign $P = \sqrt{\Lambda}Q$, then the expression $A = P^TP$ still holds true, and it's simpler. – Ly Minh Hoang Apr 22 at 4:56 | 2019-05-23T15:08:36 | {
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http://blog.travelkarnataka.net/362ia/530c0b-curl-formula-2d | In 2D, the dual to a bivector is a scalar. If you’ve seen a current sketch giving the direction and magnitude of a flow of a fluid or the direction and magnitude of the winds then you’ve seen a … Again, we let and compute Not surprisingly, the curl is a vector quantity. The curl of F is the new vector field This can be remembered by writing the curl as a "determinant" Theorem: Let F be a three dimensional differentiable vector field with continuous partial derivatives. 2. and N = x, so curl F = 1 − 2x y3. (Or a two-form; I'm not sure which. Its gradient $$\nabla f(x,y,z)$$ is a vector field. Notes. A curl is always the same type of beast in any number of dimensions. Divergence and Curl calculator. Then Curl F = 0, if and only if F is conservative. Next: Physical Interpretation of the Up: The Curl of a Previous: The Curl of a The Curl in Cartesian Coordinates. 3 2. i + x j then M = x y3. The remaining answer is: - The term in parenthesis is the curl of a vector function, which is also a vector. - The gradient of a scalar function is a vector. Discover Resources. Let $$f(x,y,z)$$ be a (scalar-valued) function, and assume that $$f(x,y,z)$$ is infinitely differentiable. And the value of that third component will be exactly the 2D curl. Just “plug and chug,” as they say. $\endgroup$ – Stephen Montgomery-Smith Mar 5 '15 at 17:46 It's neither a vector nor a scalar; it's a bivector. Given a vector field F(x, y, z) = Pi + Qj + Rk in space. This article uses the standard notation ISO 80000-2, which supersedes ISO 31-11, for spherical coordinates (other sources may reverse the definitions of θ and φ): . Notice that F(x, y) is a vector valued function and its curl … That may not make a lot of sense, but most people do know what a vector field is, or at least they’ve seen a sketch of a vector field. An alternative notation for divergence and curl may be easier to memorize than these formulas by themselves. It also will generally be a (vector valued) function. By definition, if F = (M, N) then the two dimensional curl of F is curl F = N x − M y Example: If F = x y. Then the 3D curl will have only one non-zero component, which will be parallel to the third axis. You'll see fancier equations for curl where the surface shrinks to zero (such as in wikipedia), but recognize the basic intuition -- curl is … The point is that it's an intrinsically two-dimensional object.) The polar angle is denoted by θ: it is the angle between the z-axis and the radial vector connecting the origin to the point in question. The Curl The curl of a vector function is the vector product of the del operator with a vector function: where i,j,k are unit vectors in the x, y, z directions. On the other hand, we can also compute the curl in Cartesian coordinates. In 3D, the dual to a bivector is a vector. Taking the divergence of the term in parenthesis, we get the divergence of a vector, which is a scalar. It can also be expressed in determinant form: Curl in cylindrical and sphericalcoordinate systems Practice; Classic Net of Cuboid; Definite Integral Illustrator (I) Thus, the curl of the term in parenthesis is also a vector. Using these facts, we can create the formula for curl: Where (S) is the surface we are considering; the direction of the curl is the normal to the surface. Given these formulas, there isn't a whole lot to computing the divergence and curl. What is the curl of the gradient? Two Dimensional Curl We have learned about the curl for two dimensional vector fields. Can you come up with a formula for a two-dimensional vector field $$\vec r(xy)$$ with constant nonzero curl at every point? So in that sense, the 2D curl could be considered to be precisely the same as the 3D curl. | 2022-06-29T03:19:46 | {
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http://linear.ups.edu/fcla/section-B.html | A basis of a vector space is one of the most useful concepts in linear algebra. It often provides a concise, finite description of an infinite vector space.
We now have all the tools in place to define a basis of a vector space.
##### DefinitionBBasis
Suppose $V$ is a vector space. Then a subset $S\subseteq V$ is a basis of $V$ if it is linearly independent and spans $V\text{.}$
So, a basis is a linearly independent spanning set for a vector space. The requirement that the set spans $V$ insures that $S$ has enough raw material to build $V\text{,}$ while the linear independence requirement insures that we do not have any more raw material than we need. As we shall see soon in Section D, a basis is a minimal spanning set.
You may have noticed that we used the term basis for some of the titles of previous theorems (e.g. Theorem BNS, Theorem BCS, Theorem BRS) and if you review each of these theorems you will see that their conclusions provide linearly independent spanning sets for sets that we now recognize as subspaces of $\complex{m}\text{.}$ Examples associated with these theorems include Example NSLIL, Example CSOCD and Example IAS. As we will see, these three theorems will continue to be powerful tools, even in the setting of more general vector spaces.
Furthermore, the archetypes contain an abundance of bases. For each coefficient matrix of a system of equations, and for each archetype defined simply as a matrix, there is a basis for the null space, three bases for the column space, and a basis for the row space. For this reason, our subsequent examples will concentrate on bases for vector spaces other than $\complex{m}\text{.}$
Notice that Definition B does not preclude a vector space from having many bases, and this is the case, as hinted above by the statement that the archetypes contain three bases for the column space of a matrix. More generally, we can grab any basis for a vector space, multiply any one basis vector by a nonzero scalar and create a slightly different set that is still a basis. For “important” vector spaces, it will be convenient to have a collection of “nice” bases. When a vector space has a single particularly nice basis, it is sometimes called the standard basis though there is nothing precise enough about this term to allow us to define it formally — it is a question of style. Here are some nice bases for important vector spaces.
##### Proof
The bases described above will often be convenient ones to work with. However a basis does not have to obviously look like a basis.
We have seen that several of the sets associated with a matrix are subspaces of vector spaces of column vectors. Specifically these are the null space (Theorem NSMS), column space (Theorem CSMS), row space (Theorem RSMS) and left null space (Theorem LNSMS). As subspaces they are vector spaces (Definition S) and it is natural to ask about bases for these vector spaces. Theorem BNS, Theorem BCS, Theorem BRS each have conclusions that provide linearly independent spanning sets for (respectively) the null space, column space, and row space. Notice that each of these theorems contains the word “basis” in its title, even though we did not know the precise meaning of the word at the time. To find a basis for a left null space we can use the definition of this subspace as a null space (Definition LNS) and apply Theorem BNS. Or Theorem FS tells us that the left null space can be expressed as a row space and we can then use Theorem BRS.
Theorem BS is another early result that provides a linearly independent spanning set (i.e. a basis) as its conclusion. If a vector space of column vectors can be expressed as a span of a set of column vectors, then Theorem BS can be employed in a straightforward manner to quickly yield a basis.
# SubsectionBSCVBases for Spans of Column Vectors¶ permalink
We have seen several examples of bases in different vector spaces. In this subsection, and the next (Subsection B.BNM), we will consider building bases for $\complex{m}$ and its subspaces.
Suppose we have a subspace of $\complex{m}$ that is expressed as the span of a set of vectors, $S\text{,}$ and $S$ is not necessarily linearly independent, or perhaps not very attractive. Theorem REMRS says that row-equivalent matrices have identical row spaces, while Theorem BRS says the nonzero rows of a matrix in reduced row-echelon form are a basis for the row space. These theorems together give us a great computational tool for quickly finding a basis for a subspace that is expressed originally as a span.
Example IAS provides another example of this flavor, though now we can notice that $X$ is a subspace, and that the resulting set of three vectors is a basis. This is such a powerful technique that we should do one more example.
Click to open
# SubsectionBNMBases and Nonsingular Matrices¶ permalink
A quick source of diverse bases for $\complex{m}$ is the set of columns of a nonsingular matrix.
##### Proof
Perhaps we should view the fact that the standard unit vectors are a basis (Theorem SUVB) as just a simple corollary of Theorem CNMB? (See Proof Technique LC.)
With a new equivalence for a nonsingular matrix, we can update our list of equivalences.
Click to open
# SubsectionOBCOrthonormal Bases and Coordinates¶ permalink
We learned about orthogonal sets of vectors in $\complex{m}$ back in Section O, and we also learned that orthogonal sets are automatically linearly independent (Theorem OSLI). When an orthogonal set also spans a subspace of $\complex{m}\text{,}$ then the set is a basis. And when the set is orthonormal, then the set is an incredibly nice basis. We will back up this claim with a theorem, but first consider how you might manufacture such a set.
Suppose that $W$ is a subspace of $\complex{m}$ with basis $B\text{.}$ Then $B$ spans $W$ and is a linearly independent set of nonzero vectors. We can apply the Gram-Schmidt Procedure (Theorem GSP) and obtain a linearly independent set $T$ such that $\spn{T}=\spn{B}=W$ and $T$ is orthogonal. In other words, $T$ is a basis for $W\text{,}$ and is an orthogonal set. By scaling each vector of $T$ to norm 1, we can convert $T$ into an orthonormal set, without destroying the properties that make it a basis of $W\text{.}$ In short, we can convert any basis into an orthonormal basis. Example GSTV, followed by Example ONTV, illustrates this process.
Unitary matrices (Definition UM) are another good source of orthonormal bases (and vice versa). Suppose that $Q$ is a unitary matrix of size $n\text{.}$ Then the $n$ columns of $Q$ form an orthonormal set (Theorem CUMOS) that is therefore linearly independent (Theorem OSLI). Since $Q$ is invertible (Theorem UMI), we know $Q$ is nonsingular (Theorem NI), and then the columns of $Q$ span $\complex{n}$ (Theorem CSNM). So the columns of a unitary matrix of size $n$ are an orthonormal basis for $\complex{n}\text{.}$
Why all the fuss about orthonormal bases? Theorem VRRB told us that any vector in a vector space could be written, uniquely, as a linear combination of basis vectors. For an orthonormal basis, finding the scalars for this linear combination is extremely easy, and this is the content of the next theorem. Furthermore, with vectors written this way (as linear combinations of the elements of an orthonormal set) certain computations and analysis become much easier. Here is the promised theorem.
##### Proof
A slightly less intimidating example follows, in three dimensions and with just real numbers.
Not only do the columns of a unitary matrix form an orthonormal basis, but there is a deeper connection between orthonormal bases and unitary matrices. Informally, the next theorem says that if we transform each vector of an orthonormal basis by multiplying it by a unitary matrix, then the resulting set will be another orthonormal basis. And more remarkably, any matrix with this property must be unitary! As an equivalence (Proof Technique E) we could take this as our defining property of a unitary matrix, though it might not have the same utility as Definition UM.
##### 1
The matrix below is nonsingular. What can you now say about its columns? \begin{equation*} A=\begin{bmatrix} -3 & 0 & 1\\ 1 & 2 & 1\\ 5 & 1 & 6 \end{bmatrix} \end{equation*}
##### 2
Write the vector $\vect{w}=\colvector{6\\6\\15}$ as a linear combination of the columns of the matrix $A$ above. How many ways are there to answer this question?
##### 3
Why is an orthonormal basis desirable?
# SubsectionExercises
##### C10
Find a basis for $\spn{S}\text{,}$ where \begin{align*} S &= \set{ \colvector{1\\3\\2\\1}, \colvector{1\\2\\1\\1}, \colvector{1\\1\\0\\1}, \colvector{1\\2\\2\\1}, \colvector{3\\4\\1\\3} }\text{.} \end{align*}
Solution
##### C11
Find a basis for the subspace $W$ of $\complex{4}\text{,}$ \begin{align*} W &= \setparts{\colvector{a + b - 2c\\a + b - 2c + d\\ -2a + 2b + 4c - d\\ b + d}} {a, b, c, d \in\complexes}\text{.} \end{align*}
Solution
##### C12
Find a basis for the vector space $T$ of lower triangular $3 \times 3$ matrices; that is, matrices of the form \begin{align*} \begin{bmatrix} * & 0 & 0\\ * & * & 0\\ * & * & *\end{bmatrix} \end{align*} where an asterisk represents any complex number.
Solution
##### C13
Find a basis for the subspace $Q$ of $P_2\text{,}$ $Q = \setparts{p(x) = a + bx + cx^2}{p(0) = 0}\text{.}$
Solution
##### C14
Find a basis for the subspace $R$ of $P_2\text{,}$ $R = \setparts{p(x) = a + bx + cx^2}{p'(0) = 0}\text{,}$ where $p'$ denotes the derivative.
Solution
##### C40
From Example RSB, form an arbitrary (and nontrivial) linear combination of the four vectors in the original spanning set for $W\text{.}$ So the result of this computation is of course an element of $W\text{.}$ As such, this vector should be a linear combination of the basis vectors in $B\text{.}$ Find the (unique) scalars that provide this linear combination. Repeat with another linear combination of the original four vectors.
Solution
##### C80
Prove that $\set{(1,\,2),\,(2,\,3)}$ is a basis for the crazy vector space $C$ (Example CVS).
##### M20
In Example BM provide the verifications (linear independence and spanning) to show that $B$ is a basis of $M_{mn}\text{.}$
Solution
##### T50
Theorem UMCOB says that unitary matrices are characterized as those matrices that “carry” orthonormal bases to orthonormal bases. This problem asks you to prove a similar result: nonsingular matrices are characterized as those matrices that “carry” bases to bases.
More precisely, suppose that $A$ is a square matrix of size $n$ and $B=\set{\vectorlist{x}{n}}$ is a basis of $\complex{n}\text{.}$ Prove that $A$ is nonsingular if and only if $C=\set{A\vect{x}_1,\,A\vect{x}_2,\,A\vect{x}_3,\,\dots,\,A\vect{x}_n}$ is a basis of $\complex{n}\text{.}$ (See also Exercise PD.T33, Exercise MR.T20.)
Solution
##### T51
Use the result of Exercise B.T50 to build a very concise proof of Theorem CNMB. (Hint: make a judicious choice for the basis $B\text{.}$)
Solution | 2020-05-28T07:41:00 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=0f6jdnc8eanf2pk5uf3ta0o7l0&action=printpage;topic=1292.0 | # Toronto Math Forum
## MAT334-2018F => MAT334--Lectures & Home Assignments => Topic started by: Nikita Dua on October 01, 2018, 08:02:20 PM
Title: Section 1.4 Question 1
Post by: Nikita Dua on October 01, 2018, 08:02:20 PM
lim n -> infinity zn = ((1 +i)/(sqrt(3))n.
I used polar coordinates to solving giving r = sqrt(2/3)
xn =sqrt(2/3) cos(n * pi /4)
yn =sqrt(2/3) sin(n * pi /4)
zn = sqrt(2/3) cos(n * pi /4) + isqrt(2/3) sin(n * pi /4)
From this how can show that it converges to 0?
Title: Re: Section 1.4 Question 1
Post by: oighea on October 02, 2018, 01:27:44 AM
This sequence converges to 0 because the absolute value converges to 0. Note ${\sqrt{\frac{2}{3}}} < 1$, so powers of ${\sqrt{\frac{2}{3}}}^n$
Proof:
The Arg of $\frac{(1+i)}{\sqrt{3}}$ is $\frac{\pi}{4}$. That can be verified as $(1+i)$ "points northeast", and the $\sqrt{3}$ denominator is irrelevant to the Arg.
The magnitude $|\frac{(1+i)}{\sqrt{3}}|$ is ${\sqrt{\frac{2}{3}}}$. That can be verified as $|1+i|$ = $\sqrt2$, and $\frac{\sqrt2}{\sqrt3}$ = ${\sqrt{\frac{2}{3}}}$.
Therefore, $\frac{(1+i)}{\sqrt{3}} = {\sqrt{\frac{2}{3}}}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})$
And by DeMoivre's law, $[\frac{(1+i)}{\sqrt{3}}]^n$ = ${\sqrt{\frac{2}{3}}}^n(\cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4})$
Intuitively magnitude of $[\frac{(1+i)}{\sqrt{3}}]^n$ can only spiral down counterclockwise as $n$ increases, and eventually approaches 0.
Title: Re: Section 1.4 Question 1
Post by: Victor Ivrii on October 02, 2018, 09:54:22 AM
In this problem argument is irrelevant, only modulus matters.
Title: Re: Section 1.4 Question 1
Post by: Min Gyu Woo on October 02, 2018, 02:27:13 PM
"This sequence converges to 0 because the absolute value converges to 0."
Where did you get that information?
The textbook states on pg 34 that,
If $z_n \rightarrow A$ then $|z_n| \rightarrow |A|$.
Does the converse work as well?
Title: Re: Section 1.4 Question 1
Post by: Victor Ivrii on October 02, 2018, 02:40:42 PM
Look at the definition of $z_n\to A$ . | 2022-12-03T12:55:00 | {
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https://electronics.stackexchange.com/questions/373326/phase-difference-between-two-signals-problem | # Phase Difference Between Two Signals (Problem)
I am asked to calculate the phase difference between the two signals below.
$$i_1 = -4\sin(377t + 55^\circ) \hspace{0.2cm}\quad \mathrm{and}\quad \hspace{0.2cm} i_2 = 5\cos(377t - 65^\circ)$$
What I did was to convert $$i_1 = -4\sin(377t + 55^\circ)$$ to $$i_1 = 4\cos(377t + 145^\circ)$$
Then find the difference by $$145 - (-65)=210$$ Then we conclude $$i_1\hspace{0.2cm} leads \hspace{0.2cm} i_2 \hspace{0.2cm}by \hspace{0.2cm}210^\circ$$ But, let's add 360 to (since the result won't change) $$i_2 = 5\cos(377t - 65^\circ)$$ We get $$i_2 = 5\cos(377t + 295^\circ)$$
Now comparing the two we get difference to be 150 degrees,which means $$i_2\hspace{0.2cm} leads \hspace{0.2cm} i_1 \hspace{0.2cm}by \hspace{0.2cm}150^\circ$$
My questions are that,
• What is the difference between these two answers? Is the latter wrong?
• How could we rewrite the second answer in terms of i1 leading i2?
• It is not a coincidence that 210 + 150 = 360 – evildemonic May 9 '18 at 14:30
Both phase differences are correct and, in fact, indistinguishable. There is an infinite number of ways writing the same sine wave, just by adding any integer multiple of 2*pi (or 360 degrees) to the phase.
The problem is with trying to apply the concept of "leading" and "lagging" to sine waves. Strictly speaking a sine wave has infinite length, i.e. it has no start and no end. They have been going forever, so it's pointless to ask "which one comes first". Every peak of the first sine wave is preceded by a peak of the second one, which in turn is again preceded by another peak of the first one and so forth ad infinitum.
In practice, if the phase difference is reasonably small (say 90 degrees or less), you could call the one with the larger phase "leading" because the peaks are relatively close together and the one with the larger phase visually appears to come first. However with larger phase differences and especially 180 degrees that makes no sense any more.
Both of the results you got are right. You feel it confusing, because you imagine it as a running race, where one has to be first and the other the second, but in fact these are periodic signals. Which means that the leading is always relative.
Try to sketch the signals and select a local maximum of i1. The i1 leads i2 by the distance between your chosen i1 maximum and the closest i2 maximum to the left. But there is another i2 maximum to the right from the chosen i1 maximum, so i2 leads i1 by that distance. Its always true that if you add these two phase differences, you always get 360 degrees, what you can use to check your answer
Usually when you specify leading vs lagging phase difference you limit yourself to a "normalized" phase angle of ±180°.
More specifically, to be a mathematically nitpicker, any phase angle should be reduced to the normalized interval
$$(-180°,180°]$$
so that leading vs. lagging would be uniquely defined. | 2019-08-24T15:51:22 | {
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https://math.stackexchange.com/questions/2879022/discriminant-in-the-context-of-pde-classification-b2-4ac-or-b2-ac | # Discriminant (in the context of PDE classification): $b^2 - 4ac$ or $b^2 - ac$?
I'm reading two textbooks on partial differential equations. In their respective sections on classification of PDEs (hyperbolic, parabolic, elliptical), they differ in what they describe as being the discriminant. One textbook says that the discriminant is $b^2 - 4ac$, while the other describes it being $b^2 - ac$. Are these both correct, or is one correct and the other incorrect?
EDIT: What are the $a$, $b$, and $c$? The second-order linear PDE in two independent variables is $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants.
• What are $a$, $b$ and $c$? – Angina Seng Aug 11 '18 at 3:45
• @LordSharktheUnknown Edited with clarification. – user423167 Aug 11 '18 at 3:54
• Still no $a$, $b$ and $c$: the discriminant is $B^2-4AC$ here. – Angina Seng Aug 11 '18 at 7:11
Some authors/texts like to write the middle term as $2Bu_{xy}$ as opposed to $Bu_{xy}.$ In that case, it is $B^2-AC$ and in your case, it should be $B^2-4AC.$
In short, either can be correct, depending on how the original equation is expressed.
Assuming that the mixed partials of $u$ are equal, we can write your differential equation in operator form: $$\left( A\partial_x^2+B\partial_x\partial_y+C\partial_y^2+D\partial_x+E\partial_y+F \right)u = 0.$$ The parenthesized operator is reminiscent of the left-hand side of the general conic equation $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$, which can be written in matrix form as $$\begin{bmatrix}x&y\end{bmatrix} \begin{bmatrix} A & \frac B2 \\ \frac B2 & C \end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} + \begin{bmatrix}D&E\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}+F = 0.$$ The discriminant of this equation, which indicates the type of conic that the equation represents, is the negative determinant of the $2\times2$ matrix of the quadratic part of this equation, i.e., $B^2/4-AC$. This expression conventionally gets multiplied by $4$ for convenience in writing, if nothing else: $B^2-4AC$.† On the other hand, it can also be convenient not to have those stray factors of two floating around. They can be eliminated by using $2B$ instead of $B$ as the coefficient of the mixed term, with corresponding matrix $$\begin{bmatrix}A&B\\B&C\end{bmatrix},$$ in which case the discriminant becomes $B^2-AC$. If you remember that it’s related to the determinant of the quadratic part of the equation, you’ll get the right version.
† In fact, multiplying by $4$ gives you the negative determinant of the Hessian of the left-hand side of the conic equation, which is another way to arrive at the discriminant.
• Nice!!, actually, I am curious to know how the above PDE corresponds to the general conic equation, well it's intuitive but any rigorous treatment, any references? – BAYMAX Aug 11 '18 at 4:45
• @BAYMAX There’s a brief gloss in this Wikipedia article. – amd Aug 11 '18 at 6:50
A few points -
*) If we are discussing the second order linear PDE of two independent variables, which say is of the form - $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants, then we must be careful about the coefficients of the PDE, they are $A,B,C,D,E,F,G$ and not $a,b,c,d,e,f,g$, (very simple yet much important).
*) If we are considering the PDE $Au_{xx} + Bu_{xy} + Cu_{yy} + Du_x + Eu_y + Fu = G$, where $A$, $B$, $C$, $D$, $E$, $F$, $G$ are functions of $x$ and $y$ and could be constants, then their classification is as follows -
We check the Discriminant which is $d = B^2(x_{0},y_{0}) - 4 A(x_{0},y_{0})C(x_{0},y_{0})$.
At $(x_{0},y_{0})$, the equation is said to be -
1) Elliptic if $d <0$
2) Parabolic if $d = 0$
3) Hyperbolic if $d>0$
If this is true for all points $(x_{0},y_{0}) \in$ domain $\Omega$, then the equation is said to be Elliptic, Parabolic or Hyperbolic in that domain.
Extra point -
Now if there is the case of $n$ independent variables like $x_{1},x_{2},...,x_{n}$and second order linear PDE (of the form $\sum \sum a_{i,j} u_{x_{i},x_{j}}+$ lower order terms $= 0$ then the classification depends on the signature of the eigenvalues of the coefficient matrix.
*) Elliptic if the eigenvalues are all positive or all negative.
*) Parabolic - The eigenvalues are all positive or all negative, save one which is zero
*) Hyperbolic - There is only one negative eigenvalue and all the rest are positive, or there is only one positive eigenvalue and all the rest are negative.
Few references similar to this question -
Perhaps by checking the second reference above and the answers to your questions above you will know the difference of usage of $B^2 -4AC$ and $B^2 - AC$! | 2021-01-18T10:29:16 | {
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https://math.stackexchange.com/questions/2493037/probability-of-rolling-a-double-6-with-two-dice | # Probability of rolling a double $6$ with two dice
Two dice (with numbers 1 to 6 on the faces) are rolled.
One die rolls a 6.
What is the probability of rolling a double 6?
One solution is to say that P(2 sixes) = $\frac{1}{6}$ since the first die gives a 6, so the only way to get a double six is by rolling a six on the other die (which has a 1 in 6 chance).
Another solution is to say that there are 11 possible combinations if one die rolls a six i.e. (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2) and (6, 1). So the probability of rolling a double six if one six has already been rolled is $\frac{1}{11}$.
Which answer is correct and why?
• The second one is correct. The exercise says that one rolls a 6. This roll can be the first one or the second one. Given that someone roll $a$ six, what is the probability to roll a double 6? – callculus Oct 27 '17 at 22:39
• This is related to the Two Child Problem or Boy-or-Girl Paradox en.wikipedia.org/wiki/Boy_or_Girl_paradox – Jay Oct 28 '17 at 1:32
There is confusion between two questions.
1. You have rolled a six. Now for a double-six, you need to get a 6 on the second roll. The second roll is independent of the first. So the probability of a 6 again on the second roll is $1/6.$
2. If both dice have already been rolled out of your sight, and you are told that there is at least one 6, then conditional on that information, what is the probability that the dice actually show a double-6. Then the analysis of @Shayne2020 (+1) leading to the answer $1/11$ is correct.
• Dear Teacher, Dear Professor, I'm so sorry for this comment..I asked mathoverflow for help. But the question was downvoted. I had to delete the question. I really need help. But unfortunately, no one helped. If I ask you for help, can you take a look? If you have a few minutes. Then, I will delete this comment..Thank you very much.. mathoverflow.net/q/298997/123863 – Mathematics is Life Apr 29 '18 at 17:40
• @MathematicsIsLife: This seems to be an entirely different question. If you could explain what you are doing and summarize your data, someone might be able to help. For example, your first vector $P_1$ when multiplied by 42 can be summarized as: values 0, 1, 2, 3, 4, 5 with respective frequencies 25, 4, 5, 5, 2, 1 . – BruceET Apr 29 '18 at 18:01
• Teacher, the sum of all values are equal to $1$. For any $P(x).$ Because, this is probabilistic distribution...(English is my second language..Sorry for wrong words) – Mathematics is Life Apr 29 '18 at 18:11
• Teacher, I can not do anything because I do not have a computer and I can not use mathematical softwares..:( – Mathematics is Life Apr 29 '18 at 18:42
This can be phrased as
Given that at least one die rolls a 6, what is the probability of rolling a double 6?
We can use the conditional probability formula: $$P(A|B) = \frac{P(A\ \mathrm{and}\ B)}{P(B)}$$ This means "the probability of event A given event B, is the probability of A and B divided by the probability of B".
\begin{align} P(\mathrm{double\ 6}|\mathrm{at\ least\ one\ 6}) &= \frac{P(\mathrm{double\ 6}\ \mathrm{and}\ \mathrm{at\ least\ one\ 6})}{P(\mathrm{at\ least\ one\ 6})}\\ &= \frac{P(\mathrm{double\ 6})}{P(\mathrm{at\ least\ one\ 6})}\\ &= \frac{1/36}{11/36}\\ &= \frac{1}{11} \end{align}
One die rolls a 6.
This doesn't clarify which one. Was it the first die or the second one? (Note that the dice are distinguishable by the turn of their throws.)
So, the latter logic is correct. The conditional probability reduces the sample space $S$ into ${(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6), (6, 5), (6, 4), (6, 3), (6, 2) (6, 1)}$
Now, $|S|= 11$
And the only favorable outcome (rolling two 6's ) is $(6,6)$.
Thus, the probability is $\frac{1}{11}$
The odds are $$1$$ in $$6$$ in both cases (seeing one die vs not seeing the dice) because the arguments given for all the $$11$$ permutations neglect to address doubles for the $$5$$ (which equal $$5$$ combinations) that are not $$(6,6)$$. Independently, if one die is known to be six, there is a $$1$$ in $$6$$ chance of the other being six.
The answer is $$1$$ in $$6$$. | 2020-12-03T23:32:52 | {
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