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1969615
|
Evaluate [imath]\sum^{n-1}_{k=1} {n\choose k}\frac{kn^{n-k}}{k+1}[/imath]
How can we evaluate: [imath]\sum^{n-1}_{k=1} {n\choose k}\frac{kn^{n-k}}{k+1}[/imath] I observe that the expression has a similar format with the following formula: [imath]\sum^{n}_{j=0} {n\choose j}\frac{u^j}{j+1}=\frac{1}{n+1}\sum^n_{k=0}(u+1)^k[/imath] but I have no idea how to process further. A blind evaluation using Wolfram Alpha gives us [imath]\frac{n(n^n-1)}{n+1}[/imath]
|
1968896
|
Let [imath]n \in \mathbb{Z}^+[/imath], prove the identity [imath] \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}[/imath]
Let [imath]n \in \mathbb{Z}^+[/imath], prove the identity [imath] \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}[/imath] First of all [imath] \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)[/imath] [imath]=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)[/imath] [imath]=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k} \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)[/imath] We have for the first sum [imath](1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.[/imath] For the second sum [imath](1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.[/imath] Integrating both sides from [imath]0[/imath] to [imath]x[/imath], we see that [imath]\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.[/imath] Putting [imath]x=1[/imath] yields [imath]\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}[/imath] Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
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1969803
|
Prove that if [imath]X[/imath] is and infinite set and [imath]Y[/imath] is countable or finite, then [imath]|X \cup Y| = |X|[/imath]
Let [imath]X[/imath] be an infinite set and [imath]Y[/imath] a finite or countable set. Prove that [imath]|X \cup Y| = |X|[/imath]. The form to prove this is by finding a bijection between both sets [imath]X \cup Y[/imath] and [imath]X[/imath]. I got a hint to use the property that every infinite set has a finite or countable subset. My problem is how to define this bijection. Any ideas?
|
1967678
|
If A is an infinite set and B is at most countable set, prove that A and [imath]A \cup B[/imath] have the same cardinality
I am trying to prove: If A is an infinite set and B is at most countable set, prove that A and [imath]A \cup B[/imath] have the same cardinality. Set definitions here are from baby Rudin. Attempt: There are 4 possible cases: 1.A is countable and B is finite. 2.A is countable and B is countable. 3.A is uncountable and B is finite. 4.A is uncountable and B is countable. Then for each case I need to show that [imath]A \cup B \sim A[/imath]. I can prove 1-3, but I am unsure how to prove case 4. And in general, it seems that my proof strategy is inefficient and there exist much faster proof then proof by cases.
|
1969954
|
If [imath]X=f^{-1}(Y)[/imath], is it true that [imath]f(X)=Y[/imath]?
Let [imath]f:S\to T[/imath]. Let [imath]S\subset X[/imath] and [imath]Y\subset T[/imath]. If [imath]X=f^{-1}(Y)[/imath], is it true that [imath]f(X)\subset Y[/imath]? Is it true that [imath]f(X)=Y[/imath]? The first claim is true. Pick some [imath]y_0\in f(X)[/imath]. Then there exists some [imath]x_0\in X[/imath] such that [imath]f(x_0)=y_0[/imath]. By definition of inverse image, [imath]X = \{s\in S: f(x)\in Y\}[/imath]. Thus, we have [imath]f(x_0)\in Y[/imath]. That is, [imath]y_0\in Y[/imath]. This implies that [imath]f(X)\subset Y[/imath]. But I'm not sure about the second claim?
|
914957
|
What is [imath]f(f^{-1}(A))[/imath]?
Suppose that [imath]f : E \rightarrow F[/imath]. What is [imath]f(f^{-1}(A))[/imath]? Is it always [imath]A[/imath]? [imath]f^{-1}[/imath] is the inverse function. This is not a homework, I'm confused by this statement.
|
1970087
|
Quick bijection between [imath]\mathbb{Q}\cap (a,\ b)[/imath] and [imath]\mathbb{Q}[/imath]
I need a really quick way of showing there's a bijection from [imath]\mathbb{Q}\cap(a,\ b)[/imath] to [imath]\mathbb{Q}[/imath] for any real numbers [imath]a < b[/imath]. I attempted a few ways but I'm drawing a blank right now .Nothing I've worked on is fruitful (it either goes nowhere or is much too complicated) so I'm omitting it from the question. Any simple ideas? Mainly I'm looking for something that can be rigorously justified and explained in a matter of no more than three to four lines. Clarification: I don't need to construct a bijection, I just need to show that there is one. Clarification 2: The context I'm working in doesn't have a definition of "countable", so I can't just say both sets are countable unfortunately. Clarification 3: We know that there exists a bijection from [imath]\mathbb{Q}[/imath] to [imath]\mathbb{N}[/imath]. Our construction was essentially that you can list all elements of [imath]\mathbb{Q}[/imath] in a grid and spiral outwards from the origin, ignoring duplicates, and assigning the next natural to the next unique rational in the spiral path.
|
17568
|
Is there a bijection between [imath](0,1)[/imath] and [imath]\mathbb{R}[/imath] that preserves rationality?
While reading about cardinality, I've seen a few examples of bijections from the open unit interval [imath](0,1)[/imath] to [imath]\mathbb{R}[/imath], one example being the function defined by [imath]f(x)=\tan\pi(2x-1)/2[/imath]. Another geometric example is found by bending the unit interval into a semicircle with center [imath]P[/imath], and mapping a point to its projection from [imath]P[/imath] onto the real line. My question is, is there a bijection between the open unit interval [imath](0,1)[/imath] and [imath]\mathbb{R}[/imath] such that rationals are mapped to rationals and irrationals are mapped to irrationals? I played around with mappings similar to [imath]x\mapsto 1/x[/imath], but found that this never really had the right range, and using google didn't yield any examples, at least none which I could find. Any examples would be most appreciated, thanks!
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1969368
|
[imath]{f_n}[/imath] : seq of measurable functions with common domain E, show that each of the following functions is measurable: inf{[imath]f_n[/imath]},sup{[imath]f_n[/imath]}
For a sequence [imath]{f_n}[/imath] of measurable functions with common domain E, show that each of the following functions is measurable: inf{[imath]f_n[/imath]},sup{[imath]f_n[/imath]},lim inf{[imath]f_n[/imath]},lim sup{[imath]f_n[/imath]} At first, I want to know what inf{[imath]f_n[/imath]} means. If {[imath]f_n[/imath]} is increasing function, then I understand {[imath]f_1[/imath]} is the infimum of sequence of function {[imath]f_n[/imath]}. But what if it doesn't? If two candidates have a crossed point, then what is infimum of {[imath]f_n[/imath]}? Secondly proving the statement : Since {[imath]f_n[/imath]} is measurable on E, for any real number [imath]c[/imath], {[imath]x\in E |f_n(x) >c[/imath]}(Or [imath]\gt[/imath] can be [imath]\lt,\le,\ge[/imath]) is measurable. Then {[imath]x\in E[/imath]| inf{[imath]f_n[/imath]} [imath]\ge c[/imath]} , since for any [imath]n[/imath],[imath]f_n[/imath] [imath]\ge[/imath] inf{[imath]f_n[/imath]} so, done. Is it right?
|
1456350
|
The Supremum and Infimum of a sequence of measurable functions is measurable
I am reading through Folland's Real Analysis: Modern Techniques and Their Applications, and they have the following proposition and proof: Proposition: If [imath]\{f_{j}\}[/imath] is a sequence of [imath]\bar{\mathbb{R}}[/imath]- valued measurable functions on [imath](X, \mathcal{M})[/imath], then the functions: [imath]g_{1}(x)=\sup_{j} f_{j}(x)[/imath] [imath]g_{2}(x)=\inf_{j} f_{j}(x)[/imath] are measurable. Proof: We have [imath]g_{1}^{-1}((a,\infty])= \bigcup_{j=1}^{\infty} f_{j}^{-1}((a,\infty])[/imath] [imath]g_{2}^{-1}([-\infty,a))= \bigcup_{j=1}^{\infty} f_{j}^{-1}([-\infty,a))[/imath] Where the result follows from the fact that [imath]g_{1}^{-1}((a,\infty])\in \mathcal{M}[/imath] for all [imath]a \in \mathbb{R} \iff g[/imath] is measurable, and [imath]g_{2}^{-1}([-\infty,a))\in \mathcal{M}[/imath] for all [imath]a \in \mathbb{R} \iff g[/imath] is measurable. I understand the last part of the proof, but how do we know [imath]g_{1}^{-1}((a,\infty])= \bigcup_{j=1}^{\infty}f_{j}^{-1}((a,\infty])[/imath] and [imath]g_{2}^{-1}([-\infty,a))=\bigcup_{j=1}^{\infty} f_{j}^{-1}([-\infty,a))[/imath]? I would appreciate any help with this.
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1968655
|
Finding nth power of a 4 state Markov Chain Matrix
I am required to find P[imath]_{ij}^{(n)}[/imath] for the P matrix of a 4 state Markov chain with states 0,1,2,3 as given below [imath]\begin{matrix} 1-p&&p&&0&&0\\0&&1-p&&p&&0\\0&&0&&1-p&&p\\0&&0&&0&&1\end{matrix}[/imath] There were two options. One was to solve it by finding eigenvalues and then left and right eigenvectors...etc. Second choice was to use method of generating function P(s) = I+sP+s[imath]^2[/imath]P[imath]^2[/imath]+s[imath]^3[/imath]P[imath]^3[/imath]........... I chose to use the second method and got the (I-sP)[imath]^{-1}[/imath] as under: \begin{matrix}\frac{1}{ps-s+1}&&\frac{-ps}{(ps-s+1)^2}&&\frac{p^2s^2}{(ps-s+1)^3}&&\frac{p^3s^3}{(s-1)(ps-s+1)^3}\\0&&\frac{1}{ps-s+1}&&\frac{-ps}{(ps-s+1)^2}&&\frac{-p^2s^2}{(s-1)(ps-s+1)^2}\\0&&0&&\frac{1}{ps-s+1}&&\frac{ps}{(s-1)(ps-s+1)}\\0&&0&&0&&\frac{-1}{(s-1)}\end{matrix} Now to gather the coefficients of [imath]s^n[/imath] in each of the terms, I could do it for all terms of the inverse matrix except terms [imath]a_{14}[/imath], [imath]a_{24}[/imath] and [imath]a_{34}[/imath]. (a) Could I please get guidance to do this? (b) Is there any other easier method to calculate the P[imath]_{ij}^{(n)}[/imath]?
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1930680
|
Markov Chain - method of generating function -to find nth power
P= [imath]\left(\begin {matrix}q&p&0&0\\0&q&p&0\\0&0&q&p\\0&0&0&1\end{matrix}\right)[/imath] We are to find [imath]P^n[/imath] using method of generating function. I have understood the steps up to finding the inverse as inv (I-sP) = [imath]\frac{adj(I-sP)}{|I-sP|}[/imath] This matrix works out to P= [imath]\left(\begin {matrix}(1-sq)^{-1}&sp(q-sq)^{-2}&s^2p^2(1-sq)^{-3}&s^3p^3(1-s)^{-1}(1-sq)^{-3}\\0&(1-sq)^{-1}&sp(1-sq)^2&s^2p^2(1-s)^{-1}(1-sq)^{-2}\\0&0&(1-sq)^{-1}&sp(1-s)^{-1}(1-sq)^{-1}\\0&0&0&(1-s)^{-1}\end{matrix}\right)[/imath] I have not understood the next step in the textbook which says now collect coefficients of [imath]s^n[/imath] by expanding each element of [imath](I-sP)^{-1}[/imath] in powers of s to get the matrix. The first row of this matrix is [imath]\left(\begin{matrix} q^n&& npq^{n-1}&&\frac{n(n-1)p^2q^{n-2}}{2}&&p^3(1+3q+6q^2+.......)\end{matrix}\right)[/imath] Request guide as to how is this done for the first row of the matrix. I will work out the rest based on this
|
1969764
|
Trigonometric identity between sines of multiples of [imath]\pi/70[/imath]
Prove that: [imath]\left(\sin\frac {9\pi}{70}+ \sin\frac {29\pi}{70} - \sin\frac {31\pi}{70}\right) \left(\sin\frac {\pi}{70}-\sin\frac {11\pi}{70} - \sin\frac {19\pi}{70}\right) =\frac {\sqrt {5} -4}{4}[/imath] I Could not get any idea to solve. Please help
|
1895204
|
Help with this trigonometry problem
Prove the given identity. [imath]\left(\sin\frac {9\pi}{70}+ \sin\frac {29\pi}{70} - \sin\frac {31\pi}{70}\right) \left(\sin\frac {\pi}{70}-\sin\frac {11\pi}{70} - \sin\frac {19\pi}{70}\right) =\frac {\sqrt {5} -4}{4}[/imath] Please help, I could not gather enough ideas, even on how to get to the first step. I thought of using the transformation formula (from sum to product) but did not find a fruitful result. UPDATE 1.: [imath]\left(-\frac{1 +\sqrt{5}}{8} + \frac{1}{8} \sqrt{14\left(5 - \sqrt{5} \right)}\right)\left(-\frac{ 1+ \sqrt{5}}{8} -\frac{1}{8} \sqrt{14 \left(5 -\sqrt{5} \right)}\right) [/imath], I noticed that Alexis had got to this point. Can anyone explain me how did he get here, with calculations. Please help. Thanks in Advance.
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1969588
|
Why are there exactly [imath]2^{\aleph_0}[/imath] Turing degrees?
I know that there are exactly [imath]2^{\aleph_0}[/imath] Turing degrees, but I am trying to realize the reason: I am convinced there can't be more than [imath]2^{\aleph_0}[/imath] of them, but why can't there be less? I know it should be trivial...
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1702688
|
How many Turing degrees are there?
So I know there are precisely [imath]2^{\aleph_0} [/imath] Turing degrees, but is there a proof of this somewhere?
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1970835
|
Proof of inequality [imath]n!\leq(\frac{n+1}{2})^n[/imath]
I need to prove the following inequality: [imath]n!\leq(\frac{n+1}{2})^n[/imath] Thank you very much for your help.
|
1967405
|
Proof that [imath]n!\leq {(\frac{n+1}{2})}^{n}[/imath]
I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision [imath]n!\leq {\left(\frac{n+1}{2}\right)}^{n}[/imath]
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1970454
|
How to solve a second order but nonlinear differential equation?
I am a physics student and I am trying to solve a central potential problem and came to an equation like this: [imath]y^{\prime\prime}+y=\frac{1}{y^3}[/imath] and I don't know how to solve this equation. Can anyone help me out?
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1970552
|
How to solve [imath]y'' + y = y^{-3}[/imath]
I am trying to solve a central force problem and came to an equation like this: [imath]\frac{d^2 y}{dx^2} + y = \frac{1}{y^3}[/imath] I can't find a decent method to solve it.
|
1966283
|
How do I find a flaw in this false proof that [imath]7n = 0[/imath] for all natural numbers?
This is my last homework problem and I've been looking at it for a while. I cannot nail down what is wrong with this proof even though its obvious it is wrong based on its conclusion. Here it is: Find the flaw in the following bogus proof by strong induction that for all [imath]n \in \Bbb N[/imath], [imath]7n = 0[/imath]. Let [imath]P(n)[/imath] denote the statement that [imath]7n = 0[/imath]. Base case: Show [imath]P(0)[/imath] holds. Since [imath]7 \cdot 0 = 0[/imath], [imath]P(0)[/imath] holds. Inductive step: Assume [imath]7·j = 0[/imath] for all natural numbers [imath]j[/imath] where [imath]0 \le j \le k[/imath] (induction hypothesis). Show [imath]P(k + 1)[/imath]: [imath]7(k + 1) = 0[/imath]. Write [imath]k + 1 = i + j[/imath], where [imath]i[/imath] and [imath]j[/imath] are natural numbers less than [imath]k + 1[/imath]. Then, using the induction hypothesis, we get [imath]7(k + 1) = 7(i + j) = 7i + 7j = 0 + 0 = 0[/imath]. So [imath]P(k + 1)[/imath] holds. Therefore by strong induction, [imath]P(n)[/imath] holds for all [imath]n \in \Bbb N[/imath]. So the base case is true and I would be surprised if that's where the issue is. The inductive step is likely where the flaw is. I don't see anything wrong with the strong induction declaration and hypothesis though and the math adds up! I feel like its so obvious that I'm just jumping over it in my head.
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1991348
|
A wrong mathematical induction
Where is the mistake? Statement: For [imath]n\in \mathbb{N}_{0}[/imath] is [imath]2n=0[/imath]. Bais: Show that the basis holds for [imath]n=0[/imath] : [imath]2*0=0[/imath]. Assumption: The statement is valid for all [imath]k\leq n[/imath] : [imath]2*k=0[/imath] for all [imath]k \leq n[/imath] Inductive Step: For [imath]k = n+1[/imath] is [imath]k=a+b[/imath] for two natural numbers [imath]a,b \leq n[/imath]. It is [imath]2(n+1) = 2a + 2b = 0+0=0[/imath]. It is obviously wrong. I have a few ideas but I would like to be sure where the exact mistake is. The first thing I noticed was that in this equoation [imath]2(n+1) = 2a + 2b = 0+0=0[/imath] the left side [imath]2(n+1)[/imath] is always positive but the right side isn't. I cant find a mathematical mistake here so I thought the Assumption is wrong. I always thought the Assumption is valid for a specific n. It is is indeed freely selectable but specific. Is that right and is that the problem with this induction? If it isn't - maybe it might be the fact that there aren't two natural numbers below 1 that [imath] a+ b =k[/imath] and thet you cant use the assumption two times?
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1970705
|
Does it mean we can never know the exact area of circle?
I performed three calculations for a circle of radius [imath]10^7[/imath] metre [imath](\pi-3.14)(10^7)^{2}=1.59265359\times10^{11}[/imath] [imath](\pi-3.1415926535897)(10^{7})^2=9.31[/imath] [imath](\pi-3.141592653589793238462)(10^{7})^2=0.00000006433832795028842[/imath] It implies large difference in the calculated area of the circle and it's dependence on the accuracy of value of [imath]\pi[/imath] used. Since we may never know the actual value of [imath]\pi[/imath],Does it imply that it is impossible to calculate the exact value of the area of a circle?
|
918658
|
Finding the exact area of a circle?
Background: I recently began taking calculus and it has come to alter the way I look at circles, and curves. The equation of a circle is [imath]\pi r^2[/imath], traditionally in school we have always left the answer in terms of pi (i.e.) if the radius [imath]r=2[/imath] then the area [imath]A = 4\pi[/imath]. Question: If one were to attempt to write the area of a circle in decimal form (i.e.) if the radius=2 then the area [imath]A = 4\pi[/imath], but [imath]\pi[/imath] doesn't have an end, it has (per what I have learned in school) an infinite number of decimal places so it is [imath]3.14159\ldots[/imath] therefore if one multiplied [imath]4\cdot 3.14159\ldots[/imath] one would have to approximate ones answer. Does that mean that it is impossible to calculate the exact (without approximation) area of a circle? Thanks for any responses, Joel
|
1971404
|
Convergence of rearranged infinite series
Suppose that [imath]\sum_{k=1}^{\infty}a_k[/imath] is nonabsolutely convergent. Then does [imath] \sum_{k=1}^{\infty}a_k = a_2+a_1+a_4+a_3...? [/imath] I'm not sure how to determine whether this rearrangement changes the value of the sum. Would it depend on the fact that the original series does not converge absolutety? How would I prove this?
|
1970427
|
Rearranging infinite series
Is the computation [imath]\sum_{k=1}^{\infty}a_k = a_2+a_1+a_4+a_3...[/imath] valid? I think that it is a valid rearrangment because the permutation function is bijective, and the computation is valid if the original series is unconditionally convergent. Is my thinking correct?
|
1966489
|
Explain if I understand correctly.
For a formula [imath] \gamma = r \leftrightarrow (p_1 \vee p_2) [/imath] there is true that: [imath]\rho \models \gamma [/imath] iff [imath]\rho(r) = \max(\rho(p_1), \rho(p_2))[/imath] Is there exists such set of formulas [imath]\Gamma[/imath] that [imath]\rho \models \Gamma[/imath] iff [imath]\rho(r) = \max_{n \in \mathbb{N}} (\rho(p_n))[/imath] where [imath]\rho[/imath] is a vaulation. Do I understand correctly that I should prove that there exists expected set that it is true [imath]\rho(r) = \max_{n \in \mathbb{N}} (\rho(p_n))[/imath] for any [imath]r[/imath]? [imath]\rho \models \gamma [/imath] means that [imath]\rho[/imath] satisfies [imath]\gamma[/imath] If yes, please hint me If no, please explain what I should do.
|
1965591
|
True for every formula.
For a formula [imath] \gamma = r \iff (p_1 \vee p_2) [/imath] there is true that: [imath]\rho \models \gamma \iff \rho(r) = \max(\rho(p_1), \rho(p_2))[/imath] Is it a true for every formula r?
|
1969946
|
Strongly complete completion of profinite group
Let [imath]G[/imath] be à profinite group and [imath]\widehat{G}[/imath] his profinite completion. I want to show (I hope it's true) that [imath]\widehat{G}[/imath] is strongly complete that is all normal subgroups of finite index are open sets. For that I take [imath]\widehat{N}[/imath] a finite index normal subgroup of [imath]\widehat{G}[/imath] and try to find [imath]N[/imath] normal subgroup of finite index of [imath]G[/imath] such that [imath]\widehat{N}=\ker\left(\varphi_N:\widehat{G}\to G/N\right)[/imath]. I tried [imath]N=\theta^{-1}\left(\widehat{N}\right)[/imath] where [imath]\theta:G\to\widehat{G}[/imath] but could not conclude. Not rely with this question I think: I don't want a different definition of strongly complete but I want to check that somewath ([imath]\widehat{G}[/imath] here) is strongly complete.
|
180280
|
Strongly complete profinite group
Let [imath]G[/imath] be a profinite group (or equivalently a compact and totally disconnected topological group ) with the property that all of its normal subgroups of finite index are open sets. Does this imply that all of its subgroups of finite index are open sets ? (if all subgroups of finite index from [imath]G[/imath] are open sets, than [imath]G[/imath] is called strongly complete ; this motivates the title of this post)
|
246855
|
Probability Problem with [imath]n[/imath] keys
A woman has [imath]n[/imath] keys, one of which will open a door. a)If she tries the keys at random, discarding those that do not work, what is the probability that she will open the door on her [imath]k^{\mathrm{th}}[/imath] try? Attempt: On her first try, she will have the correct key with probability [imath]\frac1n[/imath]. If this does not work, she will throw it away and on her second attempt, she will have the correct key with probability [imath]\frac1{(n-1)}[/imath]. So on her [imath]k^{\mathrm{th}}[/imath] try, the probability is [imath]\frac1{(n-(k-1))}[/imath] This does not agree with my solutions. b)The same as above but this time she does not discard the keys if they do not work. Attempt: We want the probability on her [imath]k^{\mathrm{th}}[/imath] try. So we want to consider the probability that she must fail on her [imath]k-1[/imath] attempts. Since she keeps all her keys, the correct one is chosen with probability [imath]\frac1n[/imath] for each trial. So the desired probability is [imath](1-\frac{1}{n})^{k-1} (\frac1n)^k[/imath]. Again, does not agree with solutions. I can't really see any mistake in my logic. Can anyone offer any advice? Many thanks
|
2742167
|
Probability that a key open a door after some trials
Suppose we have [imath]n[/imath] keys. We try each key at random to open a door. If we [imath]\bf discard[/imath] the ones that don't work, what is the probability that we open the door in the kth try? What if we dont discard the previously tried keys? [imath]\bf \underline{Attempt \; to \; the \; solution}[/imath] We can call [imath]E_i[/imath] to be event that the ith key opens the door. Clearly, since we have [imath]n[/imath] keys [imath]P(E_i) = \frac{1}{n}[/imath]. Here is where I am having trouble trying to express our outcome in terms of events [imath]E_i[/imath]. I feel like the answer is just [imath]P(E_k) = \frac{1}{n}[/imath] but perhaps the phrasing of the problem is confusing. Now, if we dont discard. Suppose we put first key into door. there are [imath]n-1[/imath] possible ways that it wont work cause [imath]1[/imath] of them is the one that works.same holds for the second and third and so forth until we get to the [imath]k-1[/imath]th case. The next case the one that opens the door can be done in 1 way of course. Thus, [imath] P = \frac{(n-1)^{k-1} }{n} [/imath] Is this a correct approach?
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1972482
|
Months containing different birthday
If we are given [imath]20[/imath] people , what is the probability that among the [imath]12[/imath] months there are [imath]4[/imath] months containing exactly two birthdays and [imath]4[/imath] months containing exactly [imath]3[/imath] birthdays . I tried , that I even not able to start it .
|
504187
|
Combinatorics problem on [imath]20[/imath] people, [imath]12[/imath] months, and distinguishable groups
Given [imath]20[/imath] people, what is the probability that, among the [imath]12[/imath] months in the year, there are [imath]4[/imath] months containing exactly [imath]2[/imath] birthdays and [imath]4[/imath] months containing exactly [imath]3[/imath] birthdays? I am less concerned about knowing the answer than making sure I have thought about the problem correctly. The general problem seems to be finding the distinct integer-valued vectors that satisfy [imath]x_1 + x_2 + ... x_{12} = 20\ \ \ \ \ \ \ \ \ x_i \ge 0, i = 1, ..., 12[/imath] It can be proven that there are [imath]\binom{20+12-1}{12-1}[/imath] such vectors. I am presuming this to be the sample space. (Please correct me if I'm wrong) [imath]4\cdot 2 + 4\cdot 3 + 4 \cdot 0= 20[/imath], so there are [imath]3[/imath] distinct groups of [imath]4[/imath] months. Therefore we have [imath]\binom{12}{4\ 4\ 4}[/imath] possible ways to get the desired distribution of months. The probability I get is [imath]\frac{\binom{12}{4\ 4\ 4}}{\binom{20+12-1}{12-1}} [/imath]. The answer in the back of the textbook is [imath]1.0604 \times 10^{-3}[/imath] so my answer is off. I just want to know where I went wrong.
|
1964950
|
Why can't be ([imath]\mathbb{Z}_5,+) [/imath] isomorphic to [imath]U(R)[/imath] for some rings [imath]R[/imath] where [imath]U(R)[/imath] is the unity groups of [imath]R[/imath]?
Why can't be [imath](\mathbb{Z}_5,+) [/imath] isomorphic to [imath]U(R)[/imath] for some rings [imath]R[/imath] where [imath]U(R)[/imath] is the unity groups of [imath]R[/imath] ? Is important that [imath](\mathbb{Z}_5,+)[/imath] is cyclic?
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384362
|
Group of invertible elements of a ring has never order [imath]5[/imath]
Let [imath]R[/imath] be a ring with unity. How can I prove that group of invertible elements of [imath]R[/imath] is never of order [imath]5[/imath]? My teacher told me and my colleagues that problem is very hard to solve. I would be glad if someone can provide me even a small hint because, at this point, I have no clue how to attack the problem.
|
1972947
|
Let G be a simple group and let [imath]{n_p}[/imath] be the number of Sylow p-subgroups, where p is prime. Show |G| divides [imath]({n_p})![/imath].
Using this or otherwise, prove groups of order 48 cannot be simple. could someone help me about this problem? Do we use the Sylow-Theorem?
|
813091
|
trouble applying Sylow's theorems
Let [imath]G[/imath] be a simple group and let [imath]n_p[/imath] be the number of Sylow [imath]p[/imath]-subgroups, [imath]p[/imath] prime. Show that [imath]|G|[/imath] divides [imath](n_p)![/imath] (factorial). If i start off by assuming G is abelian then G is isomorphic to Z/pZ. So |G| = p, by Sylow's third theorem n_p is of the form n_p = 1+pk, k some integer. n_p must divide order of G witch implies k=0 and n_p = 1. But this would mean that the Sylow-p-subgroup is normal in G, contradicting the simplicity of G. Worse, |G|=p does not divide 1. So i guess G is not abelian, not really sure how to continue from here.
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1971282
|
Inexpressibility of [imath]\vee[/imath] in term of [imath]\wedge, \to[/imath] and [imath]\bot[/imath] in intuitionistic logic.
Show that it is not possible to express [imath]\vee[/imath] using [imath]\wedge, \to, \bot[/imath]. I have no idea how to show it. My system is intuitionistic logic.
|
1128646
|
Is [imath]\lor[/imath] definable in intuitionistic logic?
The Wikipedia page mentions that [imath]\{\lor,\leftrightarrow,\bot\}[/imath] and [imath]\{\lor,\leftrightarrow,\neg\}[/imath] are complete sets of operators for intuitionistic logic, and also gives a few equivalences for other operators in terms of [imath]\lor[/imath], but there are no equivalences listed for [imath]\lor[/imath] itself. So my question is whether it is possible to define an expression in [imath]\{\land,\to,\neg\}[/imath] (which also includes [imath]\leftrightarrow,\bot[/imath]) that is intuitionistically equivalent to [imath]\lor[/imath] in the sense that it satisfies the "definition" [imath]((A\lor B)\to C)\iff((A\to C)\land(B\to C)).\tag{1}[/imath] (I suspect the answer is no, but I'd like a proof.) This question extends to the classification of all minimal complete operator subsets of [imath]\{\land,\lor,\to,\leftrightarrow,\neg,\top,\bot\}[/imath]. Depending on the answer to this question, it seems that we have [imath]\{\lor,\leftrightarrow,\neg\},\{\lor,\leftrightarrow,\bot\},\{\lor,\land,\to,\neg\},\{\lor,\land,\to,\bot\}[/imath] if it is true, otherwise [imath]\{\land,\to,\neg\},\{\land,\to,\bot\},\{\land,\leftrightarrow,\neg\},\{\land,\leftrightarrow,\bot\},\{\lor,\leftrightarrow,\neg\},\{\lor,\leftrightarrow,\bot\},\{\to,\leftrightarrow,\neg\},\{\to,\leftrightarrow,\bot\}.[/imath]
|
1973337
|
Infinite composition of Logs with infinite argument
If you were to compose an infinite number of ln's what is the result? Is it finite? Does it exist? Perhaps the most natural way to think about writing down this limit would be as follows: [imath]\lim_{a->\infty}\log_a\circ\log_{a-1}\circ...\circ\log_2(a)[/imath] (starting at 1 is obviously not going to work since log(1)=0, leads to the log(log(1)) being divergent.) The problem with this definition is that the logs kill the 2 too quickly, i.e we get complex evaluations of the function, which is not necessarily bad. There are various other ways to think about this sort of limit as well. Suppose you already had an infinite composition of logs, then took a limit as x went to infinity. i.e: [imath]\lim_{x->\infty}...\log\circ\log\circ...\circ\log(x)[/imath] who wins, x or the logs. Numerically I have found that composing 6 logs with 10^10,000,000 results in -2.165, which seems to suggest that the logs would win, but things, as we know, tend to act very differently in the limit. Is there another way to think about this limit, or does it not exist, since taking the limit two ways leads to two different results? Thanks.
|
1587322
|
Limit to infinity and infinite logarithms?
When trying to evaluate[imath]\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))[/imath]I noticed that the answer was bound to be complex for any [imath]x[/imath]. Plugging in a very, very large real number in for [imath]x[/imath] will eventually become complex, and the [imath]\log[/imath] of a complex number is always complex or imaginary. But what happens when we try to evaluate [imath]\lim_{x\to\infty}\ln(\ln(\ln(\ln(\cdots\ln(x)\cdots))))[/imath] We see that [imath]\lim_{x\to\infty}\ln(x)=\infty[/imath]Meaning that if we try to do this repeatedly, we should still get infinite as our final result. But we also see that for arbitrarily large [imath]x[/imath] where [imath]x[/imath] is real and finite, the result is a complex answer. Which creates a sort of contradiction. What does it evaluate to?!?!
|
1973421
|
Why [imath]\lim_{n\to\infty}\sqrt{n^2+100n} - n=50[/imath]
[imath]\lim_{n\to\infty}\sqrt{n^2+100n} - n=50[/imath] Why is it [imath]50[/imath]? I'm unsure how to solve this. Is there some transformation to apply first? How do you know?
|
1972095
|
Big-O proof showing that t(n) is O(1)
Let t(n) = [imath]\sqrt{n^2+100n} - n[/imath] Show that [imath]t(n)[/imath] is [imath]O(1).[/imath] I tried solving this using the formal definition of O( ) but am not getting anywhere. Quite confused on how to tackle this question.
|
1340021
|
Find the sign of the following integral: [imath]\int_{0}^{2\pi}{\sin x\over x}[/imath].
Am I required to know if it is positive or negative? This expression is not really integrable, not within the boundaries of the course at least. I tried comparing it to other integrals but I get zero all the time. I am not sure I have fully understood what I should do. I would really appreciate any help in this. An attempt using an advice here: [imath]\sin x[/imath] between [imath]0[/imath] to [imath]\pi[/imath] equals [imath]-\sin x[/imath] between [imath]\pi [/imath] to [imath]2\pi[/imath]. Since we are looking at [imath]\sin x\over x[/imath], [imath]\int_{\pi}^{2\pi}{\sin x\over x}[/imath] might be negative, but still is smaller than [imath]\int_{0}^{\pi}{\sin x\over x}[/imath], therefore, the sign is +1. Is that sufficient?
|
107113
|
Find the sign of [imath]\int_{0}^{2 \pi}\frac{\sin x}{x} dx[/imath]
I'd love your help with finding the sign of the following integral: [imath]\int_{0}^{2 \pi}\frac{\sin x}{x} dx[/imath],I know that computing it is impossible. I tried to use integration by parts and maybe to learn about the sign of each part and conclude something but It didn't work for me. Any suggestions?
|
989043
|
Rigorous proof that surjectivity implies injectivity for finite sets
I'm trying to prove that, for a finite set [imath]A[/imath], if the map [imath]f: A\rightarrow A[/imath] is a surjection, then it's an injection. I've looked at this post: Surjectivity implies injectivity but the arguments there seem very round-about, proving that injectivity implies surjectivity first. I think I've seen other proofs, too, but they all sound kind of flimsy. They tend to go something like "If the function is surjective then no two arguments can go to the same image, since that would require some element of the range not being the image of any element." While that's true and I see what the argument has in mind, it's no more convincing than the statement of the claim that we're trying to prove! Without a more rigorous proof, both seem to rely on the intuition that everything has to go to one and only one thing, but we're supposed to be proving that. But I've been trying to develop a more thorough and direct proof, and haven't been able to. I've tried induction: The theorem holds trivially if [imath]|A|=0[/imath]. Suppose the theorem holds up to [imath]n=|X|[/imath], and let [imath]|A|=n+1[/imath] and [imath]f:A\rightarrow A[/imath] be a surjection. Then let [imath]x\in A[/imath] and [imath]f^{-1}(x)[/imath] be the set of all elements [imath]y\in A[/imath] such that [imath]f(y)=x[/imath]... But at this stage I want to restrict my function, but that has the disadvantage that if [imath]f(y)[/imath] has more than one element in it, when I pluck it out of [imath]A[/imath] as the domain and pluck [imath]x[/imath] out of [imath]A[/imath] as the range, then I may not be dealing with a function mapping from a set to itself and so I don't get to use the inductive hypothesis. Any other proofs you know about?
|
2628829
|
Proving a function is bijection
My problem is: Let [imath]A[/imath] and [imath]B[/imath] finite sets of equal cardinality. Show that if a function [imath]f : A \to B[/imath] is injective then it is also surjective. Similarly show that if [imath]f[/imath] is surjective, then it is also injective. My thought is since the number of elements in [imath]|A| = |B|[/imath], function [imath]f: A \to B[/imath] is bijection. But my instructor said I have to explain more. Please help me with this. I don't know how a injective function leads to a surjective function and vice versa. Thank you.
|
1973495
|
Prove a binomial identity [imath]\sum^{n}_{k=0}\binom{n+k}{k}2^{-k} = 2^n[/imath]
I was asked to show that [imath]\sum^{n}_{k=0}\binom{n+k}{k}2^{-k} = 2^n[/imath] using pascal identity. I don't know how to do it. Whenever I was trying to use [imath]\binom{n+k+1}{k} = \binom{n+k}{k}+\binom{n+k}{k-1}[/imath], my index is always off by 1.
|
389099
|
Proof of the identity [imath]2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}[/imath]
I just found this identity but without any proof, could you just give me an hint how I could prove it? [imath]2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}[/imath] I know that [imath]2^n = \sum\limits_{k=0}^n \binom{n}{k}[/imath] but that didn't help me
|
1974702
|
Index of the subgroup [imath]\operatorname{Im}(f)[/imath]
In a course of group theory an exercise is left with no indications : Let [imath]A[/imath] be an abelian group of finite type and [imath]a_1,\dots ,a_n[/imath] a [imath]\mathbb{Z}[/imath]-basis of [imath]A[/imath]. Let [imath]f : A \to A[/imath] be a group homomorphism. Les [imath]M=(m_{ij})[/imath] be the matrix (invertible) defined by [imath]\displaystyle f(a_j)=\sum_{i=1}^n m_{ij}a_i[/imath] with [imath]\det (M)\neq 0[/imath]. Show that [imath][A:\operatorname{Im}(f)]=|\det M |[/imath], i.e the index of [imath]\operatorname{Im}(f)[/imath] in [imath]A[/imath] as a subgroup is precisely [imath]\det (M)[/imath]. I am not able to prove this fact, and need a bit of help. In fact I do not see where I can make [imath]\det (M)[/imath] appear. Can you give me some hints ? Thank you
|
245733
|
Index of a sublattice in a lattice and a homomorphism between them
I am asked to show that if [imath]\phi_A[/imath] is the homomorphism from [imath]\mathbb{Z}^k \rightarrow \mathbb{Z}^k[/imath] given by [imath]\phi_A(x)=xA[/imath] then the index of [imath]\phi(\mathbb{Z}^k)[/imath] in [imath]\mathbb{Z}^k[/imath] is finite if and only if [imath]A[/imath] is nonsingular. While this seems intuitive to me, one obstacle to proving this is that I am not quite sure of how to represent the index of a sublattice in a lattice! The way I understand it, [imath]j\mathbb{Z}^k[/imath] has index [imath]j^k[/imath] in [imath]\mathbb{Z}^k[/imath], for an integer [imath]j[/imath]. I am used to indices in finite groups but not infinite groups, so is the trick to show this sort of correspondence? Furthermore, I would like to know if this is a valid sketch of a proof: If [imath]A[/imath] nonsingular, [imath]A'[/imath] is in Smith Normal Form ([imath]PAQ=A'[/imath]) and [imath]\det A'[/imath] is nonzero, so the columns of [imath]P^{-1}[/imath] form a basis for [imath]\mathbb{Z}^k[/imath] and the columns of [imath]AQ[/imath] form a basis for the sublattice such that each column in the basis for the sublattice is an integral multiple of the columns in the basis for the lattice [imath]\mathbb{Z}^k[/imath], and since these multiples are given by the diagonal entries of SNF, we have that for every one point in the sublattice, there are [imath]\prod a_i=\det A'=\det A[/imath] points in the main lattice. If [imath]A[/imath] singular, we have the same as (1) except now at least one of these columns is a multiple [imath]0[/imath] times the column for the basis in the main lattice, so there are infinitely many points in the main lattice for each point in the sublattice.
|
1974347
|
Prove the following equality: [imath]\int_{0}^{\pi} e^{4\cos(t)}\cos(4\sin(t))\;\mathrm{d}t = \pi[/imath]
Prove the following equality: [imath] \int_{0}^{\pi} e^{4\cos(t)}\cos(4\sin(t))\;\mathrm{d}t = \pi [/imath]
|
1938543
|
Evaluate [imath]\int_0^{\pi} e^{a\cos(t)}\cos(a\sin t)dt[/imath]
What is the value of [imath]\int_0^{\pi} e^{a\cos(t)}\cos(a\sin t)dt?[/imath]
|
1403184
|
Bijective Conformal Mapping onto the Open Unit Disc [imath]\mathbb{D}[/imath]
What is the explicit bijective conformal mapping [imath]f(z):G_n\to\mathbb{D}[/imath], [imath]z\in\mathbb{C}[/imath] for the following domain transformations: [imath]G_1=\{x+iy~|~x>1/2,y>0\}[/imath] is the open region of the first quadrant. [imath]G_2=\{x+iy~|~x<0,y>0\}[/imath] is the open second quadrant. [imath]G=_3\{x+iy~|~x<0,y<0\}[/imath] is the open third quadrant. [imath]G_4=\{x+iy~|~x>0,y<0\}[/imath] is the open fourth quadrant. Where it is noted that [imath]\mathbb{D}[/imath] is the open unit disc defined by [imath]\mathbb{D}=\{x+iy~|\sqrt{~x^2+y^2}<1\}[/imath]. I am aware that the aforementioned transformations have the specific form [imath]f(z)=Ke^{cz}[/imath], [imath]\forall K\in\mathbb{C}[/imath] or [imath]f(z)=z^2[/imath], where such mappings exist by the Riemann Mapping Theorem.
|
52338
|
Conformal Maps onto the Unit Disc in [imath]\mathbb{C}[/imath]
I am interested in finding explicit formulae for (better yet characterizing) conformal functions from various domains onto the open unit disc [imath]\mathbb{D}\subset\mathbb{C}[/imath], and in understanding the key ideas necessary to establish such functions. Specifically, what can [imath]f[/imath] look like when [imath]f:G\to\mathbb{D}[/imath] is conformal and (1) [imath]G=\{x+iy~|~x,y>0\}[/imath] is the open first quadrant. (2) [imath]G=\{x+iy~|~x>0,~0<y<1\}[/imath] is an open horizontal strip in the first quadrant. (3) [imath]G=\{z\in\mathbb{C}~|~\frac{1}{2}<|z|<1\}[/imath] is an annulus. (4) [imath]G=\mathbb{D}\cap\{|z-\frac{1}{2}|>\frac{1}{2}\}[/imath] is something else (torus?).
|
1974649
|
Series with non-negative terms: [imath]\sum_{n=1}^{\infty} \frac{|\cos n|^n}{n}[/imath]
Given the following series: [imath]\sum_{n=1}^{+\infty} \frac{|\cos n|^n}{n}[/imath] (being a series with non-negative terms), does it converge or diverge? Unfortunately, I can't prove in any way or convergence or divergence of this series. Any ideas?
|
823816
|
Is [imath] \sum\limits_{n=1}^\infty \frac{|\sin n|^n}n[/imath] convergent？
Is the series [imath] \sum_{n=1}^\infty \frac{|\sin n|^n}n\tag{1}[/imath] convergent？ If one want to use Abel's test, is [imath] \sum_{n=1}^\infty |\sin n|^n\tag{2}[/imath] convergent？ Thank you very much
|
1975893
|
Degree of a field extension [imath]F[x]/(g(x))[/imath] over the field [imath]F[/imath]
Let [imath]F[/imath] be a field and let [imath]F[x][/imath] be the ring of polynomials in [imath]x[/imath] over [imath]F[/imath]. Let [imath]g(x)[/imath], of degree [imath]n[/imath], be in [imath]F[x][/imath] and let [imath]V=(g(x))[/imath] be the ideal generated by [imath]g(x)[/imath] in [imath]F[x][/imath]. I have to prove that [imath]F[x]/V[/imath] is an [imath]n[/imath]-dimensional vector space over [imath]F[/imath]. Help me with the proof.
|
547861
|
Establishing a Basis for [imath]F[x]/(f(x))[/imath]
Hypothesis: [imath]f(x) = a_0 + a_1 x + \ldots + a_n x^n \in F[x][/imath] s.t. [imath]f(x)[/imath] irreducible over field [imath]F[/imath]. [imath]deg(f(x)) = n \ge 1[/imath]. Let [imath]V = (f(x)) \subset F[x][/imath]. Goal: Show that [imath]\mathcal{B} = \{1 + V, x + V, x^2 + V, \ldots , x^{n-1} + V\}[/imath] forms a basis for [imath]E = F[x]/V[/imath] Attempt: Linear Independence: If this set weren't linearly independent, then a non-trivial [imath]b_i \in F[/imath] would satisfy [imath] b_0 + b_1x + \ldots + b_{n-1} x^{n-1} + V = 0 + V [/imath] But this would then imply that [imath]b_0 + b_1x + \ldots + b_{n-1} x^{n-1} \in 0 + V[/imath]. But this would contradict [imath]V = (f(x))[/imath] since it is impossible that [imath]f(x)\,\,|\,\,b_0 + b_1x + \ldots + b_{n-1} x^{n-1}[/imath] of strictly lower degree. Now how would I show that this set spans [imath]F[x]/V[/imath]?
|
1975986
|
For [imath]c \in \mathbb{R}[/imath], the equation [imath]\dfrac{1}{x-a_1} + \dfrac{1}{x-a_2} + \ldots + \dfrac{1}{x-a_n} = c[/imath] has [imath]n-1[/imath] real solutions
For [imath]c \in \mathbb{R}[/imath], the equation [imath]\dfrac{1}{x-a_1} + \dfrac{1}{x-a_2} + \ldots + \dfrac{1}{x-a_n} = c[/imath] has [imath]n-1[/imath] real solutions The follow up to this is that prove that the equation has [imath]n[/imath] solutions if [imath]c \neq 0[/imath]. Here [imath]a_1 < a_2 < \ldots < a_n[/imath] and [imath]c \in \mathbb{R}[/imath]. How does this equation even have solutions, I'm really confused with this!
|
549060
|
Roots to an equation using analysis
Suppose that [imath]a_{1}<a_{2}<…<a_{n}[/imath]. Prove that the equation [imath] \frac{1}{x-a_{1}}+\frac{1}{x-a_{2}} +…+\frac{1}{x-a_{n}}=c[/imath] has exactly [imath]n-1[/imath] roots if [imath]c=0[/imath] and [imath]n[/imath] roots if [imath]c\neq 0[/imath]. I am absolutely not sure but I think I need to use Rolle's theorem on this. Multiply both sides by [imath]x-a_{1}[/imath] so we get: [imath]1+\frac{x-a_{1}}{x-a_{2}}+…+\frac{x-a_{1}}{x-a_{n}}-c(x-a_{1})=0[/imath]. Now it is defined for [imath]a_{1}[/imath] but not for [imath]a_{2},…,a_{n}.[/imath]. If we consider the interval [imath](a_{1}-\varepsilon,a_{n}+\varepsilon), \varepsilon>0[/imath] we get..nowhere. I want to create an interval with endpoints that are solutions to the equation. Yet I don't know how to prove that there are exactly [imath]n[/imath] roots(Fundamental theorem of algebra can't be used).
|
1976110
|
Is there a quicker way to work out highest common factors in [imath]\Bbb{Z}[i][/imath]?
I wanted to find HCF of [imath]5+i[/imath] and [imath]3+5i[/imath]. So I considered norms and found that the factors of the units in [imath]\Bbb{Z}[i][/imath], and the associates of [imath]5+i[/imath] for the first. And units in [imath]\Bbb{Z}[i][/imath], the associates of [imath]1+i[/imath], the associates of [imath]4+i[/imath], and the associates of [imath]3+5i[/imath] in the second. So the only common factors are the units. So I concluded that HCF was [imath]\{\pm1,\pm i\}[/imath]. However this approach to quite a while of working and I wondered if there was a smarter approach rather than just directly calculating the factors in each case and comparing? Thanks!
|
82350
|
GCD of gaussian integers
Let [imath]\mathbb Z [i] =\{a+bi: a,b \in \mathbb Z\}[/imath]. What is the gcd of [imath]11+7i[/imath] and [imath]18-i[/imath] in [imath]\mathbb Z [i][/imath]?
|
1977020
|
How to give a consistent mathematical explanation to a paradoxical divergent/convergent phenomena.
How can be explained, in a "clean" mathematical way, the "stabilization" of the different digits in the following divergent sequence? [imath]\tan(89°)=57.28996163...[/imath] [imath]\tan(89.9°)=572.9572134...[/imath] [imath]\tan(89.99°)=5729.577893...[/imath] [imath]\tan(89.999°)=57295.77951...[/imath] [imath]\tan(89.9999°)=572957.7951...[/imath] etc...
|
418077
|
Why does the tangent of numbers very close to [imath]\frac{\pi}{2}[/imath] resemble the number of degrees in a radian?
Testing with my calculator in degree mode, I have found the following to be true: [imath]\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}[/imath] Why is this? What is the proof or explanation?
|
1976779
|
Show that if the smallest prime factor [imath]p[/imath] of the positive integer [imath]n[/imath] exceeds [imath]n^{\frac{1}{3}}[/imath] then [imath]\frac{n}{p}[/imath] must be a prime or 1.
I am really having a problem understanding the proof of this question. Let me write the proof and i will explain my question Let [imath]p[/imath] be the smallest prime factor of [imath]n[/imath]. Assume that [imath]p > n^{\frac{1}{3}}[/imath] Then [imath]p > n^{\frac{1}{3}} \Rightarrow p^{-1} < n^{-\frac{1}{3}} \Rightarrow np^{-1} < nn^{-\frac{1}{3}} \Rightarrow \frac{n}{p} < n^{\frac{2}{3}}[/imath]. Let [imath]m = \frac{n}{p}[/imath], and assume that it is greater than [imath]1[/imath]. Notice that [imath]m[/imath] cannot have a prime factor less than [imath]n^{\frac{1}{3}}[/imath]. However [imath]n^{\frac{1}{3}} = \sqrt{n^{\frac{2}{3}}} > \sqrt{m}[/imath] Therefore, [imath]m[/imath] does not have a prime factor less than [imath]\sqrt{m}[/imath]. Hence, [imath]m[/imath] is a prime. I really dont understand this proof and Why does [imath]m[/imath] not have a prime factor less than [imath]n^{\frac{1}{3}}[/imath]? If someone can explain to me this proof i would really appreciate it ! thanks
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1946444
|
Prove this number must be a prime number or 1.
Show that if the smallest prime factor [imath]p[/imath] of the positive integer [imath]n[/imath] exceeds [imath]\sqrt[3]{n}[/imath], then [imath]\frac{n}{p}[/imath] must be prime or 1. I'm stuck trying to prove this. I tried this using contradiction, but I haven't been able to prove it. My approach: Let [imath]p[/imath] be the smallest prime factor of the positive integer [imath]n[/imath]. Let [imath]p > \sqrt[3]{n}[/imath] Then we must show that [imath]\frac{n}{p}[/imath] is either a prime number or 1. We know that [imath]n = pe[/imath] for some [imath]e \in \mathbb{Z}[/imath]. Thus, [imath]\frac{n}{p} = e[/imath] We also know that [imath]p[/imath] > [imath]\sqrt[3]{n}[/imath], so [imath]p^3 > n[/imath], and [imath]p^2 > \frac{n}{p}[/imath] So, assume [imath]\frac{n}{p} = e[/imath] is not prime nor 1, meaning it's a composite number. Let's try to get a contradiction. If [imath]\frac{n}{p}[/imath] is composite, then [imath]\exists a,b \neq 1: \frac{n}{p} = ab [/imath] Now, here's where I get stuck. What can I do with this info? [imath]\frac{n}{p} = ab [/imath] and [imath]p^2 > \frac{n}{p}[/imath].
|
1977318
|
Exponential inequality(Proof)
Given [imath]\forall x,y>0[/imath] Prove that [imath]x^y+y^x\ge1[/imath] I have tried weighted inequalities and Jensen's but unfortunately ended up no where. Please help me. (I know this is a basic inequality).
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827754
|
Proving the exponential inequality: [imath]x^y+y^x\gt1[/imath]
How can the following inequality be proven? [imath]x^y+y^x\gt1[/imath] for [imath](x\gt0,y\lt1)[/imath]
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1977560
|
If [imath]f[/imath] is analytic and [imath]3\operatorname{Re}f+2\operatorname{Im}f[/imath] is constant, is [imath]f[/imath] also constant?
Let [imath]f\colon U\to\mathbb{C}[/imath] be analytic on a region [imath]U[/imath]. If [imath]f=u+iv[/imath] and [imath]3u+2v=c\in\mathbb{R}[/imath] is constant, is it true that [imath]f[/imath] is also constant?
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1756099
|
Show a given analytic function is constant
Suppose that [imath]f[/imath] is analytic on some region [imath]R\in\mathbb{C}[/imath]. If Im[imath](f)[/imath] = [imath]k\cdot[/imath]Re[imath](f)[/imath] for some nonzero constant [imath]k\in\mathbb{C}[/imath], then show that [imath]f[/imath] is constant on [imath]R[/imath]. I know that if [imath]f'(z)=0[/imath] for all [imath]z[/imath] in some region [imath]R[/imath], then [imath]f(z)[/imath] is constant. However, I'm not sure how to apply this to the question at hand. I also think the Cauchy-Riemann equations could be helpful, but again, I'm not sure how to apply them to the question. Any guidance would be greatly appreciated. Thank you!
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1977094
|
Lie algebra of [imath]Gl(V)[/imath]
What is the mean of bijective is an open condition ? And why the Lie algebra of [imath]Gl(V)[/imath] can be identified with [imath]gl(V)[/imath] ?
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65784
|
Lie algebra of [imath]GL_n(\mathbb{C})[/imath]
I would like to do Tao's exercise 6 (i) but before I can even attempt it I need to be clear about his terminology. Exercise 6 Show that the Lie algebra [imath]gl_n(\mathbb{C})[/imath] of the general linear group [imath]GL_n(\mathbb{C})[/imath] can be identified with the space [imath]M_n(\mathbb{C})[/imath] of [imath]n \times n[/imath] complex matrices, with the Lie bracket [imath][A,B] := AB - BA[/imath]. When he writes "identified", does he just mean there is an isomorphism? So the exercise would be to find an explicit isomorphism from [imath]M_n(\mathbb{C})[/imath] to [imath]gl_n(\mathbb{C})[/imath]? Is that correct? Many thanks for your help.
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1978063
|
What does it mean [imath]2^\mathbb{N}[/imath]?
I have an exercise that involves this set: [imath]2^\mathbb{N}[/imath], but I don't know what set it is. I know that the elements of the set are functions whose domain is [imath]\mathbb{N}[/imath], but no idea which type of functions they are. Thanks in advance.
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104522
|
Strange set notation (a set as a power of 2)?
I am not very well versed on set theory or syntax, but I thought I knew the basics. However, in a book about databases I am reading now, the author uses [imath]2^x[/imath] to signify "a set of [imath]x[/imath]." For example, [imath]2^{\text{dogs}}[/imath] is a set of [imath]\text {dogs}[/imath], etc. The author never really explained this or why he does it, I just picked up the meaning from context. I am not sure why the exponent operator is used, nor am I sure what the number [imath]2[/imath] has to do with it. The sets being represented are NOT powers of [imath]2[/imath] (in size)... they come in all sizes. Is this a valid notation? I have not seen it anywhere before...
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1977040
|
Prove that [imath]1\cdot 1! + 2\cdot 2! +\dots+n\cdot n! = (n + 1)! - 1[/imath]
(whenever [imath]n[/imath] is a non-negative integer) I did the basic step [imath]P(1)[/imath] and found the statment [imath]P(n+1)[/imath] I now have [imath](n+1)! - 1 + (n+1)\cdot(n+1)![/imath] This should equal [imath](n+2)! - 1[/imath], but how do I show that?
|
326105
|
[imath]\sum_{i=1}^n i\cdot i! = (n+1)!-1[/imath] By Induction
I am trying to prove the following by Mathematical Induction: [imath]\sum_{i=1}^n i\cdot i! = (n+1)!-1\quad\text{for all integers $n\ge 1$}[/imath] My proof by Induction follows: First prove [imath]P(1)[/imath] is true, [imath]\sum_{i=1}^1 i\cdot i! = (1+1)! - 1[/imath] Then, for all integers [imath]k >= 1[/imath], if [imath]P(k)[/imath] is true then [imath]P(k+1)[/imath] is also true, [imath] \sum_{i=1}^k i\cdot i! = (k+1)! - 1[/imath] Then, we must show that [imath]P(k+1)[/imath] is true [imath] \sum_{i=1}^{k+1} i\cdot i! = ((k+1)+1)! - 1\\ \sum_{i=1}^{k+1} i\cdot i! = (k+2)! - 1[/imath] I am currently struggling with the next step to show that [imath]P(k)[/imath] is equal to [imath]P(k+1)[/imath]. How would I finish this prove by induction?
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1397224
|
Show that [imath]n[/imath] does not divide [imath]2^n - 1[/imath] where [imath]n[/imath] is an integer greater than [imath]1[/imath]?
Clearly [imath]2^n - 1[/imath] is an odd integer, therefore, let [imath]n[/imath] be an odd integer and it divides [imath]2^n - 1[/imath]. We can write [imath]2^n - 1 = (2-1)(2^{n-1} + \cdots \cdots + 1) = (2^{n-1} + \cdots \cdots + 1)[/imath] From here on I don't know how to proceed?
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20487
|
For [imath]n \geq 2[/imath], show that [imath]n \nmid 2^{n}-1[/imath]
Here is a problem which i have not been able to do for quite sometime. For [imath]n \geq 2[/imath], show that [imath]n \nmid 2^{n}-1[/imath]. I have thought of proving this in two ways: One by using induction which didn't actually work. Next by Fermat's little theorem we have [imath]2^{p-1} \equiv 1 \ (\text{mod} \ p)[/imath], which actually says that for [imath]p \mid 2^{p}-2[/imath]. But i couldn't proceed more than this. Any ideas by which i can actually solve the problem?
|
1978487
|
Proof of an existance of a finite atlas for a smooth manifold
In couple of places on that site we can find the following statement: A smooth manifold [imath]M[/imath] admits a finite atlas. Look for example here Definition of manifolds with finite atlas or here Topology of a manifold, yet with no reference for a proof. I would appreciate if someone give any since I could'n found.
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92881
|
Upper bound on the number of charts needed to cover a topological manifold
If [imath]M^n[/imath] is a compact topological manifold (not necessarily with additional structure), is there an upper bound on the number of charts needed to cover [imath]M[/imath] ? Does this bound depend on the dimension of [imath]M[/imath] ? Thanks in advance... Cheers
|
1978494
|
Solve [imath]x^2 + x + 47 ≡ 0 \pmod {7^3}[/imath]
Solve [imath]x^2 + x + 47 ≡ 0 \pmod {7^3}[/imath]. I don't know how to go about doing this question. I've tried completing the square and other routes but I always seem to end up with horrible answers for [imath]x[/imath] involving surds. Any help or hints are appreciated.
|
30778
|
Is there a formula for solving the congruence equation [imath]ax^2 + bx + c=0[/imath]?
Using the quadratic formula, we have either 0, 1, or 2 solutions. I wonder if we could use it this formula for congruence? Or is there a formula to solve quadratic equation for congruence? Edit Assume that [imath]ax^2 + bx + c \equiv 0 \pmod{p}[/imath], where [imath]p[/imath] is prime with [imath](a, p) = 1[/imath], then is there a formula for this equation? Thanks,
|
1979059
|
Decomposing a permutation into multiplication of transpositions
I have a permutation in cyclic notation, for example [imath](132)[/imath], and i want to represent it as multiplication of transpositions. What is the fastest way to do it?
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499613
|
Logic for decomposing a permutation into different products composed of transpositions
I know that any permutation cycle can be decomposed into transpositions as follows: [imath](a_1,a_2...,a_n) = (a_1,a_{n-1})...(a_1,a_2)[/imath] But in my book there is an example of the following form [imath](1,2,3,4,5) = (5,4)(5,2)(2,1)(2,5)(2,3)(1,3)[/imath] I verified that it holds true. But I cant seem to figure out how the author came up with that. Also, What is the generic algorithm to produce all decompositions of permutation in terms of transpositions.
|
1978536
|
How to prove formally that these two definitions of Complex projective space are same?
Definition: [imath]\mathbb CP^n[/imath] Complex Projective Space is defined as the space of lines through origin in [imath]\mathbb C^{n+1}[/imath]. Definition: [imath]\mathbb CP^n= \frac {S^{2n+1}}{S^1}[/imath] How to prove formally that above two definitions of [imath]\mathbb C P^n[/imath] are same? Intuitively i can see that both definitions are same but i am struggling in proving formally.Any ideas?
|
1018078
|
How to prove that [imath]S^{2n+1}/S^1[/imath] is homeomorphic to [imath]\mathbb CP^n[/imath] under a given identification
We represent an element [imath](x_1,y_1,...,x_n,y_n,x_{n+1},y_{n+1})\in S^{2n+1}[/imath] as an element [imath](z_1,...,z_{n+1})\in \mathbb C^{n+1}[/imath] where [imath]z_k = x_k+iy_k[/imath]. Now I'm considering the action of [imath]S^1[/imath] on [imath]S^{2n+1}[/imath] as [imath]e^{iθ}·(z_1,...,z_{n+1}) = (e^{iθ}z_1,...,e^{iθ}z_{n+1})[/imath]. Under this action we identify any two elements as [imath](z_1,...,z_{n+1}) ∼ e^{iθ} ·(z_1,...,z_{n+1})\forall (z_1,...,z_{n+1})\in S^{2n+1}\subseteq \mathbb C^{n+1}[/imath] and [imath]e^{iθ}\in S^1[/imath]. For the resulting quotient space [imath]S^{2n+1}/S^1[/imath], I'm trying to prove that [imath]S^{2n+1}/S^1[/imath] is homeomorphic to [imath]\mathbb CP^n[/imath]. How do I prove this? Any solution will be highly appreciated.
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1975343
|
prime numbers and greatest common divisor
Question: For any prime numbers [imath]p[/imath] and [imath]q[/imath], one has [imath]\text{gcd}(p+q, q) = \text{gcd}(p,q)[/imath]. Is this statement true? Answer: True. can someone explain why? my reasoning: for any prime numbers, if an integer [imath]q[/imath] is added to it, the [imath]\text{gcd}[/imath] between [imath]p+q[/imath] and [imath]q[/imath] will still be [imath]1[/imath], since [imath]q[/imath] is still a prime number (prime = [imath]1[/imath](prime))
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644252
|
Intuition of why [imath]\gcd(a,b) = \gcd(b, a \pmod b)[/imath]?
Does anyone have a intuition or argument or sketch proof of why [imath]\gcd(a,b) = \gcd(b, a \pmod b)[/imath]? I do have a proof and I understand it, so an intuition would be more helpful. The proof that I already have: I show [imath]\gcd(a,b) \mid \gcd(b, a \pmod b)[/imath] and [imath]\gcd(b, a \pmod b) \mid \gcd(a, b)[/imath] which implies [imath]\gcd(a,b) = \gcd(b, a \pmod b)[/imath] and stuff is non-negative. WLOG [imath]a \geq b[/imath] [imath]\gcd(a,b) \mid a[/imath] [imath]\gcd(a,b) \mid b[/imath] so it divides any linear combination of a and b Since [imath]a \pmod b = a - qb[/imath] then: [imath]\gcd(b, a - qb) = bx + (a-qb)y[/imath] [imath]\gcd(b, a - qb) = bx + ay - qby [/imath] [imath]\gcd(b, a \pmod b) = b(x-qy) + ay[/imath] which is a LC of [imath]a[/imath] and [imath]b[/imath]. So [imath]\gcd(a,b) \mid \gcd(b, a \pmod b)[/imath]. Other direction is nearly identical.
|
1979510
|
A combinatorial question: Let [imath]S = \{1,2,3,4\}[/imath] and [imath]X = \{f:S\rightarrow S | x Find |X|.[/imath]
Let [imath]S = {1,2,3,4}[/imath] and [imath]X = \{f:S\rightarrow S \ | \ x<y \implies f(x)\leq f(y)\}.[/imath] Find [imath]|X|[/imath]. The solution is [imath]\binom{7}{3}[/imath], and the explanation was using stars and bars. I don't see how it could possibly be connected to a stars and bars problem. Can someone help me with this.
|
1978224
|
Let [imath]S=\{1,2,3,4,\dotsc,N\}[/imath] and [imath]X=\{ f: S \rightarrow S \mid x < y \Rightarrow f(x)\leqslant f(y)\}[/imath]. Then [imath]|X|[/imath] is equal to what?
A trick question about combinators (Stars and Bars)- Need some explanations. Let [imath]S=\{1,2,3,4,\dotsc,N\}[/imath] and [imath]X=\{ f: S \rightarrow S\mid x < y \Rightarrow f(x)\leqslant f(y)\}[/imath]. Then [imath]|X|[/imath] is equal to what? How to use Stars and Bars here? Thank you very much.
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1974447
|
Proof of this property of greatest common divisor: [imath]\gcd( n! + 1 , (n+1)! + 1 ) = 1[/imath]
Show that for any strictly positive integer [imath]n \in \mathbb{Z}^{+}[/imath] , one has [imath]\gcd( n! + 1 , (n+1)! + 1 ) = 1[/imath] The hint given to me was : Recall that the greatest commond divisor of any two integers [imath]a, b \in \mathbb{Z}[/imath] is defined as the unique non-negative integer [imath]gcd(a, b) \in \mathbb{Z}^{+}[/imath] such that [imath]\{ ar + bs \in \mathbb{Z} | r, s \in \mathbb{Z} \} = \gcd(a, b) · \mathbb{Z}[/imath] I don't understand the equation [imath]\{ ar + bs \in \mathbb{Z} | r, s \in \mathbb{Z} \} = \gcd(a, b) · \mathbb{Z}[/imath]. help?
|
25688
|
Prove that [imath]\gcd(n!+1,(n+1)!+1)=1[/imath]
I'd like to solve this one similarly to my previous question: Is this a Valid proof for [imath](2n+1,3n+1)=1[/imath]? I did find a somewhat related post that uses a different method: How to show that [imath]\gcd(n! + 1, (n + 1)! + 1) \mid n[/imath]? So how would I go about this? I can write the above like: [imath]\exists \ d \ \in \mathbb{Z}[/imath] [imath]n!+1 \equiv 0[/imath] (mod [imath]d[/imath]) [imath](n+1)!+1 \equiv 0[/imath] (mod [imath]d[/imath]) Not sure what to do next. Any hints? Thanks!
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1980432
|
Property of short exact sequences.
If [imath]0 \to L \to M \to N \to 0[/imath] is a short exact sequence of [imath]A[/imath]-modules then the following statements are equivalent: There is an isomorphism [imath]M \cong L \oplus N[/imath] under which [imath]L \to M[/imath] is given by [imath]m \mapsto (m, 0)[/imath] and [imath]M \to N[/imath] by [imath](m, n) \mapsto n[/imath]. There exists a section of [imath]M \to N[/imath]. I understand that the condition "there is an isomorphism [imath]M \cong L \oplus N[/imath]" is much weaker than the condition 1. above, so I wonder if there are some simple [imath]M, L[/imath] and [imath]N[/imath] such that [imath]M \cong L \oplus N[/imath], but there is no section of [imath]M \to N[/imath].
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1082283
|
Example of a non-splitting exact sequence [imath]0 → M → M\oplus N → N → 0[/imath]
Recently, someone stated that every short exact sequence (of, say, modules) of the form [imath]0 → M → M \oplus N → N → 0[/imath] splits. I think this is false in general because the arrow [imath]M → M \oplus N[/imath] might not be the natural inclusion. (The difference pointwise isomorphic/diagramwise isomorphic.) Maybe one can realize an arrow [imath]M\oplus N → N[/imath] with a kernel isomorphic to [imath]M[/imath], but not being an (internal) direct summand of [imath]M \oplus N[/imath]? I tried something along the lines of [imath]ℚ[/imath] and [imath]ℚ/ℤ[/imath], but I have been unsuccessful so far. Can anyone provide me with a counterexample?
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1979521
|
how to show [imath]*[/imath] gives a binary operation
Set [imath]S[/imath] be the set of all real number except -1. Define [imath]*[/imath] on [imath]S[/imath] by [imath]a*b=a+b+ab[/imath] a) Show that [imath]*[/imath] gives a binary operation on [imath]S[/imath]. Answer from my lecturer was; Suppose that [imath]a*b=a+b+ab=-1[/imath]. Then we obtain that [imath]a=-1[/imath] or [imath]b=-1[/imath]. Therefore, [imath]S[/imath] is closed under *. I not really clear how [imath]a[/imath] or [imath]b[/imath] be -1 since it is not belong to [imath]S[/imath] and how it be close? I need an explanation. Thanks in advanced.
|
1909266
|
How to prove the set [imath]S[/imath] is closed under the binary operation [imath]a*b=a+b+kab[/imath]?
Let [imath]k[/imath] be any nonzero real number and let [imath]S[/imath] be all real numbers except [imath]{-1\over k}[/imath]. Let [imath]a,b \in S[/imath]. Define * on S by [imath]a*b=a+b+kab[/imath]. Prove that [imath]S[/imath] is closed under *. All I can think to do is set [imath]a*b[/imath] equal to [imath]{-1\over k}[/imath], but I don't know where to go from there. Is that the extent of the proof?
|
1980442
|
1 = e^2π. Where did I make a mistake?
I seem to have proven that [imath]e^{2\pi} = 1[/imath]. What is my mistake? See here for proof.
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1347504
|
For which complex [imath]a,\,b,\,c[/imath] does [imath](a^b)^c=a^{bc}[/imath] hold?
Wolfram Mathematica simplifies [imath](a^b)^c[/imath] to [imath]a^{bc}[/imath] only for positive real [imath]a, b[/imath] and [imath]c[/imath]. See W|A output. I've previously been struggling to understand why does [imath]\dfrac{\log(a^b)}{\log(a)}=b[/imath] and [imath]\log(a^b)=b\log(a)[/imath] not always hold (while I was always thinking of logarithms of positive reals as [imath]\log_a(a^b)=\dfrac{\log(a^b)}{\log(a)}=\dfrac{b\log(a)}{\log(a)}=b[/imath] before I've started to self-study complexes), but then learned about branch cutting of multifunctions, so that natural logarithm can be defined to be a function by first finding a set of [imath]w[/imath] for each [imath]z[/imath], so that [imath]z=e^w[/imath] (inverse natural exponential), and then somehow (no matter how) selecting a unique solution [imath]w[/imath] from that set, so that [imath]\forall{z}\exists!{w}[/imath], yielding a function [imath]z\mapsto{w}[/imath]. This results in some sort of discontinuity where solutions are being dropped: This is done because functions are generally more easy to deal with than multifunctions. Wolfram's convention is to define [imath]\log(z)[/imath] as the inverse of [imath]e^z[/imath] such that [imath]\log(1)=0[/imath] and such that the branch cut discontinuity is at [imath]{(-\infty;0]}[/imath]. I know this is somewhat unpopular to think of natural logarithm as a single-valued function, but I am going to follow this convention (I personally think it is well-justified, at least that's what happens with [imath]\operatorname{arcsin}[/imath], [imath]\operatorname{sqrt}[/imath], etc, so in fact I am comfortable with it). Then it is clear for me why is [imath]\dfrac{\log\left((-2)^{-3}\right)}{\log(-2)}\approx{0.814319+0.841574i}\neq{-3}[/imath], while [imath]-3[/imath] perfectly satisfies the equation [imath](-2)^x=(-2)^{-3}[/imath]. Given that [imath]\log(a^b)=b\log(a)[/imath] holds for positive real [imath]a, b[/imath] only, I thought how would I then solve equations of form [imath]a^x=b[/imath], where [imath]a,b[/imath] are complexes and not necessarily positive reals, since my first step was always to rewrite everything to base [imath]e[/imath]: [imath]e^{\log(a^x)}=e^{\log(b)}[/imath] and then carry the exponent out of the logarithm: [imath]e^{x\log(a)}=e^{\log(b)}[/imath] (adding [imath]2\pi i k,\,k\in\mathbb{Z}[/imath] to any exponent and then eliminating the exponentials yields the result). So I've asked on ##math, where I've been pointed out that another way to rewrite [imath]a^b[/imath] to base [imath]e[/imath] is using the definition of the logarithm: [imath]a^b=(a)^b=(e^{\log(a)})^b=e^{b\log(a)}[/imath]. I was happy with that (and even solved a [imath](-2)^x=-3[/imath] for [imath]x[/imath] just for fun using this approach, see here) but some time later I have realized that this neither does not actually work for arbitrary complex [imath]a[/imath] and [imath]b[/imath]! [imath]1=1^{1/2}=((-1)^2)^{1/2}=(-1)^{2\cdot\frac{1}{2}}=(-1)^1=-1[/imath] The rule [imath](a^b)^c[/imath], as I was told by W|A, requires [imath]a, b, c[/imath] to be positive reals. When I pointed that out on ##math, I was told that [imath]a^b=e^{b\log(a)}[/imath] is by definition of complex exponentiation. I've checked, and Wolfram Mathematica agreed with this identity. Too good! But then I realized from these something must not be true: [imath](a^b)^c=a^{bc}[/imath] holds only for positive real [imath]a,b,c[/imath] [imath]\forall{a,b\in\mathbb{C}}:a^b=e^{b\log(a)}[/imath] [imath]\forall{a\in\mathbb{C}}:a=e^{log(a)}[/imath] The latter is the definition of the logarithm, so should be true. The second also must be true, otherwise I do not know how to solve equations. Hence: [imath]a^c=e^{c\log(a)}[/imath] is by the definition of complex exponentiation, just as I was told. Then, rewriting [imath]a[/imath] in the LHS using the definition of logarithm: [imath]a=e^{log(a)}[/imath] [imath]\left(e^{log(a)}\right)^c=e^{c\log(a)}[/imath] Now, since that for every complex [imath]b[/imath] there exists a [imath]z[/imath] such that [imath]b=\log(z)[/imath], we can rewrite [imath]\log(a)=b[/imath]: [imath](e^b)^c=e^{bc}[/imath] for all complex [imath]b,c[/imath]! So, what is that? I've made a mistake? Or is base [imath]a=e[/imath] that special? Or is in fact (I suspect) one just needs to require [imath]a[/imath] to be positive real, and [imath]b,c[/imath] are in fact irrelevant? [imath]\forall{a}\in\mathbb{R}\forall{b,c}\in\mathbb{C}:a>0\implies{(a^b)^c=a^{bc}}[/imath] Have I found a bug in Mathematica and W|A or made a huge stupid mistake leading myself to drastic misunderstanding? P. S. This is my first post at MSE, I am not a math major, just a hobbyist, so sorry if I am struggling at basics here. Also sorry for my English: it is not my native language. Edit: thank you @Andrew for your answer. [imath]\forall{a,b,c}:-\pi<\Im(b\log(a))\leq\pi\implies(a^b)^c=a^{bc}[/imath] Very clear and straightforward, works flawlessly. But it appears that though the implication is obviously true, there are more cases (read "values of [imath]a,\,b,\,c[/imath]") from which [imath](a^b)^c=a^{bc}[/imath] does follow, i.e. I found that it is true for [imath]c\in\mathbb{Z}[/imath] and arbitrary complex [imath]a,\,b[/imath], for example: [imath]\big((-2)^{-3}\big)^{-2}=(-2)^{(-3)\times(-2)};[/imath] [imath]\left(\frac{1}{(-2)^3}\right)^{-2}=(-2)^{6};[/imath] [imath]\left(-\frac{1}{8}\right)^{-2}=64;[/imath] [imath]\left(-8\right)^2=64;[/imath] [imath]64=64.[/imath] For this case, [imath]b\log(a)=-3\log(-2)=-3(\log(2)+i\pi)=-3\log(2)-3i\pi[/imath], and hence [imath]\Im(-3\log(2)-3i\pi)=-3\pi\notin{({-\pi;\pi}]}[/imath], therefore @Andrew's implication does not cover all cases. So, is there any more solutions of [imath](a^b)^c=a^{bc}[/imath]?
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1981192
|
What is the cartesian product of a nonempty set and an empty set
Let [imath]A=\emptyset[/imath] and [imath]B=[a,b][/imath] where [imath]R⊃A,B[/imath] What is [imath]A \times B [/imath] where [imath]\times[/imath] denotes the cartesian product? How can we explain this [imath]A\times B[/imath] in [imath]R^2[/imath], what does it look like?
|
937245
|
Why is the Cartesian product of [imath]A[/imath] and [imath]\varnothing[/imath] empty?
Given two sets [imath]A[/imath] and [imath]\varnothing[/imath], where [imath]\varnothing[/imath] is the empty set. The Cartesian product is defined by [imath]A \times \varnothing = \{ (x,y) \mid x \in A, y \in \varnothing \}[/imath] Why is [imath] A \times \varnothing = \varnothing[/imath]?
|
1981570
|
How to find fundamental group in next space
Question is same as title and Space is [imath]R^3\(x-axis U y-axis)[/imath].
|
189524
|
The fundamental group of some space
I have been solving some past exam questions and I came across the following question. Let [imath]X=R^3-[/imath]{(x-axis)U (y-axis)} be the complement of the x and y-axes in [imath]R^3[/imath]. Compute the fundamental group of [imath]X[/imath]. I am not really sure how to go about this. My guess is that if I could find a retract of this space whose fundamental group is known then the question is solved. I am also wondering if the Seifert-van Kampen theorem would work but I have no clue. Any help would greatly be appreciated.
|
1980684
|
Prove that [imath]\nu(n) \le \nu(2^{n}-1)[/imath] where [imath]\nu(n)[/imath] is the number of positive divisors of n
Prove that [imath]\nu(n) \le \nu(2^{n}-1)[/imath] Ok so I have very few ideas for this question. I thought if I could find the prime decomposition of [imath]2^{n}-1[/imath], I could go somewhere with the formula for [imath]\nu[/imath], but I've had little success. Any help would be much appreciated.
|
503502
|
Prove that [imath]\tau(2^n-1) \geq \tau(n)[/imath] for all positive integers [imath]n[/imath].
Prove that [imath]\tau(2^n-1) \geq \tau(n)[/imath] for all positive integers [imath]n[/imath]. Note that [imath]\tau(2^n-1)=\sum_{d|2^n-1}{1}[/imath]. I try to prove by induction. Base case: When [imath]n=1[/imath], we have [imath]\tau(2-1)=\tau(1)[/imath]. Hence, the inequality is true. Fix an [imath]n[/imath]. Assume that the inequality [imath]\tau(2^n-1) \geq \tau(n)[/imath], which is [imath]\sum_{d|2^n-1}{1} \geq \sum_{d|n}{1}[/imath] is true for [imath]n[/imath]. Note that [imath]\sum_{d|2^{n+1}-1}{1}=\sum_{d|2^{n}-1}{1}+1 \geq \sum_{d|n}{1}+1=\sum_{d|n+1}{1}[/imath] I don know whether the equality above hold or not. Can anyone guide me?
|
1980702
|
why this sequence of functions does not uniformly converges?
Show that the sequence: [imath]h_n(x)=\begin{cases}\frac{x}{n}\left(1+\frac{1}{n}\right) \mbox{ if } x=0\mbox{ or }x\notin\mathbb{Q}\\ x(b+\frac{1}{n})(1+\frac{1}{n})\mbox{ if } x\in\mathbb{Q}. \end{cases}[/imath] Does not uniformly converges at any bounded interval. This excercise was taken from the book ``Mathemathical Analysis'' of Tom Apostol.
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1981458
|
Help with counter-example on uniformly convergence.
Show that the sequence: [imath]h_n(x)=\begin{cases}\frac{x}{n}\left(1+\frac{1}{n}\right) \mbox{ if } x=0\mbox{ or }x\notin\mathbb{Q}\\ x(b+\frac{1}{n})(1+\frac{1}{n})\mbox{ if } x\in\mathbb{Q}. \end{cases}[/imath] Does not uniformly converges at any bounded interval.
|
535961
|
Prove that if [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)[/imath].
Let [imath]a,b,c \in \mathbb{Z}[/imath], prove that if [imath]\gcd(a,b)=1[/imath], then [imath]\gcd(a\cdot b,c) = \gcd(a,c)\cdot \gcd(b,c)[/imath].
|
536179
|
Prove that if [imath]\gcd(a,b)=1[/imath] then [imath]\gcd(ab,c) = \gcd(a,c) \gcd(b,c)[/imath]
Let [imath]a, b, c[/imath] be integers. Prove that if [imath]\gcd(a,b)=1[/imath] then [imath]\gcd(ab,c) = \gcd(a,c) \gcd(b,c)[/imath] First time asking here. I'm not sure what your policies are on general homework help but I truly am stuck. So far I have shown [imath]\gcd(a,c) \gcd(b,c)[/imath] as an integer combination of [imath]ab[/imath] and [imath]c[/imath]. So if I can show that [imath]\gcd(a,c) \gcd(b,c)[/imath] divides [imath]ab[/imath] and [imath]c[/imath] I can use the proof that if an integer [imath]d[/imath] is a common divisor of [imath]a[/imath] and [imath]b[/imath], and [imath]d=ax+by[/imath] for some [imath]x[/imath] and [imath]y[/imath], that [imath]d=\gcd(a,b)[/imath]. However I don't really know where to start with this. Any help would be appreciated.
|
1221283
|
Let [imath]\pi[/imath] be a prime element in [imath]\Bbb{Z}[i][/imath]. Then [imath]N(\pi)=2[/imath] or [imath]N(\pi)=p[/imath] s.t. [imath]p[/imath] is prime and [imath]p\equiv 1\pmod 4[/imath]
Let [imath]\pi[/imath] denote a prime element in [imath]\Bbb{Z}[i][/imath] such that [imath]\pi\not\in \Bbb{Z},i\Bbb{Z}[/imath]. Prove that [imath]N(\pi)=2[/imath] or [imath]N(\pi)=p[/imath], where [imath]p[/imath] is a prime number [imath]\equiv 1\pmod4[/imath]. Give a complete classification of the prime elements of [imath]\Bbb{Z}[i][/imath] using the prime numbers in [imath]\Bbb{Z}[/imath]. This exact question has been asked here but I do not understand the answers given, so I will ask it again in hope of further explanation. Here's everything I know that will probably help me with the problem: We know that [imath]\pi=a+bi[/imath] where [imath]a,b\in \Bbb{Z}\setminus \{ 0\}[/imath]. Fermat's Two Square theorem: A prime number [imath]p\equiv 1 \mod4[/imath] satisfies [imath]p=a^2+b^2[/imath] where [imath]a\neq b[/imath] and [imath]a,b\in \Bbb{N}\setminus \{0\}[/imath]. [imath]\pi[/imath] is irreducible in [imath]\Bbb{Z}[i][/imath] If [imath]\pi=xy[/imath] then [imath]\pi \mid x[/imath] or [imath]\pi \mid y[/imath] If [imath]N(\pi)=2[/imath] then [imath]a=\pm 1, b=\pm 1[/imath]. If [imath]N(\pi)=p[/imath] where prime [imath]p\equiv 1\pmod 4[/imath] then we know [imath]p=a+bi[/imath] for some [imath]a,b\in \Bbb{Z}\setminus \{0\}[/imath] by Fermat's theorem. I can't figure out where to start. I was thinking of starting with this: Let [imath]x,y\in \Bbb{Z}[i][/imath] such that [imath]x=x_1+x_2 i[/imath] and [imath]y=y_1+y_2i[/imath]. Then [imath]\pi\mid x[/imath] or [imath]\pi\mid y[/imath]. Suppose [imath]\pi\mid x[/imath]..... Can I have a hint on how to begin this?
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577152
|
Let [imath]\pi[/imath] denote a prime element in [imath]\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z[/imath]. Prove that [imath]N(\pi)=2[/imath] or [imath]N(\pi)=p[/imath], [imath]p \equiv 1 \pmod 4[/imath]
Let [imath]\pi[/imath] denote a prime element in [imath]\mathbb Z[i], \pi \notin \mathbb Z, i \mathbb Z[/imath]. Prove that [imath]N(\pi)=2[/imath] or [imath]N(\pi)=p[/imath], [imath]p \equiv 1 \pmod 4, p[/imath] is a prime. I know that [imath]\pi[/imath] is prime in [imath]\mathbb Z[i][/imath] implies it is irreducible. Also if [imath]\pi[/imath] has a prime norm, that is [imath]N(\pi) = p[/imath] then [imath]\pi[/imath] is irreducible. For any [imath]z \in \mathbb Z[i][/imath] we have [imath]N(z) = a^2 + b^2[/imath]. Since [imath]\pi \notin \mathbb Z, i \mathbb Z[/imath] both [imath]a[/imath] and [imath]b[/imath] must be non-zero in the case of [imath]\pi[/imath]. By Fermat's Two Square theorem every prime number [imath]\equiv 1 \pmod 4[/imath] can be written as a unique sum of two squares. I have some idea I should utilize this theorem here, but no success so far. However I've come to a dead end. I don't know how to go on from here. Could someone help me out ? Thanks.
|
1982243
|
Suppose [imath]p[/imath] is prime and [imath]G[/imath] is a group and [imath]|G| = p(p + 1)[/imath]. Prove [imath]G[/imath] has normal subgroup of order [imath]p[/imath] or a normal subgroup of order [imath]p + 1[/imath].
I am having a huge issue that I can't fix in my argument. I will present my argument first then I will mention why I am having troubles fixing it. Let [imath]p[/imath] be a sylow [imath]p[/imath]-subgroup of [imath]G[/imath]. Then [imath]n_p \mid (p + 1)[/imath] and [imath]n_p = 1 \pmod p[/imath]. If [imath]n_p = 1[/imath], then [imath]p[/imath] is normal and we are done. Otherwise we have [imath]n_p = kp + 1[/imath]. Since [imath](kp + 1) \mid (p + 1)[/imath] so this imply that [imath]n_p = p + 1[/imath]. Therefore there are [imath]p + 1[/imath] sylow [imath]p[/imath] each having size [imath]p[/imath], so that means each distinct [imath]p[/imath] subgroups have intersection [imath]\{e\}[/imath]. Therefore the union of all [imath]p[/imath] subgroups gives [imath](p + 1)(p - 1) + 1 = p^2[/imath] elements. Therefore this leave [imath]|G| - p^2 = p[/imath] elements remaining outside all [imath]p[/imath]-sylows. Let [imath]A[/imath] be the set of elements that don't lie in any [imath]p[/imath]-subgroup of [imath]G[/imath]. [imath]|A| = p[/imath]. Let [imath]q[/imath] be any prime that divides [imath]p + 1[/imath]. Then A must contain some element [imath]g[/imath] of order [imath]q[/imath]. Since [imath]n_p = p + 1[/imath], so this means that [imath]N_G(P) = P[/imath], therefore [imath]g[/imath] isn't in [imath]N_G(P)[/imath]. Let [imath]x[/imath] be a generator of [imath]P[/imath]. Then all powers [imath]x^i[/imath] for [imath]i \in \{1,\ldots,p - 1\}[/imath] are generators of [imath]P[/imath]. Since [imath]g[/imath] isn't in [imath]N_G(P)[/imath] so none of the [imath]x^{i}[/imath] centralizes [imath]g[/imath]. Therefore [imath]\{g,\ldots,x^{p - 1}gx^{-(p - 1)}\}[/imath] are all distinict. All of those elements have order [imath]q[/imath]. Since each of them must lie in [imath]A[/imath], so [imath]A = \{g,\ldots,x^{p - 1}gx^{-(p - 1)}\}[/imath]. I can't really now deduce that [imath]p + 1[/imath] is a power of [imath]q[/imath] (for example consider [imath]|G| = (5)(6)[/imath]). Any suggestions would be nice.
|
1953593
|
[imath]|G|=p(p+1)[/imath] for [imath]p[/imath] prime, then [imath]G[/imath] has a normal subgroup of order [imath]p[/imath] or [imath]p+1[/imath]
I am trying to solve the above question, as an application of Sylow's theorem. Let [imath]P[/imath] be the p-Sylow subgroup. Then [imath]n_p | (p+1)[/imath] and [imath]n_p \equiv 1 \pmod{p}[/imath]. If [imath]n_p =1[/imath], [imath]P[/imath] is normal and we are done, else [imath]n_p = p+1[/imath]. Now, \begin{equation} 1+n_p(p-1) = 1 + (p+1)(p-1) = p^2, \end{equation} is the total number of elements in the p-Sylow subgroups. So if [imath]n_p = p+1[/imath], that means the number of elements not in any p-Sylow subgroup is [imath]|G|-p^2=p[/imath]. If these [imath]p[/imath] elements and the identity form a subgroup then its a subgroup of the smallest prime index, so we are done. But how do I show that all the elements not in the p-Sylow subgroups form a subgroup, i.e. the subgroup generated by these elements has trivial intersection with the [imath]p-[/imath]Sylow subgroups?
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1982355
|
If n is an integer with n>1, then n has a prime factor. How do you do this using strong induction??
If [imath]n[/imath] is an integer with [imath]n>1[/imath], then [imath]n[/imath] has a prime factor. How do you do this using strong induction??
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934660
|
Proof that every number has at least one prime factor
Prove that for [imath] n \geq 2[/imath], n has at least one prime factor. I'm trying to use induction. For n = 2, 2 = 1 x 2. For n > 2, n = n x 1, where 1 is a prime factor. Is this sufficient to prove the result? I feel like I may be mistaken here.
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1981838
|
Proof-method: Show that if [imath]n[/imath] is not prime then [imath]\Bbb Z /n\Bbb Z[/imath] is not a field
The question pertains to the question in the title. That is, Show that if [imath]n[/imath] is not prime then [imath]\Bbb Z /n\Bbb Z[/imath] is not a field. I realise that this has many duplicates, but my question is rather this; Why does it not suffice to say that if [imath]n[/imath] is not prime then [imath]\lvert\Bbb Z /n\Bbb Z \rvert = n \neq p^m[/imath] for some prime [imath]p[/imath] and integer [imath]m[/imath] and so is not a field by definition? I realise this somewhat misses the reason why it's not a field, but would it suffice as an answer to this question?
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522921
|
The ring [imath]ℤ/nℤ[/imath] is a field if and only if [imath]n[/imath] is prime
Let [imath]n \in ℕ[/imath]. Show that the ring [imath]ℤ/nℤ[/imath] is a field if and only if [imath]n[/imath] is prime. Let [imath]n[/imath] prime. I need to show that if [imath]\bar{a} \neq 0[/imath] then [imath]∃\bar b: \bar{a} \cdot \bar{b} = \bar{1}[/imath]. Any hints for this ? Suppose [imath]ℤ/nℤ[/imath] is a field. Therefore: for every [imath]\bar{a} \neq 0[/imath] [imath]∃\bar b: \bar{a}\cdot \bar{b}=1[/imath]. How can I show that [imath]n[/imath] must be prime ?
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1983193
|
[imath]f\in K[t][/imath] is irreducible if and only if [imath]G[/imath] [imath](\Gamma(L:K))[/imath] acts transitively on the roots of [imath]f[/imath].
Let [imath]K[/imath] be a field, [imath]f\in K[t][/imath] separable over [imath]K[/imath], [imath]L[/imath] the splitting field of [imath]f[/imath] over [imath]K[/imath], and [imath]G=\Gamma(L:K)[/imath]. I need to prove that [imath]f[/imath] is irreducible if and only if [imath]G[/imath] acts transitively on the roots of [imath]f[/imath]. I was able to show one direction, that [imath]f[/imath] irreducible implies [imath]G[/imath] acts transitively on the roots. I'm stuck on the other direction of implication; supposing [imath]G[/imath] acts transitively on the roots of [imath]f[/imath]. Any guidance or a point in the right direction would be much appreciated!
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1561512
|
[imath]f[/imath] is irreducible [imath]\iff[/imath] [imath]G[/imath] act transitively on the roots
Let [imath]F[/imath] a field and [imath]f\in F[X][/imath] a separable polynomial. Let [imath]K_f[/imath] the splitting field of [imath]f[/imath] and [imath]G_f=\text{Gal}(K_f/F)[/imath] his group of Galois. Show that [imath]f[/imath] is irreducible [imath]\iff[/imath] [imath]G_f[/imath] act transitively on the roots on [imath]f[/imath]. My work [imath]\implies :[/imath] Let [imath]f\in F[x][/imath] irreducible, [imath]\alpha \in K_f[/imath] a root and [imath]\alpha _1,...,\alpha _n[/imath] the other roots of [imath]f[/imath] (so [imath]K_f=F(\alpha ,\alpha _1,...,\alpha _n)[/imath]). What I have to show (I think) is that for all [imath]i[/imath] there is a [imath]\sigma \in G_f[/imath] s.t. [imath]\sigma \alpha =\alpha _i.[/imath] Q1) Is it right ? Q2) Does [imath]f[/imath] is the minimal polynomial of [imath]\alpha [/imath] ? (To me it is, but my teacher said that it can be not the case, but I don't understand why). Since [imath]\alpha [/imath] and [imath]\alpha _1[/imath] has the same minimal polynomial, there is an isomorphism [imath]\sigma :\alpha \longmapsto \alpha _1[/imath]. We can observe that [imath]\sigma : F(\alpha )\longrightarrow K_f[/imath]. Now [imath]F(\alpha ,\alpha _1)/F(\alpha )[/imath] is algebraic, then we can extend [imath]\sigma [/imath] to a morphism [imath]F(\alpha ,\alpha _1)\longrightarrow K_f[/imath]. Now, we can continue like that an extend [imath]\sigma [/imath] as [imath]\psi: K_f\longrightarrow K_f[/imath] s.t. [imath]\psi(\alpha )=\alpha _1[/imath]. Q3) The fact that [imath]\psi(\alpha )=\alpha _1[/imath] comes from the fact that [imath]\psi|_{F(\alpha )}=\sigma [/imath], right ? The bijectivity is obvious since it's a field homomorphism and that [imath]\sigma (K_f)=K_f[/imath]. Q4) How conclude that [imath]G_f[/imath] act transitively on the roots ? I don't have particular problem for the converse.
|
1979186
|
Examples of Intervals
Claim. Let [imath]a[/imath] and [imath]b[/imath] be real numbers. If [imath]a\geq b[/imath] then [imath](a,b), [a,b)[/imath] and [imath](a,b][/imath] are empty set. My is question is that why?
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2000183
|
[imath](a,a]=[/imath]? Where [imath]a \in \Bbb R[/imath].
We were taught that [imath][a,a]=\{a\}[/imath] and [imath](a,a)=\emptyset [/imath], For [imath]a \in \Bbb R[/imath]. So I wonder what will be the result in the case [imath](a,a][/imath]? Let [imath]A=(a,a][/imath]. Then [imath]a \in A[/imath] and [imath]a \notin A[/imath]. This is absurd. So I guess [imath]A[/imath] is undefined.
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1983507
|
surjective function and identity
Let [imath]f:M\to N[/imath] be a function. For all [imath]Y[/imath] subset of [imath]N[/imath], [imath]f(f^{-1}(Y))[/imath] is a subset of [imath]Y[/imath]. When [imath]f[/imath] is surjective, [imath]f(f^{-1}(Y))[/imath] equals [imath]Y[/imath]. Is this true or false? I guess because of [imath]f(f^{-1}(Y)) = \mathrm{id}(Y)[/imath] this statement is true. Can give some one a better explanation?
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557764
|
Demonstrate that if [imath]f[/imath] is surjective then [imath]X = f(f^{-1}(X))[/imath]
I haven't been able to do this exercise: Let [imath]f: A \rightarrow B[/imath] be any function. [imath]f^{-1}(X)[/imath] is the inverse image of [imath]X[/imath]. Demonstrate that if [imath]f[/imath] is surjective then [imath]X = f(f^{-1}(X))[/imath] where [imath]X \subseteq B[/imath]. Since [imath]X \subseteq B[/imath], all the elements in [imath]X[/imath] belong to the codomain of [imath]f[/imath]. Since [imath]f[/imath] is surjective, it means that all elements in the codomain [imath]B[/imath] have some preimage in [imath]A[/imath]. Given that [imath]X \subseteq B[/imath], all elements in [imath]X[/imath] must also have a preimage in [imath]A[/imath]. Have [imath]\triangle = f^{-1}(X)[/imath], [imath]\triangle[/imath] is now a set containing the preimages of the elements in [imath]X[/imath]. Because of this, [imath]\triangle \subseteq A[/imath]. If we evaluate [imath]f(\triangle)[/imath], we...... nope, I don't know what I'm doing now. What do you think?
|
1983812
|
Prove for any ideal [imath]J[/imath], [imath]\operatorname{rad} J[/imath] is an ideal of [imath]A[/imath], if [imath]J[/imath] is ideal of [imath]A[/imath].
Let [imath]A[/imath] be a ring, and let [imath]J[/imath], [imath]K[/imath] be ideals of [imath]A[/imath]. The radical of [imath]J[/imath] is the set [imath]\operatorname{rad}J=\{a \in A:a^n \in J \text{ for some } n \in \mathbb{Z}\}.[/imath] Prove for any ideal [imath]J[/imath], [imath]\operatorname{rad}J[/imath] is an ideal of [imath]A[/imath]. I know that I will start the proof supposing that [imath]a^n \in J[/imath] and [imath]b^m \in J[/imath], and will work to show [imath](a+b)^{m+n} \in J.[/imath] How would I go about this? and how do I use this to prove the ideal?
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1127291
|
[imath]Rad(I)[/imath] is an ideal of [imath]R[/imath]
[imath]Rad(I)=\{a \in R | \exists n \in \mathbb{N} \text{ such that } a^n \in I\}[/imath] R is a commutative ring, I is an ideal. To show that [imath]Rad(I)[/imath] is an ideal of [imath]I[/imath], we have to show that for [imath]a,b \in Rad(I)[/imath] it stands that [imath]a-b \in Rad(I)[/imath] and for [imath]r\in R[/imath] it stands that [imath]a\cdot r\in Rad(I)[/imath], right?? When [imath]a,b \in Rad(I)[/imath] we have that [imath]a,b \in R[/imath] and [imath]a^n, b^m \in I[/imath]. [imath]a-b \in R[/imath]. But how can we show that [imath](a-b)^x \in I, x \in \mathbb{N}[/imath] ??
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1981054
|
Prove the form [imath]P_1P_2\cdots P_n + 1[/imath] where [imath]P_k[/imath] is kth prime is not necessary a prime.
Consider the sequence [imath]\{a_n\}[/imath] such that [imath]a_n = \prod_{k=1}^n P_k +1[/imath] where [imath]P_k[/imath] is kth prime. How can I prove exist composite number in this sequence without by examples like [imath]2\times3\times5\times7\times11\times13 +1=59\times 509[/imath]?
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1103357
|
How often is [imath]1+\prod_{k=1}^n p_k[/imath] not prime?
How often is [imath]1+\prod_{k=1}^n p_k[/imath] is not prime?,wkere [imath]p_k[/imath] is k'th prime consider that [imath]2\times3\times5\times7\times11\times13+1=59\times509 = 30031[/imath] is this one off or, are there infinitely many composites of the form [imath]1+\prod_{k=1}^n p_k[/imath]?
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1985376
|
How many elements of order 5 are there in [imath]S_7[/imath]
This seems relatively simple, since the lowest common multiple of the length of the disjoint cycles must be the order of the permutation. Writing this I got: [imath](5)(1)(1)[/imath] Is the only cycle in [imath]S_7[/imath] whose lowest common multiple that I could find. The number of ways to write this are [imath]\frac{7!}{5}[/imath], but my book is instead doing this: [imath]\frac{7!}{2!\cdot 5}[/imath] Why is this the right answer for the combinations? And why dividing by [imath]2![/imath]? It seems arbitrary to me, could anyone englighten me on this?
|
301791
|
Elements of order 5 in [imath]S_7[/imath], odd permutations of order 4 in [imath]S_4[/imath], and find a specific permutation in [imath]S_7[/imath]
[imath](1)[/imath] How many elements of order [imath]5[/imath] are in [imath]S_7[/imath](symmetric group)? [imath](2)[/imath] How many odd permutation of order [imath]4[/imath] does [imath]S_4[/imath] have? [imath](3)[/imath] If [imath]\beta \in S_7[/imath] and [imath]\beta^4=(2143567)[/imath] then find [imath]\beta[/imath] Give me some hints for these three questions.please help me. Thanks in advance.
|
1985595
|
Strong Induction - Any integer greater than 1 is divisible by a prime number
Prove any integer greater than 1 is divisible by a prime number(strong induction) Let P(n) be an integer divisible by a prime number, where n>=2. Base Case: Show true for [imath]P(2)[/imath] [imath]P(2)[/imath] is true because 2 is a prime number itself. Inductive Case: Show that for integers [imath]k>2[/imath], if true for all [imath]i[/imath] with [imath]2<=i<k[/imath], then it's true for k. Suppose that for all integers i with [imath]2<=i<k[/imath], i is divisible by a prime number (inductive hypothesis) If someone could explain how the inductive case above works that would be appreciated
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1039424
|
Use mathematical induction to prove that any integer [imath]n\ge2[/imath] is either a prime or a product of primes.
Use strong mathematical induction to prove that any integer [imath]n\ge2[/imath] is either a prime or a product of primes. I know the steps of weak mathematical induction... Basis step: [imath]p(n)[/imath] for [imath]n=1[/imath] or any arbitrary [imath]n_0[/imath] ... show that it is true Inductive hypothesis: [imath]p(n)[/imath] for [imath]n=k[/imath] ... Assume that it is true for [imath]n=k[/imath] Inductive step: [imath]P(n)[/imath] for [imath]n=k+1[/imath] ... Show that this is true for [imath]n=k+1[/imath]
|
1986033
|
For all natural numbers [imath]x[/imath] and [imath]y[/imath] if [imath]S(x)=S(y)[/imath] then [imath]x=y[/imath].
Definition. [imath]S(x):=x\cup[/imath]{[imath]x[/imath]}. Theorem1. For all natural numbers [imath]x[/imath] and [imath]y[/imath], if [imath]x\in y[/imath] then [imath]x\subseteq y[/imath]. Theorem2. For all natural numbers [imath]x[/imath] and [imath]y[/imath] if [imath]S(x)=S(y)[/imath] then [imath]x=y[/imath]. Note that we will use Theorem1 for proof of Theorem2. Proof of Theorem2. Assume not. i.e., if [imath]S(x)=S(y)[/imath] then [imath]x\neq y[/imath]. We know that [imath]x\in x\cup[/imath] {[imath]x[/imath]}. Also, by the assumption [imath]x\in y\cup[/imath] {[imath]y[/imath]}. Hence, either [imath]x\in y[/imath] or [imath]x\in[/imath] {[imath]y[/imath]}. So,... Can you help? Answer: Show that for any natural numbers, if [imath]S(x)=S(y)[/imath] then [imath]x=y[/imath].
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1976631
|
Show that for any natural numbers, if [imath]S(x)=S(y)[/imath] then [imath]x=y[/imath].
Definition. [imath]0=\emptyset[/imath]. Definition. [imath]S(x):=x\cup\{x\}[/imath], [imath]S(y):=y\cup\{y\}[/imath]. Definition. [imath]S(x):=x\cup[/imath]{[imath]x[/imath]}. Theorem1. For all natural numbers [imath]x[/imath] and [imath]y[/imath], if [imath]x\in y[/imath] then [imath]x\subseteq y[/imath]. Theorem2. For all natural numbers [imath]x[/imath] and [imath]y[/imath] if [imath]S(x)=S(y)[/imath] then [imath]x=y[/imath]. Note that we will use Theorem1 for proof of Theorem2.
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1985948
|
GRE Subject question [imath]\log x=c x^4[/imath]
For what positive value of [imath]c[/imath] does the equation [imath]\log x=cx^4[/imath] have exactly one real solution for [imath]x[/imath]?. Thank you.
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523319
|
[imath]\log x =Cx^4[/imath] has only one root. Find C
[imath]\log x =Cx^4[/imath] has only one root. Find C. I don't know how to solve this problem. Do you take derivative on both sides? I am thinking C equals 0. Am I correct on that?
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1987286
|
for triamgle [imath]ABC[/imath] prove that [imath]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2[/imath]
Let [imath] a,b,c[/imath] be lengths of sides triangle [imath] ABC[/imath]. Prove that: [imath]\displaystyle \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2[/imath] [imath]\bf{My\; Try::}[/imath] For a [imath]\triangle [/imath] with sides [imath]a,b,c[/imath] [imath]a+b>c[/imath] and [imath]b+c>a[/imath] and [imath]a+b>c[/imath] Now let [imath]x=a+b-c>0[/imath] and [imath]y=b+c-a>0[/imath] and [imath]z=c+a-b>0[/imath] So [imath]\displaystyle 2a=x+z[/imath] and [imath]2b=x+y[/imath] and [imath]2c=y+z[/imath] So equality convert into [imath]\frac{x+z}{x+2y+z}+\frac{x+y}{x+y+2z}+\frac{y+z}{2x+y+z}<2[/imath] How can i solve above inequality, Help required, Thanks
|
1903775
|
Triangle Inequality [imath]\sum_{\text{cyc}}\frac{a}{b+c}<2[/imath]
Let [imath]a,b,c[/imath] be the lengths of the sides of a triangle. Prove [imath]\sum_{\text{cyc}}\frac{a}{b+c}<2[/imath]
|
1987452
|
How many distinct values can be obtained for the expression [imath]1/2/3/5/7/11/13/17[/imath]
What is a simple way to solve this problem? I can solve the problem by counting all possibilities. How many distinct values can be obtained for the expression [imath]1/2/3/5/7/11/13/17[/imath] if an unlimited number of parentheses may be placed in the expression?
|
1795909
|
how many distinct values does it have?
I solved this problem by manually adding parentheses and counting them, and got correct answer of 32. Is there a simple to find the answer? Thanks. The value of the expression [imath]1÷2÷3÷5÷7÷11÷13[/imath] can be altered by including parentheses. If we are allowed to place as many parentheses as we want, how many distinct values can be obtained for this expression?
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1987223
|
Prove that [imath]a^2[/imath] when divided by [imath]5[/imath] cannot have a remainder of [imath]3[/imath].
For any natural number [imath]a[/imath], prove that [imath]a^2[/imath] when divided by [imath]5[/imath] cannot have a remainder of [imath]3[/imath]. (Hint: What are the possible values of the remainder when [imath]a[/imath] is divided by [imath]5[/imath]? Using the hint, I found that the possible values of the remainder are [imath]1, 2, 3, 4[/imath], and [imath]0[/imath], and [imath]3[/imath] is only a remainder when the last digit of [imath]a[/imath] is [imath]3[/imath] or [imath]8[/imath]. But I'm not sure how to explain this in a proof, and I don't know how to extend it to [imath]a^2[/imath]. Any help would be much appreciated!
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1975116
|
Prove no solution to [imath]x^2\equiv3\bmod5[/imath]
Show that there is no solution to the congruence [imath]x^2\equiv3\bmod5[/imath]. I'm not sure if the fact that the only numbers divisible by [imath]5[/imath] are those ending in [imath]0[/imath] or [imath]5[/imath] will help…
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1986980
|
Prove that the series [imath] \sum_{n=1}^{\infty} \frac{\sin n}{n} [/imath] is convergent
I know I can use Dirichlet's Convergence Test by writing the series as [imath]\sum_{n=1}^{\infty} \frac{1}{n}\cdot\sin(n)[/imath] The [imath]\frac{1}{n}[/imath] is easy but how can I prove that [imath]\sin(n)[/imath] is bounded ?
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1046026
|
Prove [imath]\sum_{k=1}^{\infty} \frac{\sin(kx)}{k} [/imath] converges
How to prove [imath]\sum_{k=1}^{\infty} \frac{\sin(kx)}{k}[/imath] converges without using integral test?
|
1987807
|
Show that group of order 105 with normal 3-sylow is abelian.
I'm having trouble solving this question. What i've managed so far is. Let G be a group and [imath]|G|=105=3*5*7[/imath], [imath]n_p=|Syl_p(G)|[/imath] and let [imath]P\in Syl_3(G)[/imath], [imath]Q\in Syl_5(G)[/imath] and [imath]R\in Syl_7(G)[/imath]. We have that [imath]n_3=1 \text{ (from assumption.)}\\ n_5=1|21,\\ n_7=1|15.[/imath] Since the 7-Sylow and 5-Sylow groups are generated by a single element if two 5-Sylow or 7-Sylow must have a trivial intersection, otherwise they are the same group. By a counting the elements we find that at least two of the p-Sylow in G must be normal. This reduces the problem into three cases as follows. Case 1: [imath]n_7= 15[/imath] and [imath]n_5=1[/imath]: Case 2: [imath]n_7= 1[/imath] and [imath]n_5=21[/imath]: Case 3: [imath]n_7= 1[/imath] and [imath]n_5=1[/imath]: Any suggestions on how to proceed from here?
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257016
|
Suppose [imath]|G| = 105[/imath]. Show that it is abelian if it has a normal [imath]3[/imath]-Sylow subgroup.
Since [imath]105 = 3\cdot5\cdot7[/imath], another result implies that there must exist a normal [imath]7[/imath]-Sylow subgroup. Does this help prove that [imath]G[/imath] is abelian? Also, can this result be generalized (If [imath]|G| = p\cdot q\cdot r[/imath], and [imath]G[/imath] has a normal [imath]p[/imath]-Sylow subgroup, then [imath]G[/imath] is abelian (where [imath]p<q<r[/imath]).
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1986878
|
Prove that [imath]\cos^2(z)+\sin^2(z)=1[/imath]
Can anybody show me how to prove that [imath]\cos^2(z)+\sin^2(z)=1[/imath] for [imath]z\in \mathbb{C}[/imath]? I can prove it for the case where [imath]z[/imath] is real, but can't seem to find a way to prove it for complex numbers in general.
|
1598436
|
Is [imath]\sin^2(z) + \cos^2(z)=1[/imath] still true for [imath]z \in \Bbb{C}[/imath]?
Is [imath]\sin^2(z) + \cos^2(z)=1[/imath] still true for all [imath]z \in \Bbb{C}[/imath]? I've tried rewriting it using complex definitions of [imath]\sin[/imath] and [imath]\cos[/imath], and I don't see why it wouldn't, but the text I'm reading asks this question as if it shouldn't hold?
|
1988163
|
Find and justify the Supremum of the following set
Find the supremum of the following set: [imath]A:=\left\{{(n-1)\over(2n+3)} : n \in\mathbb{N}\right\}[/imath] So I have as my answer that sup(A) = ½ but need to justify. We were taught to justify in two steps, first show our answer is an upper bound for the set then show it is a least upper bound. For step 1: if a is a member of A then a = (n-1)/(2n+3) for some natural number n and (n-1)/(2n+3) Step 2 is where I am uncertain. I think I need to show that for any y<½ there exists a member of the set between y and ½ but not sure how to do this. Thanks for all the answers so far!
|
1987690
|
Find the supremum of the following set
Find the supremum of the following set: [imath]A:=\left\{{(n-1)\over(2n+3)} : n \in\mathbb{N}\right\}[/imath] So I have as my answer that [imath]\sup(A) = \frac12[/imath] but need to justify. We were taught to justify in two steps, first show our answer is an upper bound for the set then show it is a least upper bound. For step [imath]1[/imath]: if [imath]a[/imath] is a member of [imath]A[/imath] then [imath]a = (n-1)/(2n+3)[/imath] for some natural number [imath]n[/imath] and [imath]\frac{n-1}{2n+3} \leq \frac{n}{2n} = \frac12[/imath] so [imath]\frac12[/imath] is an upper bound for the set Step 2 is where I am uncertain. I think I need to show that for any [imath]y<\frac12[/imath] there exists a member of the set between [imath]y[/imath] and [imath]\frac12[/imath] but not sure how to do this. Thanks for all the answers so far!
|
1988636
|
Least fixed point of a function
Consider the function [imath]f:\mathcal{P}(\{0,1\}^*)\to\mathcal{P}(\{0,1\}^*)[/imath] defined as: [imath]f(X)=X\cup\{w01:w\in X\}\cup\{\epsilon\}[/imath] Where [imath]\epsilon[/imath] denotes a word of length 0. Find the least fixed point of [imath]f[/imath] on [imath]\mathcal{P}(\{0,1\}^*)[/imath] ordered by inclusion. I cannot convince myself that this function even has a fixed point, since feeding any set of words to it would cause it to output twice as many words (the original input, plus every word from the original with 01 appended to the end). Is my reasoning incorrect?
|
1986549
|
Confused about language and words in logic
I am very confused about how to interpret the following. If we write [imath]\{0,1\}^{+}[/imath]=[imath]\cup_{n \in \mathbb{N}} \{0,1\}^{n}[/imath] then I am told that elements of [imath]\{0,1\}^{+}[/imath] are called words and subsists of it are called languages, Then I was asked, if we consider [imath]g: P( \{0,1\}^{+}) \to P( \{0,1\}^{+})[/imath] defined as [imath]g(X)=X \cup \{w01:w \in X\} \cup \{\tau\}[/imath] where [imath]\tau[/imath] represents the word of zero length, then find the least fixed point of g ordered by inclusion. I have been trying to make sense of it but cant make progress. I know that a fixed point is an element [imath]X[/imath] such that [imath]g(X)=X[/imath] I know that since we want the least, the fixed point has to be less then any other fixed points ordered by inclusion to. Can anyone help to make sense of this? Thank you
|
1987949
|
What is the result of this equation?
Is there any sensitive equation solver which will not show the result as approximately [imath]0[/imath] for this equation: [imath]\cos x=e^{-\Large\frac{3}{10^{45}}}[/imath] or how can I calculate it?
|
1987534
|
How can I take a sensitive result for this equation?
Is there any sensitive equation solver which will not show the result as approximately 0 for this equation: [imath]cosx=e^{-\frac{x}{10^{45}}}[/imath] being x is variable, or how can I calculate it?
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1987081
|
Show that if [imath]G[/imath] is connected and it has two spanning trees with differente edges, then [imath]G[/imath] has a spanning graph eulerian
i need help with this: Show that if [imath]G[/imath] is connected and it has two spanning trees with differente edges, then [imath]G[/imath] has a spanning subgraph eulerian spanning tree: spanning subgraph of [imath]G[/imath] and it's a tree I really need to solve it, can you give a clue, please? thansk!
|
1149059
|
Two disjoint spanning trees, spanning subgraph with all even degrees
Show that if a graph has two edge-disjoint spanning trees then it has a connected, spanning subgraph with all degrees even. I start by looking at the union of the two spanning trees. I know it has [imath]2n-2[/imath] edges and is connected. But I don't know where to go from there. Perhaps it has to do with Euler circuits? If I prove the union has a Euler circuit I can just take the circuit and then the degrees are all [imath]2[/imath].
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1983260
|
Exterior derivative of wedge product
How can I show that [imath] \mathrm d(a\wedge b)=(\mathrm d\,a)\wedge b + (-1)^{q}a\wedge(\mathrm d\,b) [/imath] for a [imath]q[/imath]-form [imath]a[/imath] and an [imath]r[/imath]-form [imath]b[/imath]?
|
1164823
|
Differential Geometry-Wedge product
How can we prove the following relation for differentiating the wedge product of a p-form [imath]\alpha_p[/imath] and a q-form $\beta_q[imath]$d(\alpha_p\wedge\beta _q)=d\alpha_p\wedge\beta_q+(-1)^{p}\alpha_p\wedge d\beta_q[/imath]
|
1986810
|
proving that formulas are theorems using deduction theorem
I'm working in [imath]L_0[/imath] I know I have to use the three axioms and inference rule. However, I'm finding it difficult to prove that [imath](\neg \alpha \rightarrow (\alpha \rightarrow \beta))[/imath] is a theorem. I find it very fiddly and unintuitive - as if I just have to sort of plug in random formulae and combine them to get what I need. Any advice would be appreciated.
|
1928232
|
How to prove [imath]\vdash\neg P\to (P\to Q)[/imath]?
I am trying to prove, [imath]\vdash\neg P\to (P\to Q)[/imath] where [imath]P[/imath] and [imath]Q[/imath] are arbitrary but otherwise fixed formulas. The only axioms and rule of inference that I can use are, [imath]\color{crimson}{\text{Axiom 1.}}\ P\to (Q\to P)[/imath] [imath]\color{crimson}{\text{Axiom 2.}}\ (S\to (P\to Q))\to((S\to P)\to (S\to Q))[/imath] [imath]\color{crimson}{\text{Axiom 3.}}\ (\neg Q\to\neg P)\to(P\to Q)[/imath] [imath]\color{crimson}{\text{Rule of Inference.}}[/imath] Modus Ponens. I know that in Angelo Margaris's book First Order Mathematical Logic there is a proof (page 53) but as I have mentioned here, the dot notation seems very confusing and translating them to the nested parentheses notation seems very difficult to me. So, I am trying on my own to prove the result. Can anyone help?
|
1988950
|
Prob. 7, Chap. 3 in Baby Rudin: If [imath]a_n \geq 0[/imath], then how does convergence of [imath]\sum a_n[/imath] imply convergence of [imath]\sum \frac{\sqrt{a_n}}{n}[/imath]?
Here's Prob. 7 in the Exercises of Chapter 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: Prove that the convergence of [imath]\sum a_n[/imath] implies the convergence of [imath] \sum \frac{\sqrt{a_n}}{n}[/imath] if [imath]a_n \geq 0[/imath]. How to show this? I have no clue! Can anybody here please be of any help? Is this result a special case (or application) of a more general result about infinite series?
|
1352843
|
Series convergence proof review (Baby Rudin)
Ch.3 #7. Prove that the convergence of [imath]\sum a_n[/imath] implies the convergence of [imath]\sum \frac{\sqrt{a_n}}{n},[/imath] if [imath]a_n \geq 0[/imath]. My attempt. If [imath]\sum a_n[/imath] is convergent, then by the root test, [imath]\lim_{n \rightarrow \infty} sup \sqrt[n]{|a_n|} < 1[/imath]. This means [imath]|a_n|^{1/n} < 1[/imath]. Because [imath]a_n \geq 0[/imath], this means [imath]a_n ^{1/n} < 1[/imath]. Now apply the root test to [imath]\sum \frac{\sqrt{a_n}}{n}[/imath] by considering [imath]\lim_{n \to \infty} sup \sqrt[n]{|\frac{\sqrt{a_n}}{n}|}.[/imath] [imath]\sqrt[n]{|\frac{\sqrt{a_n}}{n}|} = \frac{\sqrt{{a_n}^{\frac{1}{n}}}}{\sqrt[n]n}.[/imath] Because [imath]a_n ^{1/n} < 1[/imath], certainly the numerator on the left-hand side of the equation above is [imath]<1[/imath]. And because [imath]n[/imath] tends to [imath]\infty[/imath], and [imath]n > 1 \rightarrow \sqrt[n]n > 1,[/imath] the denominator is [imath]>1[/imath]. Therefore the limit is [imath]<1[/imath] and the root test implies that [imath]\sum \frac{\sqrt{a_n}}{n}[/imath] converges. If there are any errors in my proof, I think they come from 1) my assumption of the converse of the root test theorem and 2) my misunderstanding of [imath]\lim_{n\to\infty} sup[/imath]. Thanks in advance for your help.
|
1989580
|
Describe a partition of [imath]\mathbb{N}[/imath] that divides [imath]\mathbb{N}[/imath] into [imath]\aleph_0[/imath] countably infinite subsets.
Does anyone have any hints on how to describe the following partition? Describe a partition of [imath]\mathbb{N}[/imath] that divides [imath]\mathbb{N}[/imath] into [imath]\aleph_0[/imath] countably infinite subsets.
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961294
|
Infinite partition of [imath]\mathbb N[/imath] by infinite subsets
I am looking for an explicit partition of [imath]\mathbb N[/imath] with the following condition: [imath]\mathbb N=\bigsqcup_{i\in\mathbb N}A_i[/imath] where all the [imath]A_i[/imath]'s are infinite. What I mean by explicit is a formula for each [imath]A_i[/imath] (I will have to do computation with the elements of [imath]A_i[/imath]). So, I can not use the choice axiom. If someone has an idea to build such a partition...
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1989775
|
How to prove the determinant of this matrix invoving two sets of numbers is greater than zero?
Suppose [imath]0<a_1<a_2<\cdots <a_n[/imath],[imath]\lambda_1<\lambda_2<\cdots<\lambda_n[/imath]and \begin{align*} A=\left({a_i}^{\lambda_j}\right)= \begin{pmatrix} {a_1}^{\lambda_1}&{a_1}^{\lambda_2}&\ldots &{a_1}^{\lambda_n}\\ {a_2}^{\lambda_1}&{a_2}^{\lambda_2}&\ldots &{a_2}^{\lambda_n}\\ \vdots &\vdots&\ldots&\vdots\\ {a_n}^{\lambda_1}&{a_n}^{\lambda_2}&\ldots &{a_n}^{\lambda_n} \end{pmatrix} \end{align*} Prove that the determinant of A is greater than zero,[imath]\mathrm{i,e.}[/imath] [imath]\det A>0[/imath]. It is easy to check that the assertion is true when [imath]n=2[/imath],for when [imath]n=2[/imath], \begin{align*} \det A&=\begin{vmatrix} {a_1}^{\lambda_1}&{a_1}^{\lambda_2}\\ {a_2}^{\lambda_1}&{a_2}^{\lambda_2} \end{vmatrix} ={a_1}^{\lambda_1}{a_2}^{\lambda_2}-{a_1}^{\lambda_2}{a_2}^{\lambda_1}\\ &=\exp\{\lambda_1\ln a_1+\lambda_2 \ln a_2\}-\exp\{\lambda_2\ln a_1+\lambda_1 \ln a_2\} \end{align*} Since [imath](\lambda_1\ln a_1+\lambda_2 \ln a_2)-(\lambda_2\ln a_1+\lambda_1 \ln a_2)=(\lambda_2-\lambda_1) (\ln a_2-\ln a_1)>0[/imath],then the case for [imath]n=2[/imath]holds.While about the general case for [imath]n>2[/imath],I have no idea. Can anyone help me?It's a homework.
|
150131
|
Positivity of a determinant
I'm stuck to prove the following exercise : Given real numbers [imath]x_1,\ldots,x_n[/imath] and [imath]y_1,\ldots,y_n[/imath], show that [imath] \det(e^{\large{x_iy_j}})_{i,j=1}^n>0 [/imath] provided that [imath]x_1<\cdots<x_n[/imath] and [imath]y_1<\cdots<y_n[/imath]. Any idea ?
|
1989822
|
How can I show that [imath]\{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \}[/imath] is connected?
I am having a lot of trouble showing that this set is connected. [imath]\{(x,y,z) \in \mathbb{R}^3 : x^2 + y^2 + z^2 = 1 \}[/imath] This set contains all of the points on the boundary of a sphere (or the surface, if you like) with a empty core/inside. What I have tried doing is proving path connectedness, by picking two arbitrary points a and b and defining a map [imath] [0, 1] [/imath] that maps on S, but I am not sure how to construct such a path since I need to go around the circle. Though a technique I have also tried is going through the sphere and simply projecting every vector on the line onto the sphere (by dividing every point on the line with its norm resulting in a unit vector) - however I am not looking for "tricks" and a rather more robust solution - something that I can learn from. :) Any help appreciated!
|
364379
|
Proving connectedness of the [imath]n[/imath]-sphere
My homework assignment contains the following question: Prove that [imath]S^n = \{x \mid x \in \mathbb R^{n+1}, d(x,0)=1\}[/imath] is connected. Can you give me a hint please? What I can do is nothing..
|
1990167
|
How is the limit infimum of sets different from the limit infimum of a sequence of real numbers?
I am trying to understand whether the limit infimum of a set is related or a generalization of the limit infimum of a sequence of real numbers. Suppose that [imath]X_n[/imath] is a sequence of sets, and so the limit infimum of [imath]X_n[/imath] is defined as saying that from some stage onwards, all of the [imath]X_n[/imath] occur. Concretely: [imath] \liminf_{n \to \infty}X_n = \bigcup_{n=1}^{\infty}\bigcap_{k\geq n}X_k [/imath] Now, suppose that we let [imath]X_n(\omega) = (-1)^n[/imath] and let [imath]X(\omega) = 1[/imath]. Then, we have that: [imath] \liminf_{n \to \infty}X_n = -1 [/imath] Here it appears that after some time, only [imath]-1[/imath] occurs. But, I know that both [imath]1[/imath] and [imath]-1[/imath] alternate, and so there shouldn't be a time after which [imath]-1[/imath] only occurs. I am wondering if I am somehow confusing something here between the limit of sets and the limit of a sequence? It also seems that the definition of the limit infimum/supremum of sets has no direct connection to that of a sequence and is an arbitrary definition. Am I wrong here?
|
1605312
|
How to prove the following properties of infimum and supremum involving the union and intersection of the sets [imath]A_k[/imath]
I am reading a book on probability theory and I have troubles understanding why the following holds [imath] \sup_{k \ge n} A_k = \bigcup_{k\ge n} A_k [/imath] [imath] \inf_{k \ge n} A_k = \bigcap_{k\ge n} A_k [/imath] I know how to define infimum and supremum, but I have troubles proving the above expressions. Unfortunately, the book states those expressions as a definition without proving them. A graphical explanation in terms of a Venn diagram will be very helpful. Thank you in advance
|
1990262
|
equation of lines which intersect another line at given angle.
Find the equation of [imath]2[/imath] lines through the origin which intersect the line [imath]\displaystyle \frac{x-3}{2} = \frac{y-3}{1} = \frac{z}{1}[/imath] at an angle of [imath]\displaystyle \frac{\pi}{3}[/imath] [imath]\bf{My\; Try::}[/imath] Let equation of line be [imath]\displaystyle \frac{x-0}{a} = \frac{y-0}{b} = \frac{z-0}{c}[/imath] Where [imath]<a,b,c>[/imath] be the direction cosine of line parallell to that line. and Given Line is [imath]\displaystyle \frac{x-3}{2} = \frac{y-3}{1} = \frac{z}{1}[/imath] Where [imath]<2,1,1>[/imath] be the direction cosine of line parallell to that line and Given [imath]\displaystyle \frac{\pi}{3}[/imath] be the angle between [imath]<a,b,c>[/imath] and [imath]<2,1,1>[/imath] So [imath]\displaystyle \cos \frac{\pi}{3} = \frac{2a+b+c}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{6}}\Rightarrow \frac{1}{2}=\frac{2a+b+c}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{6}}[/imath] Now how can i solve it, Help required, Thanks
|
1530416
|
Equations of the two lines through the origin which intersect the line [imath]\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}[/imath] at an angle of [imath]\frac{\pi}{3}[/imath]
Find the equations of the two lines through the origin which intersect the line [imath]\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}[/imath] at an angle of [imath]\frac{\pi}{3}[/imath]. Let the direction ratios of the two required lines be [imath](a_1,b_1,c_1)[/imath] and [imath](a_2,b_2,c_2)[/imath]. Therefore the two equations are [imath]\frac{x-0}{a_1}=\frac{y-0}{b_1}=\frac{z-0}{c_1}[/imath] and [imath]\frac{x-0}{a_2}=\frac{y-0}{b_2}=\frac{z-0}{c_2}[/imath] As these lines are making an angle of [imath]\frac{\pi}{3}[/imath] with [imath]\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}[/imath]. So,[imath]\cos\frac{\pi}{3}=\frac{2a_1+b_1+c_1}{\sqrt6\sqrt{a_1^2+b_1^2+c_1^2}}[/imath] And [imath]\cos\frac{\pi}{3}=\frac{2a_2+b_2+c_2}{\sqrt6\sqrt{a_2^2+b_2^2+c_2^2}}[/imath] But i am stuck here.How i can solve three variables with one equation.The book gives answer as [imath]\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}[/imath] and [imath]\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}[/imath] Please help me.
|
1990194
|
Integration with Fractional Part (Calculus) [imath]\lim\limits_{n\to \infty} \int\limits_0^1{e^{\{nx\}}\cdot x^{2016}dx} [/imath]
[imath]\lim\limits_{n\to \infty} \int\limits_0^1{e^{\{nx\}}\cdot x^{2016}dx} = \text{ } ? [/imath] [imath]\{nx\}[/imath] - Fractional Part Number. Example: [imath]\{-2/3\}=\{4/3\}=1/3[/imath] My solve: [imath]\lim\limits_{n\to \infty} \frac{1}{n^{2017}}\int\limits_0^n{e^{\{t\}}\cdot t^{2016}dt} , t=nx[/imath] Please give me a clue what to do next
|
1847922
|
Compute: [imath]\lim\limits_{n\to+\infty}\int\limits_{0}^1 e^{\{nx\}}x^{100}dx[/imath]
Can anyone help to compute this limit, please? [imath]\{nx\}[/imath] denotes the fractional part of [imath]nx[/imath]. Tried to replace [imath]nx[/imath] by another variable and write an entire integral as a sum of integrals: [imath]\sum\limits_{i=0}^{n-1}\int\limits_{i}^{i+1} [/imath] but it seems not to work out.
|
1989691
|
Proof that [imath]2^k+3^m=5^n[/imath] has only a finite number of solutions?
[imath]2^1+3^1=5^1[/imath] [imath]2^4+3^2=5^2[/imath] In these cases a power of [imath]2[/imath] plus a power of [imath]3[/imath] is a power of [imath]5[/imath]. If I understand the abc conjecture correctly, this is only finitely often the case. Question 1: Is there an easy proof for this? Question 2: Are there other known solutions of [imath]2^k+3^m=5^n[/imath]?
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348538
|
Finding all integer solutions of [imath]5^x+7^y=2^z[/imath]
Find all integers [imath]x,y,z[/imath] such that [imath]5^x+7^y=2^z[/imath]. This one comes from an online contest that I arranged some years ago, and I can assure that a completely elementary solution exists.
|
1989408
|
Help with proving that [imath]\phi(n)[/imath] [imath]\ge[/imath] [imath]{\sqrt n} \over 2[/imath]
So far I have [imath]{{\phi(n)^2} \over n} = n \prod_{i = 1}^k \left(1 - {1 \over p_i}\right)^2[/imath] We know that [imath]\left(1 - \frac{1}{p_i}\right) \geq \frac{1}{2}[/imath] So [imath]n \prod_{i = 1}^k \left(1 - {1 \over p_i}\right)^2 \geq \frac{n}{4^k} = \frac{n}{2^{2k}}[/imath] Want to show that [imath]\frac{n}{2^{2k}} \geq \frac{1}{2}[/imath] [imath]n \geq 2^{2k - 1}[/imath] Got stuck here. Any help to prove this? EDIT: Ok so I looked at the answer in the thread provided and I don't understand how [imath]\frac{\phi(n)^2}{n}= \prod_{p|n \ \text{prime}} \frac{(p^{a_p-1}(p-1))^2}{p^{a_p}}[/imath] There's a factor of n missing since the [imath]\phi[/imath] function is being squared
|
527946
|
Prove that [imath]\phi(n) \geq \sqrt{n}/2[/imath]
So I'm trying to prove the following two inequalities: [imath]\frac{\sqrt{n}}{2} \leq \phi(n) \leq n.[/imath] The upper bound we get from simply noting that [imath]\phi(n) = n \prod_{p | n}\left( 1 - \frac{1}{p}\right)[/imath] and the fact that [imath](1 - \frac{1}{p}) \leq 1[/imath]. But how can we get the lower bound? I tried using the same expression for [imath]\phi(n)[/imath] but it seems to not really give me the inequality. Can you help?
|
1990934
|
Show that the following matrix has positive determinant
The following is an interview question. Show that any matrix of the form [imath]\begin{pmatrix} 1 & 1/2 \\ 1/2 & 1/3\end{pmatrix}[/imath] [imath]\begin{pmatrix} 1 & 1/2 & 1/3 \\ 1/2 & 1/3 & 1/4 \\ 1/3 & 1/4 & 1/5 \end{pmatrix}[/imath] etc. is positive definite. By induction, it suffices to show that they have positive determinant. But how? Things I've considered: Use multilinearity/row operations Factor into [imath]A^t A[/imath] Do the Cholesky decomposition and notice a pattern Guess the minimimal polynomial and observe that all the roots are positive I can't seem to figure it out though. I'm looking for a solution that you could plausibly come up with in an interview Note: I have verified for sizes up to [imath]13\times13[/imath] that the determinant is positive using Python, but the matrix starts getting pretty close to singular as it increases in size. (Of course, you wouldn't know that in an interview)
|
783164
|
Prove positive definiteness
I want to prove that the matrix [imath]\begin{pmatrix} 1 &\cfrac{1}{2} &\cfrac{1}{3} &\cdots &\cfrac{1}{n} \\ \cfrac{1}{2} &\cfrac{1}{3} &\cfrac{1}{4} &\cdots &\cfrac{1}{n+1} \\ &\vdots &&\ddots &\vdots \\ \cfrac{1}{n} &\cfrac{1}{n+1} &\cfrac{1}{n+2} &\cdots &\cfrac{1}{2n-1} \end{pmatrix}[/imath] is positive definite. Using mathematical induction, I only need to show that its determinant is positive. But I can't find the way out.
|
1990916
|
Equivalent metrics on [imath]\Bbb N[/imath]
Show that [imath](\Bbb N,d_1)[/imath] and [imath](\Bbb N,d_2)[/imath] are equivalent where [imath]d_1[/imath] is the discrete metric and [imath]d_3(m,n)=|\frac{1}{m}-\frac{1}{n}|[/imath]. My account In order to show that [imath](\Bbb N,d_1)[/imath] and [imath](\Bbb N,d_2)[/imath] are equivalent,take any [imath]x\in \Bbb N[/imath] and consider [imath]B_{d_1}(x,r)[/imath] .We need to find [imath]s>0[/imath] such that [imath]x\in B_{d_3}(x,s)\subset B_{d_1}(x,r)[/imath] and vice versa. If we choose [imath]s=r[/imath] in the first case then if [imath]y\in B_{d_1}(x,s)\implies d_1(x,y)<s\implies |x-y|<s\implies |\frac{1}{x}-\frac{1}{y}|<s\implies y\in B_{d_3}(x,r)[/imath]. But I am stuck in the other subset inequality.Please help . Is there any other way to solve this ?
|
1164560
|
[imath]\mathbb N[/imath] when given the metric [imath]d(m,n)=\frac{1}{m}-\frac{1}{n}[/imath]
How to show that [imath]\mathbb N[/imath] when given the metric [imath]d(m,n)=\dfrac{1}{m}-\dfrac{1}{n}[/imath] and when given the subspace topology as inherited from [imath]\mathbb R[/imath] are equivalent. When [imath]\mathbb N[/imath] is given the subspace topology then [imath]\mathbb N[/imath] becomes the discrete space.But I cant show that in the first case single point sets are open. I know to show equivalent we have to proceed as take any [imath]x\in B(x,r)[/imath] and then find [imath]s>0[/imath] such that [imath]x\in B(x,s)\subset B(x,r)[/imath] But I cant find the [imath]s>0[/imath] in the metric given .Please help
|
1991321
|
Prove [imath]\sqrt n\leq \sqrt[n]{n!} \leq \frac{n+1}{2}[/imath]
Prove [imath]\sqrt n\leq \sqrt[n]{n!} \leq \frac{n+1}{2}[/imath]. I thought I would use induction to prove this, only I don't know how to use the assumption in the inductive step. Base: [imath]n=1[/imath] [imath]\Rightarrow \sqrt 1\leq \sqrt[1]{1!}\leq \frac{2}{2}[/imath]. So for [imath]n=1[/imath] this inequality holds. Assumption: [imath]\sqrt k\leq \sqrt[k]{k!}\leq \frac{k+1}{2}[/imath] Inductive step: [imath]\sqrt{k+1}\leq \sqrt[k+1]{(k+1)!}\leq \frac{k+1+1}{2}[/imath] Is it correct so far and how do I continue from here?
|
1572094
|
Estimate of n factorial: [imath]n^{\frac{n}{2}} \le n! \le \left(\frac{n+1}{2}\right)^{n}[/imath]
on our lesson at our university, our professsor told that factorial has these estimates [imath]n^{\frac{n}{2}} \le n! \le \left(\dfrac{n+1}{2}\right)^{n}[/imath] and during proof he did this [imath](n!)^{2}=\underbrace{n\cdot(n-1)\dotsm 2\cdot 1}_{n!} \cdot \underbrace{n\cdot(n-1) \dotsm 2\cdot 1}_{n!}[/imath] and then: [imath](1 \cdot n) \cdot (2 \cdot (n-1)) \dotsm ((n-1) \cdot 2) \cdot (n \cdot 1)[/imath] and it is equal to this [imath](n+1)(n+1) \dotsm (n+1)[/imath] why it is equal, I didn't catch it. Do you have any idea? :)
|
248573
|
Show that [imath]rank(A)+rank(B) \leq n[/imath], when [imath]A,B[/imath] are [imath]2[/imath] matrices of size [imath]n \times n[/imath], and [imath]AB=0[/imath]
Question from homework in Linear Algebra: Let [imath]A,B[/imath] be two matrices of size [imath]n \times n[/imath] such that [imath]AB=0[/imath]. Show that: [imath]rank(A) + rank(B) \le n[/imath] . It probably has something to do with the dim of the null space or column space but I can't put things together from what we've learned... Please help.. Thanks. :)
|
16339
|
Proof of: [imath]AB=0 \Rightarrow Rank(A)+Rank(B) \leq n[/imath]
As the title says, am searching for a proof of If [imath]A,B \in \mathbb{R}^{n\times n}[/imath] and [imath]AB=0[/imath] then [imath]\mathrm{rank}(A)+\mathrm{rank}(B) \leq n[/imath] I am doing this as preparation for an upcoming exam and can't figure a way to start. Please just post small hints as answers. I will try to go from there. Thank you ftiaronsem
|
1992916
|
[imath]\sum_{k= 2}^\infty (\zeta(k) - 1)[/imath]
Show that [imath]\sum_{k=2}^\infty (\zeta(k) - 1)[/imath] converges and find the limit. (Hint : Deduce that [imath]\zeta(s) = 1 + O(2^{-\sigma})[/imath] where [imath]s = \sigma + it[/imath], [imath]\sigma \geq 2[/imath].) [imath]\textbf{Proof}[/imath] Let [imath]s = \sigma + it[/imath] with [imath]\sigma \geq 2.[/imath] Then [imath]\zeta(s) = 1 + \sum_{n \geq 2} \frac{1}{n^s} = 1 + \sum_{n \geq 2} \frac{1}{n^\sigma e^{it \log n}}.[/imath] Since [imath]s[/imath] is complex and complex number cannot be compared, is it a bit vague to prove [imath]\zeta(s) = 1 + O(2^{-\sigma}).[/imath] Even if I ignore the imaginary part, it is not clear why [imath]\sum_{n \geq 2} \frac{1}{n^\sigma}\leq \frac{C}{2^\sigma}[/imath] for some constant [imath]C[/imath].
|
1433797
|
Why does the [imath]\sum_{n>1}(\zeta(n)-1)=1?[/imath]
While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus [imath]1[/imath], converge to [imath]1[/imath]. Better put: [imath]\sum_{n=2}^{\infty} \zeta(n)-1 = 1 [/imath] Furthermore, if you use only the even numbers for the zeta function, the sum will converge to [imath]\frac{3}{4}[/imath], or [imath]\sum_{n=1}^{\infty} \zeta(2n)-1 = \frac{3}{4}[/imath] Leaving [imath]\sum_{n=2}^{\infty} \zeta(2n-1)-1 = \frac{1}{4}[/imath] This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so?
|
1993331
|
How did Ramanujan calculate these interesting values for [imath]e[/imath]?
While reading a mini-bio about Ramanujan, there was an equation that I found somewhat intriguing:[imath]e^{\pi\sqrt{58}}=24591257751.99999982\ldots\tag1[/imath] This is interesting in the way that [imath]e^{\pi\sqrt{58}}[/imath] is very close to [imath]24591257752[/imath], and how Ramanujan managed to calculate such a large number back when computers didn't exist! Others include \begin{equation}35\sqrt{\pi}\ln 2=42.9999986\ldots\tag2\\ 51\ln(36\pi)-2e^\pi+\sqrt[4]{21}=196.99999991695\ldots\\ e^{{\frac \pi 4\sqrt{102}}}=800\sqrt3+196\sqrt{51}\\\vdots\end{equation} And there are much more. So, I'm wondering: Main Question: Is there some sort of formula, trick or follow-through that you can use that enables you to calculate other fascinating [imath]e[/imath] equations that are very close to an integer? Note: I'm not the best at mathematics. In fact, I haven't even started Calculus yet. I would love it if the answer also had a detailed description of how you solved this.
|
101376
|
How did Hermite calculate [imath]e^{\pi\sqrt{163}}[/imath] in 1859?
Pretend you are in 1859. What is a fast, efficient, and accurate way to numerically evaluate constants like that to, say, 20 decimal places, using ONLY pen and paper?
|
1988680
|
Every separable Banach space is a quotient of [imath]\ell_1[/imath]
I'm really confused with this question... Let [imath]X[/imath] be a separable Banach space. Prove that [imath]X[/imath] is quotient space [imath]\ell_1[/imath]. I have some denotations. [imath]\{ \xi_i \}[/imath] - convergent dense sequence in [imath]K_1(0)[/imath] = [imath]\{ x \in X \mid \|x\| \leq 1 \}[/imath] - closed balls with center [imath]0[/imath]. [imath] \alpha : \{ x_i \} \mapsto \sum_i (x_i \xi_i),\{ x_i \} \in \ell_1 [/imath] How can these denotations help me to prove statement? Thanks in advance.
|
1538920
|
Every separable Banach space is isomorphic to [imath]\ell_1/A[/imath] for some closed [imath]A\subset \ell_1[/imath]
How to prove the following mind-blowing fact? Let [imath]X[/imath] be a separable Banach space and let [imath]\ell_1[/imath] be the space of all absolutely summable scalar sequences. Then there exists such closed subspace [imath]A\subset \ell_1[/imath] that factor space [imath]\ell_1/A[/imath] and [imath]X[/imath] are isomorphic as normed spaces. Edit: So what, this is like a classification up to isomorphism of all separable Banach spaces? Each separable Banach space corresponds to some closed subspace of [imath]\ell_1[/imath]?
|
1994202
|
Independence and addition?
Suppose that [imath]X[/imath] and [imath]Y[/imath] are independent random variables. Is [imath]X + Y[/imath] independent from [imath]X[/imath]? Intuitively I think it should be so - because if information about [imath]X[/imath] gave information about [imath]X + Y[/imath], it should also give information about [imath]Y[/imath] by subtracting X - which then rephrases as, if [imath]X[/imath] and [imath]Y[/imath] are not independent, are [imath]X + Y[/imath] and [imath]Y[/imath] dependent? To answer this I would need to say something about [imath]\sigma(X + Y)[/imath] (in terms of [imath]\sigma(X)[/imath] and [imath]\sigma(Y))[/imath], but I am not sure how to do so. Or is it false? I don't know. Edit: Duplicate of Are [imath]X[/imath] and [imath]X+Y[/imath] independent, if [imath]X[/imath] and [imath]Y[/imath] are independent?
|
1224083
|
Are [imath]X[/imath] and [imath]X+Y[/imath] independent, if [imath]X[/imath] and [imath]Y[/imath] are independent?
As asked in the title? Does the independence of two random variables [imath]X[/imath] and [imath]Y[/imath] imply the independence of [imath]X[/imath] and [imath]X+Y[/imath]? If so, what's the easiest way to prove that?
|
1994258
|
[imath]a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}[/imath] implies [imath]a_n/n^2[/imath] converges?
Let [imath]a_0=\pi[/imath], [imath]a_1=\pi^2[/imath], [imath]a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}[/imath] How can we prove that [imath]a_n/n^2[/imath] converges? It is easy to see that [imath]a_n-a_1=\sum_{k=1}^{n-1}\frac{2a_{k-1}}{k+1}\geq 2a_0\sum_{k=1}^{n-1}\frac{1}{k+1}\to\infty.[/imath] Then how to do?
|
1981219
|
How to calculate the limit of [imath]\frac{a_n}{n^2}[/imath] for the sequence [imath]a_{n+1}=a_n+\frac{2 a_{n-1}}{n+1}[/imath]?
Assume [imath]\{a_n\}[/imath]is a sequence which satisfies the following recursion formula: [imath]a_{n+1}=a_n+\frac{2a_{n-1}}{n+1}~(n\geqslant 1).[/imath] and [imath]a_0=\pi,a_1=\pi^2 [/imath]. How to compute [imath]\displaystyle\lim_{n\to \infty}\frac{a_n}{n^2}[/imath]?
|
1994363
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Ways to evaluate [imath]\int_0^1 \int_0^1 \frac{1}{1-xy}dxdy = \frac{\pi^2}{6}[/imath]
A classmate told me this, but he didn't tell me how to evaluate the integral. [imath]\int_0^1 \int_0^1 \frac{1}{1-xy}dxdy=\int_0^1 \int_0^1\sum_{n=0}^\infty (xy)^n dxdy=\sum_{n=1}^\infty \frac{1}{n^2}=\zeta(2)[/imath] So if you can evaluate that integral this might be an easy way to solve the Basel problem. I tried substituting [imath]\frac{1}{x}=v[/imath] and [imath]\frac{1}{y}=t[/imath] to get [imath]\zeta(2)=\int_1^\infty\int_1^\infty\frac{1}{vt-1}-\frac{1}{vt}dvdt=\lim_{N\to\infty}\left(\int_1^N\int_1^N\frac{1}{vt-1}dvdt-\ln(N)^2\right)[/imath] The areas of the cross sections of [imath]\frac{1}{vt-1}[/imath] are going to be infinite near [imath](1,1)[/imath] so this seems like a dead end.
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1884418
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How to evaluate [imath]\int_0^1\int_0^1 \frac{1}{1-xy} \, dy \, dx[/imath] to prove [imath]\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}[/imath].
I've read somewhere on this site that if you consider: [imath]\int_0^1 \int_0^1 \frac{1}{1-xy} \,dy\,dx[/imath] Then using the power series, we have this is equal to [imath]\sum_{n=1}^{\infty} \frac{1}{n^2}[/imath] which I decided to try and was able to show. Apparently we can show this is equal to [imath]\frac{\pi^2}{6}[/imath], and using what little I know about double integrals from a few khan academy videos (I haven't taken multivariable calculus yet), I tried to evaluate this double integral by techniques of single variable calculus [imath]u=xy[/imath]..and I got this: [imath]-\int_{0}^{1} \frac{\ln (1-x)}{x}dx[/imath] The usual way I would evaluate this is with a Taylor series, but that just that just leads us in circles. So I want to know how can I evaluate this, so we can prove [imath]\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}[/imath].
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1993860
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How to solve for matrix of the linear transformation?? (Coordinates)
Apologies for a simple question but my textbook is horrible. I don't seem to understand what this notation means. [imath] \left[ \begin{array}{c} T \end{array} \right]_b^a [/imath] I was looking online for resources to help me understand and found the following question: Suppose a basis for V = R3 is the set C = {(1, 0, 0), (1, 1, 0), (1, 1, 1)}. And the coordinate vector of an element (x, y, z) is [imath] \left[ \begin{array}{c} x-y\\ y-z\\ z \end{array} \right] [/imath] How would you get the matrix representation of T. The solution is [imath] \left[ \begin{array}{c} 3&4&4\\ 1&1&2\\ 1&1&0 \end{array} \right] [/imath] but I don't seem to understand how to get there so can someone provide a brief explanation of what is going on?
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1992188
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How to find the matrix representation of a linear tranformation
How does one find the matrix representation of a linear transformation [imath]T:V\to W[/imath] with respect to the basis [imath]B[/imath] for [imath]V[/imath] and [imath]D[/imath] for [imath]W[/imath]?
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