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769221 | Resolve [imath]\lim_{x \to 0} \frac {\tan 3x}{\tan 5x}[/imath]
I started resolving this limit: [imath]\lim_{x \to 0} \dfrac{\tan 3x}{\tan 5x}[/imath] so: [imath]\begin{align}\lim_{x \to 0} \dfrac{\tan 3x}{ \tan 5x} & = \lim_{x \to 0} \dfrac{\frac{\sin 3x}{\cos 3x }}{\frac{\sin 5x}{\cos 5x}} \\ \\ & = \lim_{x \to 0} \dfrac{ \sin (3x) \cdot \cos (5x)}{ \sin(5x) \cdot \cos (3x)}\\ \\ & = \quad?\end{align}[/imath] Could someone help me please | 905745 | How to find [imath]\lim_{x\to 0}\frac{\tan 3x}{\tan 5x}[/imath]?
I am asked to find the following limit: [imath] \lim_{x \to 0} \frac{\tan 3x}{\tan 5x}[/imath] My problem is in simplifying the function. I followed two different approaches to solve the problem. But both seems incorrect. Apprach 1) Since [imath]\tan \theta = \frac{sin \theta}{cos \theta}[/imath] and [imath]\cot \theta = \frac{\cos \theta}{\sin \theta}[/imath], we have: [imath]\frac{\tan 3x}{\tan 5x} = \frac{\sin 3x \times \cos 5x}{\sin 5x \times \cos 3x}[/imath] This approach does not work well, because I cannot simplify more. Approach 2) Since [imath]\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \times \tan y)}[/imath], we have: [imath]\frac{\tan 3x}{\tan 5x} = \frac{\tan 3x}{\tan 3x + \tan 2x} = \frac{\tan 3x}{\frac{\tan 2 + \tan3}{1 - \tan 2 \times \tan 3}}[/imath] What am I doing wrong? |
924185 | [imath](x+y)^c\le x^c+y^c[/imath] for [imath]0[/imath]
The statement I'm trying to prove is: [imath](x+y)^c\le x^c+y^c[/imath] whenever [imath]0\le x,y[/imath] and [imath]0\le c\le1[/imath]. This comes up in the proof that [imath]|x|_*^c[/imath] is an absolute value whenever [imath]0<c\le1[/imath] and [imath]|x|_*[/imath] is an absolute value over some integral domain [imath]D[/imath]. The given statement, however, is simply a question about a real inequality. How do you prove this one? The case [imath]x=0[/imath] or [imath]y=0[/imath] is trivial, so we can assume that [imath]x,y>0[/imath]. Then expanding the definitions we have [imath]e^{c\log(x+y)}\le e^{c\log x}+e^{c\log y}[/imath], but the addition doesn't play well with the exponentials and logs, so I don't see what else can be done. This feels like a convexity result, but I'm not seeing exactly how to make the connection (not to mention that I still haven't proved that [imath]\exp[/imath] is convex and [imath]\log[/imath] is concave, so if I can avoid that I'd prefer to). | 264156 | Prove that [imath](p+q)^m \leq p^m+q^m[/imath]
If [imath]p,q[/imath] are positive quantities and [imath]0 \leq m\leq 1[/imath] then Prove that [imath](p+q)^m \leq p^m+q^m[/imath] Trial: For [imath]m=0[/imath], [imath](p+q)^0=1 < 2= p^0+q^0[/imath] and for [imath]m=1[/imath], [imath](p+q)^1=p+q =p^1+q^1[/imath]. So, For [imath]m=0,1[/imath] the inequality is true.How I show that the inequality is also true for [imath]0 < m < 1[/imath]. Please help. |
936526 | Does [imath]\sum_{n=1}^\infty \frac{\cos(\ln(n))}{n}[/imath] converge?
EDIT: the question is answered here Divergence of [imath]\sum\limits_{n=1}^{\infty} \frac{\cos(\log(n))}{n}[/imath] Using integrals, I managed to prove that [imath]\displaystyle \forall m, \sin(\ln(m+1))\leq \sum_{n=1}^m \frac{\cos(\ln(n))}{n}\leq 1+\sin(\ln(m))[/imath] and I noticed that [imath]\sin(\ln(m+1)) - \sin(\ln(m)) \to 0[/imath] I tend to believe that the series diverges, but I can't prove it with the previous inequality. | 533014 | Divergence of [imath]\sum\limits_{n=1}^{\infty} \frac{\cos(\log(n))}{n}[/imath]
I'm struggling to prove that [imath]\sum \limits_{n=1}^{\infty} \frac{\cos(\log(n))}{n}[/imath] diverges. Does anyone have any idea how to prove it? Breaking it into smaller pieces din't work. Maybe I should bound it with another series? But how? |
936744 | Recurrence problem with a game of probability
Fair coin flipping (50% on both sides) [imath]P_1[/imath] and [imath]P_2[/imath] plays a few games of fair coin flipping. Assume player [imath]A[/imath] starts with [imath]x[/imath] coins and player [imath]B[/imath] with [imath]y[/imath] coins. Let [imath]P_n[/imath] denote the probability of player [imath]A[/imath] winning all coins. Find [imath]P_0[/imath] and [imath]P_{x+y}[/imath], then write a difference equation for [imath]P_n[/imath]. Solve the difference equation Find the odds of player [imath]A[/imath] winning over player [imath]B[/imath] if [imath]A[/imath] starts with, e.g. half the coins [imath]B[/imath] starts with. How far I've come I've been able to produce a recurrence relation [imath]P_n = \frac{1}{2}P_{n+1} + \frac{1}{2}P_{n-1}[/imath]. And I assume [imath]P_0 = 0, P_{x+y}=1[/imath], since when player [imath]A[/imath] has 0 coins, he also has [imath]0\%[/imath] chance of winning, and when player [imath]A[/imath] has [imath]x+y[/imath] coins, he's already won, therefore it would be [imath]100\%[/imath] Correct me if I'm wrong. All help appreciated, thanks a lot. | 936549 | Gambler's ruin and coin toss
Edit 3. Fixed question to be more clear and include current solution Problem Two players player 1 and player 2 plays a game of fair coin flipping. Player 1 starts with [imath]A[/imath] coins and Player 2 with [imath]B[/imath] coins. Let [imath]P_n[/imath] defines the probability of player 1 winning all coins in a game when player 1 has [imath]n[/imath] coins. Find the recurrence relation of [imath]P_n[/imath] and then find its closed formula. This formula can then be used to know the probability of player 1 winning if he has, i.e. twice the coins player 2 has. Solution? I know that [imath]P_0 = 0[/imath], because if she has [imath]0[/imath] coins, then she can't win, what about [imath]P_{A+B}?[/imath] [imath]P_n = 0.5P_{n+1} + 0.5P_{n-1}[/imath] Which gives the characteristic equation: [imath]x^2-2x+1 = (x-1)^2=0 \implies x=1\ (double\ root)[/imath] This can then be put into the general solution for linear homogenous recurrence relations: [imath]P_{hn} = C_1+nC_2[/imath] and here I'm stuck once again. To find these two values I'd need two boundary points |
936891 | Non Maximal Prime ideal!
Assume [imath]S[/imath] to be all continuous functions from [imath][0,1][/imath] to [imath]\mathbb R[/imath]. I know by compactness of [imath][0,1][/imath] it follows that all maximal ideals of [imath]S[/imath] have the form [imath]M_{x_0}=\{f\in S \mid f(x_0)=0\}[/imath].Does there exist any non maximal prime ideal of [imath]S[/imath].I tried with some ideals of S but could not find any.please give me some idea how can I construct this? | 527658 | Prime ideals in [imath]C[0,1][/imath]
Are there any prime ideals in the ring [imath]C[0,1][/imath] of continuous functions [imath][0,1]\rightarrow \mathbb{R}[/imath], which are not maximal? Perhaps, I duplicate smb's question, but this is an interesting problem! Could you give me any hint or give a link to some literature? |
936955 | a problem on functional equations
if fuction is defined from [imath]N to N[/imath] then can we say that it is not continuous as it is not defined for all [imath]x[/imath]??a simple statement ,but this is stopping me from giving the solution to a question..pls help | 326709 | Possible to have a continuous sequence?
I'm wondering if it's possible to have a continuous sequence [imath]f: \mathbb{N} \to \mathbb{R}[/imath]? My intuition is telling me no because logically it would be impossible to map the natural numbers onto the real numbers but I'm not sure how to formalize it. What do you think? How would you prove that no function over the natural numbers (no sequence) is continuous? For example, how would you prove that [imath]f: \mathbb{N} \to \mathbb{R}[/imath] such that [imath]f(n) = n^{2}[/imath] is not continuous over the natural numbers? |
936924 | Limit of the sequence [imath]\frac{1^k+2^k+...+n^k}{n^{k+1}}[/imath]
How would someone find the limit of the sequence [imath]a_n = \frac{1^k+2^k+...+n^k}{n^{k+1}}, k \in \mathbb{N}[/imath] as [imath]n[/imath] goes to Infinity? Can someone give me maybe a hint where to start? | 150391 | Evaluate [imath]\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}}[/imath]
By considering: [imath]\lim_{n\to\infty}\frac{\sum_{k=1}^n k^1}{n^{2}} = \frac 1 2[/imath] [imath]\lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^{3}} = \frac 1 3[/imath] [imath]\lim_{n\to\infty}\frac{\sum_{k=1}^n k^3}{n^{4}} = \frac 1 4[/imath] Determine if this is true: [imath]\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac 1 {{m+1}}[/imath] If it is, prove it. If it is not, evaluate [imath]\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{{m+1}}[/imath]. |
936027 | What is the value of [imath]\csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}[/imath]?
How to compute [imath]S=\csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}[/imath] I tried to rewrite it in terms of [imath]\sin[/imath] [imath] \csc^2\frac{\pi}{14}+\csc^2\frac{3\pi}{14}+\csc^2\frac{5\pi}{14}= \frac{1}{\sin^2\frac{\pi}{14}}+\frac{1}{\sin^2\frac{3\pi}{14}}+\frac{1}{\sin^2\frac{5\pi}{14}}=\\ \\ \frac{\sin^2\frac{3\pi}{14}\sin^2\frac{5\pi}{14}+\sin^2\frac{\pi}{14}\sin^2\frac{5\pi}{14}+\sin^2\frac{\pi}{14}\sin^2\frac{3\pi}{14}}{\sin^2\frac{\pi}{14}\sin^2\frac{3\pi}{14}\sin^2\frac{5\pi}{14}}[/imath] then i used [imath]2\cos x\cos y=\cos(x-y)+\cos(x+y)\\ 2\sin x\sin y=\cos(x-y)-\cos(x+y)[/imath] then i found \begin{align} S &= 2\frac{\left(\cos\frac{\pi}{7}-\cos\frac{4\pi}{7}\right)^2+\left(\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)^2+\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}\right)^2}{\left(\cos\frac{\pi}{7}-\cos\frac{4\pi}{7}\right)\left(\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}\right)} \end{align} i made a simplification and used again the transformation of product on sum and arrived at \begin{align} S &= 4\frac{6-5\cos\frac{\pi}{7}+2\cos\frac{2\pi}{7}-4\cos\frac{3\pi}{7}+2\cos\frac{4\pi}{7}-4\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}}{5-8\cos\frac{\pi}{7}+6\cos\frac{2\pi}{7}-5\cos\frac{3\pi}{7}+4\cos\frac{4\pi}{7}-3\cos\frac{5\pi}{7}+\cos\frac{6\pi}{7}} \\ & \cos\frac{6\pi}{7}=-\cos\frac{\pi}{7} \hspace{5mm} \cos\frac{5\pi}{7}=-\cos\frac{2\pi}{7} \hspace{5mm} \cos\frac{4\pi}{7}=-\cos\frac{3\pi}{7} \\ S &= 4\frac{6+6\left(-\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)}{5+9\left(-\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}\right)} \end{align} [imath]-\cos\frac{\pi}{7}+\cos\frac{2\pi}{7}-\cos\frac{3\pi}{7}=-\frac{1}{2}[/imath] [imath]S=4\frac{6-3}{5-\frac{9}{2}}=4\frac{3}{\frac{1}{2}}=4\cdot3\cdot2=24[/imath] | 45144 | Proving [imath]\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24[/imath]
How do I show that: [imath]\frac{1}{\sin^{2}\frac{\pi}{14}} + \frac{1}{\sin^{2}\frac{3\pi}{14}} + \frac{1}{\sin^{2}\frac{5\pi}{14}} = 24[/imath] This is actually problem B [imath]4371[/imath] given at this link. Looks like a very interesting problem. My attempts: Well, I have been thinking about this for the whole day, and I have got some insights. I don't believe my insights will lead me to a [imath]\text{complete}[/imath] solution. First, I wrote [imath]\sin\frac{5\pi}{14}[/imath] as [imath]\sin\frac{9 \pi}{14}[/imath] so that if I put [imath]A = \frac{\pi}{14}[/imath] so that the given equation becomes, [imath]\frac{1}{\sin^{2}{A}} + \frac{1}{\sin^{2}{3A}} + \frac{1}{\sin^{2}{9A}} =24[/imath] Then I tried working with this by taking [imath]\text{lcm}[/imath] and multiplying and doing something, which appeared futile. Next, I actually didn't work it out, but I think we have to look for a equation which has roots as [imath]\sin[/imath] and then use [imath]\text{sum of roots}[/imath] formulas to get [imath]24[/imath]. I think I haven't explained this clearly. [imath]\text{Thirdly, is there a trick proving such type of identities using Gauss sums ?}[/imath] One post related to this is: How to prove that: [imath]\tan(3\pi/11) + 4\sin(2\pi/11) = \sqrt{11}[/imath] I don't know how this will help as I haven't studied anything yet regarding Gauss sums. |
937815 | Proving cardinality of the reals and the cross product of the reals
I am trying to prove that [imath]\Bbb {R} \times \Bbb {R} \sim \Bbb {R}[/imath] using the Cantor-Bernstein Theorem. So then that would mean that I need to prove that [imath]|\Bbb {R}| \leq |\Bbb {R} \times \Bbb {R}|[/imath] and [imath]|\Bbb {R} \times \Bbb {R}| \leq |\Bbb {R}|[/imath]. This is my work so far: For the first part ([imath]|\Bbb {R}| \leq |\Bbb {R} \times \Bbb {R}|[/imath]) I let [imath]\Bbb {R} \rightarrow \Bbb {R} \times \Bbb {R}[/imath] be defined as [imath]f(r)=(r,r) , \forall r \in \Bbb {R}[/imath], which is the identity function. So then also let [imath]f(s) = (s,s), \forall s \in \Bbb {R}[/imath], where [imath]r \neq s[/imath]. So then since [imath](r,r) \neq (s,s)[/imath], then [imath]f(r) \neq f(s)[/imath]. Hence [imath]f[/imath] is injective and therefore [imath]|\Bbb {R}| \leq |\Bbb {R} \times \Bbb {R}|[/imath]. Is this correct so far? Am I on the right track? I'm confused about how to go about proving the second part, that [imath]|\Bbb {R} \times \Bbb {R}| \leq |\Bbb {R}|[/imath]. I know I need to show that there is an injection from [imath]\Bbb {R} \times \Bbb {R} \rightarrow \Bbb {R}[/imath] but I can't figure out how to do it. Also how would I define this function for the second part? | 935181 | Cross product of the reals question
Is [imath]\Bbb {R} \times \Bbb {R} \subseteq \Bbb {R}[/imath]? If this is the case then would it be true that [imath]|\Bbb {R} \times \Bbb {R}| \leq |\Bbb {R}|[/imath]? |
560950 | How to evaluate [imath] \int_0^1 {\log x \log(1-x) \log^2(1+x) \over x} \,dx [/imath]
Solve that the following integral: [imath] \int_0^1 {\log x \log(1-x) \log^2(1+x) \over x} \,dx. [/imath] I haven't solved it yet. | 503405 | A Challenging Logarithmic Integral [imath]\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx[/imath]
How can we prove that: [imath]\int_0^1 \frac{\log(x)\log(1-x)\log^2(1+x)}{x}dx=\frac{7\pi^2}{48}\zeta(3)-\frac{25}{16}\zeta(5)[/imath] where [imath]\zeta(z)[/imath] is the Riemann Zeta Function. The best I could do was to express it in terms of Euler Sums. Let [imath]I[/imath] denote the integral. [imath]I=-\frac{\pi^2}{24}\zeta(3)+2\sum_{r=2}^\infty \frac{(-1)^r (H_r)^2}{r^3}-2\sum_{r=2}^\infty \frac{(-1)^r H_r}{r^4}+2 \sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}H_r}{r^2}-2\sum_{r=2}^\infty \frac{(-1)^r H_r^{(2)}}{r^3}[/imath] where [imath]\displaystyle H_r^{(n)}=\sum_{n=1}^r \frac{1}{k^n}[/imath]. I am unable to simplify these sums further. Does anyone have any idea on how to solve this integral? Please see here for more details. Some time ago I was able to solve the simpler integral: \begin{align*} \int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{x}dx &=-\frac{3 \pi^4}{160}+\frac{7\log(2)}{4}\zeta(3)-\frac{\pi^2 \log^2(2)}{12} +\frac{\log^4(2)}{12} \\ &\quad+ 2 \text{Li}_4 \left(\frac{1}{2} \right) \end{align*} where [imath]\text{Li}_n(z)[/imath] is the Polylogarithm. |
938108 | Prove [imath]\frac {1}{(a-b)^{2}} + \frac {1}{(b-c)^{2}} + \frac {1}{(c-a)^{2}}=1[/imath]
Given that [imath]a,b,c[/imath] are distinct real numbers such that [imath]a^{2}*(1-b+c)+b^{2}*(1-c-a)+c^{2}*(1-a+b)=ab+bc+ca[/imath] First, I tried to expand the bottom of what we need to prove. [imath]\frac{1}{a^{2}-2ab+b^{2}}+\frac{1}{b^{2}-2bc+c^{2}}+\frac{1}{c^{2}-2ac+a^{2}}[/imath] I then tried to expand what we are given. [imath]a^{2}-a^{2}*b+a^{2}*c+b^{2}-b^{2}*c-b^{2}*a+c^{2}-c^{2}*a+c^{2}*b=ab+bc+ca[/imath] [imath]a^{2}-a^{2}*b+a^{2}*c+b^{2}-b^{2}*c-b^{2}*a+c^{2}-c^{2}*a+c^{2}*b-ab-bc-ca=0[/imath] Add 1 to both sides [imath]a^{2}-a^{2}*b+a^{2}*c+b^{2}-b^{2}*c-b^{2}*a+c^{2}-c^{2}*a+c^{2}*b-ab-bc-ca+1=1[/imath] Help will be appreciated. | 937093 | An inequality with a weird condition
[imath]a,b,c[/imath] are distinct real numbers that [imath](a^2)(1-b+c)+(b^2)(1-c+a)+(c^2)(1-a+b)=ab+bc+ca.[/imath] Prove [imath]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}=1.[/imath] I have tried two different approaches to this problem but am stuck. First I started by expanding the bottom of each fraction. [imath]\frac{1}{(a^2)-2ab+(b^2)}+\frac{1}{(b^2)-2bc+(c^2)}+\frac{1}{(c^2)-2ac+(a^2)}=1[/imath] I also tried to expand [imath](a^2)(1-b+c)+(b^2)(1-c-a)+(c^2)(1-a+b)=ab+bc+ca[/imath] [imath](a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b=ab+bc+ca[/imath] [imath](a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca=0[/imath] Then I added one to both sides [imath](a^2)-(a^2)b+(a^2)c+(b^2)-(b^2)c-(b^2)a+(c^2)-(c^2)a+(c^2)b-ab-bc-ca+1=0+1[/imath] Help will be appreciated. |
938262 | What is the Fourier Transform of [imath]f'(x)/x[/imath]?
What is the Fourier Transform of [imath]f'(x)/x[/imath]? Is it even possible to find? It's deceptively simple looking. What about [imath]f(x)/x[/imath]? | 938194 | What is the Fourier Transform of [imath]f'(x)/x[/imath]
What is the Fourier Transform of [imath]f'(x)/x[/imath]? Is this even possible to find? It's deceptively simple looking. |
937453 | [imath][\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}][/imath]
What is [imath][\mathbb{Q}(\sqrt [3] {2}+\sqrt {5}):\mathbb{Q}][/imath]? A straight forward way will be to just set [imath]x=\sqrt [3] {2}+\sqrt {5}[/imath], take powers and reach at a polynomial in [imath]x[/imath] and show the polynomial is irreducible. But the method is very tedious and seems to be hopeless. Any idea? | 937071 | [imath]\mathbf{Q}[\sqrt 5+\sqrt[3] 2]=\mathbf{Q}[\sqrt 5,\sqrt[3] 2][/imath]?
Is there a general or elegant way to approach this problem? One can show that [imath]\sqrt 5+\sqrt[3] 2[/imath] is a root of the hexic [imath]x^6-6x^4-10x^3+12x^2-60x+17[/imath], which should then be its minimal polynomial to get the result. I see no straightforward way to show this is irreducible, though. It has a root in [imath]\mathbf{F}_4[/imath], and to show it's irreducible over [imath]\mathbf{F}_3[/imath] already seems to require a lot of grunt work. One thing I can do is rule out the possibility that [imath][\mathbf{Q}[\sqrt 5+\sqrt[3] 2]:\mathbf{Q}]=2[/imath], because [imath][\mathbf{Q}(\sqrt 5,\sqrt[3] 2):\mathbf{Q}]=6[/imath] and [imath]\mathbf{Q}(\sqrt 5,\sqrt[3] 2)[/imath] is the compositum of [imath]\mathbf{Q}[\sqrt 5+\sqrt[3] 2][/imath] and [imath]\mathbf{Q}[\sqrt 5][/imath], where the latter has degree two. But I don't seem to be able to play an analogous trick to rule out [imath]3[/imath], because I know of no general lower bound on the degree of a compositum. Could there be some way to bring Galois theory into the picture? These extension aren't Galois, and so it's by no means clear to me how to reason from their normal closures, but perhaps that's the direction to go in. Suggestions are very much appreciated. |
939108 | Verify that [imath](I−XY)^{(-1)}*X=X*(I−YX)^{(-1)}[/imath]
Verify that [imath](I_n−XY)^{-1}\cdot X=X\cdot (I_m−YX)^{-1}[/imath] The first [imath]I[/imath] is of order [imath]n[/imath] and the second is of order [imath]m[/imath]. [imath]X[/imath] is [imath]n\times m[/imath] [imath]Y[/imath] is [imath]m\times n[/imath] | 938317 | Linear algebra proof regarding matrices
I'd like a hint rather than a full solution. The problem I am considering is the following: [imath]X[/imath] is an [imath]n\times m[/imath] matrix [imath]Y[/imath] is [imath]m\times n[/imath] Show that [imath](I - XY)^{-1}\cdot X = X\cdot(I - YX)^{-1}[/imath] The first [imath]I[/imath] is of dimension [imath]n[/imath] and the second [imath]I[/imath] is of dimension [imath]m[/imath]. |
939021 | Find [imath][\mathbb{Q}(\sqrt [3] {3},\sqrt [3] {2}):\mathbb{Q}][/imath]
I am trying to find [imath][\mathbb{Q}(\sqrt [3] {3},\sqrt [3] {2}):\mathbb{Q}][/imath]. My guess is it is [imath]9[/imath]. There are 3 possibilities-3,6,9. If it is not 9 then [imath]X^{3}-3[/imath] is not irreducible over [imath]\mathbb{Q}(\sqrt [3] {2})[/imath] hence it has a root in [imath]\mathbb{Q}(\sqrt [3] {2})[/imath]. But [imath]\mathbb{Q}(\sqrt [3] {2})[/imath] is real field so the only possibility is that it contains [imath]\sqrt [3] {3}[/imath]. To show this is not the case I assumed that [imath]\sqrt [3] {3}=a+b\sqrt [3] {2}+c\sqrt [3] {4}[/imath] and cubed both side. I have not yet completed it , but seems this methode is very long. Is there a better approach? I WANT A SOLUTION WHICH DOES NOT USE ANY GALOIS THEORY/ALGEBRAIC NUMBER THEORY. | 367013 | Prove [imath]\sqrt[3]{3} \notin \mathbb{Q}(\sqrt[3]{2})[/imath]
I've tried solving [imath]\sqrt[3]{3} =a + b* \sqrt[3]{2}+c* \sqrt[3]{4}[/imath], but there is no obvious contradiction, even when taking the norms/traces of both sides. I can't think of another approach. This is part of a proof, it seems obvious, but it's really bothersome I can't prove it. |
939679 | The number of elements in [imath]\mathbb{Z}_{11}[/imath] satisfies [imath]x^{12}-x^{10}=2[/imath].
The number of elements in [imath]\mathbb{Z}_{11}[/imath] satisfies [imath]x^{12}-x^{10}=2[/imath]. I don't know how to start it. | 464750 | How many elements [imath]x[/imath] in the field [imath]\mathbb{ Z}_{11}[/imath] satisfy the equation [imath]x^{12} - x^{10} = 2[/imath]?
How many elements [imath]x[/imath] in the field [imath]\mathbb{ Z}_{11}[/imath] satisfy the equation [imath]x^{12} - x^{10} = 2[/imath]? |
231145 | Prove [imath]f(S \cap T) \subseteq f(S) \cap f(T)[/imath]
[imath]f(S \cap T) \subseteq f(S) \cap f(T)[/imath] [imath]x[/imath] lies in ([imath]S \cap T[/imath]), which means the domain has fewer elements than the domain of [imath]S[/imath] and [imath]T[/imath], since [imath]x[/imath] must be in [imath]S[/imath] and [imath]T[/imath]. All [imath]f(x)[/imath] values of [imath]x[/imath], which resides in ([imath]S \cap T[/imath]) is also a member of [imath]f(S) \cap f(T)[/imath], because [imath]f(S)[/imath] encompasses all [imath]x[/imath] in [imath]S[/imath] even those in ([imath]S \cap T[/imath]) and the same can be said about [imath]f(T)[/imath]. Can you give me the solution? | 2236828 | Proof of [imath]f(A∩B)⊆f(A)∩f(B)[/imath]
Let there be a function [imath]f:R→R[/imath]. Let [imath]A[/imath] & [imath]B[/imath] be two subsets of [imath]R[/imath]. From my experience, I know that [imath]f(A∩B)⊆f(A)∩f(B)[/imath]. I don't know how to prove it. Also, are there certain conditions that the function [imath]f[/imath] or the sets [imath]A[/imath] & [imath]B[/imath] should follow for the above condition to be true? |
939811 | Trascendental extension
Let [imath]p[/imath] a prime number and let [imath]\mathbf{F}_p(X)[/imath] the field of rational functions over [imath]\mathbf{F}_p[/imath]. The degree [imath][\mathbf{F}_p(X):\mathbf{F}_p(X^p)][/imath] is [imath]p[/imath]? I'm a little confused; thanks. | 924329 | [imath][F(t):F(t^n)]=n[/imath] where [imath]t[/imath] is trascendental
Let [imath]F[/imath] be a field and let [imath]t[/imath] be trascendental over [imath]F[/imath]. Prove that [imath][F(t):F(t^n)]=n[/imath]. Obviously [imath][F(t):F(t^n)]\le n[/imath] since the polynomial [imath]f(x)=x^n-t^n \in F(t^n)[/imath] has [imath]t[/imath] as a root. But I don't know how to prove that it's irreducible over [imath]F(t^n)[/imath]. I tried to consider that [imath]t[/imath] is trascendental to deduce some algebraic relations over [imath]F[/imath] on [imath]t[/imath]. But it did not work. |
940618 | How do I create an injection here?
I am trying to show that [imath]|\Bbb {R} \times \Bbb {R}| \leq |\Bbb {R}|[/imath]. I don't know how to define [imath]f:\Bbb {R} \times\Bbb {R} \rightarrow \Bbb {R}[/imath] in a way that would make [imath]f[/imath] injective. My professor gave us the hint that [imath](0,1) \sim \Bbb {R}[/imath] would imply that [imath]|(0,1) \times (0,1)| \leq |(0,1)|[/imath], but I don't understand how that is helpful here. | 197735 | Injective function from [imath]\mathbb{R}^2[/imath] to [imath]\mathbb{R}[/imath]?
Is there an injective function [imath]f :\mathbb{R}^2\rightarrow\mathbb{R}[/imath]? I approached this problem from the perspective of [imath]f^{-1}[/imath], from which I showed there exists a surjective function [imath]f^{-1}:\mathbb{R}\rightarrow\mathbb{R}^2[/imath]. Would this imply that there exists an injective function for the inverse mapping? |
940822 | Proving a matrix is always symmetric
[imath]B[/imath] is a square matrix of real numbers. Show that the matrix [imath]BB^T[/imath] is always symmetric. | 465799 | [imath]A^TA[/imath] is always a symmetric matrix?
Through experience I've seen that the following statement holds true: "[imath]A^TA[/imath] is always a symmetric matrix?", where [imath]A[/imath] is any matrix. However can this statement be proven/falsefied? |
940785 | A proof involving nested integrals and induction
Prove that [imath]\int_0^x dx_1 \int_0^{x_1}dx_2 \cdots \int_0^{x_{n-1}}f(x_n) \, dx_n =\frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t) \, dt[/imath] I'm trying induction over [imath]n[/imath]. The case [imath]n=1[/imath] is trivial. When [imath]n=2[/imath] \begin{align}\int_0^x dx_1 \int_0^{x_1}f(x_{2})\,dx_2 = & \int_0^x\int_0^{x_1}f(x_2) \, dx_2 \, dx_1 \\ = & \int_0^x \int_{x_2}^{x} f(x_2) \, dx_1 \, dx_2 \\ =& \int_0^x f(x_2)(x_1-x_2) \, dx_2 = \frac{1}{(2-1)!}\int_0^x (x-t)f(t) \, dt\end{align} I couldn't take the induction step though. Any thoughts? I appreciate the help. | 932351 | Evaluate a Nested Integral
According to two questions [1] and [2] asked on this site earlier there exists a nice relation: [imath]\frac1{n!} \left(\int_{0}^t\mathrm dt \; f(t)\right)^n = \int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_1)\,f(t_2) \ldots f(t_n). [/imath] Unfrotunatly I am more interested in integrals of the form [imath]\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; f(t_n). [/imath] Can it be somehow brought to an unary integral? All I have figured out by now is that [imath]\int_{0}^t\mathrm dt_1 \int_{0}^{t_1}\mathrm dt_2 \cdots \int_{0}^{t_{n-1}}\mathrm dt_n\; t_n^m = \frac{m!}{(m+n)!} t^{m+n} .[/imath] |
941022 | Differentiable functions with two variables
I'm reposting this question because the original didn't receive answers. Thank you. I struggling with some problems. Thank you for any help: 1) This function is given : [imath] f(x,y)=(e^x-1)\frac y{(x^2+y^2)^\alpha}\;[/imath] , and they ask the values of [imath]\;\alpha\;[/imath] for which f is can be defined in origin and is differentiable at [imath]\;(0,0)\;[/imath] . 2) Function [imath]\;g(x,y)\;[/imath] is given, for which a) [imath]\;g(0,0)=7\;[/imath] b) [imath]\;g(t+2t^2,\sin3t+4t^2)=5e^t\;[/imath] c) [imath]\;\lim_{t\to 0}\frac{g(t,2t)-g(3t,4t)}t=10\;[/imath] They ask to show there's a point for which [imath]\;g\;[/imath] isn't differentiable and what is that point. I tried in (1) to check when is [imath]\;f\;[/imath] continuous at [imath]\;(0,0)\;[/imath], and so the limit must exist, so if we take for example [imath]\;y=x\;[/imath] and let [imath]\;x\to 0\;[/imath] we have [imath]\;(e^x-1)\frac x{2^\alpha x^{2\alpha}}\to0\;[/imath] if [imath]\;1-2\alpha\ge 0\implies \alpha\le\frac12\;[/imath] and we can put [imath]\;f(0,0)=0\;[/imath] . But I know there can be functions continuous but not differentiable so I'm stucked because the condition above is perhaps not enough. For (2) I use b) to have that [imath]g(t+2t^2,\sin3t+4t^2)=5e^t\to 5\;[/imath] with [imath]\;t\to 0\;[/imath], and then it can't be continuous at origin since [imath]\;g(0,0)=7\;[/imath] and then not differentiable, but then I don't get what condition c) tells. | 940393 | Advanced Real Calculus - Differentiability
I struggling with some problems. Thank you for any help: This function is given : [imath] f(x,y)=(e^x-1)\frac y{(x^2+y^2)^\alpha}\;[/imath] , and they ask the values of [imath]\;\alpha\;[/imath] for which f is can be defined in origin and is differentiable at [imath]\;(0,0)\;[/imath] . I tried in (1) to check when is [imath]\;f\;[/imath] continuous at [imath]\;(0,0)\;[/imath], and so the limit must exist, so if we take for example [imath]\;y=x\;[/imath] and let [imath]\;x\to 0\;[/imath] we have [imath]\;(e^x-1)\frac x{2^\alpha x^{2\alpha}}\to0\;[/imath] if [imath]\;1-2\alpha\ge 0\implies \alpha\le\frac12\;[/imath] and we can put [imath]\;f(0,0)=0\;[/imath] . But I know there can be functions continuous but not differentiable so I'm stucked because the condition above is perhaps not enough. |
941759 | Diophantine equation [imath]x^2+y^2+1=xyz[/imath]
Let [imath]x,y,z[/imath] be positive integers such that [imath]x^2+y^2+1=xyz[/imath]. Show that [imath]z=3.[/imath] | 115272 | Let [imath]x[/imath] and [imath]y[/imath] be positive integers such that [imath]xy \mid x^2+y^2+1[/imath].
Let [imath]x[/imath] and [imath]y[/imath] be positive integers such that [imath]xy \mid x^2+y^2+1[/imath]. Show that [imath] \frac{x^2+y^2+1}{xy}= 3 \;.[/imath] I have been solving this for a week and I do not know how to prove the statement. I saw this in a book and I am greatly challenged. Can anyone give me a hint on how to attack the problem? thanks |
941739 | Convergence of a series: TV show problem
I came across the following video on youtube where a kid was asked to show that [imath]\sum_{n = 1}^{\infty} \frac{\sin{(2n)}}{1 + \cos^4{n}}[/imath] is convergent. He tried to use the integral test but wasn't allowed to finish his argument. I have two questions. (1) Can you apply integral test in this situation since it is not clear if the terms of the series are decreasing? (2) How do you solve this problem? | 925471 | Is [imath]\sum_{n=1}^\infty \frac{\sin(2n)}{1+\cos^4(n)}[/imath] convergent?
As I was just checking this 'child prodigy' out on Youtube, I stumbled upon this video, in which Glenn Beck asks the kid to do the following proof: Further on, the kid starts sketching a proof (without a shadow of a doubt regarding the accuracy of his solution ) including the Integral Test. I don't know much about improper integrals since I just finished highschool, but this integral approach seemed, intuitively, pretty inaccurate to me since this is not a strictly decreasing function. Then I found this out from Wikipedia ! Conditions for the Integral Test. Furthermore, the blunt assessment that the series are convergent seems dubious as well..Upon a few computations of my own(mostly partial sums) , I tend to believe that the series are, in fact, divergent . Can anyone suggest a rigurous take on this problem (easy as it may seem to some amongst you) ? |
941968 | Derivative help
Use the definition of the derivative to find [imath]f'(x)[/imath] for [imath]f(x)=\sqrt{x-2}[/imath]. The answer that I got was [imath]\frac{1}{2(x-2)^.5}[/imath]. Is this correct? The second part asks use your answer from part 1 to find the equation of the line tangent to [imath]f(x)=\sqrt{x-2}[/imath] at the point [imath](6,2)[/imath]. How do I do this? | 941864 | Use the definition of the derivative to find [imath]f'(x)[/imath] for [imath]f(x)=\sqrt{x-2}[/imath]
Use the definition of the derivative to find [imath]f'(x)[/imath] for [imath]f(x)=\sqrt{x-2}[/imath]. I don't even know where to start with this. I have done [imath]\sqrt{x-2}= (x-2)^.5[/imath]. Is this correct? |
942335 | Why does shifting the integration line from the real-axis not affect the integral [imath] \int_{-\infty}^\infty e^{-x^2} dx [/imath]?
I know that [imath] \int_{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}, [/imath] and clearly this integral is invariant under translation along the real axis. But today I learned that [imath] \int_{-\infty}^\infty e^{-(x+ik)^2} dx = \sqrt{\pi} [/imath] holds for any choice of constant [imath] k \in \mathbb{R}[/imath], so we even have translation invariance along the imaginary axis. Why is that? | 801163 | a generalization of normal distribution to the complex case: complex integral over the real line
How to prove [imath]\int_{\mathbb{R}} e^{-\frac{(x+it)^2}{2}}dx=\sqrt{2\pi}[/imath] for any [imath]t\in \mathbb{R}[/imath]? I only obtained the case that [imath]t=0[/imath], [imath]\int_{\mathbb{R}} e^{-\frac{x^2}{2}}dx=\sqrt{2\pi}[/imath]. Thanks. |
942631 | [imath]\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}?[/imath]
How do I show that for [imath]n \geq 0[/imath], [imath]\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}?[/imath] I know that [imath]\sum_{k=0}^n \binom{n}{k} = 2^n[/imath], but does this really tell me anything? Thanks. | 942969 | Prove [imath]{2n\choose n}=\sum\limits_{k=0}^n {n\choose k}^2[/imath]
Prove [imath]{2n\choose n}=\sum\limits_{k=0}^n {n\choose k}^2[/imath] My Approach: I will be making use of [imath]\tag 1\quad{m+n\choose r} = {m\choose 0}{n \choose r} + {m\choose 1}{n\choose r- 1} + \cdot\cdot\cdot + {m\choose r}{n \choose 0}[/imath] and [imath]\tag 2{a \choose b}={a\choose a-b}[/imath] By (1) [imath]{2n\choose n} = {n+n\choose n}= {n\choose 0}{n\choose n}+{n\choose 1}{n\choose n-1}+\cdot\cdot\cdot + {n\choose n}{n\choose 0}[/imath] By (2) [imath]{2n\choose n} = {n\choose 0}^2 + {n\choose 1}^2 + \cdot\cdot\cdot + {n\choose n}^2[/imath] Then [imath]{2n\choose n}=\sum\limits_{k=0}^n {n\choose k}^2[/imath] Please have a look at my solution and give me any hints and\or suggestions you may have. |
226559 | Examples of non-measurable sets in [imath]\mathbb{R}[/imath]
I'm a newcomer in real analysis. I am learning the concept of measurable by myself using Royden's book "Real Analysis". I have a question regarding measurable sets. The following definition comes from Royden's book (page 35). Definition: A set [imath]E[/imath] is said to be measureable provided for any set [imath]A[/imath], [imath]m^*(A)=m^*(A\cap E)+m^*(A\cap E^C)[/imath] where [imath]m^*(\cdot)[/imath] denotes the outer measure of a set. To me, intuitively the above equation holds for all sets. In [imath]\mathbb{R}[/imath], I think a set can either be an interval or a series of isolated points (right?). It seems these kinds of sets are all measurable by the definition. Can anybody give me an example of non-measureable sets so that I can have an intuitive understanding regarding this concept? Thanks. | 2109894 | Does any subset of [imath][0,1][/imath] have a well-defined Lebesgue Measure? No.
New comer to measure theory! I'm struggling to wrap my head around this concept... Referring to Lebesgue Measures! Could someone provide a concrete example of a set in the interval [imath][0,1][/imath] such that the probability of choosing a point in that set is not well defined? |
944135 | How to find a symmetric matrix [imath]B[/imath] where [imath]x^T Ax = x^T Bx.[/imath]
Find a symmetric matrix [imath]B[/imath] such that for every [imath]3\times 1[/imath] matrix [imath]x[/imath]. [imath]x^T Ax = x^T Bx\ .[/imath] Let [imath]A = \begin{pmatrix}2& 1& -1\\3& 0& 1\\-2& 5& 3\end{pmatrix}[/imath] | 943574 | Find a symmetric matrix [imath]B[/imath] such that for every [imath]3\times 1[/imath] matrix [imath]\text{x}[/imath]
Let [imath]A=[/imath] [imath]\begin{pmatrix} 2 & 1 & -1\\ 3 & 0 & 1\\ -2 & 5 & 3 \end{pmatrix}.[/imath] Find a symmetric matrix [imath]B[/imath] such that for every [imath]3\times1[/imath] matrix [imath]\text{x}[/imath], [imath]\text{x}^TA\text{x}=\text{x}^TB\text{x}[/imath] I've tried to take the inverse of [imath]A[/imath] and [imath]\text{x}[/imath] to cancel out, but I get nowhere. Thanks. |
944115 | Show that two consistent systems are equivalent to each other
[imath]A: n \times n[/imath], [imath]B: n \times m[/imath] and [imath]A[/imath] is invertible. Show that "[imath]\forall \vec{b} \in \mathbb{R}^n, B \vec{x} = \vec{b}[/imath] is consistent" is equivalent to "[imath]\forall \vec{b} \in \mathbb{R}^n, (AB) \vec{x} = \vec{b}[/imath] is consistent." I know that [imath](AB)\vec{x}=\vec{b}[/imath] can be written as [imath]B\vec{x}=A^{-1}\vec{b}[/imath] since we know that [imath]A[/imath] is invertible. The form [imath]B\vec{x}=A^{-1}\vec{b}[/imath] then allows us to equate the first and second expressions so that [imath]\vec{b} = A^{-1}\vec{b}[/imath]. This means that [imath]A^{-1} = I_n[/imath] so that [imath]A = I_n[/imath] as well (since [imath]A^{-1}(AB)\vec{x}=A^{-1}\vec{b} \Rightarrow B\vec{x} = A^{-1}\vec{b}[/imath], and [imath]BI_n = B[/imath]). So the expression [imath]B\vec{x}=\vec{b} \equiv (AB)\vec{x}=\vec{b}[/imath]. Can someone help me check my work or tell me where I've gone wrong here? | 943913 | [imath]\forall \vec{b} \in \mathbb{R}^n, B \vec{x} = \vec{b}[/imath] is consistent is equivalent to....
Suppose [imath]A: n \times n[/imath] and [imath]B: n \times m[/imath] and that [imath]A[/imath] is invertible. Prove that [imath]\forall \vec{b} \in \mathbb{R}^n, B \vec{x} = \vec{b}[/imath] is consistent is equivalent to [imath]\forall \vec{b} \in \mathbb{R}^n, (AB) \vec{x} = \vec{b}[/imath] is consistent. I'm assuming this boils down to proving that [imath]B[/imath] and [imath]AB[/imath] are equivalent expressions, but I can't see why this would be the case. Why does [imath]A[/imath] being invertible make the two statements equivalent (that is, why is [imath]A[/imath] invertible required)? |
918625 | Cube of an integer
[imath]\frac{x}{y}+\frac{y}{z}+\frac{z}{x}=k[/imath] and [imath]x, y, z, k[/imath] are integers. Prove that [imath]xyz[/imath] is cube of some integer number. I was wondering about giving a parametrization for the rational points on the elliptic curve in [imath]\mathbb{P}^2(\mathbb{R})[/imath] [imath] x^2 y + y^2 z + z^2 x = k xyz[/imath] but I am not too confident in the topic, so I am asking you to shed some light. Thank you for your help. | 948565 | Proving from (a/b + b/c + c/a) being natural that abc is some natural number cubed
[imath]a, b, c \in \Bbb Z[/imath] [imath]a, b, c, \gt 0[/imath] [imath](\frac {a}{b} + \frac {b}{c} +\frac {c}{a}) \in \Bbb Z[/imath] How to prove that [imath]abc[/imath] is an integer cubed? |
943141 | Integrating [imath]x^2e^{-x}[/imath] using Feynman's trick?
In the second episode of season [imath]8[/imath] of "The Big Bang Theory," which aired yesterday night, it is stated that one can integrate [imath]x^2e^{-x}[/imath] by using Feynman's trick of differentiating under the integral. Is this actually true, and if so, how to do it? And is it "better," in any sense, than the usual way of doing it by integration by parts? | 205797 | Differentiation under integral sign (Gamma function)
This might be a silly question, but I'm reading this article about differentiation under the integral sign, and I'm stumped by something that's written early on. The author is giving a derivation of the formula for [imath]n![/imath] in terms of the gamma function. He shows how you can get [imath]\frac{n!}{t^{n+1}} = \int_0^{\infty}x^ne^{-tx}\,dx[/imath] by differentating under the integral sign of [imath]\int_0^{\infty}e^{-tx}dx[/imath]. He then says that the above "immediately implies" the formula [imath]n! = \int_0^\infty x^ne^{-x}\,dx.[/imath] However, I can't for the life of me see how this follows. Multiplying the first equation by [imath]t^{n+1}[/imath] gives [imath]n! = t^{n+1}\int_0^{\infty}x^ne^{-tx}[/imath], so apparently [imath]t^{n+1}\int_0^\infty x^ne^{-tx} \, dx = \int_0^\infty x^ne^{-x} \, dx,[/imath] but I don't see how this is true. Can anyone explain this? |
944326 | Find the all possible real solutions of [imath]x^y=y^x[/imath]
Find the all possible real solutions of [imath]x^y=y^x[/imath] [imath]x,y[/imath] both are real numbers. My attempt:I observed the following solutions [imath]x=2,y=4[/imath] [imath]x=4,y=2[/imath] [imath]y=x[/imath] Is there any other possible solutions? | 2282466 | Solutions for [imath]a^b = b^a[/imath]?
How to solve the equation [imath]a^b = b^a[/imath] if [imath]a ≠ b[/imath]. It doesn't have any whole solutions, but are there no solutions at all? And how to prove that? Any help is appreciated. Edit: Thanks to Henry & Bernard. It is correct that (2,4) is a solution, but is it the only one? Irracional solutions maybe? |
944366 | [imath]a^2b^2=b^2a^2 , a^3b^3=b^3a^3[/imath] , then [imath]G[/imath] is abelian
If in a group [imath]G[/imath] , [imath]a^2b^2=b^2a^2 , a^3b^3=b^3a^3 , \forall a.b\in G[/imath] , then how to prove that [imath]G[/imath] is abelian ? It is supposed to be some elementary manipulation but I can't figure it out , please help . | 539049 | The conditions [imath]a^2b^2=b^2a^2[/imath] and [imath]a^3b^3=b^3a^3[/imath] make the group [imath]G[/imath] abelian?
Let [imath]G[/imath] be a group. Assume that [imath]a^2b^2=b^2a^2[/imath] and [imath]a^3b^3=b^3a^3[/imath] for all [imath]$a, b \in G$[/imath]. Prove that the group [imath]$G$[/imath] is abelian. Also please tell me whether there is any standard approach in proving commutativity of groups like this problem. |
943359 | A positive polynomial is the sum of two squares in [imath]\mathbb{R}[X][/imath]
Let [imath]P\in\mathbb{R}[X][/imath] be a positive polynomial. I want to show that there exists [imath]A,B\in\mathbb{R}[X][/imath] so that [imath]P=A^2+B^2[/imath] [imath]\displaystyle P=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^{2p}(x-b_k)=a\prod_{k=0}^q(x-a_k)\prod_{k=0}^p(x^2-2\Re(b_k)+|b_k|^2)[/imath] where [imath]a[/imath] is the leading coefficient, [imath]a_k[/imath] are the real rooks, [imath]b_k[/imath] are the complex roots. What can i do now ? I know that the identity [imath](a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2[/imath] should come in handy, but I can't figure out : 1) How to show that the product real roots is a square 2) How to use the above identity for the complex roots. If I can solve those two points, then I can solve the problem. | 823627 | Prove that [imath]p \in \mathbb{R}[x][/imath] can be represented as a sum of squares of polynomials from [imath]\mathbb{R}[x][/imath]
[imath]p \in \mathbb{R}[x][/imath] and [imath] \forall x\in\mathbb{R} \ \ p(x) > 0[/imath]. Prove that [imath]p = \sum_{k=1}^{n}p^2_k, \ \ p_k \in \mathbb{R}[x][/imath]. I have noticed that [imath]deg(p)[/imath] must be even, because [imath]p(x)[/imath] doesn't intersect [imath]\{x=0\}[/imath]. How can I proceed with the proof? |
927289 | Demonstration that 0 = 1
I have been proposed this enigma, but can't solve it. So here it is: [imath]\begin{align} e^{2 \pi i n} &= 1 \quad \forall n \in \mathbb{N} && (\times e) \tag{0} \\ e^{2 \pi i n + 1} &= e &&(^{1 + 2 \pi i n})\ \text{(raising both sides to the $2\pi in+1$ power)} \tag{1} \\ e^{(2 \pi i n + 1)(2 \pi i n + 1)} &= e^{(2 \pi i n + 1)} = e &&(\text{because of (1)}) \tag{2} \\ e^{1 + 4 \pi i n - 4 \pi^2 n^2} &= e && (\div e) \tag{3} \\ e^{4 \pi i n - 4 \pi^2 n^2} &= 1 &&(n \rightarrow +\infty) \tag{4} \\ 0 &= 1 &&(?) \tag{5} \end{align}[/imath] So the question is: where is the error? | 178763 | Why this proof [imath]0=1[/imath] is wrong?(breakfast joke)
We have [imath]e^{2\pi i n}=1[/imath] So we have [imath]e^{2\pi in+1}=e[/imath] which implies [imath](e^{2\pi in+1})^{2\pi in+1}=e^{2\pi in+1}=e[/imath] Thus we have [imath]e^{-4\pi^{2}n^{2}+4\pi in+1}=e[/imath] This implies [imath]e^{-4\pi^{2}n^{2}}=1[/imath] Taking the limit when [imath]n\rightarrow \infty[/imath] gives [imath]0=1[/imath]. |
945026 | Order of Group with Elements of Order 2
Let G be a finite group such that every element in G which isn't the identity has order of 2. Show that [imath]|G| = 2^{n}[/imath] for some [imath]n \in \mathbb{N}[/imath]. I know that G is necessarily going to be abelian. But what is a good method to prove the order of the group? | 943429 | Cardinal of a group [imath]G[/imath] such that for all [imath]x\in G[/imath] we have [imath]x^2=e[/imath]
Let [imath]G[/imath] be a group such that for all [imath]x\in G[/imath] we have [imath]x^2=e[/imath]. Show that if [imath]G[/imath] is finite then the order of [imath]G[/imath] is [imath]2^n[/imath]. Here is the solution I have seen in a book. If G is finite, it can be considered as a vector space over the field [imath]\Bbb{Z}/2\Bbb{Z}[/imath], and then necessarily is finite dimensional, which gives [imath]G[/imath] as a vector space isomorphic to [imath](\Bbb{Z}/2\Bbb{Z})^n[/imath]. I don't understand why [imath]G[/imath] can be considered as a vector space over the field [imath]\Bbb{Z}/2\Bbb{Z}[/imath]. Can, someone, explain this? Please. |
945281 | Solving this summation: [imath]\sum_{i=1}^k i\cdot2^i[/imath]
[imath]\sum_{i=1}^k i\cdot2^i[/imath] I'm working on a recurrence relation question and now I'm stuck at this point. I have no idea how to simplify this down to something I can work with. Can I seperate the terms into [imath]\sum_{i=1}^k i \cdot \sum_{i=1}^k 2^i[/imath] and then just use the geometric series? | 2219925 | Trick to solving this Summation?
I am currently following some slides I found to try and learn how to find the average case complexity for some algorithms. I got stuck when having to handle this summation though: [imath]\sum_{i=1}^k i*2^i [/imath] According to the slides: it can be shown that the the summation is equal to: [imath](k - 1)·({2^k+1}) + 2 [/imath] Can somebody explain how to get to this answer? |
945511 | Combinatorics Summation Simplification
I am at a loss for how to approach simplifying the following combinatorics summation. I assume some method using summation formulas and combinatorial identities is required. [imath]\sum_{c = 0}^{3b}{\sum_{a = 0}^{c}{(-1)^{c}{n \choose 3b - c}{n \choose a}{a \choose c - a}}}[/imath] | 944035 | Binomial theorem / Generating function
I had trouble figuring out if the following equality holds by applying the binomial theorem and using generating functions. Could anyone please shed some light? Any help is greatly appreciated. [imath]{n \choose k} = \sum_{x = 0}^{3k} \sum_{y=0}^{x} (-1)^{x} {n \choose {3k-x}} {n \choose y} {y \choose {x -y}} [/imath] |
666843 | If [imath]E[X|Y]=Y[/imath] almost surely and [imath]E[Y|X]=X[/imath] almost surely then [imath]X=Y[/imath] almost surely
Assume that [imath]X[/imath] and [imath]Y[/imath] are two random variables such that [imath]Y=E[X|Y][/imath] almost surely and [imath]X= E[Y|X][/imath] almost surely. Prove that [imath]X=Y[/imath] almost surely. The hint I was given is to evaluate: [imath]E[X-Y;X>a,Y\leq a] + E[X-Y;X\leq a,Y\leq a][/imath] which I can write as: [imath]\int_A(X-Y)dP +\int_B(X-Y)dP[/imath] where [imath]A=\{X>a, Y\leq a\}[/imath] and [imath]B=\{X\leq a,Y\leq a\}[/imath]. But I need some more hints. | 978423 | Equality of conditional expectations
Let [imath]X,Y[/imath] be integrable random variables such that [imath]\mathbb E(X|\sigma(Y))=Y[/imath] and [imath]\mathbb E (Y|\sigma(X))=X[/imath], where [imath]\sigma(X)[/imath] is the smallest sigma algebra such that [imath]X[/imath] is measurable. Show that [imath]X=Y[/imath] almost surely. The claim seems plausible since I think of the equality [imath]\mathbb E(X|\sigma(Y))=Y[/imath] as [imath]X[/imath] being finer than [imath]Y[/imath]. But still, I didn't manage to formally prove the claim. Any suggestions? |
934090 | Prove the congruence [imath]pB_{p-1} \equiv -1 \pmod p[/imath] for Bernoulli numbers.
I need to prove that: If [imath]p[/imath] is prime greater than or equal to five, then [imath]pB_{p-1}[/imath] belongs to the p-integers and more over: [imath]pB_{p-1} \equiv -1 \pmod p[/imath] Hint:Put [imath]N=p[/imath] in the Faulhaber´s formula: [imath]S_{j}(N-1)= \displaystyle\sum_{k=o}^j \frac{1}{(j-k+1)}{j \choose k}N^{j-k+1}B_{k}[/imath] and reduce everything mod p and use the Fermat´s little theorem. But I do not know how to reduce this: [imath]S_{j}(p-1)= \displaystyle\sum_{k=o}^j \frac{1}{(j-k+1)}{j \choose k}p^{j-k+1}B_{k}[/imath] to mod p, and I have problems trying to prove that [imath]pB_{p-1}[/imath] belongs to the p-integers, Can you help to prove the theorem please, Thank you. | 933666 | Prove the von Staudt-Clausen congruence of the Bernoulli numbers
I need to prove that: If [imath]p[/imath] is prime greater than or equal to five, then [imath]pB_{p-1}[/imath] belongs to the p-integers and more over: [imath]pB_{p-1} \equiv -1 \pmod p[/imath] Hint:Put [imath]N=p[/imath] in the Faulhaber´s formula: [imath]S_{j}(N-1)= \displaystyle\sum_{k=o}^j \frac{1}{(j-k+1)}{j \choose k}N^{j-k+1}B_{k}[/imath] and reduce everything mod p and use the Fermat´s little theorem. But I do not know how to reduce this: [imath]S_{j}(p-1)= \displaystyle\sum_{k=o}^j \frac{1}{(j-k+1)}{j \choose k}p^{j-k+1}B_{k}[/imath] to mod p,and I have problems trying to prove that [imath]pB_{p-1}[/imath] belongs to the p-integers, Can you help to prove the theorem please, Thank you. |
946468 | Calc question about inequality and electric circuit theory
In electric circuit theory, the combined resistance [imath]R[/imath] of two resistors [imath]R_1 > 0[/imath] and [imath]R_2 > 0[/imath] connected in parallel obeys [imath]\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2}[/imath] Show that [imath]R < \sqrt{\frac{R_1 R_2}{2}}[/imath]. This proves that [imath]R[/imath] cannot be large if the individual resistances [imath]R_1[/imath] and [imath]R_2[/imath] are small. Can anyone show me how to solve this questions? | 944693 | Inequality for the combined resistance of two resistors connected in parallel
Exact question: In electric circuit theory, the combined resistance [imath]R[/imath] of two resistors [imath]R_1>0[/imath] and [imath]R_2>0[/imath] connected in parallel obeys [imath]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}[/imath] Show that [imath]R<\sqrt{\frac{R_1R_2}{2}}[/imath]. This proves that [imath]R[/imath] cannot be large if the individual resistances [imath]R_1[/imath] and [imath]R_2[/imath] are small. |
946745 | Integrating a function wrt different measures
Suppose that [imath](\Omega, \mathcal E, P)[/imath] is a probability space and [imath]X\colon \Omega \to \mathbb R[/imath] is a RV defined on [imath]\Omega[/imath]. Denote as [imath]\mu\colon \mathcal B \to [0,1][/imath] the probability measure on [imath]\mathbb R[/imath] defined as [imath]\mu(I)=P(X^{-1}(I)).[/imath] Suppose that [imath]\mu << dx[/imath], by Radon-Nikodym theorem there exists [imath]f \colon \mathbb R\to [0,+\infty][/imath] ([imath]f[/imath] is exactly the PDF of [imath]X[/imath]) such that [imath]\mu(I)=\int_I f(x)dx.[/imath] Is it correct that, given [imath]g\colon \mathbb R \to \mathbb R[/imath], [imath]\int_\mathbb R g d\mu=\int _\mathbb R g(x)f(x)dx?[/imath] And if it is, why? | 243529 | What is the name of this theorem, and are there any caveats?
For random variable [imath]X[/imath] that follows some distribution, [imath]f(x)[/imath] is the probability density function of that distribution if and only if [imath]\mathbb{E}[\phi(X)] = \int_{-\infty}^\infty \phi(x) f(x)dx[/imath] for all functions [imath]\phi[/imath]. Context: My professor used this in lecture to demonstrate a way to find the distribution of [imath]cX[/imath] given random variable [imath]X[/imath] that follows a specific distribution. What is the name of this theorem? Have I missed any qualifications/caveats? I am in particular curious about whether "for all functions" is correct. My professor mentioned "positive, bounded, deterministic functions," but I am not sure what he meant by that. Where can I find a proof of this theorem? |
946650 | Possible usage of Fubini-Tonelli when measures aren't [imath]\sigma[/imath]-finite?
Let [imath](X,\mathcal{A},\mu)[/imath] and [imath](Y,\mathcal{B}, \nu)[/imath] be measure spaces such that [imath]\mu(X)>0[/imath] and [imath]\nu(Y)>0[/imath]. Let [imath]f:X \rightarrow \mathbb{R}[/imath] and [imath]g:Y \rightarrow \mathbb{R}[/imath] be measurable functions (with respect to [imath]\mathcal{A}[/imath] and [imath]\mathcal{B}[/imath] respectively) such that [imath]f(x)=g(y) \text{ $\mu\times\nu$-almost everywhere on $X\times Y$}[/imath] Show that there exists a constant [imath]\lambda[/imath] such that [imath]f(x)=\lambda[/imath] for [imath]\mu[/imath]-a.e. [imath]x[/imath] and [imath]g(y)=\lambda[/imath] for [imath]\nu[/imath]-a.e. [imath]y[/imath]. Here is what I was working on: Consider [imath]E = \{ (x,y) : f(x)=g(y) \}[/imath]. Then, [imath]\mu \otimes \nu (E^c)=0[/imath]. Thus, [imath] 0 = \iint_{X \times Y} 1_{E^c} \, d\mu \otimes d\nu = \int_Y \int_X 1_{E^c} \, d\mu \, d\nu,[/imath] by using Fubini (since [imath]1_{E^c}[/imath] is integrable) or Tonneli (since [imath]1_{E^c}[/imath] is nonnegative). Now, this yields that [imath]\int_X 1_{E^c} \, d\mu = 0[/imath] [imath]\nu[/imath]-a.e. Hence, we can pick [imath]y \in Y[/imath] such that [imath]f(x)=g(y)=:\lambda[/imath] [imath]\mu[/imath]-a.e. You can do a similar argument to show that [imath]g(y) = \lambda[/imath] [imath]\nu[/imath]-a.e. Now, I realize that I cannot use Fubini or Tonneli unless the measures are [imath]\sigma[/imath]-finite, and I haven't used the fact that [imath]\mu(X)>0[/imath] or [imath]\nu(Y)>0[/imath]. What am I missing? | 938580 | When [imath]f(x) = g(y)[/imath] for almost every [imath](x,y)[/imath], must [imath]f[/imath] and [imath]g[/imath] be constant almost everywhere?
Consider two measure spaces [imath](X,\mathcal{A},\mu)[/imath] and [imath](Y,\mathcal{B},\nu)[/imath], where [imath]\mu\times\nu(X\times Y)>0[/imath]. Given two measurable functions [imath]f:X\to \mathbb{R}[/imath] and [imath]g:Y\to\mathbb{R}[/imath] such that [imath]f(x) = g(y) \qquad\mu\times\nu \,\,\text{a.e,}[/imath] does it follow that there exists a constant [imath]\lambda[/imath] such that [imath]f(x)=\lambda[/imath] for [imath]\mu[/imath]-a.e. [imath]x[/imath] and [imath]g(y)=\lambda[/imath] for [imath]\nu[/imath]-a.e. [imath]y[/imath]? Clarification: The displayed statement means that [imath]\mu\times\nu\big(\{(x,y):f(x)\neq g(y)\}\big)=0[/imath]. I know that this is true when you assume [imath]\mu[/imath] and [imath]\nu[/imath] are [imath]\sigma[/imath]-finite, by applying Fubini's theorem to [imath]\int |f(x)-g(y)|\,d(\mu\times\nu)[/imath]. In the general case, the only first step I can think of is to find a countable set of rectangles with arbitrarily small total area where [imath]f(x)=g(y)[/imath] on the complement of their union. But there I'm stuck. |
293244 | Compute [imath]\sum_{k=1}^{n} \frac 1 {k(k + 1)} [/imath]
More specifically, I'm supposed to compute [imath]\displaystyle\sum_{k=1}^{n} \frac 1 {k(k + 1)} [/imath] by using the equality [imath]\frac 1 {k(k + 1)} = \frac 1 k - \frac 1 {k + 1}[/imath] and the problem before which just says that, [imath]\displaystyle\sum_{j=1}^{n} a_j - a_{j - 1} = a_n - a_0[/imath]. I can add up the sum for any [imath]n[/imath] but I'm not sure what they mean by "compute". Thanks! | 2946183 | [imath]\sum_{k=1}^{519}\frac{1}{k(k+2)}[/imath], how do I solve this?
[imath]\sum_{k=1}^{519}\frac{1}{k(k+2)}[/imath] I've been trying to figure out this summation for a while now but I can't seem to get it. I've been looking at my notes and the only trick I've been given for summations like this one is [imath]\sum_{k=1}^n\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}[/imath]. This doesn't help me to my knowledge because this kind of fraction decomposition doesn't work on my question. I have premium Symbolab and that says it doesn't support this kind of question. WolframAlpha gives me the decimal approximation of 0.7480 but no support work. I'm sorry if this question has been posted here before but I couldn't find this question anywhere online. Thank you for the help, -jjleahy |
509719 | Proof of equivalence of algebraic and geometric dot product?
Geometrically the dot product of two vectors gives the angle between them (or the cosine of the angle to be precise). Algebraically, the dot product is a sum of products of the vector components between the two vectors. However, both formulae look quite different but compute the same result. What is an easy/intuitive proof showing the equivalence between the two definitions? I.e., demonstrate that [imath]a_x b_x + a_y b_y + a_z b_z = \lvert a \rvert \lvert b \rvert \cos{\theta}[/imath] | 116133 | How to understand dot product is the angle's cosine?
How can one see that a dot product gives the angle's cosine between two vectors. (assuming they are normalized) Thinking about how to prove this in the most intuitive way resulted in proving a trigonometric identity: [imath]\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)[/imath]. But even after proving this successfully, the connection between and cosine and dot product does not immediately stick out and instead I rely on remembering that this is valid while taking comfort in the fact that I've seen the proof in the past. My questions are: How do you see this connection? How do you extend the notion of dot product vs. angle to higher dimensions - 4 and higher? |
947528 | Trigonometry curiosity
How prove that: [imath]-\tan[(\frac{10\pi}{41}+4[\sin[(\frac{2\pi}{41}+\sin[(\frac{4\pi}{41}+\sin[(\frac{12\pi}{41}+\sin[(\frac{20\pi}{41}-\sin[(\frac{26\pi}{41})]-\sin[(\frac{30\pi}{41})])])])])]])]=\cot[(\frac{2\pi}{41}-\cot[(\frac{8\pi}{41}+\cot[(\frac{20\pi}{41}+\cot[(\frac{32\pi}{41}+\cot[(\frac{36\pi}{41})])])])][/imath] | 947496 | Trigonometric curiosity
How prove this [imath]-\tan\frac{10\pi}{41}+4\left(\sin\frac{2\pi}{41}+\sin\frac{4\pi}{41}+\sin\frac{12\pi}{41}+\sin\frac{20\pi}{41}-\sin\frac{26\pi}{41}-\sin \frac{30\pi}{41}\right)= \\\cot\frac{2\pi}{41}-\cot\frac{8\pi}{41}+\cot\frac{20\pi}{41}+\cot\frac{32\pi}{41}+\cot\frac{36\pi}{41}[/imath] |
371492 | Linear Algebra - Elementary Canonical Forms
Let [imath]N[/imath] be a [imath]2\times2[/imath] complex matrix such that [imath]N^2=0[/imath]. Prove that either [imath]N=0[/imath] or [imath]N[/imath] is similar over [imath]\mathbb{C}[/imath] to [imath]\begin{pmatrix}0&0\\1&0\end{pmatrix}[/imath] | 197512 | Matrix is either null or similar to the elementary matrix [imath]E_{2,1}[/imath]
Let [imath]A[/imath] be a [imath]2\times 2[/imath] complex matrix such that [imath]A^2=0[/imath]. Prove that either [imath]A=0[/imath] or [imath]A[/imath] is similar over [imath]\mathbb{C}[/imath] to [imath]\left(\begin{array}{cc} 0 & 0 \\1 & 0 \end{array}\right) [/imath] |
526055 | Linear Algebra : Check similarity
If [imath]A[/imath] is a [imath]2 \times 2[/imath] matrix with complex entries, then [imath]A[/imath] is similar over C to a matrix of one of the two types [imath] \left[ {\begin{array}{cc} a & 0 \\ 0 & b \\ \end{array} } \right] [/imath] [imath] \left[ {\begin{array}{cc} a & 0 \\ 1 & a \\ \end{array} } \right] [/imath] When For [imath]2 \times 2[/imath] matrix [imath]N[/imath] such that [imath]N^{2} = 0[/imath], [imath]N[/imath] is similar to [imath] \left[ {\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array} } \right] [/imath] The source of the problem is by Hoffman and Kunze. | 549674 | Classifying complex [imath]2\times 2[/imath] matrices up to similarity
I would like to prove the following proposition, which is given as an exercise in Hoffman and Kunze: If [imath]A[/imath] is a [imath]2\times 2[/imath] matrix with coefficients in [imath]\mathbb{C}[/imath], then [imath]A[/imath] is similar either to a matrix of the form [imath]\begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix}[/imath] or to a matrix of the form [imath]\begin{pmatrix} a & 0 \\ 1 & a \end{pmatrix}[/imath]. A hint directs the reader to prove that if [imath]N[/imath] is an nilpotent matrix (also in [imath]M_{2}(\mathbb{C})[/imath]), then either [imath]N=0[/imath] or [imath]N[/imath] is similar to [imath]\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}[/imath]. I have proven this claim (by supposing that [imath]N \neq 0[/imath] and showing that the transformation induced by multiplication by [imath]N[/imath] has matrix representation [imath]\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}[/imath] under a particular basis). However, I'm not really sure how to use this knowledge to prove the proposition in question. Any steps in the right direction would be appreciated. Also, I'd love to see any other proofs of the proposition. Thanks in advance! |
947822 | If [imath]\limsup x_n = x[/imath], [imath]\lim y_n = y[/imath], [imath]x_n, y_n > 0[/imath], then does [imath]\limsup (x_n y_n)= xy[/imath]?
I have to prove the following statement, but I can't. If [imath]\limsup x_{ n }=\, x,\lim y_{ n }=\, y, \, x_{ n },y_{ n }>0[/imath], then [imath]\limsup (x_{n}y_{n})=xy[/imath]. Will you give me some hint or solution? | 768011 | Product of lim sups
Suppose lim sup [imath]a_n[/imath] is finite, and [imath]c_n \to c[/imath] Prove that if [imath]c \geq 0[/imath] lim sup [imath]a_n c_n[/imath] = c lim sum [imath]a_n[/imath] and find a counterexample to this if [imath]c <0[/imath]. Is there a rule that the product of lim sups is equal to the lim sup of the product? Also, what counterexample will work here? |
947729 | What can be said about [imath]\pi+e[/imath] and [imath]\pi e[/imath]? Are these numbers rational or irrational?
"homework" What can be said about [imath]\pi+e[/imath] and [imath]\pi e[/imath]? Are these numbers rational or irrational? I know that both [imath]\pi[/imath] and [imath]e[/imath] are irrational. What can be said about [imath]\pi+e[/imath], and [imath]\pi e[/imath]? | 159350 | Why is it hard to prove whether [imath]\pi+e[/imath] is an irrational number?
From this list I came to know that it is hard to conclude [imath]\pi+e[/imath] is an irrational? Can somebody discuss with reference "Why this is hard ?" Is it still an open problem ? If yes it will be helpful to any student what kind ideas already used but ultimately failed to conclude this. |
947710 | Implications from [imath]f(z)\in\mathbb{R} \Longleftrightarrow z\in \mathbb{R}[/imath]
Let [imath]f:D(0,1)\longrightarrow \mathbb{C}[/imath] be a holomorphic function such that [imath]f(z)\in\mathbb{R} \Longleftrightarrow z\in \mathbb{R}[/imath]. How to prove that [imath]f[/imath] has at most one zero on the disk. By hypothesis the zeros of [imath]f(z)[/imath] are real and [imath]\displaystyle f(z)=\sum_{n=0}^\infty a_nz^n[/imath] , [imath]a_n\in\mathbb{R}[/imath] since [imath]f(z)=\overline{f(\overline{z})}[/imath]. Any help would be appreciated. | 913339 | Show that an entire function that is real only on the real axis has at most one zero, without the argument principle
Could someone advise me on how to approach this problem: Suppose an entire function [imath]f[/imath] is real if and only if [imath]z[/imath] is real. Prove that [imath]f[/imath] has at most [imath]1[/imath] zero. without the use of argument principle ? Here is my attempt: Suppose [imath]f(z)[/imath] has two zeroes at [imath]z=a.[/imath] Let [imath]f(z) = \sum^{\infty}_{n=0}a_n(z-a)^n, \forall z. \ [/imath] Then [imath]f(z)=(z-a)^2 \left(\dfrac{a_{0}}{(z-a)^2} +\dfrac{a_1}{z-a}+a_2+a_3(z-a)+...\right)[/imath] [imath]\implies a_0=a_1=0.[/imath] [imath]\implies f(z)= (z-a)^2\left(a_2+a_3(z-a)+a_4(z-a)^2+...\right)[/imath] [imath]\implies .... ?[/imath] Thank you. |
947677 | the need for the formal definition of a limit
The intuitive definition for [imath]\lim\limits_{x \to a} f(x) = L[/imath] is the value of [imath]f ( x )[/imath] can be made arbitrarily close to [imath]L[/imath] by making [imath]x[/imath] sufficiently close, but not equal to, [imath]a[/imath]. the formal definition of limit:For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies | f(x) − L | < ε. Why we need the formal definition of a limit? Does the intuitive definition has some flaw ? I understand the intuitive definition quite well, but I don't understand much about the formal one. P.S.I must declare I only have some basic knowledge of limit ,I started to learn calculus a few days ago . Understanding of the formal and intuitive definition of a limit, This post is not good ,because it covered more than one question , I picked one question from it ,so we can discuss it more targeted . | 945755 | Understanding of the formal and intuitive definition of a limit
The intuitive definition for [imath]\lim\limits_{x \to a} f(x) = L[/imath] is the value of [imath]f ( x )[/imath] can be made arbitrarily close to [imath]L[/imath] by making [imath]x[/imath] sufficiently close, but not equal to, [imath]a[/imath] . I can easily understand this ,but for the (ε, δ)-definition of limit:For every real ε > 0, there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ implies | f(x) − L | < ε. Oh, god, I cannot understand it completely. As it is the formal definition of limit, I think it should be precise but somewhat should include the mean of the above intuitive definition, so as for "[imath]f ( x )[/imath] can be made arbitrarily close to [imath]L[/imath]" in the intuitive definition correspond to “For every real ε > 0,| f(x) − L | < ε” in the formal definition, it's fine! but does “making [imath]x[/imath] sufficiently close, but not equal to, [imath]a[/imath] .” correspond to “there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ” ? This is the point I cannot understand, because I am not sure if “there exists a real δ > 0 such that for all real x, 0 < | x − a | < δ” shows "x close enough, but not equal, to [imath]a[/imath]". The other question is: does the biggest δ also get smaller as ε is getting smaller? Why? (Exclude the case when f(x) is a constant function.) Why do we need the formal definition of a limit? Does the intuitive definition have some flaw? P.S. Thank you everyone, but I must declare I only have some basic knowledge of limit, I only started to learn calculus a few days ago. |
948128 | What is the proof of [imath]n^2 = 1 + 3 + 5 ... (2*n - 1)[/imath]
What is the proof of [imath]n^2 = 1 + 3 + 5 + ... + (2\times n - 1)[/imath]? While I verified that this is true for small numbers, I am looking for a mathematical proof for all Natural Numbers . | 727774 | How to prove for each positive integer [imath]n[/imath], the sum of the first [imath]n[/imath] odd positive integers is [imath]n^2[/imath]?
I'm new to induction so please bear with me. How can I prove using induction that, for each positive integer [imath]n[/imath], the sum of the first [imath]n[/imath] odd positive integers is [imath]n^2[/imath]? I think [imath]9[/imath] can be an example since the sum of the first [imath]9[/imath] positive odd numbers is [imath]1,3,5,7,9,11,13,15,17 = 81 = 9^2[/imath], but where do I go from here. |
74682 | Injective functions also surjective?
Is it true that for each set [imath]M[/imath] a given injective function [imath]f: M \rightarrow M[/imath] is surjective, too? Can someone explain why it is true or not and give an example? | 1851033 | Prove surjectivity using the injective property.
Let's say that we have a function [imath]f:\mathbb{M} \rightarrow \mathbb{M}[/imath] where [imath]\mathbb{M}[/imath] is an arbitrary set. And let's say that the function f is injective.Does that mean that the function is surjective too? More generally: If a function's domain and codomain coincide and the function is injective does that imply that the function is surjective?My way of thinking this: Suppose that f is not surjective [imath]\Rightarrow \exists y \in \mathbb{M}[/imath] such that [imath]f(x) \neq y, \forall x \in \mathbb{M}[/imath] And since the domain and the codomain coincide this mean that two elements from domain must go in the exact element from the codomain which implies that f is not injective, which is a contradiction. Therefore f is surjective. Am I right? Thanks |
948402 | Finding the cumulative distribution
How can I find the cumulative distribution function for the following prob density. [imath] f(x) = \begin{cases} x & \text{if } 0<x<1 \\ 2-x & \text{if } 1 \leq x <2 \\ 0 & \text{otherwise} \end{cases} [/imath] This is what I did First I did 0 for [imath]x\le0[/imath] if [imath]0<x<1[/imath] then [imath]\int_0^xf(y)\,dy=\int_0^xy\,dy=\frac{x^2}{2}[/imath] if [imath]1\le x <2[/imath] [imath]\int_0^1y\,dy+\int_1^x(2-y)\,dy=\frac{1}{2}+\big(2x-\frac{x^2}{2}-({2}-\frac{1}{2})\big)[/imath] if [imath]x\ge2[/imath] then 1 But would my cumulative distribution be correct? | 943233 | Finding Distribution function from probability density function
Given [imath]f(x) = \begin{cases} x, \; 0 < x <1\\ 2-x, \; 1 \leq x < 2\\ 0 \text{ everywhere else} \end{cases}[/imath] as our P.D.F, I must find the corresponding distribution function. I know that [imath]F(x) = P(X \leq x) = \int_{-\infty}^{x}f(t) dt[/imath] is the distribution function, but I don't know how to apply it to this particular probability density function. The P.D.F has three different cases. How do I handle that? A sum of integrals with the appropriate bounds for each case of [imath]f(x)[/imath]? I'm kind of confused on how to create those bounds. How do we get rid the of the [imath]-\infty[/imath]? I'm thinking: For [imath]x > 0[/imath]: [imath]F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)dt = \int_{0}^{1} x dx + \int_{1}^{2} 2-x dx[/imath] Eh.... |
887968 | Prove that [imath](a+b\sqrt[3]{2}+c\sqrt[3]{4})^{-1}[/imath] with a,b,c∈Q is a number of the form [imath]d+e\sqrt[3]{2}+f\sqrt[3]{4}[/imath] with [imath]d,e,f∈Q[/imath]
Prove that [imath](a+b\sqrt[3]{2}+c\sqrt[3]{4})^{-1}[/imath] with [imath]a,b,c∈Q[/imath] is a number of the form [imath]d+e\sqrt[3]{2}+f\sqrt[3]{4}[/imath] with [imath]d,e,f \in Q[/imath] I'd like to do this without using too much fancy field-theoretic machinery. | 734636 | Prove the ring [imath]a+b\sqrt[3]{2}+c\sqrt[3]{4}[/imath] has inverse and is a field
How can I prove that [imath]\frac{1}{a+b\sqrt[3]{2}+c\sqrt[3]{4}}[/imath] is of the form [imath]a+b\sqrt[3]{2}+c\sqrt[3]{4}[/imath] (i.e. that [imath]a+b\sqrt[3]{2}+c\sqrt[3]{4}[/imath] is a field) for all rational [imath]a,b,c[/imath] and [imath]a+b\sqrt[3]{2}+c\sqrt[3]{4} \neq 0[/imath]? |
584498 | Prove [imath]|e^{i\theta} -1| \leq |\theta|[/imath]
Could you help me to prove [imath] |e^{i\theta} -1| \leq |\theta| [/imath] I am studying the proof of differentiability of Fourier Series, and my book used this lemma. How does it work? | 2680456 | Explain why [imath]| e^{iwx} - 1 | \le 2 |w| \, |x|[/imath]
A paper stated [imath]| e^{iwx} - 1 | \le 2 |w| \, |x|[/imath] without any further explanation. I struggle to find an explanation because Euler's formula doesn't help. Can you give me a give why this statement might be true? |
950236 | Trig limit without L'Hospital Rule: [imath]\lim\limits_{x\to0}\frac{\tan x-\sin x}{x^3}[/imath]
I'm really getting stuck on this and would appreciate some help: [imath] \lim_{x\ \to\ 0}\left[\,\tan\left(\,x\,\right) - \sin\left(\,x\,\right) \over x^{3}\,\right] [/imath] I know I need to change [imath]\tan\left(\,x\,\right)[/imath] into [imath]\sin\left(\,x\,\right)/\cos\left(\,x\,\right)[/imath] and turn [imath]\sin\left(\,x\,\right)[/imath] into [imath]1 - \cos^{2}\left(\,x\,\right)[/imath]. But then I get stuck. | 818000 | Limits of trig functions
How can I find the following problems using elementary trigonometry? [imath]\lim_{x\to 0}\frac{1−\cos x}{x^2}.[/imath] [imath]\lim_{x\to0}\frac{\tan x−\sin x}{x^3}. [/imath] Have attempted trig identities, didn't help. |
552604 | mapping properties of [imath](1-z)^i.[/imath]
There is a question on the complex analysis of Conway "Functions of one complex variable" (pg.57 #28) about the mapping properties of [imath](1-z)^i[/imath]. Can anyone describe me what does the exponent '[imath]i[/imath]' mean? What does it do to [imath](1-z)[/imath]? I am confused. Can i write in the following way? [imath](1-z)^i=e^{i \log (1-z)}=\cos (\log (1-z))+i \sin (\log (1-z))[/imath] | 950306 | mapping properties of [imath](1−z)^i[/imath]
There is a question on the complex analysis of Conway "Functions of one complex variable" (pg.[imath]57\ \#28[/imath]) about the mapping properties of [imath](1−z)^i[/imath]. Can anyone describe me what does the exponent '[imath]i[/imath]' mean? What does it do to [imath](1−z)[/imath]? I am confused. Can i write in the following way? [imath](1−z)^i=e^{i\cdot\log(1−z)}=\cos(\log(1−z))+i\sin(\log(1−z))[/imath] |
898096 | How to prove divisibility test for [imath]4[/imath]?
Let [imath]n[/imath] be an integer, [imath]4|n[/imath] if and only if the last two digit of [imath]n[/imath] are divisible by [imath]4[/imath]. I tried to use [imath]4|n[/imath] implies that [imath]n\equiv 0 \pmod4[/imath] | 676431 | Divisibility test for [imath]4[/imath]
Claim: A number is divisible by [imath]4[/imath] if and only if the number formed by the last two digits is divisible by [imath]4[/imath]. Here's where I've gotten so far. Let [imath]x[/imath] be an [imath](n+1)[/imath]-digit number. So [imath]x= a_na_{n-1} \dots a_2a_1a_0[/imath]. If [imath]a_1 = 0[/imath] and [imath]a_0 =0[/imath], then [imath]x[/imath] is a multiple of [imath]100[/imath] and therefore clearly divisible by [imath]4[/imath]. So we must deal with the case when [imath](a_1 \neq 0 \lor a_0 \neq 0)[/imath]. Then if [imath]10a_1 + a_0 \equiv 0 \mod 4[/imath] is true, then [imath]x[/imath] is divisible by [imath]4[/imath]. Do I need to do anything else or is this done? I feel like it's not quite complete, but I'm not sure how to proceed. |
103491 | Using Vieta's theorem for cubic equations to derive the cubic discriminant
Background: Vieta's Theorem for cubic equations says that if a cubic equation [imath]x^3 + px^2 + qx + r = 0[/imath] has three different roots [imath]x_1, x_2, x_3[/imath], then [imath]\begin{eqnarray*} -p &=& x_1 + x_2 + x_3 \\ q &=& x_1x_2 + x_1x_3 + x_2x_3 \\ -r &=& x_1x_2x_3 \end{eqnarray*}[/imath] The exercise is: A cubic equation [imath]x^3 + px^2 + qx + r = 0[/imath] has three different roots [imath]x_1, x_2, x_3[/imath]. Find [imath](x_1 - x_2)^2 (x_2 - x_3)^2 (x_1 - x_3)^2[/imath] as an expression containing [imath]p, q, r[/imath]. Spoiler alert: the answer is [imath]-4p^3r - 4q^3 + p^2q - 27r^2 - 18pqr[/imath]. My question is: how am I supposed to find that without using a computer? | 2714376 | Show that [imath]\dfrac{b}{a} = -(A+B+C)[/imath] and [imath]\dfrac{c}{a} = (AB+AC + BC)[/imath]
Let´s assume we have the polynomial: [imath]ax^3+bx^2+cx+d[/imath] This polynomial has three zero points: [imath]A[/imath], [imath]B[/imath] and [imath]C[/imath]. How can I show that [imath]\dfrac{b}{a} = -(A+B+C)[/imath] and [imath]\dfrac{c}{a} = (AB + AC + BC)[/imath]? Thanks for your help! |
951291 | If [imath]\sum a_n<\infty[/imath] then [imath]\exists (b_n)[/imath]
If [imath]\displaystyle\sum_{n=1}^\infty a_n<\infty[/imath] and [imath]a_n>0[/imath], then [imath]\exists (b_n)[/imath] such that [imath]b_n\geq1[/imath], [imath]b_n\to+\infty[/imath] and [imath]\displaystyle\sum_{n=1}^\infty a_nb_n<\infty[/imath]. | 884782 | Prove that [imath]\exists \{c_n\}[/imath] monotonically increasing to [imath]\infty[/imath] such that [imath]\sum_{i=1}^\infty a_nc_n[/imath] coverges.
Given a real positive sequence [imath]\{a_n\}[/imath] such that [imath]\sum_{i=1}^n a_i[/imath] converges. Prove that there exists a real sequence [imath]\{c_n\}[/imath] monotonically increasing to [imath]\infty[/imath] such that [imath]\sum_{i=1}^\infty a_nc_n[/imath] converges. What I have done: I have written [imath]\sum_{i=1}^{N+1}a_ic_i=c_{N+1}\sum_{i=1}^{N+1}a_i-\sum_{i=1}^N(c_{n+1}-c_n)S_n[/imath] and then try to prove the convergence by the Cauchy criterion. However my evaluation is still not good enough. I'm really stuck. |
951504 | Prove: [imath]\sum_{k=1}^{n-1} k·k!=n!−1[/imath]
Show: [imath]\sum_{k=1}^{n-1} k·k!=n!−1[/imath] via combinatoric proof. I know how to prove via induction but not via combinatorial method. | 737402 | Combinatorial proof of [imath]\sum_{k=0}^n k \cdot k! = (n+1)! -1[/imath]
Is there a nice combinatorial proof of the following identity? (That is, by showing that both sides count the same thing.) [imath]\sum_{k=0}^n k \cdot k! = (n+1)! -1 [/imath] I was searching Wikipedia for nice identities to assign to my students for a homework on combinatorial proof, and thought this one looked innocent enough, but then realized I couldn't solve it myself. Perhaps we should take the set of permutations of [imath]n+1[/imath] letters and partition it in some clever way (maybe the [imath]-1[/imath] suggests that the identity should be set aside), but I can't see what that might be. Of course it is very easy to prove by induction; that's not what I'm looking for. |
949640 | If [imath]\mathcal{E}[/imath] are subsets of [imath]X[/imath] and [imath]A \subset X[/imath]. Show that [imath]\mathcal{A}\mathcal{(E \cap}[/imath] A)=[imath]\mathcal{A}\mathcal{(E)} \cap[/imath] A
If [imath]\mathcal{E}[/imath] is collections of subsets of a set [imath]X[/imath] and let [imath]A \subset X[/imath] be a subset. Show that the generated [imath]\sigma[/imath]-algebra of [imath]\mathcal{(E \cap}[/imath] A) =the generated [imath]\sigma[/imath]-algebra [imath]\mathcal{E}[/imath] [imath] \cap A[/imath], i.e., [imath]\mathcal{A}\mathcal{(E \cap}[/imath] A)=[imath]\mathcal{A}\mathcal{(E)} \cap[/imath] A | 1037690 | Measure Theory Problems
Let [imath]\mathcal{E}[/imath] be an arbitrary collection of subsets of a set [imath]X[/imath], let [imath]A[/imath] be a nonempty subset of [imath]X[/imath], and let [imath] \mathcal{E}\cap A:=\{E\cap A:E\in \mathcal{E}\}. [/imath] Show that the [imath]\sigma[/imath]-algebra [imath]\sigma_A(\mathcal{E}\cap A)[/imath] on [imath]A[/imath] generated by [imath]\mathcal{E}\cap A[/imath] is equal to [imath]\sigma_X(\mathcal{E})\cap A[/imath], where [imath]\sigma_X(\mathcal{E})[/imath] is the [imath]\sigma[/imath]-algebra on [imath]X[/imath] generated by [imath]\mathcal{E}[/imath]. |
949718 | Relationship between 2 L-p spaces
Here is the question: Given function [imath]f[/imath] is bounded on a measurable set [imath]E[/imath]. Show that if [imath]f \in L^{p_1}(E)[/imath] then [imath]f \in L^{p_2}(E)[/imath] whenever [imath]p_1<p_2[/imath] . I know that I need to show [imath]|f|_{p_1}\geq c |f|_{p_2}[/imath] for a constant c I tried to let [imath]p=\frac{p_1}{p_2}[/imath] and then try to use Holder inequality but it did not get anywhere. I also try to let [imath]g=f^{p_1}[/imath] and try to look at [imath]g^p[/imath] but it did not work . | 21460 | How to show that [imath]L^p[/imath] spaces are nested?
Suppose [imath]1<p_1<p_2<\infty[/imath], then show [imath]L^{p_1}[a,b] \supset L^{p_2}[a,b][/imath]. I was able to show [imath]\|f\|_{p_1} \le \|f\|_{p_2} (b-a)^{1/p_1 - 1/p_2}[/imath] but I'm not sure how to proceed from here. |
199680 | A function has at least 4 critical points on [imath][0,1]\times[0,1][/imath]
Define a smooth function [imath]f(x,y)[/imath] on [imath]\mathbb{R}^2[/imath],which satisfies the following condition:[imath]f(x,y)=f(x+1,y);\quad f(x,y)=f(x,y+1)[/imath] (i.e.[imath]f[/imath] can be defined on a torus)then hotw to proof that there are at least 4 critical points([imath]f_{x}=f_{y}=0[/imath]) on the unit cube [imath][0,1]\times[0,1][/imath] (obviously,it has at least 2 critical point)? I'm also like to know other examples of thhis kind(the topology implies some properties of the function defined on that space) *Edit*I think there are 2 different proffs of this question,one only need some use of calculus techniques(which I tried but failed),another one may include some topological statements like Neal said. | 930103 | A smooth function [imath]f:S^1\times S^1\to \mathbb R[/imath] must have more than two critical points.
I am trying to show that a smooth function [imath]f:S^1\times S^1\to \mathbb R[/imath] must have more than two critical points. Since [imath]f[/imath] attains maximum and minimum, it must have at least two critical points. How would one show that they can't be two? If one considers the gradient vector field [imath]\nabla f[/imath] then by the Poincare Hopf index theorem its index is equal to the Euler characteristic of the torus, i.e. it is [imath]0[/imath]. Therefore [imath]\nabla f[/imath] has an even number of zeros. This question has been posted here but it hasn't been answered. |
952406 | Prove that a sum of squares is a CSR modulo prime
How can I prove that a sum of two integer squares, namely [imath] x^2 + y^2 [/imath] (ranging from [imath] x = 0 \to p, \; y = 0 \to p [/imath]) is a complete system of residues (CSR) modulo [imath] p [/imath] (prime)? Or, how can I prove that [imath] x^2 + y^2 \equiv \left \{ 0, 1, 2, \cdots, p-1 \right \} \pmod p [/imath], for some integers [imath] x,y [/imath] and all prime values of [imath] p [/imath]? I've failed to prove [imath] \left (0, 1, 4, 9, 16, \cdots \right ) [/imath] is a CSR modulo [imath] p [/imath] , and I feel that this is an important step to solve my question. I've also tried showing that Lagrange's four-square theorem solves the question, but it obviously gets stuck when [imath] x^2 + y^2 \equiv z \pmod p [/imath], and [imath] z [/imath] cannot be expressed as a sum of squares. | 891861 | Sum of two squares modulo p
I have heard somewhere that for all primes [imath]p[/imath], for all [imath]k[/imath], there exist [imath]x, y[/imath] s.t. [imath]x^2 + y^2\equiv k \pmod{p}[/imath]? I recall that the proof is very elementary, but I can't remember such a proof. How would one prove this? One way is to use Cauchy-Davenport, but I don't think that this is the simplest solution. |
952716 | Give a combinatorial proof
[imath]\sum_{k=1}^n k{n\choose k}=n\cdot 2^{n-1}[/imath] I have to prove the identity using a combinatorial proof: I think this should be my combinatorial proof. We want to form a committee of [imath]k[/imath] people from a total of [imath]n[/imath] people. | 509238 | Combinatorial Proof with Summation Identity
Must show: [imath]\sum_{i=0}^n i{n \choose i} = n2^{n-1}[/imath] using a combinatorial argument. I have tried picking the problem apart by determining that the summation is equal to 1/2 of the size of all subsets of a set of size n times n. But I am having trouble relating that to what each term of the summation represents. |
953079 | Using Newton's binomial theorem to prove that a sum evaluates to [imath]36^n-26^n[/imath]
Using Newton's binomial theorem to argue that: [imath]n \ge 1[/imath] [imath]36^n - 26^n = \sum_{k=1}^{n}\binom{n}{k}10^k \cdot 26^{n-k}[/imath] my argument [imath](26+10)^n = \sum_{k=1}^{n}\binom{n}{k}10^k \cdot 26^{n-k} +26^n [/imath] [imath](26+10)^n = \sum_{k=0}^{n}\binom{n}{k}10^k \cdot 26^{n-k} +26^n [/imath] I'm stuck at the part on what to do with the [imath]26^n[/imath] After I moved it over a made the sum [imath]\sum_{k=1}^{n}[/imath] to [imath]\sum_{k=1}^{n}[/imath] | 948779 | Prove this equality by using Newton's Binomial Theorem
Let [imath] n \ge 1 [/imath] be an integer. Use newton's Binomial Theorem to argue that [imath]36^n -26^n = \sum_{k=1}^{n}\binom{n}{k}10^k\cdot26^{n-k}[/imath] I do not know how to make the LHS = RHS. I have tried [imath](36^n-26^n) = 10^n [/imath] which is [imath]x[/imath] in the RHS, but I don't know what to do with the [imath]26^{n-k}[/imath] after I have gotten rid of the [imath]26^n[/imath] on the right. I also know I might have to use [imath]\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}[/imath] pascal's identity in this question. Maybe I am approaching it from a completely wrong point of view, If someone can help point me in the right direction. It would be much appreciated!!! |
953294 | If [imath]H[/imath] is a subgroup of [imath]G[/imath] and [imath]K[/imath] is a subgroup of [imath]H[/imath], then [imath]|G:K|=|G:H||H:K|[/imath]
I would like to prove that : If [imath]H[/imath] is a subgroup of [imath]G[/imath] and [imath]K[/imath] is a subgroup of [imath]H[/imath], then [imath]|G:K|=|G:H||H:K|[/imath] I appreciate any clear explanation of this. Thanks ! | 730728 | If [imath]K \leq H \leq G[/imath], show that [imath][G:K] = [G:H][H:K][/imath].
This is not for homework. (I am a grader for a class.) The case in which [imath]G[/imath] is finite is trivial. (That is, use a corollary to Lagrange's Theorem, and set [imath][G:H] = \dfrac{|G|}{|H|}[/imath], and similarly for [imath][H:K][/imath].) How do you prove this for when [imath]G[/imath] is infinite? The proof of this statement that I know well tries to show that [imath][G:K][/imath] is finite iff [imath][G:H][/imath] and [imath][H:K][/imath] are finite, but I am not a fan of that proof. For pedagogical purposes, I'm looking for a better proof. (The course uses right cosets by convention.) The course has just started covering normal subgroups, so things like the Isomorphism Theorems are not known yet. |
953208 | Show that [imath]f'_n \rightarrow f'[/imath] uniformly on every compact set.
Assume that [imath]f_n,f:D(0,1)\rightarrow C[/imath] are holomorphic. and [imath]f_n \rightarrow f[/imath] uniformly on every compact set. we need to show that [imath]f'_n \rightarrow f'[/imath] uniformly on every compact set. I kind of have no idea how to approach this one, its from a test I attended and I'm going for a second go at it and I want to understand the first one, first. | 888719 | For holomorphic functions, if [imath]\{f_n\}\to f[/imath] uniformly on compact sets, then the same is true for the derivatives.
Let [imath]\Omega[/imath] be an open subset in [imath]\mathbb{C}[/imath]. Let [imath]\{f_n\}[/imath] be a sequence of holomorphic functions on [imath]\Omega[/imath] such that [imath]f_n\to f[/imath] pointwise and converges uniformly on any compact subset [imath]K\subseteq \Omega[/imath]. Then by Cauchy Theorem and Morera Theorem, [imath]f[/imath] is holomorphic. Let [imath]f_n[/imath] and [imath]f'[/imath] be the derivatives of [imath]f_n[/imath] and [imath]f[/imath] respectively. Prove that [imath]f_n'\to f'[/imath] uniformly on any compact subset [imath]K\subseteq \Omega[/imath]. How to prove? |
953545 | How can I find the distribution of a stochastic variable X^2 if X is normal standard distributed?
I am considering a stochastic variable X that is standard normal distributed i.e. [imath] F_X(x) = \int_{-\infty}^x\frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt [/imath] How do I find out the distribution of [imath]X^2[/imath]? /Gus | 925662 | Proof that if [imath]Z[/imath] is standard normal, then [imath]Z^2[/imath] is distributed Chi-Square (1).
Suppose that [imath]Z\sim N(0,1)[/imath] and let [imath]V=Z^2[/imath]. Prove that [imath]V\sim \chi^2(1)[/imath]. I want to use the method of moment generating functions, because I already understand the proof using the method of distribution functions. I will show my work, and then where I got stuck. Since [imath]Z\sim N(0,1)[/imath], then [imath]\mu=0[/imath] and [imath]\sigma^2=1[/imath], and we have [imath]M_V(t) =E[e^{tV}]=E[e^{tZ^2}]=\int_{-\infty}^\infty e^{tz^2}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}z^2}dz=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{z^2(t-\frac{1}{2})}dz.[/imath] At this point, I'm out of ideas. I want to eventually get something that looks like [imath]\frac{1}{(1-2t)^{\frac{1}{2}}}[/imath]. Could I get a hint please? |
954077 | To prove [imath]\det (xy^t)=0[/imath]
Let [imath]x,y[/imath] be arbitrary non-zero column vectors in [imath]\mathbb R^n[/imath] , then how do we prove that [imath]\det (xy^t)=0[/imath] ? | 480854 | [imath]AB[/imath] is not invertible
Is it true that if [imath]A[/imath] is an [imath]m \times n[/imath] matrix and [imath]B[/imath] is an [imath]n \times m[/imath] matrix, with [imath]m > n [/imath] then [imath]\det(AB)=0[/imath]? |
954438 | Google Interview Question about a town where if a couple has a girl born, they can't have more children...
Today I was reading about Google Interview math puzzles and I couldn't solve the following puzzle. Imagine a town where there is a law: If a couple have a girl born, then they can't have more children. If they have a boy born, then they can have more children. They keep having children until a girl is born. Question is: What is the portion of girls to boys in town. I'm trying to solve it using mathematics. Here is my approach but I'm not getting anywhere. I'm using probabilities to model this world. Probability that a new couple has 1 boy is 1/2. Probability that a new couple has 1 girl is 1/2. Probability that a couple with 1 boy has 2 boys is 1/4. Probability that a couple with 1 boy has 2 girls is 0. (You can only have 1 girl by law.) Probability that a couple with 1 boy has 1 boy and 1 girl is 1/4. Probability that a couple with 2 boys has 3 boys is 1/8. Probability that a couple with 2 boys has 3 girls is 0. (You can only have 1 girl by law.) Probability that a couple with 2 boys has 2 girls is 0. (You can only have 1 girl by law.) Probability that a couple with 2 boys has 2 boys and 1 girl is 1/8. Probability that a couple with 3 boys has 4 boys is 1/16. Probability that a couple with 3 boys has 4 girls is 0. (You can only have 1 girl by law.) Probability that a couple with 3 boys has 3 girls is 0. (You can only have 1 girl by law.) Probability that a couple with 3 boys has 2 girls is 0. (You can only have 1 girl by law.) Probability that a couple with 3 boys has 3 boys and 1 girl is 1/16. At this moment I'm starting to see a pattern. If couple has a boy, then with equal probability they can have a girl as well. So I sum all the probabilities of all couples in town (infinite number of couples), and I multiply probability by number of boys: [imath]Boys in Town = 1*1/2 + 2*1/4 + 3*1/8 + 4*1/16 + ...[/imath] I get infinite series: [imath] 1/2 + 1/2 + 3/8 + 4/16 + 5/32 + ...[/imath] I sum it up and I get: [imath] 1 + 3/8 + 4/16 + 5/32 + ...[/imath] (I don't know how to sum this) So there will be more than 1 boy for sure! So I think in this town the ratio will be that there are more boys than girls. In particula there wil be [imath] 3/8 + 4/16 + 5/32 + ...[/imath] boys more than girls in town. There will be 1 girl and 1 + [imath] 3/8 + 4/16 + 5/32 + ...[/imath] boys. Please let me know what you think of my analysis. Thanks! | 20426 | Famous puzzle: Girl/Boy proportion problem (Sum of infinite series)
Puzzle In a country in which people only want boys, every family continues to have children until they have a boy. If they have a girl, they have another child. If they have a boy, they stop. What is the proportion of boys to girls in the country? My solution (not finished) If we assume that the probability of having a girl is 50%, the set of possible cases are: Boy (50%) Girl, Boy (25%) Girl, Girl, Boy (12.5%) ... So, if we call G the number of girls that a family had and B the number of boys that a family had, we have: [imath]B = 1[/imath] [imath]P(G = x) = (1/2)^{x+1}*x[/imath] So [imath]G = \Sigma (1/2)^{x+1}*x[/imath] I feel like the sum of this infinite serie is 1 and that the proportion of girls/boys in this country will be 50%, but I don't know how to prove it! Thanks! |
954905 | Showing a particular part of toy contour tends to 0?
I am trying to solve: [imath]\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^2 + a^2}dx[/imath] I translate it to its complex equivalent: [imath]\lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz - \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz[/imath] where [imath]C[/imath] is a half circle with radius [imath]R[/imath], above the real axis, and [imath]\gamma[/imath] is only the curved (top part) of the half circle. I am trying to show that [imath]\int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz[/imath] tends to 0 as [imath]R \rightarrow \infty[/imath]. I have that: [imath]\left| \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz \right|\leq \pi R \cdot \sup_{\xi \in \gamma} \mid \frac{z\sin{z}}{z^2 + a^2} \mid[/imath] That is, the integral is upper bounded by the length of the contour multiplied by the supremium of the contour. This leads me to: [imath]\pi R \cdot \sup_{\xi \in \gamma} \left| \frac{z\sin{z}}{z^2 + a^2} \right| \leq \pi R \cdot \mid \frac{R}{R^2 + a^2} \mid[/imath] as [imath]\sin{z}[/imath] has a maximum of 1. But what other inequalities can I produce to show that this tends to 0 as [imath]R \rightarrow \infty[/imath]? Thanks. | 954916 | Contour integration with poles
I am trying to solve: [imath]\int_{-\infty}^{\infty}\frac{x\sin{x}}{x^2 + a^2}dx[/imath] and show that it equals [imath]\pi e^{-a}[/imath] for all [imath]a > 0[/imath]. I translate it to its complex equivalent: [imath]lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz - \int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz[/imath] where C is a half circle with radius R, above the real axis, and [imath]\gamma[/imath] is only the curved (top part) of the half circle. I've shown that [imath]\int_{\gamma}\frac{z\sin{z}}{z^2 + a^2}dz[/imath] tends to 0 as [imath]R \rightarrow \infty[/imath]. So now: [imath]lim_{R \rightarrow \infty} \int_{-R}^{R}\frac{z\sin{z}}{z^2 + a^2}dz = \int_{C}\frac{z\sin{z}}{z^2 + a^2}dz[/imath] By residue theorem, [imath]\int_{C}\frac{z\sin{z}}{z^2 + a^2}dz = 2\pi i \cdot res_{ai}f[/imath] as [imath]ai[/imath] is the pole of [imath]f[/imath] that lies above the real axis. [imath]2\pi i \cdot res_{ai}f = 2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z sin{z}}{z^2 + a^2}[/imath] By L'Hospital's rule, this is equivalent to: [imath]2\pi i \cdot lim_{z \rightarrow ai} (z-ai)\frac{z^2\cos{z} + 2z\sin{z} - zai\cos{z} - ai\sin{z}}{2z} = 2\pi i \cdot \frac{\sin{ai}}{2} = \pi i \sin{ai}[/imath] But how is [imath]\pi i \sin{ai}[/imath] equivalent to [imath]\pi e^{-a}[/imath]? Did I make a mistake? |
689111 | Find the eigenvalues of a matrix with ones in the diagonal, and all the other elements equal
Let [imath]A[/imath] be a real [imath]n\times n[/imath] matrix, with ones in the diagonal, and all of the other elements equal to [imath]r[/imath] with [imath]0<r<1[/imath]. How can I prove that the eigenvalues of [imath]A[/imath] are [imath]1+(n-1)r[/imath] and [imath]1-r[/imath], with multiplicity [imath]n-1[/imath]? | 904926 | Determinant of a rank [imath]1[/imath] update of a scalar matrix, or characteristic polynomial of a rank [imath]1[/imath] matrix
This question aims to create an "abstract duplicate" of numerous questions that ask about determinants of specific matrices (I may have missed a few): Eigenvalues of a matrix of [imath]1[/imath]'s Eigenvalues for the rank one matrix [imath]uv^T[/imath] Calculating [imath]\det(A+I)[/imath] for matrix [imath]A[/imath] defined by products How to calculate the following determinants (all ones, minus [imath]I[/imath]) Determinant of a specially structured matrix ([imath]a[/imath]'s on the diagonal, all other entries equal to [imath]b[/imath]) Determinant of a special [imath]n\times n[/imath] matrix Find the eigenvalues of a matrix with ones in the diagonal, and all the other elements equal Determinant of a matrix with [imath]t[/imath] in all off-diagonal entries. Characteristic polynomial - using rank? Determinant of rank-one perturbations of (invertible) matrices The general question of this type is Let [imath]A[/imath] be a square matrix of rank[imath]~1[/imath], let [imath]I[/imath] the identity matrix of the same size, and [imath]\lambda[/imath] a scalar. What is the determinant of [imath]A+\lambda I[/imath]? A clearly very closely related question is What is the characteristic polynomial of a matrix [imath]A[/imath] of rank[imath]~1[/imath]? |
955123 | how to prove the eigenvalues of tridiagonal matrix
Assume the tridiagonal matrix [imath]T[/imath] in this form: [imath] T = \begin{bmatrix} a & c & & & \\ b & a & c & & \\ & b & a & \ddots & \\ & & b & a & \\ & & & b & \\ \end{bmatrix} [/imath] we must show the eagen valus have form [imath] a+2\sqrt{bc}\, \cos(k \pi/{(n+1)}) [/imath] | 955168 | How to find the eigenvalues of tridiagonal Toeplitz matrix?
Assume the tridiagonal matrix [imath]T[/imath] is in this form: [imath] T = \begin{bmatrix} a & c & & & &\\ b & a & c & & &\\ & b & a & c & &\\ & & &\ddots & &\\ & & & b & a & c\\ & & & & b & a\\ \end{bmatrix} [/imath] we must show that its eigenvalues are of the form [imath]a + 2 \sqrt{bc} \, \cos \left( \frac{k \pi}{n+1} \right)[/imath] where [imath]a=qh^2−1, ~~ b=1- \frac{ph}{2}, ~~ c =1+\frac{ph}{2} , ~~q \leq 0.[/imath] |
954922 | Series question,related to telescopic series, 1/2*4+ 1*3/2*4*6+ 1*3*5/2*4*6*8 ...infinity
The series is [imath]\frac{1}{2*4}+ \frac{1*3}{2*4*6}+ \frac{1*3*5}{2*4*6*8}+....[/imath] It continues to infinity.I tried multiplying with [imath]2[/imath] and dividing each term by[imath](3-1)[/imath],[imath](5-3)[/imath] etc,starting from the second term which gives me [imath]\frac{1}{8} -\frac{1}{4*6} -\frac{1}{4*6*8}-\frac{1}{4*6*8*10}[/imath]. Also if anybody is wondering the answer 0.5 | 936236 | Sum of the series [imath]\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\dots[/imath]
I am recently struck upon this question that asks to find the sum until infinite terms [imath]\frac{1}{2\cdot 4}+\frac{1\cdot3}{2\cdot4\cdot6}+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8}+.....∞[/imath] I tried my best to get something telescoping or something useful, but I failed. I even made a recurrence as [imath]t_n=t_{n-1}\frac{2n-1}{2n+2}[/imath], but this question was expected to be done with simple logic of series (Also that the recurrence on solving gives a higher order charasteristic polynomial, which might be difficult to solve without calculator). So, thus I ended up being confused with this question. So, can anyone prode a small solution to this (might be easy) problem . |
955758 | Calculate the sum of infinite series with general term [imath]\dfrac{n^2}{2^n}[/imath].
Please explain different methods to calculate the sum of infinite series with [imath]\dfrac{n^2}{2^n}[/imath] as it's general term i.e. Calculate [imath]\sum_{n=0}^\infty \dfrac {n^2}{2^n}[/imath] Please avoid the method used for summation of arithmetic geometric series. It is very tedious approach. Does any other simpler approach exist? | 439062 | Summation of series [imath]\sum_{n=1}^\infty \frac{n^a}{b^n}[/imath]?
How can we evaluate this series [imath]\sum_{n=1}^\infty \frac{n^a}{b^n}?[/imath] Here [imath]a[/imath] and [imath]b[/imath] are positive integers. If [imath]b=1[/imath] then series will be diverging, in other cases, it will be converging, but how to find this sum? Isn't there a simple solution which doesn't involve Stirling numbers? |
955806 | Finding [imath]E(X \mid X > Y )[/imath] when [imath]X, Y \sim N(0,1)[/imath]
This is the problem: The random variables [imath]X[/imath] and [imath]Y[/imath] are independent and [imath]N(0,1)[/imath]-distributed. Determine [imath]E(X \mid X > Y )[/imath], [imath]E(X + Y \mid X > Y )[/imath]. I go by the definition of [imath]f_{X \mid X>Y}(x) = \frac{\int_x^\infty f_{X,Y}(x,t)dt}{f_X(x)}= \cdots = (2\pi)^{-1/2} \int_x^\infty e^{-t^2/2} dt[/imath] but that function does not have a primitive. I thought about "getting" a Gamma under the int but I'm not sure on how to proceed here. Edit 1. I got the following tip. If [imath](X,Y)[/imath] is normal centered, there exists some nonnegative [imath](\sigma,\tau,\varrho)[/imath] and [imath](U,V)[/imath] i.i.d. standard normal such that [imath]X-Y=\sigma U[/imath] and [imath]X=\tau U+\varrho V[/imath]. Thus, [imath]E(X\mid X\gt Y)=\tau E(U\mid U\gt0)+\varrho E(V\mid U\gt0)==\frac{2\tau}{\sqrt{2\pi}}.[/imath] But I do not see why the == is true. I also don't know how this would help me, do I pick [imath]\tau[/imath] and [imath]\varrho[/imath] to be 1? | 417926 | conditional expectation under bivariate normal
When [imath]X[/imath], [imath]Y[/imath] are bivariate normal, how can we calculate the conditional expectation [imath]E(X\mid X>Y)[/imath]? A more concrete setting of this general problem can be referred A hard problem on conditional expectation. |
957297 | Using the chain rule to finding the second order partial derivative
[imath]z=f(x,y)[/imath] where [imath]x=r\cos\theta [/imath] and [imath]y=r\sin\theta[/imath]. Find [imath] \frac{\partial z}{\partial x}[/imath] and [imath] \frac{\partial^2 z}{\partial x^2}[/imath] I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have... [imath] \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}[/imath]. So I'm guessing that [imath]\frac{\partial z}{\partial r}[/imath] and [imath]\frac{\partial z}{\partial θ}[/imath] are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question. Full Problem: | 957434 | Express partial derivatives of second order (and the Laplacian) in polar coordinates
[imath]z=f(x,y)[/imath] where [imath]x=rcosθ[/imath] and [imath]y=rsinθ[/imath] Find [imath] \frac{\partial z}{\partial x}[/imath] and [imath] \frac{\partial^2 z}{\partial x^2}[/imath] I'm having big troubles with using chain rule, in particularly the second derivative. Spent 2 hours on this already. What I have... [imath] \frac{\partial z}{\partial x}=\frac{\partial r}{\partial x}\frac{\partial z}{\partial r}+\frac{\partial z}{\partial θ}\frac{\partial θ}{\partial x}[/imath]. So I'm guessing that [imath]\frac{\partial z}{\partial r}[/imath] and [imath]\frac{\partial z}{\partial θ}[/imath] are unkown? And I will need to use them to differentiate again. Any help is really welcome. This is simply only part of a bigger question. Full Question: |
957159 | Prove that a logarithm is irrational
I’m stuck with the following problem: Prove that [imath]\log_{2} 3 \in \mathbb{R} - \mathbb{Q} [/imath] . Thanks in advance! | 225277 | Use the unique factorization for integers theorem and the definition of logarithm to prove that [imath]\log_3 (7)[/imath] is irrational.
Use the unique factorization for integers theorem and the definition of logarithms to prove that [imath]\log_3 (7)[/imath] is irrational. I am taking a beginners fundamental mathematics module, no advanced stuff please. Thanks! My attempt. Suppose for a contradiction that it is rational, that is [imath]\log_3(7)=\frac{a}{b}[/imath] for some [imath]a,b\in R[/imath] where [imath]b\neq0[/imath]. Therefore, by the definition of logarithms, [imath]7=3^{\frac{a}{b}}[/imath]. By the theorem of unique factorization, [imath]7=1*7[/imath] is unique. Ok I'm plain stuck! Any help please? |
956297 | Related rates with respect to time
The sun is shining and a spherical snowball of volume 340 ft[imath]^3[/imath] is melting at a rate of 10 cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after 2.5 hours? Please help I'm really confused 340-10*2.5 = 315 so (-10) = 4pi/3 * ? | 956406 | Related rates with snowball
The sun is shining and a spherical snowball of volume 340 ft[imath]^3[/imath] is melting at a rate of [imath]10[/imath] cubic feet per hour. As it melts, it remains spherical. At what rate is the radius changing after [imath]2.5[/imath] hours? And this answer is wrong? please help [imath]10\times 2.5 =25[/imath] [imath]340 - 25 = 315[/imath] so [imath]r^3 = \dfrac{315\times 3 }{ 4\pi}[/imath] [imath]r= 4.2[/imath] ft |
957587 | Van Der Pol ODE
Hi I'm having troubles linearising and solving for the Van Der Pol equation [imath]\frac{d^2x}{dt^2} - (1-x^2)\frac{dx}{dt}+x =0[/imath] with initial cond [imath]x(0)=1[/imath] and [imath]x'(0)=1[/imath] Any hints or help would be greatly appreciated thank you. | 800782 | van der pol equation
Consider the van der Pol equation below: [imath](x'')+a(x^2-1)(x')+(x)=0[/imath] I need to : Find an equilibrium point and linearize this equation near it Find solutions of the linearized equation depending on a |
958006 | Is there a general method for solving a cubic polynomial?
I'm doing a course in linear algebra at the moment, and whenever I need to find the eigenvalues of a 3x3 matrix, I'm faced with the issue that I don't know a general method for solving a cubic equation, or I don't remember enough from high school algebra to be able to solve any given cubic equation. For instance, if I have the characteristic equation [imath]-\lambda^3+8\lambda^2-19\lambda+12=0[/imath] then I can't factor it easily because 19 is a prime number, but a computer was able to somehow find it's solutions for me. Is there a general, algorithmic method for solving any given cubic equation without factorization? If not, how would you solve an equation like the above? And is there a similar kind of general method for solving quartic or higher degree polynomials? | 427585 | How does one derive these solutions to the cubic equation?
In the first major studies into the magical properties of the complex number and the application to solving the cubic equation, Cardano did the following in around 1539: He reduced the cubic equation [imath]ax^3+bx^2+cx+d[/imath] to the form [imath]x^3=3px+2q[/imath] where [imath]p[/imath] and [imath]q[/imath] are real numbers. (1). How did Cardano get to this? There was also another solution to the cubic, discovered before 1926, and it is often referred to as the (del Ferro-)Cardano solution, and is perhaps where Cardano went from in his first reduction. This solution is: [imath]x=(q+w)^{1/3}+(q-w)^{1/3}[/imath] where [imath]w=\sqrt{(q^2-p^3)}[/imath]. (2). How is the del Ferro-Cardano solution derived? |
958077 | Is locally Lipshitz mapping on a compact set globally Lipschitz?
Let [imath]f: K \rightarrow \mathbb R[/imath] be a locally Lipschitz mapping on a compact subset [imath]K[/imath] of [imath]\mathbb R^n[/imath]. Is it then [imath]f[/imath] Lipschitz? | 154721 | In [imath]\mathbb{R}^n[/imath], locally lipschitz on compact set implies lipschitz
I need to prove: Let [imath]A[/imath] be open in [imath]\mathbb{R}^m[/imath], [imath]g:A \longrightarrow \mathbb{R}^n[/imath] a locally lipschitz function and [imath]C[/imath] a compact subset of [imath]A[/imath]. Show that [imath]g[/imath] is lipschitz on [imath]C[/imath]. Can anyone help me? |
958593 | Matrix Problem about commutative multiplication
The problem is: "Find all [imath]2\times 2[/imath] matrices A that have the property that for any [imath]2\times 2[/imath] matrix B, AB = BA." Given hint: "The given equation must hold for all B. Try matrices B that have lots of zero entries." I tried bashing by letting [imath]A=\begin{pmatrix}a & b\\c & d\end{pmatrix}[/imath] and [imath]B=\begin{pmatrix}w & x\\y & z\end{pmatrix}[/imath], but as you can tell, this got messy quickly. Any help would be appreciated, thank you. | 147370 | Commutative matrices are multiples of the identity
Assume M is a [imath]2 \times 2[/imath] real matrix such that [imath]MX = XM[/imath] for all real [imath]2\times2[/imath] matrices [imath]X[/imath]. Show that [imath]M[/imath] must be some real multiple [imath]q[/imath] of [imath]I[/imath]. I can see that this is logical and have tried a few examples where I have multiplied [imath]qI[/imath] by some random [imath]2\times2[/imath] matrix from both sides and get the same matrix in both cases. How would I go about showing this though? Surely a few examples aren't actually showing anything for the general case, right? Also, does the property shown in the question hold for all square matrices, or just [imath]2\times2[/imath] ones? |
959201 | The intersection of an empty family of sets
I am confused about the following. Could you explain me why if [imath]A=\varnothing[/imath],then [imath]\cap A[/imath] is the set of all sets? Definition of [imath]\cap A[/imath]: For [imath]A \neq \varnothing[/imath]: [imath]x \in \cap A \leftrightarrow (\forall b \in A )x \in b[/imath] EDIT: I want to prove that [imath]\cap \varnothing[/imath] is not a set. To do that, do I have to begin, supposing that it is a set? | 6613 | Unary intersection of the empty set
In MK (Morse-Kelley) set theory life is easy: [imath]\forall X\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)[/imath]. If [imath]X=\left\{\right\}[/imath] then [imath]\bigcap X=U[/imath], where [imath]U[/imath] is the universal class. So the (unary) intersection of the empty set is the class that contains all sets as elements. In ZF (Zermelo-Fraenkel) set theory, instead, proper classes are not allowed. So, how can I define [imath]\bigcap X[/imath] in ZF? I tried with the following definitions: [imath]\forall X\left(X\not=\left\{\right\}\rightarrow\forall y\left(y\in\bigcap X\leftrightarrow\forall x\left(x\in X\rightarrow y\in x\right)\right)\right)[/imath]. This means that [imath]\bigcap\left\{\right\}[/imath] is undefined, which is not that good. [imath]\forall X\forall y\left(y\in\bigcap X\leftrightarrow X\not=\left\{\right\}\land\forall x\left(x\in X\rightarrow y\in x\right)\right)[/imath]. This means that [imath]\bigcap\left\{\right\}=\left\{\right\}[/imath], which is the opposite of MK. I couldn't find any other valuable definition. Any ideas? Thank you. |
959054 | Show [imath]\mathcal{L}\left\{\frac{1}{t}f(t)\right\} = \int_{s}^{\infty}F(u)du[/imath]
Show for [imath]\mathcal{L}[/imath], the Laplace transform, that [imath]\mathcal{L}\left\{\frac{1}{t}f(t)\right\} = \int_{s}^{\infty}F(u)du.[/imath] I know that [imath]\mathcal{L}\left\{ t^n f(t) \right \} = (-1)^n \frac{d^n}{ds^n} F(s)[/imath] and this is sort of suggestive for [imath]n=-1[/imath], but I can't seem to show the result. Some help would be appreciated. | 393610 | Proof [imath]L{\rm{[}}\frac{{x(t)}}{t}{\rm{] = }}\int_s^\infty {X(u)du} [/imath]
I see that we usually use the theorem to solve the Laplace transform, however i want to proof the theorem, who could give me some details!!! [imath]L{\rm{[}}\frac{{x(t)}}{t}{\rm{] = }}\int_s^\infty {X(u)du} [/imath] where: [imath]L{\rm{[x(t)] = X(s)}}[/imath] |
959642 | Finding solutions of equation without using inequalities.
The problem is as follows: Find the unique ordered triple of positive integers [imath](a,b,c)[/imath] with [imath]a\leq b\leq c[/imath] such that [imath]\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}=\frac{25}{84}[/imath] The problem is easy to solve by first bounding [imath]a[/imath], then bounding [imath]b[/imath] and finally [imath]c[/imath]. This is the solution I was given: It's obvious that [imath]a\geq 4[/imath], otherwise [imath]\frac{1}{ab}+\frac{1}{abc}[/imath] would have to be negative, which we don't want. We also have [imath]\frac{25}{84}=\frac{1}{a}+\frac{1}{ab}+\frac{1}{abc}\leq \frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}=\frac{1+a+a^2}{a^3}[/imath] [imath]\frac{25}{84}\leq \frac{1+a+a^2}{a^3}[/imath] so [imath]a\leq 4[/imath] in order to satisfy the inequality, hence [imath]a=4[/imath]. We can proceed in a similar manner to solve the rest of the problem. Yes, this solution is simple and beautiful, but suppose I just want to solve this problem using a rigorous number-theoretic method (e.g. without inequalities). How would I go about doing this? *EDIT* By the way, this problem was from part C of the 2013 COMC. | 895556 | Determine variables that fit this criterion...
There is a unique triplet of positive integers [imath](a, b, c)[/imath] such that [imath]a ≤ b ≤ c[/imath]. [imath] \frac{25}{84} = \frac{1}{a} + \frac{1}{ab} + \frac{1}{abc} [/imath] Just having trouble with this Canadian Math Olympiad question. My thought process going into this, is: Could we solve for [imath]\frac{1}{a}[/imath] in terms of the other variables? Then substitute that value in for each occurrence of [imath]a[/imath], to solve for [imath]a[/imath]? That's all I can really think of right now. It's a question I'm not exactly used to... It's sort of the first of these kinds that I've faced. Thanks. |
960289 | Show that every number is the limit of a sequence of irrational numbers
I am not even sure where to start with this one. I understand for something to be a limit then that means if we have a sequence [imath]\{a_n\}[/imath] of irrational numbers then for any number [imath]a[/imath] in the reals, the [imath]|a_n - a| < \epsilon[/imath] for [imath]\epsilon > 0[/imath]. But how would you show that any number fulfils this statement? | 338328 | Irrational and rational sequence proof
Show that every irrational number in [imath]\mathbb{R}[/imath] is the limit of a sequence of rational numbers. Every rational number in [imath]\mathbb{R}[/imath] is the limit of a sequence of irrational numbers. How can I prove this? |
959791 | Expression for the Fourier transform of [imath]f(x) = \frac{1}{1 +\|x\|^2}[/imath]
I'm having troubles with the Fourier transform of [imath]f(x) = \frac{1}{1 +\|x\|^2} \in L^2(\mathbb{R}^{n})[/imath]. For the case [imath]n=1[/imath] I got [imath]\hat{f}(\xi) = \pi e^{-2\pi |\xi|}[/imath] using residues. Does the general case have a nice expression? How is that expression obtained? | 662056 | Integral in [imath]n-[/imath]dimensional euclidean space
I want to calculate this integral in [imath]n[/imath]-dimensional euclidean space. [imath]I(x)=\int_{\mathbb{R}^n}\frac{d^n k}{(2\pi)^n}\frac{e^{i(k\cdot x)}}{k^2+a^2},[/imath] where [imath]k^2=(k\cdot k)[/imath], [imath]k=(k_1,\ldots,k_n)\in\mathbb{R}^n[/imath], [imath]x=(x_1,\ldots,x_n)\in\mathbb{R}^n[/imath],[imath]a\in \mathbb{R}[/imath]. I've done this integral for [imath]n=3[/imath] by spherical coordinates and residue theorem. I have [imath]I(r)=\frac{1}{4\pi r}e^{-ar},[/imath] where [imath]r=|x|[/imath] But in [imath]n[/imath]- dimensions I failed in using spherical coordinates, because I have never done it before. Also I see that this integral is Fourier transform of [imath]\frac{1}{k^2+a^2}[/imath], but I failed here too, because I can't find Fourier pair in my reference books. If someone could guide me in this integration it would be great. |
960571 | Subset of a Cartesian product?
Is it true the following [imath] R \subset R^2[/imath] ? (If yes, I would like to see a rigorous proof.) | 897775 | Is [imath]\mathbb{R}[/imath] a subspace of [imath]\mathbb{R}^2[/imath]?
I think I've been confusing myself about the language of subspaces and so on. This is a rather basic question, so please bare with me. I'm wondering why we do not (or perhaps "we" do, and I just don't know about it) say that [imath]\mathbb{ R } [/imath] is a subspace of [imath]\mathbb{ R }^2 [/imath]. It's elementary to prove that the set [imath] S:= \left\{ c \cdot \mathbf{x} \mid c \in \mathbb{ R }, \mathbf{x} \in \mathbb{ R }^2 \right\}[/imath] is a vector subspace of [imath]\mathbb{ R } ^2[/imath]. What is confusing me is that there seems to be an isomorphism between the set [imath]S[/imath] and [imath]\mathbb{ R } [/imath]: \begin{align*} \varphi: S &\rightarrow \mathbb{ R } \\ c \cdot \mathbf{x} &\mapsto c \\ \end{align*} If this is indeed true, as I believe it is having checked that [imath]\varphi[/imath] gives an isomorphism, wouldn't we say that [imath]\mathbb{ R } [/imath] is a subspace of [imath]\mathbb{ R } ^2[/imath]? Any help sorting out this (language) problem will be greatly appreciated! |
960640 | Find all positive integer that [imath]2^{2^n}+5 [/imath] is a prime number.
Find all nonnegative integer that [imath]2^{2^n}+5 [/imath] is a prime number. For [imath]n=0[/imath] we have [imath]7[/imath] - correct For [imath]n=1[/imath] we have 9 - false For [imath]n=2[/imath] we have 21 - false For [imath]n=3[/imath] we have 259 ... Maybe any ideas how to do it in general? | 952942 | Find all primes of the form [imath]2^{2^n} + 5[/imath] for a nonnegative integer n
I'm a little lost on how to do this problem. It looks a lot like the definition for the Fermat numbers: [imath]F_n = 2^{2^n} + 1[/imath], however I'm not sure how to use that in order to find all of the primes of the form: [imath]2^{2^n} + 5[/imath]. |
960269 | Proving that [imath]\lim y_i=\lim x_i[/imath]
Let [imath]P_{ij}[/imath], where [imath]P_{ij}[/imath] is a real number and [imath]i, j[/imath] natural number, allow the three statements to hold: [imath]P_{ij}\ge 0[/imath] for all [imath]i, j[/imath] [imath]\lim_{i\to\infty}P_{ij}=0[/imath] for all [imath]j[/imath] [imath]\sum_{j=1}^i P_{ij}[/imath] =1 for all [imath]i[/imath] Let [imath](x_j)[/imath] be a convergent sequence and let a sequence [imath](y_i)[/imath] be defined by [imath]y_i=\sum_{j=1}^i P_{ij}x_j[/imath] Prove [imath](y_i)[/imath] is a convergent sequence, and prove [imath]\lim y_i=\lim x_i[/imath] I don't really know how to approach the problem, thanks in advance | 961003 | Proving Two Limits Are Equal to Each Other
Let [imath]P_{ij}[/imath], where [imath]P_{ij}[/imath] is a real number and [imath]i, j[/imath] natural number, allow the three statements to hold: [imath]P_{ij}\ge 0[/imath] for all [imath]i, j[/imath] [imath]\lim_{i\to\infty}P_{ij}=0[/imath] for all [imath]j[/imath] [imath]\sum_{j=1}^i P_{ij}[/imath] =1 for all [imath]i[/imath] Let [imath](x_i)[/imath] be a convergent sequence and let a sequence [imath](y_i)[/imath] be defined by [imath]y_i=\sum_{j=1}^i P_{ij}x_j[/imath] Prove [imath](y_i)[/imath] is a convergent sequence, and prove [imath]\lim y_i=\lim x_i[/imath] Ok so here's what I have: Fix arbitrary [imath]\epsilon>0[/imath], there exists [imath]N[/imath] such that [imath]|x_i-x|<\epsilon [/imath] for all [imath]i>N[/imath], so [imath]\lim x_n = x[/imath], which exists since ([imath]x_n[/imath]) is a convergent sequence. We may assume that [imath]|x_i-x|<M[/imath] holds always, for a constant M. Then [imath]y_i=\sum_{j=1}^i P_{ij}x_j [/imath], so by triangle inequality and since [imath]\sum_{j=1}^i P_{ij} =1[/imath] for all [imath]i[/imath], the following occurs: \begin{align*} |y_i-x|\leq {}&\sum_{j=1}^i P_{ij}|x_j- x|\\ ={}&\sum_{j\leq N} P_{ij}|x_j- x| +\sum_{j> N} P_{ij}|x_j- x|\\ <{}&\sum_{j\leq N} P_{ij}M +\sum_{j> N} P_{ij}\epsilon\\ \leq{}&\sum_{j\leq N} P_{ij}M +\epsilon \end{align*} Now I don't know what to do with the sum that remains, if I factor out the [imath]M[/imath] and make the sum that remains equal to [imath]1[/imath], then I don't really get anywhere. Thanks in advance. |
793755 | Given that [imath]x^y=y^x[/imath], what could [imath]x[/imath] and [imath]y[/imath] be?
It's not too difficult to figure out that [imath]x[/imath] and [imath]y[/imath] can both be 1, and also [imath]x[/imath] can be 2 and [imath]y[/imath] can be 4 (and vice versa). But I can't rule out if there are other solutions. Does it have anything to do with inverse functions? Is there a way to see the solutions by graphing, or algebraically? | 871169 | Find all solutions of [imath]a^b = b^a[/imath]
Find all solutions to [imath]a^b = b^a[/imath] where [imath]a, b[/imath] are natural numbers with [imath]a<b[/imath]. So far I've been able to conclude that this is equivalent to [imath]\frac{log(a)}{a} = \frac{log(b)}{b}[/imath] but I'm not sure how to proceed. Can someone please give me a hint? |
961841 | Proving that [imath]\left|\Re\left( \frac{1+i\sqrt{7}}{2}\right)^n\right| \to \infty[/imath]
Let [imath]u_n=\displaystyle\Re\left( \frac{1+i\sqrt{7}}{2}\right)^n[/imath] Prove that [imath]|u_n| \to \infty[/imath] This appeared in a recent issue of French Revue de la Filière Mathématiques, as it was reportedly asked during an oral exam. The editor deems it is an "extremely difficult problem", and uses Skolem–Mahler–Lech theorem to prove the result. They're also looking for a proof an undergraduate could find. Do you know one ? | 705877 | An exotic sequence
Let [imath]a=\frac{1+i\sqrt 7}{2}[/imath] and [imath]u_n=\Re(a^n)[/imath] show that [imath](|u_n|)\to +\infty[/imath] I think basics method does not works here. Any ideas ? |
578957 | Definite Integral [imath]\int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(3+x)}}dx[/imath]
How can I find the value of this following definite integral? [imath]\int_2^4\frac{\sqrt{\log(9-x)}}{\sqrt{\log(9-x)}+\sqrt{\log(3+x)}}dx[/imath] | 957510 | A Putnam Integral [imath]\int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}.[/imath]
This is a Putnam Problem that I have been trying to solve (on and off) for two years, but I have failed. I am in Calculus BC. This problem comes from the book "Calculus Eighth Edition by Larson, Hostetler, and Edwards". This problem is at the end of the first section of the chapter 8 exercises. Here's the problem: Evaluate [imath]\int_2^4 \frac{\sqrt{\ln(9-x)}\,dx}{\sqrt{\ln(9-x)} + \sqrt{\ln(x+3)}}.[/imath] Please. Any help is very much appreciated. So are solutions. Thank you! Edit: I like the solution given, but I was interested to see if there is any other way of doing the problem? I'm excited to see the results. |
964125 | Set of zeros of derivate - Lebesgue measure
I'm currently struggeling with the following: Let [imath]\lambda[/imath] be the Lebesgue measure and [imath]f \in C^2[0, 1][/imath]. Show: If [imath]\lambda(\{x \in [0, 1]; f(x) = 0 \}) > 0[/imath], then [imath]\lambda(\{ x \in [0, 1]; f'(x) = 0\}) > 0[/imath]. Although the statement seems to be very obvious and the proof is trivial if [imath]\{x \in [0, 1]; f(x)=0\}[/imath] has an inner point, I don't really know how to approach the general case. I would be very grateful for a little hint. | 963663 | [imath]m\{x\in [0,1]:f'(x)=0\}>0[/imath]
Let [imath]f\in C^2\{[0,1],\mathbb{R})[/imath] and [imath]m\{x\in [0,1]:f(x)=0\}>0[/imath]. Prove that [imath]m\{x\in [0,1]:f'(x)=0\}>0,[/imath] where [imath]m[/imath] denotes Lebesgue measure. I don't have any clue to solve this exercise.. |
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