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Show that $ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, [\arcsin{\frac{x}{a}} / \arccos{\frac{x}{a}}]= \frac{\pi^2}{8} $? I found this very interesting result: $$ \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}} = \int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arccos{\frac{x}{a}} = \frac{\pi^2}{8}, $$ for $a>0.$ I wond...
We have that $$\int_0^a \frac{dx}{\sqrt{a^2-x^2}} \, \arcsin{\frac{x}{a}}=\int_0^1 \frac{dt}{\sqrt{1-t^2}} \, \arcsin{t}\\ =\int_0^1 \arcsin(t)\ d(\arcsin{t})=\frac{1}{2}\left[\arcsin^2{t}\right]_0^1=\frac{\pi^2}{8}.$$ For the other one you can integrate in a similar way or use the identity $\arcsin(t)+\arccos(t)=...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1893735", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0 }
Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$? $\color{red}{\mathbf{EDIT}}$ The question was misinterpreted - it was actually: 'what is the absolute value of $z/\bar{z}$?'; I'am grateful for the answers given on the original problem though and will keep this up as is in case someone els...
https://ernstchan.com/b/src/1457375466-129.pdf The question is actually "what is the value of $z /\overline{z}$. Which is easy. By theorem 2: $|z /\overline{z}| = |z|/|\overline{z}| = \sqrt{x^2 + y^2}/\sqrt{x^2 + (-y)^2} = 1$ ;I don't see if Lang ever stated this in the text but it should be obvious $|z| = |\overline ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1894307", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 1 }
Examining convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$ with mean value theorem $$\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$$ I would like to examine covergence of this series using mean value theorem. I would like to check my solution and optionally ...
It is OK. Alternatively, you may write, by the Taylor series expansion, as $n \to \infty$, $$ \begin{align} \frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}&=\frac{1}{n}\sqrt{\left(1+O\left(\frac{1}{n}\right)\right)-\left(1+O\left(\frac{1}{n+1}\right)\right)} \\\\&=\frac{1}{n}\sqrt{O\left(\frac{1}{n}\right)} \\\\&=O\...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1895278", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? Is the series $\sqrt{1} - \sqrt{2} + \sqrt{3} - \sqrt{4} + \dots$ summable? I think it diverges although: $$ \sqrt{n+1} - \sqrt{n} \approx \frac{1}{2\sqrt{n}}$$ for example by the Mean Value Theorem $f(x+1)-f(x) \approx f'(x)$ and then I might...
This sum can be done with some form of zeta function regularization. For $\Re s >1$, define: $$\eta(s) = \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = (1-2^{1-s})\zeta(s) $$ Then, by analytic continuation, we can calculate: $$\sum_{n=1}^\infty (-1)^{n-1} \sqrt{n} \to \eta\left(-\frac{1}{2}\right) = (1-2\sqrt{2})\zeta\le...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1896464", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "9", "answer_count": 5, "answer_id": 0 }
Smallest positive integer that gives remainder 5 when divided by 6, remainder 2 when divided by 11, and remainder 31 when divided by 35? What is the smallest positive integer that gives the remainder 5 when divided by 6, gives the remainder 2 when divided by 11 and 31 when divided by 35? Also, are there any standard m...
The system of equations $$ \left\{\begin{array}{ll} n \equiv 5 &\pmod{6} \\ n \equiv 2&\pmod{11} \\ n\equiv 31 &\pmod{35}\end{array}\right. $$ is equivalent to $n\equiv m\pmod{2\cdot 3\cdot 5\cdot 7\cdot 11}$ by the Chinese remainder theorem. If $m_6, m_{11}, m_{35}$ are the smallest integer solutions of the systems $$...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1897521", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 8, "answer_id": 0 }
Find large power of a non-diagonalisable matrix If $A = \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$, then find $A^{30}$. The problem here is that it has only two eigenvectors, $\begin{bmatrix}0\\1\\1\end{bmatrix}$ corresponding to eigenvalue $1$ and $\begin{bmatrix}0\\1\\-1\end{bmatrix}$ corresp...
Since the given matrix is rather simple, you could also compute a few powers: \begin{align*} A^1&= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}\\ A^2&= \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{pmatrix}\\ A^3&= \begin{pmatrix} 1 & 0 & 0 \\ 2 & 0 & 1 \\ 1 & 1 & 0 \en...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1898710", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "14", "answer_count": 5, "answer_id": 3 }
Parabolic Representation The points $(-1/2, 0)$, $\left(0, \frac{\sqrt3}{2}\right)$, $(1/2, 0)$ are the vertices of an equilateral triangle. A parabolic equation that contains the three points is $$y= (-2 \sqrt3)x^2+\frac{\sqrt{3}}{2} \,.$$ How can the other two parabolas, that are congruent to the first parabola and ...
Since you are given the three points, there is a straight vectorial approach that you can follow. Premised that the parabola $y=x^2$, the parabola with vertex in $O=(0,0)$ and passing through $U_{-1}=(-1,1)$ and $U_{1}=(1,1)$, can be written as: $$ \mathop {OP}\limits^ \to \cdot \mathbf{u}_y = \left( {\mathop {O...
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How can I prove that $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$? How can I prove that $\binom{3n}{n,n,n}$ divided by $6$ with no remainder for all $n>0$?
Referring to the standard result, the number of factors of $p$ in $\frac{(3n)!}{n!n!n!}$ is given by $$ \sum_{i \ge 1} \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \sum_{i \ge 1} \left\lfloor \frac{n}{p^i} \right\rfloor = \sum_{i \ge 1} \left( \left\lfloor \frac{3n}{p^i} \right\rfloor - 3 \left\lfloor \frac{n}{p^i} \r...
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Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting $$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$ but to no avail. Could someone point me in the right direction?
If you have trouble approaching this, try some examples: Note that (for instance) $$ 1 \times \color{red}{2 \times 3} \times 4 = 24 = 5^2-1 $$ $$ 2 \times \color{red}{3 \times 4} \times 5 = 120 = 11^2-1 $$ $$ 3 \times \color{red}{4 \times 5} \times 6 = 360 = 19^2-1 $$ and observe that $$ 5 = \color{red}{2 \times 3} - 1...
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Multi-index sum property Exercise 1.2.3.29 in Donald Knuth's The Art of Computer Programming (3e) states the following property of a multi-indexed sum: $$ \sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k = \frac{1}{3}S_3 + \frac{1}{2}S_1S_2 + \frac{1}{6}S_1^3, $$ where $S_r = \sum_{i=0}^n a_i^r$. I tried to prove it an...
Here is a rather elementary approach based upon two aspects: * *Iterative calculation We look at first at the simpler double sum \begin{align*} A_2&:=\sum_{i=0}^n\sum_{j=0}^ia_ia_j=\sum_{0\leq j\leq i\leq n}a_ia_j\\ &=\sum_{0\leq i\leq j\leq n}a_ia_j \end{align*} and obtain an expression in $S_1=\sum_...
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What is $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? I encountered the integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ while researching the evaluation of harmonic sums. Mathematica 11 is not able to evaluate this integral, and is not able to e...
As an addendum to the previous answer, it can be shown (through the FL-expansion of $\frac{\arcsin\sqrt{x}}{\sqrt{x}}$) that $$\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right)=4\text{ Im } \text{Li}_3\left(\frac{1+i}{2}\right)-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2).$$...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1900704", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "19", "answer_count": 1, "answer_id": 0 }
Craft Inequality Originally the inequality looks like : $$ \frac{1}{4}(\sqrt{\sum_{cyc}\sin(u)\sin(v)} - \sum_{cyc}\sin(u)\sin(v)) \geq\sum_{cyc}\cos(u)-(\sum_{cyc}\sin(\frac{3u}{2})) $$ with $$u+v+w=\pi$$ after many transformations the inequality looks like : $$ \sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2C+\sum_{\math...
I begin with a transformation : $$x^2=1-C$$ $$y^2=1-B$$ $$z^2=1-A$$ We obtain : $$ \sqrt{\frac{1}{2}\left(\sum_{\mathrm{cyc}}2(1-x^2)+\sum_{\mathrm{cyc}}2(1-x^2)(1-y^2)\right)}-\frac{4}{2}\sum_{\mathrm{cyc}}((1-x^2))-\frac{1}{2}\sum_{\mathrm{cyc}}2((1-x^2)(1-y^2))\geq-\frac{1}{2}\sum_{\mathrm{cyc}}\sqrt{\frac{x^2}{2}}(...
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If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$. Let $a,b,c$ be the lengths of the sides of a triangle. Prove that $$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$ Attempt. By clearing the denominators, the required inequality is equiv...
As @user236182 pointed out, we can write $a=p+q$, $b=q+r$, and $c=r+p$, with $p,q,r>0$. Letting $s = p+q+r$, we have $$\frac{a}{b+c} = \frac{s-r}{s+r} = 1 - \frac{2r}{s+r} $$ and hence $$\sum\limits_{\text{cyc}}{\frac{a}{b+c}}< 2 \iff \sum\limits_{\text{cyc}}{\left(1-\frac{2r}{s+r}\right)}< 2 \iff \sum\limits_{\text{cy...
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Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows: Let $a,b,c$ be positive numbers with ...
$$\frac{a + \frac 1 3 + \frac 1 3 + \frac 1 3}{4} \ge \sqrt[4]{a\cdot \frac 1 3 \cdot \frac 1 3 \cdot \frac 1 3} $$ and similarly for b and c. So $(a+1)(b+1)(c+1) \ge 64\sqrt[4]{\frac{abc}{27^3}}$. You must show this is $\ge 64 abc$, which follows from your inequality (1).
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prove $abc \ge 8$ for $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$ I am reading this book about inequalities and the chapter about AM-GM inequalities includes this problem: Let $a,b,c$ be positive numbers for which $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1$, prove that $$abc \ge 8$$ The book does not provide full ...
Define \begin{align*} x&=\frac{1}{1+a}\\ y&=\frac{1}{1+b}\\ z&=\frac{1}{1+c} \end{align*} Then your problem transforms into: given $x+y+z=1$ prove $(\frac{1}{x}-1)(\frac{1}{y}-1)(\frac{1}{z}-1) \geq 8$. I believe you know the rest.
{ "language": "en", "url": "https://math.stackexchange.com/questions/1906492", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0 }
If $a, b, c, d> 0$ then I try to prove the following inequality.. If $a, b, c, d> 0$ then prove that $$a^2 b^2 c^2 + b^2 c^2 d^2 + a^2 b^2 d^2 + a^2 c^2 d^2 \geq ab^2 c^2 d + ab^2 cd^2 + a^2 bcd^2 + a^2bc^2d$$ I'm using AM GM relation but I'm not getting the answer, probably due to a conceptual mistake?
We may write $a^2b^2c^2 + b^2c^2d^2 + a^2b^2d^2 + a^2c^2d^2$ as $$\frac{a^2b^2c^2 + b^2c^2d^2}{2} + \frac{b^2c^2d^2 + a^2b^2d^2}{2} + \frac{a^2b^2d^2 + a^2c^2d^2}{2} + \frac{a^2c^2d^2 + a^2b^2c^2}{2}$$ Apply the AM-GM inequality to each term to obtain the result.
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How can it be proven that $\frac{x}{y}+\frac{y}{x}\geq2$, with $x$ and $y$ positive? So I realized that I have to prove it with the fact that $(x-y)^2+2xy=x^2+y^2$ So $\frac{(x+y)^2}{xy}+2=\frac{x}{y}+\frac{y}{x}$ $\Leftrightarrow$ $\frac{(x+y)^2}{xy}=\frac{x}{y}+\frac{y}{x}-2$ Due to the fact that $(x+y)^2$ is a sq...
Let $a = x/y$, then this is equivalent to proving $a + 1/a \geq 2$. $a$ is positive so multiply both sides by $a$ we get $$a^2 - 2a + 1 = (a - 1)^2 \geq 0.$$
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Infinitude of solutions to Bézout's equation If $c=\gcd(a,b)$ for $a,b\in\Bbb Z$ it is possible to write $ax + by = c$ with $x,y\in\Bbb Z$. Is it possible to prove that there are infinitely many integer solutions for $x$ and $y$ without using Diophantine equations, and if so how? (The particular problem I have is $1947...
Writing $ax + by = c$, you get infinitely many solutions as follows: Let $k\in \mathbb Z$ be arbitrary, then \begin{align*} c &= ax + by = ax + \frac{ab}{c}\cdot k + by - \frac{ab}{c}\cdot k\\ &= a\cdot \left(x+\frac{b}{c}\cdot k\right) + b\cdot \left(y-\frac{a}{c}\cdot k\right) \tag{1} \end{align*} is another solution...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1911772", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
limit of product of $(a_1a_2.\dots a_n)^{\frac{1}{n}}$ How to calculate the following limit $$ \lim_{n\rightarrow \infty} \left[ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\cdots \left(1+\frac{n}{n}\right) \right]^\frac{1}{n} .$$ I was trying this by taking the $\log $ of the product and then limit but I am ...
This answer uses Sterling's approximation to get the limit: \begin{align} l=\;&\left[ \left( 1+\frac { 1 }{ n } \right) \left( 1+\frac { 2 }{ n } \right) \cdots \left( 1+\frac { n }{ n } \right) \right]^{ \frac { 1 }{ n } } \\ =\; & \left[ \frac{1}{n^n}(n+1)(n+2)\dots(n+n) \right]^{ \frac { 1 }{ n } } \\ = \;& \fra...
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Why isn't there an $x$ such that $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$? If $\tan\left(\frac{\sin^{-1} x}{5}\right)=1$, then $x$ is equal to $$\frac{-\pi}{2}\le \sin^{-1}x \le \frac{\pi}{2}$$ $$\frac{-\pi}{10}\le \frac{\sin^{-1}x}{5} \le \frac{\pi}{10}$$ This show that no value of such $x$ exist. I want to know wh...
$\sin^{-1} x = \frac{5 \pi}{4}$ gives $x=\sin \frac{5 \pi}{4} = \sin (\pi + \frac{\pi}{4})$, then $x = -\sin \frac{\pi}{4} = -\frac{1}{\sqrt 2}$. Your error is restricting $\sin^{-1} x$ to the values $-\frac{\pi}{2} \le \sin^{-1} x \le \frac{\pi}{2}$ (the principal value is within this range).
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A faster way than tedious multiplying Show that $(a+b+c)(a+b\epsilon +c\epsilon^2)(a+b\epsilon^2 + c\epsilon) = a^3 + b^3 + c^3 - 3abc$ If $$\epsilon^2 + \epsilon + 1 =0$$ The solution in the back of the book is given as Proved by a direct check, taking into consideration that $\epsilon^2 = -\epsilon -1 \: \:$ and ...
One way would be by factorizing $a^3+b^3+c^3-3abc$. $$a^3+b^3+c^3-3abc=(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$$ where $\omega$ is the cube root of unity. More on the factorization can be found here. Then, clearly, $\epsilon=\omega$, and thus, $$\epsilon^3=1$$ Factorizing this gives, $$(\epsilon-1)(\epsilon^...
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$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$, for all integers $n\ge 1$. Let $x_1,\ldots, x_n$ be positive integers. Use mathematical induction to prove that $$\left(\sum_{i=1}^n x_i\right) \cdot \left(\sum_{i=1}^n \frac{1}{x_i}\right) \ge n^2$$ for all integers $n \ge 1$. (Give...
Suppose the given inequality holds for an integer $n(\geq 1)$. Then for $n+1$, $$\left(\sum_{i=1}^{n+1} x_i\right)\left(\sum_{i=1}^{n+1} \frac{1}{x_i}\right)=\left(\sum_{i=1}^{n} x_i+x_{n+1}\right)\left(\sum_{i=1}^{n} \frac{1}{x_i}+\frac{1}{x_{n+1}}\right)$$ Expanding this gives us $$=\left(\sum_{i=1}^{n} x_i\right)\le...
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Mistake in basic algebra, I think? Problem prove $(2n+1)+(2n+3)+(2n+5)+...+(4n-1) =3n^2$ Induction proof: base case $n=1$ assume true for all $n$ prove for $n+1$. The $n$th or last term becomes $(4(n+1)-1)=4n+3$. We also sub $n+1$ in for all $n$ the $n-1$ term is $(4n-1)$ and the first term is $2(n+1)+1=2n+3$ The right...
The left hand side is the sum of the odd numbers from $2n+1$ up to $4n-1$. Thus when going $n\to n+1$, the first summand $2n+1$ is dropped, and two summands $4n+1$ and $4n+3$ are appended a the end.
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Prove: $\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$ Prove that: $$\cos^3{A} + \cos^3{(120°+A)} + \cos^3{(240°+A)}=\frac {3}{4} \cos{3A}$$ My Approach: $$\mathrm{R.H.S.}=\frac {3}{4} \cos{3A}$$ $$=\frac {3}{4} (4 \cos^3{A}-3\cos{A})$$ $$=\frac {12\cos^3{A} - 9\cos{A}}{4}$$ Now, please help m...
First note that $\cos^3\theta=\frac34\cos\theta+\frac14\cos3\theta$. Observing that $$\cos(3A+3\cdot\tfrac23\pi)=\cos(3A+3\cdot\tfrac43\pi)=\cos3A,$$and adding the results of $\cos^3\theta$ for $\theta=A$, $\theta=A+\frac23\pi$, and $\theta=A+\frac43\pi$, using the fact that $\cos A+\cos(A+\frac43\pi)=2\cos(A+\frac23\...
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Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$ I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions? $$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\thet...
\begin{align*} \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} &= \tan^2\theta + \cot^2\theta \\ &= \sec^2\theta - 1+ \csc^2\theta - 1\\ &= \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} - 2\\ &= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta \cos^2\theta} -2\\ &= \sec^2\theta \csc^2\theta - 2 \...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1923555", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 6, "answer_id": 0 }
Let $p,q$ be irrational numbers, such that, $p^2$ and $q^2$ are relatively prime. Show that $\sqrt{pq}$ is also irrational. Progress: Since, $p,q$ are irrationals and $p^2$ and $q^2$ are relatively prime, thus, $p^2\cdot{q^2}$ cannot be a proper square, so, $pq$ is also irrational. Suppose, $pq=k$, then: $$\sqrt{k}\cd...
Suppose $\sqrt{pq}$ is rational, say $\sqrt{pq} = \dfrac mn$ where $m$ and $m$ are positive, relatively prime integers. Then $p^2 q^2 = \dfrac{m^4}{n^4}$ where $p^2q^2$ is an integer. Since $m$ and $n$ are relatively prime integers, then $m^4$ and $n^4$ are also relatively prime integers. It follows that $n^4=1$ and he...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1924527", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 2 }
Show by induction that $n! > n \cdot F_n$ for all $n > 3$ Show by induction that $n! > n \cdot F_n$ for all $n> 3$, where $F_n$ is the $n$th fibonacci number. For the basis step, I put $n=4$ and got: \begin{align*} &4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24 \\ &4 \cdot F_4 = 4 \cdot 3 = 12 \end{align*} So the statement hold...
How about the following way? It is sufficient to prove that $(n-1)!\gt F_n$ for $n\gt 3$. Inductive step : we assume that $(k-1)!\gt F_k$. Then, multiplying the both sides by $k$ gives $$k!\gt kF_k\gt F_k+F_k\gt F_k+F_{k-1}=F_{k+1}$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/1928184", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Partition of a prime $p$ into primes raised to a power to be dvisible by $p$ Find all the partitions of prime $p$ in primes less than $p$. Raise each term in each partition by some power $k > 1$ to see if the sum of these terms will be divisible by $p$. Of course, $k$ can differ for each partition. For example, for $5$...
As McFry has already answered, the first question is answered by Fermat's little theorem: Since $a^p+b^p+\cdots+z^p\equiv a+b+\cdots+z$, there will always be a least power satisfying the OP's condition, and that least power will be no greater than $p$. As for the second question, $p=7=2+2+3$ is an example where the le...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1930650", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
Simplifying inverse trigonometric expressions such as $\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{2x}{x^2-1}\right)$ Okay so I'm just looking for a short cut method or a method that is not so long for simplifying expressions like this $$\cos^{-1}\left(\frac{x^2-1}{x^2+1}\right)+\tan^{-1}\left(\frac{...
Revised for the last part In your specific case $$ \cos ^{ - 1} \left( {\frac{{x^2 - 1}} {{x^2 + 1}}} \right) + \tan ^{ - 1} \left( {\frac{{2x}} {{x^2 - 1}}} \right) $$ you may note that: $$ \left\{ \begin{gathered} x^2 - 1 = \operatorname{Re} \left( {\left( {x + i} \right)^2 } \right) \hfill \\ 2x = \operator...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1931605", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 1 }
Zeros of $a(1-x)x^b-(1-a)x(1-x)^b = 0$ What are the zeros of this equation? $$a(1-x)x^b-(1-a)x(1-x)^b = 0$$ I was told that there were 3 zeros, however I've only been able to find 2 of them. Namely, $x=0$ and $x=1$. I can't think of any way to solve it algebraically and I'm not seeing any other obvious solutions. I a...
The equation has at either $3$ or $4$ roots, assuming that $0 < a < 1$. In particular, the third root they were probably looking for is $$ x = \frac{1}{1 + \sqrt[(b-1)]{r}}, $$ where $r = \frac{a}{1-a}$. This root is between $0$ and $1$. If $b$ is odd and $a \ne \frac12$, there is a second root, $$ x = \frac{1}{1 - \sq...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1933476", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Show that $\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$ is a subgroup of $S_4$ Show that {$1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)$} is a subgroup of $S_4$ My attempt: Let $H = \{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, where $1$ is the identity permutation. Then, $1\circ(1, 2)(3,4) = (1, 2)(3,4) $ $1\circ(1, 3...
Let $S=\{1, (1, 2)(3,4), (1, 3)(2,4), (1,4)(2, 3)\}$, and relabel the elements $\sigma_1=1$, $\sigma_2=(1, 2)(3,4)$, $\sigma_3=(1, 3)(2,4)$, $\sigma_4=(1,4)(2, 3)$, where we note $\sigma_1=1=(1)(2)(3)(4)$ is the identity permutation. Consider the product $(1,3)(2,4){\circ}(1,3)(2,4)$ in $S_4$. Remember when computing p...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1934863", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 1 }
How to find all solutions of $x^2-y^2=x+y$? For this equation $$x^2-y^2=x+y$$ I know there are 2 functional solutions, $y=x-1$, and $y=-x$. The second solution we can see immediately because both sides will always be $0$. The first solution is also easy to get:$$(x-y)(x+y)=x+y\ \ ///\div(x+y)\ \ \ [y\neq-x]\\x-y=1\\y=x...
$x^2 - y^2 = (x-y)(x+y) = x+y$ If $x+y=0$ then all $\{(x, -x)|x \in \mathbb R\}$ are solutions. If $x+y\ne 0$ then $(x-y) = 1$ and $y = x -1$ so all $\{(x, x-1)|x \in \mathbb R\}$ are solutions. So solutions are $\{(x, -x)|x \in \mathbb R\}\cup\{(x, x-1)|x \in \mathbb R\}$. .... or $x^2 - y^2 = x+y$ $x^2 - y^2 -(x+y) =...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1935369", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Arc Circle Intersection in Euclidean 3D While programming on a GPS based game I came upon the following problem for 3D euclidean space: Assume we have an arc (given by a sphere center coordinate C and two euclidean coordinates $A$ and $B$ on that sphere, with $AC = AB$ and $A !=B$) and another sphere (given by an arbit...
I thought about it, and the second answer I proposed earlier is not all that bad. I'm going to treat the points $A$, $B$, $C$, $Q$, as coordinate triples, so that I can add and subtract them as vectors. I'm also going to write $r_2$ for $R$, the radius of the second sphere, and $r_1$ for the radius of the first. Here g...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1937019", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
How to prove $\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$? How to prove this identity? $$\sin 10^\circ = \frac{-1+\sqrt{9-8\sin 50^\circ}}{4}$$ Is this a particular case of a more general identity? Also, is it possible to give a geometric proof of this equality?
The way you would derive this identity is as follows: Note that $\sin A + \sin B = 2\sin\big(\frac{A+B}{2}\big)\cos\big(\frac{A-B}{2}\big)$ Putting $A=50$ and $B=10$, $$ \sin 50 + \sin 10 = 2\sin\big(30\big)\cos\big(20\big) = \cos 20 $$ Now, we know that $\cos 20 = 1- 2\sin^2 10$. So we substitute: $$ \sin 50 + \sin 1...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1939066", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0 }
Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$. Suppose $0< a,b,c < 1$ and $ab + bc + ca = 1$. Find the minimum value of $a + b + c + abc$. How can I use the first two equations to help solve the third? I'm stuck. Any solutions are greatly appreciated!
When $a = b = c = \frac{1}{\sqrt3}$, we have $ab + bc + ca = 1$ and $a + b + c + abc = \frac{10\sqrt3}{9}$. Indeed, the minimum of $a + b + c + abc$ is $\frac{10\sqrt3}{9}$. It suffices to prove that, for all $a, b, c\in (0, 1)$ with $ab + bc + ca = 1$, $$a + b + c + abc\ge \frac{10\sqrt3}{9}.$$ Let $p = a + b + c, q =...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1939973", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 2, "answer_id": 1 }
Find limit of sequence sum Let us have sum of sequence (I'm not sure how this properly called in English): $$X(n) = \frac{1}{2} + \frac{3}{4}+\frac{5}{8}+...+\frac{2n-1}{2^n}$$ We need $$\lim_{n \to\infty }X(n)$$ I have a solution, but was unable to find right answer or solution on the internet. My idea: This can be re...
$$\frac12+\frac34+\ldots+\frac{2n-1}{2^n}=\left(1-\frac12\right)+\left(1-\frac14\right)+\left(\frac34-\frac18\right)+\ldots\left(\frac n{2^{n-1}}-\frac1{2^n}\right)=$$ $$1+1+\frac34+\ldots+\frac n{2^{n-1}}-\frac12-\frac14-\ldots-\frac1{2^n}\xrightarrow[n\to\infty]{}$$ $$\to\sum_{k=1}^\infty\frac k{2^{k-1}}-\sum_{k=1}^\...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1947082", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
Weighted sum of two dice such that the result is a random integer between $0$ and $35$ The numbers shown by 2 dice are labelled $d$ and $e$. $A, B$ and $C$ are constants, giving a score $S=Ad + Be + C$. Find $A, B$ and $C$ such that the range of possible values for $S$ covers all integers from $0$ to $35$, with an equ...
The maximum value of $S$ (assuming $A,B\geqslant0$) is $6(A+B)+C$, and the minimum value $A+B+C$. This gives us the system of equations \begin{align} 6A+6B+C &= 35\\ A+B+C &= 0. \end{align} Gaussian elimination yields $C=-7$, and therefore $A+B=7$. So $(A,B,C)=(6,1,-7)$ and $(1,6,-7)$. Taking $(A,B,C)=(6,1,-7)$ yields ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1949330", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 3, "answer_id": 1 }
Why does $\sum_{n=0}^k \cos^{2k}\left(x + \frac{n \pi}{k+1}\right) = \frac{(k+1)\cdot(2k)!}{2^{2k} \cdot k!^2}$? In the paper "A Parametric Texture Model based on Joint Statistics of Complex Wavelet Coefficients", the authors use this equation for the angular part of the filter in polar coordinates: $$\sum_{n=0}^k \cos...
First note that $$ \left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} = \frac{1}{2^{2k}} \exp\left(-i2k\left(x+\frac{n\pi}{k+1}\right)\right)\sum_{j=0}^{2k} {2k \choose j} \exp\left(i2j\left(x+\frac{n\pi}{k+1}\right)\right) $$ Then, \begin{align} \sum_{n=0}^k \left[\cos\left(x+\frac{n\pi}{k+1}\right)\right]^{2k} &= ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1951708", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 3, "answer_id": 2 }
$y'=$ ${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ Prove that the solutions $y_1$ and $y_2$ for the initial value problem $y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$ are: $y_1=1-x$ $y_2={{-x^2}\over 4}$ And explain why the two solutions not contradiction with the theory of existence and uniqueness. ( sorry I don't spe...
Let $u=\sqrt{x^2+4y}$ , Then $y=\dfrac{u^2-x^2}{4}$ $\dfrac{dy}{dx}=\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}$ $\therefore\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$ $\dfrac{u}{2}\dfrac{du}{dx}-\dfrac{x}{2}=\dfrac{u-x}{2}$ with $u(2)=0$ $\dfrac{u}{2}\dfrac{du}{dx}=\dfrac{u}{2}$ with $u(2)=0$ $\d...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1952299", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Greatest open ball in the unit $n$-cell Let $n\geq2$ be an integer, let $C=[-1,1]^n$ and let $A$ be the set of all real numbers $r$ such that $r$ is the radius of some open ball $V$ such that $V$ is contained in $C$ and $V$ is disjoint to the open ball whose center is $\bf 0$ and radius is $1.$ Find $\sup A.$ I think...
Consider a point $p ∈ [-1, 1]^n \setminus B(\mathbf{0}, 1)$. Observe that $B(p, r) ⊆ [-1, 1]^n \setminus B(\mathbf{0}, 1)$ if and only if $r$ is not greater than the distance from $p$ to any of the faces of the cube and also not greater than the distance from $p$ to $B(\mathbf{0}, 1)$. But we can calculate those! For $...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1954214", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 1, "answer_id": 0 }
How to find the determinant of a 5x5 matrix How do I find the determinant of this? $$\begin{bmatrix} 0& 6& −2& −1& 5\\ 0& 0& 0& −9& −7\\ 0& 15& 35& 0& 0\\ 0 &−1 &−11& −2& 1\\ −2 &−2& 3& 0& −2\end{bmatrix}$$ I tried doing row reductions but every time I get a $0$ and I gain a number. I'm not really sure how to do this b...
By using a Laplace expansion along the first column the problem immediately boils down to computing $R=-2\cdot\det(M)$ with $$ \det M=\det\begin{pmatrix}6&-2&-1& 5 \\ 0 & 0 & -9 & -7 \\ 15 & 35 & 0 & 0 \\ -1&-11&-2&1\end{pmatrix}=-5\cdot\det\begin{pmatrix}6&-2&1& 5 \\ 0 & 0 & 9 & -7 \\ 3 & 7 & 0 & 0 \\ -1&-11&2&1\end{p...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1955784", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Find $\Pr(Y\le 2X)$ for two iid random variables of a given pdf Consider two independent random variables $X$ and $Y$. Let both PDFs are defined by $$f_X(x)=\begin{cases}\displaystyle\frac{4-2x}{3}&\text{if }x\in(0,1),\\0&\text{otherwise.}\end{cases}$$ Find $\Pr(Y \le 2X)$. I solved this the following process. * *I...
Yes, indeed, that's the right integral. But not the right value. $$\begin{align}\mathsf P(Y\leq 2X) ~=~& \int_0^{1/2}\frac{4-2x}3\int_{0}^{2x}\frac{4-2y}3\,\mathsf d y\,\mathsf d x +\int_{1/2}^1\frac{4-2x}3\int_0^1 \frac{4-2y}3\,\mathsf d y\,\mathsf d x \\[1ex] ~=~& \int_0^{1/2}\frac{4-2x}3\frac{4(2x-x^2)}3\,\mathsf d...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1957502", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Show that $x^3 \equiv 0,1,8 \mod{9}$ (Critique my solution) My solution is based on the idea that it is sufficient to consider the cubes less than 9. Ie $0^3 \equiv 0, 1^3\equiv 1, 2^3 \equiv 8$. Since these are the only cubes in the residue class$\mod{9}$ these are the only possibilities for $x^3 \mod{9}$ Is this a su...
According to that reasoning, since $1^3\equiv1,2^3\equiv8,3^3\equiv0\pmod{27}$, we have that $x^3\equiv0,1,8\pmod{27}$ for any $x$. But $4^3\equiv10\pmod{27}$.
{ "language": "en", "url": "https://math.stackexchange.com/questions/1959276", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 1, "answer_id": 0 }
How to prove statement about sum We have such sum: $$\sum_{n=1}^{\infty}\frac{(-1)^{(n+1)}}{n^p}, 0<p<1$$ How to prove that sum is between 1/2 and 1?
Let $ \quad f(x) = 1/x^{p} \quad \colon \{ x \ge 1 \,, \space 0 \lt p \lt 1 \} \space\Rightarrow\space 1/x \lt 1/x^{p} \lt 1 $ $ \Rightarrow \quad f^{\prime}(x) = -p/x^{p+1} \lt 0 \space\Rightarrow\space f^{\prime\prime}(x) = p(p+1)/x^{p+2} \gt 0 \space\Rightarrow\space f(x) \space $ is convex $$ \begin{align} & \\ &...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1960920", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$, then what is $\int_{-1}^{1} p(x)dx$? Q.If $p(x)$ is a cubical polynomial with $p(1)=3,p(0)=2,p(-1)=4$,Then $\int_{-1}^{1} p(x)dx$=__? My attempt: Let $p(x)$ be $ax^3+bx^2+cx+d$ $p(0)=d=2$ $p(1)=a+b+c+d=3$ $p(-1)=-a+b-c+d=4$ From them,we get $b=3.5$,$d=2$...
It could be helpful to do the integration step as well: \begin{align} \int_{-1}^1p(x)dx&=\int_{-1}^1\left(ax^3+bx^2+cx+d\right)\,\text{d}x\\ &=\color{blue}{\frac{a}{4}x^4\big|_{x=-1}^{x=1}}+\color{red}{\frac b3x^3 \big|_{x=-1}^{x=1}}+\color{green}{\frac c2 x^2\big|_{x=-1}^{x=1}}+\color{black}{dx\big|_{x=-1}^{x=1}}\\ &=...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1962163", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 2 }
Probability that a number is divisible by 11 The digits $1, 2, \cdots, 9$ are written in random order to form a nine digit number. Then, the probability that the number is divisible by $11$ is $\ldots$ I know the condition for divisibility by $11$ but I couldn't guess how to apply it here. Please help me in this regard...
A number is divisible by 11 if and only if the sum of the odd-position digits and the even-position digits differ by a multiple of 11. Now, the sum of the digits from 1 to 9 is odd, so the difference in any such number must be either 11 or 33. The only subsets of $\{1,2,3,4,5,6,7,8,9\}$ where the difference between the...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1967378", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "22", "answer_count": 5, "answer_id": 4 }
How to show $\tanh^{-1}(3i)=i\tan^{-1}(3)$? I want to find all the complex values associated with $\tanh^{-1}(3i)$. On WolframAlpha, it computes $\tanh^{-1}(3i)=i\tan^{-1}(3)$. I know that $\tanh^{-1}(z)=\frac{1}{2} \log{\frac{1+z}{1-z}}$ and so $$\begin{align} \tanh^{-1}(3i)&=\frac{1}{2} \log{\frac{1+3i}{1-3i}}\\ &=\f...
put $\arg(-4+3i)=2\theta$ $\displaystyle\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}=-\frac{3}{4} $ $\displaystyle\tan\theta=3$ $\displaystyle\theta=i\tan^{-1}3$
{ "language": "en", "url": "https://math.stackexchange.com/questions/1968000", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 0 }
Equation which has to be solved with logarithms I need some help how to solve these equations for $x$. I think I have to use logarithms but still not sure how to do it and would be really grateful if someone could explain me. $x^2 \cdot 2^{x + 1} +2 ^{\lvert x - 3\rvert + 2} = x^2 \cdot 2^{\lvert x - 3\rvert + 4} + 2^...
1) Making $2^{x-1}=a$ and $2^{|x-3|+2}=b$ you have $$4ax^2+b=4bx^2+a\iff(4x^2-1)(a-b)=0$$ This gives $x=\frac 12$ and $x\ge3$ 2) You have two independent possibilities $$x^2-7x+5=1\\x^2-2x-15=0$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/1976125", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 5, "answer_id": 1 }
Supremum and infimum of asinx + bcosx I am trying to draw conclusions about the supremum and infimum of $F=\{a\sin x+b\cos x: x \in \mathbb{R}\}$ where $a,b \in \mathbb{R}$ are parameters. I am able to rewrite the function as $$\sqrt{a^2+b^2}\left(\frac{a}{\sqrt{a^2+b^2}}sinx+\frac{b}{\sqrt{a^2+b^2}}cosx\right).$$ Noti...
Checking with calculus, we have the extreme point of \begin{align} f(x) = a \cos x + b\sin x \ \ \Rightarrow \ \ f'(x) = -a\sin x + b\cos x. \end{align} Solving for the critical points \begin{align} b\cos x - a\sin x = 0 \end{align} leads to the cases Case 1: $a\neq 0$ Then it follows \begin{align} \tan x = \frac{\sin ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1976228", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
combinatorics: sum of product of integer compositions I am trying to solve a problem from Stanley's book, it says: Fix $k,n \in \mathbb{P}$. Show that: \begin{align} \sum a_1 a_2 \cdots a_k = \binom{n+k-1}{2k-1} \end{align} where the sum ranges over all compositions $(a_1 , a_2 , \ldots , a_k)$ of $n$ into $k$ parts. I...
This is a supplement to @MarkoRiedel's answer which should clarify OPs question. At first we look at the compositions of $n=4$ which consists of two terms ($k=2$) and look at the corresponding terms of the generating function. \begin{array}{crl} 1+3\qquad&\qquad x^1\cdot3x^3=&3x^4\\ 2+2\qquad&\qquad2 x^2\cdot 2 x^2=...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1976347", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 3, "answer_id": 2 }
Difficult limit problem involving sine and tangent I encountered the following problem: $$\lim_{x\to 0} \left(\frac 1{\sin^2 x} + \frac 1{\tan^2x} -\frac 2{x^2} \right)$$ I have tried to separate it into two limits (one with sine and the other with tangent) and applied L'Hôpital's rule, but even third derivative doesn'...
Since $\sin x$ and $\tan x$ are “almost equal” to $x$, separating into a sum of two limits seems like an interesting approach. The first limit is $$ \lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)= \lim_{x\to0}\frac{x^2-\sin^2x}{x^2\sin^2x} $$ Finding a suitable Taylor expansion of the numerator is easy: $$ x^...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1976711", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 3, "answer_id": 2 }
Solution to System of Linear Differential Equations with Variable Coefficients I'm stuck trying to solve the system $\frac{dx(t)}{dt} = A(t)x(t)$ with $A(t) =\frac{1}{2}\begin{pmatrix} ln(t+1) & ln(t-1)\\ ln(t-1) & ln(t+1) \end{pmatrix}$. I think I could try computing $e^{\int A(z) dz}$ and that should give me a fundam...
We shall consider our initial time at $t_0 = 1$ for this post. Since \begin{align} A(t) = \frac{1}{2} \begin{pmatrix} \ln (t+1) & \ln (t-1)\\ \ln (t-1) & \ln (t+1) \end{pmatrix} \end{align} then it follows \begin{align} \int^t_{1} A(s)\ ds = \frac{1}{2} \begin{pmatrix} (t+1)\ln (t+1) +(t+1) & (t-1)\ln(t-1) + (t-1)\\ (t...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1979528", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
The value of following integral $$\frac {x^2\cos (x)}{1+e^x} dx$$ from $[-\pi,\pi] $ ? Now I converted to $$R\frac{e^{ln (x^2)+ix}}{1+e^x}dx $$ wherw R is real part now it isnt an even or odd so those manipulations are useless. $1+e^x=u $ thus $e^xdx=du $ but from here I can go nowhere.
Following @Sangchul Lee's comment (although the idea is pretty straight forward to someone who is aware of basic calculus technics) we have: \begin{align*} \int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^x+1} \, {\rm d}x &\overset{u=-x}{=\! =\! =\! =\!} \int_{-\pi}^{\pi} \frac{x^2 \cos x}{e^{-x} +1} \, {\rm d}x \\ &= \int_{-\...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1979669", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Find the coefficient of $x^{80}$ in the power series expansion $\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}$ Find the coefficient of $x^{80}$ in the power series expansion $$\dfrac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}.$$ I don't know how to find coefficients in power series, solutions are greatly appreciated!
$$\frac{x^2(1+x^2+x^5)}{(1-x)^2(1+x)}=x^2(1+x^2+x^5)\frac{1}{4}\left(\frac{2}{1-x^2}+\frac{1}{(1-x)^2}\right)$$ then use the following $${\frac {1}{1-x^2}}=\sum _{n=0}^{\infty }x^{2n}$$ $${\frac {1}{(1-x)^{2}}}=\sum _{n=1}^{\infty }nx^{n-1}$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/1985027", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2 }
Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$. Maybe a very stupid question but I am stuck. Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$. Is it enough to simply multiply, i.e. $z\bar{z} = \frac{1\times z}{z} = 1$? Showhow I feel this is not correct. I know that if $z = \pm 1$ or $z \pm i$ the...
$z = a + bi$ for $\frac{1}{z} = \bar{z}$ $\frac{1} {a + bi} = a - bi$ $\frac{1} {(a + bi)} \frac{a - bi}{a - bi} = a - bi$ $\frac{a - bi}{a^2 + b^2} = a - bi$ cancel and rearrange $(a^2 + b^2) = 1$ recalling the formula for |z| $|z| = \sqrt{a^2 + b^2} = \sqrt 1 = 1$ the other way if |z| = 1 then $a^2 + b^2 = 1$ $z = a...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1985829", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 6, "answer_id": 3 }
What is the curved asymptote of $\frac{x+1}{\sqrt{3x-2}}$? Please excuse any non-technical language I use... I'm confused by the process for finding the curved asymptote (as $x$ gets really large) of the graph $y=\frac{x+1}{\sqrt{3x-2}}$. I first looked at the (naive?) approach of separating the fraction up into two p...
Using equivalents, when $x$ is large $$x+1\sim x \qquad , \qquad \sqrt{3x-2}\sim \sqrt{3x}\qquad \implies y \sim \frac{x}{\sqrt{3x}}=\frac{\sqrt x}{\sqrt{3}}$$ Letting $x=\frac 1t$, you also could write $$y=\frac{1+\frac 1t}{\sqrt{\frac 3t -2}}=\frac 1{\sqrt t}\frac{1+t}{\sqrt{ 3 -2t}}$$ Now, use Taylor expansion ar...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1987344", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 0 }
Inverse of function $f : \mathbb{N}\times\mathbb{N}\to\mathbb{N}$ Matthew Szudzik's mapping function is another approach on mapping $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$, however, I have trouble finding the inverse of it. The function is $$f(a, b)=\left\{\begin{array}{ll}a^2+a+b&\mbox{, if }a\geq b\\a+b^2&\mbox{...
$f(a,b) = a^2 + a + b=c; a\ge b$ $f(a,b) = a+b^2=c; a< b$. So make wild guesses. If $a' = -b'$ we have $f(a',b) = a'^2 = c$ so $g(c) = (\sqrt{c}, -\sqrt{c})$ will work if $c$ is a perfect square. But obviously $c$ needn't be a perfect square. But we can be close. Suppose $n^2 \le c < (n+1)^2$ or in other words $n ...
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Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$ Find all Pairs of Integers $(x,y)$ , $x \gt y \ge 2$ such that $x^y=y^{x-y}$ My Try: if $y=2$ then $$x^2=2^{x-2}$$ and Taking log on both sides we get $$2 \log x=(x-2) \log 2$$ i.e., $$2 \log_2 x=x-2$$ if $x=2^k$ then we get $$2k=2^k-2$$ or $...
For $x^y=y^{x-y} $ I find that $x=8, y=2$ and $x=9, y=3$ are the only solutions. Must have $x > y$. $x^y =y^{x-y} $ so $(xy)^y = y^x $. Let $x = ry$ where $r > 1$. $(ry^2)^y = y^{ry}$ so $ry^2 = y^r$ or $r = y^{r-2}$ or $y = r^{1/(r-2)}$. If $x = 2y$ then $(2y)^y = y^y$ which has no solution. If $1 < r < 2$, then $r-2 ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1992626", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
How To Write A Formula For The First N Terms of a Sequence? So I have the sequence $$\{5, 15, 45,\cdots\}$$ and I figured out that the formula to find a particular term is $$S_n = 5 \times 3^{n-1}$$ but how do I use this to find the sum of the first $n$ terms?
Geometric Summation: Given the sequence$$a,ar,ar^2,ar^3,\ldots,ar^{n-1}\tag1$$ We have the sum as$$S=a\left(\frac {1-r^n}{1-r}\right)\tag2$$ Where $S$ is the total sum and $a$ is the first term of the sequence. Proof: Multiplying $S$ by $r$, we obtain another equation$$Sr=ar+ar^2+ar^3+\ldots+ar^n\tag3$$ And subtr...
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If $f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$, find the required value We have $$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$, then find the value of $\sum_{k=1}^{3} k f(k)$. Could someone give me slight hint as how to find $f(x)$ here.
$$f(x)= \lim_ {n \to \infty} \sum_{r=1}^{n} \frac{n}{n^2+x^2r^2}$$ $$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{n^2}{n^2+x^2r^2}$$ $$= \lim_ {n \to \infty} \frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+x^2\cdot \frac{r^2}{n^2}}$$ $$= \lim_ {h \to 0} \text{h} \sum_{r=1}^{n} \frac{1}{1+x^2(rh)^2}$$ Then using Riemann ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/1999572", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
Prove that, for all non-negative real numbers $x,y,z$ that satisfy $x+y+z=1, x^2y+y^2z+z^2x≤4/27$ Prove that, for all non-negative real numbers $x, y, z$ that satisfy $x + y + z = 1$, $$x^2 y + y^2 z + z^2 x \leq \frac {4}{27} $$ I'm having trouble with this question. I suspect it may have a fairly simple proof using t...
Let $\{x,y,z\}=\{a,b,c\}$, where $a\geq b\geq c$. Hence, by Rearrangement and AM-GM we obtain: $$x^2y+y^2z+z^2x=x\cdot xy+y\cdot yz+z\cdot zx\leq a\cdot ab+b\cdot ac+c\cdot bc=$$ $$=b(a^2+ac+c^2)\leq b(a+c)^2=4b\left(\frac{a+c}{2}\right)^2\leq4\left(\frac{b+\frac{a+c}{2}+\frac{a+c}{2}}{3}\right)^3=\frac{4}{27}$$
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An Identity for Pell-numbers The Pell-numbers are defined recursively by: $P_0 = 0, P_1 = 1$ and $ P_{n+2} = 2P_{n+1} + P_n$ I am stuck trying to prove the identity: $P_{2n+1}^2 - 1 = 4P_n P_{n+1}(P_{n+1}^2 - P_n^2)$ A proof would be great, otherwise a list of known Pell-identies would also help
You know $P_{n+2} = 2 P_{n+1} + P_{n}$ $\Rightarrow$ $P_{n+2}-(1+\sqrt{2})P_{n+1}= (1-\sqrt{2})(P_{n+1}-(1+\sqrt{2})P_{n})$. In other word, we have \begin{align*} \frac{P_{n+2}-(1+\sqrt{2})P_{n+1}}{P_{n+1}-(1+\sqrt{2})P_{n}} = (1-\sqrt{2}). \end{align*} Since $P_0 = 0, P_1 = 2, P_2 = 2$, we deduce that \begin{align*}...
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Calculate a determinant with pattern A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is: $$ \begin{vmatrix} 0 & 1 & \cdots &n-2&n-1 \\ n-1 & 0 & \cdots &n-3&n-2 \\ ...
I take it this is a Toeplitz matrix (diagonals are constant). Subtract row $n-1$ from row $n$, then row $n-2$ from row $n-1$, ... row $1$ from row $2$. Then subtract column $1$ from each other column. Then add $1/n$ times the second, third, ..., last columns to the first column. The result should be upper triangula...
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Show that $(1-\beta)(p+\beta)=1$ If $$(7+4\sqrt{3})^n = p+\beta,$$ where $n$ and $p$ are positive integers and $\beta$ is a proper fraction, then show that $$(1-\beta)(p+\beta)=1.$$ I cant even understand how to express the term in a positive number and a proper fraction. I would appreciate any hint.
Fun question! The key realization is that $$ (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n \in \mathbb{Z} $$ (do you see why?) and moreover, that $$ 0 < (7 - 4\sqrt{3}) < 1, $$ so that $$ 0 < (7 - 4\sqrt{3})^n < 1, $$ for all natural numbers $n$. It follows from here that \begin{align*} p &= (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})...
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Solve for $x$ $2\log_4(x+1) \le 1+\log_4x$ $2\log_4(x+1) \le 1+\log_4x$ I did: $$2\log_4(x+1) \le 1+\log_4x \Leftrightarrow \log_4(x^2+1) \le 1+\log_4(x) \Leftrightarrow \log_4(\frac{x^2+1}{x}) \le 1 \Leftrightarrow 4 \ge \frac{x^2+1}{x} \Leftrightarrow 4x \ge n^2 +1 \Leftrightarrow 0\ge x^2 -4x +1$$ Using the quadrati...
write your inequation like that: $$\log_4(x+1)^2\le \log_4 4+\log_4 x$$ thus $$\log_4 (x+1)^2\le \log_4 4x$$ this is equivalent to $$(x+1)^2\le 4x$$ can you proceed?
{ "language": "en", "url": "https://math.stackexchange.com/questions/2006674", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2 }
Evaluate $\lim_\limits{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$ $$\lim_{x \to +\infty}\left(x\ln (1+x)-x\ln x + \arctan\frac{1}{2x}\right)^{x^2\arctan x}$$ My attempt \begin{align*} &=\exp \lim_\limits{x \to +\infty} x^2\arctan x \cdot\ln\left[x\ln (1+x)-x\ln x + \arctan\frac{1...
Let $f(x)$ be your function. we have $x\ln(1+\frac{1}{x})+\arctan(\frac{1}{2x})=$ $=x(\frac{1}{x}-\frac{1}{2x^2}+\frac{1}{3x^3})+(\frac{1}{2x}-\frac{1}{24x^3})+\frac{1}{x^3}\epsilon(x)$ $=1+\frac{1}{3x^2}(1+\epsilon(x))$ thus $\ln(f(x))\sim \frac{1}{3}\arctan(x)\;\; (x\to+\infty)$ and $$\lim_{x\to+\infty}f(x)=e^{\frac{...
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interesting scalene triangle trigonometry problem the problem is to find the minimum value of cos(2α)+cos(2β)+cos(2γ) in a scalene triangle. So what should the value of α,β, and γ be? I already transformed the expression using γ=180∘−(α+β) to cos(2α)+cos(2β)+cos(2α+2β) so how can I finish the problem if this is a good ...
Given $A+B+C = \pi$ and $$\cos 2A+\cos 2 B+\cos 2C = 2\cos(A+B)\cos(A-B)+\cos 2C$$ So $$\cos 2A+\cos 2B+\cos 2C = -2\cos C\cdot \cos(A-B)+2\cos^2 C-1$$ $$ = -2\cos C\left[\cos(A-B)-\cos(C)\right]-1=-2\cos C\left[\cos(A-B)+\cos(A-B)\right]$$ $$ = -4\cos A \cos B\cos C-1$$ So $$\cos 2A+\cos 2B+\cos 2C = -1-4\cos A\cos B\...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2009502", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Manipulation of conditions of roots of a quadratic equation. How do I write $a^5+b^5$ in terms of $a+b$ and $ab$. Also is there any general way of writing $a^n+b^n$ in terms of $a+b$ and $ab$?
This is what I found when I first came across this question. Now let me see if I can find the steps to get Lozenges' expression. First of all, Dr. Sonhard's factorization is correct: $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4).$$ Were Zlatan's sign doubt correct, we would get, I suppose, either $ab$ or $a-b$ as the first...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2010573", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 0 }
How does one solve a bivariate normal density function? If the exponent of $e$ of a bivariate normal density is $$\frac{-1}{54} *(x^2+4y^2+2xy+2x+8y+4) \\\text{find } \sigma_{1},\sigma_{2} \text{ and } p \text{ given that } \mu_{1} =0 \text{ and } \mu_{2}=-1. $$ One must use this definition to solve. A pair of rando...
If we substitute $\mu_1=0$ and $\mu_2=-1$ in the exponent of $e $ of the joint probability density formula, we get $$\displaystyle -\frac 1{2(1-\rho^2)}\left[\left(\frac{x}{\sigma_1}\right)^2 -2\rho \left(\frac{x}{\sigma_1}\right) \left(\frac{y+1}{\sigma_2}\right) +\left(\frac{y+1}{\sigma_2}\right)^2\right]$$...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2011353", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
If the matrices $A^3 = 0$, $B^3=0$ and $AB=BA$ then show this: The question: If $A$ and $B$ are square matrices of the same type such that $A^3=0$, $B^3=0$ and $AB=BA$. Show that $$\left(I+A+\frac{A^2}{2!}\right)\left(I+B+\frac{B^2}{2!}\right)=I+(A+B)+\frac{(A+B)^2}{2!}+\frac{(A+B)^3}{3!}+\frac{(A+B)^4}{4!}$$ This is h...
I would do it without brute force: * *Since $A^3=0$ we have $$ e^A=I+A+\frac{A^2}{2!}. $$ Similar for $B$. *The LHS of your expression is then $e^Ae^B$. For commuting matrices it is true that $$ e^Ae^B=e^{A+B}. $$ *Finally notice that if $A^3=B^3=0$ and $AB=BA$ then $(A+B)^5=0$ (e.g. do the binomial expansion and ...
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Disproving a ring Homomorphism Problem My question is: $\varphi=\left\{ \begin{array}{c l} T &\mbox{$\longrightarrow R$} \\ \begin{pmatrix} a & 0\\ b & c \end{pmatrix} & \mbox{$\longrightarrow$ $\begin{pmatrix} a & b\\ 0 & c \end{pmatrix}$} \end{array}\right.$ Show that this function...
I will call $f$ to your $\varphi$. Look that if you have: $$ A= \begin{pmatrix} 1 & 0\\ 1 & 0 \end{pmatrix}, B= \begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix} $$ Then, $$\begin{pmatrix} 1 & 0\\ 1 & 0 \end{pmatrix}*\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix}= \begin{pmatrix} 0 & 0\\ 0 & 0 ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2013078", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
How does one prove that this function is injective? As the title suggests, I am trying to prove that the following function is injective but at the moment I am stuck. $$f(x)=-x^5-16x -1 $$ $$-x_1^5 - 16x_1 -1 = -x_2^5 - 16x_2 -1$$ $$x_1^5 + 16x_1= x_2^5 + 16x_2$$ What should be the next step? Could you also tell me if ...
I suppose that your function is defined on $\mathbb{R}$. Assume that $f(x)=f(y)$. This means that $-x^{5}-16x-1=-y^{5}-16y-1$, so $x^5+16x=y^5+16y$. From this we get $$(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4+16)=0.$$ Then, $x=y$ or $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$. If $x=y$ we're done. Otherwise, $x^4+x^3y+x^2y^2+xy^3+y^4+16=0$....
{ "language": "en", "url": "https://math.stackexchange.com/questions/2013839", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$ In a $\triangle ABC,\angle B=60^{0}\;,$ Then range of $\sin A\sin C$ $\bf{My\; Attempt:}$ Using Sin formula: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ So $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin 60^0}\Rightarrow \sin A = \fr...
It is sufficient to consider the function $$f(x)=\sin (x)\sin(\frac{2\pi}{3}-x)$$ restreint to the domain $D=\{x|\space 0\lt x\lt\dfrac{2\pi}{3}\}$. It is easy to find $f$ is positive and has a maximun at the point $x=\dfrac{\pi}{3}\in D$; furthermore the infimum of $f(x)$ is equal to $0$ taken to the neighborhood of ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2015166", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 3 }
Show that the set of numbers whose Euler function is less than a certain value has an upper bound Let $S = \{n \in \mathbb{Z} \mid \varphi(n) \le 12 \}$ be the set of number whose Euler function is less or equal 12. Prove that $S$ has an upper bound. In the example $S= \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 1...
The Fundamental theorem of arithmetic allows to write $k = p_1^{\alpha_1} \cdot \ldots \cdot p_n^{\alpha_n}$. As $\varphi(p) = p-1$, $p$ must be less than 13. So the only primes allowed are $\{2,3,5,7,11,13\}$. Consider every possibile combination: * *$p_1 = 13$ gives $13$ and $13 \cdot 2 = 26$ *$p_1 = 11$ gives $...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2015685", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$. Prove that $a,b$ are both divisible by $p$. [Solution verification] Problem:Let $p$ be a prime of the form $3k+2$ that divides $a^2+ab+b^2$ for some $a,b\in \mathbb{Z}$. Prove that $a,b$ are both divisible by $p$. My Attempt: $a^2+ab+b^2\equiv 0 \pmod p\R...
Here's a correct different solution. If $p=3k+2$ is an odd prime, $p\mid a^2+ab+b^2$, where $a,b\in\mathbb Z$, then $$a^2+ab+b^2\equiv 0\pmod{p}$$ $$\stackrel{\cdot 4}\iff (a+2b)^2\equiv -3a^2\pmod{p}$$ $$\iff (2a+b)^2\equiv -3b^2\pmod{p}$$ For contradiction, let either $p\nmid a$ or $p\nmid b$. Then either $$\left((a+...
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if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) i want to show that if a,b,c,d are integers such that 5 divides ($a^4+b^4+c^4+d^4$) then 5 also divides (a+b+c+d) but im not quite sure how to aproach this problem.
Mod $5$, we have that the only squares are $0,1,4$. The squares of these are $0,1$. So, we have each of $a^4,b^4,c^4,d^4\in\lbrace 0,1\rbrace $, and their sum is $0$ (becaue $5$ divides them). Because of this, each of $a^4,b^4,c^4,d^4\equiv 0\pmod{5}$. Because of this, we have that $a,b,c,d\equiv 0\pmod{5}$, so the...
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How can I derive what is $1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$ ?? I'd like to evaluate the series $$1\cdot 2\cdot 3\cdot 4 + 2\cdot 3\cdot 4\cdot 5+ 3\cdot 4\cdot 5\cdot 6+\cdots + (n-3)(n-2)(n-1)(n)$$ Since I am a high school student, I only know how to p...
As shown in this post, $$ \sum_{k=1}^n x^k = x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1}$$ Differentiate five times: $$ \sum_{k=1}^n k (k-1)(k-2)(k-3) x^{k-4} = \frac{d^5}{dx^5} \left[ x \sum_{k=1}^{n} \binom{n}{k} (x-1)^{k-1} \right] \tag{1}$$ For left side, $$ \sum_{k=1}^n k (k-1)(k-2)(k-3)(k-4) x^{k-4}$$ We can sta...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2021231", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "8", "answer_count": 10, "answer_id": 9 }
If $ \sin \alpha + \sin \beta = a $ and $ \cos \alpha + \cos \beta = b $ , then show that $\sin(\alpha + \beta) = \frac {2ab } { a^2 + b^2} $ I've been able to do this, but I had to calculate $ \cos (\alpha + \beta) $ first. Is there a way to do this WITHOUT calculating $\cos(\alpha+\beta)$ first ? Here's how I did it ...
How about: $$\begin{array}{} a&=\sin\alpha+\sin\beta&=2\sin\frac 12(\alpha+\beta)\cos\frac 12(\alpha -\beta) \\ b&=\cos\alpha+\cos\beta&=2\cos\frac 12(\alpha+\beta)\cos\frac 12(\alpha -\beta) \\ ab&=4\sin\frac 12(\alpha+\beta)\cos\frac 12(\alpha+\beta)\cos^2\frac 12(\alpha -\beta)&=2\sin(\alpha+\beta)\cos^2\frac 12(\al...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2021356", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 3, "answer_id": 1 }
Real values of $a$ for $x^4+(a-1)x^3+x^2+(a-1)x+1=0$ to have at least two negative roots Problem Statement:- Find all the values of $a$ for which the equation $$x^4+(a-1)x^3+x^2+(a-1)x+1=0$$ possess at least two negative roots. I know that there has been a post regarding this same problem here, but I have a dif...
Just posting an alternative solution, as the comments already helped you to debug your solution: For quartic polynomials, there is a closed-form analytical solution for the zeroes. However horrible that formula may seem, you don't need it here. You only need the discriminant. A polynomial $ax^4+bx^3+cx^2+dx+e=0$ has tw...
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Show that $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ by induction I have some with proving by induction. I cannot find a solution for the inductive step: $1^3 + 2^3 + ... + n^3 = (n(n+1)/2)^2$ I already did the induction steps: Basis: P(1) = $1^3 = (1(1+1)/2)^2$ (This is true) Inductive step: Assume $P(k) = ((k)(k+1)/2)^2$...
First, show that this is true for $n=1$: $\sum\limits_{k=1}^{1}k^3=(1(1+1)/2)^2$ Second, assume that this is true for $n$: $\sum\limits_{k=1}^{n}k^3=(n(n+1)/2)^2$ Third, prove that this is true for $n+1$: $\sum\limits_{k=1}^{n+1}k^3=$ $\color\red{\sum\limits_{k=1}^{n}k^3}+(n+1)^3=$ $\color\red{(n(n+1)/2)^2}+(n+1)^3=$ $...
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How can $\frac{x^3-4x^2+4x}{x^2-4}$ be both $0$ and "undefined" when $x = 2$? Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$ Now I want to find the value of $F(2)$. I can do it in 2 ways: * *Put $x=2$ and solve the function. It will give: $$F(2)=\frac{0}{0}$$ which is not defined. *Solve $...
What you have found is a simplification for $F(x)$, provided $x\neq 2, x\neq -2$. The denominator $(x^2 - 4)$ of the original function makes it undefined at $x = 2, \;x=-2:$ $(2^2-4) = (-2)^2 - 4 = 0$ So your simplification $$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$ is valid, $\forall x \in \mathbb R \s...
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How can simultaneous sinusoidal equations be solved? I have come across a set of simultaneous equations which I can't figure out how to solve. I have 3 equations and only two unknowns, but they are angular quantities and feature in the equations as sinusoidal functions of the angular quantities. The system of equations...
Note that $$ \begin{array}{l} \left( {\begin{array}{*{20}c} x \\ y \\ z \\ \end{array}} \right) = \left( {\begin{array}{*{20}c} {\cos \psi \sin \theta + \cos \theta \sin \phi \sin \psi } \\ {\sin \psi \sin \theta - \cos \psi \cos \theta \sin \phi } \\ {\cos \phi \cos \theta } \\ \end{array}} ...
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Advice on proof in linear algebra. I just wrote my first proof in linear algebra so I'd love some advice on the things that go well and what could be improved upon. It's a proof by induction. Theorem: Let $A_n$ be a $n\times n$ matrix of the form: $\begin{pmatrix} 2 & 1 & 0 & 0 && & \cdots & 0\\ 1 & 2 & 1 & 0 && & \cdo...
Let $d_n = \det A_n$. Note that $d_1 = 2, d_2 = 3$. For the induction step, look at the (clunky) picture below: Note that $d_n = 2 d_{n-1} -d_{n-2}$. Now show that $n \mapsto 3(n-1)$ satisfies the equation with the same initial conditions.
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Solving inequality involving floor function We have this inequality (over real numbers) : $$x^2-2x\le \frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x \rfloor + \lfloor -x \rfloor}$$ How we can solve it using both of algebraic and geometric methods ?
$$\begin{cases}x= \lfloor x\rfloor+\{x\}\\-x= \lfloor -x\rfloor+1-\{x\}\end{cases}\Rightarrow\lfloor x\rfloor+\lfloor -x\rfloor=-1$$ $$\sqrt{1-\lfloor x\rfloor^2}=\begin{cases}0 \text{ if } -1\le x\lt0\\1\text{ if } 0\le x\lt1\\0 \text{ if } 1\le x\lt2\end{cases}$$ Hence $$\frac{\sqrt{1-\lfloor x\rfloor^2}}{\lfloor x\...
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Differential equation with integral that is hard to compute Task and my attempt to solve it $$xy'=\sqrt{2x^2+y^2}-y$$ I have tried substitution like $y=z·x$ but finally got very difficult integral $$ \int \frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{dx}{x}$$ at the left side. Maybe i have to do something different to avoid di...
* *Multiply by $\frac{\sqrt{2+z^2}+2z}{\sqrt{2+z^2}+2z}$: $$ \int\frac{dz}{\sqrt{2+z^2}-2z}=\int\frac{\sqrt{2+z^2}+2z}{(2+z^2)-4z^2}\,dz=\int\frac{\sqrt{2+z^2}}{2-3z^2}\,dz+\int\frac{2z}{2-3z^2}\,dz. $$ The second integral is simple. *In the first integral do the change of variables $t=z+\sqrt{z^2+2}$: $$ \int\frac{\...
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Solve system of equation using matrices (4 variables) The Question: Solve using matrices. $$2w-2x-2y+2z=10\\w+x+y+z=-5\\3w+x-y+4z=-2\\w+3x-2y+2z=-6$$ My work: $$ \begin{bmatrix} 2&-2&-2&2&10\\ 1&1&1&1&-5\\ 3&1&-1&4&-2\\ 1&3&-2&2&-6\\ \end{bmatrix} \rightarrow \begin{bmatrix} 1&-1&-1&1&5\\ 0&2&2&0&-10\\ 0&4&-1&1&-1...
In the second matrix of your work, entry $a_{3,3}$ should be $2$, not $-1$, and entry $a_{3,5}$ should be $-17$, not $-11$. Also, the whole fourth row seems wrong. It appears that the four row operations performed in that step should have been: \begin{align} &1.\text{ Replace R1 with $\frac12\times$ R1.} \\ &2.\text{ R...
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How to show that $f(x,y)=x^4+y^4-3xy$ is coercive? How to show that $f(x,y)=x^4+y^4-3xy$ is coercive ? This is my attempt : $$f(x,y)=x^4+y^4-3xy$$ $$f(x,y)=x^4+y^4\left(1-\frac{3xy}{x^4+y^4}\right)$$ As $||(x,y)|| \to \infty $ , $\frac{3xy}{x^4+y^4} \to 0$ So $||(x,y)|| \to \infty $ , $f(x,y)=x^4+y^4-3xy \to \infty$. I...
Your answer looks right. Here's another way. Observe \begin{align} xy \le \frac{x^2+y^2}{2} \ \ \text{ and }\ \ x^4+y^4\geq \frac{1}{2}(x^2+y^2)^2 \end{align} which means \begin{align} x^4+y^4-3xy \geq&\ x^4+y^4-\frac{3}{2}(x^2+y^2) \\ \geq&\ \frac{1}{2}(x^2+y^2)^2-\frac{3}{2}(x^2+y^2)\\ =&\ \frac{1}{2}\left(x^2+y^2...
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find $g$ such that $g\circ f=h$ Let $$f(x)=\dfrac{2x+3}{x-1} \quad \mbox{and}\quad h(x)=\dfrac{6x^2+8x+11}{(x-1)^2} $$ find polynom $g$ such that $g\circ f=h$ Indeed, $$g\circ f=h \iff \forall x\in \mathbb{R}\quad g\circ f(x)=h(x) $$ \begin{align} g\circ f(x)&=h(x)\\ g\left(f(x)\right)&=h(x)\\ g\left(\dfrac{2x+3}{x-1}...
It is wrong. After you have written the expression of $g(y)$ we should have $$ g(y) = \frac{6(3+y)^{2} + 8(3+y)(y-2) + 11(y-2)^{2}}{25} $$ instead of your expression. On simplifying we get $g(y) = y^{2}+2$, i.e, $$ g(x) = x^{2} + 2$$
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Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$ Solve for $x$ : $x^4 = 5(x-1)(x^2 - x +1)$ I am having difficulty factoring the whole equation so that I can equate each factor to $0$ one by one and get the roots accordingly.
Divide both sides by $x^4$ and change variable to $y = \frac{x-1}{x^2}$, we have $$\begin{align}1 = 5y(1-y) \iff & 4y(y-1) = -\frac45 \iff (2y-1)^2 = \frac15\\ \implies & y = \frac12\left(1 \pm \frac{1}{\sqrt{5}}\right) = \frac{\sqrt{5}\pm 1}{2\sqrt{5}} = \frac{2}{5 \mp \sqrt{5}} \end{align} $$ Substitute $y$ back by ...
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Computation of eigenvalues/vectors of a $9\times 9$ matrix I have a symmetric matrix (with real coefficients) and I need to compute its eigenvalues and eigenvectors. My matrix depends on 3 parameters $(\nu_1,\nu_2,\nu_3)$ that are not independent (in fact we have $\nu_1^2+\nu_2^2+\nu_3^2=1$). If I am using Maple to com...
One can do this with most of the CASs, by taking the condition $a^2+b^2+c^2-1:=0$ as another polynomial equation for computing a Gröbner basis. It depends on your specific matrix, whether or not the complexity of the system of polynomial equations is still manageable. Edit: the characteristic polynomial of $B$ is given...
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Evaluating $\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$ How to evaluate this integral? $$\int_{ - \infty }^\infty {\frac{{{e^{7\pi x}}}}{{{{\left( {{e^{3\pi x}} + {e^{ - 3\pi x}}} \right)}^3}\left( {1 + {x^2}} \right)}}dx}$$ M...
$$I=\frac{1}{8} \int_{-\infty}^{\infty} \frac{\cosh(7\pi z)}{(1+z^2)\cosh^3(3\pi z)} \text{d}z$$ Now we're considering the function $$f(z)=\frac{\cosh(7\pi z)\psi^{(0)}\left ( 1-iz\right ) }{\cosh^3(3\pi z)}$$ From the residue theorem and calculate the residues at $$z=\frac{i}{6},\frac{i}{2},\frac{5i}{6}$$ We finally ...
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Determine the vectors of components For the polynomial vector space $\mathbb{R}[x]$ of degree $\leq 3$ we have the following three bases: $$B_1 = \{1 - X^2 + X^3, X - X^2, 1 - X + X^2, 1 - X\} , \\ B_2 = \{1 - X^3, 1 - X^2, 1 - X, 1 + X^2 - X^3\}, \\ B_3 = \{1, X, X^2, X^3\}$$ How can we determine the following vec...
Hint: $$ B_1= \begin{bmatrix} 1&X&X^2&X^3 \end{bmatrix} \begin{bmatrix} 1&0&1&1\\ 0&1&-1&-1\\ -1&-1&1&0\\ 1&0&0&0 \end{bmatrix} $$ $$ B_2= \begin{bmatrix} 1&X&X^2&X^3 \end{bmatrix} \begin{bmatrix} 1&1&1&1\\ 0&0&-1&0\\ 0&-1&0&1\\ -1&0&0&-1 \end{bmatrix} $$ $$ B_3= \begin{bmatrix} 1&X&X^2&X^3 \end{bmatrix} \begin{bmatrix...
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The Most general solution satisfying equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$ The most general value of $x$ satisfying the equations $\tan x=-1$ and $\cos x=1/\sqrt{2}$, is found to be $x=2n\pi+\frac{7\pi}{4}$. My approach: $$ \frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}=\si...
Here is what you got wrong: $$\frac{\sin x}{\cos x}=-1\implies \sqrt{2}\sin x=-1\implies\sin x=\frac{-1}{\sqrt{2}}$$ So you got $\sin x<0$ and $\cos x>0$. Then you are in the $4$th quadrant. $$\sin x=\frac{-1}{\sqrt{2}}=\sin \left(\frac{7\pi}{4}\right) \Rightarrow x=\frac{7\pi}{4}+2n\pi$$ P.S.: You got wrong because $\...
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General solution to $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ $(\sqrt{3}-1)\cos x+(\sqrt{3}+1)\sin x=2$ is said to have a general solution of $x=2n\pi\pm\frac{\pi}{4}+\frac{\pi}{12}$. My Approach: Considering the equation as $$ a\cos x+b\sin x=\sqrt{a^2+b^2}\Big(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\...
converting the given equation in $$\tan$$ we get $${\frac { \left( 1+\sqrt {3} \right) \left( 1- \left( \tan \left( x/2 \right) \right) ^{2} \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}+2\,{\frac { \left( \sqrt {3}-1 \right) \tan \left( x/2 \right) }{1+ \left( \tan \left( x/2 \right) \right) ^{2}}}...
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Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$ Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$. We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$. How can we continue?
You have that $3y-2$ divides $y^2$. Notice that $(3y-2,y^2)| (3y-2,y)^2$. On the other hand, if $d$ divides $3y-2$ and $y$ we have $d|2$. So $(3y-2,y^2)| 4$. Therefore $3y-2|4$ Therefore $y=0,1$ or $2$. If $y=0$ we have $x=0$. If $y=1$ we have $x=\pm 1$ If $y=2$ we have $x=\pm 1$
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How to factorize $a^2-b^2-a+b+(a+b-1)^2$? The answer is $(a+b-1)(2a-1)$ but I have no idea how to get this answer.
$$(a+b-1)^2=a^2+b^2+1-2a-2b+2ab$$ $$\implies a^2-b^2-a+b+(a+b-1)^2=2a(a+b)-\{2a+(a+b)\}+1$$ $$ =2a\{a+b-1\}-\{a+b-1\}=?$$
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How to integrate $\int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ in a faster way? $\displaystyle \int_a^b (x-a)(x-b)\,dx=-\frac{1}{6}(b-a)^3$ $\displaystyle \int_a^{(a+b)/2} (x-a)(x-\frac{a+b}{2})(x-b)\, dx=\frac{1}{64}(b-a)^4$ Instead of expanding the integrand, or doing integration by part, is there any faster way to ...
You can do this with substitution and Cavalieri's formula: $$ \int_0^1 u^n \,du = \frac{1}{n+1} $$ For the first one, let $u= \frac{x-a}{b-a}$. Then $x = (b-a)u +a$, which means $x-a = (b-a)u$ and $x-b = (b-a)(u-1)$. Also $dx=(b-a)\,du$. So \begin{align*} \int_a^b (x-a)(x-b)\,dx &= (b-a)^3 \int_0^1 u (u...
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How to see there exists const $C$ such that $\frac{d^{n-m}}{dx^{n-m}}(1-x^2)^n=C(1-x^2)^m\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^n$ This comes up in relation to Legendre functions. The claim is made that for $n =0,1,2,3,\cdots$ and $m=0,1,2,3,\cdots,n$, there is a constant $C_{n,m}$ such that $$ \frac{d^{n-m}}{dx^{n-m}}(1...
Suppose we seek to determine the constant $Q$ in the equality $$ Q_{n,m} \left(\frac{d}{dz}\right)^{n-m} (1-z^2)^n = (1-z^2)^m \left(\frac{d}{dz}\right)^{n+m} (1-z^2)^n$$ where $n\ge m.$ We will compute the coefficients on $[z^q]$ on the LHS and the RHS. Writing $1-z^2 = (1+z)(1-z)$ we get for the LHS $$\sum_{p=0}^{n-m...
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Help verify the proof that $\min \left({\lfloor{N/2}\rfloor, \lfloor{(N + 2)/p}\rfloor}\right) = \lfloor{(N + 2)/p}\rfloor$ for $N \ge p$ Well, I think that I have this correct, however my proof seams cumbersome, so can someone check these results or suggest a simplified proof. Prove that ($N \ge p$) \begin{equation*} ...
Show that \begin{equation*} \left\lfloor{\frac{N}{2}}\right\rfloor < \left\lfloor{\frac{N + 2}{p}}\right\rfloor \end{equation*} is not valid. Let $N = k\, p + m$ where $k \in \mathbb{Z}^{+}$ and $m \in \left\{{0}\right.$, $1$, $\cdots$, $\left.{p - 1}\right\}$ by the Quotient Remainder Theorem. Then show this holds f...
{ "language": "en", "url": "https://math.stackexchange.com/questions/2066562", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step: $$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$ This is how I proved it...
Expanding $(1+x)(1+x^2)(1+x^4)\cdots(1+x^{2^n})$ gives $1+x+x^2+\cdots +x^{2^{n+1}-1}$ because every integer $k$ with $0 \le k \le 2^{n+1}-1$ has a unique binary representation (as a sum of powers of $2$).
{ "language": "en", "url": "https://math.stackexchange.com/questions/2067075", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "6", "answer_count": 4, "answer_id": 1 }