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Multiplying square roots How do I simplify the following types of question: $ \sqrt{x^2+5} \times \sqrt{x^2+20}$ Do I need to get both answer out of their roots first or not? This is how I would do it: $ \sqrt{x^2+5} \times \sqrt{x^2+20} = x^2 + x\sqrt{20} + x\sqrt{5} + \sqrt{100}$ but I'm very certain this isn't the ...
For positive real $a,b,c,d$ $\sqrt{a+b}\cdot\sqrt{c+d}=\sqrt{(a+b)(c+d)}=\sqrt{ac+ad+bc+bd}$ which is $\ne \sqrt{ac}+\sqrt{ad}+\sqrt{bc}+\sqrt{bd}$
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Calculating the following limit: $\lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}} $ I am trying to calculate this limit: $$ \lim_{x \to 0} \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}} $$ I've tried using conjugate of both denominator and numerator but I can't get the right result.
$$ \frac{\sqrt{x^2+1}-\sqrt{x+1}}{1-\sqrt{x+1}}=\frac{x^2-x}{-x}\cdot\frac{1+\sqrt{x+1}}{\sqrt{x^2+1}+\sqrt{x+1}}\sim1\cdot\frac{2}{2} $$
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Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$ How can I solve integral of types (1) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4+1}}dx$ (2) $\displaystyle \int\dfrac{1}{\sqrt[4]{x^4-1}}dx$
(2) $$\int \frac{dx}{(x^4-1)^\frac14}=\int\frac{dx}{x(1-\frac1{x^4})^\frac14}=\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$ Let $1-\frac1{x^4}=y^4,4y^3dy=-4\frac{dx}{x^5}\implies \frac{dx}{x^5}=-y^3dy$ and $\frac1{x^4}=1-y^4\implies x^4=\frac1{1-y^4}$ $$\int\frac{x^4dx}{x^5(1-\frac1{x^4})^\frac14}$$ $$=\int\frac{-y^3...
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Showing that $ |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$ For every nonnegative integer $n$ and every real number $ x$ prove the inequality: $$\sum_{k=0}^n|\cos(2^kx)|= |\cos x|+|\cos 2x|+\cdots+|\cos 2^nx|\geq \dfrac{n}{2\sqrt{2}}$$
This is terse and skips some important steps, but I speculate that it can be turned nicely into a clean proof! We show: $$|\cos 2^n x | + |\cos 2^{n+1} x| \geq \frac{1}{\sqrt{2}}$$ We have $$ \sqrt{\cos^2 2^n x } + \sqrt{\cos^2 2^{n+1} x}. $$ And since $\cos^2 x = \frac{1}{2}(\cos(2x)+1)$ we get $$ \frac{1}{\sqrt{2}}\s...
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Does $\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$ Converges? $$\frac{3}{1\cdot 2} - \frac{5}{2\cdot 3} + \frac{7}{3\cdot 4} - ...$$ Do you have an idea about this serie? If it converges what is the sum?
Use the equality $$ \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1} $$
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Finding the derivative of $x$ to the power something that is a function of $x$ if $y = x^{(x+1)^\frac12}$ then how can I get the first derivative of $y$?
$$y=x^{(x+1)^{1/2}}=x^{\sqrt{x+1}}=e^{\sqrt{x+1}\log x}$$ and since $$(\sqrt{x+1}\log x)'=\frac{\log x}{2\sqrt{x+1}}+\frac{\sqrt{x+1}}{x}=\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$ we get, applying as suggested the chaing rule, that $$y'=y\frac{x\log x+2x+2}{2x\sqrt{x+1}}=x^{\sqrt{x+1}}\cdot\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$...
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How does this sum go to $0$? http://www.math.chalmers.se/Math/Grundutb/CTH/tma401/0304/handinsolutions.pdf In problem (2), at the very end it says $$\left(\sum_{k = n+1}^{\infty} \frac{1}{k^2}\right)^{1/2} \to 0$$ I don't see how that is accomplished. I understand the sequence might, but how does the sum $$\left ( \fr...
One can forget about the square root part for a while. Note that $\dfrac{1}{k^2}\lt \dfrac{1}{(k-1)k}$. But $\dfrac{1}{(k-1)k}=\dfrac{1}{k-1}-\dfrac{1}{k}$. Thus your sum is less than $$\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+2}-\frac{1}{n+3}+\cdots.$$ Note the wholesale cancellation: the abov...
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Integrating a Rational Function I am studying for a test and I am trying to evauate the integral below. I know how to simplify it with partial fractions, but when I try to solve it, I cannot seem to find a substitution that will simplify it enough to solve in reasonably quick . I plugged it into wolfram and as usual it...
I think the integral can be solved in a very easy way. $$ \begin{align} \int\frac{x^3+x+2}{x^4+2x^2+1}dx&=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{x^4+2x^2+1}dx\\ &=\int\frac{x^3+x}{x^4+2x^2+1}dx+\int\frac{2}{(x^2+1)^2}dx. \end{align} $$ In the RHS part, for the left integral uses substitution $u=x^4+2x^2+1\;\Righta...
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Simple AM-GM inequality Let $a,b,c$ positive real numbers such that $a+b+c=3$, using only AM-GM inequalities show that $$ a+b+c \geq ab+bc+ca $$ I was able to prove that $$ \begin{align} a^2+b^2+c^2 &=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{a^2+c^2}{2} \geq \\ &\ge \frac{2\sqrt{a^2b^2}}{2}+\frac{2\sqrt{b^2c^2}}{2}+\...
Update: Upps, that is the same as that of @Vincent, sorry didn't see that first. However, it's a bit more explicte A nicer way to use your results and proceed frm there is the following. You've got that $$ a^2+b^2+c^2 \ge ab+bc+ca $$ Now use the fact that $(a+b+c)=S=3$ and multiply each side of your original inequalit...
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Cramer von Mises test statistic I am trying to derive the Cramer von Mises test statistic $$nC_{n}=\frac{1}{12n}+\sum_{i=1}^{n}\left(U_{(i)}-\frac{2i-1}{2n}\right)^2$$ where $U_{(i)}=F_{0}(X_{(i)})$ the order statistics from the original $$C_{n}:=\int (\hat{F}_{n}(t)-F_{0}(t))^2dF_{0}(t)$$ Could anyone help me with t...
Let $X_1,\dots,X_n$ be a random sample of size $n$ from $f(x)$. The empirical CDF is \begin{equation} \hat{F}_n(x) = \frac{1}{n} \sum_{i=1}^n I(X_i \leq x) = \begin{cases} 0, & x < X_{1:n},\\ i/n, & X_{i:n} \leq x < X_{i+1:n},\\ 1, & X_{n:n} \leq x. \end{cases} \end{equation} Let $U_...
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Problem involving permutation matrices from Michael Artin's book. Let $p$ be the permutation $(3 4 2 1)$ of the four indices. The permutation matrix associated with it is $$ P = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \end{bmatrix} $$ This is the matrix that permutes the comp...
The permutation is indeed $(1342)$, but this decomposes as $(13)(14)(12)$, so you should compute $$ P_{(13)} P_{(14)} P_{(12)} = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 0 \end{bm...
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Find $\lim\limits_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$. Find $\displaystyle \lim_{x\to 1}\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}$. I tried to rationalize it, but doesn't help either. Please give me some hints. Thank you.
Hint: $$ \frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}=\left(\frac {\sqrt{x+3}-2}{\sqrt{x+8}-3}\frac {\sqrt{x+3}+2}{\sqrt{x+8}+3}\right)\frac {\sqrt{x+8}+3}{\sqrt{x+3}+2} $$
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$2005|(a^3+b^3) , 2005|(a^4+b^4 ) \implies2005|a^5+b^5$ How can I show that if $$2005|a^3+b^3 , 2005|a^4+b^4$$ then $$2005|a^5+b^5$$ I'm trying to solve them from $a^{2k+1} + b^{2k+1}=...$ but I'm not getting anywhere. Can you please point in me the correct direction? Thanks in advance
$(a^5+b^5) = (a+b)(a^4+b^4) - ab(a^3+b^3)$. So for any $n$, if $n \mid a^3+b^3$ and $n \mid a^4+b^4$ then $n \mid a^5+b^5$.
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Finding the area of a triangle using fractions? To find the area of the triangle do you use Pythagorean theorem from what you have? Could this use similar triangles.
The area of $\triangle PST$ is the sum of the areas of $\triangle PSV$ and $\triangle VST$. $$\begin{align} \text{The area of }\triangle PSV &= \frac12 PV\cdot SV\\ &=\frac12 PV\cdot\frac23 QV\\ &=\frac13 PV\cdot QV, \end{align}$$ $$\begin{align} \text{and the area of }\triangle VST &= \frac12VT\cdot SV\\ &=\frac12 VT\...
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Find the points on the given curve where the tangent line is horizontal or vertical $r=e^θ $ $($ Assume $0 ≤ θ ≤ 2π.)$ Apparently I keep getting this answer wrong. I dont know if i need to use $n $ in the answer or not...
As $r=\sqrt{x^2+y^2}$ and $\theta=\arctan \frac yx$ Differentiating $$\sqrt{x^2+y^2}=e^{\arctan \frac yx}$$ wrt $x,$ $$\frac{x+y\frac{dy}{dx}}{\sqrt{x^2+y^2}}$$ $$=e^{\arctan \frac yx}\cdot\frac1{1+\left(\frac yx\right)^2}\cdot\frac{\left(x\frac{dy}{dx}-y\right)}{x^2}$$ $$=\sqrt{x^2+y^2}\frac{\left(x\frac{dy}{dx}-y\ri...
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Commuting Skew-symmetric Nilpotent 4x4 Matrices Suppose $A$ and $B$ are nonzero, commuting, skew-symmetric, nilpotent matrices in $M_4(k)$, $k$ a field (char $k\ne 2$). Must $A=\lambda B$ for some $\lambda\in k$? I have shown that this is true for $3\times 3$ matrices, and I believe it should also be true for $4\time...
It's not true. Consider $$ A = \left[ \begin {array}{cccc} 0&0&1&-i\\ 0&0&i&1\\ -1&-i&0&0\\ i&-1&0&0 \end {array} \right],\ B = \left[ \begin {array}{cccc} 0&1&-i&0\\ -1&0&0&i\\ i&0&0&1\\ 0&-i&-1&0\end {array} \right]$$ where $i$ is a square root of $-1$. EDIT: Or, a bit more generally, with the same $A$, $$ B = \le...
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Further clarification needed on proof invovling generating functions and partitions (or alternative proof) Show with generating functions that every positive integer can be written as a unique sum of distinct powers of $2$. There are 2 parts to the proof that I don't understand. I will point them out as I outline the...
For your second question, first note that $$g^*(x)=\prod_{k\ge 0}\left(1+x^{2^k}\right)=(1+x)(1+x^2)(1+x^4)(1+x^8)\dots$$ makes sense: if $n\le 2^m$, the coefficient of $x^n$ in $g^*(x)$ is its coefficient in the polynomial $$\prod_{k=0}^m\left(1+x^{2^k}\right)\;,$$ because the contribution of any factor $1+x^{2^k}$ w...
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Evaluation of definite integral using residue theorem $$ \int^{+\infty}_{-\infty} \frac{x-1}{x^3-1} dx$$ I need to evaluate the above integral . My idea is to consider the same integral but with the $x$'s as $z$'s, over the complex plane, have a closed contour integral over $\gamma$, and then use the residue theorem. ...
Note that $\frac{x - 1}{x^3 - 1} = \frac{1}{x^2 + x + 1}$. So the integral is $$ \int_{-\infty}^{+\infty} \frac{1}{x^2 + x + 1} \, dx. $$ First take the integral from $-a$ to $a$ and then take the limit after. Just complete the square in the denominator and use substitution to get an integral in terms of $\arctan$. $$ ...
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General term of $a_{n}= \sqrt{ a_{n-1}+6}$ What is the general term of $a_{n}= \sqrt{ a_{n-1}+6}$,$a_{1}=4$?
I doubt if such general closed form exists. We have $\sqrt{a_{n-1}+6}$ to be $\sqrt{10}$ in the first step, and $\sqrt{6+\sqrt{10}}$ is hard to simplify as if one assume it has the form $a+\sqrt{b}$, then we must have $a^{2}+b=6,2a\sqrt{b}=\sqrt{10}$. This force $a^{2}b=5/2$, and one has to solve $x^{2}-6x+5/2$. This ...
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"Fat" sets of integers and Fibonacci numbers Let us call a set of integers "fat" if each of its elements is at least as large as its cardinality. For example, the set $\{10,4,5\}$ is fat, $\{1,562,13,2\}$ is not. Define $f(n)$ to count the number of fat sets of a set of integers $\{1...n\}$ where we count the empty set...
This problem has a simple solution using ordinary generating functions. Observe that the generating function of subsets of $\{k, n\}$ indexed by number of elements ($u$) and total sum ($x$) is by inspection seen to be $$\prod_{m=k}^n (1+ux^m).$$ Therefore the generating function of the sets being considered is $$\sum...
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Simplify $ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $ Please help me find the sum $$ \frac{1}{x-y}+\frac{1}{x+y}+\frac{2x}{x^2+y^2}+\frac{4x^3}{x^4+y^4}+\frac{8x^7}{x^8+y^8}+\frac{16x^{15}}{x^{16}+y^{16}} $$
Note that each term is of the form $$\dfrac{df_k/dx}{f_k} = \dfrac{d(\log(f_k)))}{dx}$$ Hence, $$\sum_{k=0}^n \dfrac{df_k/dx}{f} = \sum_{k=1}^n \dfrac{d(\log(f_k)))}{dx} = \dfrac{d(\sum_{k=0}^n \log(f_k)))}{dx} = \dfrac{d(\log(f_0 \cdot f_1 \cdot f_2 \cdot f_3 \cdots f_n))}{dx}$$In your case, $f_0 f_1 f_2 \cdots f_n$ r...
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Logarithm simplification Simplify: $\log_4(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}})$ Can we use the formula to solve this: $\sqrt{a+\sqrt{b}}= \sqrt{\frac{{a+\sqrt{a^2-b}}}{2}}$ Therefore first term will become: $\sqrt{\frac{3}{2}}$ + $\sqrt{\frac{1}{2}}$ $\log_4$ can be written as $\frac{1}{2}\log_2$ Please guide further....
Let $x=\sqrt{2+\sqrt 3}+\sqrt{2-\sqrt 3}$ $x^2=(2+\sqrt 3)+(2-\sqrt 3)+2\sqrt{(2+\sqrt 3)(2-\sqrt 3)}=6$ I think you get it from here.
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Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am hav...
Your work looks good so far. Here is a hint: $$ \frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n} $$ To elaborate, apply the hint to get: $$ \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(...
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Proof $ab-cd \ge 3$ $a$, $b$, $c$, and $d$ are real number $a\ge b\ge c\ge d$ $a+b+c+d = 13$ $a^2+b^2+c^2+d^2 =43$ Proof that $ab-cd\ge 3$
Hint $$ (a + b + c + d)^2 = 169 = \sum a^2 + 2 ( ab + bc + cd + da + ac + bd) $$ This gives: $ ab + bc + cd + da + ac + bd = 63 $. We already had $ ab + cd \leq \frac{43}{2} $ from $ (a - b)^2 + (c - d)^2 \geq 0 $. Use the above results with $$ \left[ (a + b) - (c + d) \right]^2 \geq 0 $$
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How to prove that $\frac{(5m)!(5n)!}{(m!)(n!)(3m+n)!(3n+m)!}$ is a natural number? How to prove that $$\frac{(5m)! \cdot (5n)!}{m! \cdot n! \cdot (3m+n)! \cdot (3n+m)!}$$ is a natural number $\forall m,n\in\mathbb N$ , $m\geqslant 1$ and $n\geqslant 1$? If $p$ is a prime, then the number of times $p$ divides $N!$ is...
I became aware of this question due to some recent edits. The shorter proofs seem to rely on $$ \lfloor5x\rfloor+\lfloor5y\rfloor\ge\lfloor x\rfloor+\lfloor y\rfloor+\lfloor3x+y\rfloor+\lfloor3y+x\rfloor $$ My answer also uses this inequality. I have tried to provide a complete, yet shorter, proof of it. Define $$ f(x...
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$\lim_{n \to \infty}\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$ Let $n=1,2,...$ and define $f(n)$ by $$f(n)=\int_0^\infty\left ( \log \left ( {\frac{x+n}{x+\frac{1}{n}}}\right )\right )^2dx$$ For some values of $n$, Matehamtica's shows that $f(n)$ is finite and seems to converge t...
$$ \frac{x + n}{x + n^{-1}} = 1 + \frac{n-1}{nx + 1}$$ Using the expansion of $\log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + O(x^4)$ We see that the resulting expansion of the integrand is $$\frac{n - 1}{nx + 1} - \frac{(n - 1)^2}{2(n x+ 1)^2} + O\left (\left(\frac{n-1}{nx + 1} \right)\right)^3$$ Squaring this again,...
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Find a polynomial of degree $4$ with the coefficient $x^2$ equal to $6$, and zeros $-3$, $ 2$, $ -1$, $-2$ I started off with: $$f(x)= a(x-(-3)) (x-(2)) (x-(-1)) (x-(-2))$$ $$f(x)= a(x+3) (x-2) (x+1) (x+2)$$
You are right that $f(x)=a(x+3)(x+2)(x+1)(x-2)$ Expanding gives $$f(x)=a(x^{2}+3x+2x+6)(x^{2}+x-2x-2)=a(x^{2}+5x+6)(x^{2}-x-2)$$ Multiplying the quadratics, $$f(x)=a(x^{4}+5x^{3}+6x^{2}-x^{3}-5x^{2}-6x-2x^{2}-10x-12)$$ Collecting terms gives $f(x)=a(x^{4}+4x^{3}-x^{2}-16x-12)$ Hence, the coefficient on $x^{2}$ is $...
{ "language": "en", "url": "https://math.stackexchange.com/questions/338205", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 0 }
What may be the ratio of the perimeter of the trapezium to its midline? The diagonals of a symmetric trapezium are perpendicular to each other. What may be the ratio of the perimeter of the trapezium to its midline?
Let the perimeter $p = a+b+2c$, where $c$ are the sides and set $d = d_a + d_b$ to be the length of the diagonals with $d_a$ and $d_b$ being their parts. From the Pythagorean theorem we know that $d_a = \frac{a}{\sqrt{2}}$ and $d_b = \frac{b}{\sqrt{2}}$. Also, $c^2 = d_a^2+d_b^2$, hence $c = \frac{1}{\sqrt{2}}\sqrt{a^2...
{ "language": "en", "url": "https://math.stackexchange.com/questions/338529", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
How prove this summation prove that: $$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\dfrac{n}{2}$$ I think can prove by the probability my idea: $$\dfrac{n}{n+1}+\dfrac{2n(n-1)}{(n+1)(n+2)}+\dfrac{3n(n-1)(n-2)}{(n+1)(n+2)(n+3)}+\cdots=\displaystyle\sum_{k=1}^{n}\dfrac{(n!)^...
Here is another method using the convolution of two generating functions. Observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\...
{ "language": "en", "url": "https://math.stackexchange.com/questions/338850", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 1, "answer_id": 0 }
When are both fractions integers? The sum of absolute values of all real numbers $x$, such that both of the fractions $\displaystyle \frac{x^2+4x−17}{x^2−6x−5}$ and $\displaystyle \frac{1−x}{1+x}$ are integers, can be written as $\displaystyle \frac{a}{b}$, where $a$ and $b$ are coprime positive integers. What is the v...
Note that $$\frac{1+x}{1-x}=\frac{2}{1+x}-1$$ and $$\frac{x^2+4x-17}{x^2-6x-5} = \frac{10x-12}{x^2-6x-5}+1.$$ We need both of these to be integers. So $\frac{2}{1+x}$ must be an integer, and happens when, for example, $x=-3,-2,0,1$ (you also need to check rational values). You can plug these into the second fraction t...
{ "language": "en", "url": "https://math.stackexchange.com/questions/339042", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
Sum of the series $\sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} $ for $n>3$, The sum of the series $\displaystyle \sum_{k=0}^{r}(-1)^k.(k+1).(k+2).\binom{n}{r-k} = $ where $\displaystyle \binom{n}{r} = \frac{n!}{r!.(n-r)!}$ My try:: I have expand the expression $\displaystyle 1.2.\binom{n}{r}-2.3.\binom{n}{r-1}+3.4...
Felix Marin's approach is nice and short, but I thought I'd add another approach that uses standard Binomial tools: Vandermonde's Identity, the symmetry of Pascal's Triangle, and $\binom{n}{k}=\frac nk\binom{n-1}{k-1}$. $$ \hspace{-12pt}\begin{align} &\phantom{={}}\sum_{k=0}^r(-1)^k(k+1)(k+2)\binom{n}{r-k}\\ &=\sum_{k=...
{ "language": "en", "url": "https://math.stackexchange.com/questions/339213", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 5, "answer_id": 4 }
$x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$. How do I prove that the diophantine equation $x^4-4y^4=z^2$ has no solution in positive integers $x$, $y$, $z$.
We have $$x^4-4y^4=z^2\iff\left(2y^2\right)^2+z^2=\left(x^2\right)^2.$$ Therefore if a primitive solution exists then $2y^2=2nm, \ z=m^2-n^2, \ x^2=m^2+n^2.\;$ [Note: Primitive here means one such that $x^2, 2y^2$ and $z$ are coprime which has as a consequence that $n$ and $m$ are coprime.] From $2y^2=2nm$ and because...
{ "language": "en", "url": "https://math.stackexchange.com/questions/339413", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
proving or disproving that two tangent lines are parallel to a curve Im trying to prove or disprove that given the function, $f(x)=0.5\sqrt{1-x^{2}}$, There are two different tangent lines to $f(x)$ so they are parallel. I tried to derivative but with no success.
$y=f(x)=0.5\sqrt{1-x^2}\implies 4y^2=1-x^2$ with $y\ge0$ Let us find the condition of $y=mx+c$ being a tangent to the curve So, $4y^2=1-\left(\frac{y-c}m\right)^2\implies y^2(1-4m^2)+2cy+(m^2-c^2)=0$ This is a Quadratic Equation in $y$ For tangency, both the root must be same i..e, the discriminant must be $0$ So, $$...
{ "language": "en", "url": "https://math.stackexchange.com/questions/339842", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 0 }
Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$? Let $N$ = $11^2 \times 13^4 \times 17^6$. How many positive factors of $N^2$ are less than $N$ but not a factor of $N$? $Approach$: $N$=$11^2$.$13^4$.$17^6$ $N^2$=$11^4$.$13^8$.$17^{12}$ This means $N...
N= 11^2 * 13^4 * 17^6 No of Factors of N = 3*5*7 = 105 There will be one factor which is N, So remaining = 105-1 = 104 No of factors of N^2= 5*9*13 = 585, of these N will be one of the factors. Out of the remaining 584 factors, half will be less than N and half will be grater than N , So no. of factors of N^2 which are...
{ "language": "en", "url": "https://math.stackexchange.com/questions/340533", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 1 }
Proof by induction for a recursive function $F(n) = F(n–1)+F(n–2)$ I'm having a lot of trouble with the following homework question: Consider the following recursive function: Base Case: $F(0) = 0,F(1) = 1$. Recursive Step: $F(n)=F(n−1)+F(n−2)$ for all $n≥2$. Prove that $F(n)\le \left ( \cfrac{1+ \sqrt5}{2} \right) ^{...
Hint: $\displaystyle \frac{1+\sqrt{5}}{2}+1= \frac{3+\sqrt{5}}{2}=(\frac{1+\sqrt{5}}{2})^2$ The above implies that, $$\displaystyle F(k+1)=F(k)+F(k-1)\le(\frac{1+\sqrt{5}}{2})^{k-1}+(\frac{1+\sqrt{5}}{2})^{k-2} \text{ (By induction hypothesis)}$$ $$\displaystyle \Rightarrow (\frac{1+\sqrt{5}}{2})^{k-1}+(\frac{1+\sqrt{...
{ "language": "en", "url": "https://math.stackexchange.com/questions/341737", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 0 }
How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$ Is there an easy way to prove the identity? $$\cos \left ( \frac{\pi}{7} \right ) - \cos \left ( \frac{2\pi}{7} \right ) + \cos \left ( \frac{3\pi}{7} \right ) = \cos \left (\frac{\pi}{3}...
Because $$\cos \frac{\pi}{7} - \cos\frac{2\pi}{7} + \cos\frac{3\pi}{7} = \frac{2\sin\frac{\pi}{7}\cos \frac{\pi}{7} - 2\sin\frac{\pi}{7}\cos\frac{2\pi}{7} + 2\sin\frac{\pi}{7}\cos\frac{3\pi}{7}}{2\sin\frac{\pi}{7}}=$$ $$=\frac{\sin\frac{2\pi}{7}- \sin\frac{3\pi}{7}+\sin\frac{\pi}{7} + \sin\frac{4\pi}{7}-\sin\frac{2\pi}...
{ "language": "en", "url": "https://math.stackexchange.com/questions/347112", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "11", "answer_count": 4, "answer_id": 2 }
Integer solutions for $x^3+2=y^2$? I've heard a famous result that $26$ is the only integer, such that $26-1=25$ is a square number and $26+1=27$ is a cubic number.In other words, $(x,y)=(5,3)$ is the only solution for $x^2+2=y^3$. How if we make it like this $x^3+2=y^2$? Are there any integral solutions? If so, finite...
Edit: Misread the question. Will leave the answer for a while, since the writeup of the solution to the wrong problem may be interesting to MSE users. We need to take a moderately lengthy detour through the arithmetic of $\mathbb{Z}[\sqrt{-2}]$, that is, the arithmetic of numbers of the form $a+b\sqrt{-2}$, where $a$...
{ "language": "en", "url": "https://math.stackexchange.com/questions/347425", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "11", "answer_count": 3, "answer_id": 1 }
Proving $x\equiv 2^{11} \mod{97}$ In $\mathbb{N}$ we suppose the equation $(F): x^{35} \equiv 2 \mod{97}$, and $(97,x)=1$. How could I prove that $x\equiv 2^{11} \mod{97}$ ?
Here we look for relations ('tricks') that will solve the problem without using Fermat's Little Theorem. We interpret the OP's question as a 'how to calculate' ${2^{11}}^{35} \pmod{97}$. Well $2^{11} \equiv 2^8 \times 2^3 \equiv 62 \times 8 \equiv 11 \pmod{97}$ so we've found our first relation, $\tag 1 {2}^{11} {11}^{...
{ "language": "en", "url": "https://math.stackexchange.com/questions/348553", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 4, "answer_id": 3 }
Proving $1-\frac{1}{2}\sin(2x)=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}$ without factoring Is there a way to prove this identity without factoring? $$1-\frac{1}{2}\sin(2x)=\frac{\sin^3x+\cos^3x}{\sin x+\cos x}$$
Motivated by the fact that $(a^2 + b^2 - ab)(a+b) =a^3 + b^3$, we see LHS = $1 - \frac{1}{2} \sin(2x) = \sin(x)^2 + \cos(x)^2 - \sin(x) \cos(x)$ and we are done.
{ "language": "en", "url": "https://math.stackexchange.com/questions/349611", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 2 }
Finding roots of complex equation? Determine all roots of the equation $x^6+(3+i)x^3 + 3i = 0$ in $\mathbb{C}$ Express answers in standard form
Let $x^3 = z$. We then have that $$z^2 + (3+i)z + 3i = 0 \implies (z+3)(z+i) = 0 \implies z = -3, -i \implies x^3 = -3,-i$$ Hence, $$x = \sqrt[3]{-3},\sqrt[3]{-3} \omega, \sqrt[3]{-3} \omega^2, i, i\omega, i\omega^2$$where $\omega$ is the complex cube-root of $1$ and $\sqrt[3]{-3}$ is the real number $y$ such that $y^3...
{ "language": "en", "url": "https://math.stackexchange.com/questions/350727", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
Find Rational Roots and Factor $x^4 + x^3 - 4x^2 + x + 1$ Find all rational roots of the polynomial $x^4 + x^3 - 4x^2 + x + 1$. Factor this polynomial into a product of monic irreducible polynomials over the rationals. I am confused about this question, but isn't the rational root theorem tell us the roots are only $\...
\begin{align} x^4+x^3-4x^2+x+1 & = (x^4 - 2x^2 + 1) + (x^3 - 2x^2 + x) = (x^2-1)^2 + x(x-1)^2\\ & = (x-1)^2 \left( (x+1)^2 + x \right) = (x-1)^2(x^2 + 3x +1)\\ & = (x-1)^2 \left( \left(x+\dfrac32\right)^2 + 1 - \dfrac94\right)\\ & = (x-1)^2 \left( \left(x+\dfrac32\right)^2 - \dfrac54\right) = (x-1)^2 \left( \left(x+\df...
{ "language": "en", "url": "https://math.stackexchange.com/questions/350790", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 1 }
$2\int_0^\pi \sin x + \sin x \cos x\,dx$. Where am I going wrong? $2\displaystyle\int_0^\pi \sin x + \sin x \cos x\,dx$ I let $u=\sin x \implies \dfrac{du}{dx}=\cos x \implies dx=\dfrac{du}{\cos x}$ $2\displaystyle\int_{x=0}^{x=\pi} 2u\,du$ When I try to change the limits I just get 0 and 0: Lower limit $=\sin 0 = 0$...
As $$\int_a^b f(x)dx=\int_a^b f(a+b-x)dx$$ If $f(x)=\sin x+\sin x\cos x,$ $f(\pi+0-x)=\sin(\pi+0-x)+\sin(\pi+0-x)\cos(\pi+0-x)=\sin x-\sin x\cos x$ as $\sin(\pi-x)=\sin x,\cos(\pi-x)=-\cos x$ So, $$\int_0^\pi (\sin x + \sin x \cos x)dx=\int_0^\pi (\sin x - \sin x \cos x)dx=I\text{ say}$$ So, $$2I=\int_0^\pi (\sin x + \...
{ "language": "en", "url": "https://math.stackexchange.com/questions/350936", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 1 }
How many solutions does $x^2 \equiv {-1} \pmod {365}$ have? How many solutions does $x^2 \equiv {-1} \pmod {365}$ have? My thought: $365 = 5 \times 73$ where $5$ and $73$ are prime numbers. So we can obtain $x^2 \equiv {-1} \pmod 5$ and $x^2 \equiv {-1} \pmod {73}$. For $x^2 \equiv {-1} \pmod 5$, we checked $5 \equ...
Because $5$ and $73$ are relatively prime, we have $x^2\equiv -1\pmod{5\cdot 73}$ if and only if $x^2\equiv -1\pmod{5}$ and $x^2\equiv -1\pmod{73}$. Because $5$ is a prime of the form $4k+1$, the congruence $x^2\equiv -1 \pmod{5}$ has a solution, and hence two solutions. The same applies to $x^2\equiv -1\pmod{73}$. Su...
{ "language": "en", "url": "https://math.stackexchange.com/questions/352585", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "7", "answer_count": 2, "answer_id": 0 }
Determine the definite limit The following limit $$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$ evaluates to 1/12. This is my progress so far: $$\lim_{x\to 1}\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}$$ $$\lim_{x\to 1}\frac{1 + \sqrt{x}}{2(1 - x)} - \frac{1 + \sqrt[3]{x} + \sq...
If we pose $x=1-h$ we find $$\frac{1}{2(1 - \sqrt{x})} - \frac{1}{3(1 - \sqrt[3]{x})}=\frac{1}{2(1 - \sqrt{1-h})} - \frac{1}{3(1 - \sqrt[3]{1-h})}$$ and we have $$\sqrt{1-h}=1-\frac{1}{2}h-\frac{1}{8}h^2+o(h^2)$$ and $$(1-h)^{1/3}=1-\frac{1}{3}h-\frac{1}{9}h^2+o(h^2)$$ hence we find $$\frac{1}{2(1 - \sqrt{x})} - \frac{...
{ "language": "en", "url": "https://math.stackexchange.com/questions/354311", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 2, "answer_id": 1 }
Use Cauchy product to find a power series represenitation of $1 \over {(1-x)^3}$ Use Cauchy product to find a power series represenitation of $$1 \over {(1-x)^3}$$ which is valid in the interval $(-1,1)$. Is it right to use the product of $1 \over {1-x}$ and $1 \over {(1-x)^2}$ if I know thier expantion (HOW).. Is ...
This is most easily done with the binomial theorem and the identity $$ \binom{-n}{k}=(-1)^k\binom{n+k-1}{k} $$ which gives $$ \begin{align} \frac1{(1-x)^3} &=\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\ &=\sum_{k=0}^\infty\binom{k+2}{k}x^k\\ &=\sum_{k=0}^\infty\frac{(k+2)(k+1)}{2}x^k \end{align} $$ To use the Cauchy product...
{ "language": "en", "url": "https://math.stackexchange.com/questions/355041", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 5, "answer_id": 0 }
How to find $\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$? Integration by parts is of no success. What else to try? $$\int_0^\pi (\log(1 - 2a \cos(x) + a^2))^2 \mathrm{d}x, \quad a>1$$
Let $z=e^{ix}$ and then $\cos x=\frac{1}{2}(e^{ix}+e^{-ix})=\frac{1}{2}\left(z+\frac{1}{z}\right),dz=izdx$. We have \begin{eqnarray*} \int_0^\pi(\log(1-2a\cos x+a^2))^2dx&=&\frac{1}{2}\int_{-\pi}^{\pi}(\log(1-2a\cos x+a^2))^2dx\\ &=&\frac{1}{2}\int_{|z|=1}\left[\log(1-a\left(z+\frac{1}{z}\right)+a^2)\right]^2\frac{dx}{...
{ "language": "en", "url": "https://math.stackexchange.com/questions/355139", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "14", "answer_count": 6, "answer_id": 5 }
How to derive a closed form of a simple recursion? Consider: $$T(n) = 2 T(n-1) + 1$$ with $T(1)$ a positive integer constant $a$. I just stuck in finding a closed form for this simple recursion function. I would appreciate it, if someone gives me a hint.
A proof with generating functions: Let $$f(X) := \sum_{n=1}^{\infty} T(n) X^n$$ be a formal power series (we don't care about convergence); then you have $$f(X) = aX +\sum_{n=2}^{\infty} (2T(n-1) + 1)X^n$$ $$= aX + 2Xf(X) + \sum_{n=2}^{\infty} X^n$$ $$=2Xf(X) + \frac{X^2}{1-X} +aX.$$ Manipulate this to $$f(X) = \frac{X...
{ "language": "en", "url": "https://math.stackexchange.com/questions/355963", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 2, "answer_id": 0 }
Finding functional equation for generating function I'm given $$ a_n = \sum_{i=2}^{n-2} a_ia_{n-i} \quad (n\geq 3), a_0 = a_1=a_2 = 1 $$ and I need to find the functional equation for the generating function satisfying the above equality. I obtained the correct answer, but I don't understand why some of the steps are a...
For this problem, I would go in the other direction: $\begin{align} f^2(x) &= (\sum_{n=0}^{\infty} a_n x^n) (\sum_{m=0}^{\infty} a_m x^m)\\ &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n x^n a_m x^m\\ &= \sum_{n=0}^{\infty} \sum_{m=0}^{\infty}a_n a_m x^{n+m}\\ &= \sum_{i=0}^{\infty} \sum_{j=0}^{i}a_j a_{i-j} x^{i}\\ &...
{ "language": "en", "url": "https://math.stackexchange.com/questions/356509", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Find all values x, y and z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes. Find all positive integers x, y, z which satisfy the equation $(x^2 + 1)(y^2 + 1) = z^2 + 1$ given that $(x^2 + 1)$ and $(y^2 + 1)$ are both primes. It seems trivial that the only...
Odd primes: We first take care of the case $x^2+1$, $y^2+1$ both odd primes. If $x^2+1$ and $y^2+1$ are odd, then $x$, $y$, and $z$ are even. Factor in the Gaussian integers. We get $$(x-i)(x+i)(y-i)(y+i)=(z-i)(z+i).$$ Note that $x\pm i$ and $y\pm i$ are Gaussian primes. And since $z$ is even, the Gaussian integers...
{ "language": "en", "url": "https://math.stackexchange.com/questions/357970", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "9", "answer_count": 3, "answer_id": 1 }
Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ Show that if $n$ is prime then $2^n - 1$ is not divisible by $7$ for any $n > 3$ Hint: Follow the example in lectures to show that $2^n -1$ is not divisible by 3. In lectures, the example showed the $3 \mid 2^n -1 \iff n$ is even. So wha...
Considering $7|2^n-1$. $2^n-1= (2-1)(2^{n-1}+2^{n-2}+ \dots1)$ $7k=(2^{3m}-1)=(2-1)(2^{3m-1}+2^{3m-2}+ \dots1)$ Let $7k=2^n-1=(2-1)(2^{n-1}+2^{n-2}+ \dots1)=(2-1)(2^{3m-1}+2^{3m-2}+ \dots1)$ $(2^{n-1}+2^{n-2}+ \dots1)=(2^{3m-1}+2^{3m-2}+ \dots1) \implies n=3m$. But $n$ is a prime greater $3$, a contradiction.
{ "language": "en", "url": "https://math.stackexchange.com/questions/359732", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 2, "answer_id": 1 }
What is "prime factorisation" of polynomials? I have the following question: Find the prime factorization in $\mathbb{Z}[x]$ of $x^3 - 1, x^4 - 1, x^6 - 1$ and $x^{12} - 1$. You will need to check the irreducibility in $\mathbb{Z}[x]$, of three quadratic polynomials and of one quartic. In the case of the quadratic, yo...
For quadratics and cubics, irreducibility is easy to test, for if there are no rational roots, the polynomial must be irreducible. And the polynomials $x^2+x+1$ and $x^2-x+1$ don't even have real roots. In more complicated cases, the Rational Roots Theorem could be helpful. Let's deal with $x^{12}-1$. This immediately...
{ "language": "en", "url": "https://math.stackexchange.com/questions/361599", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 2, "answer_id": 0 }
Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that $$4S \sqrt{3}\le a^2+b^2+c^2$$
You can use the Law of cosine and the area formula of a triangle to solve this problem. Suppose the three angles are $A,B,C$ opposite to the sides $a,b,c$, respectively. Then $$ c^2=a^2+b^2-2ab\cos C, S=\frac{1}{2}ab\sin C $$ and hence \begin{eqnarray*} a^2+b^2+c^2-4S\sqrt{3}&=&2[a^2+b^2-ab(\cos C+\sqrt{3}\sin C)]\\ &=...
{ "language": "en", "url": "https://math.stackexchange.com/questions/362361", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 5, "answer_id": 2 }
Summation of sequence $a_n - a_{n-1} = 2n$ $(a_1,a_2,a_3,..)$ be a sequence such that $a_1$ =2 and $a_n - a_{n-1} = 2n$ for $n \geq 2$. Then $a_1 + a_2 + .. + a_{20}$ is equal to? $a_1$ = 2 $a_2$ = 2 + 2x2 $a_3$ = 6 + 2x3 $a_4$ = 12 + 2x4 $a_5$ = 20 + 2x5 $a_n$ = $b_n$ + $g_n$ ,here $g_n$=2xn $b_3$-$b_2$=4 $b_4$-$b_...
I have 2 methods of doing this: Method one: $a_2-a_1=4$, i.e. $a_2=6$ $a_3-a_2=6$, i.e. $a_3=12$ $a_4-a_3=8$, i.e. $a_4=20$ You will notice that $a_k = 2+4+6+8+ \cdots +2k = \frac{k}{2}(2k+2)=k^2 + k$ (using summation formula) $a_1+a_2+ \cdots + a_{20} = \displaystyle\sum_{k=1}^{20} (k^2+k)=\frac{(20)(20+1)(2\times20+1...
{ "language": "en", "url": "https://math.stackexchange.com/questions/363242", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 1 }
Elementary Diophantine equation Solve $(x+y)(xy+1)=2^z$ in positive integres. My attempts is to use $x+y=2^a$, $xy=2^b-1$ and therefore $x,y$ are the roots of the quadratic equation $w^2-2^aw+2^b-1=0$. I try to analyze its dicriminant but it seems to be a dead end...
The idea of discriminant works nicely, as @Jyrki, @Andre and @Mike answered. Yet, I think of another (perhaps more "number theory"ish :) way. First, the equation is symmetric in $x$ and $y$, hence without loss of generality, we assume $x\ge y$. Since $xy+1$ divides $2^z$ and is not equal to $1$, then it should be eve...
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In how many ways i can write 12? In how many ways i can write 12 as an ordered sum of integers where the smallest of that integers is 2? for example 2+10 ; 10+2 ; 2+5+2+3 ; 5+2+2+3; 2+2+2+2+2+2;2+4+6; and many more
Following answer is for general case. Denote by $c_m(1,n)=\binom{n-1}{m-1}$ the number of composition of $n$ into $m$ positive parts, and by $c_m(2,n)$ number of compositions of $n$ into $m$ parts greater than $1$. Each composition of $n$ into $m$ parts greater than $1$, if all parts decrease by 1, becomes a compositi...
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Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$ The question is basically in the title: Prove $7\mid x^2+y^2$ iff $7\mid x$ and $7\mid y$ I get how to do it from $7\mid $ and $7\mid y$ to $7\mid x^2+y^2$, but not the other way around. Help is appreciated! Thanks.
$x^2+y^2 \equiv \mod 7 \implies x^2 \equiv k \mod 7$ and $y^2 \equiv 7-k \mod 7$ And any $a^2 \equiv 0,1,4,2\mod 7$(Why?) $\implies k=0$
{ "language": "en", "url": "https://math.stackexchange.com/questions/365496", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 6, "answer_id": 1 }
Determine the values of $a,b,c$ for which the function is continuous at $x=0$ Determine the values of $a,b,c$ for which the function is continuous at $x=0$ $$ f(x) = \begin{cases} \frac{\sin(a+1)x+sinx}{x} \qquad \text{if} \ x<0 \ ; \\ \\ c \qquad\quad\qquad \text{if} \ x=0 \ ; \\ \\ \frac{\sqrt{x+bx^2}-\sqrt{x}}{b...
Hint: $$\frac{\sqrt{1+bx}-1}{bx}=\frac{\sqrt{1+bx}-1}{bx}\cdot\frac{\sqrt{1+bx}+1}{\sqrt{1+bx}+1}=\frac{1}{\sqrt{1+bx}+1}$$
{ "language": "en", "url": "https://math.stackexchange.com/questions/366357", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 1, "answer_id": 0 }
Evaluate $\int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Evaluate $\displaystyle \int_0^{2\pi} \frac{d\theta}{(1-a\cos(\theta)+a^2)}$ Super general. I get to a step: $\displaystyle \frac{2}{i}$ multiplied by Path integral $\displaystyle \frac{z}{[(2-a)z^2 + 2(a^2 z) + a]}.$ No idea if I'm on the right track. May...
Take a Weierstrass substitution $u = \tan \frac{\theta}{2}$ after reducing the integration range using $\cos(2\pi - \theta) = \cos(\theta)$ in the second integral below: $$ \int_0^{2\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} = \int_0^{\pi} \frac{\mathrm{d}\theta}{a^2+1-a \cos\theta} + \int_{\pi}^{2\pi} \frac{\...
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For positive numbers a,b,c ,$abc=1$ prove$ \frac{1}{a+b} +\frac{1}{a+c}+\frac{1}{c+b} \leq\frac{3}{2}$ For positive numbers $a$,$b$,$c$ , with $abc=1$ prove $$\frac{1}{a+b} +\frac{1}{a+c}+\frac{1}{c+b} \leq\frac{3}{2}$$
The problem as stated is false. Take $a=b=0.1$, $c=100$. $\frac{1}{a+b}=5$. However, if $\min(a,b,c)\ge 1$ (for example, if they are integers), then it's true, by the other posted solution.
{ "language": "en", "url": "https://math.stackexchange.com/questions/371093", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Finding the locus of the midpoint of chord that subtends a right angle at $(\alpha,\beta)$ There is a circle $x^2+y^2=a^2$. On any line that cuts the circle in two distinct points(it is a secant), the points of intersection with circle are taken and at those two points I draw the tangents that intersect at some point, ...
Parameterize circle as $(a \cos \alpha, a \sin \alpha)$ with $\alpha \in \left[ 0 , 2 \pi \right]$, now the tangent vector is given as $(-a \sin \alpha , a \cos \alpha)$. Let the two points correspond to end of chord be at angle $\alpha$ and at angle $\beta$, then: $$ (a \cos \alpha, a \sin \alpha) \cdot ( a \cos \beta...
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Finding a coefficient using generating functions I need to find a coefficient of $x^{21}$ inside the following expression: $$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}$$ I think the only way (using generating functions) is to express the parentheses content with a generating function. The generating function for the $...
$$(x^{2} + x^{3} + x^{4} + x^{5} + x^{6})^{8}=(x^2(1+x+x^2+x^3+x^4))^8=$$ $$=x^{16}\left(\frac{1-x^5}{1-x}\right)^8=x^{16}(1-x^5)^8(1-x)^{-8}=$$ $$=x^{16}\sum_{k=0}^{8}\binom{8}{k}(-x)^{5k}\sum_{l=0}^{\infty}\binom{8+l-1}{l}x^l=$$ $$=\sum_{l=0}^{\infty}\sum_{k=0}^{8}(-1)^{5k}\binom{8+l-1}{l}\binom{8}{k}x^{16+5k+l}$$ $$...
{ "language": "en", "url": "https://math.stackexchange.com/questions/371914", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0 }
Prove That $x=y=z$ If $x, y,z \in \mathbb{R}$, and if $$ \left ( \frac{x}{y} \right )^2+\left ( \frac{y}{z} \right )^2+\left ( \frac{z}{x} \right )^2=\left ( \frac{x}{y} \right )+\left ( \frac{y}{z} \right )+\left ( \frac{z}{x} \right ) $$ Prove that $$x=y=z$$
Let $p = a+b+c$, then $ab+bc+ca = \frac{1}{2}(p^2 - p)$ and $abc=1$. So $a,b,c$ are solutions to the equation $$x^3 - px^2 + \frac{1}{2} (p^2-p)x - 1 = 0$$ The discriminant is $(\frac{1}{2}(p^2-p))^2 - 4(\frac{1}{2}(p^2-p))^3 - 4p^3 - 27 + 18 p (\frac{1}{2}(p^2-p)) \ge 0$, since the roots are real. This is equivalent t...
{ "language": "en", "url": "https://math.stackexchange.com/questions/372143", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "10", "answer_count": 4, "answer_id": 0 }
The number $ \frac{(m)^{(k)}(m)_k}{(1/2)^{(k)} k!}$ For a real number $a$ and a positive integer $k$, denote by $(a)^{(k)}$ the number $a(a+1)\cdots (a+k-1)$ and $(a)_k$ the number $a(a-1)\cdots (a-k+1)$. Let $m$ be a positive integer $\ge k$. Can anyone show me, or point me to a reference, why the number $$ \frac{(m)...
Using Knuth's notation: $$ x^{\underline{k}} = x (x - 1) \ldots (x - k + 1) = (x)_{(k)} \qquad x^{\overline{k}} = x (x + 1) \ldots (x + k - 1) = (x)^{(k)} $$ What you are looking for is: $$ \frac{m^{\overline{k}} m^{\underline{k}}}{(1/2)^{\overline{k}} k!} $$ As if $k > m$ then $m^{\underline{k}} = 0$, while all other ...
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How to solve a polynomial with power fractions like $a-ax+x^{0.8}-x^{0.2}=0$ I have something like $a-ax+x^{0.8}-x^{0.2}=0$ with parameter a>0 and variable x>0. I know by trial and error that the equation has three real roots for parameter a greater than certain value, otherwise it has only one root. x=1 is one obviou...
Let $y=x^{0.2}>0$, then the problem reduces to finding all $a>0$ such that the polynomial equation $a-ay^5+y^4-y=0$ has only 1 positive real root $y=1$. For $y=1$ to be a double root, we must have $-5a(1)^4+4(1)^3-(1)=0$, so $a=\frac{3}{5}$. Putting this into the equation gives $$0=\frac{3}{5}-\frac{3}{5}y^5+y^4-y=-\f...
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Differentiating $\;y = x a^x$ My attempt: $$\eqalign{ y &= x{a^x} \cr \ln y &= \ln x + \ln {a^x} \cr \ln y &= \ln x + x\ln a \cr {1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left(x \times {1 \over a} + \ln a \times 1\right) \cr {1 \over y}{{dy} \over {dx}} &= {1 \over x} + \left({x \over a} + \ln a\...
When going from line $$\ln y = \ln x + x \ln a$$ to $$\frac{1}{y} \frac{dy}{dx} = \frac{1}{x} + (x \times \frac{1}{a} + \ln a \times 1)$$ you made a mistake: when you tried to differentiate $x \ln a$ with respect to x via the product rule, you differentiated $\ln a$ with respect to a, turning it into $\frac{1}{a}$, whe...
{ "language": "en", "url": "https://math.stackexchange.com/questions/376642", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "5", "answer_count": 4, "answer_id": 2 }
Write the expression in terms of $\log{a}$ and $\log{b}$ Write in terms of $\log{a}$ and $\log{b}$ : $\log{\frac{a^3}{b^4}}$? using log laws $log(\frac{x}{y})$ = $\log{x} - \log{y}$ $\log{\frac{a^3}{b^4}} = \log{a^3} - \log{b^4}$ and using log law that $\log{x^y} = y\log{x}$ $\log{a^3} = 3 \log{a}$ & $\log{b^4} = 4\log...
It seems correct, you used well the logarithm laws.
{ "language": "en", "url": "https://math.stackexchange.com/questions/376857", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
finding a primitive root. It says for part A to Find a primitive root r of 38? Im not sure if I did it right. I first calculated $\phi(38)=\phi(19*2)=18$. So there are 18 numbers that are relatively prime to 38. Listing them out we get 1,3,5,7,9,11,13,15,17,21,... so on. So I decided to test out 1. But $ord_1 38$ doe...
As $\phi(19)=18$ and $ord_{19}2$ must divide $\phi(19),$ we only need to test for the powers $1,2,3,6,9,18$ $3^2=9;3^3=27\equiv-11\pmod{38};3^6\equiv(-11)^2\equiv121\equiv7\pmod{38}$ So, $3^9=3^3\cdot3^6\equiv(-11)(7)\equiv-77\equiv-1\pmod{38}$ $\implies 3 $ is a primitive root $\pmod{38}$ Alternatively, let's start w...
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Proving/disproving $\{a^{k_1}\}=\{a^{k_2}\}=\{a^{k_3}\}$ Let $a\in\mathbb{R}\setminus\mathbb{Z}$. Prove or disprove that there do not exist three distinct $k_1, k_2, k_3\in\mathbb{N}$ such that $\{a^{k_1}\}=\{a^{k_2}\}=\{a^{k_3}\}\neq 0$, where $\{x\}=x-\lfloor x \rfloor$.
This is not a complete answer, but I thought the approach might help someone find a solution. We can ask when the following condition holds for positive integers $a>b>c$ and $a>d\ge c$: $$ x^a - x^d - m = P(x)(x^b - x^c - n), $$ where $P(x)$ is some polynomial over $x$, and $m,n\in\mathbb{Z}$. If it does, then $\{x^a...
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Prove by mathematical induction that $1 + 1/4 +\ldots + 1/4^n \to 4/3$ Please help. I haven't found any text on how to prove by induction this sort of problem: $$ \lim_{n\to +\infty}1 + \frac{1}{4} + \frac{1}{4^2} + \cdots+ \frac{1}{4^n} = \frac{4}{3} $$ I can't quite get how one can prove such. I can prove basic di...
i have a solution, but not seem mathematical induction $s = 1 + \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \cdots$ $\dfrac1{4}s = \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \dfrac1{4^5} + \cdots$ $\dfrac1{4}s = - 1 + (1 + \dfrac1{4} + \dfrac1{4^2} + \dfrac1{4^3} + \dfrac1{4^4} + \dfrac1{4^...
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Distance between point and a line - problems simplifying the minimised distance equation Someone asked how to prove the distance between a point $(x_1,y_1)$ and a line $Ax + By + C = 0$ is:$$\text{Distance} = \frac{\left | Ax_{1} + By_{1} + C\right |}{\sqrt{A^2 + B^2} }$$ The currently accepted answer shows that a poin...
Since you wanted to see how this is done algebraically, we'll need to look at a particular property of quadratic functions. As I've said in a comment, if $r > 0$, then $ f(x) = rx^2 + sx + t \ $ represents an "upward-opening" parabola. By completing the square, we can write this as $$f(x) \ = \ r \cdot (x^2 \ + \ \f...
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Find integral $\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$ (most likely substitution) $$\int_0^1 \frac{\ln(1+x^2)}{1+x^2} \ dx$$ I tried letting $x^2=\tan \theta$ but it didn't work. What should I do? Please don't give full solution, just a hint and I will continue.
Integrate as follows \begin{align} &\int_0^1\frac{\ln(1+x^2)}{1+x^2} dx\\ =&\int_0^1\int_0^1 \frac{2x^2y}{(1+x^2)(1+x^2y^2)}dy\ dx=\int_0^1\frac{\frac\pi2 y-2\tan^{-1}y}{y^2-1} \ \overset{y\to \frac{1-y}{1+y}}{dy}\\ = &\ \frac\pi2\int_0^1 \frac{1}{1+y}dy -\int_0^1\frac{\tan^{-1}y}{y}dy=\frac\pi2\ln2-G \end{align}
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To find the x and y-intercepts of the line $ax+by+c=0$ Please check if I've solved the problem in the correct way: The problem is as follows: Find the points at which the line $ax+by+c=0$ crosses the x and y-axes. (Assume that $a \neq 0$ and $b \neq 0$. My solution: We have to find the x and y-intercepts of the line....
easier way is this equation of line intercept form $$ \frac xa+\frac yb=1$$ so write the given eqn in this form $$ax+by+c=0$$ $$ax+by=-c$$ $$\frac {ax}{-c}+\frac {by}{-c}=1$$ $$\frac {x}{\frac{-c}{a}}+\frac {y}{\frac {-c}{b}}=1$$ so intercept at X axis is (-c/a,0) and Y axis is (0,-c/b)
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Reducibility of $x^{2n} + x^{2n-2} + \cdots + x^{2} + 1$ Just for fun I am experimenting with irreducibility of certain polynomials over the integers. Since $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$, I thought perhaps $x^6+x^4+x^2+1$ is also reducible. Indeed: $$x^6+x^4+x^2+1=(x^2+1)(x^4+1)$$ Let $f_n(x)=x^{2n}+x^{2n-2}+\cdots + ...
Note that $f_n(x)(x^2-1) = x^{2n+2} - 1 = (x^{n+1}-1)(x^{n+1}+1)$. Now use unique factorization...
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Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$ Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me? Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$ Isolate one of the square roots: $\sqrt{(2x-5...
I suppose you know this relation: $(a+b)^2=a^2+2ab+b^2$. In the step that you don't understand exactly this relation is used with $a:=1$ and $b:= \sqrt{1-x}$.
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Class Group of $\mathbb Q(\sqrt{-35})$ As an exercise I am trying to compute the class group of $\mathbb Q(\sqrt{-35})$. We have $-35\equiv 1$ mod $4$, so the Minkowski bound is $\frac{4}{\pi}\frac12 \sqrt{35}<\frac23\cdot 6=4$. So we only need to look at the prime numbers $2$ and $3$. $-35\equiv 5$ mod $8$, so $2$ is ...
Note that the ring of integers is $\mathbb Z[(1+\sqrt{-35})/2]$. If you compute $(3, 1 + \sqrt{-35})^2$, you get $$(9,3 + 3\sqrt{-35}, -34 + 2 \sqrt{-35} ) = (9, 1 + \sqrt{-35}) = ( \dfrac{1-\sqrt{-35}}{2} \dfrac{1 + \sqrt{-35}}{2}, 2 \dfrac{1+\sqrt{-35}}{2}) = ((1 + \sqrt{-35})/2 )$$ (because $\dfrac{1-\sqrt{-35}}{2}...
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How to prove the existence of odd numbers $a$ and $b$ in $2m\equiv a^{20}+b^{11}\pmod{2^n}$ Show that for any natural numbers $m,n$, there exist odd numbers $a,b$ such that $$2m\equiv a^{20}+b^{11}\pmod{2^n}$$ Thank you everyone.
We proceed by induction on $n$. When $n=1$, we have $1^{20}+1^{11} \equiv 0 \equiv 2m \pmod{2^1}$. Suppose that the statement holds for $n=k$, so $\exists a, b \in \mathbb{Z}$, $a, b$ odd, such that $2m \equiv a^{20}+b^{11} \pmod{2^k}$. Thus $a^{20}+b^{11} \equiv 2m, 2m+2^k \pmod{2^{k+1}}$. If $a^{20}+b^{11} \equiv 2m ...
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finding nth term Let 3,8,17,32,57 . . . . . be a pattern.How do we find the nth number?My brains are completely jammed,I am tired.I do not even recognize the pattern.I calculated a few ways,but all I want is a little hint,not the whole solution.
Note: I just added a derivation of an explicit formula for the terms as the OP requested. From 3,8,17,32,57, if the term is n, the next term is 2n+k where k = 2, 1, -2, -7. The differences of k are -1, -3, -5. If we assume that the next difference is -7, the next k is -7-7=-14 and the next term is 2*57-14 = 100. To get...
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Simplify $\sqrt n+\frac {1}{\sqrt n}$ for $n=7+4\sqrt3$ If $n=7+4\sqrt3$,then what is the simplified value of $$\sqrt n+\frac {1}{\sqrt n}$$ I was taking LCM but how to get rid of $\sqrt n$ in denominator
You could as OP suggests, add 2, and take the square root. For example, $\dfrac{1}{(7+4\sqrt3)}$ is $(7-4 \sqrt3)$, so the sum of these numbers is $14$. You add $2$ to it to get $16$, and take the square root. The new number is $4 = x + \dfrac {1}x$, which leads to $x = \dfrac{\sqrt{4+2} + \sqrt{4-2}}{2}$ = 1.931...
{ "language": "en", "url": "https://math.stackexchange.com/questions/394041", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 4, "answer_id": 3 }
Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ Find the closed solution of $s_{n} = 3s_{n-1} + 2^{n-2} - 1$ if $s_1 = 0, s_2 = 0, s_3 = 1$ I have attempted to use $p_n = c2^{n-2} - d$ [where $h_n = A(3)^n$, but to no avail] - i ended up with $c=-1$ and $d=-\frac{1}{2}$, which is incorrect. Any help is app...
A general technique is taught by Wilf's "generatingfunctionology". Define $S(z) = \sum_{n \ge 0} s_n z^n$ and write ($s_0$ you get from the recurrence "backwards", mostly for not having to mess around with indices): $$ s_{n + 1} = 3 s_n + 2^{n - 1} - 1 \qquad s_0 = \frac{1}{6} $$ Multiply the recurrence by $z^n$, add o...
{ "language": "en", "url": "https://math.stackexchange.com/questions/395282", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 2, "answer_id": 1 }
Formulas for calculating pythagorean triples I'm looking for formulas or methods to find pythagorean triples. I only know one formula for calculating a pythagorean triple and that is euclid's which is: $$\begin{align} &a = m^2-n^2 \\ &b = 2mn\\ &c = m^2+n^2 \end{align}$$ With numerous parameters. So are there other fo...
Here is the method I use: * *Take a number $c >= 1$ *Find all pairs $a, b$ such that $ab = 2c^2$ *For every pair of $a, b$ you get an unique solution $x = a + 2c, y = b + 2c, z = a + b + 2c$. It can be easily proven that $x^2 + y^2 = z^2$ is reduced to $4c^2 = 2ab$ which is satisfied at point 2. The advantage of...
{ "language": "en", "url": "https://math.stackexchange.com/questions/395543", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "3", "answer_count": 9, "answer_id": 4 }
Evaluating $\int_0^\infty \frac{\log (1+x)}{1+x^2}dx$ Can this integral be solved with contour integral or by some application of residue theorem? $$\int_0^\infty \frac{\log (1+x)}{1+x^2}dx = \frac{\pi}{4}\log 2 + \text{Catalan constant}$$ It has two poles at $\pm i$ and branch point of $-1$ while the integral is to b...
In this answer, the substitution $x=\frac{1-y}{1+y}$ is used to get $$ \int_0^1\frac{\log(1+x)}{1+x^2}\mathrm{d}x=\frac\pi8\log(2)\tag{1} $$ We can use the substitution $x\mapsto1/x$ to get $$ \begin{align} \int_1^\infty\frac{\log(1+x)}{1+x^2}\mathrm{d}x =\int_0^1\frac{\log(1+x)-\log(x)}{1+x^2}\mathrm{d}x\tag{2} \end{a...
{ "language": "en", "url": "https://math.stackexchange.com/questions/396170", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "16", "answer_count": 9, "answer_id": 2 }
linear Transformation of polynomial with degrees less than or equal to 2 I would like to determine if the following map $T$ is a linear transformation: \begin{align*} T: P_{2} &\to P_{2}\\ A_{0} + A_{1}x + A_{2}x^{2} &\mapsto A_{0} + A_{1}(x+1) + A_{2}(x+1)^{2} \end{align*} My attempt at solving: \begin{align} T(p + q)...
Note that you're hiding the core of the proof, which is simply that if $p,q$ are polynomials, we define their sum, which is also a polynomial, and $(p+q)(x)=\sum_{i=1}^n (a_i+b_i)x^i=p(x)+q(x)$. Thus, if $T$ is your transformation, $$T((p+q)(x))=(p+q)(x+1)=p(x+1)+q(x+1)=T(p(x))+T(q(x))$$ and $$T(\lambda p(x))=(\lambda ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/397748", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 3, "answer_id": 0 }
Solve for $x$, $3\sqrt{x+13} = x+9$ Solve equation: $3\sqrt{x+13} = x+9$ I squared both sides and got $9 + x + 13 = x^2 + 18x + 81$ I then combined like terms $x^2 + 17x + 59 = 0$ I then used the quadratic equation $x= -\frac{17}2 \pm \sqrt{\left(-\frac{17}2\right)^2-59}$ However, the answer is 3
3 √x+13 = x + 9 Squaring both sides, we get 9 (x + 13) = x^2 + 81 + 18x 9x + 117 = x^2 + 81 + 18x x^2 + 18x - 9x + 81 - 117 = 0 x^2 + 9x - 36 = 0 x^2 + 12x - 3x - 36 = 0 x(x + 12) - 3(x + 12) = 0 (x + 12)(x - 3) = 0 Therefore, x = -12 or 3 However, when the problem equation does not get satisfied when x is substituted ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/398051", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 3 }
Behavior of $|\Gamma(z)|$ as $\text{Im} (z) \to \pm \infty$ Let $\Gamma(z)$ be the gamma function. In a paper I'm reading, the author states that $$ |\Gamma(z)| = |\Gamma(a+ib)| \sim \sqrt{2 \pi} |b|^{a-\frac{1}{2}} e^{-|b|\frac{\pi}{2}}$$ as $|b| \to \infty$. Can this asymptotic behavior be derived from Stirling's fo...
Assume $|y|\gg|x|$. Note that $$ \begin{align} y\arctan\!2(y,x) &=|y|\arctan\!2(|y|,x)\\[6pt] &=\frac\pi2|y|-|y|\arctan\!2(x,|y|)\\ &\sim\frac\pi2|y|-x \end{align} $$ and $$ x^2+y^2\sim|y|^2 $$ Therefore, $$ \begin{align} \left|\sqrt{2\pi}\,\frac{\color{#C00}{z^{z-1/2}}}{\color{#090}{e^z}}\right| &=\sqrt{2\pi}\color{#C...
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About the property of $m$: if $n < m$ is co-prime to $m$, then $n$ is prime The number $30$ has a curious property: All numbers co-prime to it, which are between $1$ and $30$ (non-inclusive) are all prime numbers! I tried searching(limited search, of course) for numbers $\gt 30$ that have this property, but could not f...
Let $n>30$ be such a number. Observe that $p^2<n$ implies $p|n$. Thus $2|n$, $3|n$, $5|n$ follow directly from $n>25>9>4$. Thus $n$ is a multiple of $30$ and $n>30$, which implies $n\ge 60>49$ and hence $7|n$. Up to now, $n$ is divisible by the four smallest primes. Lemma. The product of four consecutive primes is gre...
{ "language": "en", "url": "https://math.stackexchange.com/questions/398399", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "26", "answer_count": 4, "answer_id": 2 }
how to solve this multivariate quadratic equation? Any hope to acquire an analytic solution to such equations: Solve: $$\sum_{j=1}^n a_{ij} x_i x_j = b_i$$ for $i=1,\ldots,n$, where $a_{ij}$'s and $b_i$'s are known constants and $x_i$'s are unknowns to be solved. Thanks a lot! P.S. Thanks for alex.jordan's interesting ...
Let $A=\left( a_{ij} \right)_{n \times n} $, $B=\mathrm{diag}\{b_1, b_2, \cdots, b_n\}=\begin{pmatrix} b_1 \\ & \ddots \\ & & b_n \end{pmatrix}$, $C=A^{-1}B$. There is a possible numerical solution when $A$ is invertible and $b_i\neq 0$ ($i$ = 1, 2, $\cdots$, $n$ ). From the conditions, it will result $$ \begin{pmatr...
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Trigonometric function integration: $\int_0^{\pi/2}\frac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$ How to integrate $$\int_0^{\pi/2}\dfrac{dx}{(a^2\cos^2x+b^2 \sin^2x)^2}$$ What's the approach to it? Being a high school student , I don't know things like counter integration.(Atleast not taught in India in high school education...
You want to integrate $$J = \int_0^{\pi/2} \dfrac{dx}{\left(a^2 +(b^2-a^2) \sin^2(x) \right)^2} = \dfrac1{a^4}\int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$$ where $c = \dfrac{b^2-a^2}{a^2}$. We now want to integrate $I = \displaystyle \int_0^{\pi/2} \dfrac{dx}{\left(1 +c \sin^2(x) \right)^2}$. From Taylor ...
{ "language": "en", "url": "https://math.stackexchange.com/questions/402223", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "7", "answer_count": 4, "answer_id": 1 }
Proving that sequence $a_n = \sqrt{x \sqrt{x \sqrt{x \sqrt{\cdots}}}} = x^{1-2^{-n}}$ Let $x>0$. For sequence $a_n$, such that $n$ denotes the $n$th term: $$\begin{align} a_1&= \sqrt{x}\\ a_2&= \sqrt{x \sqrt{x}}\\ a_3&= \sqrt{x \sqrt{x \sqrt{x}}}\\ a_4&= \sqrt{x \sqrt{x \sqrt{x \sqrt{x}}}}\\ &\vdots\\ a_{n-1}&= \sqrt{...
Hint: \begin{align} a_4&=\sqrt{x \sqrt{x\sqrt{{x \sqrt{x}}}}}\\&=\sqrt{x \sqrt{x\sqrt{{x^{\frac{3}{2}} }}}}\\&=\sqrt{x \sqrt{x^{\frac{7}{4}}}}\\&=\sqrt{x^{\frac{15}{8}}}\\&=x^{\frac{15}{16}}\\&=x^{1-2^{-4}} \end{align} Can you use induction in these footsteps?
{ "language": "en", "url": "https://math.stackexchange.com/questions/403279", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 1, "answer_id": 0 }
Simplified form of $\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$. Tried this one a couple of times but can't seem to figure it out. I am trying to simplify the expression: $$\left(6-\frac{2}{x}\right)\div\left(9-\frac{1}{x^2}\right)$$ So my attempt at this is: $$=\bigg(\dfrac{6x}{x}-\dfrac{2}{x}\bigg)\di...
$$\dfrac{(6x^3-2x^2)}{x(9x^3-x)} = \dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} $$ Cancel common factor of $x$ in numerator and denominator gives us: $$\dfrac{2x^2(3x - 1)}{x(9x^2 - 1)} = \frac{2x(3x-1)}{\left[9x^2 - 1\right]}$$ Now we have a difference of squares in the denominator, and can factor it: $$\frac{2x(3x-1)}{\color{bl...
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Subtracting roots of unity. Specifically $\omega^3 - \omega^2$ This is question that came up in one of the past papers I have been doing for my exams. Its says that if $\omega=\cos(\pi/5)+i\sin(\pi/5)$. What is $\omega^3-\omega^2$. I can find $\omega^3$ and $\omega^2$ by De Moivre's Theorem. But I cant make much headwa...
$$\omega=\cos(\pi/5)+i\sin(\pi/5)$$ $$\omega^3-\omega^2=\cos(3\pi/5)+i\sin(3\pi/5)-(\cos(2\pi/5)+i\sin(2\pi/5))$$ $$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$ $$\omega^3-\omega^2=\cos(3\pi/5)-\cos(2\pi/5)+i(\sin(3\pi/5)-\sin(2\pi/5))$$ $$\omega^3-\omega^2=-2\sin(\pi/2)\cdot\sin(\pi/10)-2...
{ "language": "en", "url": "https://math.stackexchange.com/questions/405183", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 3, "answer_id": 1 }
$R_n = 3(2^n)-4(5^n)$, $n \geq0$, prove $R_n$ satisfies $R_n = 7R_{n-1}-10R_{n-2}$ So the question is: $R_n=3(2^n)-4(5^n)$ for $n\ge 0$; prove that $R_n$ satisfies $R_n=7R_{n-1}-10R_{n-2}$. I don't really know what to do from here. If I substitute $$R_n = 3(2^n)-4(5^n)$$ into $$Rn = 7R_{n-1}-10R_{n-2}$$ I end up ge...
Having rewritten the recurrence, we get an equation: $$10R_{n-2}-7R_{n-1}+R_n=0$$ that can easily be solved. The roots of a characteristical polymomial $10-7k+k^2=0$ are $k_1=2,k_2=5$. So general solution is $R_n=c_12^{n}+c_25^{n}$, where $c_1$ and $c_2$ are arbitrary constants.
{ "language": "en", "url": "https://math.stackexchange.com/questions/405384", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "2", "answer_count": 4, "answer_id": 1 }
$1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1 = 12$ or $1$? Does the expression $1+1+1+1+1+1+1+1+1+1+1+1 \cdot 0+1$ equal $1$ or $12$ ? I thought it would be 12 this as per pemdas rule: $$(1+1+1+1+1+1+1+1+1+1+1+1)\cdot (0+1) = 12 \cdot 1 = 12$$ Wanted to confirm the right answer from you guys. Thanks for your help.
If you type $1+1+1+1+1+1+1+1+1+1+1+1⋅0+1$ into most pocket calculators you will get the answer $1$. This is because the calculator applies the next operation to the result of the previous one, effectively computing$$(1+1+1+1+1+1+1+1+1+1+1+1)⋅0+1=0+1=1$$ When using a calculator you have to respect the logic of the machi...
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Convergence of $1+\frac{1}{2}\frac{1}{3}+\frac{1\cdot 3}{2\cdot 4}\frac{1}{5}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\frac{1}{7}+\cdots$ Is it possible to test the convergence of $1+\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1\cdot 3}{2\cdot 4}\dfrac{1}{5}+\dfrac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\dfrac{1}{7}+\cdots$ by Gauss test...
Ratio test is not enough. Example 1: $$S=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\cdots$$ $$\dfrac{u_n}{u_{n+1}}=\dfrac{\left(\dfrac{1}{n}\right)}{\left(\dfrac{1}{n+1}\right)}=\frac{n+1}{n}=1+\frac{1}{n}$$ Ratio will be $1$ if $n--->\infty$ But we know that S series is divergent. So the ratio test says us If the ratio...
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How to simplify $a^n - b^n$? How to simplify $a^n - b^n$? If it would be $(a+b)^n$, then I could use the binomial theorem, but it's a bit different, and I have no idea how to solve it. Thanks in advance.
$$ a^n - b^n = (a-b+b)^n - b^n = \sum_{k=0}^n {n \choose k}(a-b)^{k}b^{n-k} - b^n = (a-b)b^{n-1} + {n \choose 1}(a-b)^2b^{n-2} + ... + {n \choose n-1} (a-b)^{n-1}b + (a-b)^n = (a-b)(b^{n-1} + {n \choose 1}(a-b)b^{n-2} + {n \choose 2}(a-b)^2b^{n-3} + {n \choose 3}(a-b)^3b^{n-4} + ... + {n \choose n-1} (a-b)^{n-2}b + (a...
{ "language": "en", "url": "https://math.stackexchange.com/questions/406703", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "12", "answer_count": 3, "answer_id": 0 }
What does $\lim\limits_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$ evaluate to? What does $$\lim_{x\to\pi/6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ evaluate to? This very likely needs substitution.
without L-hospital law $$\lim_{x\to\dfrac \pi6}\frac{1-\sqrt{3}\tan x}{\pi-6x}$$ $$\lim_{(x-\dfrac \pi6)\to 0}\frac{1-\sqrt{3}\dfrac {\sin x}{\cos x}}{6\left(\dfrac\pi 6-x\right)}$$ $$\lim_{(x-\dfrac \pi6)\to 0}\dfrac{1\cdot\cos x-\sqrt{3} \cdot{\sin x}}{6\left(\dfrac\pi 6-x\right)\cos x}$$ $$\lim_{(x-\dfrac \pi6)\to 0...
{ "language": "en", "url": "https://math.stackexchange.com/questions/408315", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 6, "answer_id": 3 }
Why dividing by zero still works Today, I was at a class. There was a question: If $x = 2 +i$, find the value of $x^3 - 3x^2 + 2x - 1$. What my teacher did was this: $x = 2 + i \;\Rightarrow \; x - 2 = i \; \Rightarrow \; (x - 2)^2 = -1 \; \Rightarrow \; x^2 - 4x + 4 = -1 \; \Rightarrow \; x^2 - 4x + 5 = 0 $. Now he...
The division your teacher did is polynomial division. He did not divide by zero; he divided by $x^2 - 4x + 5$. The long division he did is just an algorithm that allows you to get the following identity: $$x^3 - 3x^2 + 2x - 1 = (x^2 - 4x + 5)(x+1) + x-6$$ This is an identity involving multiplication, not division. Now,...
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How find $\int(x^7/8+x^5/4+x^3/2+x)\big((1-x^2/2)^2-x^2\big)^{-\frac{3}{2}}dx$ How can I compute the following integral: $$\int \dfrac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{\left(\left(1-\frac{x^2}{2}\right)^2-x^2\right)^{\frac{3}{2}}}dx$$ According to Wolfram Alpha, the answer is $$\frac{x^4 - 32x^2 + 20}{2\sqr...
Perform the substitutions $y=\frac{x^2}{2}, y=2+\sqrt{3}\sec{\theta}$ to get: \begin{align} &\int{\frac{\frac{x^7}{8}+\frac{x^5}{4}+\frac{x^3}{2}+x}{((1-\frac{x^2}{2})^2-x^2)^{\frac{3}{2}}} dx} \\ & =\int{\frac{y^3+y^2+y+1}{(y^2-4y+1)^{\frac{3}{2}}} dy} \\ &=\int{\left(\frac{y+5}{\sqrt{(y-2)^2-3}}+\frac{20y-4}{\sqrt{((...
{ "language": "en", "url": "https://math.stackexchange.com/questions/408995", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "1", "answer_count": 1, "answer_id": 0 }
If $\frac{\cos^4\theta}{\cos^2\phi}+\frac{\sin^4\theta}{\sin^2\phi}=1$, show $\frac{\cos^4\phi}{\cos^2\theta} +\frac{\sin^4\phi}{\sin^2\theta}=1$ If $\dfrac{\cos^4\theta}{\cos^2\phi}+\dfrac{\sin^4\theta}{\sin^2\phi}=1$, prove that $\dfrac{\cos^4\phi}{\cos^2\theta} +\dfrac{\sin^4\phi}{\sin^2\theta}=1$. Unable to move ...
Hint: Let $ x = \cos \theta$, $y = \cos \phi$. Show by expansion (and clearing denominators) that both equations are equivalent to $x^4 - 2x^2 y^2 + y^4 =0$, hence these statements are equivalent to each other. Note: This shows that the condition is satisfied iff $x = \pm y$. This is not required, but very strongly hin...
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Integrate: $\int \frac{dx}{x \sqrt{(x+a) ^2- b^2}}$ How to evaluate $$\int \frac{dx}{x \sqrt{(x+a) ^2- b^2}}$$ I tried trigonometric substitution $x + a = b \sec \theta$ and I encountered $$\int \frac{\tan \theta}{ (b - a \cos \theta) \sqrt{\tan^2 \theta}}d\theta$$ how to handle this term $\displaystyle \frac{\tan \the...
I substituted $x=1/y$ and got $$-\int \frac{dy}{\sqrt{1+2 a y+(a^2-b^2) y^2}}$$ I then completed the square in the square root to get $$-\frac{1}{\sqrt{a^2-b^2}} \int \frac{dy}{\displaystyle\sqrt{\left(y+\frac{a}{a^2-b^2}\right)^2-\left(\frac{a}{a^2-b^2}\right)^2}}$$ Now let $$y=\frac{a}{a^2-b^2} (\cosh{u}-1)$$ Then t...
{ "language": "en", "url": "https://math.stackexchange.com/questions/412085", "timestamp": "2023-03-29T00:00:00", "source": "stackexchange", "question_score": "4", "answer_count": 2, "answer_id": 0 }
What are the odds of rolling 3 pairs with six dice? Given a roll of six standard six-side dice, how do you calculate the odds of rolling 3 pairs? This should included non-unique pairs like 2,2,2,2,3,3 or even 5,5,5,5,5,5.
Let's split into $3$ cases: Case 1: All dice show the same number. There are $6$ ways this can happen. Case 2: $4$ dice show the same number, and $2$ dice show a different number. There are $\binom{6}{2}=15$ different ways to choose the $2$ dice that show a different number, and $\binom{6}{2}=15$ ways to choose which...
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