Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to determine equation for $\sum_{k=1}^n k^3$ How do you find an algebraic formula for $\sum_{k=1}^n k^3$? I am able to find one for $\sum_{k=1}^n k^2$, but not $k^3$. Any hints would be appreciated.
| One classic method is
$$\begin{align} \sum_{k=1}^{n} k^4 &= \left( \sum_{k=1}^n (k+1)^4 \right) + 1 - (n+1)^4
\\&= \left( \sum_{k=1}^n k^4 + 4 k^3 + 6 k^2 + 4 k + 1 \right) + 1 - (n+1)^4
\\&= \left( \sum_{k=1}^n k^4 \right)
+ 4 \left( \sum_{k=1}^n k^3 \right)
+ 6 \left( \sum_{k=1}^n k^2 \right)
+ 4 \left( \sum_{k=1}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/320985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 10,
"answer_id": 2
} |
Find the value of $\int \frac {du}{(a-u^2)^2}$ I am stuck on the following integration problem:
$\int \frac {du}{(a-u^2)^2}; a$ being a constant.
Can someone point me in the right direction? Thanks in advance for your time.
| Use $u = \sqrt{a} \sin{x}$; we get:
$$
\begin{align}
\int{\dfrac{du}{ (a - u^2)^2}} &= \int{\dfrac{\cos{x} dx}{ (a - a\sin^2{x})^2}} \\
&= \dfrac{1}{a^2}\int{\dfrac{\cos{x} dx}{ \cos^4{x}}} \\
&= \dfrac{1}{a^2}\int{\sec^3{x} dx} \\
&= \dfrac{1}{a^2} \cdot \left( \dfrac{1}{2} \sec{x} \tan{x} + \dfrac{1}{2} \ln{|\sec{x} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/322663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Questions regarding p-adic expansion and numbers
As opposed to real number expansions which extend to the right as sums
of ever smaller, increasingly negative powers of the base $p$,
$p$-adic numbers may expand to the left forever, a property that can
often be true for the $p$-adic integers. For example, conside... | Because is says that if $x=\cdots1313_5$ then $3x+1\equiv 0 \text{ mod } 5^n$ for all $n$, which is precisely what it means to be $0$ in $\mathbb{Q}_5$. Thus, you see that $3x+1=0$ so that $\displaystyle x=\frac{-1}{3}$.
EDIT:
Now that I have more time, let me be less glib about this response.
Whenever possible, we wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Definite Integral $\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$ I want to prove that
$$\int_0^{\pi/2} \frac{\log(\cos x)}{x^2+\log^2(\cos x)}dx = \frac{\pi}{2}\left(1-\frac{1}{\log 2}\right)$$
| Note that for $|x| < \frac{\pi}{2}$, we have
$$ \frac{\log\cos x}{\log^2 \cos x + x^2} = \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)}. $$
Thus if $I$ denotes the given integral, we have
\begin{align*}
I
&= \frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \Re \frac{1}{\log\left( \frac{1+e^{2ix}}{2} \right)} \, d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/325651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 2
} |
Divisibility criteria for $7,11,13,17,19$ A number is divisible by $2$ if it ends in $0,2,4,6,8$. It is divisible by $3$ if sum of ciphers is divisible by $3$. It is divisible by $5$ if it ends $0$ or $5$. These are simple criteria for divisibility.
I am interested in simple criteria for divisibility by $7,11,13,17,19$... | $(1)$
The formulae for $2,3,5,9,11$ can be derived from $\sum_{0\le r\le n}{a_r10^r}$
Observe that $\sum_{0\le r\le n}{a_r10^r}\equiv a_0\pmod 2$
$\sum_{0\le r\le n}{a_r10^r}\equiv a_0\pmod 5$
$\sum_{0\le r\le n}a_r10^r\equiv \sum_{0\le r\le n}a_r\pmod 3$ as $9\mid(10^r-1)$
$\sum_{0\le r\le n}a_r10^r\equiv \sum_{0\le ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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"answer_id": 1
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Generating Functions: Solving a Second-Order Recurrence I'm self-studying generating functions (using GeneratingFunctionology as a text). I came across this programming problem, which I immediately recognized as a modification of the Fibonacci sequence. I wanted to place my newly found generating function techniques ... | Something went wrong between
$$A(x)= \frac{1}{1 - x - kx^2}$$
And
$$A(x) = \frac{1}{(x-r_1)(x-r_2)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/330292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Wolframalpha step-by-step of $\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$ I wonder, where the minus sign goes after the first $u$-substitution of integral $\displaystyle\int\frac{\cos(x)}{\sqrt{2+\cos(2x)}}dx$?
| Integrate
$$\int \frac{\cos x}{\sqrt{2+\cos(2x)}}dx$$
Known Identity
$$\cos(2x)=1-2\sin^2(x)$$
Replacing
$$\int \frac{\cos x}{\sqrt{3-2\sin^2(x)}}dx$$
Let $u = \sin(x)$, $du = \cos(x)dx$
Substituting
$$\int \frac{u}{\sqrt{3-2u^2}}du$$
Let $z = \sqrt{3-2u^2}$
$$dz = \frac{1}{2}\frac{-4u}{\sqrt{3-2u^2}}du$$
$$dz = -\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ \sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}. $ The series is:
$$
\sum_{n=2}^\infty \frac{1}{n^3(n^3+1)}.
$$
I tried splitting the whole thing into simple fractions but I don't seem to get anywhere.
Any ideas?
| $$
\begin{align}
\sum_{n=2}^\infty\frac1{n^3(n^3+1)}
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3+1}\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3(1+1/n^3)}\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^3}-\frac1{n^3}\left(1-\frac1{n^3}+\frac1{n^6}-\dots\right)\right)\\
&=\sum_{n=2}^\infty\left(\frac1{n^6}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/331850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Power series involving complex variable. Show that
$1 + \binom{m+1}{1}z + \binom{m+2}{2}z^2 +...+ \binom{m+n}{n}z^n +... = \frac{1}{(1-z)^{m+1}}$
for non-negative integers $m$ and $|z| < 1$.
| Hint: start with the geometric series:
$$1+z+z^2+\ldots = \frac{1}{1-z}$$
and differentiate. For example,
$$1+2 z+3 z^2+\ldots=\frac{1}{(1-z)^2}$$
$$2+(3)(2)z+(4)(3)z^2+\ldots=\frac{2}{(1-z)^3} \implies 1+\binom{3}{1} z+\binom{4}{2}z^2+\ldots=\frac{1}{(1-z)^3}$$
Differentiate $m$ times to get your result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factoring $x^8-x^4+1$ over $GF(7)$ Could anyone suggest any good way to do it? (The only way I can think of is by looking for roots (There are none), checking a factorization into the product of a 6 and a 2 polynomial (Many unknowns for the coefficients), checking a factorization of a 5 and 3, and finally one of two 4 ... | First one substitutes $z=x^4$ and arrives at $z^2-z+1=0$. Quadratic polynomials are easy to factor (over any field of characteristic $\neq 2$), here we get $(z-3)(z-5)$. Using a bit of Galois theory (see below) one can find $x^4-3=(x^2+2x+2)(x^2+5x+2)$ and $x^4-5=(x^2+x+4)(x^2+6x+4)$ and these polynomials of degree $2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/332406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Computing $\int \frac{6x}{\sqrt{x^2+4x+8}} dx$ I am trying to compute an indefinite integral. Thanks!
$$\int \frac{6x}{\sqrt{x^2+4x+8}} dx.$$
| $$\int \frac{6x}{\sqrt{x^2+4x+8}}dx=\int\frac{6(x+2)-12}{\sqrt{(x+2)^2+2}}dx$$
$$=6\int\frac{x}{\sqrt{x^2+2}}dx-12\int\frac{1}{\sqrt{x^2+2}}dx$$
$$=6\sqrt{x^2+2}-12\sinh^{-1}\left(\frac{x+2}{2}\right)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/332473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Balls of 3 colours in a bag - Probability There are $12$ balls in a bag. $3$ of them are red, $4$ of them are green, and $5$ of them are blue. We randomly take out $3$ balls from the bag at the same time. What is the probability that all three balls are of the same colour?
My answer: $(3/12)^3 + (4/12)^3 + (5/12)^3$. I... | For $3$ balls of red color $$\mathrm{Prob}=\frac{3}{12}\cdot\frac{2}{11}\cdot\frac{1}{10}$$
For $3$ balls of green color $$\mathrm{Prob}=\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}$$
For $3$ balls of blue color $$\mathrm{Prob}=\frac{5}{12}\cdot\frac{4}{11}\cdot\frac{3}{10}$$
Total probability of all ball same color ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof that there exists an integer $n \geqslant2$ such that $n^2$ divides $2^n + 3^n$ Proof that there exists an integer $n \geqslant 2$ such that $n^2$ divides $2^n + 3^n$. I came up with this problem and I don't have a clue how to start, or even if it is not trivial at all.
| Suppose there is an $n\in\mathbb{N}$ such that $2^n+3^n$ is divisible by $n^2$. Then, $2^n+3^n\equiv 0 \bmod n^2$. In particular, $2^n+3^n\equiv 0 \bmod n$. Suppose for convenience that $n=p\neq 2,3$ is a prime number. Then, by Fermat's little theorem, $2^p\equiv 2 \bmod p$ and $3^p\equiv 3 \bmod p$, so
$$2^p+3^p\equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/334982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6$
If $a,b,c$ are non-negative numbers and $a+b+c=3$, prove that:
$$\sqrt{3a + b^3} + \sqrt{3b + c^3} + \sqrt{3c + a^3} \ge 6.$$
Here's what I've tried:
Using Cauchy-Schawrz I proved that:
$$(3a + b^3)(3 + 1) \ge (3\sqrt{a} + \sqrt{b^3})^2$$
$$\sqrt{(3a ... | I think your first method can still do something, but you probably have to use it in some other way, since using Cauchy-Schwarz directly will fail.
This is a suggestion.
Suppose
$\sqrt{3a+b^3} = u$
$\sqrt{3b+c^3} = v$
$\sqrt{3c+a^3} = w$
Then
$u^2+v^2+w^2 = 3(a+b+c)+a^3+b^3+c^3 = 9 + \sum a^3$
we are going to prove $u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/336367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 5
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Evaluate $\lim\limits_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
*
*Evaluate $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$.
*Examine whether $x^{1/x}$ possesses a maximum ... | $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3}{n^4}+\frac{2^3}{n^4}+\frac{3^3}{n^4}+\dots +\frac{n^3}{n^4}\right\}$
= $\displaystyle\lim_{n\rightarrow \infty}\left\{\frac{1^3+2^3+3^3+\dots +n^3}{n^4}\right\}$ =
$\displaystyle\lim_{n\rightarrow \infty}\left\{\left(\frac{n(n+1)}{2n^2}\right)^2\right\}$
=$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/338121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Divisibility problem involving the floor function I am trying to prove the following:
$$2^{m+1}\mid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$$ but $2^{m+2}\nmid\lfloor(1+\sqrt{3})^{2m+1} \rfloor$ for every natural number $m$.
I tried to use induction on $m$ but am not sure how to use the induction hypothesis. (The base case... | Consider the recurrence $x_0=2, x_1=20$ and $x_n=8x_{n-1}-4x_{n-2}$ for $n \geq 2$. It has characteristic polynomial $x^2-8x+4$, with roots $4 \pm 2\sqrt{3}=(1 \pm \sqrt{3})^2$, so $x_n=a(1+\sqrt{3})^{2n}+b(1-\sqrt{3})^{2n}$ for some contants $a, b$.
Using the values for $x_0, x_1$, we get:
$$2=a+b$$
$$20=a(1+\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/339314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the locus of a complex number Find the locus of $\arg\left(\frac{z-3}{z}\right) = \frac{\pi}{4}$ where $z$ represent complex number.
Working: $\arg\left(\frac{z-3}{z}\right) $ can be written as $\arg(z-3)-\arg(z) = \frac{\pi}{4}$, or $\arg\left((x-3)+iy\right) - \arg(x+iy)=\frac{\pi}{4}$.
If we take tangent to ... | $\arg(\dfrac{z-3}{z})=\dfrac{\pi}{4}\implies\Re(\dfrac{z-3}{z})=\Im(\dfrac{z-3}{z})$ and$\Re(\dfrac{z-3}{z})>0,z\neq0 $$\implies\Re(\dfrac{|z|^2-3\bar z}{|z|^2} )=\Im(\dfrac{|z|^2-3\bar z}{|z|^2})$$\implies x^2+y^2-3x=3y,y>0$(As$\Im(\dfrac{z-3}{z})=\dfrac{3y}{x^2+y^2}>0 $ if and only if $y>0$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/340799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x > 1$, prove that $f(x) = \dfrac{1}{\sqrt{x^2+1}}\log(x+\sqrt{x^2-1})$ $$f(x) = 2\int_{0}^{1}\dfrac{du}{u^2(1-x)+1+x}$$
I have used partial fractions but solves nothing.
| I don't see a logarithm but an arctangent. Check this:
$$(1-x)u^2+1+x=(1+x)\left(\frac{1-x}{1+x}u^2+1\right)=(1+x)\left(1+\left(\sqrt\frac{1-x}{1+x}\;u\right)^2\right)\Longrightarrow$$
$$2\int\limits_0^1\frac{du}{(1-x)u^2+1+x}=\frac{2}{\sqrt{(1-x)(1+x)}}\;\int\limits_0^1\frac{\sqrt{\frac{1-x}{1+x}}\;du}{\left(1+\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/341144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stu... | $$ \frac{4}{x+4} < 2 $$
(We'll do a case-by-case analysis here, which I feel is slightly more illuminating.)
Case 1: $x+4 > 0 \quad$ ($x > -4$)
In this case, we just multiply both sides by $x+4$ to get:
$$ 4 \le 2(x+4) $$
$$ 4 \le 2x + 8 $$
$$ -4 \le 2x $$
$$ -2 \le x $$
So this solution occurs when both $-4 < x$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solving $x+2y+5z=100$ in nonnegative integers I have not done combinatorics since high school, so this is an embarrassingly simple question.
We can solve the diophantine equation $x+y+z=100$ in nonnegative integers using the "bars and boxes" combinatorial method. We have $100$ dots, and we want place 2 partition marker... | You can solve this problem by using the idea of generating functions; specifically, for the example above, let $c_n$ be the number of positive integer solutions to the equation
$$x+2y+5z=n$$
Then the generating function $f(a)$ for the sequence $a_i$ is
$$f(a)=c_0+c_1a+c_2a^2+c_3a^3+...+c_na^n+...$$
We have
$$f(a)=(1+a... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Please, I need a more detailed explanation of the particular solution of the problem with vectors Here is the problem and its solution (link to the source if you are interested):
Two different points $A$ and $B$ are given. Find a set of such points $M$, that $\overrightarrow{MA}\cdot\overrightarrow{MB}=k^2$, where $k$ ... | Are you familiar with how to "complete the square?" To get from
$$
x_a x_b - (x_a+x_b)x + x^2 + y_a y_b - (y_a+y_b)y + y^2 = k^2
$$
to
$$
\left( x - \frac{x_a + x_b}{2} \right)^2 - \frac{(x_a - x_b)^2}{4} +
\left( y - \frac{y_a + y_b}{2} \right)^2 - \frac{(y_a - y_b)^2}{4} = k^2 \\
$$
consider the following:
$$
\be... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3... | You're so close, but you have simply multiplied incorrectly.
Note that if $u' = -\frac{4}{3}x(x^2 - 1)^{-5/3}$ and $v = \frac{2}{3}x$, then $$u'v = -\frac{8}{9}x^2(x^2-1)^{-5/3}$$
This will make your answer correct. To match W|A, you just need to combine fractions carefully.
| {
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"url": "https://math.stackexchange.com/questions/346797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Find the floor value of a finite continued surd Given $x=20062007$, and let
$$A=\sqrt{x^2+\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}}.$$
Find the greatest integer not exceeding $A$.
| $$10x+1<\sqrt{100x^2+39x+\sqrt{3}}<10x+2$$
$$4x+1<\sqrt{16x^2+10x+1}<\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}<\sqrt{16x^2+10x+2}<4x+2$$
$$2x+1=\sqrt{4x^2+4x+1}<\sqrt{4x^2+\sqrt{16x^2+\sqrt{100x^2+39x+\sqrt{3}}}}<\sqrt{4x^2+4x+2}<2x+2$$
$$x+1=\sqrt{x^2+2x+1}<A<\sqrt{x^2+2x+2}<x+2$$
Thus $\lfloor A \rfloor=x+1=20062008$.
| {
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"url": "https://math.stackexchange.com/questions/347126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to factor $2b^2c^2 + 2c^2a^2 + 2a^2b^2 -a^4-b^4-c^2$? The term is: $2b^2c^2 + 2c^2a^2 + 2a^2b^2 -a^4-b^4-c^2$
And the answer is : $(a+b+c)(b+c-a)(c+a-b)(a+b-c)$
I have tried a lot, but could't accomplish. Please don't bring up any complex method, it is just a high school math problem. But in vain I just can't do i... | $$2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$$
$$=(2bc)^2-\{(a^2)^2+(b^2)^2+(c^2)^2-2a^2b^2+2b^2c^2-2a^2c^2\}$$
$$=(2bc)^2-(a^2-b^2-c^2)^2$$
$$=(2bc+a^2-b^2-c^2)(2bc-a^2+b^2+c^2)$$
$$=\{a^2-(b-c)^2\}\{(b+c)^2-a^2\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/349260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Residue Integral
Verify the integral with the aid of residues:
$$\int^{\infty}_0\frac{x^2+1}{x^4+1}dx=\frac{\pi}{\sqrt 2}$$
I got:
$f(z)=\frac{z^2+1}{z^4+1}$ and now I must find the residues for $f(z)$ and I got that the poles are: $e^{i\frac{\pi}{4}}$ and $e^{i\frac{3\pi}{4}}$. But I do not know how to finish.
| This answer, which uses residues, says that
$$
\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x
$$
Using $m=4$ and $n=0$ and $n=2$ yields
$$
\begin{align}
\int_0^\infty\frac{x^2+1}{z^4+1}\,\mathrm{d}x
&=\frac\pi4\csc\left(\frac14\pi\right)+\frac\pi4\csc\left(\frac34\pi\right)\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353429",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving Recursions Without Initial Conditions I am trying to solve the following recursion but it does not appear that I can use characteristic equations or generating functions since I do not have initial conditions. Is there another way I am missing or can I assume initial conditions to use the above methods?
$$a_n ... | A general way of solving such, as expounded in Wilf's "generatingfunctionology":
$$
a_{n + 2} = 3 a_{n + 1} - 2 a_n + 4 \cdot 2^n + (n + 2)^2
$$
Define the ordinary generating function:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
By the properties of generating functions, with the operator $D = \dfrac{d}{d z}$:
$$
\frac{A(z) -... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$k-$ Subsets of $\{1,\cdots,n\}$ with no consecutive integers This is a practice-exam question in discrete mathematics.
Denote this number (as described in the title) with $f(n,k)$ where $f(n,0)= 1$. I figured out that the recurrence $f(n+2,k) = f(n+1,k) + f(n,k-1)$ holds if $k \leq \left \lfloor \frac{n+1}{2}\right \r... | OK, get at this with generating functions, à la Wilf's "generatingfunctionology", but in two indices. Define:
$$
F(x, y) = \sum_{n, k \ge 0} f(n, k) x^n y^k
$$
Write the recurrence as:
$$
f(n + 2, k + 1) = f(n + 1, k + 1) + f(n, k)
$$
Clearly $f(n, 0) = 1$ for all $n$, $f(0, k) = [k = 0]$, $f(1, k) = [0 \le k \le 1]$ (... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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limit with $\arctan$ I have to find the limit
and want ask about a hint:
$$\lim_{n \to \infty} n^{\frac{3}{2}}[\arctan((n+1)^{\frac{1}{2}})- \arctan(n^{\frac{1}{2}})]$$
I dont have idea what to do. Derivatives and L'Hôpital's rule are so hard
| For $x>0$ we have $$\arctan x+\arctan\frac{1}{x}=\frac{\pi}{2}$$
so
\begin{align}(u_n=\arctan^{\frac{1}{2}}(n+1)- \arctan^{\frac{1}{2}}(n)&=\sqrt{\frac{\pi}{2}}\left[(1-\frac{2}{\pi}\arctan\frac{1}{n+1})^{1/2}-(1-\frac{2}{\pi}\arctan\frac{1}{n})^{1/2}\right]\\&=\sqrt{\frac{\pi}{2}}\left(\frac{1}{\pi}\frac{1}{n(n+1)}+O(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Asymptotics of sum of binomials How can you compute the asymptotics of
$$S=n + m - \sum_{k=1}^{n} k^{k-1} \binom{n}{k} \frac{(n-k)^{n+m-k}}{n^{n+m-1}}\;?$$
We have that $n \geq m$ and $n,m \geq 1$.
A simple application of Stirling's approximation gives
$$S \approx T = n + m - \frac{n^{3/2-m}}{\sqrt{2\pi}} \sum_{k=1}... | Computation for constant $m$ positive integer.
This uses the same method as in Estimate $\sum_{k=1}^{n} k^{k-1} \binom{n}{k} (n-k)^{n+1-k}$ ,
further explanation is there.
Write
\begin{equation*}
u_{-1}(z) = \sum_{n=1}^\infty \frac{n^{n-1}}{n!} z^n
\tag{1}\end{equation*}
Then the unique singularity nearest to the orig... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Looking for help understanding the asymptotic expansion of the digamma function I was recently given an example using this asymptotic expansion of the digamma function where:
$$\frac{d}{dx}(\ln\Gamma(x)) = \psi(x) \sim \ln(x) - \frac{1}{2x} - \frac{1}{12x^2}$$
Here's the example:
$$\frac {\psi\left(\frac x4\right)}4 - ... | The $\log(x)$ terms cancel because $\dfrac{1}{4} - \dfrac{1}{5} - \dfrac{1}{20} = 0$.
In somewhat more detail,
$$ \eqalign{\frac{\psi(x/4)}{4} &= -\frac{1}{2}\,\ln \left( 2 \right) +\frac{1}{4}\,\ln \left( x \right) -\frac{1}{2\,x}
-\frac{1}{3\, x^2}+O \left( {x}^{-4} \right)\cr
\frac{\psi(x/5+1/2)}{5} &= -\frac{1}{5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem related to a clock I faced the following problem:
At what time after 4 o'clock, the hour and the minute hand will lie opposite to each other?
*
*$\quad$ 4-50'-31"
*$\quad$ 4-52'-51"
*$\quad$ 4-53'-23"
*$\quad$ 4-54'-33"
Can someone point me in the right direction?
| At $4$ o'clock the angle between the hour & the minute hand is $-\frac4{12}\cdot360^\circ=-120^\circ$
To be at $180^\circ,$ the difference of angle to be generated will be $180^\circ-(-120^\circ)=300^\circ$
Now, the hour hand moves in $12$ hours $360^\circ$
So, it moves in $1$ hour $=60$ minutes $\frac{360^\circ}{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/362334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Can someone check the solution to this recurrence relation? Here's the recurrence relation: $a_n = 4a_{n−1} − 3a_{n−2} + 2^n + n + 3$ with $a_0 = 1$ and $a_1 = 4$
Here's the solution:Write:
$$
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^n + n + 3 \quad a_0 = 1, a_1 = 4
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$. If you multiply... | You substituted $n+2$ instead of $n$ in your first step, but forgot to change the powers of $2$, etc. In other words, you should have
\begin{align}
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^{n+2} + (n+2) + 3 \quad a_0 = 1, a_1 = 4
\end{align}
instead of
\begin{align}
a_{n + 2} = 4 a_{n + 1} - 3 a_n + 2^n + n + 3 \quad a_0 = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find the last two digit of $7^{81}$? Could any one tell me how to find the last two digit of $7^{81}$? I have succeeded in finding the last digit only which is $7$.
Any group theoretic approach or any other approach is welcome.
| An alternative method of determining the value that does not rely on calculating $\phi(100)$ is to perform the calculation in modular arithmetic using efficient exponentiation.
$81=64+16+1$, so here's the process (all lines are mod 100):
$$7 \equiv 7\\7^2 \equiv 49\\7^4 \equiv 49^2 \equiv 1\\7^5 \equiv 1\cdot 7 \equiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/365248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Computing Legendre symbol value using quadratic reciprocity I have an example in my lecture notes that says:
Check
$$\left( \frac{31}{1019} \right) = -1$$
where $\left( \frac{a}{b} \right)$ is the Legendre symbol.
I said, because $1019$ and $31%$ are both primes, we can use quadratic reciprocity and get:
$$ \left( \f... | The Question seems to be wrong
$$\left(\frac{27}{31}\right)=\left(\frac{-4}{31}\right)=\left(\frac{2^2}{31}\right)\left(\frac{-1}{31}\right)=\left(\frac{-1}{31}\right)\text{ as } (\pm2)^2\equiv 4\pmod {31}$$
Again, we know, $\left(\frac{-1}p\right)=1\iff $ prime $p\equiv1\pmod 4$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to factor $x^4-7x^2-18$ I am not sure how I would factor this. The $x^4$ and $x^2$ are really throwing me off. Can someone explain how I would factor this?
| Solution 1.
\begin{eqnarray*}
x^4-7x^2-18&=&(x^4+2x^2)-(9x^2+18)\\
&=&x^2(x^2+2)-9(x^2+2)\\
&=&(x^2+2)(x^2-9)\\
&=&(x^2+2)(x-3)(x+3)
\end{eqnarray*}
Solution 2.
\begin{eqnarray*}
x^4-7x^2-18&=&(x^4-9x^2)+(2x^2-18)\\
&=&x^2(x^2-9)+2(x^2-9)\\
&=&(x^2-9)(x^2+2)\\
&=&(x-3)(x+3)(x^2+2)
\end{eqnarray*}
Solution 3.
\begin{eqn... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating Sums Algebraically or Combinatorially Consider
(1) $$\sum_{k=0}^{n}\binom{n}{k}2^{k-n}$$
(2) $$\sum_{k=0}^{n}\binom{n}{k}\frac{k!}{(n+k+1)!}$$
These sums appear too difficult (in my mind) to evaluate combinatorially. What are some good methods to attack these problems algebraically?
| The second sum
$$ \sum_{k=0}^{n} \binom{n}{k} \frac{k!}{(n+k+1)!} $$
can be rewritten as
$$\frac{n!}{(2n+1)!} \sum_{k=0}^{n} \binom{2n+1}{n-k} = \frac{2^{2n} n!}{(2n+1)!}$$
expand $\binom{n}{k} = \frac{n!}{(n-k)!k!}$, cancel the $k!$ and multiply and divide by $(2n+1)!$ and,
$\sum_{k=0}^{n} \binom{2n+1}{n-k} = \binom{2... | {
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"source": "stackexchange",
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How to find limit of this function- with summation notation i.e Sigma notation Find the value of $\displaystyle\lim_{n \rightarrow \infty} \dfrac 1{n^4}\left[1\left(\sum^n_{k=1}k\right)+2\left(\sum^{n-1}_{k=1}k\right)+3\left(\sum^{n-2}_{k=1}k\right)+\dots+n.1\right]$
Please guide how to proceed in this case .....Thanks... | $$\displaystyle\lim_{n \rightarrow \infty} \dfrac 1{n^4}\left[1\left(\sum^n_{k=1}k\right)+2\left(\sum^{n-1}_{k=1}k\right)+3\left(\sum^{n-2}_{k=1}k\right)+\dots+n.1\right]=$$
$$\lim\limits_{n\rightarrow\infty}\dfrac{1}{n^4}\left[1\dfrac{n(n+1)}{2}+2\dfrac{(n-1)(n)}{2}+\dots +k\dfrac{(n-k+1)(n-k+2)}{2}+\dots+n\cdot1\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368332",
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"source": "stackexchange",
"question_score": "2",
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How to prove that $\lim_{n\to \infty} (n^k/2^n) = 0$? I'm having a hard time trying to prove this statement.
$\lim_{n\to \infty} (n^k/2^n) = 0$
k is a positive number.
Please, help me.
Thanks in advance.
| Let us prove that $\lim_{n \to \infty}\frac{n^k}{a^n}=0,$ for $a>1$. Let $m \in Z$ and $m>k$, then
$$ 0<\frac{n^k}{a^n}\le \frac{n^m}{a^n}=\left(\frac{n}{\sqrt[m]{a^n}}\right)^m=\left(\frac{n}{b^n} \right)^m $$
where $b=\sqrt[m]{a}>1$.
Now,
\begin{align*} 0<\frac{n}{b^n}&=\frac{n}{(1+(b-1))^n}=\frac{n}{\sum\limits_{k=0... | {
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Find expansion around $x_0=0$ into power series and find a radius of convergence My task is as in the topic, I've given function $$f(x)=\frac{1}{1+x+x^2+x^3}$$
My solution is following (when $|x|<1$):$$\frac{1}{1+x+x^2+x^3}=\frac{1}{(x+1)+(x^2+1)}=\frac{1}{1-(-x)}\cdot\frac{1}{1-(-x^2)}=$$$$=\sum_{k=0}^{\infty}(-x)^k\c... | If I recall, if you have two power series, based at the same point, then the radius of convergence of their product is at least the smaller of the two radii of convergence. Generally it will be the smaller of the two radii of convergence. The at least part comes from the possibility of numerators in one cancelling with... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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What's the Maclaurin Series of $f(x)=\frac{1}{(1-x)^2}$? This function seemed to be pretty much straight forward, but my solution is incorrect.
I have two questions:
1. Where did I make a mistake?
2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / in... | As far as shortcuts go, there are two you might find interesting.
You've probably seen the geometric series several times:
$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots $$
Now the first idea is that your function $f(x)$ is the square of this one, so:
$$ f(x) = (1 + x + x^2 + x^3 + \ldots )^2 $$
If the indicated multipl... | {
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"url": "https://math.stackexchange.com/questions/369889",
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"source": "stackexchange",
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mathematical induction ($(1+x)^n\ge1+nx+n(n-1)x^2/2$) Suppose that $x > 0$ and let $n \geq 2$ be a positive integer. Prove that $(1 + x)^n \geq 1 + nx + \frac{n(n-1)}{2}x^2$
So for the base case, I have $x=1$, but that really is not getting me anywhere.
Would it be better to manipulate $n$? Is there a way to know whic... | HINT:
Let $(1+x)^m\ge \{1+mx+\frac{m(m-1)}2x^2\}$
So, $(1+x)^{m+1}=(1+x)(1+x)^m \ge (1+x)\{1+mx+\frac{m(m-1)}2x^2\}$
$=1+(m+1)x+x^2\{\frac{m(m-1)}2+m\}+x^3\frac{m(m-1)}2\ge \{1+(m+1)x+\frac{m(m+1)}2x^2\}$ as $x>0$ and $m\ge2$
Now, for $m=2,1+2x+x^2-\{1+2x+x^2\}=0\implies (1+x)^2\ge \{1+2x+\frac{2(2-1)}2x^2\}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this resu... | There is an infinite family of solutions coming from the idea $(2+\sqrt 3)^3=26+15\sqrt 3$. We can form $(2+\sqrt 3)^{3n}$ and find another solution. The next one is $(2+\sqrt 3)^6=1351+780 \sqrt 3$ and $(1351+780\sqrt 3)^{(1/3)}+(1351-780\sqrt 3)^{(1/3)}=14$ There is a recurrence, if $(a,b)$ is a solution, the next... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
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I cannot find the last factor of this expression? I'm supposed to factor $x^8-y^8$ (the exponents are 8 for both if it is too difficult to see) as completely as possible. It is easy to factor this to $(x+y)(x-y)(x^2+y^2)(x^4+y^4)$. However, the book says "(Hint: there are 5 factors. Note that we sat real coefficients, ... | Hint $\rm\,\ x^4 + y^4 = (x^2+y^2)^2 - (\sqrt{2}\, xy)^2 =\, \cdots\ $ (factor the difference of squares)
Remark $\ $ One can perform analogous factorizations in more exotic cases, and these prove very useful for factor factoring integers having such forms. $ $ For example, $ $ Aurifeuille, Le Lasseur and Lucas $ $ dis... | {
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Complex definite integrals/residue I am trying to evaluate this integral:
$$
\int_0^\pi \frac{d\theta}{(a + b \cos \theta)^2}
$$
for $0 < b < a$. I feel like the way I'm trying to solve [using $\cos t=(e^{it}+e^{-it})/2$ and $z=e^{it}$, but it doesn't seem to come out right. Can anyone help me step by step?
| From Taylor series of $\dfrac1{(a+bx)^2}$, we have
$$\dfrac1{(a+b \cos(t))^2} = \sum_{k=0}^{\infty} \dfrac{(k+1)(-b)^k}{a^{k+2}} \cos^{k}(t)$$
Now $$\int_0^{2\pi} \cos^k(t) dt = 0 \text{ if $k$ is odd}$$
We also have that
$$\color{red}{\int_0^{2\pi}\cos^{2k}(t) dt = \dfrac{(2k-1)!!}{(2k)!!} \times \pi = \pi \dfrac{\dbi... | {
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Why is $\;n^2-\frac{n^2}{2} =\frac{n^2}{2}\;$? Could someone please expand on how to get from $\;\displaystyle\left( n^2-\frac{n^2}{2}\right)\;$ to $\;\left(\dfrac{n^2}{2}\right)\;?\;$
I can't seem to wrap my head around that.
| We proceed by induction on $n$. The base case for $n = 0$ is $0^2 - \frac{0^2}{2} = 0 =\frac{0^2}{2}$, which follows from the fact that $0$ is the additive identity in $\mathbb{Q}$. Suppose the claim holds for some $n = k$. Now
\begin{aligned}
(k+1)^2 - \frac{(k+1)^2}{2} &= k^2 + 2k + 1 - \frac{k^2}{2} - \frac{2k}{2} -... | {
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Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2... | There is a remarkable series converging to $\ln(2)$ on Wikipedia: $$\ln(2)=\sum_{k=0}^\infty\frac{2}{2k+1}\left(\frac{7}{31^{2k+1}}+\frac{3}{161^{2k+1}}+\frac{5}{49^{2k+1}}\right)$$ although I don't see a source for why this is true.
Summing just the $k=0$ and $k=1$ terms, gives
$$\begin{align}
\ln(2)&>\frac{2955848868... | {
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"timestamp": "2023-03-29T00:00:00",
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Integral of $\sin x \cdot \cos x$ I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?
$ \int \sin x \cos x dx $
1) via subsitution $ u = \sin x $
$ u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Righta... | Antiderivatives are only unique up to adding a constant ('of integration'). If you were to stick limits in your integrals then you'd always get the same number.
| {
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Mellin transform of $\sin x$ aka $\int^{\infty}_0 x^{s-1}\sin x dx $ I am trying to find the Mellin transform of $\sin x $, put in other words to solve:
$$\int^{\infty}_0 x^{s-1}\sin x \mathrm{d} x $$
And I know that the answer is:
$$\Gamma(s) \sin \left(\frac{\pi s}{2}\right)$$
From several tables on the internet but... | Alternatively, the Mellin transform for $\sin x$ can be found by employing the following useful property for the Laplace transform:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$
and
$$\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
if $ab=cd$ then $a+b+c+d $ is composite Let $a,b,c,d$ be natural numbers with $ab=cd$.
Prove that $a+b+c+d$ is composite.
I have my own solution for this (As posted) and i want to see if there is any other good proofs.
| From $ab=cd$ you have $$(a+b)^2-(a-b)^2=(c+d)^2-(c-d)^2\Rightarrow(a+b)^2-(c+d)^2=(a-b)^2-(c-d)^2$$ Hence we have $$(a+b+c+d)(a+b-c-d)=(a-b+c-d)(a-b-c+d)$$ Now note that $|a+b+c+d|>|a-b+c-d|$ and $|a-b-c+d|$. If $(a+b+c+d)$ was prime then it must divide one of $(a-b+c-d)$ or $(a-b-c+d)$, which is not possible.
| {
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"url": "https://math.stackexchange.com/questions/383394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 8,
"answer_id": 5
} |
Integral Of $\int\sqrt{\frac{x}{x+1}}dx$ I want to solve this integral
$$\int\sqrt{\frac{x}{x+1}}dx$$
And think about:
1) $t=\frac{x}{x+1}$
2) $dt = (\frac{1}{x+1} - \frac{x}{(x+1)^2})dx$
Now I need your advice! Thanks!
| put
$$\sqrt{\frac{x}{x+1}}=t,\quad x=\frac{t^2}{1-t^2},\quad dx=\frac{2t}{(1-t^2)^2}\,\,dt$$
So
\begin{align*}
\int\sqrt{\frac{x}{x+1}}\,\,dx &=\int t\cdot \frac{2t}{(1-t^2)^2}\,\,dt=2\int \frac{ t^2}{(1-t^2)^2}\,\,dt\\
&=2\int \frac{ t^2+1-1}{(1-t)^2(1+t)^2}\,\,dt\\
&=2\int \frac{ (t-1)(t+1)}{(1-t)^2(1+t)^2}\,\,dt-... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Another trigonometric proof...? ...sigh..another problem how shall I prove the following?
$$ {\cot A\over1- \tan A} + {\tan A \over 1- \cot A} = 1 + \tan A + \cot A$$
so what now? the following's what I've done:
$$\cot A - \cot^2 A + \tan A- \tan^2 A \over 2 - \tan A - \cot A$$
| Let $\tan A=a\implies \cot A=\frac1a$
So, the problem reduces to
$$\frac{\frac1a}{1-a}+\frac a{1-\frac1a}=\frac1{a(1-a)}+\frac{a^2}{a-1}$$
$$=\frac1{a(1-a)}-\frac{a^2}{1-a}=\frac{1-a^3}{a(1-a)}=\frac{1+a+a^2}a=a+\frac1a+1$$
Alternatively,
$$\frac{\tan A}{1-\cot A}=\frac{\tan^2A}{\tan A-1} (\text{ multiplying the nume... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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Evaluate a sum with binomial coefficients: $\sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$ $$\text{Find} \ \ \sum_{k=0}^{n} (-1)^k k \binom{n}{k}^2$$
I expanded the binomial coefficients within the sum and got $$\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \dots + \binom{n}{n}^2$$
What does this equal to? I think this c... | This can also be done using a basic complex variables technique.
Start as in @robjohn's answer. Suppose we seek to evaluate
$$\sum_{k=0}^n (-1)^k k {n\choose k}^2
= \sum_{k=1}^n {n\choose k} (-1)^k k {n\choose k}
\\= \sum_{k=1}^n {n\choose k} (-1)^k k \frac{n}{k} {n-1\choose k-1}
= n \sum_{k=1}^n {n\choose k} (-1)^k {... | {
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"source": "stackexchange",
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"answer_count": 3,
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Integrating a partial fraction with multiple quadratic denominators When integrating a real rational fraction, you first break it into partial fractions. You then end up with fractions with linear denominators $\frac{A}{(x-b)^n}$, which are easy. You also end up with quadratic denominators $\frac{Ax+B}{(x+ax+b)^n}$. I ... | As André Nicolas describes, the end result will come down to generating some sort of reduction formula. With the techniques we learn a little way into integral calculus,
$$\int \frac{Ax+B}{(x^2 + ax + b)^n} \ dx $$
can be "stripped down" using completion of squares to
$$ \frac{A}{2}\int \ \frac{[2x + a] }{( \ [x + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How can this $T(n) = T(n-1)+T(n-2)+3n+1$ non homogenous recurrence relation be solved How are can the above recurrence relation be solved?
I've reached here:
$(x^{2}-x-1)(x-3)^2(x-1)$
And then here:
$$a_n = l_1 \cdot (x_1)^n+l_2 \cdot (x_2)^n+l_3 \cdot (x_3)^n+l_4\cdot n \cdot (x_3)^n+l_5\cdot (x_4)^n$$
And we are giv... | Modify $T$ as follows:
$$
\begin{aligned}
T(n) &= T(n-1) + T(n-2) + 3n + 1\\
T(n) + 3n &= T(n-1) + T(n-2) + 6n + 1\\
T(n) + 3n &= T(n-1) + 3n + T(n-2) + 3n + 1\\
T(n) + 3n &= T(n-1) + 3(n-1) + T(n-2) + 3(n-2) + 10\\
T(n) + 3n + 10 &= T(n-1) + [3(n-1) + 10] + T(n-2) + [3(n-2) + 10 ]
\end{aligned}
$$
So, $S(n) = T(n) + 3... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Solving recurrence relation, $a_n=6a_{n-1} - 5a_{n-2} + 1$ I'm trying to solve this recurrence relation:
$$
a_n = \begin{cases}
0 & \mbox{for } n = 0 \\
5 & \mbox{for } n = 1 \\
6a_{n-1} - 5a_{n-2} + 1 & \mbox{for } n > 1
\end{cases}
$$
I calculated generator function as:
$$
A = \frac{31x - 24x^2}{1 - ... | Use Wilf's technique from "generatingfunctionology". Define $A(z) = \sum_{n \ge 0} a_n z^n$, and write:
$$
a_{n + 2} = 6 a_{n + 1} - 5 a_n + 1 \qquad a_0 = 0, a_1 = 5
$$
Multiply by $z^n$ and add for $n \ge 0$, which gives:
$$
\frac{A(z) - a_0 - a_1 z}{z^2} = 6 \frac{A(z) - a_0}{z} - 5 A(z) + \frac{1}{1 - z}
$$
Solving... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Integral of $\int \frac{x^4+2x+4}{x^4-1}dx$ I am trying to solve this integral and I need your suggestions.
$$\int \frac{x^4+2x+4}{x^4-1}dx$$
Thanks
| welcome to math.stackexchange this question were answered already.
Here is the link
use polynomial division, we get $$\int \frac{x^4+2x+4}{x^4-1} dx = \int 1 + \frac{2x+5}{(x^2 - 1)(x^2 + 1)}dx
= \int 1 + \frac{2x+5}{(x+1)(x-1)(x^2+1)} dx $$
Expressing this as partial fractions, we need only find $A, B, C$
$$= \int... | {
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"source": "stackexchange",
"question_score": "2",
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Solve $(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$ I got the square root of 14 and 11 but the answer book states that these answers are wrong. Can someone help me? I used this formula to find the individual roots
$x = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 - q}$
| Given:
$(x^2 + 5)^2 - 15(x^2 + 5) + 54 = 0$
Let a = $(x^2+5)$ which gives us a simple quadratic trinomial of the form $ax^2+bx+c=0.$
$a^2-15a+54=0$
Factoring this we get:
$(a-9)(a-6)=0$
$a-9=0$
$a-6=0$
$\implies a=9$ or $a=6$
$x^2+5=9$
$x^2+5=6$
$\implies x = \pm2$ or $\pm 1$
You can check that these solutions are co... | {
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"source": "stackexchange",
"question_score": "1",
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What type of Hypergeometric series is this? I am trying to find a closed form for the series
$$ \sum^\infty_{n=0} \frac{1}{n!} \frac{1}{n+1}(-z)^n {}_2F_2\left(m,n+1;\frac{1}{2},n+2; b z\right)$$
$m$ is a nonzero positive integer, and $b$, $z$ are positive real numbers. I to rewrite the sum as
$$ \sum^\infty_{n=0} \sum... | $\sum\limits_{n=0}^\infty\dfrac{1}{n!}\dfrac{1}{n+1}(-z)^n{}_2F_2\left(m,n+1;\dfrac{1}{2},n+2;bz\right)$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(n+1)_k(-1)^nb^kz^{n+k}}{\left(\dfrac{1}{2}\right)_k(n+2)_k(n+1)!k!}$
$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(m)_k(-1)^nb^kz^{n+k}}{\le... | {
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"source": "stackexchange",
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"answer_count": 3,
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Evaluation of a specific determinant.
Evaluate $\det{A}$, where $A$ is the $n \times n$ matrix defined by $a_{ij} = \min\{i, j\}$, for all $i,j\in \{1, \ldots, n\}$.
$$A_2
\begin{pmatrix} 1& 1\\
1& 2
\end{pmatrix};
A_3 = \begin{pmatrix} 1& 1& 1\\
1& 2& 2\\
1& 2& 3
\end{pmatrix};
A_4 = \begin{pmatrix} ... | If you expand along the bottom row, then most of the $(n-1)\times(n-1)$ subdeterminants are $0$ because the $(n-1)\times(n-1)$ submatrices have their two right-most columns identical.
So expansion along the bottom row leaves only the last two terms: $$(-1)(n-1)\det(A_{n-1})+n\det(A_{n-1})$$ which is just $\det(A_{n-1})... | {
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"source": "stackexchange",
"question_score": "1",
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Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over... | Starting from your second to last line (your integration was fine, minus a few $dx$'s in you integrals):
$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C \tag{1}$$
Good, up to this point... $\uparrow$.
So the error was in your last equality at the very end:
You made an error by ignoring the fact that the f... | {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Find the value of $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$ I want to know an objective approach to solve these type of expression in a quick time
Which of the expression equals to
$$\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}$$
a)$1-\tan\theta-\cot\theta$
b)$1+\tan\theta-... | I'm not sure where you got stuck after writing the expression in terms of $ \sin \theta $ and $ \cos \theta $ but here's how I would do it: $$ \begin{align*}\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} &= \frac{\sin \theta \tan \theta}{\sin \theta - \cos \theta} + \frac{\cos\theta \cot \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$ I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$
Is it OK how I want to do?
$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\fra... | Assuming you have the Mean Value Theorem at your dispoeal, that's the easiest way to show that the limit equals $0$: with $g(x) = \sin\sqrt x$, we have
$$
\sin\sqrt{x+1}-\sin\sqrt x = g(x+1)-g(x) = ((x+1)-x)g'(\xi) = \frac{\cos\sqrt\xi}{2\sqrt\xi}
$$
for some $x\le\xi\le x+1$. In particular,
$$
|\sin\sqrt{x+1}-\sin\sqr... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "15",
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"answer_id": 3
} |
quadratic equation precalculus from Stewart, Precalculus, 5th, p56, Q. 79
Find all real solutions of the equation
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{x^2-4}$$
my solution
$$\dfrac{x+5}{x-2}=\dfrac{5}{x+2}+\dfrac{28}{(x+2)(x-2)}$$
$$(x+2)(x+5)=5(x-2)+28$$
$$x^2+2x-8=0$$
$$\dfrac{-2\pm\sqrt{4+32}}{2}$$
$$\dfrac{... | $$\frac{x+5}{x-2}=\frac{5}{x+2}+\frac{28}{x^2-4}$$
multiply both sides with $(x-2)(x+2)\neq 0$ or $x\neq 2,x\neq -2$
we get
$$(x+5)(x+2)=5(x-2)+28,x\neq 2,-2$$
$$x^2+2x-8=0,x\neq 2,-2$$
$$x^2-2x+4x-8=0,x\neq 2,-2$$
$$x(x-2)+4(x-2)=0,x\neq 2,-2$$
$$(x-2)(x+4)=0,x\neq 2,-2$$
so $x+4=0$ or $x=-4\neq 2,-2$ is unique soluti... | {
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"source": "stackexchange",
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How do I evaluate the following expression? How to evaluate the following expression:
$\displaystyle \frac{1}{\sqrt{2}+1}+ \frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}} +\cdots +\frac{1}{\sqrt{9}+\sqrt{8}}$
| Observe that:
$(\sqrt{2} + 1)(\sqrt{2} - 1) = (\sqrt{2})^2 - 1^2 = 2 - 1 = 1.$
Therefore,
$\dfrac1{\sqrt{2} + 1} = \sqrt{2} - 1.$
Similarly,
$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 2 - 1 = 1.$
And therefore,
$\dfrac1{\sqrt{3} + \sqrt{2}} = \sqrt{3} - \sqrt{2}.$
We keep doing this u... | {
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"url": "https://math.stackexchange.com/questions/396511",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
solution to a root inequality I have the inequality
$$\sqrt{a^2+b^2+c^2}+2\sqrt{ab+ac+bc} \geq \sqrt{a^2+2bc}+\sqrt{b^2+2ac}+\sqrt{c^2+2ab}.$$I tried to do $u=a^2+b^2+c^2$ and $v=ab+ac+bc$ and $x=a^2+2bc$, $y=b^2+2ac$, $z=c^2+2ab$ ...but I did not find any solution. Any help is appreciated.
| I will assume that $a, b, c \geq 0$. Change to cylindrical coordinates with axis in the direction of $(1, 1, 1)^T$ as follows:
$$\left(\begin{matrix} a \\ b \\ c \end{matrix}\right) = \frac{z}{\sqrt{3}}\left(\begin{matrix} 1 \\ 1 \\ 1\end{matrix}\right) + \rho\left[\frac{\cos\phi}{\sqrt{6}}\left(\begin{matrix} 1 \\ 1 \... | {
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"source": "stackexchange",
"question_score": "7",
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How to resolve this algebra equation? $$f = X^3 - 12X + 8$$ $a $- complex number, $a$ is a root for $f$
$b = a^2/2 - 4 $.
Show that $f(b) = 0$
This is one of my theme exercises ... Some explanations will be appreciated ! Thank you all for your time .
| As $a$ is root of $f = X^3 - 12X + 8=0, a^3-12a+8=0$
Now, $b=\frac{a^2}2-4\implies a^2=2(b+4)=2b+8$
So, from $a^3-12a+8=0, a(2b+8)-12a+8=0\implies a=-\frac8{2b+4}=-\frac4{b+2}$
So, $$\left(-\frac4{b+2}\right)^3-12\cdot \left(-\frac4{b+2}\right)+8=0$$
On simplification we get , $$b^3-12b+8=0$$
| {
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"source": "stackexchange",
"question_score": "1",
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Basis of kernel and image of a linear transformation - verification The transformation matrix I found is: $$\begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 0 & 0\end{pmatrix}$$
Is this how a basis for $\ker$ and $\mathrm{im}$ is calculated?
$$\begin{pmatrix} 1 & -1 \\ 1 & 1 \\ 0 & 0\end{pmatrix} \sim \begin{pmatrix} 1 & 0 \\ 0 & 1 ... | This CW answer intends to remove the question from the unanswered queue.
As DonAntonio already noted, your calculation is correct.
| {
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Prove with integration the inequality $e(\frac{n}{e})^n < n! < n \times e(\frac{n}{e})^n$ Prove with integration the inequality, I need some advice about how to start prove it.
I know that if function is Monotonically increasing function so :
$$ f(1)+\int^n_1f(x)dx\leq f(1)+f(2)+....+f(n)\leq f(n)+\int^n_1f(x)dx$$
$$e... | Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
$$\begin{align}
a_n ~~~&:=~~~ \left ( 1 + \frac{1}{n} \right ) ^n \\ ~ \\
b_n ~~~&:=~~~ \left ( 1 - \frac{1}{n} \right ) ^{-n}
\end{align}$$
and
$$\underset{n \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Recurrence Relations with single roots I have the following recurrence: $a_{n+3}=3a_{n+2}-3a_{n+1}+a_n$
with initial values $a_1 = 1, a_2 = 4, a_3 = 9$
I have found the characteristic equation to be $x^3-3x^2+3x-1$ and the root to be 1.
My text book is not helpful in how I should go about solving this when I have a sin... | Use Wilf's "generatingfunctionology". Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply your recurrence by $z^n$ and sum over $n \ge 0$ to get:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
= 3 \frac{A(z) - a_0 - a_1 z}{z^2} - 3 \frac{A(z) - a_0}{z} + A(z)
$$
Using the recurrence "backwards" gives $a_0 = 0$. Solving for... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d
$$a+b+c+d=2$$
$$a^2+b^2+c^2+d^2=30$$
$$a^3+b^3+c^3+d^3=44$$
$$a^4+b^4+c^4+d^4=354$$
so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$
If the qusetion i... | Here's an approach which involves AM-GM inequality(To bound the numbers).
I have considered the positive values of $(a,b,c,d)$. I can consider $a^2,b^2,c^2$ and $d^2$, since they are all strictly positive.
$\dfrac{a^2+b^2+c^2+d^2}{4} \ge \sqrt{abcd}$
$\dfrac{30}{4} \ge \sqrt{abcd} \implies 56.25 \ge abcd$
Since $a^2+b^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/402856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 0
} |
Finding the correct statements about $(5+2\sqrt{6})^{2n+1} = S + t$ with $S$ integer and $0 \leq t < 1$ Problem:
If $n$ is a positive integer and $(5+2\sqrt{6})^{2n+1} = S + t$, where $S$ is an integer and $0 \leq t < 1$, then
(a) $S$ is an odd integer
(b) $S + 1$ is not divisible by $9$
(c) The integer next above $(5... | Everything flows from the hint by Ross Millikan.
Let $(5+2\sqrt{3})^{2n+1}=a_n+b_n\sqrt{3}$. (we can imagine computing $a_n$ and $b_n$ by expanding using the Binomial Theorem). Note that $a_n$ and $b_n$ are integers. Now imagine expanding $(5-2\sqrt{3})^{2n+1}$. We get $a_n-b_n\sqrt{3}$. It follows that
$$(5+2\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/406928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Not so easy optimization of variables? What is the maximum value of $x^2+y^2$, where $(x,y)$ are solutions to $2x^2+5xy+3y^2=2$ and $6x^2+8xy+4y^2=3$. (calculus is not allowed). I tried everything I could but whenever I got for example $or$ $x^2+y^2=f(y)$ or $f(x)$ the function $f$ would always be a concave up parabola... | Since my other entry is already rather long, I thought I'd separately add a couple of other methods that work (at least in principle), but which involve too much effort to be practical "contest math" answers.
The approach I discussed already can also be carried out in polar coordinates, with the equations
$$ 2 \cos^2 \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Integrality of $\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$ I've been asked to provide a proof for
If $n$ is an integer then
$$\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$$
is also an integer.
Any help would be appreciated
| Hint:$$\frac{n}{6}(n^2+3n+2)=\frac{n}{6}(n+1)(n+2)$$
What can you say about the divisibility of $n,(n+1),(n+2)$ by $2$ and $3$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/410069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Two roots of polynomial If a polynomial with rational coefficients has a root $1 + \cos(2\pi/9) + \cos^2(2\pi/9)$, then the one also has a root $1+\cos(8\pi/9)+\cos^2(8\pi/9).$ How to prove it?
| Take $\omega=e^{i2\pi/9}$.
Let $\alpha=1 + \cos(2\pi/9) + \cos^2(2\pi/9)$.
Then $4\alpha = 6+2\omega+ \omega^2 + \omega^7 + 2\omega^8$, by using $2\cos(2\pi/9)=\omega+\bar\omega$ and $\omega^9=1$.
Let $\beta = 1+\cos(8\pi/9)+\cos^2(8\pi/9)$. Then $4\beta = 6 + \omega+ + 2\omega^4 + 2\omega^5 + \omega^8$, by using $2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/410952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit with roots I have to evaluate the following limit:
$$ \lim_{x\to 1}\dfrac{\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}}{\sqrt{x-1}+\sqrt{x^2+1}-\sqrt{x^4+1}} . $$
I rationalized both the numerator and the denominator two times, and still got nowhere. Also I tried change of variable and it didn't work.
Any help is gratefu... | For every $x>1$ we have
\begin{eqnarray}
P(x):&=&\sqrt{x+1}+\sqrt{x^2-1}-\sqrt{x^3+1}=\sqrt{x^2-1}+\frac{x+1-(x^3+1)}{\sqrt{x+1}+\sqrt{x^3+1}}\\
&=&\sqrt{x^2-1}-\frac{x(x^2-1)}{\sqrt{x+1}+\sqrt{x^3+1}}=\sqrt{x^2-1}\left(1-\frac{x\sqrt{x^2-1}}{\sqrt{x+1}+\sqrt{x^3+1}}\right)\\
&=&\sqrt{x-1}p(x),
\end{eqnarray}
with
$$
p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/411676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Diagonalising quadratic form Given the quadratic form $$Q(x) = \alpha\alpha_1\alpha_2 + 2\alpha^2\alpha_1\alpha_3$$ on $\mathbb{R}^2$ where $x = (\alpha_1,\alpha_2,\alpha_3)$ in some basis I want to find the signature of $Q$ dependent on $\alpha$ and the diagonal basis of $Q$.
I was trying to rewrite $Q$ into a sum of ... | As in your question, let
$$Q(x) = \alpha \alpha_1 \alpha_2 + 2 \alpha^2 \alpha_1 \alpha_3$$
be a quadratic form on $\mathbb{R}^3$ in some basis. We know what the signature is $(3,0,0)$ if $\alpha = 0$, so let us assume that $\alpha \neq 0$.
Something went wrong in your diagonalization process because you managed to di... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/412564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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There are no integers $x,y$ such that $x^2-6y^2=7$ How to show that there is no here are no integers $x,y$ such that $x^2-6y^2=7$?
Help me. I'm clueless.
| First show that since $a^2 \equiv 0,1,4$
$$7 \vert x^2 + y^2 \implies 7 \vert x \text{ and } 7 \vert y$$
We have
$$x^2 - 6y^2 = 7 \implies (x^2+y^2) - 7y^2 = 7 \implies 7 \vert (x^2+y^2) \implies 7 \vert x \text{ and } 7 \vert y$$
Hence, we have $x = 7x_1$ and $y=7y_1$, which gives us
$$7(x_1^2 - 6y_1^2) = 1$$
which is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/414870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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How to simplify $\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$? $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}}$$
I have been staring at it for ages and know that it simplifies to $x$, but have been unable to make any significant progress.
I have tried doing $(\frac{1-x}{1-2x})(\frac{1+2x}{1+2x})$ but that doesn't h... | $$\frac{1-\frac{1-x}{1-2x}}{1-2\frac{1-x}{1-2x}} = \frac{\frac{-x}{1-2x}}{\frac{-1}{1-2x}} = \frac{-x}{1-2x} \frac{1-2x}{-1} = \frac{-x(1-2x)}{-1(1-2x)} = \frac{-x}{-1} = x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/415304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
How to find the integral of $\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$ $$\frac{1}{2}\int^\pi_0\sin^6\alpha \,d\alpha$$
What is the method to find an integral like this?
| Note that $\displaystyle \int_{0}^{\pi} \sin^n(x) dx = 2 \int_0^{\pi/2} \sin^n(x) dx$. Let $I_n = \displaystyle \int_{0}^{\pi/2} \sin^n(x) dx$.
$I_n = \displaystyle \int_{0}^{\pi/2} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\pi/2} (n-1) \sin^{n-2}(x) \cos^2(x) dx$
The first ex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/417601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Why does the tangent of numbers very close to $\frac{\pi}{2}$ resemble the number of degrees in a radian? Testing with my calculator in degree mode, I have found the following to be true:
$$\tan \left(90 - \frac{1}{10^n}\right) \approx \frac{180}{\pi} \times 10^n, n \in \mathbb{N}$$
Why is this? What is the proof or ex... | The power series for cotangent (for $x$ in radians) is
$$\cot x = \frac{1}{x} - \frac{1}{3}x - \frac{1}{45}x^3 - \frac{2}{945}x^5 - \cdots \approx \frac{1}{x} \; \text{for $x$ small}$$
So,
$$\tan\left( 90^\circ - \frac{1^\circ}{10^n} \right)
= \cot\frac{1^\circ}{10^n}
= \cot\frac{\pi/180}{10^n}
\approx \frac{10^n}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 2
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Show that the following matrix is diagonalizable My question is related to this question discussed in MSE.
$J$ be a $3\times 3$ matrix with all entries $1\,\,$. Then prove that $J$ is
diagonalizable.
Can someone explain it in terms of A.M. and G.M. (algebraic and geometric multiplicity) concept ? Thanks in advance ... | So we have
$$ J = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix} $$
Let's compute the characteristic polynomial $\chi_J(X) = \det(J - X)$:
\begin{align*}
\chi_J(X) &= \det \begin{pmatrix} 1-X & 1 & 1 \\ 1 & 1-X & 1 \\ 1 & 1 & 1-X \end{pmatrix}\\
&= (1-X)^3 + 2 - 3(1-X)\\
&= 1 - 3X + 3X^2 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/418414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the range of $ f(x)=9^x - 3^x+1$
Problem:-Find range of function $ f(x)=9^x - 3^x+1$, here the domain of $f$ is $\mathbb R$.
Solution: $ f(x)=9^x - 3^x+1$. Let $f(x)=y$. Then
$$ \begin{split}y&=9^x - 3^x+1\\
y&=3^{2x} - 3^x+1
\end{split}$$
Let $3^x= u$. Then $ y=u^2 - u+1$, so
$$ u^2 - u+1-y=0.$$
Am I do... | After some computation, you will get $$f(x)=(3^x-1/2)^2+3/4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/419271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this equality $\frac{x}{y^2+5}+\frac{y}{z^2+5}+\frac{z}{x^2+5}\le\frac{1}{2}$ let $x^3+y^3+z^3=3,x,y,z>0$ show that
$$\dfrac{x}{y^2+5}+\dfrac{y}{z^2+5}+\dfrac{z}{x^2+5}\le\dfrac{1}{2}$$
I have show that
let $x,y,z$ be positive numbers,such that $x+y+z=3$,prove that
$$\dfrac{x}{1+y^3}+\dfrac{y}{1+z^3}+\dfrac{z}{1+... | By Cauchy-Schwarz inequality
$$\frac{x}{y^2 + 5} + \frac{y}{z^2 + 5} + \frac{z}{x^2+5} \le \sqrt{x^2 + y^2 + z^2}\sqrt{\frac{1}{(x^2+5)^2} + \frac{1}{(y^2 + 5)^2} + \frac{1}{(z^2 + 5)^2}}.$$
By AM-GM the RHS is less than
$$\frac{\frac{x^2 + y^2 + z^2}{6} + \frac{6}{(x^2 +5)^2} + \frac{6}{(y^2+5)^2} + \frac{6}{(z^2+5)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/419334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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There are infinitely many $N$ such that $\frac{N}{2}$ is a perfect square, $\frac{N}{3}$ is a perfect cube, and $\frac{N}{5}$ is a perfect fifth power
Show that there are infinitely many $N$ such that $\frac{N}{2}$ is a perfect square, $\frac{N}{3}$ is a perfect cube, and $\frac{N}{5}$ is a perfect fifth power.
A hi... | The general form of this kind of number is $2^{15}3^{10}5^{6}x^{30}$, for all values of $x$.
When $x=1$, this number is the product of powers of 2, 3, and 5 just under some integer, eg $32768$, $59049$ and $15625$ are the largest powers under, say $60,000$.
The general working for such a number is to treat the indece... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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Proving the relation $\det(I + xy^T ) = 1 + x^Ty$
Let $x$ and $y$ denote two length-$n$ column vectors. Prove that $$\det(I + xy^T ) = 1 + x^Ty$$
Is Sylvester's determinant theorem an extension of the problem? Is the approach the same?
| Hint: Decomposing $$ \begin{pmatrix} 1 & -y^T\\ x & I \end{pmatrix} $$
as lower $\cdot$ upper and upper $\cdot$ lower gives
$$ \begin{pmatrix} 1 & 0\\ x & I\end{pmatrix} \cdot \begin{pmatrix} 1 & -y^T\\ 0 & I + xy^T \end{pmatrix} = \begin{pmatrix} 1 + x^Ty & -y^T\\ 0 & I\end{pmatrix} \cdot \begin{pmatrix} 1 & 0\\ x & I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 7,
"answer_id": 5
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How to show that T is a projection operator For $x ∈ [0, 2π]$ let $G(x) = π^{−1}\cos x$, and define an operator $T$ on $L^2([0, 2π])$ as follows:
$$(Tf)(x) = ∫_0^{2π}G(x − x')f(x') \,dx'. $$
Show that $T$ is a projection operator.
I guess I must show that $T$ is selfadjoint and idempotent, in other words that:
$T=T^*$ ... | To show $T^2 = T$, just compute $T^2$. We have for $f \in L^2([0,2\pi])$:
\begin{align*}
(T^2 f)(x) &= T(Tf)(x)\\
&= \int_0^{2\pi} G(x-x')(Tf)(x')\, dx'\\
&= \int_0^{2\pi} G(x-x')\int_0^{2\pi} G(x'-x'')f(x'')\, dx''\, dx'\\
&= \int_0^{2\pi} \int_0^{2\pi} G(x-x')G(x'-x'')\,dx'\, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/420279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find a minimal polynomial (field theory) I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over Q.
I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2... | To begin, clear denominators:
$$(1 - \sqrt{5} + \sqrt{7}) \alpha = 3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35}$$
We need to make the coefficient of $\alpha$ rational, so use a difference-of-squares trick to get rid of the $\sqrt{7}$ on the LHS (i.e. multiply both sides by $1 - \sqrt{5} - \sqrt{7}$),
$$((1 - \sqrt{5})^2 - 7) \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/422233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 5,
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Induction to prove that $1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}$ Have I started this right? I know I have to add $(k+1)$ but why?
Use mathematical induction to prove that $$1^2+ 3^2 + 5^2+ \cdots +(2n + 1)^2=\frac{(n+1)(2n+1)(2n + 3)}{3}.$$
BASIS STEP: $P(0)$ is true since $$\begin{align}(2(... | You've proved that the statement is true for $k=0$ and now supposing that is true to an arbitrary $k$,
$$1^2+2^2+\cdots+(2k+1)^2 = \frac{(k+1)(2k+1)(2k+3)}{3}$$
let's see the case $k+1$. Then we use the hypothesis of induction and we get
$$\begin{align}1^2+2^2+\cdots+(2k+1)^2+(2(k+1)+1)^2 &= \frac{(k+1)(2k+1)(2k+3)}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/422359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can i show, this is a fundamental system for a given differential equation? My Problem is: i have a given differential equation: $$y^{\prime\prime\prime}-\frac{x^2}{x^2-2x+2}\cdot y^{\prime\prime}+\frac{2x}{x^2-2x+2}\cdot y^{\prime}-\frac{2}{x^2-2x+2}\cdot y= 0$$and the given functions :
$$y_{1}=x \quad y_{2}=x^2 \... | Your proof is missing two parts:
*
*That every function of the form $\sum a_i y_i$ is also a solution.
*That every solution is of the form $\sum a_i y_i$.
$1$ is satisfied since this is a linear ODE. $2$ is satisfied since the solutions are linearly independent and the order of the ODE is equal to the number of s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ I am trying to find a closed form for
$$\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx = 0.094561677526995723016 \cdots$$
It seems that the answer is
$$\frac{\pi^2}{12}\left( 1-\sqrt{3}\right)+\log(2) \log \left(1+\sqrt{3}... | Unfortunately, the following generalization works for only positive integer $a$:
$$\int_0^1\frac{\ln(1+x^{2a})}{1+x}dx=\ln^2(2)-\frac{2a^2-1}{8a}\zeta(2)+\frac12\sum_{j=0}^{2a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{4a}\right)\right)$$
$$-\frac12\sum_{j=0}^{a-1}\ln^2\left(2\sin\left(\frac{(2j+1)\pi}{2a}\right)\right).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/426325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "169",
"answer_count": 3,
"answer_id": 2
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Polynomials - Solutions How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial ... what methods occupy for this "problem"?
$$x^3+3x^2-7x+1=0$$
Thanks for your help.
|
How I can find the exact solutions of this polynomial?
I can not get to the exact roots of the polynomial
You have a real polynomial of the third degree with real coefficients. There are exact formulas to find the roots of any polynomial of this kind.
A general cubic equation of the form
$$
\begin{equation*}
ax^{3}+... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of inequality $\sum\limits_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$ Show that
$$\sum_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$$
where $$\binom n k=\frac{n!}{k!(n-k)!}$$
| It seems that we can prove the inequality as follows: $$S=\sum_{k=0}^{n}\binom n k\dfrac{5^k}{5^k+1}=$$
$$\frac 12+\sum_{k=1}^{n}\binom n k\dfrac{1}{1+5^{-k}}=$$
$$\frac 12+\sum_{k=1}^{n}\binom n k\sum_{i=0}^\infty (-1)^i(5^{-k})^i=$$
$$\frac 12+\sum_{i=0}^\infty \sum_{k=1}^{n}\binom n k (-1)^i5^{-ki}=$$
$$\frac 12+\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/429007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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If $A = \tan6^{\circ} \tan42^{\circ},~~B = \cot 66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$ My trigonometric problem is:
If $A = \tan6^{\circ} \tan42^{\circ}$ B = cot$66^{\circ} \cot78^{\circ}$ find the relation between $A$ and $B$.
Working :
$$B = \cot 66^{\circ} \cot78^{\circ} = 1- \frac{\tan... | First, note that $A \approx 0.0946362785$, $\ \ B \approx 0.0946362785$.
Now, we will prove that
$\ \ \ \Large{A=B.}$
a).
$$
\dfrac{A}{B} =
\dfrac
{\sin 6^\circ \sin 42^\circ}
{\cos 6^\circ \cos 42^\circ}
\cdot \dfrac
{\sin 66^\circ \sin 78^\circ}
{\cos 66^\circ \cos 78^\circ}
=
\dfrac
{\bigl( 2 \sin 6^\circ \sin 66^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Solution of Lagrangian Could you give me advice how to solve the following Lagrangian?
$$L=x^3+y^3 - \lambda (x^2-xy+y^2-5)$$
$$ \left\{ \begin{array}{c}
\frac{\partial L}{\partial x} = 3x^2 - \lambda (2x-y) = 0
\\\frac{\partial L}{\partial y} = 3y^2 - \lambda (-x+2y) = 0
\\ \frac{\partial L}{\partial \lambda} = x^2-x... | First
$$\lambda = \frac{3x^2}{2x-y} = \frac{3y^2}{-x+2y}\implies -x^3 +2x^2y = 2y^2x - y^3.$$
Using the factoring formula for $x^n -y^n$ when $n$ is an odd number:
$$
x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}).
$$
Hence we have
$$
x^3 -y^3 + 2y^2x - 2x^2y = 0\implies (x-y)(x^2-xy+y^2) = 0.
$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/432749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $\frac{dx}{dt} = x^3 + x$ for $x$ This is a seemingly simple first order separable differential equation that I'm getting stuck on. This is what I have so far:
$$\frac{dx}{dt} = x^3+x$$
goes to
$$\frac{dx}{x(1+x^2)} = dt$$
Now using partial fractions to integrate the left-hand side:
$$\frac{1}{x(1+x^2)} = \frac{... | You should get
$$ C^2e^{2t}=\frac{x^2}{1+x^2}=1-\frac1{1+x^2},$$
hence
$$ x=\pm\sqrt{\frac{1}{1-C^2e^{2t}}-1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/433966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Transforming trigonometric identities The problem goes like this:
If
$$N=2\sec^4x-3\sec^2x+2=\frac{\cos^2x}{\cos^2y}$$
Calculate the equivalent of
$$M=2\tan^4x+3\tan^2x+2$$
The alternaties I have are:
$$\frac{\tan^2x}{\tan^2y},\mbox{ }\frac{\tan^2y}{\tan^2x},\mbox{ }\frac{\tan^2y}{\sec^2x},\mbox{ }\frac{\sec^2y}{\tan^2... | Some ideas:
$$1+\tan^2x=\sec^2x\implies (1+\tan^2x)^2=\sec^4x$$
But
$$(1+\tan^2x)^2=1+2\tan^2x+\tan^4x$$
So
$$2\tan^4x+3\tan^2x+2=2(\tan^4x+2\tan^2x+1)-\tan^2x=2\sec^4x-\tan^2x=$$
$$=2\sec^4x-\sec^2x+1$$
Try now to end the exercise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/434034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Plotting graphs using numerical/mathematica method From the author's equation 13, 14 We can write by inserting V''(A)=0,
Solving for R we get,
$$R= \frac{6^{D/4} \sqrt{D}}{\sqrt{-2^{1+\frac{D}{2}} 3^{D/2}+3 2^{1+D} A-3^{1+\frac{D}{2}} A^2}}$$
Now inserting the V into the article equation (11)$$E= \left(\frac{\pi }{2... | Mathematica is more powerful that you give it credit for. You don't have to define any quantities explicitly. It is much more efficient to keep things in symbolic terms and numerically substitute values only when you need them. The below works for $d=3$ and gives results consistent with the paper.
Briefly:
*
*D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/434768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $\cos(6^\circ)$? Do you know any method to calculate $\cos(6^\circ)$ ?
I tried lots of trigonometric equations, but not found any suitable one for this problem.
| If you grant the use of $\cos 18^\circ$ = $\dfrac{\sqrt{10+2\sqrt5}}{4}$, you may proceed as follows:
First, calculate $\sin 18^\circ$.
Then, calculate $\cos 36^\circ$ and $\sin 36^\circ$ using $\cos 2\theta = 2\cos^2 \theta - 1$ and $\sin 2\theta = 2\sin \theta \cos \theta$ for $\theta = 18^\circ$.
Finally, use $\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/438387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
Finding $a + b + c$ given that $\;a + \frac{1}{b+\large\frac 1c} = \frac{37}{16}$ Please help me to find the needed sum:
If $a,b,c$ are positive integers such that $\;a + \dfrac{1}{b+\large \frac 1c} = \dfrac{37}{16},\;$
find the value of $\;(a+b+c)$.
Thanks!
| This is asking for the Continued Fraction expansion of $\frac{37}{16}$. We can use the Euclid-Wallis Algorithm to compute the expansion:
$$
\begin{array}{r}
&&2&3&5\\\hline
1&0&1&-3&16\\
0&1&-2&7&-37\\
37&16&5&1&0\\
\end{array}
$$
The continued fraction is above the horizontal line:
$$
\frac{37}{16}=2+\cfrac1{3+\cfrac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/441403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The least value of $4x^2-4ax +a^2-2a+2$ on $[0,2]$ is $3$. What is the integer part of $a$? The least value of $4x^2-4ax +a^2-2a+2$ on $[0,2]$ is $3$. What is the integer part of $a$?
We know that minimum value of a quadratic is $-\cfrac{b}{2a}$.
We will get one condition from here and $-\cfrac{b}{2a}$ should be equal ... | Hint: there are three possibilities, given the shape of the graph of a quadratic
Let $f(x)=4x^2-4ax +a^2-2a+2$, then
either $f(x)$ is increasing on $[0,2]$, in which case the minimum value occurs at $x=0$
or $f(x)$ is decreasing on $[0,2]$, in which case the minimum value occurs at $x=2$
or $f(x)$ has minimum value wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/442654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.