Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$
My question. Does there exist a closed-form expression of $I(a)$?
Some comments. Since $a-2<a+\cos(x)+\cos(y... | We use the identities
$$
2\sin \phi \sin \psi=\cos(\phi-\psi)-\cos(\phi+\psi)
$$
and
$$
\cos \phi+\cos \psi=2\cos\left(\frac{\phi+\psi}{2}\right)\cos\left(\frac{\phi-\psi}{2}\right)
$$
to get
$$
I=\int^{\pi}_{0}\int^{\pi}_{0}\frac{\sin^2 x \sin^2 y}{a+\cos x+\cos y}dxdy=
$$
$$
\frac{1}{4}\int^{\pi}_{0}\int^{\pi}_{0}\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Show that: $\prod_{n=1}^{\infty}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1}=\phi$ How to show that:
$$P=\prod_{n=1}^{\infty}\frac{\sqrt{5}F_{2n+2}+1}{\sqrt{5}F_{2n+2}-1}=\phi$$
Where $\phi=\frac{1+\sqrt{5}}{2}$
$$P=\frac{\sqrt{5}F_{4}+1}{\sqrt{5}F_{4}-1}\cdot \frac{\sqrt{5}F_{6}+1}{\sqrt{5}F_{6}-1}\cdot \frac{\sqrt{5... | Let $F_n$ denote the Fibonacci sequence and $L_n$ denote the Lucas sequence. Note the following identities
\begin{eqnarray*}
5F_{2n}^2-1&=&L_{2n-1}L_{2n+1} \\
2F_{2n+2}-F_{2n+1}&=&L_{2n+1} \\
5F_{2n+1}F_{2n+2}-2&=&L_{2n+1}L_{2n+2} \\
5F_{2n+2}-L_{2n+2}&=&2L_{2n+1} \\
L_{2n+2}F_{2n+2}-1&=&L_{2n+1}F_{2n+3}.
\end{eqnarra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+... | Hint:
$a+b+c=3 \equiv f(1)=3$. Using recurrence relation, $f(2)=2f(1)+1=7$, $f(3)=12$, $f(4)=18$. Notice the pattern.
$$\begin{aligned}a+b+c&=3\\4a+2b+c&=7\\9a+3b+c&=12\end{aligned}$$
This gives $a=1/2$, $b=5/2$, $c=0$. Can you proceed?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}... | Try multiplying both sides by $n^{1/3}$, expanding $(n(n+1)^2)^{1/3}$ to $(n^3+2n^2+n)^{1/3}$, and moving $n$ from the left hand side to the right hand side, which rewrites the inequality as
$$(n^3+2n^2+n)^{1/3}\lt n+{2\over3}$$
Cubing both sides (which is OK since $x^3\lt y^3\iff x\lt y$) turns the inequality to prove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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3 x 3 Grid Square For a question that I need to answer, we have to show that there is only one arrangement for which, in a $3\times 3$ grid in which the numbers from $1$ to $9$ are placed and $5$ is the centre number, the sum of the $4$ numbers in every $2\times 2$ grid found within the $3\times 3$ grid all add up to t... | The four sums being equal yields four linear equations; for the grid
$$\begin{matrix}a&b&c\\d&5&e\\f&g&h\end{matrix},$$
with $\{a,b,c,d,e,f,g,h\}=\{1,2,3,4,6,7,8,9\}$ you get the equations
\begin{eqnarray*}
a+b+d+5&=&x,\\
b+c+5+e&=&x,\\
d+5+f+g&=&x,\\
5+e+g+h&=&x.
\end{eqnarray*}
First note that $a+b+c+d+e+f+g+h=40$. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$... | There are actual computation methods for cubics but we can poke some number theory style fun into this.
$(x-1)(x-2)(x-3)=24$ are numbers whose product is 24. If we focused on integers, these numbers would be consecutive.
Oh what luck befalls us today. Watch this:
$(x-1)(x-2)(x-3)=4 \cdot 3 \cdot 2$
Let each factor pick... | {
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"url": "https://math.stackexchange.com/questions/3594574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Setting up the particular solution to $y''+4y'=e^{-4x}+3$ $y''+4y'=e^{-4x}+3$
The characteristic polynomial is $r^2+4r=0 \to r = 0,-4$
the complementary solution is: $y_c = C_1+C_2e^{-4x}$
The part that is throwing me off is the addition of 3. I can't even set that up as a polynomial?
| In this particular case, and similar cases, there is a second derivative with constant coefficients equaling a constant. This suggests a linear polynomial of the order of the highest derivative. This leads to the form (the exponential term being equal to a known solution so other rules apply):
$$f(x) = a_{0} + a_{1} \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3594730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to determine $x$ in $x^2+3x+7=4$? I'm helping my child with his homework. One of the problems is this: Determine the solution set of the equation
$x^2+3x+7=4$
Here is my attempt to determine $x_1$ and $x_2$:
$a^2+2ab+b^2=(a+b)^2$
$2ab=3x$
$2xb=3x$
$2b=3$
$b=\frac{3}{2}=1.5$
$b^2=2.25$
$x^2+3x+b^2-b^2+7=4$
$x^2+3x+2... | This is actually much simpler if you subtract the 4 to the left hand side first:
$$x^2 + 3x + 3 = 0$$
Then the solution set must satisfy the quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
where we have the quadratic $ax^2 + bx + c = 0$. In this case,
$$a=1, b=3, c = 3$$
So then $x$ must be $$x = \frac{-3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How to show that $\displaystyle\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How show that $(1)$ is:
$$\sum_{n=0}^{N}\frac{(-1)^n{2n \choose n}{N+n \choose N-n}}{[(n+1)(n+1)(n+2)(n+3)\cdots(n+N)]^2}=\frac{1}{2N!^2}\tag1$$... | Let $ N $ be a positive integer.
Observe that : $$ \left(\forall n\in\mathbb{N}\right),\ \frac{1}{\binom{N+n}{n}}=\left(n+N+1\right)\int_{0}^{1}{x^{n}\left(1-x\right)^{N}\,\mathrm{d}x} $$
Thus, \begin{aligned} \sum_{n=0}^{N}{\frac{\left(-1\right)^{n}\left(N!\right)^{2}}{\left(N+n\right)!\left(N-n\right)!}}&=\sum_{n=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$.
Question: Define $I_n=\int_0^1\frac{x^n}{\sqrt{x^2+1}}dx$ for every $n\in\mathbb{N}$. Prove that $$\lim_{n\to\infty}nI_n=\frac{1}{\sqrt 2}$$.
My approach: Given that $I_n=\int_0^1\frac{x^... | Yet another approach is to use integration by parts to obtain
\begin{equation*}
I_{n} + I_{n-2} = \int_{0}^{1}x^{n-2}\sqrt{1+x^{2}}\,\mathrm{d}x = \frac{\sqrt{2}}{n-1} - \frac{1}{n-1}I_{n}
\end{equation*}
for $n \geq 2$, from which we get
\begin{equation*}
nI_{n} = \sqrt{2} - (n-1)I_{n-2} = \sqrt{2} - (n-2)I_{n-2} - I_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$ 5.4
Can somebody verify this solution for me?
Find the area under the graph of the function $f(x)=(1-2x)^2$ between $x=-2$ and $x=2$
The area under the graph of $f(x)$ between $x=-2$ and $x=2$ is exactly equal to:
$\int_{-2}^2 (1-2x... | You would spare your time if you make a substitution $ t=2x-1$. $$\int_{-2}^2 (1-2x)^2dx = {1\over 2}\int_{-5}^3 t^2dt =...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that there do not exist any distinct natural numbers a,b,c,d such that
Show that there do not exist any distinct natural numbers a,b,c,d such that $a^3+b^3=c^3+d^3$ and $a+b=c+d$.
| Mark $a+b=x$ then $b=x-a$ and $d=x-c$. Notice that $x\ne 0$. No we have: $$a^3+(x-a)^3 = c^3+(x-c)^3$$ and thus $$-3x^2a+3xa^2 = -3x^2c+3xc^2$$
so \begin{align}xa-a^2= xc-c^2& \implies x(a-c) = (a-c)(a+c)\\ &\implies x=a+c \\& \implies a+c=a+b \\&\implies c=b\end{align}
A contradiction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is the integral $ \int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1} $ converges or diverges. Determine if the following integral diverges/converges, if it converges, is it absolutely or conditionally converges.
$$
\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1}
$$
What i tried:
In that interval we know that:
$$
0 < x
$$
Therefore ... | First, do a change of coordinates: $y=x-1$. The integral then becomes
$$\int_1^2\frac{1}{\sqrt{x^2-x+1}-1}dx=\int_0^1\frac{1}{\sqrt{y^2+y+1}-1}dy$$
Now, expand using the Taylor series around $0$ to get
$$\frac{1}{\sqrt{y^2+y+1}-1}=\frac{2}{y}-\frac{3}{2}+\frac{15 y}{8}-\frac{33 y^2}{16}+\cdots>\frac{2}{y}-\frac{3}{2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding number of integral solutions to an equation.
Find the number of integral solutions to:
$$x^2+y^2-6x-8y=0.$$
My attempt:
The equation can be rewritten as:
$$x^2+y^2-6x-8y+9+16=25,$$
basically adding 25 to both sides, or equivalently,
$$(x-3)^2+(y-4)^2=25.$$
This is a Pythagorean triplet. The only triplet of ... | There are $12$ different solutions. The square $25$ can be expressed as the sum:
$$5^2+0^2=4^2+3^2=3^2+4^2=0^2+5^2=25$$
So, the possibilities are:
$$\left\{\begin{matrix}
x-3=\pm4 \rightarrow x=7 \vee x=-1
\\ y-4=\pm3 \rightarrow y=7 \vee y=1
\end{matrix}\right.
\vee
\left\{\begin{matrix}
x-3=\pm3 \rightarrow x=6 \vee ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How to prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent?
Consider the sequence $\{x_n \}$ defined by $$x_n = e \left (\frac {n} {n+1} \right )^{n + \frac 1 2},\ n \geq 1.$$
Prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent.
My attempt: I find that \begin{align*} \sum\... | Use Taylor series to second and third order:
$$1-\left(n+\frac12\right)\log\left(1+\frac1n\right)=1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}+\frac1{3n^3}-\dots\right)$$
$$\le1-\left(n+\frac12\right)\left(\frac1n-\frac1{2n^2}\right)=\frac1{4n^2}$$
$$1-\left(n+\frac12\right)\log\left(1+\frac1n\right)\ge1-\left(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\fra... | Although this solution is mentioned in the comments to the question, I think it should be put here as an answer because of its simplicity and naturalness.
Looking at the given equation $2x^3+x^2+x-1=0$ one would quickly check possible rational roots using the Rational Root Test:
$$\text{To check: }\pm 1, \pm\frac 12$$... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why doesnt this trignometric substituition work? $$\frac 1 5\int\frac{x+1+5}{(x+1)^2+5}\,\mathrm dx$$
where i substitute $x+1 = \sqrt{5}\tan(\theta)$
After Substituition :
$$\frac15\int\frac{\sqrt5\tan(\theta)+5}{\tan^2(\theta)+1}\sqrt5\sec^2(\theta)\,\mathrm d\theta$$
but if i do $u = x +1$
and then further substitute... | As said by Doug, the substitution works. Before applying it, consider decomposing $$\frac{x+1+5}{(x+1)^2+5}=\frac12\frac{((x+1)^2+5)'}{(x+1)^2+5}+\frac{5}{(x+1)^2+5},$$ that makes life simpler.
| {
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"timestamp": "2023-03-29T00:00:00",
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$n$ is prime if $1+x+x^2+\dots+x^{n-1}$ is prime I came across this question in a number theory textbook:
Let $x$ and $n$ be positive integers such that $1+x+x^2+\dots+x^{n-1}$ is prime. Then show that $n$ is prime.
I reasoned that this proof would require Fermat's Little Theorem, and put $1+x+x^2+\dots+x^{n-1} = \fr... | If $x=1$, then $1+x+\ldots+x^{n-1}$ is simply $n$ and we are done. So assume $x\gt 1$.
Suppose that $n$ is composite, $n=ab$ with $a\geq 2,b\geq 2$. Then we have $\frac{x^n-1}{x-1}=AB$ where $A=\frac{x^n-1}{x^a-1}=1+x^a+(x^a)^2+\ldots+(x^a)^{b-1}$ and $B=\frac{x^a-1}{x-1}$. So $\frac{x^n-1}{x-1}$ is composite.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+... | We can write your idea in the following form.
Let $\sqrt[6]{\frac{x}{2}}=y.$
Thus, $$\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}=\frac{\sqrt[3]2}{\sqrt2}\lim_{x\rightarrow2}\frac{\sqrt[3]{\frac{x}{2}}-1}{\sqrt{\frac{x}{2}}-1}=\frac{\sqrt[3]2}{\sqrt2}\lim_{y\rightarrow1}\frac{y^2-1}{y^3-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624692",
"timestamp": "2023-03-29T00:00:00",
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Real Solution of $4^x + 6^x = 9^x$. So here is the problem
Solve $$4^x + 6^x = 9^x$$ for $x$.
I am trying to find its real solution.
I was trying in this way!!
$$6^x\left( \frac {4^x}{6^x} + 1\right) = 9^x$$
$$\left({\frac 23}\right)^x +1=\left({\frac 32}\right)^x$$ but I'm stuck . . .
Will anyone help ?
| $$4^x + 6^x = 9^x \implies \left(\frac{4}{9}\right)^x +\left(\frac{6}{9}\right)^x =1 \implies \left(\frac23\right)^{2x}+\left(\frac23\right)^{x}-1= 0 $$
Let $\left(\frac23\right)^x = y$
So, $$y^2+y-1 = 0 \implies y =\left(\frac23\right)^x = \frac{-1\pm \sqrt{5}}{2}$$
Considering the positive root, $$x = \frac{\ln y}{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution to a non-linear simultaneous equation I'm trying to solve the non-linear, two-variable system of equations
\begin{align*}
y-x+x^5-\frac{y^4x}{2(1+x^2)^2}-\frac{x^3}{1+y^2} &= 0 \\
-x-y+y^5-\frac{x^4y}{2(1+y^2)^2}-\frac{y^3}{1+x^2} &= 0
\end{align*}
with $x,y \in \mathbb{R}$. You might notice that the second eq... | I converted both equations into polynomials and put them into Singular to find all solutions.
LIB "solve.lib";
ring r=0,(x,y),dp;
poly p1=2x9y2+2x9+4x7y2+2x7-4x5+2x4y3+2x4y-4x3y2-6x3+4x2y3+4x2y-xy6-xy4-2xy2-2x+2y3+2y;
poly p2=-x6y-x4y-2x3y4-4x3y2-2x3+2x2y9+4x2y7-4x2y3-2x2y-2xy4-4xy2-2x+2y9+2y7-4y5-6y3-2y;
ideal i=p1,p2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x... | Attention: I suddenly found the OP requires the use of Taylor series expansion. My solution does not meet the requirement.
We have
$$f(x) = (x+3)^{1/3}(x-27) = (x+3)^{4/3} - 30(x+3)^{1/3}.$$
Thus, we have, for $k\ge 1$,
\begin{align}
f^{(k)}(x) &= \tfrac{4}{3}(\tfrac{4}{3}-1)(\tfrac{4}{3}-2) \cdots (\tfrac{4}{3}-(k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $\sum_{r=1}^{3n-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{3n}r}$
$$\sum_{r=1}^{3n-1}\dfrac{(-1)^{r-1}\cdot r}{\binom{3n}r}, n \in 2k, k\in \mathbb{Z^+}$$
Answer given (much simpler than expected)
$\dfrac{3n}{3n+2}$
I tried adding and subtracting 1 to $r$ so could use $\dfrac{\binom{n}r}{r+1}=\dfrac{\binom{n+1}... | Let's start by the following : \begin{aligned}\frac{1}{\binom{3n}{r}}=\left(3n+1\right)\int_{0}^{1}{x^{r}\left(1-x\right)^{3n-r}\,\mathrm{d}x}\end{aligned}
Since $\left(\forall x\in\left[0,1\right]\right) $, we have : $$ \sum_{r=0}^{3n-1}{r\left(-1\right)^{r-1}x^{r}\left(1-x\right)^{3n-r}}=\left(-1\right)^{n+1}x^{3n+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Calculating determinant of a symmetric matrix where the $k$th row is given by $[a_{k-1},a_k,...,a_0,a_1,...,a_{n-(k-1)}]$
For $j = 0,...,n$ set $a_{j} = a_{0} + jd$, where $a_{0}, d$ are fixed real numbers. Calculate the determinant of the $(n+1)\times (n+1)$ matrix
$$A = \begin{pmatrix}
a_{0} & a_{1} & a_{2} ... | Partial Answer: If we use the matrix determinant lemma, then it suffices to consider the case of $a_0 = 0$ and $d = 1$.
With that said, let $A_n$ denote the matrix defined with your formula, with $a_0 = 0$ and $d = 1$. Let $R$ denote an $n \times n$ identity matrix, where the first column has been replaced by $(1,0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Using that $1 + z + z^{2} + ... + z^{n} = \frac{1-z^{n+1}}{1-z}$ and taking the real parts, prove that: $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac12+\frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$
for $0 < \theta < 2\pi$.
Alright. What I have done is this, using the De Moivre's Formu... | Continue with
$$1 + e^{i\theta} + e^{2i\theta} + ... e^{ni\theta}=\frac{1 - e^{(n+1)i\theta}}{1 - e^{i\theta}}
=\frac{e^{\frac12(n+1)i\theta}}{e^{\frac12i\theta}}\cdot \frac{e^{-\frac12(n+1)i\theta} - e^{\frac12(n+1)i\theta}}{e^{-\frac12i\theta} - e^{\frac12i\theta}}
= e^{\frac12ni\theta} \frac{\sin\left(\frac{n + 1}2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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The sequence 1 + (1+1) + (1+1+2) + (1+1+2+4) + (1+1+2+4+8) + .. I am working on a brute-force algorithm for a hard problem where the number of operations $S[n]$ seem to grow as a function of $n$ as follows:
$$
\begin{align}
S[1]& = 1\\
S[2]& = 1 + (1+1)\\
S[3]& = 1 + (1+1) + (1+1+2)\\
S[4]& = 1 + (1+1) + (1+1+2) + (1+1... |
$1 + 2 + 4 + ...... + 2^k = 2^{k+1} - 1$.
So $(1+1+2+4+....... + 2^k) = 2^{k+1}$.
So $S[n] = 1 + (1+1) + (1+1+2) + (1+1+2+4) + ..... = 1 + 2^1 + 2^2 + ..... + 2^n = 2^{n+1} -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
(Different) derivatives of $f(x) = \arcsin\left(\left(5 x + 12 \sqrt{1-x^2}\right)/13\right)$ via two different substitutions? We have been given a function to differentiate:
$$f(x) = \arcsin \left(\frac{5x + 12\sqrt{1-x^2}}{13}\right)$$
My teacher told me the method to substitute $ x= \sin\vartheta$ which would sim... | I assume that the reason for the "paradox" was clarified already in comments, but possibly there is still a need to go a little bit deeper into details:
Your error has its root in the assumption:
$$
\arcsin(\sin x)=x.
$$
However the equality is valid only in the range $-\frac\pi2\le x\le \frac\pi2$ whereas the correct ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve $x^{-2} + x^{-3} = \dfrac{x^3 + x^2}{4x^6}$? For online school I was given the equation $$x^{-2} + x^{-3} = \frac{x^3 + x^2}{4x^6}.$$ The problem lies in the addition of the bases, as I do not know how to solve that. Could you help me?
| We have that
$$x^{-2} + x^{-3} = \dfrac{x^3 + x^2}{4x^6}$$
$$x^4+x^3=\frac{1}{4}(x^3+x^2)$$
$$x^4+\frac34 x^3-\frac14 x^2=0$$
$$x^2(4x^2+3x-1)=0$$
However, since $x=0$ is not a solution (it is in the denominator of the original equation), we can divide out by $x$ to get
$$4x^2+3x-1=0$$
This is a simple quadratic equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3637473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying ... | You can do it brutally with a limited developement at order 3:
Let's study the parts independantly and start with the denominator:
*
*$a^2+b^2+2ab\cos(x) = (a+b)^2-{abx^2}+O(x^4)$
*From that you have $\sqrt{a^2+b^2+2ab\cos(x)} = a+b - \frac{x^2 a b}{2 (a + b)}
+ O(x^4)$
*Hence $a+b-\sqrt{a^2+b^2+2ab\cos(x)} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $ Question -
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$... | Another way.
By C-S and P-M we obtain:
$$\sum_{cyc}\frac{a^2}{a+2b^2}-1=\sum_{cyc}\frac{a^4}{a^3+2a^2b^2}-1\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^3+2a^2b^2)}-1=$$
$$=\frac{\sum\limits_{cyc}(a^4-a^3)}{\sum\limits_{cyc}(a^3+2a^2b^2)}=\frac{9(a^4+b^4+c^4)-(a^3+b^3+c^3)(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{9\sum\limits_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Orthogonally Diagonalize the matricies The Question given was "Orthogonally diagonalize the matrices, giving an orthogonal matrix P and diagonal matrix D."
I was given eigenvalues -3 and 15
and Matrix A=
$$
\begin{bmatrix}
5 & 8 & -4 \\
8 & 5 & -4 \\
-4 & -4 & -1
\end{bmatrix}
$$
So far I've done the identity M-I(eige... | The given matrix
$$ \begin{bmatrix}
5 & 8 & -4 \\
5 & 5 & -4 \\ -4 & -4 & -1
\end{bmatrix}
,$$
has eigenvalues $-3$ and two other complex eigenvalues , so I suspect you must have miss-wrote one of the entries of the matrix. Since the matrix is $3\times 3$, by theorem it must have $3$ eigenvalues. So one of the eigenv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Generating function of $(2,2,2,4,4,4,8,8,8,\ldots)$ I know that $F(x)=\dfrac{1}{1-x}$ is a generating function of $(1,1,1,1,\ldots)$ and $F(2x)$ is a generating function of $(1,2,4,8,16,\ldots).$
Then $G(x)=\dfrac{F(2x)-1}{x}=\dfrac{2}{1-2x}$ is a generating function of $(2,4,8,16,\ldots).$
A generating function of $(2... | If the book gave the answer as $\frac{1+x+x^2}{1-x}$, that's wrong; your $H$ is correct. Indeed,
$$\begin{align*}
\frac{1+x+x^2}{1-x}&=(1+x+x^2)(1+x+x^2+x^3+\cdots)\\
&=1+2x+3x^2+3x^3+\cdots
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Proof $\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt2} + \frac{\sqrt{2 + \sqrt2}}{\sqrt2 \sqrt{2 - \sqrt{2 - \sqrt2}}} =\sqrt{2-\sqrt{2+\sqrt2}}$ I want to prove the following equation
$$\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt{2}} + \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}$$
I know this m... | $a=\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}}$
$a^2=4+\sqrt{2}$
$a=\sqrt{2(2+\sqrt{2})}$
$b=\sqrt{2-\sqrt{2-\sqrt{2}}}*\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{4+\sqrt{2}-2(\sqrt{2-\sqrt{2-\sqrt{2}}}+\sqrt{2-\sqrt{2+\sqrt{2}}})}=\sqrt{4+\sqrt{2}-2(\sqrt{2(2+\sqrt{2})})} $
$-\sqrt{2-\sqrt{2-\sqrt{2}}}/\sqrt{2}+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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In how many different ways can you prove that $\sin^2x + \cos^2x = 1$ The standard proof of the identity $\sin^2x + \cos^2x = 1$ (the one that is taught in schools) is as follows: from pythagoras theorem, we have (where $h$ is hypotenuse, $b$ is base and $p$ is perpendicular)
$$h^2 = p^2 + b^2$$
dividing by $h^2$ on bo... | 1) With identity $a^2-b^2= (a-b)(a+b)$,
$$\cos^2x+ \sin^2x
= (\cos x-i\sin x)(\cos x+i\sin x)=e^{-i x}e^{i x}=1$$
2) With double angle identities,
$$\cos^2x+ \sin^2x
=\frac12 (1+\cos 2 x)+\frac12(1-\cos 2x)=1$$
3) with half-angle subs $t= \tan\frac x2$,
$$\cos^2x+ \sin^2x
=\left(\frac{1-t^2}{1+t^2}\right)^2 +\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3644869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
What's the value of $\int \left(1-\sqrt{1-x^2} \right) dx$? $$\int \left(1-\sqrt{1-x^2} \right) dx$$
Also, What is it's value from $0$ to $(1+\sqrt7)/2$ ?
| I believe there might be missing pieces or a mistake in the inquiry, here's why:
let x = sin(u)
Thus, $\frac{dx}{du}=cos(u)$
$$\int 1-\sqrt{1-x^2}dx = \int (1-cos(u))(cos(u))du $$
Using the Double angle formula: $cos(2\theta)=2cos^2\theta -1$ , we have:
$$ sin(u) - \int\frac{cos(2u)-1}{2}du $$ which thus evaluates to (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3647455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\righ... | WLOG assume $c\neq \text{mid}\{a,b,c\}$. We have:
$$a^2 b^2 c^2 (\text{LHS-RHS}) =\left( a-b \right) ^{2} \left( ab+ca+bc-{c}^{2} \right) ^{2}+4\,ab{c}
^{2} \left( a-c \right) \left( b-c \right) \geqq 0$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Integral $\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$ $$\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$$
Here's my attempts but I'm not sure that I'm doing well :
$$\text{The integral gives :} 1-\frac1 4+\frac1 6-\frac1 9+ \frac{1}{11}-\frac{1}{14}+\dots$$
This series is :
$$ S=\sum_{k=0}^\infty \frac{3}{(5k+1)(5k+4)... | Let :
$$I=\int_0^1(1-x^3+x^5-x^8+x^{10}-x^{13}+\dots)dx$$
Let $$S=\sum_{k=0}^\infty \frac{3}{(5k+1)(5k+4)}$$
Let's compute it :
\begin{align}
S&=3\sum_{k=0}^\infty \frac{1}{(5k+1)(5k+4)} \\\\
&=\frac3 4 +\frac1 5 \sum_{k=1}^\infty\bigg(\frac{1}{k+1/5}-\frac{1}{k+4/5}\bigg) \\\
&=\frac{3}{4}+\frac{1}5\sum_{k=1}^\infty\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove $\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\ge \frac{a^2+b^2+c^2}{ab+bc+ca}$ For $a,b,c>0$. Prove that: $$\frac{a^2}{(b+c)^2}+\frac{b^2}{(c+a)^2}+\frac{c^2}{(a+b)^2}+\frac{1}{4}\geqq \frac{a^2+b^2+c^2}{ab+bc+ca}$$
NguyenHuyen gave the following expression$:$
$$\sum \frac12\, \left( 8... | Yes, SOS helps!
Indeed, we need to prove that
$$\sum_{cyc}\left(\frac{a^2}{(b+c)^2}-\frac{1}{4}\right)\geq\frac{a^2+b^2+c^2}{ab+ac+bc}-1$$ or
$$\sum_{cyc}\frac{(2a+b+c)(a-b-(c-a))}{(b+c)^2}\geq\frac{2\sum\limits_{cyc}(a-b)^2}{ab+ac+bc}$$ or
$$\sum_{cyc}(a-b)\left(\frac{2a+b+c}{(b+c)^2}-\frac{2b+a+c}{(a+c)^2}\right)\geq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting ... | I get a different answer from either of those. The probability that a professor who leaves the house with his umbrella returns with it is $$\left(\frac34\right)^3=\frac{27}{64}$$ and the probability that he loses it is $\frac{37}{64}$. There are two cases. Both A and B take their umbrellas with them (probability $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$3(a+{1\over a}) = 4(b+{1\over b}) = 5(c+{1\over c})$ and $ab+bc+ca=1$ This was the question :
$3(a+{1\over a}) = 4(b+{1\over b}) = 5(c+{1\over c})$ where $a,b,c$ are positive number and $ab+bc+ca=1$ ,
$5({1-a^2\over 1+a^2}+{1-b^2\over 1+b^2}+{1-c^2\over 1+c^2}) = ?$
I have spent a lot of time trying rearranging the... | The hint.
Let $a=\tan\frac{\alpha}{2},$ $b=\tan\frac{\beta}{2}$ and $c=\tan\frac{\gamma}{2},$ where $\{\alpha,\beta,\gamma\}\subset(0^{\circ},180^{\circ}).$
Thus, $ab+ac+bc=1$ gives $\alpha+\beta+\gamma=180^{\circ}$ and the rest gives
$$\frac{3}{\sin\alpha}=\frac{4}{\sin\beta}=\frac{5}{\sin\gamma},$$ which gives $\gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find all natural numbers $a,b$ such that $ab, 4a+b-3$ are perfect squares and $9a-4b$ is a prime number. Find all natural numbers $a,b$ such that $ab, 4a+b-3$ are perfect squares and $9a-4b$ is a prime number.
I can't find any idea for this problem :(
| If $ab$ is a perfect square, then $a$ and $b$ must be perfect squares, say $a=m^2$ and $b=n^2$. It follows that
$$
9a-4b = (3m+2n)(3m-2n),
$$
so that for $9a-4b$ to be prime it must be or $3m+2n=1$ or $3m-2n=1$. Now we consider $4a+b-3\pmod{3}$: the only squares in $\mathbb{Z}/3\mathbb{Z}$ are $0$ and $1$, thus the onl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Olympiad Inequality Question I am new to the Olympiad-style questions and I hope someone could correct my proof for this question as I do not have the answer for it. Please leave some constructive criticism if possible so I could improve. Thanks in advance!
Let $a,b,c$ be positive real numbers. Prove that:
$$a^3 +b^... | While your final inequality isn't yet what you want, you can make a small modification to get the the answer.
The idea here is that the cyclic sums (but in opposite directions) $Y$ and $Z$ are highly related to each other, so if we have several expressions involving them, we can try to cancel out a term (E.g. via Ga... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3686300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find all ordered pairs $(x,y)$ of positive integers such that the expression $x^2+y^2+xy$ is a perfect square. My approach so far:
Let $x^2+y^2+xy=n^2,$ where $n\in\mathbb Z$.
$\implies (x+y) ^2-xy=n^2$
$\implies (x+y) ^2-n^2=xy$
$\implies (x+y+n) (x+y-n) =xy$
Only one case is possible:
When $x+y+n=xy$ and $x+y-n=1$. O... | $x^2+yx+y^2-n^2=0 \implies \triangle = 4n^2-3y^2= d^2$. At this point I propose you to look up a article by L.J. Mordell or Kneser in internet and they have a formula for all the finite solutions that are bounded above by the products of the coefficients $(4,-3,-1)$ of the Diophantine equation above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to prove the following inequality with two variables How to prove the following inequality, which should hold for all $x \in \left]0; 1\right[$, $n \in \mathbb{N}$, n is even.
\begin{align}
n^2 x^{n-2} (x^{n+1}+1)^3\left[2nx^n - (n-1)(x^n-1)\right] > (n+1)^2 x^{n+1} (x^n-1)^3\left[(2(n+1)x^{n+1} - n(x^{n+1}+1)\righ... | Indeed you need to prove:
\begin{align}
\left(\frac{n}{n+1}\right)^2 \left(\frac{1+x^{n+1}}{x-x^{n+1}}\right)^3\frac{(n-1)+(n+1)x^n}{n-(n+2)x^{n+1}} &> 1
\end{align}
We have
$$ |n-(n+2)x^{n+1}|\leq n.$$
Also
$$ \left(\frac{1+x^{n+1}}{x-x^{n+1}}\right)^3>x^{-3},$$
so we need to prove that
$$(n-1)x^{-3}+(n+1)x^{n-3}>\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Packs of gum probability On each gum pack there's a prize. There's $6$ different prizes and each gum pack has the same probability to have each prize. Johnny buys a gum pack each day to collect all the different prizes and only then he will stop.
I need to calculate the standard deviation of the number of days that Jo... | Let $\mathbf{p}_n=(p_1,p_2,p_3,p_4,p_5,p_6)^T$ and $p_i$ be the probability of having $i$ distinct prizes after opening $n$ gum packs. $\mathbf{p}_1=(1,0,0,0,0,0)^T$ and $\mathbf{p}_n=A\mathbf{p}_{n-1}$ where
$$A=\begin{pmatrix}
\frac{1}{6} && \frac{5}{6} && 0 && 0 && 0 && 0\\
0 && \frac{2}{6} && \frac{4}{6} && 0 && 0 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3691575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What is the total number of positive integers <300 whose digit sum is a multiple of 5? As stated in the title
For one digit number there is $1$
For two digit numbers there are $27(=4+(4+5)+(4+5+5)) (sum
=5n, n=1,2,3)$ $since (5n<18)$
But I'm struggling to find the ones that are of three digits
$(sum)=5n, n=1,...,4)$$f... | If the first two digits are $0,0$ than the last digit must be $0$ or $5$. There are two such numbers.
If the first two digits are $0,1$ then the last digit must be $4$ or $9$. There are two such number.
If the first two digits are $2$ and $7$ that then last digit must be $1$ and $6$. There are two such numbers.
And s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3692420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
finding the characteristic polynomial of symmetric matrix So I got this symmetric matrix
$$A= \begin{pmatrix}
0 & 3 & 1 & -2 \\
3 & 0 & -2 & 1 \\
1 & -2 & 0 & 3 \\
-2 & 1 & 3 & 0\\
\end{pmatrix}$$
I've got to $$P(t)=\begin{pmatrix}
-t & 3 & 1 & -2 \\
3 & -t... | The Laplace expansion with cofactors works:
$-t\begin{vmatrix}-t&-2&1\\-2&-t&3\\1&3&-t\end{vmatrix}-3\begin{vmatrix}3&-2&1\\1&-t&3\\-2&3&-t\end{vmatrix}+\begin{vmatrix}3&-t&1\\1&-2&3\\2&1&-t\end{vmatrix}+2\begin{vmatrix}3&-t&-2\\1&-2&-t\\-2&1&3\end{vmatrix}$
$=t^4 - 28 t^2 + 48 t=t(t - 4) (t - 2) (t + 6).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3693360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$ How to prove
$$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{... | \begin{align}
J&=\int_0^1 \frac{\arctan\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx\\
K&=\int_0^1 \int_0^1 \frac{1}{(x^2+2)(y^2+2)}dxdy \\
&=\int_0^1 \int_0^1 \frac{1}{4+x^2+y^2}\left(\frac{1}{2+x^2}+\frac{1}{2+y^2}\right)dxdy\\
&=2\int_0^1 \int_0^1 \frac{1}{(4+x^2+y^2)(2+x^2)}dxdy\\
&=2 \int_0^1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3693547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proof of equation relating to vector triple product and BAC-CAB rule I am currently studying Introduction to Electrodynamics, fourth edition, by David J. Griffiths. Chapter 1.1.3 Triple Products introduces the vector triple product as follows:
(ii) Vector triple product: $\mathbf{A} \times (\mathbf{B} \times \mathbf{C... | $(A\times B)\cdot(C\times D)=(A\times B)\cdot E=A\cdot(B\times E)$
where $E=C\times D$, and by the scalar triple product identity.
Then $B\times E=B\times(C\times D)=(B\cdot D)C-(B\cdot C)D$ by
the vector triple product identity. Now dot this with $A$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How would you find the number of solutions to $m_1+2m_2+3m_3=n$? Assuming $m_i$ are nonnegative integers
I understand that we need to use $C(n+k-1,n-1)$ here but I am not sure how the coefficients of $m_i$ affect the equation?
For example to find the number of solutions to $m_1+m_2+m_2=n$, we could use the equation abo... | I'm presuming $n$ is given. The answer (call it $f(n)$) is the coefficient of $x^n$ in the Taylor expansion of the generating function
$$ g(x) = (1+x+x^2 + \ldots)(1+x^2 + x^4 +\ldots)(1 + x^3 + x^6 + \ldots) = \frac{1}{(1-x)(1-x^2)(1-x^3)}$$
Using partial fractions,
$$ g(x) = \frac{1}{6(1-x)^3} + \frac{1}{4(1-x)^2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving a nonlinear periodic ODE Is it possible to solve the below ODE for arbitrary real values of $c$?
$$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0,$$
where $\theta=\theta(x)$. Also, $\theta$ itself does not have meaning, but ($\cos(2\theta),\... | $$2 (\frac{d \theta}{dx})^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{d^2}{dx^2}\theta=0$$
Let $y=\frac{d \theta}{dx}=\frac{d \theta}{dy}\frac{dy}{dx}\quad\implies\quad \frac{dy}{dx}=y\frac{dy}{d \theta}$
$$2 y^2 [\sin(2\theta) -c \cos(2 \theta)]-[\cos(2\theta) +c \sin(2 \theta)]\frac{dy}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Second order ODE solution - help me spot a mistake Solve following ODE:
$$
(1-x)x''+2(x')^2=0; x(0)=2, x'(0)=-1
$$
$$
x''=\frac{-2(x')^2}{1-x}
$$
substitute $x'=u(x)$ and assume $u \neq 0$
$$
uu'=\frac{-2(u)^2}{1-x} \\
\frac{dx}{1-x}=-\frac{du}{2u} \\
-\ln{|1-x|}=-\frac{1}{2}\ln{|u|}+c \\
|1-x|=\sqrt(|u|)e^c
$$
undo su... | The first stage of the solution should be kept with an undetermined constant up to the point
$$
u(x)=C(x-1)^2.
$$
Then using $u(2)=-1$ gives $C=-1$ and thus
$$
x'(t)=-(x-1)^2,
$$
which has a different sign than the equation you got. The sign difference in the solution follows from here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A curious infinite product Let $R$ be the ring $\mathbb{Z[q]}/(q^2)$ whose elements $a+bq$ satisfy $q^2=0.$
Define $g_n\in R$ by $g_1=1$, $g_{2^n}=1-2^{n-1}q$ for $n>0$ and $g_{2n}=0,$ $g_{2n+1}=-q$ else.
It seems that in $R$ the following identity holds in the sense of formal power series with coefficients in $R$:
... | Write $\epsilon = -q$, which will save us some minus signs. Start by substituting $x \mapsto \left( 1 + \frac{\epsilon}{2} \right) x$ into the well-known identity, which gives
$$\prod_{k \ge 0} (1 + (1 + 2^{k-1} \epsilon)) x^{2^k} = \frac{1}{1 - \left( 1 + \frac{\epsilon}{2} \right) x} = \sum_{n \ge 0} \left( 1 + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in t... | First get rid of the annoying factors,
$$\int\frac{\sqrt{16x^2-9}}xdx=\int\frac{\sqrt{16\left(\dfrac{3y}4\right)^2-9}}{\dfrac{3y}4}d\dfrac{3y}4=3\int\frac{\sqrt{y^2-1}}{y}dy.$$
Then observe the identity
$$\left(\frac 12\left(t+\dfrac1t\right)\right)^2-1=\left(\frac 12\left(t-\dfrac1t\right)\right)^2.$$
Then with $y=\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 7
} |
$\int \frac{x^2\,dx}{(a-bx^2)^2}$ How do I integrate $\int \frac{x^2\,dx}{(a-bx^2)^2}$
I've tried substitution and partial fraction decomposition, but I'm not getting anywhere.
| You haven't placed any restrictions on ${a,b}$, and so the answer depends. We will assume ${a,b >0}$.
In this case we can actually use the hyperbolic trig substitution ${x=\sqrt{\frac{a}{b}}\tanh(t)}$, leading to
$${\left(\frac{a}{b}\right)^{\frac{3}{2}}\int\frac{\tanh^2(t)\text{sech}^2(t)}{(a-a\tanh^2(t))^2}}dt=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Eigenvectors of $\begin{pmatrix}2&-2\\ -4&-2\end{pmatrix}$ - What am I doing wrong? Eigenvalues are $2\sqrt{3}$ and $-2\sqrt{3}$, I'll calculate the eigenvector for $2\sqrt{3}$ here
We've got:
$\begin{pmatrix}2-2\sqrt{3}&-2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pma... | It is not clear that your answer disagrees with the source you cite.
\begin{align}
& \left[ \begin{array}{c} 1 \\ 1- \sqrt 3 \end{array} \right] = \frac 1 {1+\sqrt3} \left[ \begin{array}{c} (1+\sqrt3\,)\cdot1 \\[4pt] (1+\sqrt3\,) (1- \sqrt 3\,) \end{array} \right] \\[15pt]
= {} & \frac 1 {1+\sqrt3} \left[\begin{array}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove: $(\forall m, n\in\Bbb N_{>0})(\exists x\in\Bbb R)$ s. t. $2\sin n x \cos m x \ge 1$
Problem 1:
Prove that for any $m,n\in\Bbb N_{>0}$, there exists $x \in\Bbb R$ such that
$2\sin n x \cos m x \ge 1$.
Four months ago, someone asked the above question. However, when I wanted to post my answer,
the question was d... | Remarks: Here is my proof for Problem 2. I wrote it for Problem 1 originally, though Problem 1 might be easier. By the way, @mathlove gave a nice proof.
Problem 2: Prove that for any $m, n \in \mathbb{N}_{>0}$, there exists rational $r$ such that
$2\sin (n r \pi) \cos (m r\pi) \ge 1$.
Proof: Let $x = r\pi$. We split in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Solving $2\sin(x)= \cot(5x)$ without finding the intersections of the graphs of the two sides I have solved the equation $$2\sin(x)= \cot(5x)$$ by finding the intersections of the graphs of the two sides.
Is there a method to solve this equation without using graphs?
| We have that $\cos 5x-2\sin x\sin 5x=0,$ or linearising the second term, $$\cos 4x-\cos 5x-\cos 6x=0.$$ Now using the identity $$\cos z=\frac{e^{iz}+e^{-iz}}{2},$$ and with $y=e^{ix}$ we obtain $$y^4+1/y^4+y^5+1/y^5+y^6+1/y^6=0.$$ There are many ways to deal with this. You may explore the substitution $$u=y+1/y.$$ Howe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3719389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding the equation of a hyperbola, knowing the equations of its asymptotes and a point on it Find the equation of the hyperbola, given that transverse axis parallel to the $x$-axis, equations of asymptotes are $4x + y - 7 = 0$ and $3x - y - 5 = 0$ and the hyperbola passes through point $(4,4)$.
How could I solve this... | $(4x+y-7)(3x-y-5)=k,$ and it passes through $(4,4)$: $(4\cdot 4+4-7)(3\cdot 4-4-5)=k$ or $k=39,$ so your equation is $$(4x+y-7)(3x-y-5)=39.$$
Edit Expanding to the form $12x^2-xy-y^2-41x+2y-4=0,$ it is a hyperbola since $B^2-4AC=(-1)^2-4\cdot 12\cdot (-1)=37>0.$
To get the canonical form it is convenient to find the ce... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3720128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_0^1 \ln{\left(\Gamma(x)\right)}\cos^2{(\pi x)} \; {\mathrm{d}x}$ I have stumbled across the following integral and have struck a dead end...
$$\int_0^1 \ln{\left(\Gamma(x)\right)}\cos^2{(\pi x)} \; {\mathrm{d}x}$$
Where $\Gamma(x)$ is the Gamma function.
I tried expressing $\Gamma(x)$ as $(x-1)!$ then us... | The key to evaluating this integral is to utilize Euler's reflection formula, which the proof can be looked up elsewhere, by substituting $u=1-x$ so that the Gamma function "disappears":
$$I=\int_0^1 \ln{\left(\Gamma(1-u)\right)}\cos^2{(\pi u)} \; \mathrm{d}u$$
Now, add the original integral:
\begin{align*}
2I&=\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Find the sum of the series with terms given by ${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$ The given series has general term as $${T}_{r}=\frac{r}{(r+1)(r+3)(r+4)}$$
I have tried to approach this problem by making a telescopic series as follows, but I end up cancelling $r$ in the numerator, $$\frac{1}{(r+1)(r+3)}-\frac{1}{(r+3... | The partial fraction expansion of your summand is
$$-\frac{1}{6}\frac{1}{r+1} + \frac{3}{2}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
Then notice that $3/2 = 4/3+1/6$ so you have
$$-\frac{1}{6}\frac{1}{r+1} + \frac{1}{6}\frac{1}{r+3}+\frac{4}{3}\frac{1}{r+3} - \frac{4}{3}\frac{1}{r+4}.$$
And now things telescope like ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3722271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Find integers $1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$ Root numbers Problem (Math Quiz Facebook):
Consider the following equation:
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
Where $a,\,b,\,c,\,d$ are integers. Find $a+b+c+d$
I've tried it like this:
Let $w=\sqrt6,\, x=\sqrt3, \... | Expand out enough to get to
\begin{align*}
(a^2-24a+476-b)+\sqrt{2}(336-16a)+\sqrt{3}(272-12a)+\sqrt{6}(192-8a)&=\sqrt{c+\sqrt{d}}.
\end{align*}
This means, when we square the left side, we need to only have two terms with nonzero coefficient. Note that
$$(w+x\sqrt2+y\sqrt3+z\sqrt6)^2=(w^2+2x^2+3y^2+6z^2)+2\sqrt2(wx+3y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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When The Sum Of Squares Of Two Consecutive Integers Is Again A Perfect Square? Find all positive integers $n < 200$, such that $n^2 + (n + 1)^2$ is a perfect square.
Well setting this equal to $k^2$ is important. But before that, since all squares $\equiv 0$ or $1$ (mod $3$ $,4)$ so using this we get that one of the tw... | If we let $m=n+\sqrt{2n^2+(-1)^n}$, starting with $1$, we get a pair of pell numbers that feed directly into Euclid's formula for generating Pythagorean triples which are the ordered pairs (A,B,C) where $A^2+B^2=C^2$.
$$F(m,n):\qquad A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$
Examples:
$$n=1\implies m=1+\sqrt{2+(-1)^1}=1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3725577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Each positive integer is painted with some color. we know that $a+b$ and $ab$ have the same colour if $a>1, b>1$. Each positive integer is painted with some color. It is known that for all pairs $a, b$ of integers greater than $1$ the numbers $a+b$ and $ab$ have the same color. Prove that all numbers greater than $4$ a... | Are you familiar with the concept of induction?
Bear with me. Suppose $N=2K$ is a somewhat large even number. Then $N$ is the same color as $2+K$ and $2+K < N=2K$ if $K > 2$ or in other words if $N=2K> 4$
If $N=2K + 1$ is a somewhat large odd number then $N=(2K-2)+3$ so $N$ is the same color as $3(2K-2)=6(K-1)$. so $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$ Finding this minimum must be only done using ineaqualities.
$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$
Using inequalities of arithemtic and geometric means:
$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}... | Very similar to what you have done:
$$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}}{5}\geq
\sqrt[5]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[5]{\frac{1}{108}}$$
This gives us
$$x^3+\frac{1}{x^2}\geq \sqrt[5]{\frac{1}{108}}$$
Desmos screenshot:
P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving polynomials using De Moivre's theorem Given $\cos 4\theta=8\cos^4\theta−8\cos^2\theta+1$, solve $16x^4-16x^2+1=0$.
The textbook's answers are $x=\pm\cos\dfrac{\pi}{12},\pm\cos\dfrac{5\pi}{12}$. I managed to get two of the four answers and i can not figure out what i did wrong.
My Attempt
Let $x=\cos\theta$
$16\... | $$\cos4\theta=\cos\dfrac\pi3$$
$$\implies4\theta=2n\pi\pm\dfrac\pi3=\dfrac\pi3(6n\pm1)$$ where $n$ is any integer
$$\implies\theta=\dfrac\pi{12}(6n+1)$$ where $n=-2,-1,0,1$
$\cos\dfrac\pi{12}(6\cdot-2+1)=\cos\left(\pi-\dfrac\pi{12}\right)=-\cos\dfrac\pi{12}$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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For how many natural numbers(<=100) is $1111^n +2222^n+3333^n+4444^n$ divisible by 10? For how many natural numbers (0 not included) $n \leq 100$ is $1111^n +2222^n+3333^n+4444^n$ divisible by 10?
I factored out $1111^n$ and got $1111^n(1+2^n+3^n+4^n)$. So $1+2^n+3^n+4^n$ must be divisible by 10. I figured out that thi... | $$1^n+2^n+3^n+4^n\equiv0\pmod2\text{ for }n\ge0$$
As $2^3\equiv3,2^2\equiv4\pmod5,$
$$1^n+2^n+3^n+4^n$$
$$\equiv1+2^n+(2^3)^n+(2^2)^n$$
$$\equiv\begin{cases}\dfrac{2^{4n}-1}{2^n-1}\equiv 0 &\mbox{if } 2^n-1\not\equiv0\pmod5\iff4\nmid n \\
4 & \mbox{if } 4\mid n \end{cases}\pmod5$$
$$\implies1^n+2^n+3^n+4^n\equiv0\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$ Evaluate $$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} \mathop{dx}$$
I tried $u=\sqrt{2x^4-2x^2+1}$, $u=\dfrac{1}{x}$ and $u=\sqrt{x}$ but none of these worked.
A friend gave this and said it's from IIT JEE and answer is $$\dfrac{\sqrt{2x^4-2x^2+1}}{2x^2}+... | Substitute $t=\frac1{x^2}$ to get
$$\int \frac{x^2-1}{x^3 \sqrt{2x^4-2x^2+1}} {dx}
=\frac12 \int \frac{(t-1)dt}{\sqrt{2-2t+t^2}}
=\frac12 \sqrt{2-2t+t^2}+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731783",
"timestamp": "2023-03-29T00:00:00",
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$f(xy + x +y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$.
Let $f : \mathbb R \to \mathbb R$ that satisfies both 2 conditions ,
$f(xy + x + y) = f(xy) + f(x) + f(y)$ and $f(x)(y - x) + f(y)(x - y) \geq 0$ $\forall x,y \in \mathbb R$.
Determine all such $f$.
My solution.
Let $P(x,y) $ be $f(x)(y - x)... | You can show that solutions to the functional equation
$$ f ( x y + x + y ) = f ( x y ) + f ( x ) + f ( y ) \tag 0 \label 0 $$
are exactly the additive functions, i.e. those satisfying
$$ f ( x + y ) = f ( x ) + f ( y ) \text . \tag 1 \label 1 $$
It's easy to see that additive functions satisfy \eqref{0}. I try to show... | {
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"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$ Evaluate $$\int_0^{\frac{\pi}{4}} \left( \frac{\sin^2{(5x)}}{\sin^2{x}} -\frac{\cos^2{(5x)}}{\cos^2{x}} \right)\mathop{dx}$$
I tried substitutions like $u=\frac{\pi}{4}-x$, and trig identities like... | Making the problem more general for the antiderivative
$$I_{n,m}=\int\left( \frac{\sin^m{((2n+1)x)}}{\sin^m{(x)}} -\frac{\cos^m{((2+1)x)}}{\cos^m{(x)}} \right)\mathop{dx}$$ using Chebyshev formulae, just as @integrand answered,
$$\frac{\sin{((2n+1)x)}}{\sin{(x)}}$$ is a polynomial of degree $n$ in $\sin^2(x)$ that is ... | {
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"source": "stackexchange",
"question_score": "2",
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The remainder when $1^1+3^3+5^5+\dots + 1023^{1023} \pmod{1024}$ is $\dots$ The remainder when $1^1+3^3+5^5+\dots + 1023^{1023} \pmod{1024}$ is $\dots$
I've tried for the summation like this.
$\sum_{i=1}^{n} (2i-1)^{2i-1} \equiv 0 \pmod{n}$ when $n$ is even. But, I didn't get the answer when $n$ is odd and didn't yet c... | We will prove that for $n\geq 2$ we have
$$
S_n:=\sum_{k=1}^{2^{n-1}}(2k-1)^{2k-1}\equiv 0\pmod{2^n}.
$$
We will induct on $n$ (the case $n=2$ is clear). Consider $S_{n+1}$:
$$
S_{n+1}=\sum_{k=1}^{2^{n}}(2k-1)^{2k-1}=\sum_{k=1}^{2^{n-1}}\left((2k-1)^{2k-1}+(2k-1+2^n)^{2k-1+2^n}\right)\equiv 0\pmod{2^{n+1}}.
$$
Due to E... | {
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"source": "stackexchange",
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Is there a quick (hopefully elementary) way to prove that $6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0$ has only one solution? I have the Diophantine equation $$6b^2c^2 + 3c^2 - 36bc - 4b^4 - 4b^2 + 53=0.$$ Numerical calculations suggest this has only one positive integer solution, namely $(b,c)=(2,3)$. Is there a quick... | Hint
$$6b^2c^2+3c^2-36bc-4b^4-4b^2+53=(6b^2c^2-36bc+54)-(4b^4+4b^2+1)+3c^2=6(bc-3)^2-(2b^2+1)^2+3c^2=0\\\implies 6(bc-3)^2+3c^2=(2b^2+1)^2$$ Then $3\mid 2b^2+1$. Now proceed like this by checking modulo primes like $2,3$ and so on.
| {
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$\int \frac{x^3+3x+2}{(x^2+1)^2 (x+1)} \ dx$ without using partial fraction decomposition One way to evaluate the integral $$\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \ dx $$ is to rewrite it as $$ \int \frac{x^3+x+2x+2}{(x^2+1)^2(x+1)} dx \\=\int\frac{x(x^2+1) +2(x+1)}{(x^2+1)^2(x+1)}dx\\=\int\frac{x}{(x^2+1)(x+1)}dx+\int\... | Let
$$\frac{x^2+1}{(x+1)^2}=t\implies x=\frac{\sqrt{2 t-1}-t}{t-1}\implies dx=\frac{dt}{1-t \left(\sqrt{2 t-1}+2\right)}$$ to make
$$I=\int \frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx=\int\left(-\frac{1}{2 t^2}+\frac{1}{4 t}-\frac{3}{4 t \sqrt{2 t-1}}\right)\,dt$$ which does not seem too bad.
$$I=\frac{1}{2 t}+\frac 14 \log(... | {
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Simplify $\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$
Simplify:
$$\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}$$
After the substitution as $\cos(x)=a$ and $\s... | \begin{align}\frac{4\cos ^2\left(2x\right)-4\cos ^2\left(x\right)+3\sin ^2\left(x\right)}{4\sin ^2\left(x\right)-\sin ^2\left(2x\right)}&=\frac{4(1-2\sin^2x)^2-4(1-\sin^2x)+3\sin^2x}{4\sin^2x-(2\sin x\cos x)^2}\\&=\frac{4-16\sin^2x+16\sin^4x-4+4\sin^2x+3\sin^2x}{4\sin^4x}\\&=\frac{16\sin^4x-9\sin^2x}{4\sin^4x}\\&=4-\fr... | {
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prove $\sum\cos^3{A}+64\prod\cos^3{A}\ge\frac{1}{2}$ In every acute-angled triangle $ABC$,show that
$$(\cos{A})^3+(\cos{B})^3+(\cos{C})^3+64(\cos{A})^3(\cos{B})^3(\cos{C})^3\ge\dfrac{1}{2}$$
I want use Schur inequality
$$x^3+y^3+z^3+3xyz\ge xy(y+z)+yz(y+z)+zx(z+x)$$
then we have
$$x^3+y^3+z^3+6xyz\ge (x+y+z)(xy+yz+zx)$... | Let $\cos\alpha=\frac{x}{2},$ $\cos\beta=\frac{y}{2}$ and $\cos\gamma=\frac{z}{2}.$
Thus, $x$, $y$ and $z$ are positives, $x^2+y^2+z^2+xyz=4$ and we need to prove that:
$$x^3+y^3+z^3+x^3y^3z^3\geq4.$$
Indeed, let $$f(x,y,z,\lambda)=x^3+y^3+z^3+x^3y^3z^3-4+\lambda(x^2+y^2+z^2+xyz-4).$$
Thus, in the minimum point we need... | {
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Point is chosen inside the square, find angle. Inside the square $MNPK$ point $O$ in such a way that $MO:ON:OP =1:2:3$. Find $\angle MON$.
I tried to solve this problem with vectors, but I couldn't find coordinates of point $O$. Can someone explain this to me?
|
Apply the Pythagorean theorem to establish the system of three equations
\begin{align}
x^2+y^2=4,\>\>\>\>\> x^2+(a-y)^2=1,\>\>\>\>\> (a-x)^2+y^2=9
\end{align}
The 2nd and 3rd equations lead to $y=\frac{a^2+3}{2a}$ and $x=\frac{a^2-5}{2a}$, respectively. Plug them into the 1st equation to get $a^4 - 10a^2 +17=0$, whic... | {
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Computation of Definite Integral of Rational Function I am dealing with a definite integral of a rational function that seems quite hard to get a nice closed form/explicit expression for. Let $ -1 < z < 1 $, then my aim is to determine an expression for the integral $ I $ in terms of $ z $:
$$ I(z) = \int_{0}^{\infty} ... | Consider the integrand $$\frac{(1+z)t^4 + (1-z)}{(1+z)^2 t^6 + 3(1+z)(5+z)t^4 + 3(1-z)(5-z)t^2 + (1-z)^2} $$ and rewrite it as
$$\frac 1 {z+1} \frac{t^4+a}{(t^2-b)(t^2-c)(t^2-d)}$$ where $a=\frac {1-z}{1+z}$ and $(b,c,d)$ are the roots of the cubic equation in $t^2$. Now, using partial fraction decomposition
$$\frac{t^... | {
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Is there a way to find value for this integral? I am trying to solve this integral by hand and here are my steps for doing that.
The integral is:
$$A=\int_0^{2 \pi} \frac{\cos(\theta)}{\sqrt{1 - a \cos(\theta)}} d\theta$$
Using the trigonometry trick: $\cos(\theta) = 1- 2 \sin\left(\frac{\theta}{2}\right)^{2}$ and the... | Let us assume $|a|< 1$ and exploit
$$ \frac{1}{\sqrt{1-z}}=\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}z^n \qquad \text{for }|z|<1. $$
This gives
$$ I(a)=\int_{0}^{2\pi}\frac{\cos\theta}{\sqrt{1-a\cos\theta}}\,d\theta=\sum_{n\geq 0}\frac{a^n}{4^n}\binom{2n}{n}\int_{0}^{2\pi}\left(\cos\theta\right)^{n+1}\,d\theta $$
where t... | {
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"timestamp": "2023-03-29T00:00:00",
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Closed form sought for $a_1 = a_2 = 1, a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i $ where $n>2$ I've been working through a problem that I've got as far as getting a recursive answer to. I was hoping to turn this into more of a "closed form" answer, but haven't really gotten anywhere. I'm hoping that someone can help w... | After @Gary's answer, using the generating function
$$G(x) = \frac{{1 - e^{ - 2x} }}{{2(x-1)^2}}= \sum_{n = 1}^\infty {a_n x^n }$$ define
$b_n=n! \,a_n$ which gives the sequence
$$\{1,2,10,48,296,2080,16752,151424,1519744,16766208\}$$ which is $A037256$ in $OEIS$ (have a look here).
It does not seem to present any p... | {
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"timestamp": "2023-03-29T00:00:00",
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Find a general method to find particular solutions where the sum of the squares of two consecutive integers is equal to the square of another integer Question: If the sum of the squares of two consecutive integers is equal to the square of another integer, then find a general method to find particular solutions. E.g... | For $A^2+B^2=C^2:\space B=A\pm1\quad$ the $C$-values are the odd Pell numbers $(5,29,169,985\cdots)$ as shown in A000129
and the triples $T_n$ that contain them can be generate sequentially by the formula below with a seed of $\quad T_0=(0,0,1)$
$$A_{n+1}=3A_n+2C_n+1\qquad B_{n+1}=3A_n+2C_n+2\qquad C_{n+1}=4A_n+3C_n+2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find function $f(x)$ whose expansion is $\sum_{k=0}^{+\infty}k^2x^k$. I know the expansion of $\frac{1}{1-x}$ is $$1+x+x^2+\cdots+x^k+\cdots$$
So taking the derivative of $$\frac{\partial}{\partial{x}} \frac{1}{1-x}=\frac{1}{(1-x)^2}$$
And subsequently the expansion is (after taking derivative): $$1+2x+3x^2+4x^3+...+ k... |
Simplify $\displaystyle\sum_{k=0}^{+\infty}k^2x^{k}$.
Differentiate and multiply the below expression by $x$ twice
\begin{align*}
\sum_{k=0}^{+\infty}x^k&=\frac{1}{1-x}\\
\sum_{k=0}^{+\infty}kx^{k-1}&=\frac{1}{(1-x)^2}\\
\sum_{k=0}^{+\infty}kx^{k}&=\frac{x}{(1-x)^2}=-\frac{1}{(1-x)}+\frac{1}{(1-x)^2}\\
\sum_{k=0}^{+\... | {
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"source": "stackexchange",
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Evaluate: $\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx.$
Prove that:$$\int_0^{\infty}\frac{\ln x}{x^2+bx+c^2}\,dx=\frac{2\ln c}{\sqrt{4c^2-b^2}}\cot^{-1}\left(\frac{b}{\sqrt{4c^2-b^2}}\right),$$ where $4c^2-b^2>0, c>0.$
We have:
\begin{align}
4(x^2+bx+c^2)&=(2x+b)^2-\left(\underbrace{\sqrt{4c^2-b^2}}_{=k\text{ (say)}... | Hint:
$$ x^2 + bx+ c^2 = (x+b/2)^2 + (c^2 - b^2/4) $$
with $c^2 - b^2/4 > 0$.
This invites change of variables using $\arctan$ as
$$d\arctan x = \frac{1}{x^2+1}dx $$
| {
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Does $k=1$ follow from $I(5^k)+I(m^2) \leq \frac{43}{15}$, if $p^k m^2$ is an odd perfect number with special prime $p=5$? The topic of odd perfect numbers likely needs no introduction.
Denote the sum of divisors of the positive integer $x$ by $\sigma(x)$, and denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
Eu... | On OP's request, I am converting my comment into an answer.
I'm not sure if I understand your follow-up question well, but we have
$$\begin{align}\frac{\mathrm d}{\mathrm dk}\bigg(I(5^k)+\frac{2}{I(5^k)}\bigg)&=-\frac{ 7\cdot 5^{2 k} + 2\cdot 5^{k + 1} - 1}{4\cdot 5^k (5^{k + 1} - 1)^2}\log 5\lt 0
\\\\\lim_{k\to\infty}... | {
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Use residues to find $\int_0^\pi \frac{d\theta}{5+3\cos\theta}$ I am trying to use residues to find $\int_0^\pi \frac{d\theta}{5+3\cos\theta}$.
My thoughts:
Letting $z=e^{i\theta}$ we get $dz=ie^{i\theta}$. Then, $\int_0^\pi \frac{d\theta}{5+3\cos\theta}=\frac{1}{5}\int_{|z|=1}\frac{dz}{iz(1+\frac{3}{5}(\frac{z+z^{-1}... | $$I = \int_0^\pi \frac{d\theta}{5+3\cos\theta}$$
I want the range to be $0$ to $2\pi$ so I will apply the substitution $\tau = \theta / 2$.
$$I = \int_0^{2 \pi} \frac{d\tau}{10+6\cos\tau}$$
Now I can apply my Residue Theorem lemma (from Freitag):
In our case $$f(z) = \frac{1}{i z}\frac{1}{10 + \tfrac{1}{2}6(z + \frac{... | {
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Estimating $\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$. I'm trying to solve this:
Which of the following is the closest to the value of this integral?
$$\int_{0}^{1}\sqrt {1 + \frac{1}{3x}} \ dx$$
(A) 1
(B) 1.2
(C) 1.6
(D) 2
(E) The integral doesn't converge.
I've found a lower bound by manually calculating $\int_{0}... | Let's try to use integration by parts to $I = \int\limits_0^1 \sqrt{1 + \frac{1}{3x}}dx$. First, transform integral into $\frac{2}{\sqrt3}\int\limits_0^1\frac{\sqrt{1 + 3x}}{2\sqrt{x}}$. Now $u = \sqrt{3x+1}$ and $dv = \frac{dx}{2\sqrt{x}}$ and what we get after IBP is $$\frac{2}{\sqrt3}\sqrt{x(3x+1)}|_0^1 - \sqrt{3}\i... | {
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What loops are possible when doing this function to the rationals? What loops are possible when doing this function to the rationals?
Let's define this function on a simplified fraction $\frac{a}{b}$.
$$f\left(\frac{a}{b}\right)=\frac{a+b}{b+1}$$
I started this with $f(\frac{2}{3})=\frac{5}{4}$ then i did the function ... | Here is a modification of @Steven Stadnicki's proof. The novel contribution of this answer is justifying the reduction step in Steven's solution through the use of a suitable partial order on the set of lattice points.
Step 1. Settings and Useful Observations
Let $\mathbb{N}_1 = \{1, 2, 3, \dots\}$ denote the set of po... | {
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How does $e^x\cdot e^X$ equal $e^{x+X}$? I know that they equal each other, but when I'm trying to prove it, something doesn't match. Please mind the difference between the two equations, one is a lowercase $x$ and the other is an uppercase $x.$ I know that the formula to get $e^x$ is $\frac{x^n}{n!}$. So I apply on $e... | Lets look at a couple functions. $Y=e^X$ and $y=e^x$.
Then $\ln Y=X$ and $\ln y = x$. $$X+x=\ln (Y)+\ln (y) = \ln (Yy) \Longrightarrow e^{X+x}=Yy=e^Xe^x$$
Ok, so there is a little informal proof using some properties of the natural log.
$$e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$$ (Note: I am not using z as a complex ... | {
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What is the residue of $z^2 \cos(\frac{1}{z})$ Can I find it using Laurent series expansion? I tried using the Taylor series expansion of $\cos(1/z)$ but could not reach any conclusion.
| The residue is the coefficient of the term $\frac{1}{z}$ in the Laurent Series.
The Maclaurin Series of $\cos z$ is
$$\cos z = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots $$
It follows that
$$\cos\left(\frac{1}{z}\right) = 1 - \frac{1}{2!}\frac{1}{z^2} + \frac{1}{4!}\frac{1}{z^4} - \frac{1}{6!}\frac{1... | {
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Other way to evaluate $\int \frac{1}{\cos 2x+3}\ dx$? I am evaluating
$$\int \frac{1}{\cos 2x+3} dx \quad (1)$$
Using Weierstrass substitution:
$$ (1)=\int \frac{1}{\frac{1-v^2}{1+v^2}+3}\cdot \frac{2}{1+v^2}dv =\int \frac{1}{v^2+2}dv \quad (2) $$
And then $\:v=\sqrt{2}w$
$$ (2) = \int \frac{1}{\left(\sqrt{2}w\right)... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
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Is it possible to show that the fifth roots of 1 add up to 0 simply by using trigonometric identities? You can't use geometric sums, minimal polynomials, pentagon, and exact values with radicals.
All the five, fifth-roots of unity are :$1,\left(\cos \left(\frac{2 \pi}{5}\right)+i \sin \left( \frac{2\pi}{5}\right)\righ... | If you want a simple trigonometric proof, recall
$$\sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2.$$
Therefore
\begin{align}
0&=\sin(a+2\pi)-\sin a\\
&=(\sin(a+2\pi)-\sin(a+8/\pi/5))+(\sin(a+8/\pi/5)-\sin(a+6/\pi/5))\\
&+(\sin(a+6/\pi/5)-\sin(a+4/\pi/5))+(\sin(a+4/\pi/5)-\sin(a+2/\pi/5))\\
&+(\sin(a+2/\pi/5)-\sin a)\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Convergence of $\sum \limits_{n=1}^{\infty}\sqrt{n^3+1}-\sqrt{n^3-1}$ Hello I am a high school student from germany and I am starting to study math this october. I am trying to prepare myself for the analysis class which I will attend so I got some analysis problems from my older cousin who also studied maths. But I am... | Note that$$\frac{1}{n^2} \cdot \frac{2\sqrt{n}}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}=\frac{1}{n^{3/2}} \cdot\underbrace{ \frac{2}{\sqrt{1+\frac{1}{n^3}}+\sqrt{1-\frac{1}{n^3}}}}_{<\frac{2}{1+0}=2}<2n^{-3/2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Use linearisation of a certain function to approximate $\sqrt[3]{30}$ Background
Find the linearisation of the function
$$f(x)=\sqrt[3]{{{x^2}}}$$
at
$$a = 27.$$
Then, use the linearisation to find
$$\sqrt[3]{30}$$
My work so far
Applying the formula
$${f\left( x \right) \approx L\left( x \right) }={ f\left( a \right) ... | The second part of your analysis is correct. The function that you want to approximate is $f(x)=\sqrt[3]{{x}}$ at $a=27$, not $g(x)=\sqrt[3]{{x^2}}$. So
$${f'(x) = {\left( {\sqrt[\large 3\normalsize]{x}} \right)^\prime } } = {{\left( {{x^{\large\frac{1}{3}\normalsize}}} \right)^\prime } } = {\frac{1}{3}{x^{ – \large\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$ Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$
My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$
$T = {\sin ^2}{1^\circ}{\sin ^2}{3^... | As $\cos(90^\circ-A)=\sin A,$
$$\prod_{k=0}^{44}\sin(2k+1)^\circ=\prod_{k=0}^{44}\cos(2k+1)^\circ$$
As $\cos180(2k+1)^\circ=-1$
as $\cos180x=2^{180-1}c^{180}x+\cdots+(-1)^{90}$
and for $\cos180x=\cos180^\circ,180x=360^\circ n\pm180^\circ$
$x=(2 n+1)^\circ$ where $0\le n\le179$
If $\cos180x=-1,$ the roots of $$2^{180-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate $ \:\sum _{n=3}^{\infty \:}\frac{4n^2-1}{n!}\:\: $? I am trying to evaluate:
$$ \:\sum _{n=3}^{\infty \:}\:\:\frac{4n^2-1}{n!}\:\: \quad (1)$$
My attempt:
$$ \:\sum _{n=3}^{\infty \:}\:\:\frac{4n^2-1}{n!}\:\: \quad = 4\sum _{n=0}^{\infty \:} \frac{(n+3)^2}{(n+3)!} + \sum _{n=0}^{\infty \:} \frac{1}{(n... | Note that $$e= \sum _{n=0}^{\infty}\frac{1}{n!}$$
\begin{split}
\sum _{n=3}^{\infty }\frac{4n^2-1}{n!} &= 4\sum _{n=3}^{\infty }\frac{n^2}{n!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\
&=4\sum _{n=3}^{\infty }\frac{n-1+1}{(n-1)!} -\sum _{n=0}^{\infty}\frac{1}{n!}+\frac{5}{2}\\
&= 4\sum _{n=3}^{\infty }\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How do I evaluate $\sum_{m,n\geq 1}\frac{1}{m^2n+n^2m+2mn}$ I saw a problem here which state to evalute $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$
My attempt
Let $$f(m,n)=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{1}{m^2n+n^2m+2mn}$$ and interchanging $m,n$ as $n,m$ we have $$f'(n,m) = \sum_{... | Using integration trick
Note that $$\dfrac{1}{nm^2+n^2m+2mn} =\dfrac{1}{mn(m+n+2)}=\dfrac{1}{mn}\int_0^1 x^{m+n+1}\,dx$$ then we have $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{1}{mn} \int_0^1 x^{m+n+1}\,dx=\int_0^1x\left(\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \dfrac{x^{m}\cdot x^n}{mn}\right)\,dx \\= \int_0^1 x\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3772397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Why is the convergence point of $ \sum _{n=1}^{\infty }\frac{1}{2^n}-\frac{1}{2^{n+1}} $ negative? I am trying to evaluate $\frac1{2^1} - \frac1{2^2} + \frac1{2^3} - \frac1{2^4} + \cdots$
I re-wrote the sum using sigma notation as:
$$ \sum _{n=1}^{\infty } \left( \frac{1}{2^n}-\frac{1}{2^{n+1}} \right) \quad (1) $$
Hen... | In the first step
$$(1) = \sum _{n=1}^{\infty } \frac{2^{n+1} - 2^n }{2^n2^{n+1}} = \sum _{n=1}^{\infty } \frac{2^n(2-1) }{2^n2^{n+1}} = \color{red}+\sum _{n=1}^{\infty } \frac1{2^{n+1}} = \ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Prove $\cos^2(\theta)+\sin^2(\theta) = 1$ $$\cos^2(\theta) + \sin^2(\theta) = 1$$
I solved this by using right triangle,
$$\sin(\theta) = \frac{a}{c}, \quad \cos(\theta) = \frac{b}{c}$$
$$\cos^2(\theta) + \sin^2(\theta) = 1$$
$$\Bigl(\frac{b}{c}\Bigr)^2 + \Bigl(\frac{a}{c}\Bigr)^2 = 1 $$
$$\frac {a^2 + b^2} {c^2} = 1 $... | You solved this by using some right triangle but there isn't a right triangle in your solution? Remember, things like that have to be explicitly written in your proof.
Also, the proof is not very valid because you started off by asserting that:
$$\sin^2(\theta)+\cos^2(\theta) = 1$$
when really, what you should have sta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3773925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 5
} |
Find the remainder when $\sum_{n=1}^{2015}{n^2\times2^n}$is divided by 23.
Find the remainder when $\sum_{n=1}^{2015}{n^2\times2^n}$is divided by 23.
I am completely stuck at this to even start , here's the samll thing that I have noticed .
When $2^{11}$ is divided by 23 , the remainder is $1$ , so $2^{11k+r}$ is equ... | So what you said and the fact that $n^2 \equiv(n \mod 23)^2 \pmod{23}$ and $12 \equiv -11 \pmod{23}$ and $13 \equiv -10 \pmod {23}$ and so on makes a periodic sum: (let the sum be $S$)
$$S \equiv 1^2\cdot2^1+\dots+11^2\cdot2^{11}$$
$$+(-11)^2\cdot2^{1}+(-10)^2\cdot2^2+\dots+(-1)^2\cdot2^{11}+0+\dots$$
and of course th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
coordinates of focus of parabola Find the coordinates of focus of parabola $$\left(y-x\right)^{2}=16\left(x+y\right)$$
rewriting:
$(\frac{x-y}{\sqrt{2}})^2=8\sqrt2(\frac{x+y}{\sqrt{2}})$
comparing with $Y^2=4aX$
$4a=8\sqrt2,a=2\sqrt2 $
$\Rightarrow$ coordinates of focus=2,2
Is this the correct approach?
| Rewrite $$(x-y)^2=16(x+y)$$ as
$$(\frac{x-y}{\sqrt{2}})^2=8\sqrt{2}\frac{(x+y)}{\sqrt{2}}\implies Y^2=4AX,~~~ A=2\sqrt{2}$$
The co-0rdinates of focus are given by $(X=A,Y=0) \implies x+y=4,x=y \implies x=2, y=2$
So the focus is at $(2,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3775650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
integral arising in statistical mechanics The following integral arises in statistical mechanics of lattice models:
$
\displaystyle I = \int_{0}^{\pi/2}
\ln\left(\, 1 +
\sqrt{\, 1 - a^{2}\sin^{2}\left(\phi\right)\,}\,\right)\,
\mathrm{d} \phi\quad$ with $\quad a \leq 1
$.
By trial and error, the integral was shown to... | Similar to @Maxim's comment
$$I(a) = \int_{0}^{\pi/2}\log\left(\, 1 +\sqrt{\, 1 - a^{2}\sin^{2}\left(\phi\right)\,}\,\right)\,d\phi$$
$$I'(a)=- \int_{0}^{\pi/2} \frac{a \sin ^2(\phi)}{\sqrt{1-a^2 \sin ^2(\phi)} \left(1+\sqrt{1-a^2 \sin^2(\phi)}\right)}\,d\phi=\frac{\pi -2 K\left(a^2\right)}{2 a}$$
$$\int \frac{K\left(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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.