Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Express $-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$ as a product of linear factors.
Express
$$-a^3 b + a^3 c+ab^3-a c^3-b^3c+bc^3$$
as a product of linear factors.
I have tried rewriting the expression as:
$$ab^3-a^3b + a^3c-ac^3 +bc^3-b^3c$$
$$= ab(b^2-a^2)+ac(a^2-c^2)+bc(c^2-b^2)$$
$$= ab(b-a)(b+a) + ac(a-c)(a+c)+bc(... | the resultat should be $$- \left( b-c \right) \left( a-c \right) \left( a-b \right) \left( a
+c+b \right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Finding $\sin^3\frac{x}{2}\cdot\cos^7\frac{x}{3}$ periodicity I need to find the period of $\sin^3\frac{x}{2}\cdot\cos^7\frac{x}{3}$ but couldn't find a nice method how? any ideas?
| Notice that:
$$\sin\left(\frac{x}{2}\right) = \sin\left(\frac{x}{2} + 2k\pi\right) = \sin\left(\frac{x+4k\pi}{2}\right)$$
and
$$\cos\left(\frac{x}{3}\right) = \cos\left(\frac{x}{3} + 2h\pi\right) = \cos\left(\frac{x+6h\pi}{3}\right)$$
We need to find the minimum common period between $4k\pi$ and $6h\pi$ which is $12\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1521127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the center of circle touching a line pair and a point. A circle passes through the point 3,$\sqrt{\frac{7}{2}}$ and touches the line pair $x^2-y^2-2x+1=0$. The co-ordinates of the centre of the circle are:-
My attempt:-
Using quadratic formula to separate the line pair into two lines.
$$x^2-y^2-2x+1=0$$
$$x=\f... | So your two lines are $y=x-1$ and $y=-x+1$.
By symmetry we know the center of the circle must be on $x$-axis. Let the center be $(a,0)$ then the radius is $\large{a-1\over\sqrt{2}}$ because the angle between the two lines is 90 degrees and hence the two radius along with the two lines form a square where the $x$-axis i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1523541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $5$ divides $3^{3n+1}+2^{n+1}$
Prove that $5$ divides $3^{3n+1}+2^{n+1}$
I tried to prove the result by induction but I couldn't.
The result is true for $n=1$.
Suppose that the result is true for $n$ i.e $3^{3n+1}+2^{n+1}=5k$ for some $k\in \mathbb{N}$. We study the term
$$3^{3n+4}+2^{n+2}=3^{3n+1}3^3+2^{... | $$3^{3n+4}+2^{n+2}=3^{3n+1}\cdot 3^{3} +2^{n+1}\cdot 2\\
=(5k-2^{n+1})\cdot 3^{3}+2^{n+1}\cdot 2\\
=2^{n+1}(2-27)+5k\cdot 27=-2^{n+1}\cdot 25+5k\cdot 27\\
=5(-2^{n+1}\cdot 5+27k)$$ which is divisible by $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1526409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
$x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}$. Then what is the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\}$?
If the equation $x^3-3x+1=0$ has three real roots $x_{1}\;x_{2}\;,x_{3}\;,$ Where $x_{1}<x_{2}<x_{3}$.
Then the value of $\{x_{1}\}+\{x_{2}\}+\{x_{3}\} = \;,$ Where $\{x\}$ Represent fractional part of ... | $\left\{x_1\right\}+\left\{x_2\right\}+\left\{x_3\right\}=x_1+x_2+x_3-[x_1]-[x_2]-[x_3]=0-(-2+0+1)=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1527100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find polynomial $f(x)$ based on divisibility properties of $f(x)+1$ and $f(x) - 1$ $f(x)$ is a fifth degree polynomial. It is given that $f(x)+1$ is divisible by $(x-1)^3$ and $f(x)-1$ is divisible by $(x+1)^3$. Find $f(x)$.
| We can clearly see that: $f(1)+1=0$ and $f(-1)-1=0$
We can write $f(x)-1=p(x)(x+1)^3$ and $f(x)+1=q(x)(x-1)^3$
By differentiation and double differentiation, you can see that
$f'(1)=0$ and $f''(1)=0$
AND
$f'(-1)=0$ and $f''(-1)=0$
You got six conditions and six unknowns!
[assume $f(x) = x^6+a_1x^5+a_2x^4+a_3x^3+a_4x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1528699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to solve the equation $t = \sqrt{x^2 - 1} - x$ for $x$?
Let the equation $t = \sqrt {x^2 - 1} - x$. Find $x$.
So I've tried the following:
$$t^2 = x^2 - 1 -2\sqrt {x^2-1}x + x^2 = 2x^2 - 2\sqrt{x^2-1}x - 1 $$
What should I do next?
| The following should be better :
$$\sqrt{x^2-1}=t+x\quad\Rightarrow\quad x^2-1=(t+x)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I manipulate $\frac { \sqrt { x+1 } }{ \sqrt { x } +1 } $ to find $M>0$ to prove a limit? Given the following limit, find such an $M>0$ that for every $x>M$, the expression is $\frac { 1 }{ 3 }$ close to the limit. In other words find $M>0$ that for every $x>M:\left| f(x)-L \right| <\frac { 1 }{ 3 }$ for the fo... | First rationalization
i.e. $\left|\frac{\sqrt{x+1}-\sqrt{x}-1}{\sqrt{x}+1}\right|$
= $\left|\frac{(\sqrt{x+1}-\sqrt{x}-1)(\sqrt{x}-1)}{x-1}\right|$
=$\left|\frac{\sqrt{x(x+1)}-\sqrt{x+1}-x+1}{x-1}\right|$
Then, we could choose M to be a particular value to solve the problem(by taking maximum of M at last). In this case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine the minimal polynomial of $T$ with respect to $\begin{pmatrix} 0 &0& 1 \end{pmatrix}^t.$ Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(v)=\begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix}v.$ Determine the minimal polynomial of $T$ with respect to $\begin{pmatrix} 0 \\ 0 \\ 1 \en... | You got the nullspace of $A$ wrong. It is one-dimensional, since the rank of $A$ is $3$. The row-reduced form that you have indicates that the first three columns are linearly independent, and the fourth one is a certain linear combination of them. This yields
$$
\operatorname{null}(A) = \operatorname{span}( (2,-1,-2,1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1529724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of recursive series I am taking a course in algebra and this problem was on my problem set, and I had no idea how to solve it.
Suppose we have a sequence $s_n$ of real numbers such that $5s_{n+1}-s_{n}-3s_{n}s_{n+1}=1$ for $1 \leq n \leq 42$ and $s_1=s_{43}$.
What are the possible values of $s_1+s_2+ \ldots + s_{4... | Note that we have
$$s_{n+1} (5-3s_n) = 1 + s_n,$$
thus in particular $s_n \neq 5/3$ for all $n$ and so
$$\tag{1} s_{n+1} = \frac{1+s_n}{5-3s_n}.$$
The mapping
$$f(z) = \frac{z + 1}{-3z+5}$$
is a Mobius transform. One can check that
$$f^{(n)} (z) = \frac{az+ b}{cz+ d},$$
where
$$\begin{bmatrix} a & b \\ c & d \end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1531796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Calculate a limit $\lim_{x \to \pi/2} \frac{\sqrt[4]{ \sin x} - \sqrt[3]{ \sin x}}{\cos^2x}$ $$\lim_{x \to \pi/2} \frac{\sqrt[4]{ \sin x} - \sqrt[3]{ \sin x}}{\cos^2x}$$
I have an idea of replacing $\sin x$ to $n$ when $n \to 1$ but wolfram says that answer is $\frac{\pi}{48} $ so my suggestion is it's had to use trigo... | Ian Miller's answer is the nicest and most efficient solution to the problem.
Just for your curiosity, I shall give you another one using Taylor series since you will use them a lot during your studies.
First, changing variable $x=\frac \pi 2+y$ $$\frac{\sqrt[4]{ \sin (x)} - \sqrt[3]{ \sin (x)}}{\cos^2(x)}=\frac{\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1534871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Prove $ 5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3) $ if $a^2+b^2+c^2=3$ Let $a,b,c\in\mathbb{R}$ such that $a^2+b^2+c^2=3$. Prove that:
$$
5(a^4+b^4+c^4)+9≥8(a^3+b^3+c^3)
$$
I tried to homogenize the inequality to get:
$$
5(a^4+b^4+c^4)+(a^2+b^2+c^2)^2≥\frac{8}{\sqrt3}(a^3+b^3+c^3)\sqrt{(a^2+b^2+c^2)}
$$
I hoped that one don't ne... | You can prove it as follows:
The inequality is equivalent to:
$$
\sum_{cyc}5a^4-8a^3+3a^2≥0\iff\sum_{cyc}5a^2\left(\left(a-\frac{4}{5}\right)^2-\frac{1}{25}\right)≥0\iff\sum_{cyc}a^2\left(\left(5a-4\right)^2-1\right)≥0\iff\\
\sum_{cyc}a^2(5a-4)^2≥3\iff\sqrt{\frac{1}{3}\sum_{cyc}a^2(5a-4)^2}≥1
$$
Now, by AM-QM:
$$
\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Prove $\liminf(M_n = -1) \subseteq ((\lim \frac{S_n}{n}) = -1)$
Let $M_1, M_2, \dots$ be iid RVs s.t. $P(M_n = n^2 - 1) = \frac{1}{n^2} = 1 - P(M_n = - 1)$. Define $S_n = \sum_{i = 1}^{n} M_i$. Prove that $P\{(\lim \frac{S_n}{n}) = -1\} = 1$.
$$\sum_n P(M_n = n^2 - 1) = \sum_n \frac{1}{n^2} < \infty$$
By BCL1, $$\to ... | Let's denote $\liminf_n[M_n = -1]$ by $A$, by first Borel-Cantelli lemma, $P(A) = 1$. Therefore for each $\omega \in A$, there exists $N \in \mathbb{N}$, such that $M_n(\omega) = -1$ for all $n > N$. It then follows that
\begin{align}
& \left|\frac{S_n(\omega)}{n} - (-1)\right| \\
= & \left|\frac{M_1(\omega) + \cdots +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1537706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Laplace Transforms Show that ${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\frac{(2n-1)!!}{2^n}\frac{1}{s^n} \sqrt{\frac{\pi}{s}}$
My Attempt:
${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rbrace}=\int_{0}^{\infty}\mathcal e^{-st}t^{n-\frac{1}{2}}dt$
Integration by parts gives:
${\mathcal L} {\lbrace t^{n-\frac{1}{2}}\rb... | By the definition of the $\Gamma$ function,
$$\begin{eqnarray*}\mathcal{L}\left(t^{n-\frac{1}{2}}\right)&=&\int_{0}^{+\infty} t^{n-\frac{1}{2}}e^{-st}\,dt = \frac{1}{s^{n+\frac{1}{2}}}\int_{0}^{+\infty}u^{n-\frac{1}{2}}e^{-u}\,du\\&=&\frac{\Gamma\left(n+\frac{1}{2}\right)}{s^{n+\frac{1}{2}}}=\frac{\left(n-\frac{1}{2}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1541609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Alternate approaches to solve this Integral Evaluate $$I=\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\:dx$$
I have used parts taking first function as Integrand and second function as $1$ we get
$$I=x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}-\int \frac{d}{dx}\left(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\right) \times x dx$$
now $... | $$
\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{\dfrac{1-\sqrt{x}}{1+\sqrt{x}}\times \dfrac{1-\sqrt{x}}{1-\sqrt{x}}}=\dfrac{1-\sqrt{x}}{\sqrt{1-x}}
$$
thus we get
$$
\int \dfrac{1}{\sqrt{1-x}}dx -\int \sqrt{\dfrac{x}{1-x}}dx
$$
So I got to your final integrals quickly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1542797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the Maximum value.
Maximize $xy^2$ on the ellipse $b^2x^2 +a^2y^2= a^2b^2$
The steps I tried to solve:
$$\nabla f = (y^2,2yx)\lambda\qquad g = (2xb^2,2y^2a^2)\lambda$$
$$y^2= 2xb^2\lambda$$
$$2yx= 2y^2a^2\lambda$$
$$
\left.
\begin{array}{l}
\text{}&y^2= 2xb^2\lambda\\
\text{}&
\end{array}
\right\}
*a^2y
$$ ... | just out of curiosity, we know that $x^2/a^2+y^2/b^2=1$, then by AM–GM inequality
$$1=\frac{x^2}{a^2}+\frac{y^2}{2b^2}+\frac{y^2}{2b^2}\geq 3\sqrt[3]{\frac{x^2}{a^2}\cdot\frac{y^2}{2b^2}\cdot\frac{y^2}{2b^2}}$$
the equality holds when
$$\frac{x^2}{a^2}=\frac{y^2}{2b^2}=\frac{y^2}{2b^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1543654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Partial sum of $\sum \frac {1} {k^2}$ It is pretty well-known that $\sum_{k = 1}^{\infty} \frac {1} {k^2} = \frac {\pi^2} {6}$. I am interested in evaluating the partial sum $\sum_{k = 1}^{N} \frac {1} {k^2}$. Here is what I have done so far. Since we have
$$\sum_{k = 1}^{N} \frac {1} {k^2} = 1 + \sum_{k = 2}^{N} \frac... | A quick fix on the upper bound is possible.
Just change everything from $\frac{1}{k^2-1}$ to $\frac{1}{k^2-\frac{1}{4}}$.
Then, we have $$ \sum_{k=1}^N \frac{1}{k^2} = 1+ \sum_{k=2}^N \frac{1}{k^2} < 1+\sum_{k=2}^N \frac{1}{k^2-\frac{1}{4}} = 1+\sum_{k=2}^N \frac{1}{k-\frac{1}{2}} - \sum_{k=2}^N \frac{1}{k+\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Expressing as a composite function to begin differentiating with the chain rule I'm trying to differentiate $y=(x+1)(x+2)^2(x+3)^3$ using the chain rule, but I am having trouble writing it as a composite function. Any help would be great as I can finish the problem on my own once I just get started in the right directi... | When you have products such as $$y=(x+1)(x+2)^2(x+3)^3$$ logarithmic differentiation makes life simpler $$\log(y)=\log(x+1)+2\log(x+2)+3\log(x+3)$$ Differentiating $$\frac{y'}y=\frac 1{x+1}+\frac 2{x+2}+\frac 3{x+1}=\frac {6 x^2+22 x+18 }{(x+1) (x+2) (x+3)}$$ Multiply both sides by $y$ to get $$y'=\frac {6 x^2+22 x+18 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of this trigonometric expression as x approaches 0 $$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})(\cot(x)+\frac{1}{x})$$
I have computed the limits
$$\lim_{x\to 0^+} (\cot(x)-\frac{1}{x})=0$$
$$\lim_{x\to 0^+} (\cot(x)+\frac{1}{x})=\infty$$
I can't multiply the limits together.
I rewrote the initial expression:... | It is possible to do not use Taylor series. First write
\begin{eqnarray*}
\cot ^{2}x-\frac{1}{x^{2}} &=&\left( \frac{x\cos x-\sin x}{x\sin x}\right)
\left( \frac{x\cos x+\sin x}{x\sin x}\right) \\
&=&\frac{x^{2}}{\sin ^{2}x}\left( \frac{x\cos x-\sin x}{x^{3}}\right) \left(
\frac{x\cos x+\sin x}{x}\right) \\
&=&\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do I find $\liminf$ and $\limsup$ if $a_{2n}=\frac {a_{2n-1}}2$ and $a_{2n+1}=\frac12+\frac {a_{2n}}2$? Its given that $a_1=a>0$ and that for any $n>1$ two things happen:
$$a_{2n}=\frac {a_{2n-1}}2$$
$$a_{2n+1}=\frac12+\frac {a_{2n}}2$$
How do I find $\lim\inf$ and $\lim\sup$
I am trying to look at
$a_{2n+1}$ and ... | EDIT: This response was written for the recurrence equations
$$ a_{2n} = \frac{a_{2n-1}}{2}, a_{2n+1} = \frac{1}{2}+\frac{a_{2n}}{2}. $$
The exact answer may differ depending on the exact edit to the question, but the techniques remain valid, assuming the coefficients are not drastically changed.
Since there are diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Infinite series equality $\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots$ Prove the following equality ($|x|<1$).
$$\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\cdots\\
=\frac{1}{1-x}+\frac{3x^2}{1-x^3}+\frac{5x^4}{1-x^5}+\frac{7x^6}{1-x^7}+\cdots\\$$
| Let me add my own proof that I had in my mind.
$$Let\quad f(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x^5}+\frac{6x^5}{1+x^6}+\frac{7x^6}{1+x^7}+\cdots$$
$$f(x)-\frac{1}{1-x}=\color{red}{\frac{-2x}{1-x^2}}+\frac{2x}{1+x^2}+\frac{3x^2}{1+x^3}+\frac{4x^3}{1+x^4}+\frac{5x^4}{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
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Simplify $\sqrt[3]{162x^6y^7}$ The answer is $3x^2 y^2 \sqrt[3]{6y}$
How does $\sqrt[3]{162x^6y^7}$ equal $3x^2 y^2\sqrt[3]{6y}$?
| First note that 162 factors as $2\cdot 3^4$. So collecting our powers of 3 reveals
$$\sqrt[3]{162x^6y^7} = \sqrt[3]{3^3 \cdot (x^3)^2\cdot (y^3)^2 \cdot (2\cdot 3 \cdot y)} = 3x^2y^2\sqrt[3]{6y}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Integrate. $\int\frac{x+1}{(x^2+7x-3)^3}dx$ How should i solve this integral?
i know that it is the same question like here
Integrate $\int\frac{x+1}{(x^2+7x-3)^3}dx$
but I've tried solve it for more then 3 hours and i still have no idea ho to solve it. Thank for help.
$$\int\frac{x+1}{(x^2+7x-3)^3}dx$$
I tried use $$... | HELP for the integrand $\frac{1}{(x^2-a)^3}$
Use the substitution $x = \sqrt{a}\sec(u)$ and $dx = \sqrt{a}\tan(u) \sec(u) du$ so then $(x^2 - a)^3 = (a\sec^2(u) - a)^3 = a^3\tan^6(u) du$.
Moreover you'll have to take in mind that $u = \text{arcsec}\left(\frac{x}{\sqrt{a}}\right)$
Thus you have
$$\sqrt{a}\int\frac{\cot^... | {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
} |
least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
The least value of $f(x,y) = 2x^2+y^2+2xy+2x-3y+8\;,$ Where $x,y\in \mathbb{R}$
$\bf{My\; Try::}$ Let $$K = 2x^2+y^2+2xy+2x-3y+8$$
So $$\displaystyle y^2+(2x-3)y+2x^2+2x+8-K=0$$
Now For real values of $y\;,$ We have $\bf{Discriminant\geq 0}$
... | since it has calculus tag im going to try solve it with calculus
heres my try:
we are going to find some extreme point and then check it if its minimum or maximum
let $Z = 2x^2 +y^2 +2xy + 2x -3y +8 $
so we got $\frac{dz}{dx} = 4x + 2y +2 $ and $\frac{dz}{dx} = 2y +2x -3 $
now we are going to find a possible extreme... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the number of permutations in$S_9$ of the form $(a_1a_2)(a_3a_4)(a_5a_6)(a_7a_8a_9)$ How many permutations, ρ, are there in $S_9$(the group of permutations of order 9!) whose decomposition into disjoint
cycles consists of three 2-cycles (transpositions) and one 3-cycle? In other words, how
many permutations are... | *
*You need to choose the support of the first transposition : $\begin{pmatrix}9\\2\end{pmatrix}$ choices.
*You need to choose the support of the second transposition in the remaining set : $\begin{pmatrix}7\\2\end{pmatrix}$ choices.
*You need to choose the support of the third transposition in the remaining set : $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Double Angle identity??? The question asks to fully solve for $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
My question is, is this a double angle formula? And if so, how would I go about to solve it?
I interpreted it this way; $$\left(\sin{\pi \over 8}+\cos{\pi \over 8}\right)^2$$
$$=2\sin{\pi \over 4}+\le... | In general, you can write expressions of the form $a\sin \theta + b\cos\theta$ in the form $c\sin(\theta + \psi)$ for suitable choices of $c$ and $\psi$. To see this, look at the double-angle formula for sine; you obtain
$$a\sin\theta + b\cos\theta = (c\cos \psi)\sin\theta + (c\sin\psi)\cos\theta$$
Comparing coefficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 5
} |
Seemingly Simple Integral $\int_0^1\frac{x^2\ln x}{\sqrt{1-x^2}}dx$. Evaluate $$\int_0^1 f(x) dx$$ where
$$f(x) = \frac{x^2\ln x}{\sqrt{1-x^2}}$$
I started off with the substitution $x=\sin y$, which resulted in the integrand reducing to
$$\sin^2y\cdot \ln (\sin y) dy$$
Then I used the property of definite integrals th... | $$I(\alpha)=\int_0^1\dfrac{x^\alpha\ln x}{\sqrt{1-x^2}}\ dx$$
Using Beta function we have
$$\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx=\dfrac12{\bf B}\left(\frac{\alpha+1}{2},\frac{1}{2}\right)$$
then with Digamma function $\psi$
\begin{align}
I(\alpha)
= \dfrac{d}{d\alpha}\int_0^1\dfrac{x^\alpha}{\sqrt{1-x^2}}\ dx \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 2
} |
Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$
Find the coefficient of $z$ in the Laureant Series expansion of $\frac{e^z}{z-1}$ in $\{|z|>1\}$.
Ok, so for $|z|>1 \iff |\frac{1}{z}|<1$ I can write
$\frac{1}{z-1}=\frac{1}{z}\cdot \frac{1}{1-\frac{1}{z}}=\frac{1}{z}\sum_{n=0}^{\infty... | Looks like to me we can just do this,
\begin{align}
\frac{e^z}{z-1} &= \frac{1}{z-1}\sum_{n \geq 0} \frac{e(z-1)^n}{n!}.\\
&=\frac{e}{z-1}\left(1 + (z-1) + \frac{1}{2!}(z-1)^2 + \frac{1}{3!}(z-1)^2 +\dots \right) \\
&=e\left(\frac{1}{z-1} + 1 + \frac{1}{2!}(z-1) + \frac{1}{3!}(z-1)^2 +\dots \right)
\end{align}
So it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$
I can otherwise solve this limit.This is in the form of $... | Start by reducing to "known" Taylor series around $0$, and then compose them.
Here, you will rely on that of $\ln(1+x)$, $\frac{1}{1+x}$, and the fact that $a^b = e^{b\ln a}$:
$$
\left(\frac{3x^2+2}{5x^2+2}\right)^{\frac{3x^2+8}{x^2}} =
\left(\frac{1+\frac{3x^2}{2}}{1+\frac{5x^2}{2}}\right)^{\frac{3x^2+8}{x^2}}
=
e^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show that $\lim\limits_{x\rightarrow 0}f(x)=1$
Suppose a function $f:(-a,a)-\{0\}\rightarrow(0,\infty)$ satisfies $\lim\limits_{x\rightarrow 0}\left(f(x)+\frac{1}{f(x)}\right)=2$. Show that $$\lim\limits_{x\rightarrow 0}f(x)=1$$
Let $\epsilon>0$ , then there exists a $\delta>0$ such that $$\left(f(x)+\frac{1}{f(x)}... | Define $h(u) = \frac{u + \sqrt{u^2 - 4}}{2}$ (this is the inverse function of $x \mapsto x + \frac{1}{x}$). Note that $h(2) = 1$ and that $h$ is continuous at $u = 2$. Thus,
$$ 1 = \lim_{u \to 2} h(u) = \lim_{x \to 0} h \left( f(x) + \frac{1}{f(x)} \right) = \lim_{x \to 0} \frac{f(x) + \frac{1}{f(x)} + \sqrt{\left( f(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is .... Problem :
The coefficient of $x^{15}$ in the product $(1-x)(1-2x)(1-2^2x)(1-2^3x)\cdots (1-2^{15}x)$ is
(a) $2^{105}-2^{121}$
(b) $2^{121}-2^{105}$
(c) $2^{120}-2^{104}$
(d) $2^{110}-2^{108}$
My approach :
Tried to... | I'd rescale as $$2^0\cdot2^1\cdot2^2\cdots2^{15}\left[(1-x)\left(\frac12-x\right)\left(\frac14-x\right)\cdots\left(\frac1{2^{15}}-x\right)\right]$$ The coefficient of $x^{15}$ in the polynomial within the brackets is the negative of the sum of the roots (since there are $16$ linear factors). So the answer is $$-2^0\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
The coefficient of $x^3$ in $(1+x)^3 \cdot (2+x^2)^{10}$ Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.
I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.
| Let's try it straight from the binomial theorem:
$$
(1+x)^3\cdot(2+x^2)^{10} = \sum_{j=0}^3 {3 \choose j} x^{3-j} \cdot \sum_{k=0}^{10} {10 \choose k} 2^k (x^2)^{10-k}.
$$
Good so far? Now, note these terms can be arranged as
$$
\sum_{j=0}^3 \sum_{k=0}^{10} {3 \choose j} {10 \choose k}2^k x^{3-j} (x^2)^{10-k}.
$$
Note... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve for $x$, correct to two significant figures, the equation: $4^{x}-2^{x+1}-3=0$ Solve for $x$, correct to two significant figures, the equation:
$$4^{x}-2^{x+1}-3=0$$
My answer: $x\log4=\log3+(x+1)\log2 \Rightarrow 0.602x-0.301x=0.477+0.301 \Rightarrow x = 2.6$ (Conflicting with book answer)
Answer in book: $x=1... | You made a mistake in your first step. Take a look at the following
$$\eqalign{
& {4^x} - {2^{x + 1}} - 3 = 0 \cr
& {\left( {{2^x}} \right)^2} - 2\left( {{2^x}} \right) - 3 = 0 \cr
& \left( {{2^x} - 3} \right)\left( {{2^x} + 1} \right) = 0 \cr
& \left\{ \matrix{
{2^x} = 3 \hfill \cr
{2^x} = - 1,\,\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Is the inverse of an invertible circulant matrix also circulant? The circulant matrices in $M_n(F)$ ($F$ field) form a subspace $\mathcal C_n$ spanned by $I,J,J^2,\cdots,J^{n-1}$ where
$$J=\begin{bmatrix} O & I_{n-1}\\1 & O\end{bmatrix}$$
This subspace $\mathcal C_n$ is also closed under multiplication, which makes it... | By Cayley-Hamilton theorem, the inverse of every nonsingular matrix $A$ can be expressed as a polynomial in $A$ of degree $\le n-1$. Therefore, if $A$ is in the linear span of $I,J,\ldots,J^{n-1}$, then $A^{-1}$ is a polynomial in $J$ of degree $\le (n-1)^2$. Since $J^n=I$, it follows that $A^{-1}$ lies inside the line... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must... | I can't resist to post this. I think this is the easiest way.
$$(B+C)x^2+(A+2B)x+2A=5x^2+3x-2$$
\begin{align}
2A&=-2&\rightarrow A&=-1\\
A+2B&=-1+2B=3&\rightarrow B&=2\\
B+C&=2+C=5&\rightarrow C&=3
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Compute $\int \big( (4z)^5 + 4(4z)^2 \big ) \big( (4z)^3 +1 \big)^{12} dz$ by substitution The problem is $\int \big( (4z)^5 + 4(4z)^2 \big ) \big( (4z)^3 +1 \big)^{12} dz$. I know a substitution has to be used, but having the $4$ in front of the $4z$ is confusing me and I don't know how to get rid of it.
| \begin{align}
&(4z \mapsto y)\\
&\int \left((4z)^5+4(4z)^2\right)\left((4z)^3+1\right)^{12}dz\\
&=\frac14\int y^2\left(y^3+4\right)\left(y^3+1\right)^{12}dy\\
&\\
&(y^3\mapsto x)\\
&=\frac1{12}\int (x+4)(x+1)^{12}dx\\
&=\frac1{12}\int (x+1)(x+1)^{12}dx+\frac1{12}\int 3(x+1)^{12}dx\\
&=\frac1{12\cdot14}(x+1)^{14}+\frac1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Explain how to compute $\cos(2\pi/13)$ by solving quadratic and cubic equations only I know that we can express $2\pi/13$ as a root of unity on the unit circle taking $z^{13} = 1$ and $z=\cos(2\pi/13)+i\sin(2\pi/13)$ and that we should be able to find a polynomial for this over the rational numbers, and because its not... | You can use the triple angle and the quadruple angle formula for cosine.
\begin{align}
&\cos x = \cos \frac{2\pi}{13}\\
&\cos 4x = 8 \cos^4 x - 8 \cos^2 x + 1\\
&\cos 3x = 4 \cos^3 x - 3 \cos x\\
&\cos 9x = 4 (4 \cos^3 x - 3 \cos x)^3 - 3 (4 \cos^3 x - 3 \cos x)\\
&\cos (4x+9x) = \cos 4x \cos 9x - \sin 4x \sin 9x \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
What is the sum of the digits of the sum of the digits? Problem
Let $(10^{2016}+5)^2=225N$. If $S$ is the sum of the digits of N, then find the sum of the digits of S
Attempt
Let's look at some smaller cases. We have $\dfrac{(10^{3}+5)^2}{225} = 4489$,$\dfrac{(10^{4}+5)^2}{225} = 444889$, and $\dfrac{(10^{5}+5)^2}{22... | $$
{1\over225}(10^{n+1}+5)^2 =
{1\over9}(2\cdot10^{n}+1)^2=
{1\over9}(4\cdot10^{2n}+4\cdot10^{n}+1)=\\
{1\over9}(4(10^{2n}-1)+4(\cdot10^{n}-1)+9)=
4{10^{2n}-1\over9}+4{10^{n}-1\over9}+1=\\
4{10^{2n}-1\over10-1}+4{10^{n}-1\over10-1}+1=
4\sum_{k=0}^{2n-1}10^k+4\sum_{k=0}^{n-1}10^k+1=\\
\sum_{k=n}^{2n-1}4\cdot10^k+\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How can one show the inequality $\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$ How can one show the inequality
$$\frac{a^2}{\sqrt{b^2-bc+c^2}}+\frac{b^2}{\sqrt{a^2-ac+c^2}}+\frac{c^2}{\sqrt{a^2-ab+b^2}}\ge a+b+c$$
Where $a,b,c$ are real and $ab+bc+ac$ is no equal t... | Note that:
$3(a-b)^{2}\ge0$
hence $4a^{2}-4ab+4b^{2}\ge a^{2}+b^{2}+2ab$.
And so:
$a^{2}-ab+b^{2}\ge\frac{a^{2}+b^{2}+2ab}{4}$
Since both sides are positive:
$\sqrt{a^{2}-ab+b^{2}}\ge\sqrt{\frac{a^{2}+b^{2}+2ab}{4}}$
We can do the same for each of $a,c$ and $b,c$ leading us to conclude that,
$\sqrt{b^{2}-bc+c^{2}}+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Mathematical Induction question: Prove divisibility by $4$ of $5^n + 9^n + 2$ Use mathematical induction to prove that $5^n + 9^n + 2$ is divisible by $4$, where $n$ is a positive integer.
| With induction: $5^1+9^1+2=16$ is divisible by $4$. If $5^k+9^k+2$ is divisible by $4$ for some $k\ge 1$, then $5^{k+1}+9^{k+1}+2=\left(5^k+9^k+2\right)+4\left(5^k+2\cdot 9^k\right)$ is also divisible by $4$.
Without explicit induction: $$5^n+9^n+2=\left(5^n-1\right)+\left(9^n-1\right)+4$$
$$=(5-1)\left(5^{n-1}+5^{n-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1570786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ ... | First advice: when you are looking for roots of a polynomial, factorize as much as you can. You have $(x-1)^2$ and $(x-1)^3$, so you can factorize $(x-1)^2$. You have $(x+2)$ and $(x+2)^2$, so you can factorize $(x+2)$. Thus you get:
$$(x+2)(x-1)^2 (2(x-1)-3(x+2))\,. $$ So:
*
*For question 1, the $-8$ root s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
A misunderstanding concerning $\pi$ The very well-known expression
$$\frac {\pi} {4} = 1 - \frac {1} {3} + \frac {1} {5} - \frac {1} {7} + \cdots$$
puts me face to face with a contradictory position. Let
$$s_N = \sum_{k = 0}^{N} \frac {1} {4k + 1} - \sum_{k = 0}^{N} \frac {1} {4k + 3}.$$
Then it is obvious that
$$\frac... | The error surely lies in the $o\left({1\over N^2}\right)$ term(s). Look at it this way: Your use of Euler-Maclaurin would suggest
$$\sum_{k=1}^N{1\over k}=\int_1^N{dx\over x}+{1\over2}\left(1+{1\over N}\right)+o\left({1\over N^2}\right)$$
as well, which would suggest
$$\sum_{k=1}^N{1\over k}-\log N\to{1\over2}$$
ins... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to find the Maclaurin series for $f(x) = \frac{1}{1 + \sin(x)}$? I have that $\frac{1}{1 + x} = 1 - x + x^2 - x^3 + ...$
So then $\frac{1}{1 + \sin(x)}$ should be $ 1 - \sin(x) + \sin^2(x) - \sin^3(x) + ...$ but clearly this is not the case.
So how does substitution into Maclaurin series work and why does this not... | By substitution of the expansion of $\sin x$ at the required order in the expansion of $\dfrac1{1+u}$ at the same order.
Example for order 5:
$$\dfrac1{1+u}=1-u+u^2-u^3+u^4-u^5+o(u),\qquad \sin x=x-\frac{x^3}6+\frac{x^5}{120}+o(x5),$$
whence
\begin{align*}
\sin^2x&=x^2-\frac{x^4}3+o(x^5)&\sin^3x&=\Bigl(x^2-\frac{x^4}3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Let $x^2+kx=0;k$ is a real number .The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set. Let $f(x)=x^2+kx;k$ is a real number.The set of values of $k$ for which the equation $f(x)=0$ and $f(f(x))=0$ have same real solution set.
The equation $x^2+kx=0$ has solutions $x=0,... | $f(f(x))$ has as solutions, the solutions of $f(x)$, plus the solution of the equation $x^2+kx+k=0$, which are $(1/2)[-k \pm \sqrt{k^2-4k}]$. So, to have $f(x)$ and $f(f(x))$ to have the same real solution set, these two solutions must be equal to the solutions of $f(x)$, which means $0$ and $-k$. So putting the previo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Differential Equation change of variable I'm investigating how you get from this:
$$ z\frac{d^2y}{dz^2}+(1-a)\frac{dy}{dz}+a^2z^{2a-1}y=0 $$
to this:
$$ \frac{d^2y}{dx^2}+y=0 $$
with a change of variable:
$$ z=x^{1/a}, (x\ge0) $$
I'm not quite 'getting' the substitution in the derivative...
In a normal situation where ... | Note, that by the chain rule, we have
$$ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} $$
and, taking another derivative, by product and chain rules,
$$ \frac{d^2y}{dz^2} = \frac{d^2 y}{dx^2} \cdot \left(\frac{dx}{dz}\right)^2
+ \frac{dy}{dx} \cdot \frac{d^2 x}{dz^2} $$
As $x = z^a$, we have
$$ \frac{dx}{dz} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int^\ell_{-\ell} r/\sqrt{L^2+r^2}^{\,3} \, dL$ We tried to solve a magnetics problem and ended up with
$$\int^\ell_{-\ell} \frac{r}{\sqrt{L^2+r^2}^{\,3}} \, dL.$$
How do I solve this integral?
| $$\int_{-l}^{l}\frac{r}{\left(\sqrt{L^2+r^2}\right)^3}\space\text{d}L=$$
$$r\int_{-l}^{l}\frac{1}{\left(\sqrt{L^2+r^2}\right)^3}\space\text{d}L=$$
$$r\int_{-l}^{l}\frac{1}{\left(L^2+r^2\right)^{\frac{3}{2}}}\space\text{d}L=$$
For the integrand $\frac{1}{\left(L^2+r^2\right)^{\frac{3}{2}}}$, (assuming all variables are... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving a simultaneous equation How can I solve the following simultaneous equations:
$$3x^4+3x^2y^2-6xy = 0\tag 1$$
$$-2x^3y+3x^2-y^2=0\tag 2$$
I have tried rearranging for $y$ in eq(1) and plugging it into eq(2), but the result did not give me the right answer.
| Using the old Sylvester´s method of elimination, choosing $y$ (the lowest degree) to be eliminated, we have
$$\begin{vmatrix}
3x^2&-6x&3x^4&0\\
0&3x^2&-6x&3x^4\\
1&2x^3&-3x^2&0\\
0&1&2x^3&-3x^2
\end{vmatrix}=0$$ which gives $$36x^4(x^8+2x^4-3)=0$$ i.e.
$$x^4(x^4+3)(x^2+1(x+1)(x-1)=0$$
The $12$ values for $x$ are $$x=0\... | {
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"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
evaluate $\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$
$$\lim _{x\to \infty} (3x^2-x^3)^{\frac{1}{3}}+x$$
can I look at $\lim\limits_{x\to \infty} (3^{\frac{1}{3}}x^{\frac{2}{3}}-x+x)$?
| HINT:
$$\left(1+z\right)^{1/3}=1+\frac13 z+O(z^2) \tag 1$$
SPOILER ALERT: Scroll over the highlighted area to reveal the full solution
$$\begin{align}x+\left(3x^2-x^3\right)^{1/3}&=x-x\left(1-\frac{3x^2}{x^3}\right)^{1/3}\\\\&=x-x\left(1-\frac3x\right)^{1/3}\\\\&=x-x\left(1-\frac1x+O\left(\frac1{x^2}\right)\right)\\... | {
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"url": "https://math.stackexchange.com/questions/1578409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
How do you evaluate this sum of multiplied binomial coefficients: $\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} $? We have to find the value of x+y in:
$$\sum_{r=2}^9 \binom{r}{2} \binom{12-r}{3} = \binom{x}{y} $$
My approach:
I figured that the required summation is nothing but the coefficient of $x^3$ is the following e... | $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \ov... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How to use Mathematical Induction to prove $\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n + 1)} = \frac{n}{n + 1}$? $$\frac{1}{1 \cdot 2} + \frac{1}{2\cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1}$$
What I have so far in the induction is:
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + ... | Get a common denominator. Your last expression simplifies to $\dfrac{k^2 + 2k + 1}{(k+1)(k+2)}$. Now factor it.
| {
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"source": "stackexchange",
"question_score": "3",
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Way to verify a least-squares solution without actually solving for $x$ and $y$? I just found the least squares solution of the system $\mathbf{x}A = \mathbf{b} = \begin{pmatrix} x & y \end{pmatrix}\begin{pmatrix} 3 & 2 & 1 \\ 2 & 3 & 2\end{pmatrix} = \begin{pmatrix} 3 & 0 & 1\end{pmatrix}$ to be $\begin{pmatrix} x \\ ... | By construction, the least squares solution minimizes the sum of the squares of the residuals
$$
r^{2}(x,y) = \lVert A x - b \rVert_{2}^{2}.
$$
that is, the least squares solution can be defined as
$$
\left[
\begin{array}{c}
x \\ y
\end{array}
\right]_{LS}
=
\left\{
\left[
\begin{array}{c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinite sums of reciprocal power: $\sum\frac1{n^{2}}$ over odd integers The infinite series I need to solve is
$$\sum_{n=1,3,5...}^{\infty}\frac{1}{n^{2}}$$
and because the point of interest lies in the value of odd n,
the infinite series can be expressed as
$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}$$
This came up i... | by using
$$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+...)$$
$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(\frac{\pi^2}{6})$$
$$\frac{\pi^2}{6}-\frac{\pi^2}... | {
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"source": "stackexchange",
"question_score": "5",
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Integrate $\int \frac{1}{x^4+4}dx$
Integrate $\displaystyle \int \dfrac{1}{x^4+4}dx$.
I could try breaking this up into two quadratic trinomials, but that seems like it would be a lot of work. If that is the best way here how do I do it most efficiently?
| Notice
$$\frac{1}{x^4+4} = \frac{1}{(x^2+2)^2 - 4x^2}
= \frac{1}{(x^2+2x+2)(x^2-2x+2)}$$
Since
$$\begin{align}
\frac{1}{x^2-2x+2} - \frac{1}{x^2+2x+2} &= \frac{4x}{(x^2+2x+2)(x^2-2x+2)}\\
\frac{1}{x^2-2x+2} + \frac{1}{x^2+2x+2} &= \frac{2x^2+4}{(x^2+2x+2)(x^2-2x+2)}
\end{align}$$
We have
$$\frac{1}{(x^2+2x+2)(x^2-2x+2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding the vectors that form a basis of a span Let
$$\begin{pmatrix}
1 & 2\\
0 & 1
\end{pmatrix},
\begin{pmatrix}
0 & 1\\ 3 & 0
\end{pmatrix},
\begin{pmatrix}
3 & 3\\ 4 & 0
\end{pmatrix},
\begin{pmatrix}
2 & 0\\ 1 & -1
\end{pmatrix}$$
be vectors in some vector space $U$. Which vectors form a basis of $Sp(U)$?
I... | Generally, a "basis" for a vector space must both span that space and be linearly independent. If the give spanning set is already independent, then it is a basis. If not you can delete one or more vectors until it is.
Here, for example, to see if these matrices are independent, look at the equation $a\begin{pmatr... | {
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How can we find the area of the triangle which covers a finite point set in $\mathbb{R}^2$ by using the interior triangles with specified area? Suppose we have given a finite point set $X \subset \mathbb{R}^2$ in a way that any triangle made by vertices of $X$ has area at most 1. How can we prove that there is a triang... | Let $PQ$ be the segment with vertices in $X$ and let $R$ be another point of $X$ such that $PQR$ has the maximum area (of all possible triangles with $P$ and $Q$ being two of their vertices). Now let $l_R$ be the line passes through the point $R$ and parallel to the edge $PQ$, and let $H_R$ be the half plane through $l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Could you solve $x↑^n2=x↑^m2$? As my title asks, could you solve $x^2=2^x$?
But that's the worrisome part, as I noticed $x↑^n 2=x↑^m 2$ and $2↑^p x=2↑^q x$ will always have a solution at $x=2$. However, there is bound to be at least another solution.
For example:$$2x=x^2$$has a solution at $x=2,0$.
And $x^2=2^x$ has a... | As you have probably surmised, $x \uparrow^n 2 = x \uparrow^m 2$ and $2 \uparrow^p x = 2 \uparrow^q x$ have solutions at $x=1$ and $x=2$. If $m=n$ or $p=q$ then of course any $x$ will be a solution; if $m \neq n$ or $p \neq q$ then there are no other solutions.
First, I will prove the following useful theorems:
Theore... | {
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"source": "stackexchange",
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Convergence of $\sum \limits _{n=1}^{\infty} (-1)^{n} \frac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$ $\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$
How to check this? I've tried using Leibniz test, it's easy to prove that this one is monotonous, but its limit is rather no... | Rewrite each term into:
$$2\times\frac{4}{5}\times\frac{6}{7}\times\frac{8}{9}\times\frac{10}{11}\times\dots\times\frac{2n}{2n+1}\times\frac{1}{2n+3}<\frac{2}{2n+3}$$
Hence, the limit of the terms go to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Solve the functional equation $f (2x)=f (x)\cos x$ Find all $f: \mathbb R\longrightarrow \mathbb R $
such that $f $ is a continuous function at $0$ and satisfies
$$\;\forall \:x \in \mathbb R,\; f\left(2x\right) = f\left(x\right)\cos x $$
My try: I just found the $f (x)$ is periodic, i.e.
$f (2\pi / 2)= f (\pi/2) \... | We have
$$f(x) = f\left(\dfrac{x}2\right)\cos\left(\dfrac{x}2\right) = f\left(\dfrac{x}4\right)\cos\left(\dfrac{x}4\right)\cos\left(\dfrac{x}2\right)$$
Hence, we have
$$f(x) = f\left(\dfrac{x}{2^{n}}\right) \prod_{k=1}^n \cos\left(\dfrac{x}{2^k}\right) = \dfrac1{2^n} f\left(\dfrac{x}{2^n}\right)\dfrac{\sin(x)}{\sin\lef... | {
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"source": "stackexchange",
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Generating functions - deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ I would like some help with deriving a formula for the sum $1^2 + 2^2 +\cdots+n^2$ using generating functions.
I have managed to do this for $1^2 + 2^2 + 3^2 +\cdots$ by putting
$$f_0(x) = \frac{1}{1-x} = 1 + x + x^2 + x^3 +\cdots$$
$$f_1(x)... | Given a generating function $f(x)$ for $a_n$, the generating function for $\sum_{i=0}^n a_i$ is $\sum_{n\geq 0} (\sum_{i=0}^n a_i)x^n = (\sum_{n\geq 0} x^n)(\sum_{n\geq 0} a_nx^n) = \frac{1}{1-x} f(x)$.
In this case, we have $f(x) = \frac{x(1+x)}{(1-x)^3}$ as you mention, so the desired generating function is $\frac{x(... | {
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"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
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What's the method of integrating $\frac{1}{1-\ln(x)}$? How to do $\int\frac{1}{1-\ln(x)}dx$ manually? I mean I got its answer by running it through Mathematica, but is there anyway we can do it by hand?
| $\frac{1}{1 - \ln x}$ has no elementary antiderivative. If you are so inclined you could express $1 - \ln x$ as a series.
$$\ln (1+x) = x - x^2/2 + x^3/3 - ...$$
$$\to \ln (1+x) = (-1+1+x) - (-1+1+x)^2/2 + (-1+1+x)^3/3 - ...$$
$$\to \ln (x) = (-1+x) - (-1+x)^2/2 + (-1+x)^3/3 - ...$$
$$\to -\ln (x) = -(-1+x) + (-1+x)^2/... | {
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"source": "stackexchange",
"question_score": "1",
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How to prove $\sum\left(\frac{a}{b+c}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$ The question is to prove:
$$\left(\frac{a}{b+c}\right)^2+\left(\frac{b}{c+a}\right)^2+\left(\frac{c}{a+b}\right)^2\ge \frac34\left(\frac{a^2+b^2+c^2}{ab+bc+ca}\right)$$
$$a,b,c>0$$
I tried Cauchy, AM-GM, Jensen, etc. but... | This is an incomplete answer, but I think the last inequality is (correct) and easier to prove.
We have:
$$\bigg (\frac{a}{b+c} \bigg)^2 + \bigg (\frac{b}{a+c} \bigg)^2 + \bigg (\frac{c}{b+a} \bigg)^2 = \frac{a^4}{a^2(b+c)^2} + \frac{b^4}{b^2(a+c)^2} + \frac{c^4}{c^2(b+a)^2} \geq \frac{(a^2+b^2+c^2)^2}{a^2(b+c)^2 + b^2... | {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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particular solution of the second-order linear equation I'm trying to find particular solution of the second-order linear equation but I can't find $y_{1}$ and $y_{2}$ according to $y = c_{1}y_{1} + c_{2}y_{2}$
$$x^{2}y^{''}-2xy^{'}+2y=0, y(1) = 3, y'(1) = 1$$
If $r$ is used, $x^{2}r^{2}-2xr=-2$ then $xr(xr - 2) = -2... | $$y''(x)x^2+2y'(x)x+2y(x)=0\Longleftrightarrow$$
Assume a solution to this Euler-Cauchy equation will be proportional to $e^{\lambda}$ for some constant $\lambda$.
Substitute $y=x^{\lambda}$ into the differential equation:
$$x^2\frac{\text{d}^2}{\text{d}x^2}\left(x^{\lambda}\right)+2x\frac{\text{d}}{\text{d}x}\left(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590376",
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"source": "stackexchange",
"question_score": "1",
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Combinatorial Identity with Binomial Coefficients: $ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $ I got the following identity from commutative algebra.
I am curious to see elegant elementary methods.
$$ {{a+b+c-1}\choose c} = \sum_{i+j=c} {{a+i-1}\choose i}{{b+j-1}\choose j} $$
| For variety's sake here is a slightly different approach.
Suppose we seek to evaluate
$$\sum_{k=0}^c {k+a-1\choose a-1} {b-1+c-k\choose b-1}.$$
Introduce
$${b-1+c-k\choose b-1} = {b-1+c-k\choose c-k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{c-k+1}} (1+z)^{b-1+c-k} \; dz.$$
Observe that this is zero when $... | {
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"source": "stackexchange",
"question_score": "2",
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Curious combinatorial summation Let $\gamma$ denote a grid walk from the upper left corner $(1,k)$ to the lower right corner $(\ell,1)$ of the $k\times\ell$ rectangle $\{1,..,k\}\times\{1,..,\ell\}$. There are $\binom{k+\ell-2}{k-1}$ such paths. Denote
$$
X_\gamma = \prod_{(i,j)\in\gamma} \frac{1}{i+j-1}\,.
$$
Claim:... | Suppose the start point is $(a,b)$ and the end point is $(c,d)$, where $a\geq c$ and $d\geq b$.
From the general form of $F((k,1),(1,l))$, and the fact that paths from $(k,1)$ go through $(k-1,1)$ or $(k,2)$, you can deduce the general form of $F((k,2),(1,l))$.
Then $F((k,3),(1,l))$ and so on.
I got this formula:
$$F((... | {
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How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5... | A tailor-made approach by analytic combinatorics. The coefficient of $x^B$ in
$$ \frac{1}{(1-x)^N} = \left(1+x+x^2+x^3+\ldots\right)^N $$
obviously counts the number of ways of representing $B$ as a sum of $N$ non-negative integers. By stars and bars, or by the (negative) binomial theorem, such number is $\binom{N+B-1... | {
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"source": "stackexchange",
"question_score": "8",
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Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n... | Hint:
$$
5(5^n+2\cdot 3^{n-1}+1)=5^{n+1}+2\cdot 3^n+1+4(3^{n-1}+1)
$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\... | HINT:
Take $$f(x)=\ln x $$ and apply MVT in [$a,a+1$].
| {
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"question_score": "3",
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$\frac{a(a+b)}{4a^2+ab+b^2} + \frac{b(b+c)}{4b^2+bc+c^2} + \frac{c(c+a)}{4c^2+ca+a^2} \leq 1$ I've got stuck at this problem:
Let $a$, $b$, $c$ be real numbers. Prove that
$$\frac{a(a+b)}{4a^2+ab+b^2} + \frac{b(b+c)}{4b^2+bc+c^2} + \frac{c(c+a)}{4c^2+ca+a^2} \leq 1$$
Firstly, I've thought this :
$$a^2 - 2ab + b^2 \... | The inequality is equivalent to:
$$
\sum_{cyc}\frac{3a(a+b)}{4a^2+ab+b^2}≤3\iff \\
0≤\sum_{cyc}1-\frac{3a(a+b)}{4a^2+ab+b^2}=\sum_{cyc}\frac{a^2-2ab+b^2}{4a^2+ab+b^2}=\sum_{cyc}\frac{(a-b)^2}{(a-b)^2+3a(a+b)}\iff\\
0≤\sum_{cyc}\frac{(a-b)^2}{\left(\frac{7}{4}a-\frac{1}{4}b\right)^2+\frac{15}{16}(a+b)^2}
$$
Which is obv... | {
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Approximation of $\frac{x}{\sqrt{x^2+R^2}}$ How do you prove this statement?
If $x\gg R$ then
$$\frac{x}{\sqrt{x^2+R^2}}\cong 1-\frac{1}{2}\left(\frac{R}{x}\right)^2$$
I have no ideas even how to start.
| Basically, this can be seen by a Taylor series expansion around $0$. Note that $x\gg R$ "means" $\frac{R}{x} \approx 0$: so you would need to make this quantity appear, first.
$$\begin{align}
\frac{x}{\sqrt{x^2+R^2}} &= \frac{x}{x\sqrt{1+\left(\frac{R}{x}\right)^2}}
= \frac{x}{x}\left(1-\frac{1}{2}\left(\frac{R}{x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx... | HINT:
the second integral $\int\frac{1}{2x^2+x+3}dx = \int\frac{1}{2\left(x+\frac{1}{4}\right)^2+\frac{23}{8}}dx$
You should be able to use some substitutions to turn it into $C\int\frac{1}{u^2+1}du$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Solve irrational equation $x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$ Solve irrational equation
$$x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3}) = 30$$
Here is what I tried
$t^3 = 35-x^3 \implies x = \sqrt[3]{35-t^3} $
which takes me to nowhere.
| HINT:
$$x^3+35-x^3+3x \sqrt[3]{35-x^3}(x+\sqrt[3]{35-x^3})=35+3\cdot30 $$
$$(x+\sqrt[3]{35-x^3})^3=5^3$$
Assuming $x$ to be real, $$x+\sqrt[3]{35-x^3}=5\iff\sqrt[3]{35-x^3}=5-x$$
Take cube in both sides
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work... | We have:
$$ \frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca} ≥ ab + bc + ca $$
Also note that
$$ \frac{a^4}{ab}+\frac{b^4}{bc}+\frac{c^4}{ca} ≥ \frac{(a^2+b^2+c^2)^2}{ab+bc+ca}$$
(Because of the Cauchy Schwarz lemma or the Titu's Lemma)
So if we can prove that
$$\frac{(a^2+b^2+c^2)^2}{ab+bc+ca} ≥ ab + bc + ca$$
Then our w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 6
} |
how to solve the equation $\tan^2(2x)+2\tan(2x) \cdot \tan(3x)-1=0$ How to solve the equation $$\tan^2(2x)+2\tan(2x) \cdot \tan(3x)-1=0$$
Can anyone give some hints in this question ?
| Let $t$ = tan $x$. Using trig, we can see that tan $2x = \frac{2t}{1-t^2},$ and tan $3x = t\frac{3-t^2}{1-3t^2}$.
Then the equation becomes:
$$\frac{4t^2}{(1-t^2)^2} + 2\frac{2t^2(3-t^2)}{(1-t^2)(1-3t^2)} - 1 = 0$$
Multiply by the GCD, which is $(1-t^2)^2(1-3t^2)$ to get
$$(4t^2)(1-3t^2) + 4t^2(3-t^2)(1-t^2) - (1-t^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that $a\sqrt{a^2+bc}+b\sqrt{b^2+ac}+c\sqrt{c^2+ab}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}$
Let $a$, $b$ and $c$ be non-negative numbers. Prove that:
$$a\sqrt{a^2+bc}+b\sqrt{b^2+ac}+c\sqrt{c^2+ab}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}.$$
I have a proof, but my proof is very ugly:
it's enough to prove a polynomial in... | $\sum\limits_{cyc}a\sqrt{a^2+bc}\geq\sqrt{2(a^2+b^2+c^2)(ab+ac+bc)}\Leftrightarrow$
$\Leftrightarrow\sum\limits_{cyc}\left(a^4+a^2bc+2ab\sqrt{(a^2+bc)(b^2+ac)}\right)\geq\sum\limits_{cyc}(2a^3b+2a^3c+2a^2bc)\Leftrightarrow$
$\sum\limits_{cyc}(a^4-a^3b-a^3c+a^2bc)\geq\sum\limits_{cyc}\left(a^3b+a^3c+2a^2bc-2ab\sqrt{(a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 1,
"answer_id": 0
} |
Which integers can be represented as the most pair of difference of two squares? Let $f(x)$ be the number of $a,b,x\in \mathbb N$ where $a^2-b^2=x$.
For example, 1971 is not only $986^2-985^2$ but also $50^2-23^2$, $114^2-105^2$, $330^2-327^2$. So, $f(1971)=4$. Is there some sort of limit to $f(x)/x$ or $f(x)/ln(x)$ or... | We're looking for the size of the set $S$ of ordered pairs $(a, b)$ with $a, b \geq 0$ and $a^2 - b^2 = (a + b)(a - b) = x$ for some fixed $x > 0$. (Removing the 'ordered' bit is trivial, since any such pair has $a > b$. Also, we can dispense with the case $x = a^2$ separately if we want to include the case $b = 0$.) L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
solution of nested radical $\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$ This question is from my friend. he think that there is trigonometry involved to this equation.
$\sqrt{7+2\sqrt{7-2\sqrt{7-2x} } } =x$
that is from $\ \ \ x = \sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7-2\sqrt...} } }} } }$
I try to compare... | I notice that $ x^3-x^2-9 x+1=0 $ has three roots:
$ x_1=1-4 \cos \left(\frac{\pi }{7}\right)=-2.60388... $
$ x_2=1+4 \cos \left(\frac{2 \pi }{7}\right)=3.49396... $
$ x_3=1-4 \cos \left(\frac{3 \pi }{7}\right)=0.109916... $
Therefore, the above result $ \sqrt{7+2 \sqrt{7-2 \sqrt{7-2 \left(1+4 \cos \left(\frac{2 \pi }{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1596824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
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If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve:
If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer.
If I write $n=2k+1$ and $m=2l+1$ I get stuck at
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$
| I will use a method similar to yours. Letting $m,n$ respectively equal $2k+1$ and $2p+1$ and subsequently expanding the expression, we get $2k^2p^2+ 2k^2 p+ \frac{k^2}{2} + 2kp^2 +2kp + \frac{k}{2} +\frac{p^2}{2} + \frac{p}{2}$. Of course, $k$ and $p$ are both integers, so the only terms we have to worry about are the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 7
} |
How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{... | I think the simplest way is by expanding
to get terms with degrees below 4 we have to choose which combination of terms of original cos(x) series must be chosen and sum up the posibble combination
here is the answer:
$$g(x)=\left(1 - \frac{1}{2}x^2 + \frac{1}{24}x^4+\cdots\right)^6$$
$$=\left(1 - \frac{1}{2}x^2 + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
} |
How can you prove the inequality $2x^3 (x^3 + 8y^3) + 2y^3 (y^3 + 8z^3) + 2z^3 (z^3 + 8x^3) ≥ 9x^4 (y^2 + z^2) + 9y^4 (z^2 + x^2) + 9z^4 (x^2 + y^2)$ I changed the RHS as $\displaystyle \sum_{cyc} 9x^4(S-x^2) $ for $S = x^2 + y^2 + z^2$
Then I thought I could apply Jensen's inequality for $f(x) = 9x^4(S-x^2) = -9x^6 +... | In fact, we have $$ LHS-RHS=\sum\limits_{cyc}{\left(x-y\right)^4\left(x^2+4xy+y^2\right)} \ge 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Difficult Integral I have a problem with solving this integral:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx$$
I tried to use substitution but I got stuck. Can anyone help me?
| The easiest way is using undetermined coefficients:
$$\int{\frac{2x^2+3x+1}{\sqrt{x^2+1}}} dx = (Ax+B)\sqrt{x^2+1}+\lambda\int{\frac{dx}{\sqrt{x^2+1}}}$$
You must differentiate and you have:
$$\frac{2x^2+3x+1}{\sqrt{x^2+1}}=A\sqrt{x^2+1}+(Ax+B)\frac{2x}{2\sqrt{x^2+1}}+\frac{\lambda}{\sqrt{x^2+1}}$$
So you have an equat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Walk me through some basic modular arithmetic? The problem is as follows:
Positive integers $a$, $b$, and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$. What is the probability that $abc + ab + a$ is divisible by $3$?
The solution is as follows:
First we factor $abc... | You can find the solutions just by plugging the all the options ($b = 0, 1, 2$, $c = 0, 1, 2$) and seeing which satisfy $b(c+1) \equiv 2 \pmod{3}$.
It often helps me, when working with mod, to forget the definition in terms of divisibility entirely and just imagine I am working with objects called $0, 1, 2$ such that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sin 2x=-\cos x$ I'm working on solving the problem $\sin(2x)=-\cos(x)$ but I got stuck.
I got the following:
$\sin 2x=-\cos x \Leftrightarrow \sin 2x=\cos(x+\pi )\Leftrightarrow \sin 2x=\sin\left(\frac{\pi }{2}-(x+\pi)\right)\Leftrightarrow \sin 2x= \sin\left(-\frac{\pi }{2}-x\right)$
then I did
$2x =-\frac{\... | Begin by using the double-angle identity $\sin(2x) = 2\sin(x)\cos(x)$ to make the equation $2\sin(x)\cos(x) = -\cos(x)$, or $2\sin(x)\cos(x) + \cos(x) = 0$. Now we can factor to obtain $\cos(x)(2\sin(x) + 1) = 0$. This means that either $\cos(x) = 0$ or $\sin(x) = -\frac{1}{2}$. Thus, the solutions are $x = \frac{\pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Interval for area bounded by $r = 1 + 3 \sin \theta$ I'm trying to calculate the area of the region bounded by one loop of the graph for the equation
$$
r = 1 + 3 \sin \theta
$$
I first plot the graph as a limaçon with a maximum outer loop at $(4, \frac{\pi}{2})$ and a minimum inner loop at $(-2, -\frac{3 \pi}{2})$. I ... | Notice how $\arcsin(-\frac{1}{3}) = - \arcsin(\frac{1}{3})$, so your answer now looks like
$$
\frac{11 \pi}{4} + \frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2 \\
$$
That means your area is greater than the answer in your book by:
$$
2 \left(\frac{11}{2} \arcsin(\frac{1}{3}) + 3 \sqrt 2\right)
$$
This might indicate you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Where is the mistake in my solution? Trigonometry proof
To prove: $$1 + 2 \sin 70^\circ = \frac{1}{2\sin 20^\circ}$$
My attempt:
$$\begin{align}
1 + 2 \sin 70^\circ &= 1 + \frac{\sin 140^\circ}{\cos 70^\circ} \\[6pt]
&= 1 + \frac{\sin 40^\circ}{\sin 20^\circ} \\[6pt]
&= \frac{\sin 20^\circ + \sin 40^\circ}{\sin 20^\c... | As noted in the comments:
*
*The original identity is wrong (LHS $\approx 2.88$, RHS $\approx 1.46$).
*Your derivation of $1 + 2 \sin 70^\circ = \dfrac{1}{2\sin 10^\circ}$is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$?
If $A,B,A+I,A+B$ are idempotent matrices how to show that $AB=BA$ ?
MY ATTEMPT:
$A\cdot A=A$
$B\cdot B=B$
$(A+I)\cdot (A+I)=A+I$ or, $A\cdot A+A\cdot I+I\cdot A+I\cdot I=A+I$ which implies $A\cdot I+I\cdot A=0$ (using above equations)
$(A+B)\cdot (A+B... | Well, $AI=IA=A$, so your equation $AI+IA=0$ actually says $2A=0$. Over a field of characteristic $\neq 2$, this implies $A=0$, so $AB=BA=0$ trivially. Over a field of characteristic $2$, your equation $AB+BA=0$ gives $AB=-BA$, and $-BA=BA$ since the characteristic is $2$.
(Incidentally, if $A$, $B$, and $A+B$ are all... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Double Integral With a Change of Variables I need to calculate:
$$
\iint _D \frac{2y^2+x^2}{xy}~\mathrm dx~\mathrm dy
$$
over the set $D$ which is:
$$
y\leq x^2 \leq 2y , \quad 1\leq x^2 +y^2 \leq 2 , \quad x\geq 0
$$
can someone help me understand what possible change of variables can I do here?
Thanks a lot in advan... | Use this change of variables
$$\begin{align}
u&=x^2+y^2\\
v&=\frac{x^2}{y}
\end{align}$$
So we may compute the Jacobian first
$$\frac{\partial(u,v)}{\partial(x,y)}=
\begin{vmatrix}
\frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\
\frac{\partial v}{\partial x} & \frac{\partial v}{\partial y}
\end{vmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Integer solution of $ y-x = \sqrt{y+x} $ I have a problem with how to find integer solution of $\ \ \ y-x = \sqrt{y+x} \ \ \ which \ \ \ \ y>x$
$ y^2 -2xy + x^2 -y -x = 0 $
$ y^2 - (2x+1)y + (x^2 - x) = 0 $
$ y = \frac{(2x+1) \pm \sqrt{8x+1}}{2} $ and then ? I cannot solve this anymore.
because If it had a solution... | Set $a=\sqrt{y+x}$. Then $a^2=y+x$ and $a=y-x$ by the condition. Since $y>x$ in fact $a$ is a positive integer. Adding and subtracting, respectively, we have $2y=a^2+a$ and $2x=a^2-a$. Whether $a$ is even or odd, $a^2+a$ and $a^2-a$ are even, so we get integer $x,y$. Hence the solution set is given by the followin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Cantor set as an intersection I've seen that the Cantor set $C$ can be expressed as a countable intersection of $C_n$'s where $C_n=C_{n-1}-\bigcup_{k=0}^{3^{n-1}-1}(\frac{3k+1}{3^n},\frac{3k+2}{3^n})$.
Taking $C_1,C_2$ into consideration, this is clear, however, when trying to come up with this specific equation on my... | What might help is to see that the formula in the question is a piece of the inductive proof of a closed form description of $C_n$, expressed using trinary expansions of integers.
For each $n$, each integer $k \in \{0,...,3^n-1\}$ can be written as a trinary number $a_{n-1}a_{n-2}...a_0$ with entries in the set of trig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1608473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using Cauchy's integral formula to evaluate $\int_0^{2\pi} \frac{1}{a^2 \cdot {\cos}^2(t) + b^2 \cdot {\sin}^2(t)} dt$ I have to solve this integral using Cauchy's integral formula.
I tried to substitute it with several different attempts but without a solution. Can anyone help?
$$\int_0^{2\pi} \frac{1}{a^2 \cdot {\cos... | First, note that $\cos(t)=\frac{e^{it}+e^{-it}}{2}$ and $\sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Substituting this into the integral yields: $$\int_o^{2\pi}\frac{1}{a^2\cdot(\frac{e^{it}+e^{-it}}{2})^2+b^2\cdot(\frac{e^{it}-e^{-it}}{2i})^2}dt$$
Simplifying yields: $$=\int_0^{2\pi}\frac{4e^{2it}}{(a^2-b^2)e^{4it}+2(a^2+b^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all natural numbers $x,y$ such that $3^x=2y^2+1$. Find all natural numbers $x,y$ such that
$$3^x=2y^2+1$$
solutions are $(1,1)$, $(2,2)$, $(5,11)$. I found that parity of both is same and If $x$ Is odd it is of the form $4k+1$.
| There is a "standard" way to attack this using Thue's theorem, by writing the left hand side as $z^3$, $3 z^3$, or $9 z^3$, depending on the assumed residue $x \pmod 3$. I can't imagine doing a search up to any of the effective versions of Thue's bound on the size of the solutions during an Olympiad.
Some of the comme... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 2
} |
Integrating $\int \frac{u \,du}{(a^2+u^2)^{3/2}}$ How does one integrate $$\int \frac{u \,du}{(a^2+u^2)^{3/2}} ?$$
Looking at it, the substitution rule seems like method of choice. What is the strategy here for choosing a substitution?
| You can take the solution as I've presented here:
\begin{equation}
\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}, \frac{1}{1 + ax^n} \right)\right]
\end{equation}
Here $a$ is $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}... | Case $\#1:$ For $x+4\ne0$
As for real $y, a>0$
$$|y|<a\iff -a<y<a$$
$$\implies\left|\dfrac x{x+4}\right|<4\iff-4<\dfrac x{x+4}<4$$
Now $-4<\dfrac x{x+4}\iff0<\dfrac x{x+4}+4=\dfrac{5x+16}{x+4}=5\cdot\dfrac{(x+4)\left(x+\dfrac{16}5\right)}{(x+4)^2}$
$\iff(x+4)\left(x+\dfrac{16}5\right)>0$
Now if $(z-a)(z-b)>0$ with $a<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Reduction of a quadratic form to a canonical form I'm supposed to reduce following polynomial to its canonical form. But my result differs from the one given in my book, so I'm not sure if it's correct too.
$$
q = u_{xx} - u_{xy} - 2 u_{yy} + u_x + u_y = 0
$$
So, the characteristic quadratic polynomial is
$$
x^2 - xy -... | We reduce second order PDE with constant coefficients to canonic form
$ u_{xx} - u_{xy} - 2 u_{yy} + u_x + u_y = 0$
Quadratic form:
$Q=x^2 - xy -2y^2=\left(x-\frac{y}{2}\right)^2-\left(\frac{3y}{2}\right)^2$
We get transform
$\xi=x,\quad\eta=\frac{x}{3}+\frac{2y}{3}$
and canonical form of PDE:
$u_{\xi\xi}-u_{\eta\eta}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the volume of the solid bounded by $(x^2+y^2+z^2)^2=x^2+y^2-z^2$ As Title.
Find the volume of the solid bounded by $(x^2+y^2+z^2)^2=x^2+y^2-z^2$
--
By Spherical Coordinates, I got $\rho^4=\rho^2(\sin^2\phi-\cos^2\phi)$
Then $\rho^2=-\cos2\phi$
How to find the integral bounded?
| As $\rho^2>0$
It must be $-1 \leq \cos{2\phi} \leq 0$, then $\frac{\pi}{4} \leq \phi \leq \frac{3\pi}{4}$
--
$\displaystyle V=\int_{0}^{2\pi} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\int_{0}^{\sqrt{-\cos2\phi}} \rho^2 \sin\phi \, d\rho d\phi d\theta = \frac{-2\pi}{3} \int_{\frac{\pi}{3}}^{\frac{4\pi}{3}} (-\cos{2\phi})^{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate this integral using cylindrical coordinates Find the volume of the solid bounded above by the paraboloid of revolution
$z^{2}=x^{2}+y^{2}$
And below by the $xy$ plane, and on the sides by the cylinder $x^{2}+y^{2}=2ax$
We take $a>0$.
I'm struggling to understand what this would look like graphically, I under... | This is what I have right now, so correct me if I'm wrong:
I think you meant $z=x^2+y^2$ if you are trying to indicate a paraboloid of revolution. The graph of $z^2=x^2+y^2$ would be two cones.
The cylinder $x^2 + y^2 = 2ax$ can be represented as follows:
$\begin{align} x^2 + y^2 &= 2ax \\
x^2 - 2ax + y^2 &= 0 \\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
how to find all functions such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$ Find all function $f:\mathbb R\to\mathbb R$ such that $f\left( x^{2} - y^{2} \right) = ( x - y )( f(x) + f(y) )$.
My try:
If $ x=y=0$ then $f(0)=0$ and
if $x\leftarrow\frac{x+1}{2}$ and $y\leftarrow \frac{x-1}{2}$, then
$f(... | Let's investigate the special case that $f$ is continuous.
Let $k=f(1)$ and
$$S=\{\,t\in\Bbb R\mid f(t)=kt\,\}. $$
Trying $x=y=0$ we find $f(0)=0$. Then with $y=-x$ we find that $f(x)=-f(-x)$, i.e., $f$ is odd. Then also $S=-S$. So far we have $\{-1,0,1\}\subseteq S$.
With $y=0$, we have
$$\tag1f(x^2)=xf(x)$$
hence
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
To obtain the condition for vanishing of the given determinant If $a,b,c$ are distinct real numbers obtain the condition under which the following determinant vanishes.
$$\left|
\begin{array}{cc}
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{array}
\right|$$
My answer: After a little calculation I was ab... | There is a pretty easy way to obtain the determinant:
First, use the linearity of the determinant to split up the matrix: $$\det (A)=
\begin{vmatrix}
a & a^2 & 1+a^3\\
b & b^2 & 1+b^3\\
c & c^2 & 1+c^3\\
\end{vmatrix}
=\begin{vmatrix}
a & a^2 & 1\\
b & b^2 & 1\\
c & c^2 & 1\\
\end{vmatrix}+\begin{vmatrix}
a & a^2 & a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| Since $2 \equiv -1 \pmod{3}$, therefore $2^{k} \equiv (-1)^k \pmod{3}$. When $k$ is odd this becomes $2^k \equiv -1 \pmod{3}$. Thus $2^k+1 \equiv 0 \pmod{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 7
} |
Find the value of $ \int_0^{\infty} \frac{2x^2-1}{4x^4+1}\,{dx} $
Find $ \displaystyle \int_0^{\infty} \frac{2x^2-1}{4x^4+1}\,{dx}
$
I can find the primitive, but the fact that the answer is $0$ made me suspect there might be a way of getting answer without finding the primitive. I tried letting $x = 1/y$ but it did... | Instead of substituting $x = \dfrac{1}{y}$, try substituting $x = \dfrac{1}{2y}$.
This gives you:
\begin{align*}I &= \displaystyle\int_{0}^{\infty}\dfrac{2x^2-1}{4x^4+1}\,dx \\ &= \displaystyle\int_{\infty}^{0}\dfrac{\tfrac{1}{2y^2}-1}{\tfrac{1}{4y^4}+1} \cdot -\dfrac{1}{2y^2}\,dy \\ &= \displaystyle\int_{0}^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Is there a systematic way to solve in $\bf Z$: $x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$ for all $n$? Is there a systematic way to solve in $\bf Z$ $$x_1^2+x_2^3+...+x_{n}^{n+1}=z^{n+2}$$ For all $n$?
It's evident that $\vec 0$ is a solution for all $n$.
But finding more solutions becomes harder even for small $n$:
When $n... | If $n+2$ is composite, say $n+2=mk$ with $m,k>1$, take any $z$, $x_m = z^k$, all other $x_j = 0$.
EDIT:
Suppose we have a solution $(a_1, \ldots, a_{n+1})$ in positive integers with
$a_1^{d_1} + \ldots + a_n^{d_n} = a_{n+1}^{d_{n+1}}$, and positive integer $k$ so that $d_i+k$ are pairwise coprime. By the Chinese Rem... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1619089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
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.