Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluate surface integral over unit sphere
Evaluate $\iint (ax^2+by^2+cz^2)dS$, where $S$ is surface of unit sphere $x^2+y^2+z^2 = 1$.
My attempt. Using Gauss divergence theorem, $F.n = (ax^2+by^2+cz^2) \implies F=\langle ax,by,cz\rangle$
$\text{div} F = (a+b+c)$. by gauss divergence theorem, $$\iint_S F.n ds = \iiin... | Check your last evaluation. Note that
$$\int_{\phi=0}^{\pi}\int_{\theta=0}^{2\pi} \sin^3 \phi \cos^2 \theta d\theta d\phi=\int_{\phi=0}^{\pi} \sin^3 \phi d\phi\int_{\theta=0}^{2\pi}\cos^2 \theta d\theta = \frac{4}{3}\cdot \pi$$
(the same result if you replace $\cos^2 \theta$ with $\sin^2 \theta$) and
$$\int_{\phi=0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2509291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What are the possible prime factors of $3^n+2$ , where $n$ is a positive integer?
What are the possible prime factors of $3^n+2$, where $n$ is a positive integer ?
It is clear that a prime $p$ for which neither $-2$ nor $-6$ is a quadratic residue modulo $p$, cannot be a prime factor of $3^n+2$, so the primes congrue... | Just an observation not a complete answer.
For a prime $p>3$, applying Fermat's little theorem
$$2^p \equiv 2 \pmod{p}$$
and if there is a minimal $n_0 \in \mathbb{N}$ such that ($p$ - odd!):
$$3^{n_0} \equiv -2 \pmod{p} \Rightarrow 3^{n_0\cdot p} \equiv (-2)^p \equiv -2 \pmod{p}$$
Also $n_0<p-1$, otherwise $n_0=(p-1)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2512025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
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Find the Fourier Series for $f(x)=$ { $1$ from $-\frac{\pi}{2} < x < \frac{\pi}{2}$ , $-1$ from $\frac{\pi}{2} < x < \frac{3\pi}{2}$ } I'm trying to find the Fourier Series for the following function, but I'm having trouble at some point. I'm hoping someone can give me a hand...
$
f(x) =
\begin{cases}
1 \t... | Thanks to MyGlasses I was able to finish the exercise. I'll leave the full answer here for future reference.
$
f(x) =
\begin{cases}
1 \text{,} & \text{if }\quad -\frac{\pi}{2} < x < \frac{\pi}{2}\\
-1 \text{,} & \text{if }\quad \frac{\pi}{2} < x < \frac{3\pi}{2}\\
\end{cases}
$
$f(x) = a_0 + \s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For Which $a$ And $b$ Above $\mathbb{Z}_5$ There Are Solutions?
For which values of $a$ and $b$ above $\mathbb{Z_5}$ the following equations have no solution/one solution, infinite solutions
\begin{cases}
ax+4y+3z=0 \\
2y+3z=1\\
3x-bz=3
\end{cases}
So the matrix is
$$
\left[\begin{array}{rrr|r}
a & 4 & 3 &... | from the equation (III) we get
$$x=1+\frac{b}{3}z$$
plugg in (I)
$$a+z\left(\frac{ab}{3}+3\right)+4y=0$$
from (II) we have
$$4y=2-6z$$ and we obtain
$$z\left(\frac{ab}{3}-3\right)=-(2+a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A box contains a penny, two nickels, and a dime. If two coins are selected randomly from the box, without replacement, and if X is the sum... A box contains a penny (1¢), two nickels (5¢), and a dime (10¢). If two coins are selected randomly from the box, without replacement, and if $X$ is the sum of the values of the ... | A good notation for the $CDF$ would be
$$ F_{X}(x)=
\begin{cases}
1 & x \geq 15 \\
\frac{4}{6} & 11 \leq x \lt 15 \\
\frac{3}{6} & 10 \leq x \lt 11 \\
\frac{2}{6} & 6 \leq x \lt 10 \\
0 & x \lt 6
\end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2517175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $a^3+b^3=1$ and $(a+b)(a+1)(b+1)=2$, find the value of $(a+b)$
Given: $(a,b)\subset \mathbb R^2$, $a^3+b^3=1$ and $(a+b)(a+1)(b+1)=2$.
Find: The value of $(a+b)$
Question on the Brazilian Math Olympic (OBM), level 2, phase 3, 2012. No answer provided.
By inspection I can easily see that $(a,b)=(1,0)$, $(a,b)=(0... | $(a+b)^3 = a^3+b^3+3a^2b+3ab^2$
$
2=(a+b)(a+1)(b+1)
=(a^2 b + a b^2) + (a+b)^2+ (a+b)
$
Let $u=a+b$ and $v=a^2 b + a b^2$.
Then $u^3=1+3v, 2=v+u^2+u$ and so
$u^3 + 3 u^2 + 3 u - 7=0$.
Now, $u^3 + 3 u^2 + 3 u - 7 = (u+1)^3-8$. Therefore, $u+1=2$ and $u=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2518270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If polynomial is divisible by quadratic, find values of $a$ and $b$ Equation is $z^4+(a+b)z^3+4az^2+(a+b+32)z+45$ which is divisible by $z^2+6z+9$.
We're meant to find values of $a$ and $b$ and then solve the equation itself!
I figured $z^2+6z+9=(z+3)^2$, but was wondering if this automatically implies that it is a dou... | use that $${z}^{2}+ \left( a+b-6 \right) z-2\,a-6\,b+27+{\frac {4\,a+28\,b-76}{z+
3}}+{\frac {6\,a-30\,b+30}{ \left( z+3 \right) ^{2}}}=\frac{z^4+(a+b)z^3+4az^2+(a+b+32)z+45}{z^2+6z+9}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2519646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to solve this limit using Taylor formula? I have difficulties in solving this limit using Taylor formula.
$$ \lim_{x \rightarrow+ \infty} \big((x+a)^{1+\frac {1}{x}} - x^{1+\frac{1}{x+a}}\big) $$
Can you give me some advice? I will be very thankful for your help.
| Let $a = bx$,
$(x+a)^{1+\frac {1}{x}} - x^{1+\frac{1}{x+a}}=((b+1)x)^{1+\frac{1}{x}}-x^{1+\frac{1}{(1+b)x}}$
fixing $x$, expand at $b=0$
$x^{1+\frac{1}{x}}(1+b+o(b^2))-(x^{1+\frac{1}{x}}-\log(x)x^{1+\frac{1}{x}}\frac{1}{x}b+o(b^2))=x^{1+\frac{1}{x}}(1+\frac{\log(x)}{x})b+o(b^2)=ax^{\frac{1}{x}}(1+\frac{\log(x)}{x})+o(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2520343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Sum of infinite sequence Let $$T_r=\frac{rx}{(1-x)(1-2x)(1-3x)\cdots(1-rx)}$$
Can someone please tell me how to break this expression into partial fractions (because I am a bit weak at it) to find the following $$\sum_{r=2}^\infty T_r$$
| This sum is telescopic with a different decomposition as follows:
$$\begin{align}
T_r&=\frac{rx}{\prod\limits_{k=1}^r1-kx}=\frac{1}{\prod\limits_{k=1}^{r}1-kx}-\frac{1-rx}{\prod\limits_{k=1}^{r}1-kx}=\frac{1}{\prod\limits_{k=1}^{r}1-kx}-\frac{1}{\prod\limits_{k=1}^{r-1}1-kx}
\end{align}$$
It should be fairly clear that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2521841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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An atypical inequality If $a, b, c \in (0, \infty)$ with $abc=1$ prove: $$a^2+b^2+c^2+3(ab+bc+ca)\leq\frac{1}{2}(\sqrt{a}+\sqrt{b}+\sqrt{c})(a+b)(b+c)(c+a).$$
On atypical inequality (a smaller amount than a product) that resisted my attempts to prove it.
| We need to prove that
$$\prod_{cyc}(a^2+b^2)\sum_{cyc}a\geq2abc\sum_{cyc}(a^4+3a^2b^2)$$ for positives $a$, $b$ and $c$ or
$$\sum_{cyc}\left(a^4b^2+a^4c^2+\frac{2}{3}a^2b^2c^2\right)\sum_{cyc}a\geq2\sum_{cyc}(a^5bc+3a^3b^3c)$$ or
$$\sum_{cyc}(a^5b^2+a^5c^2+a^4b^3+a^4c^3+a^4b^2c+a^4c^2b+2a^3b^2c^2-2a^5bc-6a^3b^3c)\geq0,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Locating the third vertex of an equilateral triangle Two vertices of an equilateral triangle are at $A=(10,-4)$ and $B=(0,6)$.
How can one locate the third vertex?
Maybe someone could give me the easy way please.
My attempt:
*
*Find the average of $M=(x , y)$ of $A$ and $B$, which is $(10+0)/2 = 5$, $(-4+6)/2 = 1... | First, find the length of a side of the triangle. $$l=\sqrt{(10-0)^2 +(-4-6)^2}=10\sqrt{2}$$
Next, we draw two circles of radius $l$ and centers at the given points. The intersection of these circles will give us the third point (well, there will be two points so two solutions to this problem). We have two equations:
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2523286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solving $3^x = 2^y + 1$ with $x,y \in \mathbb{N}^2$ I consider the following equation
$$ 3^x = 1 + 2^y \tag{$\star$} $$
with $(x,y) \in \mathbb{N}^2$ and $y \geq 3$.
I would like to show that :
$$ 3^x \equiv 1 \; [2^y] \; \Leftrightarrow \; 2^{y-2} \mid x. $$
I assume that $3^x \equiv 1 \; [2^y]$. Writing
$$ 3^x = \... | First of all $x$ can't bigger than $y$, as then obviously $3^x > 1 + 2^y$.
Another way to solve the question is to note that if $y \ge 2$ then $x$ is even, because $4 \mid 2^y = 3^x - 1$. Therefeore if $x=2k$ we have:
$$2^y = 3^{2k} - 1 = (3^k - 1)(3^k + 1)$$
But now $\gcd(3^k - 1, 3^k + 1) = 2$ and from the above equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2528986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^... | You need to show that $\lim\limits_{n\to\infty}\left(n-n\sqrt{1+\frac1{n^4}}\right)=0$. You cannot simply use that $\lim\limits_{n\to\infty}\sqrt{1+\frac1{n^4}}=1$ since $\lim\limits_{n\to\infty}\sqrt{1+\frac1n}=1$, yet $\lim\limits_{n\to\infty}\left(n-n\sqrt{1+\frac1n}\right)=-\frac12$.
If you are looking for an alte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2531365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find limit $\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$
Find the following limit without using L'Hopital's rule:
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}$$
My attempt:
$$x-1=u\implies x=u+1$$
So we have:
$$\lim_{u\to 0} \dfrac{4(u+1)^2\sqrt{u+4}-17(u+1)+9}{(u)^2}$$
What now?
| Let $\sqrt{x+3}-2=y\implies x=y^2+4y+1$
$$\lim_{x\to 1} \dfrac{4x^2\sqrt{x+3}-17x+9}{(x-1)^2}=\lim_{y\to0}\dfrac{4(1+4y+y^2)^2(y+2)-17(1+4y+y^2)+9}{y^2(y+4)^2}$$
Now,
$$4(1+4y+y^2)^2(y+2)-17(1+4y+y^2)+9$$
$$=(8+4y)(1+8y+16y^2+2y^2+8y^3+y^4)-17(1+4y+y^2)+9$$
$$=(8-17+9)+y(64+4-68)+y^2(8\cdot18+32-17)+O(y^3)=159y^2+O(y^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2532738",
"timestamp": "2023-03-29T00:00:00",
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How do I rationalize the following fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$? As the title says I need to rationalize the fraction: $\frac{1}{9\sqrt[3]{9}-3\sqrt[3]{3}-27}$. I wrote the denominator as: $\sqrt[3]{9^4}-\sqrt[3]{9^2}-3^3$ but I do not know what to do after. Can you help me?
| Let $x=3^\frac13$ then the fraction becomes
$$-\frac{1}{x^4}\cdot\frac{1}{x^5-x^4+1}$$
What we need is to multiply nominator and denominator with some polynomial of $x$ in order to leave only terms with exponents divisible by $3$ in the denominator. How can we find such polynomial?
The answer is simple: find remained w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2535817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Finding function $f(x)$ which satisfy given functional equation
Find all function $f:\mathbb{R}-\{0,1\}$ in $$f(x)+2f\left(\frac{1}{x}\right)+3f\left(\frac{x}{x-1}\right)=x$$
Attempt: put $\displaystyle x = \frac{1}{x}$, then $$f\left(\frac{1}{x}\right)+2f(x)+3f\left(\frac{1}{1-x}\right) = \frac{1}{x}$$
could some he... | Subtract twice from what you get from the given equation to get:
$$-f(x) + f\left(\frac{x}{x-1}\right) - 2f\left(\frac{1}{1-x} \right) = \frac{x^2 - 2}{3x}$$
Then in this equation plug $x= \frac {x}{x-1}$ to get:
$$-f\left(\frac {x}{x-1}\right) + f\left(x \right) - 2f\left(1-x\right) = \frac{-x^2+4x-2}{3x(x-1)}$$
Addi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2536550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Convergence of $ \sum_{k=1}^{\infty} \frac {3^k}{5^k + 1}$
Show the convergence of the following series:
$$\sum_{k=1}^{\infty}\frac {3^k}{5^k + 1}$$
*
*a) Show the monotony of the partial sums
*b) estimate upwards
*c) remember the geometric series (I do not know how to use that here.)
The following... | We know that a geometric series $\displaystyle \sum_{k=0}^{\infty} r^k$ converges $\iff$ $|r|<1$.
Let $b_k=\left(\dfrac 35 \right)^k$.
For every $k \ge 0$, we have $0<a_k \le b_k$. Thus if $\displaystyle \sum_{k=0}^{\infty} b_k$ converges, by the comparison test $\displaystyle \sum_{k=0}^{\infty}a_k$ converges too. A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2538022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve over integers $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} = 3$ I am trying to solve $\frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} = 3$ over integers, but I have no idea what's the best way to do it. I have tried multiplying both sides by $xyz$ and then figuring out that $z$ divides $xy$, $x$ divides $yz$, etc, but ... | It appears there are no non-trivial (i.e. not $x,y,z = \pm 1$) solutions: sorry, no number theory to do here. (Just some analysis.) Dividing everything by $xyz$ puts the equation in a nice form:
$\frac{1}{x^2} + \frac{1}{y^2}+\frac{1}{z^2} = \frac{3}{xyz}$.
Staring at that for a moment, it seems like the LHS is always... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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$\sin3x-\sin2x-\sin x=0$ I have issue solving this equation.
So I wrote $$\sin3x= 3\sin x-4\sin^3x$$ and $$\sin2x = 2\sin x\cos x$$
So we have $$3\sin x-4\sin^3x-(2\sin x\cos x)-\sin x = 0$$
But now I have $$\sin x$$ and $$\cos x$$ as unknown, and I don't know how to finish this.
| $$\sin { 3x-\sin { 2x } -\sin { x } =0 } \\ \left( \sin { 3x-\sin { x } } \right) -\sin { 2x } =0\\ 2\sin { \frac { 3x-x }{ 2 } \cos { \frac { 3x+x }{ 2 } } -\sin { 2x } } =0\\ 2\sin { x\cos { 2x-2\sin { x\cos { x } } =0 } } \\ \sin { x } \left( \cos { 2x-\cos { x } } \right) =0\\ \sin { x } =0,\cos { 2x-\cos {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2540041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Generating fuctions: $\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}\binom{2n}{n}$ I'm trying to solve one of my combinatorics exercise but I struggle a bit.
Is the equality correct for all the $n\ge 0$?
$$\sum_{k=0}^\infty \binom{2k}{k}\binom{n}{k}\left(-\frac{1}{4}\right)^k=2^{-2n}... | You may consider that
$$ \frac{1}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sin^{2k}(\theta)\,d\theta \tag{A}$$
from which it follows that:
$$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{4^k}\binom{2k}{k}=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k=0}^{n}\binom{n}{k}(-\sin^2\theta)^k =\frac{2}{\pi}\int_{0}^{\pi/2}\cos^{2n}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Integral with a fixed parameter For a parameter $a\ge 1$
How to calculate
$$\int_0^1 \frac{2\sqrt2}{(\sqrt2+1)(1-u^2)+2au}\ du$$?
I note $p(u) = (\sqrt{2}+1)(1-u^2)+2au$
The roots of $p$ are
$$u=\frac{-2a-2\sqrt{a^2+(\sqrt2+1)^2}}{-2(\sqrt2+1)}=\frac{-a-\sqrt{a^2+(\sqrt2+1)^2}}{-(\sqrt2+1)}=\frac{-a-\sqrt{a^2+2\sqr... | Some Preliminary Steps:
$\require{amsmath}$
$\DeclareMathOperator\arctanh{arctanh}$
$$\begin{aligned}
&(\sqrt2+1)(1-u^2)+2au=-(\sqrt2+1)\left(u^2-\frac{2au}{\sqrt2+1}-1\right)=\\&-(\sqrt2+1)\left(u^2-\frac{2au}{\sqrt2+1}+\frac{a^2}{(\sqrt2+1)^2}-\left(1+\frac{a^2}{(\sqrt2+1)^2}\right)\right)=\\&-(\sqrt2+1)\left(\left(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Solve $|x^2+6x+8|=|x^2+4x+5|+|2x+3|$ Question : Find solution for $x \in \Bbb R$, $|x^2+6x+8|=|x^2+4x+5|+|2x+3|$
I considered 8 different cases and arrived at the answer $\big [ \frac{-3}{2}, \infty)$
I dont know if its correct. Also, considering 8 different cases is tedious. Is there any other method to solve problems... | The equation is piecewise quadratic and the pieces are delimited by the zeros of the arguments of the absolute values.
We have
$$|x^2+6x+8|=\begin{cases}x\le-4\lor x\ge-2&\to x^2+6x+8\\-4\le x\le-2&\to-(x^2+6x+8),\end{cases}$$
$$|x^2+4x+5|=x^2+4x+5,$$
$$|2x+3|=\begin{cases}x\le-\dfrac32&\to-(2x+3)\\x\ge-\dfrac32&\to2x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the real solutions for the system: $ x^3+y^3=1$,$x^2y+2xy^2+y^3=2.$
Find the real solutions for the system:
$$\left\{
\begin{array}{l}
x^3+y^3=1\\
x^2y+2xy^2+y^3=2\\
\end{array}
\right.
$$
From a book with exercises for math contests. The solutions provided are: $(x,y)=(\dfrac{1}{\sqrt[3]{2}},\dfrac{1}{\sqr... | Hint:
Observe that $xy\ne0$
Set $y=mx$
divide the resultant one equation by the other to form a cubic equation in $m$ with $m+1\ne0$ being one factor
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2544262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to show $x^{10}-2x^9+3x^8-...-10x+11=0$ has no real root? I tried to solve below equation $$x^{10}-2x^9+3x^8-...-10x+11=0$$ I plot the graph and see there is no real root , But I get stuck how to show analytical .Can some one help me or give an idea ?
Thanks in advance.
This is graph of the function $$f(x)=x^{10}-... | There is some confusion over the sign of the coefficients. But it might prove useful to consider $$x^8(x-1)^2+2x^6(x-1)^2+3x^4(x-1)^2+4x^2(x-1)^2 =$$$$=x^{10}-2x^9+x^8+2x^8-4x^7+2x^6+3x^6-6x^5+3x^4+4x^4-8x^3+4x^2 =$$$$=x^{10}-2x^9+3x^8-4x^7+5x^6-6x^5+7x^4-8x^3+4x^2$$which is clearly non-negative, or some variant of thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2547766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Exercise on elementary number theory Let $a,b,c,d$ integers, $c\not=0$ such that $ad-bc=1$ and $c\equiv 0 \pmod p$ for some prime $p>3$.
Show that if $a+d=\pm1$ then $p\equiv 1\pmod 3$
I don't know how to approach this problem because when I take the expression $ad-bc$ modulo $p$ we have that either $d-d^2\equiv 1\pm... | Given $ad-bc=1$ with $c\equiv 0 \pmod p$ for $p$ a prime exceeding $3$. If in addition we have $a+d = \pm 1$ then using $a = \pm 1 - d$ in $ad-bc=1$
$$
1 \equiv ad - bc \equiv (\pm 1 - d)d - 0 \equiv \pm d -d^2
\quad \text{or}\quad d^2 \equiv \pm d -1 \pmod p
$$
$$
d^3 \equiv \pm d^2-d \equiv \pm (\pm d - 1)-d \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Coefficient of $x^3$ in $(1-2x+3x^2-4x^3)^{1/2}$ The question is to find out the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
I tried using multinomial theorem but here the exponent is a fraction and I couldn't get how to proceed.Any ideas?
| Hint:
The coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3)^{1/2}$
$=$
the coefficient of $x^3$ in the expansion of $(1-2x+3x^2-4x^3+\cdots)^{1/2}$
Using Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $,
$$1-2x+3x^2-4... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of the series of $\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 9}+...$ Find the sum of the series
$$\frac1{1\cdot3}+\frac1{3\cdot5}+\frac1{5\cdot7}+\frac1{7\cdot9}+\frac1{9\cdot11}+\cdots$$
My attempt solution:
$$\frac13\cdot\left(1+\frac15\right)+\frac17\cdot\left(\frac15+\frac... | This is a general approach to evaluate the sum of series, like these.
First find $n^{th}$ term of series.
Let $T_n$ denote the $n^{th}$ term.
We see that,
$T_1 = \frac{1}{\color{green}{1} \cdot \color{teal}{3}} $
$T_2 = \frac{1}{\color{green}{3} \cdot \color{teal}{5}} $
And so on.
Let the numbers in $\color{green}{gree... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2556569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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How to solve a matrix equation of type $XAX=B$ This is the question:
Find out the matrix $X\in M_2(\mathbb R)$, so that
$X*A*X=B$, where $A,B \in M_2(\mathbb R)$ and $A=(a_{ij}), a_{11}=1, a_{12}=2, a_{21}=1, a_{22}=5$ and $B=(b_{ij}), b_{11}=2, b_{12}=-2, b_{21}=-2, b_{22}=2$.
How can I solve it and what is\are ... | Taking the determinant of both sides of relationship $XAX=B$, we find
$\det(X)^2 \det(A)=\det(B)=0$. As $\det(A) \neq 0$. We must have $\det(X)=0$ which is equivalent to the fact that the 2 columns of $X$ are proportional.
We can assume thus, WLOG, that $X$ can be written :
$$X=\left(
\begin{array}{rr}
a&ca \\
b&cb
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Evaluating a tricky limit I'm trying to calculate the radius of convergence for a power series and I'm stuck at the following step:
$$\lim_{n\to \infty}\Bigg\lvert\left(\frac{3^n+\left(-4\right)^n}{5}\right)^{\frac{1}{n}}\Bigg\rvert$$
How would you evaluate this limit?
| For $n$ odd, we see that
\begin{align*}
\left(\dfrac{4^{2k+1}-3^{2k+1}}{5}\right)^{1/(2k+1)}&=\dfrac{3}{5^{1/(2k+1)}}\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)},
\end{align*}
where
\begin{align*}
\lim_{k\rightarrow\infty}\log\left(\left(\dfrac{4}{3}\right)^{2k+1}-1\right)^{1/(2k+1)}&=\lim_{k\rightarrow\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prime and primary ideals in $\mathbb{Z}[\sqrt{5}]$
Let $R=\mathbb{Z}[\sqrt{5}]$. The ideal $(2, 1-\sqrt{5})$ is prime in $R$, right (as $R/(2, 1-\sqrt{5})=\mathbb{F}_2$)?
1. Is then $(2^n, 1-\sqrt{5})$ primary for some $n\geq2$?
2. Are $(2)$ and $(3)$ prime ideals and $(2^n)$ and $(3^k)$ primary ideals in $R$?
I ... | $P=(2,1-\sqrt{5})$ is a prime:
First show it is not $R$. Suppose $1=2(a+b\sqrt{5})+(1-\sqrt{5})(c+d\sqrt{5})$ for some $a,b,c,d\in \mathbb{Z}$. Simplify we have that
$$
2a+c-5d=1, \quad 2b-c+d=0,
$$
add these two we have $2a+2b-4d=1$, contradiction. (By this argument, we can show that $2\notin (4,1-\sqrt{5})$, so $(2,1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2561960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Image and kernel of a linear transformation I have a linear transformation:
$$f:\mathbb{R}^3 \rightarrow \mathbb{R}[x]_2$$
$$f((1,1,1))=2x^2-3x,\ f((1,2,3))=-3x,\ f((1,2,4))=2x^2-4x$$
I need to find kernel, image, dimension of image and dimension of kernel.
I tried to find formula describing this transformation:
$f((a,... | Your expression for $f$ is correct. Here is a slightly easier way to get it:
$$f(0,0,1) = f(1,2,4) - f(1,2,3) = 2x^2 - x$$
$$f(0,1,0) = f(1,2,3) - f(1,1,1)-2\cdot f(0,0,1) = -6x^2 + 2x$$
$$f(0,0,1) = f(1,1,1) - f(0,1,0) - f(0,0,1) = 6x^2 - 4x$$
So
$$f(a,b,c) = a\cdot f(1,0,0) + b\cdot f(0,1,0) + c\cdot f(0,0,1) = (6a-6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solution of $\int\frac{\sqrt {x^3-4}}{x}dx$ : Need Hints $$\int\frac{\sqrt {x^3-4}} x \, dx$$
My attempt: $ \displaystyle \int\frac{3x^2\sqrt {x^3-4}}{3x^3}\,dx$
Then, substituting $u=x^3$; $\displaystyle \int\frac{\sqrt {u-4}}{3u} \, du$
$$\int\frac{u-4}{3u\sqrt{u-4}} \, du$$
$$\int\frac{1}{3\sqrt{u-4}}\,du-4\int\fra... | \begin{align}
w & = \sqrt{u-4} \\
w^2 & = u-4 \\
2w\,dw & = du \\
w^2+4 & = u\\[15pt]
\int \frac{\sqrt {u-4}}{3u} \, du & = \int\frac{w}{3(w^2+4)} (2w\,dw) = \frac 2 3 \int \left( 1 - \frac 4 {w^2+4} \right) dw
\end{align}
etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2562597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Find : $\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$ with l'Hôpital's rules I want to calculate :
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right)$$
with l'Hôpital's rules.
I get
$$\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim\l... | As it turns out, you just have to use L'Hospital's rule twice and observe that on the second try the denominator is no longer zero when you use directly substitution:
\begin{align}
\lim\limits_{x \rightarrow 0}\left(\frac{1}{\sin x} - \frac{1}{x}\right)
&=\lim\limits_{x \rightarrow 0}\frac{x -\sin x}{x\sin x}\\
&=\lim\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
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Use Viete's relations to prove the roots of the equation $x^3+ax+b=0$ satisfy $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$ Use Viete's relations to prove that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3+ax+b=0$ satisfy the identity $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$.
I know that viete's relat... | Let's first prove Vieta's cubic relations.
$\underline {\text{Proof}}$: Let the roots of a cubic polynomial, $f(x)$, be $\alpha$, $\beta$, $\gamma$. Then $f(x) = (x-\alpha)(x-\beta)(x-\gamma)$. Let $f(x) = x^3 - px^2 + qx - r$.
\begin{align}
f(x) & = (x-\alpha)(x-\beta)(x-\gamma) \\
& = (x^2 - \beta x - \alpha x + \alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2564158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Convergence and divergence of an infinite series The series is
$$1 + \frac{1}{2}.\frac{x^2}{4} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6}.\frac{x^4}{8} + \frac{1\cdot3\cdot5\cdot7\cdot9}{2\cdot4\cdot6\cdot8\cdot10}.\frac{x^6}{12}+... , x\gt0$$
I just stuck over the nth term finding and once I get nth term than I can do diff... | Notice
$$
\begin{align}
\frac{1}{2\cdot 4}x^2 &= \frac{2!}{(2^1 1!)^2(2 \cdot 2)} x^2
= \frac12 \binom{2}{1} \int_0^x \left(\frac{t}{4}\right)^1 dt
\\
\frac{1\cdot 3\cdot 5}{2\cdot 4 \cdot 6 \cdot 8}x^4
&= \frac{6!}{(2^3 3!)^2 (2\cdot 4)} x^4
= \frac12 \binom{6}{3} \int_0^x \left(\frac{t}{4}\right)^3 dt\\
\frac{1\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find all real solutions for the system: $x^3=y+y^5$, $y^5=z+z^7$, $z^7=x+x^3.$
Given: $$\left\{
\begin{array}{l}
x^3=y+y^5\\
y^5=z+z^7\\
z^7=x+x^3
\end{array}
\right.
$$
Find: all real solutions for the system.
From a book on preparation for math contests. The answer states there is just one solution. My problem... | Multiplying the three equations gives $x^3 y^5 z^7 = x y z (1+x^2)(1+y^4)(1+z^6)\,$.
If $xyz \ne 0$ then it follows that $x^2 y^4 z^6 = (1+x^2)(1+y^4)(1+z^6)\,$, but the latter is not possible since $0 \le x^2 \lt 1+ x^2, 0 \le y^4 \lt 1+y^4, 0 \le z^6 \lt 1+ z^6\,$.
Therefore $xyz=0\,$, and it is easy to show that any... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Polynomial long division - intermediate steps missing I'm trying to find the intermediate step:
$$\frac{1}{(x-\frac{x^2}{2}+\frac{x^3}{12}-\frac{x^4}{144}+...)^2} = \frac{1}{x^2}+\frac{1}{x}+\frac{7}{12}+\frac{19x}{72} ...$$
Is there a quick way to find these first few terms that I'm missing here? Thanks
| The first terms of the doninator of the LHS are$$x^2-x^3+\frac{5 x^4}{12}-\frac{7 x^5}{72}+\cdots=x^2\left(1-x+\frac{5 x^2}{12}-\frac{7 x^3}{72}+\cdots\right).$$Therefore, the RHS can be written as$$\frac1{x^2}\left(b_0+b_1x+b_2x^2+b_3x^3+\cdots\right)$$and we must have$$\left(1-x+\frac{5 x^2}{12}-\frac{7 x^3}{72}+\cdo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do you go about factoring $x^3-2x-4$? How do you factor $x^3-2x-4$?
To me, this polynomial seems unfactorable.
But I check my textbook answer key, and the answer is $(x-2)(x^2+2x+2)$.
So I got to solve backward:
$$x(x^2-2)-4$$
And add some terms and subtract them later,
$$x(x^2+2x+2-2x-4)-4=0$$
$$x(x^2+2x+2)-4-2x^... | The way I factor
$x^3 - 2x - 4 \tag 1$
is to notice that $2$ is a root; this I do by intuition and a good amount of experience. $2$ looks like a good guess to me since the non-leading coefficients are divisible by $2$; once I find that $2$ is a zero of (1), I use synthetic division to find $q(x)$ such that
$x^3 - 2x -... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sums of $5$th and $7$th powers of natural numbers: $\sum\limits_{i=1}^n i^5+i^7=2\left( \sum\limits_{i=1}^ni\right)^4$? Consider the following:
$$(1^5+2^5)+(1^7+2^7)=2(1+2)^4$$
$$(1^5+2^5+3^5)+(1^7+2^7+3^7)=2(1+2+3)^4$$
$$(1^5+2^5+3^5+4^5)+(1^7+2^7+3^7+4^7)=2(1+2+3+4)^4$$
In General is it true for further increase i.e.... | Both sides are polynomials in $n$ of degree $8$. Since they coincide for $n=0,\dots,8$, they are equal.
Any $9$ points will do. Taking $n=-4,\dots,4$ is probably easier to do by hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2568157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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If $\log_{10}15=a$ and $\log_{20}50=b$, express $\log_940$ in terms of $a$ and $b$. I've done this so far: $$\log_940=\frac{\log_{10}40}{\log_{10}9}=\frac{1+\log_{10}4}{a+\log_{10}6-1}.$$ How do I proceed?
| Well, this will be little messy but here is an attempt:
$$b = \frac{\log_{10}50}{\log_{10}20} = \frac{1+\log_{10}5}{1+\log_{10}2} = \frac{2-\log_{10}2}{1+\log_{10}2} \implies \log_{10}2 = \frac{2-b}{1+b}$$
and
$$a = \log_{10}15 = \log_{10}3+\log_{10}5 = \log_{10}3 + 1 - \log_{10}2 \Rightarrow \log_{10}3 = a+\log_{10}2-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let$A$ be a $3\times3$ real symmetric matrix such that $A^6=I$ . Then $A^2=I$ Let$A$ be a $3\times3$ real symmetric matrix such $ A^6=I$ . Then $A^2=I$.
How can I prove this statement is true or false?
As it is given $A$ is symmetric so $A=A^T$. Also $ A^6=I$.
But the main problem is that I can't operate $A^{-1}$ on bo... | $A^6=I$ gives $x^6-1$ $=(x^2-1)(x^2-x+1)(x^2+x+1)$ is an annihilating polynomial of $A$. Let $m(x)$ be the minimal polynomial of $A$ then $m(x)$ divides $x^6-1$.
Since all eigenvalues of a real symmetric matrix are real and $x^2\pm x+1$ does not contain any real root, $m(x)$ cannot contain any factor of $x^2-x+1$ or $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Compare sum of radicals I am stuck in a difficult question:
Compare $18$ and
$$
A=\sqrt{7}+\sqrt{11}+\sqrt{32}+\sqrt{40}
$$
without using calculator.
Thank you for all solution.
| $\sqrt{7}$ and $\sqrt{11}$ can be written as $\sqrt{9\pm 2}$ and similarly $\sqrt{32}$ and $\sqrt{40}$ can be written as $\sqrt{36\pm 4}$.
Since $\sqrt{x}$ is a concave function on $\mathbb{R}^+$,
$$ \sqrt{9-2}+\sqrt{9+2}+\sqrt{36-4}+\sqrt{36+4} \color{red}{\leq} 2\sqrt{9}+2\sqrt{36} = 18.$$
We may also estimate the di... | {
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"url": "https://math.stackexchange.com/questions/2573332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $x$ given $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$ If $\sqrt{x+14-8\sqrt{x-2}}$ + $\sqrt{x+23-10\sqrt{x-2}} = 3$, then what is the value of $x$?
Is there an easy way to solve such equation, instead of squaring on both sides and replacing $\sqrt{x-2}$ with a different variable?
| Condition: $x\geq 2$.
You have $$LHS = \sqrt{(\sqrt{x-2}-4)^2} + \sqrt{(\sqrt{x-2}-5)^2}.$$
So, if $\sqrt{x-2} < 4$, or $x<18$, then $$LHS = 4-\sqrt{x-2} + 5-\sqrt{x-2} = 9-2\sqrt{x-2} = 3 = RHS.$$
If $4 \leq \sqrt{x-2} < 5$, or $18\leq x < 27$, then $$LHS = \sqrt{x-2} -4 + 5-\sqrt{x-2} = 1 \neq 3 = RHS.$$
If $\sqrt{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Proving whether the series $\sum_{n=1}^\infty \frac{(-1)^n}{n-(-1)^n}$ converges. I've updated my proof to be complete now, edited for proof-verification!
We know that for the partial sums with even an uneven terms, the following holds:
$S_{2N}=\sum_{n=1}^{2N} \frac{(-1)^n}{n-(-1)^n} = -\frac{1}{2} + \frac{1}{1} -\fra... | Let:
$$a_n=\frac{1}{n-(-1)^n}$$
Note that:
*
*for $n=2k-1 \implies a_{2k-1}=\frac{1}{2k}$
*for $n=2k \implies a_{2k}=\frac{1}{2k-1}$
*for $n=2k+1 \implies a_{2k+1}=\frac{1}{2k+2}$
*for $n=2k+2 \implies a_{2k+2}=\frac{1}{2k+1}$
Therefore we can reorder the series $a_n\to b_n$ in such way that $b_n$ is monolitica... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Hypergeometric Function Value Wolfram gives the following:
$$_2F_1 \left(a,b;\frac{a+b+1}{2};\frac{1}{2}\right) = \frac{\sqrt{\pi}
\Gamma \left(\frac{a+b+1}{2} \right)} {\Gamma \left(\frac{a+1}{2} \right) \Gamma \left(\frac{b+1}{2} \right) }$$ and I am struggling to rederive it.
My ultimate aim is to allow the final... | Here is a derivation of your formula using a transformation and the
Gauss's Hypergeometric Theorem. The quadratic transformation
http://dlmf.nist.gov/15.8.E18
$${_2}F_1 \left(a,b;\frac{a+b+1}{2};z\right)={_2}F_1 \left(\frac{a}{2},\frac{b}{2};\frac{a+b+1}{2};4z(1-z) \right)$$
gives
$${_2}F_1 \left(a,b;\frac{a+b+1}{2};\... | {
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"answer_count": 1,
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} |
Convergence/absolute convergence of $\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin \frac{1}{2n+1}\right)$
Does the following sum converge? Does it converge absolutely?
$$\sum_{n=1}^\infty \left(\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}\right)$$
I promise this is the last one for today:
Using Simpson's rules:
$$\sum... | As an alternative, since:
$$\sin \frac{1}{2n} - \sin
\frac{1}{2n+1}=\frac{1}{2n}-\frac{1}{2n+1}+o\left(\frac{1}{n^2}\right)=\frac{1}{2n(2n+1)}+o\left(\frac{1}{n^2}\right)$$
the series converges by limit comparison with:
$$\sum_{n=1}^\infty \frac{1}{2n(2n+1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
} |
Is it possible to compute $\int_{-\infty}^\infty {x \sin x\over x^4+1}dx$ without using complex analysis?
$$\int_{-\infty}^\infty {x\sin x\over x^4+1}dx$$
By using things like Feynman's trick (Differentiating under the integral sign) or other high school level calculus concepts? I do not know things like residue theo... | The same answer as @JackD'Aurizio with slightly more details. I was doing this after he sent his answer so I finish anyway.
Let $\alpha=\frac{1}{2\sqrt[]{2}}$. After partial fraction decomposition we get:
\begin{align}
\int^\infty_{-\infty}\frac{x\sin(x)}{x^4+1}\,dx=-\alpha\int^\infty_{-\infty}\frac{\sin(x)}{x^2+\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
$\max a^3+b^3+c^3+4abc$ sub $0\leq a,b,c \le 3/2$ and $a+b+c=3$ Let $S$ be the set of $(a,b,c) \in \mathbb{R}^3$ such that $0\leq a,b,c \leq \frac{3}{2}$ and $a+b+c=3$. Find
$$
\max_{(a,b,c) \in S} a^3+b^3+c^3+4abc.
$$
| I don't know a nice solution. However, the gradient of $f(a,b,c)=a^3+b^3+c^3+4abc-\lambda(a+b+c-3)$ is
$$
\nabla f(a,b,c,\lambda)=\left(3a^2+4bc-\lambda , 3b^2+4ac-\lambda , 3c^2+4ab-\lambda , 3-a-b-c\right).
$$
By Weierstrass' theorem, the maximum exists on, let's say, $(x,y,z,w)$. Hence, a necessary condition is $\na... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integral of rational function - which contour to use? Evaluate : $$\int_{-\infty}^{+\infty} \frac {x}{(x^2+2x+2)(x^2+4)}$$
I found that the integrand can be extended to a function on a complex plane has simple poles at $\pm 2i$ and $-1\pm i$. Now I want to compute the integral by contour integration but I am unable to... | The computation of residues leads to the partial fraction decomposition
$$ \frac{x}{(x^2+2x+2)(x^2+4)}=\frac{1}{10}\left[\frac{4-x}{x^2+4}+\frac{x-2}{x^2+2x+2}\right] $$
and now you don't have to pick any contour, since
$$\int_{\mathbb{R}}\frac{4\,dx}{x^2+4}=2\pi,\qquad \int_{\mathbb{R}}\frac{3\,dx}{(x+1)^2+1}=\int_{\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
product of terms taken $3$ at a time in polynomial expression
Finding product of terms taken $3$ at a time in $\displaystyle \prod^{100}_{r=1}(x+r)$
Try:
$$\displaystyle \prod^{100}_{r=1}(x+r)=x^{100}+(1+2+3+\cdots +100)x^{99}+(1\cdot 2+1\cdot 3+\cdots+100\cdot 99)x^{98}+(1\cdot 2\cdot 3+2\cdot 3 \cdot 4+\cdot\cdot... | If you try to expand the multiplications, you'll find that to produce terms with $x^{98}$, you'll need $2$ distinct constant factors (from the set $\{1,2,\dots,100\})$ and $98$ number of $x$ factors. So the sum of these terms, $a_{98}$, would be $$a_{98}=\sum_{n=1}^{100}\sum_{m=n+1}^{100}mn$$
How can we find this sum? ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2582647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove by induction that for every $n \ge 0$ the term $2^{2n+1} - 9n^2 + 3n -2$ is divisible by $54$ What I get is $4 \cdot 54k + 27(n(n-1)) - 2$.
I understand first two terms but $-2$ is giving me a headache. I suppose to be true every term should be divisible by $54$. Or did I make wrong conclusion?
P. S: Is there mat... | $2^{2n+1}-9n^2 +3n -2 = 54k$
$2^{2n+1} =54k +9n^2 -3n +2$
$2^{2(n+1)+1} -9(n+1)^2 +3(n+1) -2=$
$2^2(54k +9n^2 -3n +2) -9(n+1)^2 +3(n+1) -2$
After I compute everything, I get this: $4\cdot54k +27(n(n+1)) -2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 2
} |
For $a,b>0$, $c>1$, and $n=3,4,\dots$ is $|ca^n+(-b)^n|\leq(ca^2+b^2)^{n/2}$ always true? Let $a,b>0$, $c>1$, and $n=3,4,\dots$, then is
\begin{equation}
|ca^n+(-b)^n|\leq(ca^2+b^2)^{n/2},
\end{equation}
true for all possible combinations of $a$, $b$, $c$, and $n$?
| Hint: prove first that $\,\big(c+x^n\big)^2 \le \big(c+x^2\big)^n\,$ for $\,c \ge 1, x \ge 0, n \ge 2\,$, using for example the binomial expansion, noting that $\,\color{red}{c^2} \le \color{red}{c^n}\,$ and $\,\color{blue}{2cx^n}\,$ is no larger than either of the two blue terms on the RHS depending on whether $\,x \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| if $f(x)=\frac{x}{3}+3$
Then $f^2(x)=\frac{f(x)}{3}+3=\frac{x}{9}+\frac{3}{3}+3$
and $f^3(x)=\frac{f^2(x)}{3}+4=\frac{x}{27}+\frac{3}{9}+\frac{3}{3}+3$
By induction the general form is emerging as:
$$f^n(x)=3^{-n}x+3\left(\sum_{k=1}^n3^{1-k}\right)$$
Use the formula for a geometric sum and plug in $x=4, n=2018$:
$f^{20... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 8
} |
$\lim_{n\rightarrow \infty}\left[\frac{\left(1+\frac{1}{n^2}\right)\cdot \cdots\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n$
$$\lim_{n\rightarrow \infty}\Bigg[\frac{\bigg(1+\frac{1}{n^2}\bigg)\bigg(1+\frac{2}{n^2}\bigg)\cdots\cdots \bigg(1+\frac{n}{n^2}\bigg)}{\sqrt{e}}\Bigg]^n$$
Try: $$y=\lim_{n\rightarro... | We have that
$$\begin{align}
\left[\frac{\left(1+\frac{1}{n^2}\right)\cdots \left(1+\frac{n}{n^2}\right)}{\sqrt{e}}\right]^n&=\frac{\left(\frac{(n^2+n)!}{(n^2)!}\right)^n}{n^{2n^2} e^{n/2}}
\sim\frac{\left(1+\frac{1}{n}\right)^{n^3+n^2+\frac{n}{2}}}{e^{n^2+n/2}}
\\
&\sim\frac{\exp\left((n^3+n^2+\frac{n}{2})\ln\left(1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
A problem of determinants Suppose that $a,b,c,d \in \mathbb R$. Then
$$\begin{vmatrix} a& -b & -c& -d\\b& a& d& -c\\c& -d& a& b\\d& c& -b& a \end{vmatrix} = 0 \iff a＝b＝c＝d＝0$$
I can prove this by computing this determinant exactly. But I wonder if there is any brief approach since the matrix has the special structure... | In this case your matrix $M$ satisfies
$$MM^t=\pmatrix{a^2+b^2+c^2+d^2&0&0&0\\
0&a^2+b^2+c^2+d^2&0&0\\
0&0&a^2+b^2+c^2+d^2&0\\
0&0&0&a^2+b^2+c^2+d^2\\
}$$
so that $\det(M)^2=\det(MM^t)=(a^2+b^2+c^2+d^2)^2$. If
$M=\pmatrix{A&-B^t\\B&A^t}$ then
$$MM^t=\pmatrix{AA^t+B^tB&AB^t-B^tA\\BA^t-A^tB&BB^t+A^tA}$$
which will simpli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Absolute value definition Is it true that $\dots$
$$
\left| y \right| =
\begin{cases}
y \hspace{1cm} y \geq 0 \\ -y \hspace{0.7cm} y < 0
\end{cases}
$$
I'm a little bit confused with the second case, where $|y| = -y$ then $y<0$,
for example :
$$
\left| 2x-4 \right|=-(2x-4)
$$
if we assume that $ y=2x-4 $ then
$$
... | It is a bit confusing to use the same letter $x$ for your substitution.
Let set $y=2x-4$.
Then $|y|$ is $\begin{cases}+y&=2x-4&\quad\text{when}\quad y\ge0\iff 2x-4\ge 0\iff x\ge 2\\-y&=4-2x&\quad\text{when}\quad y<0\iff 2x-4<0\iff x<2\end{cases}$
Indeed for $x=1$ then $x<2$ then $|y|=|2\times 1-4|=|-2|=2=4-2\times 1$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac... | Since both $a$ and $b$ are nonnegative so are ${{1}\over{1+a}}$ and ${{b}\over{1+b}}+a$ and their multiplication.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 5
} |
Recurrent equation system How to solve this recurrent equation system?
$$\begin{cases}a_{n+1}=a_n+2b_n\\
b_{n+1}=2a_n+b_n\end{cases}$$
The possible solutions should be
\begin{cases}a_n=\frac{1}{2}3^n+\frac{1}{2}(-1)^n\\
b_n=\frac{1}{2}3^n-\frac{1}{2}(-1)^n\end{cases}
| $a_{n+1}=a_n+2(2a_{n-1}+b_{n-1})=a_n+4a_{n-1}+(a_n-a_{n-1})$
So, $a_{n+1}-2a_n-3a_{n-1}=0$.
The roots of $x^2-2x-3=0$ are $-1$ and $3$.
$a_n=\alpha (-1)^n+\beta(3)^n$ for some $\alpha$ and $\beta$.
$2b_n=a_{n+1}-a_n=\alpha(-1)^{n+1}+\beta(3)^{n+1}-\alpha (-1)^n-\beta(3)^n=-2\alpha(-1)^n+2\beta (3)^n$
$b_n=-\alpha(-1)^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Determine the Taylor expansion of $f$ defined by $f(x)=e^{-1/x^2}$ if $x \ne 0$ and $f(0)=0$ I need to find the Taylor series, at $x=0$ for the piece-wise function:
$$f(x)=\begin{cases} \displaystyle e^{\frac{-1}{x^2}} &\text{if}\, x \ne 0\\ 0 &\text{if}\, x = 0\end{cases}$$
I know that the Taylor series formula is $\s... | Yes, the Taylor series is just $0$ as you had said.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}=(m^2+2) \sqrt{m^2-1}$ Im trying to get from this expression into:
$$\frac{(m^2 + 2)(m-1)^2(m+1)^2}{(m^2-1)^{3/2}}$$
this expression:
$$(m^2+2) \sqrt{m^2-1}$$
someone know how to do it?
i tried it for hours and can't get from the first expression into the second expre... | Write $(m-1)^2(m+1)^2$ as $(m^2-1)^2$.
Then you have $$\frac{(m^2-1)^2}{(m^2-1)^\frac{3}{2}}=\frac{(m^2-1)^\frac{4}{2}}{(m^2-1)^\frac{3}{2}}=(m^2-1)^{\frac{4}{2}-\frac{3}{2}}=(m^2-1)^\frac{1}{2}=\sqrt{m^2-1}$$
The $(m^2+2)$ part stays intact.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Can we solve this without cubic formula? I'm looking for an algebraic solution for $x$.
$$ \frac{x}{x+2} -3 = \frac{5x}{x^2-4}+x$$
My first go at this involved converting this into an expression with a cubic numerator and $(x+2)(x-2)$ as the denominator. To find the roots, I then tried to divide each factor in the de... | Assuming $x^2\neq 4$, you end with the cubic equation
$$x^3+2 x^2+3 x-12=0$$ So, consider the function $$f(x)=x^3+2 x^2+3 x-12 \implies f'(x)=3 x^2+4 x+3$$ The first derivative does not cancel which means that there is only one real root.
Now, use inspection : $f(0)=-12$, $f(1)=-6$, $f(2)=10$. So, the root is between $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to find the angle between the two tangents on a parabola
Two tangents are drawn from the point $(-2,-1)$ to the parabola $y^2=4x$.
If $\alpha$ is the angle between them then find the value of $\tan \alpha$ .
My try:
Eqn of tangent $1,T_1$ say at the point $(x_1,y_1)$ is $yy_1=2a(x+x_1)$,
Eqn of tangent $2,T_2$ ... | Taking the derivative wrt $x$, $y^2=4x$ becomes $2yy'= 4$, or $y' = \dfrac{2}{y}$.
So, at the point $(\frac 14t^2,t)$ on the parabola, the slope of the tangent is
$m = \dfrac 2t$.
So the equation of the tangent line through the point $(\frac 14t^2,t)$
is $y-t = \frac 2t(x-\frac 14t^2)$.
For what values of $t$ does thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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$ABCD$ is parallelogram. $\angle ABC=105^{\circ}$. $\angle CMD=135^\circ$. Find $\angle BKC$
$ABCD$ is parallelogram. $\angle ABC=105^{\circ}$. $BMC$ is equilateral triangle and $\angle CMD=135^\circ$. $K$ is midpoint of $AB$. Find $\angle BKC$
My attempts
1) $\angle MBC=\angle MCB=\angle BMC=60^{\circ}$
2) $\angle ... | Let $N$ be the point on $CD$ such that $\angle CMN=15^\circ$.
Then $\angle CMN=15^\circ=\angle MCN$ and $\angle MDN=30^\circ=\angle MND$.
So $DM=MN=NC$.
As $BC=BM$, $MN=CN$ and $BN=BN$, $\triangle BCN\cong \triangle BMN$ and therefore, $\angle CBN=\angle MBN=30^\circ$.
So, $\angle BNC=75^\circ=\angle BCN$ and hence $BC... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
Find the angle between the two tangents drawn from the point $(1,2)$ to the ellipse $x^2+2y^2=3$.
The given ellipse is $\dfrac{x^2}{3}+\dfrac{y^2}{\frac{3}{2}}=1$
Any point on the ellipse is given by $(a\cos \theta,b\sin... | Let $m$ be the slope of a tangent to the ellipse from $(1,2)$.
The equation of the tangent is $y=mx-m+2$.
Put it into the equation of the ellipse,
\begin{align*}
x^2+2(mx-m+2)^2&=3\\
(1+2m^2)x^2-4m(m-2)x+2m^2-8m+5&=0
\end{align*}
As the tangent meets the ellipse at only one point.
\begin{align*}
[4m(m-2)]^2-4(1+2m^2)(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2602561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Which sets of base 10 digits have the property that, for every $n$, there is a $n$-digit number made up of these digits that is divisible by $5^n$? Which sets of non-zero base 10 digits have the property that, for every $n$, there is a $n$-digit number made up of these digits that is divisible by $5^n$?
This is an exte... | Partial result . . .
Claim:
If $S \subseteq \{1,2,3,4,5,6,7,8,9\}$ contains a complete residue system, mod $5$, then for all positive integers $n$, there is an $n$-digit number $x$ such that
*
*All digits of $x$ are elements of $S$.$\\[4pt]$
*$5^n{\mid}x$.
Proof:
Assume $S \subseteq \{1,2,3,4,5,6,7,8,9\}$ con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
linear system of two ODEs Consider the following system of ODE:
$$\begin{cases} \dfrac{df}{dt} = \dfrac{\alpha \beta}{1+t^2}g(t), \\ \\
\dfrac{dg}{dt} = \dfrac{\beta}{\alpha}f(t), \\ \\
t \in [-\alpha, \alpha], \\ \\
f(-\alpha) = g(-\alpha) = 1. \end{cases}$$
With $\alpha \ge 1$ and $\beta > 0$ as small as you like. ... | Another way:
$$\frac{d^2g}{dt^2}=\frac{\beta}{\alpha} \frac{df}{dt}=\frac{\beta^2}{1+t^2}g(t)$$
We obtain a 2nd order ODE:
$$(1+t^2)\frac{d^2g}{dt^2}-\beta^2g=0$$
$$g(-\alpha)=1$$
$$\frac{dg}{dt}(-\alpha)=\frac{\beta}{\alpha}$$
The general solution is (we can check with Wolfram Alpha ):
$$g(t)=C_1 \cdot {_2F_1} \left(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why does the determinant always equal zero for a square matrix of consecutive numbers? This works if the integers are listed in consecutive order either along the rows or columns. Why does the determinant of the square matrix always equal $0$ for $ n > 2 $?
| Alternatively, using arithmetic mean, for $3\times 3$ matrix determinant:
$$\begin{vmatrix}
x+1 & x+2 & x+3 \\
x+4 & x+5 & x+6 \\
x+7 & x+8 & x+9 \end{vmatrix}=0,$$
because: $C_1+C_3=2C_2$, which shows linear dependence of the column vectors.
For $4\times 4$ matrix determinant:
$$\begin{vmatrix}
x+1 & x+2 & x+3 & x+4 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Computing the series$\sum\limits_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} $ Show convergence of
$\begin{align}
\sum_{k=1}^{\infty}{\frac{3^{k-1}+(2i)^k}{5^k}} &= \sum_{k=1}^{\infty}{\frac{3^{k-1}}{5^k}}+ \sum_{k=1}^{\infty}{\frac{(2i)^k}{5^k}} \\
&= \sum_{k=1}^{\infty}{\frac{1}{3} \cdot \left( \frac{3}{5} \right) ^k... | If $i$ satisfies $i^2=-1$ then
$$\sum_{k=1}^\infty\left(\frac{2i}5\right)^k$$
converges in $\Bbb C$ to
$$\frac{2i/5}{1-2i/5}=\frac{2i}{5-2i}=\frac{-4+10i}{29}$$
(geometric series) as $|2i/5|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the indefinite integral $\int \left(\frac{\arctan x}{\arctan x - x}\right)^2 dx$. I am wondering if there is a systematic way to find the indefinite integral
$$\int \left(\frac{\arctan x}{\arctan x - x}\right)^2 dx.$$
It indeed has a clean closed form
$$\frac{x^2 + 1}{\arctan x - x} + x.$$
But I am not able to re... | The integrand is
$$ \left(1+\frac{x}{\arctan x - x}\right)^2 = 1 + \frac{2x}{\arctan x - x} + \frac{x^2}{(\arctan x - x)^2} $$
Observe that
$$ (\arctan x - x)' = \frac{1}{x^2+1} - 1 = -\frac{x^2}{x^2+1} $$
So we can perform IBP on the last term
$$ \int (x^2+1)\left(\frac{1}{(\arctan x - x)^2}\frac{x^2}{x^2+1}\right)dx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Using complete the square to determine positive definite matrices I realize this may be a basic question but I am having trouble following my notes.
I have the matrix $$\begin{bmatrix}16&12\\12&9\end{bmatrix} .$$ So I've got my equation from the matrix to be $(16x)^2 - 12xy - 12yx + 9y^2.$ Not sure where to go from her... | The direction that would usually be called completing the square is
$$ Q^T D Q = H $$
$$\left(
\begin{array}{rr}
1 & 0 \\
\frac{ 3 }{ 4 } & 1 \\
\end{array}
\right)
\left(
\begin{array}{rr}
16 & 0 \\
0 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rr}
1 & \frac{ 3 }{ 4 } \\
0 & 1 \\
\end{array}
\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate limit containing $\sum{n^6}$ Evaluate:
$$\lim_{n\to\infty}{\frac{1^6+2^6+3^6+\ldots+n^6}{(1^2+2^2+3^2+\ldots+n^2)(1^3+2^3+3^3+\ldots+n^3)}}$$
I can solve the denominator as:
$$\frac{n(n+1)(2n+1)}{6}\cdot\frac{n^2(n+1)^2}{4}$$
$$n^7\cdot\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}\cdot\frac{(1+\frac{1}{n})}{4}$$
$$... | The sum of the $k^{th}$ power of the integers is a polynomial of degree $k+1$ in $n$, with leading term $\dfrac{n^{k+1}}{k+1}$.
Indeed, by the binomial theorem,
$$n^k=\sum_{i=1}^ni^k-\sum_{i=1}^{n-1}i^k\sim\frac{n^{k+1}}{k+1}-\frac{(n-1)^{k+1}}{k+1}=n^k+\text{lower degree terms}.$$
The given limit equals that of
$$\fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Show that $Q=X^2+5X+7$ divides $P=(X+2)^m+(X+3)^{2m+3}$ for any $m\in\Bbb N$ Let $$P=(X+2)^m+(X+3)^{2m+3}$$ and $$Q=X^2+5X+7.$$ I need to show that $Q$ divides $P$ for any $m$ natural.
I said like this: let $a$ be a root of $X^2+5X+7=0$. Then $a^2+5a+7=0$.
Now, I know I need to show that $P(a)=0$, but I do not know i... | We can also prove it by induction.
$P_0(x)=1+(x+3)^3=x^3+9x^2+27x+28=(x^2+5x+7)(x+4)\quad\checkmark$
$\begin{align}
P_{m+1}(x) &=(x+2)^{m+1}+(x+3)^{2m+5}\\
&=(x+2)^m(x+2)+(x+3)^2\overbrace{\big((x^2+5x+7)Q_m(x)-(x+2)^m\big)}^{\text{induction hypothesis}}\\\\
&=(x+2)^m\underbrace{(x+2-x^2-6x-9)}_{-x^2-5x-7}+(x+3)^2(x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Real ordered pair $(a,b)$ in equation
If $$a^2+5b^2+2b=6a+2ab-10$$
then all real ordered pair of $(a,b)$ is?
Try: I am trying to convert it into $$(..)^2+(..)^2=0$$ or $$(..)^2+(..)^2+(..)^2=0.$$
So $$(a-b)^2+(2a-0.75)^2+10-(0.75)^2-6a=0.$$
Could some help me to solve it? Thanks.
| Consider the equation as a quadratic in $\,a\,$:
$$
a^2 - 2a(b+3) +5b^2+2b + 10 = 0
$$
Then its reduced discriminant is:
$$
\frac{1}{4} \Delta = (b+3)^2-(5b^2+2b +10) = -4b^2 + 4b -1 = -(2b-1)^2 \;\;\le\;\; 0
$$
It follows that $\,\Delta = 0\,$ for real roots to exist, so $\,b=\dfrac{1}{2}\,$, then substituting back $\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to find the following limit algebraically? I've been trying to answer this for a while and I know it's a simple question relative to most questions that are posted here.
$$
\lim_{x\rightarrow -2}\: \frac{x^4+5x^3+6x^2}{x^2(x+1)-4(x+1)}
$$
If we substitute -2 for $ x $ we get $ 0/0 $, an indeterminate form. I figure... | Notice that:
$$x^4+5x^3+6x^2 = x^2(x^2+5x+6) = x^2(x+2)(x+3)$$
and
$$x^2(x+1)-4(x+1) = (x^2-4)(x+1) = (x+2)(x-2)(x+1).$$
Then, you have:
$$\lim_{x\rightarrow -2}\: \frac{x^2(x+2)(x+3)}{(x+2)(x-2)(x+1)} = \\
= \lim_{x\rightarrow -2}\: \frac{x^2(x+3)}{(x-2)(x+1)} = \\
= \frac{(-2)^2(-2+3)}{(-2-2)(-2+1)} = \frac{4}{4} = ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
How to solve the integral $\int \frac{4+12x}{({1-2x-3x^2})^\frac{2}{3}}\,dx$ $$\int \frac{4+12x}{({1-2x-3x^2})^\frac{1}{3}}\,dx$$
So first I tried setting $u = 1-2x-3x^2, du = -2-6x\,dx$
Which gives
$$-2\int u^\frac{-1}{3}du = \frac{-6 u^\frac{2}{3}}{2} =-3u^\frac{2}{3}$$
And ultimately:
$$-3(1-2x-3x^2)^\frac{2}{3}$$
F... | \begin{align}
& \frac d {dx} \left( -3(1-2x-3x^2)^{2/3} \right) \\[10pt]
= {} & -3\cdot \frac 2 3(1-2x-3x^2)^{(2/3)-1}(-2-6x) & & \text{by the chain rule} \\[10pt]
= {} & -2(1-2x-3x^2)^{-1/3}(-2-6x) \\[10pt]
= {} & (1-2x-3x^2)^{-1/3} (4+12x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How do I interpret this sum? So if the sum of $n$ integers $\ge 1$ equal $\frac{n(n+1)}2$. Then my book goes on and says $1 + 2 + 3 +\ldots + 2n = \frac{2n(2n + 1) }2$.
I'm confused about what $1 + 2 + 3 + \ldots +2n$ means. If the sequence is $1, 2, 3, 4$ then where does $2n$ have to do with the $n$th number?
| The formula $1+2+3+...+n$ mean the "sum of all integers starting with $1$ and going up to $n$". You can substitute any positive integer you want in place of $n$, for instance by substituting $n=6$ you get $1 + 2 + 3 + ... + 6$, which ordinarily we would write out more explicitly as $1+2+3+4+5+6$, and them sum to get $2... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Coefficients of the stirling's series expansion for the factorial. Knowing the Stirling's approximation for the Gamma function (factorial) for integers:
$$\Gamma(n+1)=n!\approx \sqrt{2\pi n}n^ne^{-n}\bigg(1+\frac{a_1}{n}+\frac{a_2}{n^2}+\cdots\bigg)$$
Using the above approximation one can write:
$$(n+1)!=\sqrt{2\pi(n+1... | You were close. Note that1
$$
e^{-1}\left(1+\frac{1}{n}\right)^{n+1/2}=1+\frac{1}{12n^2}-\frac{1}{12n^3}+\frac{113}{1440n^4}+\cdots\tag1
$$
Therefore, you have
$$
\bigg(1+\frac{1}{12n^2}-\frac{1}{12n^3}+\cdots\bigg)\cdot\bigg(1+\frac{a_1}{n+1}+\frac{a_2}{(n+1)^2}+\frac{a_3}{(n+1)^3}+\cdots\bigg)=\\
1+\frac{a_1}{n}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Trigonometry problem, Evaluate: $\frac {1}{\sin 18°}$ My problem is,
Evaluate: $$\frac {1}{\sin 18°}$$
I tried to do something myself.
It is obvious,
$$\cos 18°= \sin 72°$$
I accept $\left\{18°=x \right\}$ for convenience and here, $\sin (x)>0$
$$\cos (x)=\sin (4x)$$
$$\cos (x)=2× \sin(2x) \cos (2x)$$
$$\cos (x)=2× ... | Take $\Delta ABC$ with $AB=AC=1$ and $BC=x,$ and with $\angle ABC=\angle ACB=2\pi/5.$ Then $\angle BAC=\pi/5$ and $x=2AB\sin \frac {1}{2}\angle BAC=2\sin \pi/10.$
Take $D$ on side $AC$ with $\angle DBC=\pi/5.$ Then $ABC$ and $BDC$ are similar triangles so $CD/x= CD / CB= CB/CA=x/1$ . So $CD=x^2.$
And $\Delta BDA$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Solving $\int(x^2 + 1)^7x^3{\mathrm d}x$ without expansion What is $\int(x^2 + 1)^7x^3{\mathrm d}x$?
I wasn't able to come up with a substitution so I attempted integration by parts:
$$u = x^2 + 1, u' = 2x, v' = x^3, v = \frac{x^4}{4}$$
$$(x^2+1)\frac{x^4}{4} - \frac{x^6}{12}$$
The derivative clearly shows that this is... | You have chosen the wrong functions for the integration by parts. Take: $$f'(x)=x(x^2+1)^7\\ f(x)=\frac{1}{16}(x^2+1)^8$$
and $$g(x)=x^2\\ g'(x)=2x$$ Hence:
\begin{align}
\int x^3(x^2+1)^7\,dx = \frac{1}{16}(x^2+1)^8 x^2-\int \frac{x}{8}(x^2+1)^8\,dx
\end{align}
And the last one is easy to do.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Showing that $\sum_{j=0}^{2n-1}{\cos^n(\frac{j\pi}{2n})(2\cos(\frac{2j\pi}n)+1)\cos(\frac{j\pi}2-\frac{2j\pi}n)}$ is never an integer for $n>10$ I want to show that
$$f(n) = \sum_{j=0}^{2n-1}{\cos^n\left( \frac{j \pi}{2n}\right) \left( 2\cos \left( \frac{2 j \pi}{n} \right) + 1\right) \cos \left( \frac{j \pi}{2} - \fr... | Claim: for $n\geq 3$,
$$
f(n)= 2^{-n}\cdot 2n\left(\binom{n}{4}+\binom{n}{2}+\binom{n}{0}\right)
$$This immediately gives that for $n\geq 16$, $0<f(n)<1$; then the only integer solutions are the ones shown in the problem statement.
Start with
$$
f(n) = \sum _{j=0}^{2 n-1} \left(2 \cos \left(\frac{2 \pi j}{n}\right)+... | {
"language": "en",
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"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Factor of a Mersenne number Why is it true that if 7 divides 91 then $(2^7-1) $ divides $(2^{91}-1)$?
1) $2^{91}-1$
$7|91 \implies (2^7-1)|(2^{91}-1)$
$\implies 2^7-1$ is factor
2) $2^{1001}-1$
$7|1001 \implies (2^7-1)|(2^{1001}-1)$
$\implies 2^7-1$ is factor
| We have $$2^{91}-1 = (2^7)^{13}-1 = (2^7-1)((2^7)^{12}+...+2^7+1)$$
More generally:
$$2^{ab}-1 = (2^a)^{b}-1 = (2^a-1)((2^a)^{b-1}+...+2^a+1)$$
so $2^a-1\mid 2^{ab}-1$
And we can replace $2$ with any other number.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 2,
"answer_id": 1
} |
Are reduced SVD and truncated SVD the same thing? Truncated SVD: http://scikit-learn.org/stable/modules/generated/sklearn.decomposition.TruncatedSVD.html
Reduced SVD, I thought this is essentially the same thing, and it appears to be actually more commonly called this way.
If you could provide reference, that'll be gre... | Suppose the shape of $A$ is $m \times n$, written as $\underset{m \times n}{A}$, with rank $r$.
In full SVD:
*
*$U$ is composed of $r$ orthonormal columns that span the column space of $A$ and $m - r$ orthonormal columns that span the left null space of $A$.
*$\Sigma$ is diagonal and composed of the square root of e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Consider polynomial $X^3-3X+1$ If $\alpha$ is a root $\alpha^3-3 \alpha+1=0 $ Consider polynomial $$ X^3-3X+1$$
If $\alpha$ is a root
$$\alpha^3-3 \alpha+1=0 $$
showing $\alpha^2-2$ is also a root
set $X=\alpha^2-2$
$$ (\alpha^2-2)^3-(\alpha^2-2)+1=\alpha^6-9\alpha^4+26 \alpha^2-24$$
Let us look at $\alpha^6$
$$\begi... | Expand and do long division to get:
$$
(x^2-2)^3-3(x^2-2)+1 = x^6 - 6 x^4 + 9 x^2 - 1 = (x^3 - 3 x - 1) (x^3 - 3 x + 1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Using Fourier series to calculate infinite series. I am asked to devolop the function $f(x)=x^2$ in a series of sine and cosine (Fourier series) in the interval $[\frac{-1}{2},\frac{1}{2}]$. And use one of these series to calculate $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2} \space\space\space \sum_{n=1}^{\infty} \fra... | The function $x^2$ is an even function in the given interval, $[\frac{-1}{2},\frac{1}{2}]$, thus the sine terms will vanish.
I calculated the Fourier series to be:
$$f(x)=x^2=\frac{1}{12}+\sum_{n=1}^\infty {\frac{(-1)^n}{\pi^2n^2}cos({2\pi}nx)}$$
NOTE: There is a mistake in your equation above where you multiplied the ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving that $\binom{2n}{n} > n2^n , \forall n \ge 4 $ I'm trying to prove the following statement: $\binom{2n}{n} > n2^n , \forall n \ge 4 $
This is my attempt at an inductive proof:
Let $P(n)$ be the following proposition: "$\binom{2n}{n} > n2^n , \forall n \ge 4 $"
Base case:
$\binom{2*4}{4} = 70 > 4*2^4 = 64$ so ... | You can actually do this without explicit induction.
$$\binom{2n}{n} = \frac{(2n)(2n-1)\dots(2n-(n-1))}{n(n-1)\dots2 \cdot 1} = \frac{2n}{n} \cdot \frac{2n-1}{n-1} \cdot \dots \cdot \frac{n+2}{2} \cdot \frac{n+1}{1}$$
Each of those $n$ fractions is bigger than or equal to $2$, thereby providing our $2^n$; but we can do... | {
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"source": "stackexchange",
"question_score": "1",
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Given a root, determine $a$ and $b$.
Given $p(x)= x^3 + ax^2 + bx -6$ with a zero at $x=1+i$. Determine the values of $a$ and $b$.
With the question given in its current form, would it be reasonable to assume that both $a$ and $b$ are real? Since, if so, one may use the fact that $x= 1-i$ is also a zero, and therefor... | Which says that $x^3+ax^2+bx-6$ is divisible by $$(x-1+i)(x-1-i)=(x-1)^2-i^2=x^2-2x+2$$ and since
$$x^3+ax^2+bx-6=x^3-2x^2+2x+(a+2)x^2+(b-2)x-6,$$ we obtain $a+2=-3$ and $b-2=6$,
which gives $$(a,b)=(-5,8).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Integral of $\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ How do we integrate the following?
$\int{\frac{\cos^4x + \sin^4x}{\sqrt{1 + \cos 4x}}dx}$ given that $\cos 2x \gt 0$
I tried to simplify this, but I cannot seem to proceed further than the below form:
$\int{\frac{\sec2x}{\sqrt{2}}dx + \sqrt{2}\int{\f... | Use $$\frac{\sin^4x+\cos^4x}{\sqrt{1+\cos4x}}=\frac{1-\frac{1}{2}\sin^22x}{\sqrt2\cos2x}=\frac{\frac{1}{2}+\frac{1}{2}\cos^22x}{\sqrt2\cos2x}=\frac{\cos2x}{2\sqrt2(1-\sin^22x)}+\frac{1}{2\sqrt2}\cos2x=$$
$$=\frac{1}{4\sqrt2}\left(\frac{\cos2x}{1+\sin2x}+\frac{\cos2x}{1-\sin2x}\right)+\frac{1}{2\sqrt2}\cos2x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Converting polar equations to cartesian equations. Where $$r=\sin(3\theta)$$ and $$y=r\sin(\theta),~x=r\cos(\theta),~r^2=x^2+y^2$$ I have started by saying that $$ r=\sin (2\theta) \cos (\theta) +\sin (\theta) \cos (2\theta) \\ r=2\sin (\theta) \cos ^2 (\theta) +\sin (\theta) (1-2\sin ^2 (\theta)) \\r=2\sin (\theta) \... | $$r=\sin (3t)=Im ((\cos (t)+i\sin (t))^3) $$
$$=3\cos^2 (t)\sin (t)-\sin^3 (t) $$
$$=3\sin (t)-4\sin^3 (t) $$
$$=3y/r-4y^3/r^3$$
thus
$$r^4=3yr^2-4y^3$$
now replace $r^2$ by $x^2+y^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Polynomials division Let $p$ be a real polynomial ($p\in\mathbb{R}[x])$, if
*
*5 is the remainder of the division $\frac{p(x)}{x-2}$, and
*2 is the remainder of the division $\frac{p(x)}{x-5}$
What is the reaminder of the division $\frac{p(x)}{(x-2)(x-5)}$?
I have tried it, but I have no idea...
Thank you
| $p(x)=(x−2)(x−5)q(x)+ax+b$, for some $a,b \in \mathbb{R}$
Since we know the remainder of $p(x)$ when divided by $x-2$ and $x-5$
we obtain the following
\begin{align}
p(2) &= (2-2)(2-5)q(2) + a(2)+b\\ &= 2a+b = 5 \label{a}
\end{align}
\begin{align}
p(5) &= (5-2)(5-5)q(5) + a(5)+b\\ &= 5a+b = 2 \label{b}
\end{align}
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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continued fraction of $\sqrt{41}$
Show that $\sqrt{41} = [6;\overline {2,2,12}]$
here's my try:
$$\sqrt{36}<\sqrt{41}<\sqrt{49}\implies6<\sqrt{41}<7\implies\lfloor\sqrt{41}\rfloor=6$$
$$\sqrt{41}=6+\sqrt{41}-6=6+\frac{1}{\frac{1}{\sqrt{41}-6}}$$
$$\frac{1}{\sqrt{41}-6}=\frac{\sqrt{41}+6}{41-36}=\frac{\sqrt{41}+6}{5}=... | Given $(\sqrt{41}+4)/5$, render $6<\sqrt{41}<7$ which you used at the start. Then your fraction lies between $(6+4)/5$ and $(7+4)/5$ showing the integral part is $2$.
You should be able to get the integer part at every stage just from $6<\sqrt{41}<7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635420",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Evaluate $\int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$ I am trying to evaluate the following integral
$$I(k) = \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right)$$
with $0< k < 1$.
My attempt
By performing the substitution
$$y=\... | Another method which gives a closed form result. Using the identity
\begin{equation}
\ln\frac{1-x}{1+x}=-2\operatorname{arctanh} x
\end{equation}
and parity of the integrand, the integral can be written as
\begin{align}
I(k) &= \int_0^1 \frac{\mathrm{d}x}{\sqrt{1-x^2}}\frac{x }{1-k^2x^2}\log\left(\frac{1-x}{1+x}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
How can I prove that: $\frac{(n+1)(n+2)(n+3)...(3n)}{(1*2)(4*5)(7*8)...(3n-2)*(3n-1)} = 3^n$ How can I prove that:
$$\frac{(n+1)(n+2)(n+3)...(3n)}{(1 * 2)(4*5)(7*8)...(3n-2)*(3n-1)} = 3^n$$
Can you help me and explain me how can I prove it? I thought to prove it by induction, but I don't have idea how to do it in fac... | Note that
*
*$(n+1)(n+2)(n+3)…(3n)=\frac{(3n)!}{n!}$
*$(1\cdot2)(4\cdot5)(7\cdot8)…(3n-2)\cdot(3n-1)=\frac{(3n-1)!}{3\cdot6\cdot9\cdot...\cdot(3n-3)}=\frac{(3n-1)!}{3^{n-1}(n-1)!}$
thus
$$\frac{(n+1)(n+2)(n+3)…(3n)}{(1\cdot2)(4\cdot5)(7\cdot8)…(3n-2)\cdot(3n-1)} = \frac{(3n)!}{n!}\frac{3^{n-1}(n-1)!}{(3n-1)!}=\\=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Limit of $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ as $x$ goes to $0$ As plugging $0$ in $\frac{x^2-\log(1+x^2)}{x^2\sin^2x}$ makes the function becomes undetermined form of $\frac{0}{0}$. I tried applying L'Hospital's rule but it became messy and did not look helpful if I do further differentiation. So I tried finding the ... | As an alternative
$$\frac{x^2-\log(1+x^2)}{x^2\sin^2x}=\frac{x^2-\log(1+x^2)}{x^4}\cdot\frac{x^2}{\sin^2x}\to \frac12$$
indeed
$\frac{x^2}{\sin^2x}\to 1$ by standard limit
and let $y=x^2\to 0$
$$\frac{x^2-\log(1+x^2)}{x^4}=\frac{y-\log(1+y)}{y^2}\stackrel{HR}\implies\frac{1-\frac1{1+y}}{2y}=\frac{1+y-1}{2y(1+y)}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Finding the area of a region bounded by two parabolas and a line
Given that the parabola $y=x^2+1$ has a tangent line at point $P$, find the area, $S$, of the region enclosed by the parabola $y=x^2+1$, the tangent line, and the parabola $y=x^2$.
The part which I do not understand is how to express the part where it a... | NOTE: In this method for finding the area enclosed, you will not need to find the actual points of intersection, see my method here.
Point $P$ can be expressed with coordinates $(a, a^2+1)$
Let $f(x)=x^2+a$
$f'(x)=2x$
$f'(a)=2a$
The equation of the tangent line is $y=2a(x-a)+a^2+1=2ax-a^2+1$
Setting this equal to $y=x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Couple of implicit differentiation equations. Check if correct. Wanted to check if I did the implied differentiation correctly:
*
*Find the tangent line at ($\pi, \pi$)
$$ \sin{(x+y)} = 2x -2y$$
$$\cos{(x+y)} \cdot (1+\frac{dy}{dx}) = 2 - (2 \cdot \frac{dy}{dx})$$
$$\cos{(x+y)} + \cos({x+y}) \frac{dy}{dx} = 2 - (2... | 1) Here you made a sign mistake
$$\frac{dy}{dx} (-\cos{(x+y)} \color{red}{+ 2}) = \cos{(x+y)} - 2$$
It should be negative not positive...
Answer is
$$\frac {dy}{dx}=\frac {2-\cos(x+y)}{2+\cos(x+y)}$$
2) $y - 2 = \frac{9}{2} \cdot x(x-1) \to y =\frac{9}{2}x^2 -\frac92x+2$
Maybe you meant $y - 2 = \frac{9}{2} \cdot (x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove the following algebraically $\left( \begin{array}{c} 2n \\ 2\ \end{array} \right) = 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) + n^2$ I came across the following proof in my textbook that was used as a end of chapter review. How can I prove the following algebraically?
$$\left( \begin{array}{c} 2n \\ ... | Since OP asks for an algebraic proof, I type this proof making use of number of roots of a degree two polynomials for fun.
Set
$f(n) = \left( \begin{array}{c} 2n \\ 2\ \end{array} \right) - 2 \left( \begin{array}{c} n \\ 2\ \end{array} \right) - n^2
$.
Since $\deg f \le 2$, $f$ has at most two distinct roots.
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer My question is related to using mathematical induction to prove that $2^{5n + 1} + 5^{n + 2} $ is divisible by 27 for any positive integer.
Work so far:
(1) For n = 1:
$2^{5(1) + 1} + 5^{(1) + 2} = 26 + 53 = 64 + 125 = 189$
Check if div... | Use the fact that $32=5+27$ to get
$$\begin{align}2^{5(k+1)+1}+5^{k+1+2}&=2^{5k+6}+5^{k+3}\\&=32\cdot2^{5k+1}+5\cdot5^{k+2}\\&=5\cdot2^{5k+1}+27\cdot2^{5k+1}+5\cdot5^{k+2}\\&=5(2^{5k+1}+5^{k+2})+27\cdot2^{5k+1}\\&=27(5t+2^{5k+1})\end{align}$$ for some positive integer $t$. The result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Determine the value of summation form. Determine the value of
$\displaystyle \sum_{k=n}^{2n} \binom{k} {n} 2^{-k}$ for $n \geq 1$
My attempt:
I have asked this question before in another math forum but it leads no solution. My friend suggested me to search for something like series-k. I also looked at my textbook in co... | This answer is based upon the Lagrange inversion formula. It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance
\begin{align*}
[z^n]\frac{1}{1-2z}=[z^n]\sum_{j=0}^\infty 2^jz^j=2^n
\end{align*}
We obtain
\begin{align*}
\color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | The coefficient of $x^n$ ($0\leq n\leq 10$) in the expansion is the number of solutions to
$$
x_{1}+x_{2}=n;\quad 0\leq x_i\leq5.
$$
We can compute this via inclusion exclusion. Let $U$ be the set of all solutions to the previous question in non-negative integers and $A_{i}$ be the set of solutions in $U$ which $x_i>5$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 5
} |
Evaluate $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$ I'm interested in finding a nice closed form expression for the sum $\sum_{k=1}^n\frac{1}{k}\binom{n}{k}$. I've tried using the Binomial Theorem to get
\begin{align*}
\sum_{k=1}^n\frac{1}{k}\binom{n}{k}x^k & =\int_0^1\frac{(1+x)^n-1}{x} \, dx\\
&=\int_1^2 (1+u+\cdots+u^{n... | $f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k $
This is equal to term by term multiplication of following series:
$∑^n_{k=1} C^n_k =∑^n_{k=0} C^n_k-1=(1+1)^n-1=2^n-1$
$∑^n_{k=1}\frac{1}{k}= 1+1/2+1/3 + . . . 1/n =H_n $
⇒ $f(n)=∑ ^n_{k=1}\frac{1}{k} C^n_k =(2^n-1)H_n=∑^n_{k=1}\frac{2^k-1}{k}$
Note that $(2^n-1)$represent$(∑ ^n_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2645464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
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