Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(... | Let $t=2-\sqrt 3$. Note that $2+\sqrt 3=\frac 1t$. Then, we have
$$\begin{align}x=t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}&\Rightarrow x^3=t+\frac 1t+3\left(t^{\frac 13}+\left(\frac 1t\right)^{\frac 13}\right)\\&\Rightarrow x^3=4+3x\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
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Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:
If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$
I tried to substitute the value of x in the expression, but that comes out to be very big.
| Hint:
$$(2-\sqrt{3})(2+\sqrt{3})=1$$
$$(2-\sqrt{3})(x)=1$$
so
$$\frac{1}{x}=2-\sqrt{3}$$
$$(x+\frac{1}{x})^2=x^2+2+\frac{1}{x^2}$$
$$x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2$$
$$x^2+\frac{1}{x^2}=(2+\sqrt{3}+2-\sqrt{3})^2-2=14$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 3
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How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that I am not able to solve this.
| The sum can be found using our favourite alternative method of converting the sum into a double integral.
Noting that
$$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^{2n} \, dy = \frac{1}{2n + 1},$$
the sum can be rewritten as
\begin{align*}
\sum_{n = 1}^\infty \frac{1}{n (2n + 1)} &= \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
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$x^2 + 3x + 7 \equiv 0 \pmod {37}$ I'm trying to solve the following
$x^2 + 3x + 7 \equiv 0 \pmod {37}$
What I've tried -
I've tried making the left side as a square and then I know how to solve
but couldn't make it as a square root..
We also learned in class that you can multiply the left side and the modulo by $4a... | In the real numbers, a method of finding a solution to a quadratic equation is to complete the square. This would involve adding and subtracting $(b/2)^2$. $b=3$ in your case, and remember that $1/2 = 19 \mod 37$.
Specifically notice: $$(x+3 \cdot 19)^2 \equiv x^2 + 2\cdot 3 \cdot 19 x + (3 \cdot 19)^2$$ $$\equiv x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Definite solution for the mean distance from an external point to the surface of a sphere.
Sphere, radius $E$, is centred at point $O$ $[0,0,0]$.
External point $Q$ is at $[D,0,0]$.
I can slice the sphere by making multiple planar cuts parallel to the $YZ$ plane to produce circular zones (quasi-discs) of equal infini... | Following suggestions from coffeemath & user2566092 I found the following standard rule:-
$$
\int (ax + b) ^{p/n} = \frac{n}{(n+p)a} (ax + b) ^{1+p/n} +C
$$
for $ p = \pm1, \pm2, \cdots p \ne -n .$
(Source: The Universal Encyclopedia of Mathematics, page 590; a similar equation is here).
Applying that to the expressio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Calculate in closed form $\sum_{n=1}^{\infty} \frac{\arctan(1/n) H_n}{n}$ Playing with Taylor series is not helpful enough. What else would you try out?
$$\sum_{n=1}^{\infty} \frac{\arctan(1/n) H_n}{n}$$
$$\approx 2.1496160413898356727147400526167103602143301206321$$
It's easy to see the series converges since $\arctan... | Let $f(z)=\dfrac{(\gamma+\psi_0(-z))^2}{z^2+x^2}$. On $z=Re^{i[0,2\pi]}$, $f(z)\sim\mathcal{O}\left(\dfrac{\ln^2{R}}{R^2}\right)$, so the residue theorem gives
\begin{align}
\sum^\infty_{n=1}\operatorname{Res}\left(f(z),n\right)+\sum_{\pm}\operatorname{Res}\left(f(z),\pm ix\right)+\operatorname{Res}\left(f(z),0\right)=... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Concluding three statements regarding $3$ real numbers.
$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$
Conclusion $I.)\ 1<b<3$
Conclusion $II.)\ 2<a<3$
Conclusion $III.)\ 0<c<1$
Options
By the given statements
$\color{green}{a.)\ \text{Only conclusion $I$ can be derived}}$.
$b.)\ $ Only conclusion $II$ can... |
$\{a,b,c\}\in \mathbb{R},\ a<b<c,\ a+b+c=6 ,\ ab+bc+ac=9$
Conclusion I.) $1<b<3$
Conclusion II.) $2<a<3$
Conclusion III.) $0<c<1$
Let $\{p,q,r\}\in \mathbb{R}: a=\min(p,q,r),\ c=\max(p,q,r),
b=\max(\min(p,q),\min(q,r),\min(r,p))$.
Let's solve a system
\begin{align}
p+q+r&=6 ,\\
pq+qr+rp&=9
\end{align}
for $q$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341548",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to get the results of this logarithmic equation? How to solve this for $x$:
$$\log_x(x^3+1)\cdot\log_{x+1}(x)>2$$
I have tried to get the same exponent by getting the second
multiplier to reciprocal and tried to simplify $(x^3+1)$.
| We have that
$$\begin{align}
\log_{x+1}x&=\frac{\log_xx}{\log_x(x+1)}\\\\
&=\frac{1}{\log_x(x+1)} \tag 1
\end{align}$$
and
$$\log_x(x^3+1)=\log_x (x+1)+\log_x(x^2-x+1) \tag 2$$
Using $(1)$ and $(2)$ reveals that
$$\begin{align}
\log_x(x^3+1)\log_{x+1}x&=\left(\log_x (x+1)+\log_x(x^2-x+1)\right)\frac{1}{\log_x(x+1)}\\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Counting points on the Klein quartic In Moreno's book "Algebraic Curves over Finite Fields", he mentions the following in passing with no further comments ($K$ denotes the Klein quartic defined by $X^3 Y + Y^3 Z + Z^3 X = 0$):
The Jacobian of $K$ is a product of three elliptic curves all isogenous to the elliptic curv... | $\def\FF{\mathbb{F}}$I'm switching to another answer to discuss relations with $X_0(49)$ and complex multiplication. Again, Elkies notes are a good reference and $\zeta$ is a primitive $7$-th root of unity.
Let $\Delta \subset PSL_2(\FF_7)$ be the group of diagonal matrices. Let $Y = \Delta \backslash K$. Then $Y$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
probability and expected value Hey I am not sure if I thinking correctly on this question? In a carnival, there is game which charges you $3$ dollars to play a game. You win $1$ dollar for every consecutive head you get and you you can play till you get tail. if you get head head tail you get back two dollars. What is ... | Taking "for every consecutive head" to mean "number of heads before a tail", your formulation is correct, and here is another way to get its sum.
A = $0\times\frac{1}{2} + 1\times\frac{1}{4} + 2\times\frac{1}{8} + 3\times\frac{1}{16} + ...$
$\frac{A}{2}$ = ............ $0\times\frac{1}{4} + 1\times\frac{1}{8} + 2\times... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$ Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$
The answer should be: $y = \frac{1}{12} x^2 -3$
But how to arrive at the answer?
I tried replacing r with $\sqrt{x^2 + y^2}$, then $\sinθ$ as $\frac{y}{r}$ but to no avail.
I also ended up... | Just multiply by denominator:
$$
r = \frac{6}{1-\sin\theta} \Longrightarrow r(1 - \sin\theta) = 6\Longrightarrow
r - y = 6
$$
Now
$$
\sqrt{x^2+y^2} = 6+y\Longrightarrow x^2 + y^2 = (6+y)^2,
$$
or
$$
x^2 = 36 + 12y \Longrightarrow y = \frac{x^2}{12} - 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
finding $a_1$ in an arithmetic progression Given an arithmetic progression such that: $$a_{n+1}=\frac{9n^2-21n+10}{a_n}$$
How can I find the value of $a_1$?
I tried using $a_{n+1}=a_1+nd$ but I think it's a loop..
Thanks.
| We are given the condition $a_{n+1}=\dfrac{9n^2-21n+10}{a_n}$ or more compactly
$a_{n} a_{n+1}=9n^2-21n+10 \tag{1}$
Solving for d
By evaluating (1) at $n-1$ we are able to get another term with an $a_n$ factor:
$a_{n-1} a_{n}=9(n-1)^2-21(n-1)+10 = 9n^2-39n+40 \tag{2}$
so by subtracting (1) - (2) : $(a_{n+1} - a_{n-1}) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Differentiate the Function: $f(x)=\ln (\sin^2x)$ $$\begin{align}f(x)&=\ln (\sin^2x)\\
f'(x)&=\frac{1}{\sin^2x}\cdot 2(\sin x)(\cos x)\\
&=\frac{2(\sin x)(\cos x)}{\sin^2x}\\
&=\frac{2\ \ (\cos x)\ }{\sin x}\\
&=2\cot x
\end{align}$$
= $2\ cot\ x$
Is this answer right?
| Not quite; you have misapplied the Chain Rule. If $f(x) = g(h(x))$, where here $g(x) = \ln x$ and $h(x) = \sin^2 x$, then $f'(x) = g'(h(x))h'(x)$, so we get
$$
\frac{1}{\sin^2 x}\cdot\frac{d}{dx}(\sin^2 x) = \frac{2\sin x\cos x}{\sin^2 x}
= \frac{2\cos x}{\sin x} = 2\cot x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
solve $|x-6|>|x^2-5x+9|$
solve $|x-6|>|x^2-5x+9|,\ \ x\in \mathbb{R}$
I have done $4$ cases.
$1.)\ x-6>x^2-5x+9\ \ ,\implies x\in \emptyset \\
2.)\ x-6<x^2-5x+9\ \ ,\implies x\in \mathbb{R} \\
3.)\ -(x-6)>x^2-5x+9\ ,\implies 1<x<3\\
4.)\ (x-6)>-(x^2-5x+9),\ \implies x>3\cup x<1 $
I am confused on how I proceed.
Or... | Another way is to note the inequality is equivalent to
$$(x-6)^2>(x^2-5x+9)^2\iff -(x-1)(x-3)(x^2-6x+15)>0$$
The quadratic is always positive, so this is the same as $(x-1)(x-3)<0$ which means $x\in (1,3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Proving the closed form for an infinite sum (related to Chebyshev polynomials) How do I prove the following identity? For $y\not= 0$, we have
$$
\sum_{n=0}^{\infty} \dfrac{1}{2y}\left( (x+y)^{n+1}-(x-y)^{n+1}\right)
= \dfrac{1}{(x+y-1)(x-y-1)}.
$$
I am trying to find the closed form for the left hand side. Thanks in ... | $$
\begin{align}
\sum_{n=0}^{\infty} \dfrac{1}{2y}\left( (x+y)^{n+1}-(x-y)^{n+1}\right)
&= \sum_{n=0}^{\infty} \sum_{k=0}^n (x+y)^{k}(x-y)^{n-k} \\
&= \sum_{k=0}^\infty \sum_{n=k}^{\infty} (x+y)^{k}(x-y)^{n-k}\\
&= \sum_{k=0}^\infty (x+y)^{k} \sum_{n=k}^{\infty} (x-y)^{n-k}\\
&= \sum_{k=0}^\infty (x+y)^{k} \sum_{m=0}^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
... | Note: This is a slightly clumsy but systematic approach. On the plus side, this allow you solving similar equations of the form
$$\begin{cases}
x^2 - Ayz &= D\\
y^2 - Bxz &= E\\
z^2 - Cxy &= F
\end{cases}$$
without knowing how to complete the squares. On the minus side, you need to factor a quartic polynomial in the mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 3
} |
How to determine congruence manually How is it possible to determine if the the following congruence is true manually?
$$
2015^{53} \equiv 8 \pmod{11}
$$
| Since $2015 = 11 \cdot 183 + 2$, then $2015 \equiv 2 \pmod{11}$.
Our problem has now been reduced to showing that $2^{53} \equiv 8 \pmod{11}$.
Since $2^{53} = \left(2^5\right)^{10} \cdot 2^3 \pmod{11}$ and $2^5 = 32\equiv -1 \pmod{11}$ then $$2^{53} \equiv (-1)^{10} \cdot 2^3 \pmod{11} = 8 \pmod{11}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Differentiate the Function $ h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$
Differentiate the function $$h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$$
My try:
$$h(z) = \frac{1}{2}\ln\left(a^2-z^2\right)-\frac{1}{2}\ln\left(a^2+z^2\right)$$
so
$$h'(z) = \frac{1}{2}\cdot\frac{2a-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2a+2z}{a^2+z^2}$$... | No, this is not correct, though you are on the good path. $a^2$ is a constant, so when differentiated, it will be gone. The correct derivative is:
$$\frac{1}{2}\cdot\frac{-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2z}{a^2+z^2}=\frac{1}{2} \frac{-2z(a^2+z^2)-2z(a^2-z^2)}{(a^2-z^2)(a^2+z^2)}=\\=\frac{1}{2} \frac{-4a^2z}{(a^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How do i evaluate this integral $ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $? Is there some one show me how do i evaluate this integral :$$ \int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx $$
Note :By mathematica,the result is :
$\frac{Gamma\left(\frac1 4\right)Gamma\left(\frac5 4\right)}{\sqrt{\pi}}-\s... | This is not an answer but it is too long for a comment.
As I wrote in comment, there is something wrong somewhere since $$\int\frac{\sqrt{\tan x}}{\sin x}dx=-2 \sqrt{\cos (x)} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\cos
^2(x)\right)$$ and $$I=\int_{\pi /4}^{\pi /3}\frac{\sqrt{\tan x}}{\sin x}dx=\frac{\Gam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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How do I prove this nice inequality $x+3\sqrt[3]{xy^2}\geq4\sqrt{xy} $?
Let $x,y\geq0$. Prove that:
$$
x+3\sqrt[3]{xy^2}\geq4\sqrt{xy}
$$
Note: It's seems easy but when I tried to show it I went to complicated formula.
| Hint :$x+3\sqrt[3]{xy^2}=x+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}+\sqrt[3]{xy^2}\ge 4\sqrt[4]{x\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}\cdot\sqrt[3]{xy^2}}=4\sqrt{xy}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to calculate $\int \frac{\sin x}{\tan x+\cos x} \, dx$ How to calculate $$\int \frac{\sin x}{\tan x+\cos x} \, dx\text{ ?}$$ I got to $$\int \frac{-u}{u^2-u-1} \, du$$ while $u=\sin x$ but can I continue from here?
| First complete the square:
$$
u^2-u-1 = \left(u^2-u+\frac 1 4\right)-\frac 5 4 = \left( u - \frac 1 2 \right)^2 -\frac 5 4 = \left( u - \frac 1 2 - \frac{\sqrt 5} 2 \right)\left( u - \frac 1 2 + \frac{\sqrt 5} 2 \right)
$$
Then use partial fractions.
$$
\frac{-u}{u^2-u-1} = \frac A {u - \frac 1 2 - \frac{\sqrt 5} 2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help w... | Let $\displaystyle\text{C}=\prod_{r=1}^{7}\cos{\left(\dfrac{r\pi}{15}\right)}$
and
$\displaystyle\text{S}=\prod_{r=1}^{7}\sin{\left(\dfrac{r\pi}{15}\right)}$
Now,
$\text{C}\cdot\text{S}=\left(\sin{\dfrac{\pi}{15}} \cdot \cos{\dfrac{\pi}{15}}\right) \cdot \left(\sin{\dfrac{2\pi}{15}}\cdot\cos{\dfrac{2\pi}{15}}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 0
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Solving for y' in a fraction Given the equation $x+xy^2 = \tan^{-1}(x^2y)$ find $y'$.
I have tried doing this but solving for $y'$ I need some help and would like your advice.
Work so far...
$$1+y^2+2xy\left(\frac{dy}{dx}\right)= \frac{2xy+x^2\left(\frac{dy}{dx}\right)}{1+x^4y^2}$$
What can be done now to solve for $y'... | Multiply the both sides by $1+x^4y^2$ to get
$$(1+y^2+2xyy')(1+x^4y^2)=2xy+x^2y'.$$
Now you'll get
$$(2xy(1+x^4y^2)-x^2)y'=2xy-(1+y^2)(1+x^4y^2).$$
So, simplify the following :
$$y'=\frac{2xy-(1+y^2)(1+x^4y^2)}{2xy(1+x^4y^2)-x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$ Any ideas fot evaluating:
$$ \lim_{x \to \infty} x(\sqrt{2x^2+1}-x\sqrt{2})$$
thanks.
| HINT:
Multiply by
$$\frac{\sqrt{2x^2+1} + x\sqrt{2}}{\sqrt{2x^2+1} + x\sqrt{2}}$$
to get
$$x(\sqrt{2x^2+1} - x\sqrt{2})\cdot \frac{\sqrt{2x^2+1} + x\sqrt{2}}{\sqrt{2x^2+1} + x\sqrt{2}}= x\cdot \frac{x}{\sqrt{2x^2+1} + x\sqrt{2}}$$
At this point, you should see that the numerator has degree 1 and (in some sense) the de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1355259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving set of 2 equations with 3 variables I'm working through an example and my answer is not coming out right.
Two equations are given and then the solution is shown.
Equations:
$$\begin{aligned}20q_{1}+15q_{2}+7.5q_{3}&=10\\
q_{1}+q_{2}+q_{3}&=1\end{aligned}$$
Solution Given:
$(X, (1/3)(1-5X), (1/3)(2+2X))$ for a... | The matrix representing the system is
$$(A|B) = \left(\begin{array}{ccc|c}
4 & 3 & 1.5 & 2\\
1 & 1 & 1 & 1
\end{array}\right) \longrightarrow \left(\begin{array}{ccc|c}
4 & 3 & 1.5 & 2\\
-1.5 & -1.5 & -1.5 & -1.5
\end{array}\right) \longrightarrow \left(\begin{array}{ccc|c}
4 & 3 & 1.5 & 2\\
2.5 & 1.5 & 0 & 0.5
\end{ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How do I prove that there is no other :$k=9,12,18$ for which this fails :$\sigma^k(114) \equiv 0\mod 6 $? let $\sigma(n)$ be the sum of divisors for a positive integer for example :
$$\sigma(6)=1+2+3+6=12$$ .
I have performed some calculations in wolfram alpha about the sum divisors of this number:
$q=114$ such that ... | Hint for fast calculation of sigma in a chain...
Let a number be written as
$$
\prod p^{n_p},
$$
where $p$ are prime numbers. Then we have
$$
\sigma\Big(\prod p^{n_p}\Big) = \prod \frac{p^{n_p+1}-1}{p-1}.
$$
Example:
$$
\sigma(114) = \sigma(2 \times 3 \times 19)
= \frac{2^2-1}{2-1} \frac{3^2-1}{3-1} \frac{19^2-1}{19-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to prove that the Fibonacci sequence is periodic mod 5 without using induction? The sequence $(F_{n})$ of Fibonacci numbers is defined by the recurrence relation
$$F_{n}=F_{n-1}+F_{n-2}$$
for all $n \geq 2$
with $F_{0} := 0$
and
$F_{1} :=1$.
Without mathematical induction,
how can I show that
$$F_{n}\equiv F_{n+2... | $$\begin{bmatrix}
F_{n+1} & F_{n} \\
F_{n} & F_{n - 1}
\end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^n$$
So
$$\begin{align}
\begin{bmatrix}
F_{n+21} & F_{n+20} \\
F_{n+20} & F_{n + 19}
\end{bmatrix} &= \begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{20 + n} &\pmod 5 \\
%
&= \begin{bmatrix} 1 & 1 \\ 1 & 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 13,
"answer_id": 9
} |
Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and... | Let us compute some figures so that you can confirm answer !
Firstly, notice that when p = q = 1/2, the binomial distribution formula simplifies to P(X) = $\dfrac{n\choose X}{2^n}$
To simply computations, we can leave the division by $2^n$ till the end.
n(A wins) = $n[A = X]\cdot n[B\le X] =1\cdot1 + 5\cdot6 + 10\cdot1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Differentiate the Function: $y=\frac{ae^x+b}{ce^x+d}$ $y=\frac{ae^x+b}{ce^x+d}$
$\frac{(ce^x+d)\cdot [ae^x+b]'-[(ae^x+b)\cdot[ce^x+d]'}{(ce^x+d)^2}$
numerator only shown (') indicates find the derivative
$(ce^x+d)\cdot(a[e^x]'+(e^x)[a]')+1)-[(ae^x+b)\cdot (c[e^x]'+e^x[c]')+1]$
$(ce^x+d)\cdot(ae^x+(e^x))+1)-[(ae^x+b)\cd... | Here are the steps
$$\frac{d}{dx}\left[\frac{ae^x+b}{ce^x+d}\right]$$
$$=\frac{\left(ce^x+d\right)\frac{d}{dx}\left[ae^x+b\right]-\left(ae^x+b\right)\frac{d}{dx}\left[ce^x+d\right]}{\left(ce^x+d\right)^2}$$
$$=\frac{\left(ce^x+d\right)\left(ae^x\right)-\left(ae^x+b\right)\left(ce^x\right)}{\left(ce^x+d\right)^2}$$
$$=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Find the roots of this 6th degree polynomial Hey guys I'm reviewing for a test and I'm getting stuck on one part, I can't remember what to do next.
$x^6+16x^3+64$
$(x^3)^2+16x^3+64$
let $x^3=w$
$w^2+16w+64$
$(w+8)^2$ now substitute again
$(x^3+8)^2$
Now what do I do?
| We can actually factor anything in the form $x^3+a^3$ into $(x+a)(x^2-ax+a^2)$.
$x^3+8$ can be factored into the following:
$$x^3+8 = x^3+2^3 =(x+2)(x^2-2x+4)$$
We actually have $(x^3+8)^2$, however we are setting this value to equal $0$ to find the roots so we can just take the square root of both sides and obtain:
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
let $0<a\le b\le c\le d$, and such $abcd=1$,show that
$$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$
it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
| Necessarily $\color{green}{d\ge 1}$ and at least one between $a,b,c$ is less or equal than $1$. We have the equivalent inequality
$$a+b^2+c^3+d^4 - \frac{1}{a}-\frac{1}{b^2}-\frac{1}{c^3}-\frac{1}{d^4}\ge 0\qquad (*)$$
Arranging otherwise to compare the larger numbers involved with smaller ones
$$(d^4-\frac 1a)+(a-\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital... | $\lim_{x \to 0}\frac{\sin(x)-x\cos(x)}{x\sin(x)}=\lim_{x \to 0}\frac{\cos(x)-\cos(x)+x\sin(x)}{\sin(x)+x\cos(x)}=\lim_{x \to 0}\frac{\sin(x)}{\frac{\sin(x)}{x}+\cos(x)}=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Find the integral: $ \int{\frac{\sin x+2\cos x}{\sin^2x+3\sin x+2}\,dx}$ I would like some help with the following integral
$$ \int{\frac{\sin x+2\cos x}{\sin^2x+3\sin x+2}\,dx}$$
I tried do split the fraction to $\displaystyle\int\frac{\sin x}{\sin^2x+3\sin x+2}\,dx + \displaystyle\int\frac{2\cos x}{\sin^2x+3\sin x+... | hint:$\dfrac{\sin x}{\sin^2x+3\sin x + 2} =\dfrac{2}{\sin x+2}-\dfrac{1}{\sin x+1}$, and for each of them use $t = \tan\left(\frac{x}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ... | Here's a solution:
$$\cos(\theta + 120^\circ) = \cos\theta \cos 120^\circ-\sin\theta \sin 120^\circ = \frac{\cos \theta}{2} - \frac{\sqrt{3}}{2}\sin \theta$$
$$ \cos(\theta + 120^\circ) = \cos\theta \cos 120^\circ+\sin\theta \sin 120^\circ = \frac{\cos \theta}{2} + \frac{\sqrt{3}}{2}\sin \theta$$
$$ \left(\frac{\cos \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| Here's the thing about this question: It is a trick.For most numbers of the form $a+b\sqrt 5$ the cube root is a mess. But some such numbers have nice cube roots and when they do we can find them.
The one thing you have to know is that if $x$ is any number of the form $a+b\sqrt5$ then $x^2, x^3$, etc. also have that fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
Maximize the area of a triangle inscribed in a semicircle. http://i.imgur.com/Q5gjaSG.png
Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.
My attempt so far, I'm new at these problems and have only don... | Almost: Since the area is
$$
A = \frac{(1+x)\sqrt{1-x^2}}{2}
$$
we want to find the zero of the derivative
$$
\frac{dA}{dx} = \frac{1-x-2x^2}{2\sqrt{1-x^2}}
$$
Since $-1 < x < 1$ (do you see why?), we can just focus on the numerator, and then
$$
2x^2+x-1 = 0
$$
or $x = -1$ or $x = 1/2$. The first solution is discarded... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\le... | There is not fast but strightforward way. $18^\circ = \pi/10$; let $s=\sin{(\pi/10)}$ so, we should evaluate
$$
a=\sin\frac\pi{10}-\cos\frac{\pi}{5}=2s^2 + s - 1.
$$
We may express $\sin5x$ in terms of $\sin x$:
$$
\sin5x=16\sin^5 x - 20\sin^3 x + 5\sin x
$$
(i.e. from $\sin5x=\sin(4x + x), \sin4x=2\sin2x\cos2x$, or by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Contradiction between integration by partial fractions and substitution Integration by substitution:
$$\int \frac {dx}{x^2-1}$$
Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$
$$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\... | On simplifying
\begin{align}
& \ln \frac{x-1}{\sqrt{x^2-1}}=\ln\frac{\sqrt{x-1}^2}{\sqrt{(x-1)(x+1)}} \\[6pt]
= {} & \ln\frac{\sqrt{x-1}}{\sqrt{x+1}} \\[6pt]
= {} & \frac 1 2 \ln \frac{x-1}{x+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Non-linear system of equations Solve following system of equations over real numbers:
$$
x-y+z-u=2\\
x^2-y^2+z^2-u^2=6\\
x^3-y^3+z^3-u^3=20\\
x^4-y^4+z^4-u^4=66
$$
This does not seem as hard problem. I have tried what is obvious here, to write $x^2-y^2$ as $(x-y)(x+y)$, $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$ etc. Problem is ... | \begin{align}
x-y+z-u&=2 \quad (1)\\
x^2-y^2+z^2-u^2&=6 \quad (2)\\
x^3-y^3+z^3-u^3&=20 \quad (3)\\
x^4-y^4+z^4-u^4&=66 \quad (4)
\end{align}
The easiest way to solve it
is to start of testing the simplest solutions first.
The first eqn suggests to check if $x-y=1$, $z-u=1$ fits.
In this case (2) becomes
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Trig Equations Using Identities How would you solve:
$2\csc^2x=3\cot^2x-1$
I said:
*
*Turn the cosecant to $1+\cot^2~x$.
*Distribute to get $3=\cot^2~x$.
*Turn it into tan. To get $\tan x=\frac{1}{\pm \sqrt3}$.
Is this correct?
| Notice, we can also solve this as follows $$2\csc^2 x=3\cot^2 x-1$$ $$\implies \frac{2}{\sin^2 x}=\frac{3\cos^2 x}{\sin^2x}-1$$ $$\implies 2=3\cos^2 x-\sin^2x$$ $$\implies 3\cos^2 x-(1-\cos^2x)=2$$ $$\implies 4\cos^2 x=3$$ $$\implies \cos^2 x=\frac{3}{4}=\left(\frac{\sqrt{3}}{2}\right)^2$$ $$\implies \cos^2 x=\left(\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the r... | Let $a=\frac{x^2}{yz},\, b=\frac{y^2}{zx},\,c=\frac{z^2}{xy}.$ Thus$,$ we need to prove$:$
$$\frac{x}{\sqrt{x^2+8yz}}+\frac{y}{\sqrt{y^2+8zx}} +\frac{z}{\sqrt{z^2+8xy}} \geqq 1$$
By AM-GM$:$ \begin{align*} \text{LHS} &=\sum\limits_{cyc} \frac{x}{\sqrt{x^2+8yz}} =\sum\limits_{cyc} \frac{x(x+y+z)}{\sqrt{(x^2+8yz)(x+y+z)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 5
} |
Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find th... | We have
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a} = \frac{a^2c+b^2a+c^2b}{abc}
$$
Since from Vieta's relations we know
$$
a+b+c =0,\quad ab+bc+ca =-3,\quad abc =-1,
$$
our goal is to calculate
$$
s = a^2c+b^2a+c^2b.
$$
Let's introduce
$$
p = ac^2+ba^2+cb^2.
$$
Than we have
$$
0 = (ab+bc+ac)(a+b+c) = p+s+3abc
$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
Find the sum: $\sum_{i=1}^{n}\dfrac{1}{4^i\cdot\cos^2\dfrac{a}{2^i}}$ Find the sum of the following :
$S=\dfrac{1}{4\cos^2\dfrac{a}{2}}+\dfrac{1}{4^2\cos^2\dfrac{a}{2^2}}+...+\dfrac{1}{4^n\cos^2\dfrac{a}{2^n}}$
| A very different approach to calculate the sum. Note that
$$
\frac{1}{4^k\cos^2\frac{x}{2^k}}=\left(-\ln\cos\frac{x}{2^k}\right)''.
$$
Let's calculate (using in the $3^\text{d}$ equality the general form of Morrie's law for $\alpha=\frac{x}{2^n}$)
$$
F(x)=\sum_{k=1}^n-\ln\cos\frac{x}{2^k}=-\ln\prod_{k=1}^n\cos\frac{x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$ without finding the integration
I don't know how to start
| $$\int_{n^2+n+1}^{n^2+1}\frac{\arctan x}{\sqrt{x}}\,dx = \frac{\pi}{2}\int_{n^2+n+1}^{n^2+1}\frac{dx}{\sqrt{x}}+\int_{n^2+1}^{n^2+n+1}\frac{\arctan\left(\frac{1}{x}\right)dx}{\sqrt{x}}$$
Now:
$$\frac{n}{\sqrt{n^2+n+1}}\leq\int_{n^2+1}^{n^2+n+1}\frac{dx}{\sqrt{x}}\leq\frac{n}{\sqrt{n^2+1}}$$
and:
$$ 0\leq \int_{n^2+1}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Evaluating the limit: $\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1)$ I need to solve this question: $$\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1)$$
I tried the graphical method and observed that the graph was approaching $0$ as $x$ approached $-1$ but I need to know if there's a way to calculate this.
| $$\lim_{x\rightarrow -1^+}\sqrt[3]{x+1}\ln(x+1) = \lim_{x\rightarrow 0^+}\sqrt[3]{x}\ln(x) = \lim_{x\rightarrow 0^+} e^{ln (x)\frac{1}{3}}\ln(x) = \lim_{x\rightarrow -\infty} e^{\frac{1}{3}x} x$$
$$= -\lim_{x\rightarrow +\infty} \frac{x}{e^{\frac{1}{3}x}} = -\lim_{x\rightarrow +\infty} \frac{x}{1 + \frac{x}{3} + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solution of Differential equation Question:
Find solution of differential equation
$$ 3e^{4x} \frac{dy}{dx} = -16\frac{x}{y^2} $$
which satisfies the initial condition y(0)=1
Solution:
I know that I have to bring it in the general form of :
$$ \frac{dy}{dx} + P(x) y = Q(x)$$
However in the equation there is no P(x)y co... | An interesting aspect:
The differential equation has the form
\begin{align}
3 \, e^{a x} \, y' = - 16 \cdot \frac{x}{y^{2}}
\end{align}
for which
\begin{align}
3 \, y^{2} \, y' = - 16 \, x \, e^{-a x}
\end{align}
where $a$ is a constant, here $a = 4$. This last equation can be seen to have the form
\begin{align}
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Cauchy-Schwarz inequality problem The problems:
*Prove that $$\frac{\sin^3 a}{\sin b} + \frac{\cos^3 a}{\cos b} \geqslant \sec (a-b),$$ for all $a,b \in \bigl(0,\frac{\pi}{2}\bigr)$.
*Prove that $$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{2\sqrt[3]{abc}} \geqslant \frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b... | One form of the Cauchy-Schwarz Inequality is
$$\sum_i^n \frac{x_i^2}{y_i}\ge \frac{\left(\sum_i^n x_i\right)^2}{\sum_i^n y_i}\tag 1$$
Let $n=2$ and $x_1=\sin^2a$, $x_2=\cos^2a$, $y_1=\sin a\sin b$, and $y_2=\cos a\cos b$
The result follows immediately after expanding $\sec (a-b)=\frac{1}{\cos a\cos b+\sin a\sin b}$
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.
I' sure that I'm overseeing a tiny little missing... | Given
$$
f(x) = \frac{ 2 x - 2 }{ x^2 - 2 x + 4 }. \tag 1
$$
Assuming you want to expand $f(x)$.
Let
$$
\phi_\pm = 1 \pm \mathtt{i} \sqrt{3}. \tag 2
$$
We can write (1) as
$$
f(x) =
\frac{ 1 }{ x - \phi_+ } +
\frac{ 1 }{ x - \phi_- }. \tag 3
$$
Whence
$$
f(x) =
- \sum_{k=0}^\infty \left(
\frac{1}{{\ph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Find a Basis $B$ of $R^2$ so that $B$ matrix of $T$ is diagonal $T([1,1]^t) = [3,7]^t$
$T([1,-1]^t) = [1,1]^t$
Here's what I get:
$T=
\left(\begin{array}{cc}3 & 1 \\7 & 1\end{array}\right)
$
The eigenvectors of $T$ is $E = \left(\begin{array}{cc} .4798 & -.2527 \\.8774 & .9675\end{array}\right)$.
$E^{-1}TE$ gives us a ... | It is convenient to work with the standard basis $\left( \begin{matrix} 1 \\ 0 \end{matrix} \right), \left( \begin{matrix} 0 \\ 1 \end{matrix} \right)$ to write the answers directly as vectors in the standard basis.
So we must solve, with $T= \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$
$$ \left( \begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to show that a polynomial does not have real roots? For example, let's take the polynomial
$$x^8-x^7+x^2-x+15$$
Here, the power ($n=8$) is even so it can have real roots or it might not have real roots.
Something which I thought was to find the minima and show that if the minima of $p(x)$ is greater than $0$ and $a... | To show that a polynomial has no real roots, we will try to write it as an equation where the sum of some positive numbers equals a strictly negative number. As the sum of positive numbers cannot be strictly negative, there is a contradiction, which means there's no real root.
$x^8-x^7+x^2-x+15 = 0$
Subtract 15: $x^8-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 4
} |
find total integer solutions for $(x-2)(x-10)=3^y$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
How many integer solutions ($x$, $y$) are there of the equation $(x-2)(x-10)=3^y$?
(A)1 (B)2 (C)3 (D)4 (E)5
If let $y=0$, we ha... | Since their product is a power of $3$, both $x-2$ and $x-10$ must be powers of $3$, perhaps with minus signs (as Andre Nicolas pointed out). Note, however, that $x-2$ and $x-10$ cannot simultaneously be divisible by $3$. The only power of $3$ which is not divisible by $3$ is $1$, hence we must have that $x-2 = \pm 1$ o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Where did I go wrong with the Gram-Schmidt orthogonalisation process? Problem: Let $\alpha = \left\{(1,2,0), (1,0,1), (2,3,1)\right\}$ be a basis vor $\mathbb{R}^3$. Apply the Gram-Schmidt orthogonalisation process to turn $\alpha$ into an orthonormal basis for $\mathbb{R}^3$ with respect to the standard innerproduct.
... | The problem is that you projected two non-orthogonal vectors, $v_1$ and $v_2$, out of $v_3$. Each of those orthogonalizations kills the other. In orthogonalizing $v_3$, you need to use the result of orthogonalizing $v_2$ against $v_1$, not $v_2$ itself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find a polynomial from an equality Find all polynomials for which
What I have done so far:
for $x=8$ we get $p(8)=0$
for $x=1$ we get $p(2)=0$
So there exists a polynomial $p(x) = (x-2)(x-8)q(x)$
This is where I get stuck. How do I continue?
UPDATE
After substituting and simplifying I get
$(x-4)(2ax+b)=4(x-2)(ax+b)$
F... | The following is essentially @drhab's solution, but uses only one idea repeatedly.
From $$ (x-8)p(2x) = 8(x-1)p(x) $$ we see $x-8$ divides $p(x)$. Let $p(x) = (x-8)p_1(x)$ and substitute, yielding $$ 2(x-8)(x-4)p_1(2x) = 8(x-1)(x-8)p_1(x) $$
From this we see $x-4$ divides $p_1(x)$. Let $p_1(x) = (x-4)p_2(x)$ and subs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Exponential function negative: $\left(\frac{81}{4}\right)^{1/4}\left(\frac{1}4\right)^{-3/4}$ This is another example. $\left(\dfrac{81}{4}\right)^{1/4}\left(\dfrac{1}4\right)^{-3/4}$
Multiply on both sides equals $\dfrac{81^{1/4}}{4^{1/4}}\cdot \dfrac{1^{-3/4}}{4^{-3/4}}$
This should be $\dfrac{3}{4^{1/4}}\cdot \dfr... | An alternative way (bring everything under a single exponent):
$$(\frac{81}{4})^\frac 14 \cdot (\frac{1}{4})^\frac {-3}{4} = (\frac{81}{4})^\frac 14 \cdot [(\frac 14)^{-3}]^{\frac 14} = (\frac{81}{4})^\frac 14 \cdot(4^{3})^{\frac 14} = (\frac{81}{4}\cdot 64)^{\frac 14} = (81 \cdot 16)^{\frac 14} = 3 \cdot 2 = 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Volume of Solid Enclosed by an Equation I'm having problems finding the triple integrals of equations. I guess it has to do with the geometry. Can someone solve the two questions below elaborately such that I can comprehend this triple integral thing once and for all:
Compute the volume of the solid enclosed by
*
*$... | for the #1 if you want to use a geometrical solution you can say that it's the volume of a pyramid with apex at $(0,0,0)$ intersecting the axes at $(a,0,0),(0,b,0),(0,0,c)$ now because of the orthogonality of axes we can say the base is a right triangle having the area $\frac{1}{2}ab$ and the length of the height is $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Eliminate the parameter of a
Eliminate the parameter to find a description of the following circles
or circular arcs in terms of $x$ and $y$. Give the center and radius,
and indicate the positive orientation.
$x=4\cos{(t)} ,\ y=3\sin{(t)} ;\ 0 \leq t \leq 2\pi$
So,
$\displaystyle x^2=4^2\cos^2{(t)} ,\ y^2=3^2\sin... | Notice, $$x=4\cos t \implies \cos t=\frac{x}{4}\tag 1$$ & $$y=3\sin t \implies \sin t=\frac{y}{3}\tag 2$$ Now, for eliminating $t$, squaring & adding (1) & (2), we get $$\cos^2t+\sin^2t=\left(\frac{x}{4}\right)^2+\left(\frac{y}{3}\right)^2$$ $$\color{blue}{\frac{x^2}{16}+\frac{y^2}{9}=1}$$ The above equation is in the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $x$ in the triangle
the triangle without point F is drawn on scale, while I made the point F is explained below
So, I have used $\sin, \cos, \tan$ to calculate it
Let $\angle ACB = \theta$, $\angle DFC = \angle BAC = 90^\circ$, and $DF$ is perpendicular to $BC$ (the reason for it is to have same $\sin, \cos, ... |
$K$ will be like that so - $DK\perp CD$
we know that $DK\perp CD$ than $DK||AB$ (because $AC\perp AB$), Also we know that $CD=AD$, because of that we can understand $DK$ is median of triangle $\Delta ABC$ so $CK=KB=6$.
We know that $CE=3$, than $CE=EK=3$. $\Delta CDK$ is a right triangle, than we can understand that -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Maximal $n$ such the the additive partition with a given product is unique. Given $n$, there are many tuples with $a + b + c = n,0 < a < b < c$. For large $n$, different tuples may give the same products. E.g. $2+8+9=19=3+4+12,2\times8\times9=144=3\times4\times12$.
What is the largest value of $n$, such that there is n... |
For sufficiently large $n$, there are always unique tuples $a,b,c, \ 0<a<b<c$ and $a',b',c', \ 0<a'<b'<c'$ such that $a+b+c=n=a'+b'+c'$ and $a \times b \times c = a' \times b' \times c'$.
Proof. Say we have $b$ divisible by $3$ and $c$ divisible by $2$. This assumption will be justified later. Relate $a,b,c$ with $a'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$ Find $\triangle ABC$ if $\angle B=2\angle C$ and $$2(\cos\angle A+1)(\cos\angle B+1)(\cos\angle C+1)=\cos(\angle A-\angle B)+\cos(\angle B-\angle C)+\cos(\angle C-\angle A)+\cos\angle A+\cos\angle B+\cos\angle C+2$$
| Re-writing the equation as
$$0 = 2\prod (\cos A + 1) - \sum \cos(B-C) - \sum \cos A - 2$$
we begin by multiplying-out the product, and carrying-on from there:
$$\begin{align}
0 &= 2\cos A \cos B \cos C + 2\sum \cos B \cos C \color{blue}{+ 2\sum \cos A} \color{red}{+ 2} \\[4pt]
&\quad-\sum\cos(B-C) \color{blue}{- \sum\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to show that this difference of products is $O \left( \frac{1}{n^2} \right) $ Let $k \leq n$. Consider the following difference of products:
$$ \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n+1} \right) - \prod_{i=1}^{k-1} \left( 1 - \frac{i}{n} \right)$$
For $n=1,2,3$, this is clearly $O \left( \frac{1}{n^2} \right) $ for... | We have
$$\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n+1}\right)-\prod_{i=1}^{k-1}\,\left(1-\frac{i}{n}\right)=\left(\prod_{i=1}^{k-2}\,\left(1-\frac{i}{n}\right)\right)\left(\left(1+\frac{1}{n}\right)^{-(k-1)}-\left(1-\frac{k-1}{n}\right)\right)\,.$$
For large $n\in\mathbb{N}$, $\prod_{i=1}^{k-2}\,\left(1-\frac{i}{n}\right)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Trying to solve a pair of trigonometric simultaneous equations I have a machine that has two shafts which are the inputs and their position is set by 2 servo motors.
Depending on the angle of these two shafts (shaft 1 has an angle designated $Ta$ degrees, shaft 2 has an angle designated $Ba$ degrees) a set of gimbals r... | I'll simplify your notation a bit, using shaft angles $T$ and $B$ (rather than $T_a$ and $B_a$), and plate angles $P$ and $Q$ (rather than $P_a$ and $P_b$). The given equations are
$$\cos P = \cos T \cos B \qquad\qquad
\tan Q = \frac{\sin T}{\tan B} \tag{$\star$}$$
Solving these for $\cos T$ and $\sin T$, we eliminate ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate a quotient with fractional power of a quadratic polynomial I need help finding the indefinite integral of
$$\int\,\frac{x}{(7x - 10 - {x^2})^{3/2}}\,\text{d}x\,.$$
| $$\int { \frac { x }{ \sqrt { { \left( 7x-10-{ x }^{ 2 } \right) }^{ 3 } } } dx } =\int { \frac { x }{ \sqrt { { -\left( { x }^{ 2 }-7x+10 \right) }^{ 3 } } } dx= } \int { \frac { x }{ \sqrt { { -\left( { x }^{ 2 }-7x+\frac { 49 }{ 4 } -\frac { 49 }{ 4 } +10 \right) }^{ 3 } } } dx= } \\ =\int { \frac { x }{ \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
find the complex number $z^4$
Let $z = a + bi$ be the complex number with $|z| = 5$ and $b > 0$ such that the distance between $(1 + 2i)z^3$ and $z^5$ is maximized, and let $z^4 = c + di$.
Find $c+d$.
I got that the distance is:
$$|z^3|\cdot|z^2 - (1 + 2i)| = 125|z^2 - (1 + 2i)|$$
So I need to maximize the distance... | You are trying to maximize the value of $|z^2-(1+2i)|$ where $z^2$ is a point on the circle, centre $0$ radius $25$
As has been pointed out by Terra Hyde, $z^2$ must be the point on the circle on the other side of the origin from $1+2i$ which would be collinear with $1+2i$ and the origin. Therefore
$$z^2=25(-\cos\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Coefficient of binomial expansion The coefficient of $x^3$ is $4$ times the coefficient of $x^2$ in the new expansion of $(1+x)^n$. Find the value of $n$.
| coefficient of $x^3$ is four times the coefficient of $x^2$
$$(1+x)^n=\\\binom{n}{0}1^{n}x^{0}+\binom{n}{1}1^{n-1}x^{1}+{\color{DarkBlue} {\binom{n}{2}1^{n-2}x^{2} }}+{\color{Red}{\binom{n}{3}1^{n-3}x^{3}} }+...+\binom{n}{n}1^{n-n}x^{n}\\$$so
$$\binom{n}{3}1^{n-3}=4*\binom{n}{2}1^{n-2}\\\binom{n}{3}=4\binom{n}{2}\\\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Math Subject GRE 1268 Question 55 If $a$ and $b$ are positive numbers, what is the value of $\displaystyle \int_0^\infty \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx$. | A slight variation of the accepted solution begins on line $3$:
\begin{align}
I &= \int_{0}^{\infty} \frac{e^{ax}-e^{bx}}{(1+e^{ax})(1+e^{bx})}dx \\
&= \int_{0}^{\infty} \frac{dx}{1 + e^{bx}} - \int_{0}^{\infty} \frac{dx}{1 + e^{ax}} \\
&= \int_{0}^{\infty} \frac{e^{-bx} dx}{(1 + e^{bx})e^{-bx}} - \int_{0}^{\infty} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 2
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find the interval of convergence of the power series like the title said i have to find the interval of convergence of this power series :
$$\sum_{n=1}^\infty{ ((-1)^n *(x-1)^{2n-1})\over 3^n}$$
I applied the ratio test and i got something like this:
$$\left|\frac{(-1)*(x-1)^2}3\right|$$
I know that I have t... | The absolute value you got was correct.
\begin{align*}
\left|\frac{(-1)(x-1)^2}{3}\right| & = \left|\frac{(x-1)^2}{3}\right|\\
&= \frac13\left|(x-1)^2\right| <1
\end{align*}
Now we have
$$\left|(x-1)^2\right| < 3$$
$$1-\sqrt3 < x < 1+\sqrt3$$
Writing out some terms: $$\sum_{n=1}^\infty{(-1)^n \cdot (x-1)^{2n-1}\over 3^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Proving $\sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n}$ with induction Problem: Prove with induction that \begin{align*} \sum_{j=1}^n \frac{1}{\sqrt{j}} > \sqrt{n} \end{align*} for every natural number $n \geq 2$.
Attempt at proof: Basic step: For $n = 2$ we have $1 + \frac{1}{\sqrt{2}} > \sqrt{2}$ which is correct.
Indu... | As Daniel Fischer points out in the comments, since you have
$$
\sum_{j=1}^{n+1} \frac{1}{\sqrt{j}} > \sqrt{n} + \frac{1}{\sqrt{n+1}}
$$
it is enough to show
$$
\sqrt{n} + \frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}
$$
or equivalently $
\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1} - \sqrt{n}
$. A way to show this final inequality ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding the length of latus rectum of an ellipse If the lines $2x+3y=10$ and $2x-3y=10$ are tangents at the extremities of its same latus rectum to an ellipse whose center is origin,then the length of the latus rectum is
$(A)\frac{110}{27}\hspace{1cm}(B)\frac{98}{27}\hspace{1cm}(C)\frac{100}{27}\hspace{1cm}(D)\frac{12... | Using standard ellipse notation and relations for
$ a, b, c, p $.
Tangent equation of ellipse
$$ \frac{x x_1}{a^2} + \frac{y y_1}{b^2} =1 \tag{1}$$
Given tangent equation
$$ \frac{x}{5} + \frac{y}{10/3} =1 \tag{2}$$
Comparing $ x, y $ coefficients,
$$ \frac{x_1}{a^2}= \frac{1}{5} \tag{3}$$
$$ \frac{y_1}{b^2}= \frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Radical under Radical expression how to find the sum of $\sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}} $ ? Is there a method to solve these kind of equations ?
| $$
x = \sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}}\\
x^2 = \frac54 + \sqrt{\frac32} + 2\sqrt{\left(\frac54\right)^2 - \left(\sqrt{\frac32}\right)^2} + \frac54 - \sqrt{\frac32} = \frac52 + 2\sqrt{\frac{25}{16} - \frac32} = \frac52 + 2\cdot\frac14 = 3\\
x = \sqrt3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to d... | Here is an "almost" solution for integer coefficients.
The zeros of $x^4+12x^3+14x^2-12x+1$ are algebraic integers. So are their reciprocals. Now in the series
$$
(1-xa)^{-1/4} = 1+\frac{a}{4} x+\frac{5 a^2}{32} x^2 +\frac{15a^3}{128} x^3 + \cdots
$$
where $a$ is an algebraic integer, the coefficients are algebraic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
How to evaluate infinite series $\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} e^{-an}$ I'm trying to evaluate an infinite series:
$$
\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} e^{-an}
$$
where $a$ and $B$ are real parameters, or equivalently:
$$\sum\limits_{n=0}^\infty\sqrt{B^2+n^2} x^n$$
where $0<x<1$.
When $n$ becomes much lar... | This series can not be evaluated directly. One approximation method is as follows. Let the Lerch transcendent be defined by
$$ \phi(z;s,\alpha) = \sum_{n=0}^{\infty} \frac{z^{n}}{(n+\alpha)^{s}}.$$
The series expansion of $\sqrt{1+x}$ is, for the first few terms,
$$\sqrt{1+x} = 1 + \frac{x}{2} - \frac{x^{2}}{8}+ \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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How do I find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17? Recently I came across a question,
Find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17?
At first I applied sum of G.P. formula but ended up with the expression $1\cdot \dfrac{4^{41}-1}{4-1}$. I couldn't figure out how ... | $$\frac{4^{41}-1}{4-1}\equiv \frac{2^{82}-1}{4-1}\equiv \frac{2^2-1}{4-1}\equiv 1\pmod{17}.$$
Fermat's little theorem was used: $2^{16}\equiv 1\pmod{17}$ implies:
$$ 2^{82} = 2^{5\cdot 16+2} = 4\cdot \left(2^{16}\right)^5 \equiv 4\pmod{17}.$$
Another approach. Let $a_n = 4^0+4^1+\ldots+4^n$. Then obviously $a_{n+1}=4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
$\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$ If $A_1A_2A_3.....A_n$ be a regular polygon and $\frac{1}{A_1A_2}=\frac{1}{A_1A_3}+\frac{1}{A_1A_4}$.Then find the value of $n$(number of vertices in the regular polygon).
I know that sides of a regular polygon are equal but i could not rel... | If you know a bit about complex numbers and roots of unity, here is another approach:
The $n^\text{th}$ roots of unity are vertices of a regular $n$-gon. Let $z=1$ denote $A_1$ and $w = e^{\frac{2\pi\iota}{n}}$ denote $A_2$ in the complex plane. Then $w^2$ denotes $A_3$ and $w^3$ denotes $A_4$. Now, the given equation ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 1
} |
Find the rank of matrix. "Find rank of matrix $\begin{bmatrix}
0&0&-3\\
9&3&5\\
3&1&1\\
\end{bmatrix}$
using echelon form?"
"I am getting an answer equal to 2."
"When I reduce this I get $\begin{bmatrix}
1&1/3&0\\... | It is indeed a rank 2 matrix.
$\begin{bmatrix}
0&0&-3\\
9&3&5\\
3&1&1\\
\end{bmatrix}\overset{r_1\leftrightarrow r_3}{\rightarrow}
\begin{bmatrix}
3&1&1\\
9&3&5\\
0&0&-3\\
\end{bmatrix}
\overset{r_2=r_2-3\times r_1}{\rightarrow}
\begin{bmatrix}
3&1&1\\
0&0&2\\
0&0&-3\\
\end{bmatrix}
\overset{r_3=r_3+\frac{3}{2}\times r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Determining the minimal polynomial over $\Bbb{Q}$ I was working on a homework assignment from Hungerford:
Find the minimal polynomial of the element $\sqrt{1+\sqrt{5}}$ over $\Bbb{Q}$.
Naturally the solution would be the polynomial with roots
$$ \pm \sqrt{1 \pm \sqrt{5}} $$
Which is found as
$$ x = \pm \sqrt{1 \pm... | It is enough to show that $(x^2-1)^2-5$, that is, $x^4-2x^2-4$ is irreducible over the rationals. There are no rational roots, so we only need to rule out factorization as a product of quadratics with integer coefficients.
Since there is no $x^3$ term. we can confine attention to factorizations $(x^2+ax+b)(x^2-ax+c)$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve the complex equation The equation is $$z^2 -4z +4+ 2i = 0$$
I know that i am supposed to use $$(a+bi)^2 = a^2 + 2abi + bi^2$$ to solve the equation but i am stuck on how to expand the equation.
Can you help out with which term to expand?
| Notice, we have $$z^2-4z+4+2i=0$$
$$(z-2)^2+2i=0$$
$$(z-2)^2=-2i\iff (z-2)^2=2i^3$$
$$z-2=\sqrt{2}i^{3/2}$$ Since, $i=\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}$, hence we get $$z-2=\sqrt{2}\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)^{3/2}=\sqrt{2}\left(\cos\left(2k\pi+\frac{\pi}{2}\right)+i\sin\left(2k\pi+\frac{\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Give the equation of a line that passes through the point (5,1) that is perpendicular and parallel to line A. The equation of line $A$ is $3x + 6y - 1 = 0$. Give the equation of a line that passes through the point $(5,1)$ that is
*
*Perpendicular to line $A$.
*Parallel to line $A$.
Attempting to find the parallel,... | Line perpendicular to A that passes through the point $(5,1)$: it is directed by the normal vector to A: $(3,6)$ \ or $(1,2)$. Hence its equation is
$$\frac{x-5}1=\frac{y-1}2\iff 2x-y-9=0. $$
Line parallel to A:it has the smame normal vector as A:
$$3x+6y=3\cdot 5+6\cdot 1\iff 3x+6y-21=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Perpendicular Bisector of Made from Two Points For a National Board Exam Review:
Find the equation of the perpendicular bisector of the line joining
(4,0) and (-6, -3)
Answer is 20x + 6y + 29 = 0
I dont know where I went wrong. This is supposed to be very easy:
Find slope between two points:
$${ m=\frac{y^2 - y^1}... | To find the midpoint, you don't need to negate the coordinates. So the midpoint is $\left(\frac{-6+4}{2}, \frac{0+(-3)}{2}\right)=\left(-1, \frac{-3}{2}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
The asymptotic of the number of integers that are sums of three nonnegative cubes Let $c(n) $ be the number of distinct integers between $0 $ and $n $ of the form $ a^3 + b^3 + c^3$, meaning the sum of $3$ nonnegative cubes.
$C(n) = O( n \space \ln(n)^x ) $
Find and prove the optimal real value of $x$.
| Maybe this could help:
The amount of cubes lower than or equal to $n$ is $\lfloor\sqrt[3]{n}\rfloor$.
If we just have $A^3+B^3$ instead of $A^3+B^3+C^3$ we get:
$$C(n)=\dfrac{1}{2}\sum_{i=1}^{\lfloor\sqrt[3]{n-1}\rfloor}\lfloor\sqrt[3]{n-i^3}\rfloor$$
And for $A^3+B^3+C^3$ (which is the result you want):
$$C(n)=\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Alternative area of a triangle formula The problem is as follows:
There is a triangle $ABC$ and I need to show that it's area is: $$\frac{1}{2} c^2 \frac{\sin A \sin B}{\sin (A+B)}$$
Since there is a half in front I decided that base*height is equivalent to $c^2 \frac{\sin A \sin B}{\sin (A+B)}$. So I made an assumptio... | Let's denote by $[ABC]$ the area of $\triangle ABC$, its known that $$[ABC]=\frac{1}{2}ab\sin C$$
From Sine Law we have $a\sin C=c\sin A$ and $b\sin C=c\sin B$, also $\sin (A+B)=\sin(\pi-C)=\sin C$, then
\begin{align*}
[ABC]&=\frac{1}{2}\frac{(a\sin C)(b\sin C)}{\sin C}\\
&=\frac{1}{2}\frac{(c\sin A)(c\sin B)}{\sin (A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find $\int\frac{x^2+x}{(e^x+x+1)^2}dx$ $\int\frac{x^2+x}{(e^x+x+1)^2}dx$
I tried solving it but could not finish.I tried putting $e^x=t$ but not getting integrable.Please help me in solving it.
| \begin{align}
\int\frac{x(x+1)}{(1+e^x+x)^2}dx&=\int\frac{(1+e^x+x)^2-(1+e^x)(1+e^x+x)-xe^x}{(1+e^x+x)^2}dx\\
&=\int1dx-\int\frac{1+e^x}{(1+e^x+x)}dx-\int\frac{xe^x}{(1+e^x+x)^2}dx\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integrate $\int \frac{4x^5-7x^4+8x^3-2x^2+4x-7}{x^2(x^2+1)^2}dx$ $\displaystyle\int \frac{4x^5-7x^4+8x^3-2x^2+4x-7}{x^2(x^2+1)^2}dx$
I attempted but in vain.
$\displaystyle\int \frac{4x^5-7x^4+8x^3-2x^2+4x-7}{x^2(x^2+1)^2}dx=\int \frac{4x(x^4+1)}{x^2(x^2+1)^2}-7\frac{(x^4+1)}{x^2(x^2+1)^2}+2\frac{x^2(4x-1)}{x^2(x^2+1)^... | By partial fractions decomposition we have $$\int\left(\frac{12}{\left(1+x^{2}\right)^{2}}-\frac{7}{x^{2}}+\frac{4}{x}\right)dx
$$ now for the first integral we can use the substitution $x=\tan\left(u\right)
$ to get $$12\int\frac{dx}{\left(1+x^{2}\right)^{2}}=12\int\cos^{2}\left(u\right)du=6\int\cos\left(2u\right)du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving a Complicated Trig Problem I am preparing for AIME and I came across this problem which I need help solving:
$$\begin{eqnarray} 10^{10^{10}} \sin\left( \frac{109}{10^{10^{10}}} \right) - 9^{9^{9}} \sin\left( \frac{101}{9^{9^{9}}} \right) - 8^{8^{8}} \sin\left( \frac{17}{8^{8^{8}}} \right) + 7^{7^{7}} \sin\left(... | Since $\sin x\approx x$ for small $x$, this should be pretty darn close to $$109−101−17+76+113=180.$$
How close? By far the largest error comes from the $6^{6^6}\sin\frac{113}{6^{6^6}}$ term, because that argument to the sine is hugely larger than the other arguments (power towers grow fast). And we can estimate the re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\infty}\frac{\ln x dx}{x^2+2x+2}$ $$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}$$
$$\int_{0}^{\infty}\frac{\ln x .dx}{x^2+2x+2}=\int_{0}^{\infty}\frac{\ln x .dx}{(x+1)^2+1}\\
=\ln x\int_{0}^{\infty}\frac{1}{(x+1)^2+1}-\int_{0}^{\infty}\frac{1}{x}\frac{1}{(x+1)^2+1}dx$$ and then lost track,answer is $\frac{\p... | Let $$\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2+2x+2}dx = \int_{0}^{\infty}\frac{\ln x}{(x+1)^2+1^2}dx$$
Put $(x+1) = \tan \theta \;,$ Then $dx = \sec^2 \theta d\theta$ and changing Limits
We get $$\displaystyle I = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\ln\left(\tan \theta - 1\right)}{1+\tan^2 \theta }\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 -... | You can use the Reduction formula:
$$I_n=\int e^{ax}\sin^n bx\mathrm. dx\\
=\frac{e^{ax}\sin^{n-1} bx (a\sin bx-nb\cos bx)}{a^2+n^2b^2}+\frac{n(n-1)b^2}{a^2+n^2b^2}I_{n-2}$$
Use $n=2,a=1,b=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
limit involving $e$, ending up without $e$. Compute the limit
$$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$
we have a bit complicated solution using Mean value theorem. Looking for others
| Clearly if $a_{n} = \left(1 + \dfrac{1}{n}\right)^{n}$ then we have
\begin{align}
L &= \lim_{n \to \infty}\sqrt{n}\{a_{n + 1} - a_{n}\}\notag\\
&= \lim_{n \to \infty}\sqrt{n}\left[\exp\left\{(n + 1)\log\left(1 + \frac{1}{n + 1}\right)\right\} - \exp\left\{n\log\left(1 + \frac{1}{n}\right)\right\}\right]\notag\\
&= \lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Linear equations (solving by substitution) I'm having trouble solving this equation with fractions.
$$\frac 23 x+\frac 16 y=\frac 23 \\
-y=12-2x$$
| Notice, we have $$\frac{2}{3}x+\frac{1}{6}y=\frac{2}{3}\tag 1$$ $$-y=12-2x\tag 2$$
$$\implies y=2x-12$$ setting this value of $y$ in the eq(1), we get $$\frac{2}{3}x+\frac{1}{6}(2x-12)=\frac{2}{3}$$
$$2x+x-6=2$$ $$3x=2+6=8$$ $$x=\frac{8}{3}$$ Setting the value of $x$ in (2), we get $$-y=12-2\frac{8}{3}=\frac{36-16}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the distance from the origin to the surface $xy^2 z^4 = 32$ using the method of Lagrange Multipliers Problem: Find the distance from the origin to the surface $xy^2z^4 = 32$.
Attempt: The Lagrange equation for this problem is $L(x,y,z, \lambda) = x^2 + y^2 + z^2 + \lambda (xy^2 z^4 - 32)$. Setting the first par... | Your solution is correct, except the last part. You should get four minimizing points. (You mistakenly assumed that $y$ and $z$ must be both positive or both negative.)
However, there is a solution without using Lagrange multipliers. Note by AM-GM that
$$x^2+y^2+z^2=x^2+2\left(\frac{y^2}{2}\right)+4\left(\frac{z^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $(a+b)(b+c)(c+a) \ge8$
Given that $a,b,c \in \mathbb{R}^{+}$ and $abc(a+b+c)=3$,
Prove that $(a+b)(b+c)(c+a)\ge8$.
My attempt: By AM-GM inequality, we have
$$\frac{a+b}{2}\ge\sqrt {ab} \tag{1}$$
and similarly
$$\frac{b+c}{2}\ge\sqrt {bc} \tag{2},$$
$$\frac{c+d}{2}\ge\sqrt {cd} \tag{3}.$$
Multiplying $(1)... | If the numbers are positive here is a solution
$$3=abc(a+b+c)\ge 3abc(abc)^{\dfrac 1 3} \Rightarrow abc\le 1 \\$$
$$3=abc(a+b+c)\le \dfrac {(a+b+c)^3} {27} \cdot (a+b+c) \Rightarrow a+b+c \ge 3 \\ $$
$$ (ab+bc+ca)^2 \ge 3abc(a+b+c)=9 \Rightarrow ab+bc+ca \ge 3$$
$$(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)-abc \ge 9-1=8$$
equal... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$ According to Mathematica, we have that
$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$
that frankly speaking looks pretty nice.
However Mathematica shows tha... | Does it look any nicer?
$$\int\frac{\arctan x}{x\,\left(x^2+1\right)^5}\, dx=\frac12\,\Im\operatorname{Li}_2\left(e^{2\,i\arctan x}\right)\\-\frac x{9216\,\left(x^2+1\right)^4}\left(3963x^6+12995x^4+14525x^2+5637\right)\\+\left[\ln\frac{2x}{\sqrt{x^2+1}}+\frac{12x^6+42x^4+52x^2+25}{24\,\left(x^2+1\right)^4}-\frac{1321}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 1
} |
Binomial sum with two parameters Let $m$ and $n$ be two integers. Evaluate
$$S_{m,n}=\sum_{j=0}^{m} (-1)^j \binom{m}{j}\binom{mn-jn}{m+1}$$
At first, for $n=2$ I got $S_{m,2}=2^{m-1}m$, for $n=3$ I obtained $S_{m,3}=3^m m$, then I tried in vain to prove by induction that
$$S_{m,n}=\frac{n^k k(n-1)}{2}$$
| Suppose we seek to evaluate
$$S(m,n) = \sum_{j=0}^m (-1)^j {m\choose j}
{mn-jn\choose m+1}.$$
Introduce
$${mn-jn\choose m+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn-jn} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{m+2}} (1+z)^{mn}
\sum_{j=0}^m
(-1)^j {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$ $\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$
Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$
$\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different ... | Let $$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{x}}{x^2-x+1}dx.............(1)$$
Now Replace $x\rightarrow (1-x)\;,$ We get
$$\displaystyle I = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{(1-x)^2-(1-x)+1}dx = \int_{0}^{1}\frac{\sin^{-1}\sqrt{1-x}}{x^2-x+1}dx$$
So we get $$\displaystyle I = \int_{0}^{1}\frac{\cos^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$ $$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt{x+x^2+x^3}}$$
My Attempt:
$$\int_{0}^{1}\frac{1-x}{1+x}.\frac{dx}{\sqrt x\sqrt{1+x(1+x)}}$$
Replacing $x$ by $1-x$,we get
$$\int_{0}^{1}\frac{x}{2-x}.\frac{dx}{\sqrt{1-x}\sqrt{1+(1-x)(2-x)}}$$
Then I got stuck. Pleas... | Hint:
$$I =\int {\frac{{1 - x}}{{1 + x}}\frac{{dx}}{{\sqrt {x + {x^2} + {x^3}} }}} = \int {\frac{{1 - x}}{{1 + x}}\frac{{dx}}{{\sqrt x \sqrt {1 + x + {x^2}} }}} $$
Making $u=\sqrt{x}$ gives
$$I=2\int {\frac{{1 - {u^2}}}{{1 + {u^2}}}\frac{{du}}{{\sqrt {1 + {u^2} + {u^4}} }}} = 2\int {\frac{{\frac{1}{{{u^2}}} - 1}}{{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the rate of change in height of water level. A cone shaped container has a diameter of $0.6m$ and height of $0.5m$. Water is poured into the container with a constant rate of $0.2m^{3}s^{-1}$.
Calculate the rate of change in height of the water level when its height reaches $0.4m$.
Ok, so $\frac {dV}{dt} = 0.2$
V ... | Using similar triangles, we have that $\frac{r}{h}=\frac{.3}{.5}$, so $r=\frac{3}{5}h$.
Then $\displaystyle V=\frac{1}{3}\pi r^2h=\frac{\pi}{3}\big(\frac{3}{5}h\big)^2h=\frac{3\pi}{25}h^3,\;\;$ so $\displaystyle\frac{dV}{dt}=\frac{9\pi}{25}h^2\frac{dh}{dt}$.
When $h=.4$, this gives $\displaystyle.2=\frac{9\pi}{25}\big(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$ $$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx$$
I tried to solve it.
$$\int\frac{1+x^2}{x^4+3x^3+3x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x^3+x^2-3x+1}dx=\int\frac{1+x^2}{(x^2+1)^2+3x(x^2-1)+x^2+1}dx$$
But this does not seem to be solving.Please help.
| Using Hermite reduction and the Rothstein-Trager Algorithm for integration of rational functions I get
$$\frac{2}{\sqrt{11}}\left(\arctan\left(\frac{3}{\sqrt{11}}+\frac{2}{\sqrt{11}}x\right)-\arctan\left(\frac{3}{\sqrt{11}}-\frac{8}{\sqrt{11}}x-\frac{6}{\sqrt{11}}x^2-\frac{2}{\sqrt{11}}x^3\right)\right)
$$
which seems ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
} |
How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I ... | Since you're just looking for the first few terms:
$$f(x)=\frac{1}{1+e^x} \implies f(0)=\frac 12$$
$$f'(x)=\frac{-e^x}{(1+e^x)^2} \implies f'(0)=-\frac 14$$
$$f''(x)=\frac{-e^x}{(1+e^x)^2}+2\frac{e^{2x}}{(1+e^x)^3}\implies f''(0)=0$$
$$f'''(x)=f''(x)+4\frac{e^{2x}}{(1+e^x)^3}-3\times 2\frac{e^x}{(1+e^x)^4}\implies f'''... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
If $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$ where $x$ and$y$ are independent variable.Find $f(x).$ If $f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$ where $x$ and$y$ are independent variable.Find $f(x).$
I tried to solve it.
$f(x)=x+\int_{0}^{1}[xy^2+x^2y]f(y)dy$
$f(x)=x+x\int_{0}^{1}y^2f(y)dy+x^2\int_{0}^{1}yf(y)dy$
I applied i... | $$
f(x)
= x + \int_{0}^{1}[xy^2+x^2y]\,f(y)\,dy
= x + x \int_{0}^{1}y^2 f(y)\,dy + x^2 \int_{0}^{1}y\,f(y)\,dy
= x + Ax + Bx^2
$$
where
$$
A = \int_{0}^{1}y^2 f(y)\,dy
\qquad\mbox{and}\qquad
B = \int_{0}^{1}y\,f(y)\,dy
$$
So,
$$
A = \int_{0}^{1}y^2 \underbrace{\left(y + Ay + By^2\right)}_{f(y)}\,\,dy
= \frac{A+1}{4} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to compute the monstrous $ \int_0^{\frac{e-1}{e}}{\frac{x(2-x)}{(1-x)}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx} $ A friend told me, that he found a closed form for the following integral:
$$
\int_0^{\frac{e-1}{e}}{\frac{x(2-x)}{(1-x)}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)... | Notice, we have $$\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(1+\frac{x^2}{2-2x}\right)\right)}{2-2x+x^2}dx$$
$$=\int_{0}^{\frac{e-1}{e}}\frac{x(2-x)}{1-x}\frac{\log\left(\log\left(\frac{2-2x+x^2}{2-2x}\right)\right)}{2-2x+x^2}dx$$
Let, $$\log\left(\frac{2-2x+x^2}{2-2x}\right)=u$$
$$\implies \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Does there exist a function that is differentiable everywhere with everywhere discontinuous partial derivatives? Does there exist a function that is differentiable everywhere with everywhere discontinuous partial derivatives?
I just had this doubt, talking about first order partials.
| Yes it does. Take $$
f\left( {x,y} \right) = \left\{ \begin{array}{l}
x^2 \sin \left( {\frac{1}{x}} \right) + y^2 \sin \left( {\frac{1}{y}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,xy \ne 0 \\
x^2 \sin \left( {\frac{1}{x}} \right),\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Does $(1^a+2^a+3^a+4^a+5^a)^b=1^c+2^c+3^c+4^c+5^c$ imply $(a,b,c)=(1,2,3)$?
Question : Is the following proposition true?
Proposition : For positive integers $a,b,c$ where $b\ge 2$, if
$$(1^a+2^a+3^a+4^a+5^a)^b=1^c+2^c+3^c+4^c+5^c$$then $(a,b,c)=(1,2,3)$.
This is an unsolved case of this question where it has been pr... | In general we have
$$
\sum_{k=1}^n k^a = \frac{1}{a+1} \prod_{\jmath=1}^{a+1} \big( n + n_\jmath\big)
\tag 1
$$
Examples
$$
\begin{eqnarray}
\sum_{k=1}^n k &=& \frac{1}{2} n \big( n + 1 \big)\\\
\sum_{k=1}^n k^2 &=& \frac{1}{3} n \big( n + 1 \big) \big( n + 1/2 \big)\\\
\sum_{k=1}^n k^3 &=& \frac{1}{4} n^2 \big( n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.