Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Can we prove that $(2^{2^{n}}+1)+(2^{2^{n-1}}+1) -1$ have at least $n$ different prime divisors My point was to prove it using induction.
Then for $n=1$ it is true, for n = k it is also true so we assume that
$ \color{Red}{ 2^{2^{k}} + 2^{2^{k-1}} + 1}$ has at least k different prime divisors
When we consider $n=k+1$ w... | Assuming $2^{2^k}+2^{2^{k-1}}+1$ has at least $k$ different prime divisors,
for the inductive step it suffices to show that $2^{2^k}-2^{2^{k-1}}+1$ has a different prime factor.
Well, if $p$ divides $2^{2^k}+2^{2^{k-1}}+1$ and $2^{2^k}-2^{2^{k-1}}+1$, then $p$ divides their difference, which is a power of $2$,
so $p$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Closed form for the integral $\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$
Somewhere, the integral
$$\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$$
has been found numerically to be real and finite, if $k \in (-\infty,3].$
Can one find a closed form for this i... | Denoting the integral by $I(k)$, let us use $\int_{-a}^{a}f(x)dx, =\int_{0}^{a} [f(x)+f(-x)] dx,$ then
$$I(k)=\int_{-1}^{1} \frac{\ln[1+x^2-x\sqrt{k+x^2}]}{\sqrt{1-x^2}} dx= \int_{0}^{1} \frac{\ln[1+(2-k)x^2]}{\sqrt{1-x^2}}dx =\frac{1}{2} \int_{0}^{\pi} \ln \left (\frac{1+r^2}{2}+\frac{1-r^2}{2} \cos \theta \right) d \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Simplify $\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$ I have to simplify the following expression:
$A =\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}$
Answer: $\sqrt{a+b}-\sqrt{a-b}$
I am trying to find the constraints of $a$ and $b$. I think that $a^2-b^2 \ge 0$ and $a+b \ge 0$. How can I simplify them? (the inequalities)
| Because $$\frac{a+b-\sqrt{a^2-b^2}}{\sqrt{a+b}}=\frac{\left(\sqrt{a+b}\right)^2-\sqrt{(a-b)(a+b)}}{\sqrt{a+b}}=\sqrt{a+b}-\sqrt{a-b}$$
The domain is $a+b>0$ and $(a-b)(a+b)\geq0,$ which gives $a+b>0$ and $a-b\geq0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$ Discovered the integral below
$$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$
which looks clean, yet challenging. Have not seen it before. Post it here in case anyone is interested.
Edit:
Here is a solution. Let $J(a)=\int_0^{\frac\pi{3}}\tan... | Let $I$ be the integral given by
$$\begin{align}
I&=\int_0^{\pi/3}\text{arctanh}(\sin(x))\,dx\\\\
&=\frac12\int_0^{\pi/3}\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\,dx\\\\
&=\frac12\int_{-\pi/3}^{\pi/3}\log\left(1+\sin(x))\right)\,dx\tag1
\end{align}$$
Then, enforcing the substitution $x\mapsto \pi/2-x$ in $(1)$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
compute the following integral in closed form : $\int_0^{\infty}\frac{\arctan x}{a^2x^2+1}\,dx$
Find :
$$\int_0^{\infty}\frac{\arctan x}{a^{2}x^2+1}\,dx,\qquad a > 0.$$
I think this integral related with polylogarithm function.
My attempt as follows:
Let $$I(b)=\int_0^{\infty}\frac{\arctan bx}{1+a^{2}x^{2}}\,dx... | Your final expression has extra $a^2$ and $b^2$. It should read $$I'(b)=\frac{1}{2(a^2-b^2)}\left.\ln\frac{1+a^2x^2}{1+b^2 x^2}\right|_{0}^{\infty}=\frac{\ln a-\ln b}{a^2-b^2}$$ which is integrated using $I(0)=0$: $$I(b)=\int_{0}^{b}\frac{\ln a-\ln x}{a^2-x^2}\,dx=\frac{1}{a}\int_{0}^{b/a}\frac{\ln(1/t)}{1-t^2}\,dt=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Inequality between $\frac{1}{\sqrt[n]{1+a}}+\frac{1}{\sqrt[n]{1+b}}+\frac{1}{\sqrt[n]{1+c}}$ and $\frac{3}{\sqrt[n]{1+(abc)^{1/3}}}$ Earlier
Exploring an inequality between $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} $ and $\frac{3}{1+(abc)^{1/3}}$ if $a,b, c>0$
the expressions $$\frac{1}{\sqrt[n]{1+a}}+\frac{1}{\sqrt[... | Take $f(x)=\frac{1}{(1+e^x)^{1/n}}, n>0$. then $f''(x)=\frac{e^x(e^x -n)}{n^2(1+e^x)^{1/n+2}}>0,~ \mbox{if}~ e^x>n.$ Then Jensen's inequality
$$\frac{f{x}+f(y)+f(z)}{3} \ge f\left(\frac{x+y+z)}{3}\right)~~~~(1)$$ holds. So we have $$\frac{1}{\sqrt[n]{1+e^x}}+\frac{1}{\sqrt[n]{1+e^y}}+
\frac{1}{\sqrt[n]{1+e^z}}\ge \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Find the sum of the even-degree coefficients I am doing practice problems from the AMC-10 math competition. I do not understand how to go about solving the problem
Let $(1+x+x^2)^n = a_0+a_1x+a_2x^2+...+a_{2n}x^{2n}$ be an identity of $x$. If we let $s = a_0 + a_2 + a_4 + ... + a_{2n}$, find $s$.
The choices are $A, 2^... | As noted by lab in a comment, for $x=1,-1$ you get
$$3^n = a_o+a_1+\ldots+a_{2n} \\ 1 = a_o-a_1+\ldots+a_{2n}$$
so $3^n+1 = 2a_0+ 2a_2+\ldots+2a_{2n}$.
Another (lazy) idea if you already have some candidates: for $n=1, a_0+a_2=2$, for $n=2, (1+x+x^2)^2 = 1+2x+3x^2+2x^3+x^4$, so $a_0+a_2+a_4 = 5$. Then the only candid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3284880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I take a fraction to a negative power? I ran into this issue during my homework. Using the rules of logarithms, I need to prove that
$$
-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln3-\ln2
$$
So here were my steps:
*
*First step:
$$
-2\ln\bigg(\frac{2}{\sqrt{6}}\bigg)=\ln\left(\bigg(\frac{2}{\sqrt{6}}\bigg)^{-2}\ri... | By definition $$a^{-k} = \frac 1{a^k}$$
So $$\left(\frac{2}{\sqrt{6}}\right)^{-2} =\frac 1{\left(\frac{2}{\sqrt{6}}\right)^{2}}=$$
$$\frac 1{\left(\frac {2^2}{\sqrt 6^2}\right)}=\frac {\sqrt 6^2}{2^2}=\frac 64=\frac 32$$
It will help to realize that $(\frac ab)^{-1} = 1/(a/b) = \frac ba$ and that $(\frac ab)^k = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
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Prove that $\sum_{k=1}^{n} \frac{(-1)^{k}}{k}B(n,k) = 0 $ for all integer $n\geq 2$ For all $ n, k\geq 1$, define $B(n,k) = \sum_{a_{1}+a_{2} +\dots+a_{k} =n} \frac{1}{a_{1}!a_{2}! \cdots a_{k}! }$, where $ a_{1},a_{2},\dots,a_{k} $ are positive integers.
For example: $B(3,2) = \frac{1}{1!2!} + \frac{1}{2!1!} =1$ be... | It is well-known that
$$
e^x = \sum_{a=0}^\infty \frac{x^a}{a!}.
$$
This implies that
$$
(e^x-1)^k = \sum_{n=0}^\infty B(n,k) x^n.
$$
It is also known that
$$
\ln (1+x) = -\sum_{k=1}^\infty (-1)^k \frac{x^k}{k}.
$$
Therefore
$$
\ln (1 + (e^x-1)) = -\sum_{k=1}^\infty (-1)^k \frac{(e^x-1)^k}{k} = -\sum_{k=1}^\infty \sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How many primes are there of the form $x^6+y^6$, where $x,y\in\mathbb{Z}$? I'm trying to find such primes that are of the form $x^6+y^6$; $x,y\in\mathbb{Z}$.
If one of $x$ and $y$ is $0$, say $y=0$ then $x^6+y^6=x^6$ which is not a prime. So assume that $(x,y)\ne(0,0).$
I factored $x^6+y^6$ as follows:
$x^6+y^6=(x^2)^3... | If $x=0$ or $y=0$ then $x^6+y^6$ is $x^6$ or $y^6$ which cannot be a prime.
If $x\ne 0\ne y$ and $x^6+y^6$ is prime then $$x^6+y^6=(x^2+y^2)(x^4-x^2y^2+y^4)=(x^2+y^2)(\,(x^2-y^2)^2+x^2y^2
)$$ and the factor $x^2+y^2\ge 2,$ so the factor $(x^2-y^2)^2+x^2y^2$ must be $1,$ which is not possible unless $|x|=|y|=1$ because ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$ Prove the following for all real $x$
i. $⌊2x⌋+⌊2y⌋≥⌊x⌋+⌊y⌋+⌊x+y⌋$
ii. $⌊x⌋-2⌊x/2⌋$ is equal to either $0$ or $1$
For ($ii$.) I attempted to split it into cases of whether the fraction part {$x$} is $≥.5$ or $<5$ but that ended up being too tedious and I know there must be a more elegant,... | Part ii:
In general: $$\lfloor z\rfloor \leq z< \lfloor z\rfloor+1.$$
For ii, set $z=x/2$ then we get:
$$\left\lfloor\frac{x}{2}\right\rfloor\leq \frac{x}{2}<\left\lfloor\frac{x}{2}\right\rfloor +1$$
Double and you get:
$$2\left\lfloor\frac{x}{2}\right\rfloor\leq x<2\left\lfloor\frac{x}{2}\right\rfloor +2$$
Taking the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ A = \begin{pmatrix}
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t ... | Observe that $(XA+A^TX^T)$ is a symmetric matrix. So the image will be a subset of the set of symmetric matrices. But the question about a basis for the image set can be resolved as follows:
$$\varphi(X)=\begin{bmatrix}
2a-2b & -a+b+c-d\\
c-d-a+b & -2c+2d\\
\end{bmatrix}=(a-b)\begin{bmatrix}2&-1\\-1&0\end{bmatrix}+(c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Evaluate $\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$ Find:
$$\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$$
I don't know how I starte & evaluate this integral
Wolfram alpha give $=1,28553$
My problem whene I use $t=\cos x$ I get $\arccos x$
Same problem w... | Let $x=\sin t$ and $a= \sin^{-1}\frac\pi4$
$$I=\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx = \int_0^a \frac{\cos ^2 t+\cos (\sin t)}{1+\sin t \sin (\sin t)} \, dt
$$
Use the shorthand $t_\pm=\frac 12 (t \pm {\sin t})$ to express the denominator and the numerator as
\begin{align}
1+ \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$? If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?
Attempt:
$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$
If we multiply the two "substraction identities" we get
$$ (\si... | HINT- $sinx-siny=2cos\frac{x+y}{2}sin\frac{x-y}{2}=-\frac{1}{3}$ ..........(1)
$cosx-cosy=-2sin\frac{x+y}{2}sin\frac{x-y}{2}=\frac{1}{2}$ ..........(2)
Divide to get $tan\frac{x+y}{2}=\frac{3}{2}$,Now use $sin\theta=2tan\frac{\theta}{2}/(1+tan^2\frac{\theta}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Calculating $\lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4}$ I can't figure out how to calculate this limit (or prove it does not exist)
$$
\lim_{(x,y) \to(0,0)} \frac{x^2y}{x^2+y^4}
$$
I've tried with restrictions on $y=mx$ and curves of the form $y=x^n$.
The limit should not exist but even with polar coordinates I can't... | Let $f(x,y)={\large{\frac{x^2y}{x^2+y^4}}}$.
Let $x^2+y^2=r^2$, with $0 < r \le 1$.
If $x\ne 0$, then
\begin{align*}
|f(x,y)|&=\left|\frac{x^2y}{x^2+y^4}\right|\\[4pt]
&\le\left|\frac{x^2y}{x^2+x^2y^4}\right|\;\;\;\;\;\text{[since $x^2\le r^2\le 1$]}\\[4pt]
&=\left|\frac{y}{1+y^4}\right|\\[4pt]
&\le |y|\\[4pt]
&\le r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Finding values of $\tan^{-1} (2i)$. I am trying to find all solutions of $\tan^{-1} (2i)$. I don't see anything that I have done wrong, my answer doesn't match the one in the textbook. Here is what I have. (The convention in Brown and Churchill is to use $\log$ for a complex number and $\ln$ for a real number.)
\begin... | Both are correct.
If $n_1$ is the integer plugged into your solution, plugging $-n_1-1$ into the textbook solution gives the same answer. $$-\left(n_1+\frac{1}{2}\right) = k$$
$$(-n_1-1)+\frac{1}{2} = k$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| $ (\sqrt u + \sqrt v)^2 = u+v + 2\sqrt{u\,v }$
So at first bring in $2$ in front of the radical sign.
$$\sqrt{2+\sqrt3}=\sqrt{\frac{4+2\sqrt3}{2}}$$
Next find two factors for $3$ whose sum is $4$
They are easy to guess. They are $3$ and $1$;
(... or else you need to solve another quadratic equation)
$$ \sqrt{\frac{(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Finding $P(X<2Y)$ given joint pdf $f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}$ for $x,y\in\mathbb R$
The joint pdf of $X$ and $Y$ is $$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2 + y^2}}\,; \quad x,y\in \mathbb{R}$$ Find $P(X<2Y)$.
I have tried this:
$$\int_{-\infty}^{\infty}\int_{-\infty}^{2y} f(x, y)\, dx\,dy$$
@S... | So this is my version that I have understood from @pre-kidney's answer please note that this is just an elaboration of his answer.
$f(x, y) = \frac {1}{2\pi} e^{-\sqrt{x^2+y^2}} \ ;\ -\infty<\ x \ <\infty\ ; -\infty < y < \infty$
$X=rcosθ,Y=rsinθ$
$f(r, \theta) = f(\theta).f(r) = (\frac {1}{2\pi} ).(re^{-r})\ \ ;\ \ 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Find a condition on real numbers $a$ and $b$ such that $\left(\frac{1+iz}{1-iz}\right)^n = a+ib$ has only real solutions I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that
$$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$
has only real solutions
This is what I got till now.
$$\... | Note that
$$
|1-iz|^2 - |1+iz|^2 = -2i(z -\bar z) = 4 \operatorname{Im}(z)
$$
so that
$$
z \in \Bbb R \iff \left | \frac{1+iz}{1-iz}\right|= 1 \, .
$$
It follows that
$$
\left(\frac{1+iz}{1-iz}\right)^n = a+ib
$$
has only real solutions $z$ if and only if $|a+ib|=1$, i.e. if $a^2+b^2=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3307060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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What is the eccentricity of $4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$? Question: What is the eccentricity of $4\sqrt{5}x^2 − 17x + y + \frac{1}{8} = 0$?
Source: James S. Rickards Invitational (Algebra II Individual)
I know how to calculate eccentricity, but I am having trouble on factoring it in to $\frac{x^2}{a^2}+\f... | Note, as given, you have: $4\sqrt{5}x^2-17x+y+\frac{1}{8}=0$.
Just rewrite this as $y=-4\sqrt{5}x^2+17x-\frac{1}{8}$.
It is very easy to see now that this is just the equation of a parabola.
Now by definition, any parabola has an eccentricity of $1$.
No need to complete the square here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac... | Let $3^a=4^b=k^{ab}\implies 3=k^b,4=k^a$
$$9^{a/b}+16^{a/b}$$
$$=(3)^{2a/b}+4^{2b/a}$$
$$=(k^b)^{2a/b}+(k^a)^{2b/a}$$
$$=k^{2b}+k^{2a}$$
$$=(k^b)^2+(k^a)^2=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
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Prove that if n is divisible by $2$ and $3$, then n is divisible by $6$
For every integer n, if $2 | n$ and $3 | n$ then $6 | n$
! Note: x | y means y is divisible by x.
!! Note: I know that there are way better ways to prove it. However, I am just curious whether the proof below, admittedly peculiar, is correct.
S... | Simpler: $\ \dfrac{n}2\in\Bbb Z,\, \dfrac{n}3\in \Bbb Z\,\Rightarrow\, \dfrac{n}6 = \dfrac{n}2-\dfrac{n}3\in\Bbb Z.\ $ Turning to your argument:
Suppose $\frac{n}{6} \notin \mathbb Z$. Since $5 \cdot \frac{n}{6}$ is an integer, $\frac{n}{6}$ can be rewritten as $\frac{n}{6} = a + 0.2$, where $a \in \mathbb Z$.
This ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Matrix rearrangement I have a matrix A in this form:
$$
A=
\left[
\begin{array}{cccc}
x_1 & x_2 & 0 & 0 \\
0 & 0& x_1 & x_2
\end{array} \right]
$$
Here, $x_1$ and $x_2$ are variables, and A is a $2 \times 4$ matrix. I
would like to rearrange the matrix product so that
$$
A^TA=BC
$$
$A^T$ is the transpose of A. $B$... | (Not really an answer but at least some development...)
Let's try it!
The left-hand side is easy to calculate:
$$
A^T A =
\left[
\begin{array}{cccc}
x_1^2 & x_1 x_2 & 0 & 0 \\
x_1 x_2 & x_2^2 & 0 & 0 \\
0 & 0 &x_1^2 & x_1 x_2 \\
0 & 0 &x_1 x_2 & x_2^2 \\
\end{array}
\right]
$$
The right-hand side should have this for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $ \sin a +\sin c =2 \sin b $, show that $ \tan\frac{a+b}{2}+\tan\frac{b+c}{2} = 2 \tan\frac{c+a}{2}$ Found the question in the textbook.
I tried many methods of manipulating the identity to be proved but I did not even see a clue of using the given condition(of different forms like
$$
\sin\frac{a+c}{2} \cos\frac{a-c... | Let $a+b=2z$ etc.
$\implies2b=a+b+b+c-(c+a)=2(z+x-y)\implies b=z+x-y$
$$\sin(y+z-x)+\sin(x+y-z)=2\sin(z+x-y)$$
$$\sin(x+y-z)-\sin(z+x-y)=\sin(z+x-y)-\sin(y+z-x)$$
$$\sin(y-z)\cos x=\sin(x-y)\cos z$$
Expand using $\sin(A-B)$ formula and divide both sides by $\cos x\cos y\cos z$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to evaluate the following integral$\int_a^{pa}\frac{ax}{\sqrt{(a-x)(x-pa)}}dx$? Can anyone help me to evaluate the definite integral $\int_a^{pa}\frac{ax}{\sqrt{(a-x)(x-pa)}}dx$?I encountered this integral while doing a problem of particle dynamics in Ganguly Saha(Applied Mathematics).Can this integral be evaluated... | Introducing integral by parts,
\begin{equation}
\begin{aligned}
I&=a \int^{pa}_a\frac{x}{\sqrt{(a-x)(p a-x)}}dx \\
&=\frac{a}{2}
\left\{
a(1+p)\int^{pa}_a\frac{1}{\sqrt{(a-x)(p a-x)}}dx
-\int^{pa}_a\frac{d[(a-x)(p a-x)]}{\sqrt{(a-x)(p a-x)}}
\right\} \\
&=\frac{a}{2}
\left\{
a(1+p)\int^{pa}_a\frac{1}{\sqrt{(a-x)(p a-x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\int _{0}^1 \sqrt {(1+x)(1+x^3)} ≤ \sqrt {15/8}$
Prove that
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)}\ dx ≤ \sqrt {15/8}$$
Now in this question, the solution that was provided me was $\int _{0}^1 \sqrt {(1+x)(1+x^3)}dx ≤ \sqrt{\int _{0}^1(1+x)dx \int _{0}^1(1+x^3)dx}$
I have never even heard or seen this in ineq... | Not an answer, just some thoughts.
In this case we can also try using the arithmetic mean - geometric mean inequality:
$$\int _{0}^1 \sqrt {(1+x)(1+x^3)} dx \leq \frac{1}{2} \int _{0}^1 (1+x) dx+\frac{1}{2} \int _{0}^1 (1+x^3) dx = \\ = \frac{3}{4}+\frac{5}{8}=\frac{11}{8}=1.375$$
Meanwhile:
$$\sqrt{\frac{15}{8}}=1.369... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Show that $7a^{2} = 13b^{2}$ Here is another question which states that -
The first and the third terms of an arithmetic sequence are
a and b respectively. The sum of the first n terms is denoted by $S_n$.
(a) Find $S_4$ in terms of a and b.
(b) Given that $S_4$, $S_5$,
$S_7$ are consecutive terms of a geomet... | It is not hard to see that $S(n)=\frac{1}{4}\left(-a n^2+5 a n+b n^2-b n\right)$. So$$\frac{S(7)}{S(5)}=\frac{S(5)}{S(4)}\iff\frac{13 b^2-7 a^2}{10 b (a+3 b)}=0\iff13b^2=7a^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
totally differentiable function $\frac{x^3}{(x^2+y^2)}$ - check my proof Consider the function: $f: \mathbb{R^2} \to \mathbb{R}$ given by $f(x,y)=\begin{cases} \frac{x^3}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y)=(0,0) \end{cases}$
I want to check if it is totally differentiable at ${0 \choose 0}$. By definition this m... | What you did is correct.
Maybe a simpler way would be to notice that for $x=y=t>0$
$$\frac{-xy^2}{(x^2+y^2)\Vert {x \choose y}\Vert}=-\frac{1}{2\sqrt 2}$$
and that consequently, above function can’t have $0$ for limit at $(0,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solve the Lagrange multiplier equations the equations :
$$\left\{\begin{array}{l}{x+6 y+4 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}\right.$$
I've done many transformations,but I still can't get the answer.
the last I did this:
$$\lambda=\left(6 \frac{x}{y}+2\right)=\left(\frac{1}{4}+\frac... | Restoring the objective function from the constraint (third) equation and partial derivatives:
$$Optimize \ \ z(x,y)=\frac12x^2+6xy+y^2 \ \ s.t. \ \ 4x^2+y^2-25=0.$$
If you take partial derivatives now:
$$\left\{\begin{array}{l}{x+6 y+\color{red}8 \lambda x=0} \\ {6 x+2 y+\lambda y=0} \\ {4 x^{2}+y^{2}-25=0}\end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solution verification on homework problem. Separable first order ODE IVP. The answer is supposedly $y^2 = 1 + \sqrt{x^2 - 16}$ I don't know where I went wrong cause I know for a fact that my substitution of $x = 4 \sec(\theta)$ is correct. I know for a fact that after substitution the integral becomes $\int \sec^2(\the... | You have to use another $4$ .
$x = 4\sec\theta \implies dx = \color{red}4\sec\theta\tan\theta
d\theta$
So, you'll have $y^2 = \sqrt{x^2-16} +c$
$2^2 = \sqrt{9} +c \implies c = 4-3= 1$
Thus,
$$y^2 =\sqrt{x^2+16} +1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means ... | We can rewrite the equation as follows:
$$4x^2 - 4x + y^2 + 2y - 7 = 0$$
By completing the squares, we obtain:
$$4\left (x^2 - x + \color{red}{\frac 1 4} \right ) + (y^2 + 2y + \color{blue}1) - 7 - \color{red}1 - \color{blue}1 = 0$$
that is,
$$\frac {\left (x - \frac 1 2 \right )^2} {\left ( \frac 3 2 \right )^2} + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Integer $N$, in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b<16$, then what is the successor of $b$?
An integer $N$, expressed in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b$ is less than $16$, then what is the successor of $b$?
I couldn't develop anything more than
$$N_b = \left(\;n\cdo... | It seems there are multiple answers.
Since $N$ can be written as $6789$ in base $b$ we know $$6b^3+7b^2+8b+9=N$$
If we divide this polynomial by $b-1$ we get $$6b^2+13b+21+\frac{30}{b-1}$$ but we know that $b-1$ divides $N$ so $\frac{30}{b-1}$ is an integer. $$b-1=1,2,3,5,6,10,15,30$$ $$b=2,3,4,5,7,11,16,31$$ but $b < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for $\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x$ It is known that $$\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x=\pi \ln \left|2 \cos \frac{a}{2}\right|+a \ln \left|\tan \frac{a}{2}\right|+2 \sum_{k=0}^{+\infty} \frac{\sin (2 k+1) a}{(2 k+1... | Another series solution, which might be helpful.
$$I(a)=\int_{0}^{+\infty} \frac{\ln \left(x^{2}+2 \sin a \cdot x+1\right)}{1+x^{2}} d x$$
$$x=\tan t$$
$$I(a)=\int_{0}^{\pi/2} \left(\ln \left(1+2 \sin a \sin t \cos t\right)-2\ln \cos t \right) d t$$
$$I(a)=\int_{0}^{\pi/2} \ln \left(1+2 \sin a \sin t \cos t\right)dt+ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
computing the Taylor series for $e^{1-\cos x}$ I was told to compute the Taylor series for $e^{1-\cos x}$ to the fourth degree.
I plugged the Taylor series for $1-\cos x$ into the Taylor series for $e$.
This got me: $$1-\frac{x^2}2+\frac{x^4}{24}+\frac{(\frac{-x^2}2)^2}{2}$$A bit of simplification results in: $1+\frac{... | My answer agrees with the answer key:
$\cos x=1-\dfrac{x^2}2+\dfrac{x^4}{24}+O(x^6),$ so $1-\cos x=\dfrac{x^2}2-\dfrac{x^4}{24}+O(x^6),$
so $\exp(1-\cos x)=1 + \left(\dfrac{x^2}2 - \dfrac{x^4}{24} \right)+ \dfrac12\left(\dfrac{x^2}2\right)^2+O(x^6)=1 + \dfrac{x^2}2 + \dfrac{x^4}{12}+O(x^6).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $x\sqrt{2}=1-x$ Solve for $x$ of the equation,
$$x\sqrt{2}=1-x$$
According to the solution, $x$ is supposed to be $\sqrt{2}-1$. The solutions to this test have proven wrong from case to case, so I wonder ...
If you solve it like this: $x\sqrt{2}+x = 1$ and pull out $x$ to get $x(\sqrt{2}+1) = 1$. Then, dividing... | $\frac 1{\sqrt 2 + 1}=\frac 1{\sqrt 2+1}\frac {\sqrt 2 -1}{\sqrt 2 -1} = \frac {\sqrt 2 -1}{\sqrt 2^2 + \sqrt 2 - \sqrt 2 - 1} = \frac {\sqrt 2 -1}{2-1} =\frac {\sqrt 2-1}1 = \sqrt 2 -1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3336537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
What is the 94th term of this sequence? $1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$
What is the 94th term of the following sequence?
$$1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$$
*
*8
*9
*10
*11
My Attempt: I found that the answer is 3rd option i.e. 94th term is 10. As every number is wr... | Using your method,
$n(n+1) \lt 94$
Easy to see $9*10 = 90$ so the $90$th term value is $9$.
And $91$st term is the beginning of value $10$.
$a(a+1)$th term has the value $a$ and its the end of that value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3339709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solve $\sin x + \cos x = \sqrt{1+k}$ for $\sin 2x$, $\sin x-\cos x$, and $\tan x$ in terms of $k$ Given that $\sin x + \cos x = \sqrt{1+k}$, $-1 \le k \le 1$
*
*Find the value of $\sin 2x$ in terms of k
*Given that $x \in (45^{\circ}, 90^{\circ})$ deduce that $\sin x - \cos x = \sqrt{1-k}$
*Hence, show that $\tan... | Here’s two more very straightforward ways to go about it:
$${\cos x+\sin x\over\sqrt 2}=\sqrt{1+k\over 2}$$
$$\implies \sin\left(x+\frac\pi4\right) = \sqrt{1+k\over 2}$$ (note that this expression is valid for all $k$ in the range). Hence, $$x=\arcsin\left(\sqrt{1+k\over 2}\right)-\frac \pi4$$
For the first part, we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the asymptotic form as $N \rightarrow \infty$ of $\sum_{a = 1}^{N} \sum_{u = 1}^{a - 2} \sum_{v = u + 1}^{a - 1} {\delta}_{N = u\, a + v}$ Here $N \ge 1$ is a positive integer and $a$, $u$, and $v$ are also integers. This triple sum arises from counting the number of reducible cubic polynomials. I am looking for ... | We can simplify the triple sum by rephrasing it to $$S = \sum_{a = 1}^{n} \sum_{u = 1}^{a - 2} [u+1 \le n-ua \le a-1]$$
Now rearranging this, we get $$\sum_{a = 3}^{n} \sum_{u = 1}^{a - 2} \left[\frac{n-a+1}{a} \le u \le \frac{n-1}{1+a} \right]$$
Now solve for $u$ in terms of $a$. For each value of $u$ there will be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is $x+y\alpha + z\alpha^2$ a subfield? Let $\alpha = \sqrt[3]{2}$. I want to Prove that the set of all numbers $x+y\alpha+z\alpha^2$, for $x,y,z \in \mathbb{Q}$, is a subfield of $\mathbb{C}$.
I have showed that this set is a subring, but I'm not sure how to show for any $a\neq 0$ in our set has a multiplicative inver... | You have to solve
$$
\begin{cases}
ax+2bz+2yc=1\\
ay+bx+2cz=0\\
az+by+cx=0
\end{cases}
$$
in the variables $a,b,c$. The notation here is somewhat of counterintuitive with respect to the usual one.
The matrix associate to the non-homogeneous linear system is
$$
A(x,y,z)=\begin{pmatrix}
x & 2z & 2y\\
y & x & 2z\\
z & y &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to find eigenvalues of the matrix This is a question from our end-semester exam:
How to find the eigenvalues of the given matrix:
M=\begin{bmatrix}
5,1,1,1,1,1\\
1,5,1,1,1,1\\
1,1,5,1,1,1\\
1,1,1,5,1,1\\
1,1,1,1,4,0\\
1,1,1,1,0,4\\
\end{bmatrix}
I know that $4$ is an eigenvalue of $M$ with multiplicity atleast $3... | For block matrices, you can use Schur complement:
$$0=\det(M-I\lambda)=\begin{vmatrix}A&B\\ C&D\end{vmatrix}=\det(D)\cdot \det(A-B\cdot D^{-1}\cdot C)= \\
\begin{vmatrix}4-\lambda&0\\ 0&4-\lambda\end{vmatrix}\cdot \det\left(A-B\cdot \begin{pmatrix}\frac1{4-\lambda}&0\\ 0&\frac1{4-\lambda}\end{pmatrix}\cdot C\right)=\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f... | There are a few ways to do this problem : one of which is a more general method which turns out to be very elaborate in this case. Let $p(x) = x^3 - x$.
The first is to note that $x^3 \equiv x \mod p(x)$, therefore $x^{2n+1} \equiv x \mod p(x)$ for all $n \geq 0$(show this by induction). Thus, the given expression simp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
If $\sqrt{k-9}$ and $\sqrt{k+36}$ are both integers, the number of possible values of $k$ is? If $\sqrt{k-9}$ and $\sqrt{k+36}$ are both integers, the number of possible values of $k$ is: A) $1$; B) $2$; C) $3$; D) more than $3$.
I started to try different values of $k$ in the first root ($13, 18, 25, 34$) and see if $... | Let's first solve $a^2-b^2=45$ with $a:=\sqrt{k+36},\,b:=\sqrt{k-9}$ both integers $\ge0$. Since $(a-b)(a+b)=3^2\times 5$ must be a factorisation into two odd positive factors with $a-b\le a+b$, either $$a-b=1,\,a+b=45\implies a=23,\,b=22\implies k=b^2+9=493,$$ or $$a-b=3,\,a+b=15\implies b=6\implies k=45,$$ or $$a-b=5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the image and inverse image of the following function $2x - x^2$ Where $f(E)$ denotes the image and $f^{-1}(E)$ denotes the inverse image.
$$E=[-2,2)$$
plugging in the values I get:
$f(-2)=-8$
$f(-1)= -3$
$f(0) = 0$
$f(1)= 1$
$f(2) = 0$
So all the numbers fall between the interval of $[-8,1)$ I think it has to be... | We can complete the square to determine the image.
\begin{align*}
f(x) & = 2x - x^2\\
& = -x^2 + 2x\\
& = -(x^2 - 2x)\\
& = -(x^2 - 2x + 1) + 1\\
& = -(x - 1)^2 + 1
\end{align*}
Thus, the graph of $f$ is a parabola with vertex $(1, 1)$ that opens downwards. Therefore, its maximum value on the inter... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3353425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$ Prove $f(x)$ has three extremum points on $(-2\pi,2\pi)$
$$f(x) = \frac{\sin x}{x}, x \not= 0 $$
$$f(x) = 1, x = 0 $$
I calculated its derivative.
$$f'(x) = \frac{x\cos x-\sin x}{x^2}, x \not= 0$$
$$f'(x) = 0, x = 0$$
I can see that $f'(x)$ has a solution for x =... | What you did is very correct and there is no explicit solution to the equation.
Discarding the trivial case $x=0$, it is better to consider that you look for the zero's of
$$f(x)=x \cos (x)-\sin (x)$$ Now, consider a Taylor expansion close to $\frac {3 \pi} 2$ (remember the symmetry) to get
$$f(x)=1+\frac{3}{2} \pi \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Minimal polynomial- fields Let $\zeta$ = $\cos(\frac{2\pi}{7} ) + i\sin(\frac{2\pi}{7} )$ , let $\alpha = \zeta +\zeta^{-1} $ note that $\zeta^{-1} =\zeta^6 $
I try to find the minimal polynomial of $\alpha$ over $\mathbb Q$.
I only managed to show that the degree of the minimal polynomial is 3.
My attempt so far:
$... | Now add the equation
$$
α^4=ζ^4+ζ^{-4}+4(ζ^2+ζ^{-2})+6=ζ^4+ζ^3+4α^2-2
$$
to your consideration to find that
$$
α^4-α^3-4α^2+3α+2=0
$$
This has a root at $2=1+1^{-1}$, the remaining factor is
$$
α^3 + α^2 - 2α - 1=0
$$
You could of course also start at
$$
0=\frac{ζ^7-1}{ζ-1}=1+ζ+ζ^2+ζ^3+ζ^4+ζ^5+ζ^6=1+α+(α^2-2)+(α^3-3α)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
if the function such $f(m+nf(m))=mf(n+1)$ find the f if $f:N^{+}\to N^{+}$,and such for any postive integer $m,n$ have
$$f(m+nf(m))=mf(n+1)$$
find the $f$
I guess this function is $f(n)=n$,But How to solve it?Thanks.
| Function $f(x)$ is defined on the countable set of points and can be considered as the countable sequence of values
$$\{f(1),f(2),f(3),\dots\},$$
with the system of the equations
$$f(nf(m)+m) = mf(n+1),\quad (m,n)\in \mathbb N^2. \tag1$$
Let us research the structure of this sequence.
The value $f(1)=p$ should satisfy ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3359892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
What is the length of a side of square $DCEF$ where $AB=3, BC=4, AC=5$ in the following diagram?
$DCEF$ is a square and $\triangle ABC$ is an internal triangle. $AB=3,$
$BC=4, AC=5$. What is the length of the square?
I tried multiple times and ended up by converting variables again and again. Can someone give me a h... | The key is $$3^2 + 4^2 = 5^2 \quad\implies\quad \angle ABC = 90^\circ$$
As a consequence,
$$\angle ABD = \angle BCF\quad\implies\quad \triangle ABD \simeq \triangle BCF$$
Let $s = CF$ be the side of the square, we have
$$BD : AB = CF : BC \iff BD = \frac{3s}{4} \implies BF = DF - BD = \frac{s}{4}$$
Apply Pythagoras th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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What did I get wrong when solving $\int\frac{\sqrt{x^2-1}}{x^4}dx$? I'm not sure that this is the problem, but I think I may not know how to find the $\theta$ value when solving an integral problem with trigonometric substitution.
I got $\frac{\sin^3(\sec^{-1}(x))}{3}+C$ for the answer, but the answer should be, $\fra... | Taking your final answer we can sub back in the original subs
$$
\cos (\theta) =\frac{1}{x}
$$
We can then use the relationship
$$
\sin^2 \theta + \cos^2\theta = 1 = \sin^2 \theta +\frac{1}{x^2}
$$
The rest is straight forward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/sma... | $$(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2= \frac {(x^2+1)^4(x^2-1)^2}{x^6}=$$
$$\frac {(x^4-1)^2(x^2+1)^2}{x^6} =\frac {(x^8-2x^4+1)(x^4+2x^2+1)}{x^6}=$$
$$\frac{x^{12} +2x^{10} -x^8-4x^6-x^4+2x^2+1}{x^6}=$$
$$x^{6} +2x^{4} -x^2-4-\frac {1}{x^2}+\frac {2}{x^4}+\frac {1}{x^6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3362964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}$ and generalize it In this post the following is proved
$$\small \sum _{n=1}^{\infty } \frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\sqrt{a^2+n^2}}=\frac{1}{2} \pi J_0(a x)-\frac{\sin (a x)}{2 a},\ \ \sum _{n=1}^{\infty... | Using the representation
\begin{equation}
\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\int_0^1 \frac{\cos \left(tx \sqrt{a^2+n^2}\right)}{a^2+n^2}\,dt\\
\end{equation}
and integrating by parts,
\begin{equation}
\frac{\sin \left(x \sqrt{a^2+n^2}\right)}{\left(a^2+n^2\right)^{3/2}}=x\frac{\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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The sum of all digits of n for which $\sum^n_{r=1} r. 2^r=2+2^{n+10}$ If we start substituting values, we get
$$S_n= 2+8+24+64......n2^n$$
The nth term of the series will be
$$T_n = (n)(2^n)$$
So the sum would be $$\sum T_n=\sum n 2^n$$
Cause that’s how usually solve it. The exponent is throwing me off, and I don’t k... | Note:
$$S_n=1\cdot 2^1+2\cdot 2^2+3\cdot 2^3+\cdots +n\cdot 2^n\\
2S_n=1\cdot 2^2+2\cdot 2^3+3\cdot 2^4+\cdots +(n-1)\cdot 2^n+n\cdot 2^{n+1}\\
2S_n-S_n=-1\cdot 2^1-1\cdot 2^2-1\cdot 2^3-\cdots -1\cdot 2^n+n\cdot 2^{n+1} \Rightarrow\\
S_n=-(2+2^2+2^3+\cdots+2^n)+n\cdot 2^{n+1}=2+2^{n+10} \Rightarrow \\
-\frac{2(2^n-1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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multiple choice question on group of matrices Consider the set of matrices $$G=\left\{ \left( \begin{array}{ll}s&b\\0&1 \end{array}\right) b \in \mathbb{Z}, s \in \{1,-1\} \right\}.$$Then which of the following are true
*
*G forms a group under addition
*G forms an abelian group under multiplication
*Every element... | *
*$1$ is false:
Your approach is correct. Since, for example, $\begin{pmatrix}1&*\\0&1 \end{pmatrix}+\begin{pmatrix}1&*\\0&1 \end{pmatrix}=\begin{pmatrix}2&*\\*&* \end{pmatrix} \notin G$
*$2$ is false:
Take $b \neq 0$.$$\begin{pmatrix}1&b\\0&1 \end{pmatrix}\begin{pmatrix}-1&b\\0&1 \end{pmatrix}=\begin{pmatrix}-1&2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3364550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove $y = z$ if $y^p = z^p$ where $p \in \mathbb{N}$ and $y, z \in \mathbb{R}_{++}$ As above. I am beginner in analysis. I don't know how rigorously you have to prove things. To me it's obvious.
| Are you familiar with induction and/or the well ordering principal?
If we assume $y \ne z$ and wolog we assume $0 < y < z$ then $y^1 < z^1$. Let's assume there is a $y^k \ge z^k$. By well ordering principal there most be so first $k$ where that is true.
But that would mean $y^{k-1} < z^{k-1}$. Byt then $y^k = y^{k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3365078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove: $\sqrt{(n^{2}-1)}= [n-1,1,2n-2,1,2n-2,1,2n-2, \ldots ]$ I'm trying to show
$$
\sqrt{n^2-1}=n-1 + \cfrac{1}{1 + \cfrac{1}{2n-2+ \cfrac{1}{1+\cdots}}}
$$
What I tried:
$$
\sqrt{n^2 - 1}- (n-1) = \frac{2n-2}{n^2-1+(n-1)}
$$
And here i got stuck on how to continue, would greatly appreciate any help
| You want to show that
$n-1+\sqrt{n^2-1}=2n-2 + \cfrac{1}{1 + \cfrac{1}{2n-2+ \cfrac{1}{1+\cfrac{1}{2n-2+\cdots}}}}$
Suppose
$y=x+\cfrac{1}{1+\cfrac{1}{x+\cfrac{1}{1+\cfrac{1}{x+\cdots}}}}$
then
$y = x + \cfrac{ 1}{1+\frac 1 y} \\\Rightarrow y=x+\frac {y}{y+1}\\\Rightarrow y^2+y = xy+x+y\\\Rightarrow y^2 - xy - x = 0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3368068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $3x^4+4y^4=19z^4$ has no integer solution If $x,y,z$ are integers such that $xyz\neq 0$, prove or disprove that
$$3x^4+4y^4=19z^4$$
has no solution.
Maybe this is can use quadratic residue to solve it, maybe this equation seem is famous? because I have solve paper Introduced this.
| COMMENT.- Just for fun. Do you have for the equation $3x^4+7y^4=19z^4$ the following irrational parameterization
$$\begin{cases}x=\left(\sqrt[4]{\dfrac{19}{3}}\sqrt{\dfrac{1-t^2}{1+t^2}}\right)z\\y=\left(\sqrt[4]{\dfrac{19}{7}}\sqrt{\dfrac{2t}{1+t^2}}\right)z\end{cases}$$ There is a value of $t$ such that both $\sqrt[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3369552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Indefinite integral of rational expression involving cubic polynomials I was given the following exercise:
$$\int \frac{x^3}{(x^2+1)^3}dx$$
As a tip, my professor suggested using the following substitution: $t=x^2+1$.
Notice that if $t=x^2+1$, then $x^2=t-1$ and therefore $x=(t-1)^\frac{1}{2}$. Then
$$x^3=x^2\cdot... | Partial integration may also work as follows:
\begin{eqnarray*} \int \frac{x^3}{(x^2+1)^3}dx
& = & \int x^2\frac{x}{(x^2+1)^3}dx \\
& = & x^2\left(-\frac{1}{4}\frac{1}{(x^2+1)^2}\right) + \frac{1}{4}\int \frac{2x}{(x^2+1)^2}dx \\
& = & -\frac{1}{4}\frac{x^2}{(x^2+1)^2} - \frac{1}{4}\frac{1}{x^2+1} (+C)
\end{eqnarray*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest ... | So let $N = 4M$ be a multiple of $4$. If you divide by $6$ and take the remainder $x$ so that $N=4M = 6k + x$ what are the possible values of $x$ if $x$ is not negative and $x < 6$.
Well, It's possible that $6$ divides into $N=4M$. If so $x = 0$.
And it's possible that $6$ doesn't divide into $N=4M$. What are the po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Evaluating $\iint_R\big(x^2+y^2\big)\,dA$
Evaluate the following double integral:
$$\iint_R\big(x^2+y^2\big)\,dA,$$
where $R$ is the region given by plane $x^2+y^2\leq a^2$.
My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\... | Whenever you have $\sqrt{a^2-x^2}$ in an integral, use the substitution $x=a \sin(\theta )$.
For information :
In case of having $\sqrt{x^2-a^2}$ use the substitution $x=a \sec(\theta )$.
In case of having $\sqrt{x^2+a^2}$ use the substitution $x=a \tan(\theta )$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3378115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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For continuous, monotonically-increasing $f$ with $f(0)=0$ and $f(1)=1$, prove $\sum_{k=1}^{10}f(k/10)+f^{-1}(k/10)\leq 99/10$ A question from Leningrad Mathematical Olympiad 1991:
Let $f$ be continuous and monotonically increasing, with $f(0)=0$ and $f(1)=1$.
Prove that:$$
\text{f}\left( \frac{1}{10} \right) +\text... | Note: In the title of the question the upper limit of $k$ needs to be 9.
When $y=f(x)$ is monotonically increasing in domain $D$ then area under the curves $y=f(x)$ and $y=f^{-1}(x)$ is more that those of the resprctive rectangles below them.
Also the sum the area integrals
$$A=\int_{x_1}^{x_2} f(x)~ dx+\int_{y_1}^{y_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3381465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find the distribution of $X|Y =k$ given the distribution of $Y|X =x$ Good night, I'm trying to find the p.d.f of $Y$ and $X|Y=k$, and $\mathbb E (X|Y=k)$ in the following exercise:
Could you please verify if my attempt is correct or contains logical mistakes? Thank you so much for your help!
My attempt:
First, we h... | I post my attempt in case $X \sim \operatorname{Beta}(a,b)$ here and is going to mark this answer as accepted to peacefully close this question.
Thank you again for your verification @Math1000 ^o^
We have $X \sim \operatorname{Beta}(a,b)$ and $(Y|X=x) \sim \mathcal B(n,x)$. Then $f_{Y|X} (k|x) = {n \choose k} x^k (1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time... | You may give a blanket to each one and and use stars and bars with remaining $7$ blanket.
You only need two bars with seven stars so the answer is $\binom {9}{2} =36$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Solutions of $1+x+x^2+...+x^n$ I was thinking about the roots of unity, the solutions of the polynomial $P(x) = x^n - 1$ which are quite easy to find, they are of the form $ \cos \frac{2k\pi}{n} + i\sin \frac{2k\pi}{n} $. I was wondering if there are any other polynomials whose solutions we know. Particularly, is ther... | Observe that
$x^{n + 1} - 1 = (x - 1)(x^n + x^{n - 1} + \ldots + 1); \tag 1$
therefore any root of
$x^{n + 1} - 1 = 0 \tag 2$
other than $x = 1$ is a root of
$x^n + x^{n - 1} + \ldots + 1 = 0; \tag 3$
thus the $n$ roots of (3) are
$x = e^{2k\pi i / (n + 1)}, \; 1 \le k \le n. \tag 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve for $x$: $\frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2} = \frac{5}{12}.$ Solve for $x$: $$\frac{1}{\log\big((x+2)^2\big)}+\frac{1}{\log\big((x-2)^2\big)} = \frac{5}{12}.$$ My Attempt: \begin{align*} & \frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2} = \frac{5}{12} \\ \implies &\> \frac{1}{2\log(x+2)}+\frac{1}{2\log(x-2... | We have that
$$f(x)=\frac{1}{\log(x+2)^2}+\frac{1}{\log(x-2)^2}$$
is even and therefore we can assume $x> 0$ with $x\neq 1$ and $x\neq 2$ ($x=0$ is not a solution).
For $0<x<1$ and $1<x<2$ we have that
$$f(x)=\frac{1}{2\log(x+2)}+\frac{1}{2\log(2-x)}\\\implies f'(x)=\frac{1}{2(2-x)\log^2(2-x)}-\frac{1}{2(x+2)\log^2(x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3391806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c| \geq |a+b|$ prove then $a+b+c = 0$
For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c|
\geq |a+b|$ prove then $a+b+c = 0$
Solution:
$$
|a| + |b| + |c| \geq |b+c| + |c+a| + |a+b| \geq 0
$$
and
$$0 \leq |a+b+c|\leq |a+b| + |c| \leq |a... | We have:
$$a^2\geq(b+c)^2,$$
$$b^2\geq(a+c)^2$$ and $$c^2\geq(a+b)^2$$ or
$$(a-b-c)(a+b+c)\geq0,$$ $$(b-a-c)(a+b+c)\geq0$$ and $$(c-a-b)(a+b+c)\geq0.$$
Now, for $a+b+c>0$ we obtain:
$$a-b-c\geq0,$$ $$b-a-c\geq0$$ and $$c-a-b\geq0,$$ which gives
$$a-b-c+b-a-c\geq0$$ or
$$c\leq0.$$
Similarly, $b\leq0$ and $a\leq0,$ which... | {
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"url": "https://math.stackexchange.com/questions/3392565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$
If $\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+2(a^2-2)x+a>-\sqrt{2}$ is satisfied for all real $x>0$ then obtain the possible values of the parameter $a$.
My attempt is as follows:
$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+... | For the first condition
($x = 0$),
you want
$a+\sqrt{2} > 0$
so
$a > -\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Ramanujan, sum of two cubes - how was it discovered? I'm looking for motivation for, and hopefully a derivation of, Ramanujan's sum of cubes formula
$$\left(x^2+7xy-9y^2\right)^3+\left(2x^2-4xy+12y^2\right)^3=\left(2x^2+10y^2\right)^3+\left(x^2-9xy-y^2\right)^3$$
I can't see where one could start to justify this with... | Observe that the discriminants of $x^2 + 7 x y -9 y^2$ and $x^2 - 9 xy - y^2$ are $85$ and the discriminants of $2x^2 - 4 x y + 12 y^2$ and $2x^2 + 10 y^2$ are $-80$. This suggests that these integral binary quadratic forms might be equivalent in pairs. Testing this, we find
$$ \left. x^2 + 7 x y -9 y^2 \right|_{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Why isn’t $f:\!R \rightarrow P : a \rightarrow ax^3 + a^2x^5$ a linear operator? Why isn’t $f:\!R \rightarrow P : a \rightarrow ax^3 + a^2x^5$ a linear operator?
Because $0 = 0$ and in my understanding it is closed in linear combinations. Isn’t it?
| Simply because $f(a+b) \neq f(a)+f(b)$.
$$f(a+b)=(a+b)x^3+(a+b)^2x^5=(a+b)x^3+(a^2+2ab+b^2)x^5$$
while
$$f(a)+f(b)=ax^3+a^2x^5+bx^3+b^2x^5=(a+b)x^3+(a^2+b^2)x^5$$
And they are different unless $ab=0$ i.e. $a=0$ or $b=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3397385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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I have to solve this initial value problem and determine where the solution attains its minimum value. This is the differential equation : $y'= 4y^2 + xy^2 , y(0)=1.$
I was able to find the solution $y$ for this equation which is :
$$y=\frac{-2}{8 x + (x^2 - 2)},$$
but I don't know how to determine where the solution a... | You can take the derivative of your solution (which is correct, by the way), and set it equal to zero to find the maximum value. Or you can work from the original DE:
\begin{align*}
y'&=0\\
4y^2+xy^2&=0\\
y^2(4+x)&=0.
\end{align*}
One solution is $y=0,$ and the other is $x=-4.$ But in looking at your solution, $y\not=0... | {
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"url": "https://math.stackexchange.com/questions/3398982",
"timestamp": "2023-03-29T00:00:00",
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The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$
$\sqrt{4^0+\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\cdots}}}}=?$
I found this problem in a book. I tried to solve this but couldn't. Using calculator, I found the value close to $2$. But how can this problem be solved with proper procedure?
| Note that:
$2^2=1+3$, $3^2=4+5$, $5^2=16+9$, $9^2=64+17$, ...
Therefore
$$2=\sqrt{4^0+3}$$
$$2=\sqrt{4^0+\sqrt{4^1+ 5}}$$
$$...$$
$$2=\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \sqrt{4^3+17}}}}$$
$$...$$
$$...$$
Let $F_n=\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \sqrt{4^3+...}}}}$ where the sequence terminates after $n$ square roots.
For ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Possible Jordan canonical forms of identity matrix plus a nilpotent matrix I am working on the following Linear algebra problem:
Suppose that $N$ is a nilpotent $5 \times 5$ real matrix (so $N^5$ is the zero matrix). List all possible Jordan canonical forms of $I + N$.
Here is where my thinking is at: I know how to li... | As you mentioned correctly- The characteristic polynomial of $N$ is $\chi_N(x)=x^5$, we know that not by computing, but by the nilpotent property and the dimension of $N$. To understand the characteristic polynomial of $N+I$ we need to go back to the definition: To any matrix $M$ it's characteristic polynomial is given... | {
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"source": "stackexchange",
"question_score": "3",
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Then find the sum of all possible values of $abc$. Let $a, b, c$ be positive integers with $0 < a, b, c < 11$. If $a, b, $ and $c$ satisfy
\begin{align*}
3a+b+c&\equiv abc\pmod{11} \\
a+3b+c&\equiv 2abc\pmod{11} \\
a+b+3c&\equiv 4abc\pmod{11} \\
\end{align*}then find the sum of all possible values of $abc$.
What I trie... | All equations below are understood to be modulo $11$.
Add the $3$ equations together
\begin{eqnarray*}
5(a+b+c)=7abc.
\end{eqnarray*}
Multiply the first equation by $5$
\begin{eqnarray*}
4a+5b+5c=5abc.
\end{eqnarray*}
Eliminate $b+c$ and divide by $2a$
\begin{eqnarray*}
10a=9abc. \\
bc=6.
\end{eqnarray*}
Similarly $... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that in the ring $Z[\sqrt2]$: $\langle3+8\sqrt2, 7\rangle$ = $\langle3+\sqrt2\rangle$ So,
I know that $\langle a\rangle$ = $aR$ and $\langle a,b\rangle$ = $aR + bR$ for any ring R.
I then multiplied each side by $\sqrt2$ and computed them.
These are the steps that I took:
$\langle3+8\sqrt2, 7\rangle$ = $\langle3... | Note that since $1,-\sqrt{2} \in\mathbb{Z}[\sqrt{2}],$
$(3+8\sqrt{2})\cdot1-7\cdot\sqrt{2}=3+\sqrt{2}\in\langle 3+8\sqrt{2},7\rangle.$
Since $\forall r\in\mathbb{Z},1\cdot r\in\mathbb{Z}$ and $-\sqrt{2}\cdot r\in\mathbb{Z},$ we have that $\forall r\in\mathbb{Z},(3+\sqrt{2})r\in\langle 3+8\sqrt{2},7\rangle\\
\Rightarr... | {
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Find all matrices $A\in \mathbb{R}^{2\times2}$ such that $A^2=\bf{0}$ Attempt:
Let's consider $A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$.
$$\begin{align}
A^2 &=
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\cdot
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix} \\&=
\begin{bmatrix}
a\cdot a+b\cdot c & a\cdot b + b... | Your proof is correct, although rather naive---in the sense that you've converted the matrix equation into a system of simultaneous equations and solved it. There's no "standard" proof per se, but there are many more conceptual ways to prove it.
One example is the eigenvalue argument provided by Bungo in the comments. ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The Calculation of an improper integral For the integral $$\int_{0}^{\infty} \frac{x \ln x}{(x^2+1)^2} dx $$
I want to verify from the convergence then to calculate the integral!
*
*For the convergence, simply we can say $$ \frac{x \ln x}{(x^2+1)^2} \sim \frac{1 }{x^3 \ln^{-1} x}$$
then the integral converge because... | Yes, the improper integral is convergent, but your computation is not correct. Note that
$$\begin{align}
\int \frac{x \ln(x)}{(x^2+1)^2} dx& = -\frac{\ln(x)}{2(x^2+1)}+ \int\frac{1}{2x(x^2+1)}dx\\
&=
-\frac{\ln(x)}{2(x^2+1)}+\frac{1}{2}\int\left(\frac{1}{x}-\frac{x}{x^2+1}\right)dx\\
&=-\frac{\ln(x)}{2(x^2+1)}+\frac{\... | {
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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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How to differentiate between external and internal angle bisectors of a triangle? I came up with this question when I was trying to figure out the coordinates of the incenter of a triangle with equations:
$4x-3y=0$, $3x-4y+12=0$, $3x+4y+2=0$.
I assumed the coordinates of the incenter to be $(h,k)$ and equated the perpe... | $4·x-3·y=0$
$3·x-4·y+12=0$
$3·x+4·y+2=0$
multiply the equations by the signs of the cofactors of the constants: 0, 12 and 2.
$M=\left [ \begin{array}{} 4 & -3 & 0 \\ 3 & -4 & 12 \\ 3 & 4 & 2\\ \end{array} \right ] $
$cofactor{M_{13}}=\left|\begin{array}{} 3 & -4 \\ 3 & 4 \\ \end{array} \right| =24$
$cofactor{M_{23}}=... | {
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Prove that the number of different ways to write an integer X as a sum of positive integers is $2^z$, if the order matters. As an example for number 5, it is equal to $2^4$.
We can prove it by brute-forcing it, since it's a small number, as follows:
$1+1+1+1+1 $
$1+1+1+2 $
$1+1+2+1 $
$1+2+1+1 $
$2+1+1+1$
$1+1+3 $
$1+3+... | Let $f(n)$ be the number you're looking for for a given $n$. Let's work by strong induction.
Consider $f(n+1)$. It could either have one term (which happens in one way) or more than one term. In this second case, the second term can be any number from $1$ through $n$. If the first term is $k$, then there are $f(n+1... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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how to find $\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$ how to find
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}$$ without L'hopital or taylor/Laurent series
I tried bu... | $$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan^{3}\left(x\right)-3\tan\left(x\right)}{\cos\left(x+\frac{\pi}{6}\right)}=$$
$$\lim_{\large x \to \frac{\pi}{3}}\frac{\tan x (\tan x - \sqrt 3)(\tan x +\sqrt 3)}{\cos x \cos (\pi /6)-\sin x \sin (\pi /6)}=$$
$$\lim_{\large x \to \frac{\pi}{3}}\frac{(2\sec x)\tan x (\tan x - ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find all pairs of prime numbers $p$ and $q$ Find all pairs of prime numbers $p$ and $q$ such that $p^3 - q^5 = (p + q)^2$. Present your answers as an ordered pair $(p, q)$.
Is this resolution correct?
It suffices to show that one of $p$, $q$ is divisible by 3, then the rest of the problem collapses.
Assume otherwise. O... | That looks correct to me. You can also establish the result by looking at $(p+q) \pmod{3}$. As you said, $p^3-q^5\equiv p-q\equiv (p+q)^2\pmod{3}$.
Case 1. $p+q\equiv 0\pmod{3}$. Then you have $3\mid p-q$ and $3\mid p+q$, which gives $3\mid p,q$.
Case 2. $p+q\equiv \pm1\pmod{3}$. Then $(p+q)^2\equiv 1\equiv p-q \pmod{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to study this sequence $u_n=\sum_{k=1}^{n}\frac{1}{n+2k}$ Please is there any way to prove that sequence is increasing ?
I do: $u_{n+1}-u_n=\sum_{k=1}^{n+1}\frac{1}{(n+1)+2k}-\sum_{k=1}^{n}\frac{1}{n+2k}=\left[\frac{1}{n+3}+\frac{1}{n+5}+\ldots+\frac{1}{3n+1}+\frac{1}{3n+3}\right]-\left[\frac{1}{n+2}+\frac{1}{n+4}+... | OP, a good start:
$$u_{n+1}-u_n=\sum_{k=1}^{n+1}\frac{1}{(n+1)+2k}-\sum_{k=1}^{n}\frac{1}{n+2k}$$
Let's continue. First,
$$ \sum_{k=1}^n
\frac 1{(n+2\!\cdot\! k-\frac 12)
\cdot(n+2\!\cdot\!k+\frac 32)}\,
\, =\,\, \frac 12\cdot\left(
\, \frac 1{n+2-\frac 12}\, -\, \frac 1{3\cdot n+\frac 32}
\right) $$
$$ =\, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3413439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all the possible values of gcd$(n^2 +2 , n^4 +4)$ if $n$ is a natural number? Is this a complete “proof”? Im new to number theory so it might be wrong. So, $$n^4 +4 = (n^2 +2+2n)(n^2 +2 -2n)\ ,$$ so if $d$ divides $n^2 +2 +2n$ and $n^2 +2$, then $d | 2n$. , so $d|2$ or $d|n$.
Suppose $d| n$. We know $d | (n^2 +2)... | We have $n^4+4 = (n^2+2)(n^2 - 2) + 8$. If $n$ is odd, then $gcd(n^2+2, 8) = 1$, and if $n$ is even then $n^2+2 = 4k^2 + 2$ is not divisible by $4$ or $8$ so $gcd(n^2+2,8) = 2$. This gives your original calculation (via a Euclidean Algorithm argument)
| {
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"question_score": "3",
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Help Understanding Proof for Inequality Problem #3 from Problem Solving Strategies The problem is the following:
Prove for $a$, $b$, $c$, $d$ that
$$\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}\geq\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}$$
I understand the proof saying
$${\frac{abc+abd+acd+bcd}{4}}=\frac{(ab)(c+d)+(cd)(a+b)}{4}$$
Apply... | Because by AM-GM and C-S we obtain:
$$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}=\sqrt[3]{\frac{ab(c+d)+cd(a+b)}{4}}\leq$$
$$\leq\sqrt[3]{\frac{\left(\frac{a+b}{2}\right)^2(c+d)+\left(\frac{c+d}{2}\right)^2(a+b)}{4}}=\sqrt[3]{\frac{a+b}{2}\cdot\frac{c+d}{2}\cdot\frac{a+b+c+d}{4}}\leq$$
$$\leq\sqrt[3]{\left(\frac{\frac{a+b}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3419524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Integration based on beta function
I am getting the value of first integration using the beta function, but I am not able to find the second one. Please help
| $\because$ $\int_0^1\dfrac{x^{a-1}(1-x)^{b-1}}{(z+x)^{a+b}}~dx=\dfrac{B(a,b)}{(z+1)^az^b}$ as https://dlmf.nist.gov/5.12 stated,
$\therefore$ $\int_0^1\dfrac{x^\frac{5}{2}(1-x)^\frac{7}{2}}{(3+x)^8}~dx=\int_0^1\dfrac{x^{\frac{7}{2}-1}(1-x)^{\frac{9}{2}-1}}{(3+x)^{\frac{7}{2}+\frac{9}{2}}}~dx=\dfrac{B\left(\frac{7}{2},\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove $1+x+x^2+\cdots+x^{n-1}=\frac{1-x^n}{1-x}$ using the induction method? I am having difficulties in solving the expression $1+x+x^2+\cdots+x^{n-1} =\frac{1-x^n}{1-x}$, $n\in \mathbb{N}$, using the induction method. How is the inductive step taken in this case?
| Assume that $1+x + \dots + x^{n-1} = \frac{1-x^n}{1-x}$.
Then
$$1+x+ \dots + x^{n} = (1+ x + \dots + x^{n-1}) + x^n = \frac{1-x^n}{1-x} + x^n$$
$$= \frac{1-x^n}{1-x} + \frac{x^n(1-x)}{1-x} = \frac{1-x^n + x^n - x^{n+1}}{1-x}= \frac{1-x^{n+1}}{1-x}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Which satisfy the system of inequalities below: For each nonnegative integer $n$, calculate the number of triples $(a, b, c)$ of nonnegative integers
which satisfy the system of inequalities below:
$ \begin{cases}
a + b \leq 2n \\
a + c \leq 2n \\
c + b \leq 2n \\
\end{cases}$
What I thought: We can solve this by plott... | Consider the triples $ (a, b, c) $ of nonnegative integers that satisfy the following relationships:
$$\begin{cases} \, a + b = x \\ a + c \leq 2n \\ b + c \leq 2n \end{cases} $$Let's divide the count of these triples into two steps. In the first we consider $ x = 2k $, while in the second we consider $ x = 2k + 1 $.
T... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(11 \cdot 31 \cdot 61) | (20^{15} - 1)$ Prove that
$$ \left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right) $$
Attempt:
I have to prove that $20^{15}-1$ is a factor of $11$, $31$, and $61$. First, I will prove
$$ 20^{15} \equiv 1 \bmod11 $$
Notice that
$$ 20^{10} \equiv 1 \bmod 11$$
$$ 20^{5} ... | Alternatively:
$$\left( 11 \cdot 31 \cdot 61 \right) | \left( 20^{15} - 1 \right)=(20^5-1)(20^{10}+20^5+1)\\
20^5\equiv (-2)^5\equiv -32\equiv 1 \pmod{11}\\
20^5\equiv 81^5\equiv 3^{20}\equiv (3^5)^4 \equiv1 \pmod{61}\\
20^5 \equiv (2^{5})^2\cdot (5^2)^2\cdot 5\equiv 1^2\cdot (-6)^2\cdot 5\equiv 25\pmod{31}\\
20^{10}\e... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Absolute difference between the first and second half of a random vector Suppose I have a zero vector of length $2n$, and I randomly choose $k$ of the entries of this vector to be $1$. The choice is taken uniformly from all $\binom{2n}{k}$ possibilities. Now let $S_1$ denote the sum of the first half of the vector, and... | TL;DR: I will show in this answer that
$$
\mathbb{P}\left(\frac{|S_1 - S_2|}{n} > \epsilon\right)
\le \frac{1}{\varepsilon^2 (2n-1)},
$$
independent of $k$ (better bounds can be given if $k$ is close to $1$ or $2n$).
In other words, the probability you ask for is bounded above by roughly $1 / (2 \varepsilon^2 n).$ In p... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $ \lim_{x\to +\infty} x\left(\frac{\pi}{4} - \arctan\left(\frac{x}{x+1}\right)\right) $
$$ \lim_{x\to +\infty} x\left(\frac{\pi}{4} - \arctan\left(\frac{x}{x+1}\right)\right) $$
I tried to do this with some kind of substitution but failed miserably. Any hints or help?
| I nice approach using triangles. Use the following figure, where $\triangle ABC$ is right-angled with $\overline{BC} = x>0$ and $\overline{AB} = 1+x$. Take $D$ on $AB$ so that $\overline{AD} = 1$ and $E$ on $AC$ so that $DE\perp AC$.
By definition
$$\angle CAB = \arctan\left(\frac{x}{x+1}\right),$$
and by the Extern... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 3
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Evaluate $\displaystyle\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}$ I have been trying to evaluate
\begin{equation*}
\lim_{x \to 0^{-}} \frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x}.
\end{equation*}
We have
\begin{equation*}
\frac{\sqrt{x^2 + x + 4} - 2}{\ln(1 + x + x^2) - x} = \frac{... | For small $x$ so $x^2+x$ is also small,$$\sqrt{x^2+x+4}-2=2\left(\sqrt{1+\frac14(x^2+x)}-1\right)\approx\frac14(x^2+x)\approx\frac14 x$$while$$\ln(1+x+x^2)-x\approx x(1+x)-\frac12 x^2(1+x)^2-x\approx\frac12 x^2,$$so the ratio $\approx\frac{1}{2x}\to-\infty$ as $x\to0^-$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$ $$\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$$
I know this limit must be equal to $\frac{1}{2}$ but I can't figure why. This is just one of the thing I tried to solve this limit:
$$\lim_{x\to\infty}\biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)$$
$$... | Her is an elementary way:
\begin{eqnarray*} \biggr(x\sqrt{\frac{x}{x-1}}-x\biggr)
& = & x \biggr(\sqrt{\frac{x}{x-1}}-1\biggr) \\
& = & x \biggr(\frac{\frac{x}{x-1}-1}{\sqrt{\frac{x}{x-1}}+1}\biggr) \\
& = & x \biggr(\frac{1}{(x-1)\sqrt{\frac{x}{x-1}}+(x-1)}\biggr) \\
& = & \frac{1}{(1-\frac{1}{x})\sqrt{1+\frac{1}{x-1}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 7
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Find the range of $f(x)=2|{\sin x}|-3|\cos x|$ Find the range of $f(x)=2|{\sin x}|-3|\cos x|$
My attempt is as follows:-
$$f(x)=\sqrt{13}\left(\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|\right)$$
Let's assume $z=\dfrac{2}{\sqrt{13}}\cdot|\sin x|-\dfrac{3}{\sqrt{13}}\cdot|\cos x|$
$$f(x)=\sqrt{13... | Your domain is 0 to $\frac{\pi}{2}$. That's where sine and cosine are positive. So at those points you are limited to the range of $[-3,2]$.
Unless you are given additional instructions, you are way over complicating it.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the unit digit of $2124^{392}+3143^{394}*7177^{392}-8818^{394}$ What I do here is I find the remainder of each parcel in mod 10 and then do the maths and mod 10 again:
$2124^{392} \equiv 4^{392} \pmod{10}$
$$4^1 \equiv 4 \pmod{10} \\
4^2 \equiv 6 \\
4^3 \equiv 4 \\
4^4 \equiv 6 \\
(...)$$
Then I do $392 \pmod{2}$... | Using Newton's binomial formula,
$$4^{392}=(10-6)^{392} $$
$$=6^{392} \mod 10 =6 \mod 10$$
Since $$6^2 = 6\mod 10$$
For the second and the third , $ 3$ and $ 7$ are coprime with $ 10$ so by Euler's Theorem
$$3^{\phi(10)}=1 \mod(10) =3^4$$
and
$$7^4 = 1 \mod 10$$
For example
$$3^{394}=3^{98.4+2}=9 \mod 10$$
For the l... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tough integral from spherical coordinates One of my friend's students attempted to find the volume of a paraboloid using spherical coordinates. I'm trying to see if there's actually a way to finish it. I have been able to find the triple integral which describes the volume of the paraboloid in spherical coordinates and... | Let $t=\cos x$, we have
\begin{align}
I&=\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{3} \cdot \left(\frac{-\cos{x} + \sqrt{\cos^2({x}) + 16\sin^2({x})}}{2\sin^2({x})}\right)^3dx \\
&= \frac{1}3\int_0^1 dt \left(\frac{-t+\sqrt{t^2+16(1-t^2)}}{2(1-t^2)}\right)^3
\end{align}
Let $z$ be the larger root of the equation $(1-t^2)z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3431869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$ What is the value of $5x^2-5x-1?$ If $$x=\sqrt{{\sqrt{5}+1\over \sqrt{5}-1}}$$ What is the value of $5x^2-5x-1?$
Efforts:
After rationalization, I got $x={\sqrt{5}+1\over 2}$ and $x^2={\sqrt{5}+1\over \sqrt{5}-1}$. Going by this method is very tedious and boring.
Is there a m... | You already have $x=\frac{\sqrt5+1}{2}$. Then $x^2=\frac{\sqrt5+3}{2}$ and so $x^2=x+1$.
Therefore $5x^2-5x-1=5x+5-5x-1=4.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3437434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Approximate \cos(x^2) using Composite Trapezoidal method Find an approximation to $$\int_2^3 \cos(x^2)dx$$ using the Composite Trapezoidal method with $r = 3$ and estimate the error.
Attempt: using the composite trapezoidal formula I calculated the approximation was equal to $$\frac 16 [\cos(4) + \cos(9) + \cos(16) + ... | Assuming your "Composite Trapezoidal method" is my Trapezoidal rule, then you've made an error in applying the approximation.
With a uniform $$\Delta x = \frac{b-a}{n} = \frac{3-2}{3} = \frac{1}{3},$$
the trapezoidal rule gives
$$\begin{align*}\int_2^3 \cos\left(x^2\right)\,dx &\approx \frac{1}{6}\Big(\cos\left(2^2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0. Prove that $5^{2n+1} - 3^{2n+1} - 2^{2n+1}$ is divisible by 30 for all integers n ≥ 0.
I have tried induction as follows.
Step 1:
Try n = 0, we get: $5 - 3 - 2 = 0$, which is divisible by 30.
Try n = 1, we get: $5^{3} - 3^{3} - 2^... | Use parity to deduce it is even. Or explicitly $\ \underbrace{105(\overbrace{2i\!+\!1}^{\textstyle 5^{N}}) - 15 (\overbrace{2j\!+\!1}^{\textstyle 3^{N}})}_{\textstyle 21\times 5^{N+1} - 5\times 3^{N+1}}\, =\, \overbrace{210 i - 30j + 90}^{\textstyle \color{#c00}{30}\,(7i -j + 3)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Evaluating $\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$ for $p \in [0,1]$ I am trying to evaluate the following sum $$\sum_{y=a}^{\infty}{y \choose a} \cdot p^{y-a}$$ for $p \in [0,1]$. This looks somewhat like the binomial theorem, but I don't know how I would go about applying it, as the index of summation is $y$... | So we have
$$S = \binom{a}{a} + \binom{a+1}{a}p + \binom{a+2}{a}p^2 + \space ...$$
$$pS = \binom{a}{a}p + \binom{a+1}{a}p^2 + \binom{a+2}{a}p^3 + \space ...$$
subtracting, we obtain
$$(1-p)S = \binom{a}{a} + \binom{a}{a-1}p + \binom{a+1}{a-1}p^2 + \space ...$$
(I've used the identity $\binom{a}{a-1} + \binom{a}{a} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the bisector in the given condition I need to find x which is bisectrix for that 60°. Given red segments 9 and 12 cm and a circle. It`s seems like school exercise, but i`m I am get stuck, please help (have tried several approaches). Please, give any advices or direction to solve it. Thanks.
↓ See the image below ... |
Let
$|AB|=c=9$,
$|AC|=b=12$,
$|AD|=d=?$,
$\angle CAB=\alpha=60^\circ$,
$\angle DAB=\angle CAD=\tfrac12\,\alpha=30^\circ$.
Then by the cosine rule we have
\begin{align}
a^2&=b^2+c^2-2bc\cos\alpha=b^2+c^2-bc=117
,\\
a&=3\sqrt{13}
.
\end{align}
The area $S$ and the radius of the circle $R$
of $\triangle ABC$ are
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx$ I am tring to obtain following formula [actually I obtained this resut via mathematica]
\begin{align}
\int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx
= \frac{\sqrt{2} \sinh(x)}{a^2 \sqrt{a^2-b^2 + (a^2+b^2) \cosh... | Using substitution: Let $u=\sinh(x)$. Then $\mathrm du$ corresponds to $\cosh(x)\,\mathrm dx$ and thus (since $a^2\sinh^2(x)+b^2\cosh^2(x)=(a^2+b^2)\sinh^2(x)+b^2$) $$I=\int \big(u^2\cdot(a^2+b^2)+b^2\big)^{-\frac32}\,\mathrm du.$$
Now we want to substitute $$u^2\cdot(a^2+b^2)=b^2\tan^2(s).$$
Then $$\mathrm du\sim \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sqrt{2a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4$ when $a+b=2$. Suppose that $a,b$ are non-negative numbers such that $a+b=2$. I want to prove $$\sqrt{2 a^2+b+1}+\sqrt{2 b^2+a+1}\geq 4.$$
My attempt: I tried proving $$2a^2+b+1\geq 4$$ but this seems wrong (try $a=0,b=1$).
How can I prove the above result?
| With $$b=2-a$$ we have to prove:
$$\sqrt{4a^2-a+3}\geq 4-\sqrt{2a^2-7a+9}$$ if $$4-\sqrt{2a^2-7a+9}\le 0$$ we have nothing to prove, in the other case we get
$$-3a+11\le 4\sqrt{2a^2-7a+9}$$ If $$-3a+11\le 0$$ then is all clear, in the other case we can square and collecting like terms we get
$$-21(a-1)^2\le 0$$ and thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Limit with radicals, $\cos$, $\ln$ and powers $\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\frac{1}{x}}\Big)}{\ln(x-1)^{\frac{1}{x}}-\ln{x^{\frac{1}{x}}}}}=\underset{x\rightarrow +\infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(2^{-\frac{1}{x}}\;-\;3^{-\f... | Let's rewrite the thing to $$L=\lim\limits_{x\to+0}\frac{\left( 1-\cos x^3\right)^\frac16\left(2^{-x}-3^{-x}\right)}{x\ln(1-x)}$$
Each of the things to first order will suffice, so let's start:
$$\lim\limits_{x\to0}\frac{e^x-1}{x}=1\Rightarrow
\lim\limits_{x\to0}\frac{a^x-1}{x}=\ln a,$$
$$1-\cos(x^3)=2sin^2\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
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.