Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Differential equations of first order but of higher degree Solve $(x-py)(px-y)=2p$ where $p=\frac{dy}{dx}$
My attempt:
I started by making it a quadratic in $p$ but after a point a could not proceed further:
$$p^2-\frac{p(x^2+y^2-2)}{xy} + 1 = 0$$
$$p = (x^2+y^2-2)\pm\frac{(x^2-y^2)^2-4(x^2+y^2-1)^{1/2}}{2xy}$$
Now I d... | We are trying to solve
$$\tag 1 (px-y)(x-py)=2p , ~\text{where}~p=\frac{dy}{dx}$$
Continuing on the path used leads to something unsolvable, so let's try another approach.
Let $x^2 = X \implies 2 x dx = dX, y^2 = Y \implies 2 y dy = dY$, so
$$P = \dfrac{dY}{dX} = \dfrac{2 y dy}{2 x dx} = \dfrac{y}{x} p \implies p = \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Closed form for $f(x)=\ _3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$
I am seeking a closed form for the function $$f(x)=\,_3F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;x\right)$$
I expect there to be one, because of this post and Wolfram. The Wolfram link produces closed forms involving... | Starting from the integral representation, we let $v = w/u$ and change the order of integration to find
\begin{align}
f(x) &= \frac{1}{4} \int \limits_0^1 \int \limits_0^u \frac{1}{u \sqrt{w(1-xw)}} \, \mathrm{d} w \, \mathrm{d} u = \frac{1}{4} \int \limits_0^1 \frac{\mathrm{d} w}{\sqrt{w(1-xw)}} \int \limits_w^1 \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 0
} |
Computing real integral using residues Let's consider the following integral
$$\int \limits_{-\infty}^{\infty} \frac{\cos2x}{(x^2+4)^2} \, dx. \tag{1}$$
I would like to compute $(1)$ using residue theory. Let's consider a complex function
$$f(z) = \frac{\cos2z}{(z^2+4)^2} = \frac{\cos2x}{(z-2i)^2(z+2i)^2}.$$
Of course ... | I want to write more details, may helps someone. Consider
$$\int_C \frac{e^{2iz}}{(z^2+4)^2} \, dz$$
where $C$ is the contour, the semicircle in upper half plane (as you have gotten I think), then the integrand has poles at $z=\pm 2i$, which it's reside at $z=2i$ is
$$\operatorname{Res}_{f}(2i) = \lim_{z\to2i} \frac{d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
If $a^b$ has $b$ digits, what is the greatest value of $b$?
For natural numbers $a$ and $b$, what is the greatest value of $b$ so that $a^b$ has $b$ digits?
I knew that the greatest value of $b$ is $21$, where $9^{21}=\underset{21 \text{ digits}}{\underbrace{109,418,989,131,512,359,209}}$, but I am not sure how to pr... | If $a^b$ has $b$ digits then
$10^{b-1} \le a^b < 10^b$.
$\log 10^{b-1} \le \log a^b < \log 10^b$ (Notation: $\log$ means $\log_{10}$)
$b-1 \le b\log a < b$
$1 - \frac 1b \le \log a <1$. so $a < 10$
The largest we can allow $b$ to be, the closer to $1$ that $\log a$ will be.
Smaller values of $a$ will result is small... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that $a_n^2 - 2b_n^2 = 1$ if $a_n+b_n\sqrt{2}=(a_{n-1}+b_{n-1}\sqrt{2})^2$
Two sequences of positive integers $a_n$ and $b_n$ are defined by $a_1 = b_1 = 1$ and $$a_n+\sqrt{2}b_n=(a_{n-1}+\sqrt{2}b_{n-1})^2$$ for $n \ge 2$. Prove that $(a_n)^2 - 2(b_n)^2 = 1$.
I found that
$$a_n+b_n\sqrt{2}=(1+\sqrt{2})^{2^{n-... | If $a_n+\sqrt{2}b_n=(a_{n-1}+\sqrt{2}b_{n-1})^2=a_{n-1}^2+2b_{n-1}^2+2\sqrt2a_{n-1}b_{n-1}$
then $a_{n}=a_{n-1}^2+2b_{n-1}^2$ and $b_n=2a_{n-1}b_{n-1},$
so you can prove that $a_n^2-2b_n^2=1$ by induction because
$a_n^2-2b_n^2=(a_{n-1}^2+2b_{n-1}^2)^2-2(2a_{n-1}b_{n-1})^2=a_{n-1}^4+4b_{n-1}^4+4a_{n-1}^2b_{n-1}^2-8a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3282545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Calculating complex roots "Find all the complex roots of the following polynomials
A) $S(x)=135x^4 -324x^3 +234x^2 -68x+7$, knowing that all its real roots belong to the interval $(0.25;1.75)$
B)$M(x)=(x^3 -1+i)(5x^3 +27x^2 -28x+6)$
"
Well, in A) I don't know how to use the given information about real roots. I mean, I... | Since $\frac{1}{3}$ is a root of $S$, we obtain:
$$S=135x^4-324x^3+234x^2-68x+7=$$
$$=135x^4-45x^3-279x^3+93x^2+141x^2-47x-21x+7=$$
$$=(3x-1)(45x^3-93x^2+47x-7)=$$
$$=(3x-1)(45x^3-15x^2-78x^2+26x+21x-7)=$$
$$=(3x-1)^2(15x^2-26x+7)=(3x-1)^2(15x^2-5x-21x+7)=(3x-1)^3(5x-7).$$
Since $\frac{3}{5}$ is a root of $5x^3+27x^2-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3283600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to prove $\int^{\pi}_0 \cos(x)\log(\tan(\frac{x}{4}))\,\mathrm dx = -2$ How to prove $\int^{\pi}_0 \cos(x)\log(\tan(\frac{x}{4}))\,\mathrm dx = -2$?
Or $\int^1_0 \log(u)\frac{(u^4-6u^2+1)}{(u^2+1)^3} \,\mathrm du = -1/2$.
(I substituted $\tan(x/4) = u$.)
| \begin{align*}
\int_0^\pi\cos x\log\left(\tan \frac x4\right)dx&=\left[\sin x\log\left(\tan\frac x4\right)\right]_0^\pi-\int_0^\pi\sin x\cdot\frac{\frac14\sec^2\frac x4}{\tan\frac x4}dx\\
&=0-\int_0^\pi\frac{\sin x}{4\sin\frac x4\cos\frac x4}dx\\
&=-\int_0^\pi\frac{\sin x}{2\sin\frac x2}dx\\
&=-\int_0^\pi\cos\frac x2dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3285568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the polynomial equation when I know the roots A polynomial of minimum degree has rational coefficients and has the roots: $x_1=-1-\sqrt5;x_2=1+2i$ so there are $x_3=-1+\sqrt5$ and $x_4=1-2i$. I need to find the polynomial equation.
I tried to use $(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ but the calculations are too "heavy" ... | Just to give a different approach, from $x_1+x_2+x_3+x_4=0$ and $x_1x_2x_3x_4=-20$, we know the answer has the form
$$P(x)=x^4+ax^2+bx-20$$
Now
$$P(1)=(2-\sqrt5)(2+\sqrt5)(-2i)(2i)=(4-5)(4)=-4$$
and
$$P(-1)=(-\sqrt5)(\sqrt5)(-2-2i)(-2+2i)=(-5)(4+4)=-40$$
so $1+a+b-20=-4$ and $1+a-b-20=-40$, or
$$\begin{align}
a+b&=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3287641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove $\sum _{n \geq 1} \left[\frac{1}{n}-\frac{2}{n+1}{_2F_1}(1,\frac{n+1}{2};\frac{n+3}{2};-1)\right]=\frac{\pi}{4}-\frac{\log 2}{2}$
Can we prove the following closed form by directly using hypergeometric series?
$$S=\sum _{n=1}^{\infty} \left[\frac{1}{n}-\frac{2 }{n+1}\, _2F_1\left(1,\frac{n+1}{2};\frac{n+3}{2};-1... | It's easy enough to work with the hypergeometric function directly, since $(n+1)/2$ and $(n+3)/2$ differ by 1. This means we have
\begin{align}_2F_1\left(1,\frac{n+1}2;\frac{n+3}2;-1\right)&=(n+1)\sum_{k=0}^\infty\frac{(-1)^k}{2k+n+1}\\&=(n+1)\sum_{k=0}^\infty\int_0^1(-x^2)^kx^n~\mathrm dx\\&=(n+1)\int_0^1\sum_{k=0}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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If $x,y,z$ are real numbers satisfying$x/(y+z) +y/(z+x) +z/(x+y) =1$ then $x^2/(y+z) +y^2/(z+x)+z^2/(x+y)=$ I have tried it a few times but I am not making any progress. Please help.
| $$\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}$$
$$=\dfrac{x^2}{y+z}+x+\dfrac{y^2}{z+x}+y+\dfrac{z^2}{x+y}+z-(x+y+z)$$
$$=(x+y+z)\left(\dfrac x{y+z}+\cdots\right)-(x+y+z)$$
$$=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3288605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Prove the following inequality $a,b,c\in R_{+}^{*}$ and $2(a+b+c)=3$ Prove :
$\sum_{cyc}\frac{4a}{b^2+2(b+1)}≤3$
Where :
$a,b,c\in R_{+}^{*}$ , $2(a+b+c)$=3$
I think we use Cauchy inequality :
$\sum_{cyc}\frac{4a}{b^2+2(b+1)}≤4\sqrt{\sum_{cyc}\frac{a^2}{(b^2+2(b+1))^2}}$
Then I use :
$a^{2}+b^{2}+c^{2}≤(a+b+c)^{... | As $b^2+2b+2\geq 2$, we have $$\sum_{cyc}\frac{4a}{b^2+2b+2}\leq \sum_{cyc} \frac{4a}{2}=3$$
Equality holds at $a=b=0$ and $c=\frac32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Taylor series at infinity I'm required to make a Taylor series expansion of a function $f(x) = \arctan(x)$ at $x = +\infty$. In order to do this I introduce new variable $z = \frac{1}{x}$, so that $x \to +\infty$ is the same as $z \to +0$. Thus I can expand $f(z)$ at $z = 0$: $$f(z) = z - \frac{z^3}{3}+\frac{z^5}{5}-..... | We could use the Taylor expansion of $\arctan(x)$ when $x \rightarrow 0$ and that
$$
\forall x \in \mathbb{R}^*_+, \arctan(x) + \arctan\bigg(\frac{1}{x}\bigg) = \frac{\pi}{2}
$$
and find the correct result.
But as @janosch points out, it's faster to use
$$
\arctan'\bigg(\frac{1}{x}\bigg)= -\frac{1}{x^2 +1}
$$
and fix t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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,... | For (i), define $\{x\} = x - \lfloor x \rfloor$. Then
\begin{align*}
\lfloor 2x \rfloor + \lfloor 2y \rfloor = 2\lfloor x\rfloor + 2\lfloor y \rfloor + \lfloor 2\{x\} \rfloor + \lfloor 2\{y\} \rfloor
\end{align*}
and
\begin{align*}
\lfloor x \rfloor + \lfloor y \rfloor + \lfloor x + y \rfloor = 2\lfloor x\rfloor + 2\l... | {
"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": 1
} |
Solve $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$ for $x\in\mathbb{R}.$ Solve $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$ for $x\in\mathbb{R}.$ There is a brute force method that relies on breaking each exponential up into $2^x$’s and $3^x$’s and substituting, but I am looking for a more elegant one line or so proof (not b... | Here's the way I did it: $$\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x)^3+(3^x)^3}{(3\cdot2^2)^x+(2\cdot3^2)^x}=\frac{(2^x)^3+(3^x)^3}{3^x\cdot(2^x)^2+2^x\cdot(3^x)^2}\tag1$$
Let $a=2^x, b=3^x$ so that $(1)$ is equivalent to $$\frac{a^3+b^3}{ba^2+ab^2}=\frac{(a+b)(a^2-ab+b^2)}{(a+b)(ab)}=\frac{7}{6}\tag2$$
Then cancelling th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding the length of a side and comparing two areas from a given figure The following figure came in two national-exam paper (different question in each paper).
First Question:
What is the length of $AE$?
Choices:
A) $\frac{5}{4}$ cm
B) $\frac{\sqrt{41}}{5}$ cm
C) $\frac{7}{5}$ cm
D) $\frac{15}{8}$ cm
Second Questi... |
Triangles $ABE$ and $CDE$ are similar, with ratio $3:5$.
Hence
$$AE=\frac{3}{8}AC=\frac{3}{8}\cdot 5=\frac{15}{8}$$
By the same similarity, if $h$ is the height of triangle $CDE$ from $E$ to $CD$, then
$$h=\frac{5}{8}BC=\frac{5}{8}\cdot 4=\frac{5}{2}$$
hence the area of triangle $CDE$ is
$$\frac{1}{2}\cdot CD\cdot h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3295703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ 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 ... | We can represent a $2\times 2$ matrix as a $1\times 4$ vector.
i.e. $\pmatrix{c_1\\c_2\\c_3\\c_4} = c_1\pmatrix{1&0\\0&0}+ c_2\pmatrix{0&1\\0&0}+c_3\pmatrix{0&0\\1&0}+c_4\pmatrix{0&0\\0&1}$
$\phi(X) = \pmatrix{2&-2&0&0\\-1&1&1&-1\\-1&1&1&-1\\0&0&2&-2}$
The linearly independent columns of the matrix above will give the ... | {
"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": 2
} |
$n$-th term of a sequence from generating function $\frac{1}{(1-x^2)(1-x^3)(1-x^7)}$ I have a generating function of a sequence such that
$$\frac{1}{(1-x^2)(1-x^3)(1-x^7)}.$$
They are the product of three geometric series with coefficient 1.
Now I want to transform this into infinite sum form so that I can get nth term... | Use partial fractions, and pry out the coefficients of interest:
$\begin{align*}
\frac{1}{(1 - x^2) (1 - x^3) (1 - x^7)}
&= \frac{x^4 + 2 x^3 + 2 x^2 + 2}
{7 (x^6 + x^5 + x^4 + x^3 + x^2 + x + 1)}
+ \frac{x + 2}{9 (x^2 + x + 1)}
+ \frac{1}{8 (x + 1)}
- \frac{17}{72 (x - 1)}
+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3299550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?
My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?
| Starting with what you have written we quickly see that $$0=x^3-18x^2+81x-8=(x-8)(x^2-10x+1)$$
It is easy to see that $x\neq 8$ so we must have $\boxed {x^2-10x+1=0}$
Now, go back to the original equation. Multiply by $x$ to get $$x^2-\sqrt {8x}=9x\implies x-\sqrt {8x}=10x-x^2$$
But the boxed equation tells us that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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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... | Since $\tan(x)=2i$
Using Euler's formula
$\tan(x)={\frac {e^{ix}-e^{-ix}}{i(e^{ix}+e^{-ix})}}$ witch must equal $2i$
Manipulating a bit we get $3e^{ix}+e^{-ix}=0$ and if $e^{ix}=y$, $3y+\frac{1}{y}=0$ thus $e^{ix}=y=±i \frac{\sqrt 3}{3}$
This, of course, means that $\ln\left(±i \frac{\sqrt 3}{3}\right)=ix$
| {
"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": 1
} |
Simplify a rational identity
Simplify:
$$\frac{\dfrac{a}{b}-\dfrac{b}{a}}{1+\dfrac{b}{a}}$$
I have a feeling the solution has to do with factoring, but I'm really not sure, and would appreciate any help.
| Well, in view of addition the rule is
$$\frac{a}{b}+\frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad+bc}{bd}$$
i.e., the fractions need to be extended first to obtain the same denominator and then the addition can be performed (same for subtraction).
In view of multiplication the rule is
$$\frac{a}{b}\cdot\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a condition on real numbers $a$ and $b$ such that $\left(\frac{1+iz}{1-iz}\right)^n = a+ib$ has only real solutions I´m new on this. I need to find a condition that relates two real numbers $a$ and $b$ such that
$$\left(\frac{1+iz}{1-iz}\right)^n = a+ib$$
has only real solutions
This is what I got till now.
$$\... | that is an awesome answer given by Martin R.
alternatively, let
$$
a+ib = \sqrt{a^2+b^2} e^{i \theta}
$$
$c$ be the real solution for satisfied $a,b$. One would obtain
$$
\label{eq}
\sqrt[n]{a^2+b^2} e^{ \frac{i (\theta + 2k\pi )}{n} } = \frac{1-c^2 +i 2c}{1+c^2} \tag1
$$
note that
$$
( \frac{1-c^2 }{1+c^2} ) ^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3307060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Exercises to see if certain series converge.
*
*Does the series $\sum \frac{2^k+1}{3^k}$ converge?
*Does the series $\sum (-1)^k\frac{1}{k^2}$ converge?
*Does the series $\sum \frac{2^k+2^{2k}}{4^k}$ converge?
*Does the series $\sum \frac{k!}{(2k)!}$ converge?
For the first question, do we use the comparis... | As regards the last one, this is an alternative way: for $k\geq 1$,
$$\frac{k!}{(2k)!}=\frac{1}{(2k)\cdot(2k-1)\cdots (k+1)}\leq \frac{1}{(2)\cdot(2)\cdots (2)}=\frac{1}{2^k}$$
and therefore the non-negative series $\sum \frac{k!}{(2k)!}$ converges by the comparison test.
Your answers for 2) and 3) are correct. For the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$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... | $$(9)^{ \displaystyle\frac{a}{b} } + (16)^{ \displaystyle\frac{b}{a} } = \left(3^{\displaystyle \color{magenta}{2}}\right)^{ \displaystyle\frac{\color{red}{a}}{b} } + \left(4^{\displaystyle \color{brown}{2}}\right)^{ \displaystyle\frac{\color{blue}{b}}{a} } =\left(3^{\displaystyle \color{red}a}\right)^{ \displaystyle\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Finding lower bounds for a function Given a function $w\left(x\right)=\sum_{n=1}^x\lfloor\frac{x}{n}\rfloor$.
And $f(x) = w(6x+1)-w(6x-2)$, prove that for $x>3; f(x) \geq 12$
EDIT: This doesn't really come from anywhere, I was just messing around with discrete functions and wondered if I could show that there are an in... | You have the $2$ functions
$$w(x) = \sum_{n=1}^x \left\lfloor \frac{x}{n} \right\rfloor \tag{1}\label{eq1}$$
\begin{align}
f(x) & = w(6x + 1) - w(6x - 2) \\
& = \sum_{n=1}^{6x+1} \left\lfloor \frac{6x+1}{n} \right\rfloor - \sum_{n=1}^{6x-2} \left\lfloor \frac{6x-2}{n} \right\rfloor \\
& = \left\lfloor \frac{6x+1}{6x+1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3309911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Matrix rearrangement I have a matrix A in this form:
$$
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
$$
$A^T$ is the transpose of A. $B$... | We have $$A^T A = M =
\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]$$
We will try to write this as $M = CB$, where $B$ is a $2*4$ matrix of constants, and $C$ is a $4*2$ matrix of constants and variables. I app... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3310228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the maximum and minimum values of $a^2\sin^2\theta + b^2\csc^2\theta$
How do I find the maximum and minimum values of the following?
$$a^2\sin^2\theta + b^2\csc^2\theta$$
Is the max value $\infty$?
I tried to find the minimum value by using A.M$\geq$G.M. inequality (is there any other way?) and got it as 2|a|... | The function is periodic with period $\pi$ so let us consider $ 0\le x \le \pi.$
Case-1: When $a>b$ in $$f(x)=a^2\sin^2 x+ b^2 \csc^2 x~~~(1)$$ then
$$f'(x)=2a^2 \sin x \cos x-2b^2 \cot x \csc^2 x.= 2 \cos x(a^2 \sin^4 -b^2)/\sin^2 x ~~~(2)$$
$f'(x)=0,$ gives $\cos x=0$ or $\sin^4 x = b^2/a^2 \Rightarrow \sin^2 x= b/a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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If $ \sin a +\sin c =2 \sin b $, show that $ \tan\frac{a+b}{2}+\tan\frac{b+c}{2} = 2 \tan\frac{c+a}{2}$ Found the question in the textbook.
I tried many methods of manipulating the identity to be proved but I did not even see a clue of using the given condition(of different forms like
$$
\sin\frac{a+c}{2} \cos\frac{a-c... | We need to prove that
$$\frac{\sin\left(\frac{a+b}{2}-\frac{c+a}{2}\right)}{\cos\frac{a+b}{2}\cos\frac{a+c}{2}}=\frac{\sin\left(\frac{c+a}{2}-\frac{b+c}{2}\right)}{\cos\frac{c+a}{2}\cos\frac{b+c}{2}}$$ or
$$\frac{\sin\frac{b-c}{2}}{\cos\frac{a+b}{2}}=\frac{\sin\frac{a-b}{2}}{\cos\frac{b+c}{2}}$$ or
$$\sin{b}-\sin{c}=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3311279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find function $f(x)$ knowing that $[f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$ is satisfied for $\forall x \in \mathbb R$.
Find function $f(x)$ knowing that $$\large [f(x^2 - x) + 1]f(x^3 + 1) = [f(x^4 - 2x^3 + x) + 1]f(x + 1)$$ is satisfied for $\forall x \in \mathbb R$ $(f(x) \ne c$ with $c$ being ... | The function
$$f(x) = -\frac 1 2 x$$
satisfies the above property.
This is how I found it. First, notice that the equation $x^2 - x = x^4 - 2x^3 + x$ has solutions $x = -1, 0, 1, 2$. If you let $x = -1, 0, 1$ in the equality you get an identity, because $-1, 0, 1$ are also solutions of the equation $x^3 + 1 = x + 1$. I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Find the equation of the circle that touches these three lines $x= 0$, $y=0$, $x = a$.
Here is my attempt:
$x = 0$ and $y = 0$, these both line go through the $x$ and $y$ axes. And also the circle touches those two lines. So the ce... | Your error is very simple: you have wronged expanding the square, in fact $\left(x - \frac{a}{2}\right)^2 + \left(y - \frac{a}{2}\right)^2 = \left(\frac{a}{2}\right)^2$ is not $x^2 + y^2 - 2ax - 2ay + \frac{a^2}{4} = 0$, but: $$x^2-ax+\frac{a^2}{4}+y^2+ay=0$$ In other words: $$4x^2+4y^2-4ax-4ay+a^2=0$$
Also, by simmetr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can the integral $\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$ be solved analytically? I'm trying to solve the following integrals:
$$I_1(a)=\int_0^\infty\ \frac{1}{\sqrt{x^2+a^2}} \cdot \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx$$
and
$$I_2(a)=\int_0^\infty\ \frac{1}{1+e^\sqrt{x^2+a^2}}\,dx.$$... | Assume $a\neq0$ for the key case.
For $\int_0^\infty\dfrac{1}{\sqrt{x^2+a^2}\left(1+e^\sqrt{x^2+a^2}\right)}~dx$ ,
$\int_0^\infty\dfrac{1}{\sqrt{x^2+a^2}\left(1+e^\sqrt{x^2+a^2}\right)}~dx$
$=\int_0^\infty\dfrac{1}{\sqrt{a^2\sinh^2t+a^2}\left(1+e^\sqrt{a^2\sinh^2t+a^2}\right)}~d(|a|\sinh t)$
$=\int_0^\infty\dfrac{1}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
A disc is cut into 12 sectors with areas in arithmetic progression. The largest angle is twice the smallest. Find the smallest angle. I was given a question which states -
A circular disc is cut into twelve sectors whose areas are in an
arithmetic sequence. The angle of the largest sector is twice the
angle of the... | You got $d = a$1$/11$.
($a$1 is angle of the smallest sector)
We can write: $11d = a$1 ........(1)
Now, we know:
Area of sector = ${\frac{1}2}.{r^2}.\theta$
And,
Sum of area of all sectors = Area of circle
(${\frac{1}2}.{r^2}.a$1) + (${\frac{1}2}.{r^2}.a$2) + ..... + (${\frac{1}2}.{r^2}.a$12) = $\pi.r^2$
${\frac{1}2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Show that $7a^{2} = 13b^{2}$ Here is another question which states that -
The first and the third terms of an arithmetic sequence are
a and b respectively. The sum of the first n terms is denoted by $S_n$.
(a) Find $S_4$ in terms of a and b.
(b) Given that $S_4$, $S_5$,
$S_7$ are consecutive terms of a geomet... | I know that:
$$\left\{\begin{matrix}
c_1=a+3b
\\c_1\cdot q=5b
\\ c_1\cdot q^2=\frac{7}{2}(3b-a)
\end{matrix}\right.$$
For simplicity, I put the first term $c_1$ equal to $S_4$.
From this I have $q=\frac{5b}{c_1}$. Substituting in the third equation: $$(a+3b)\cdot \frac{25b^2}{(a+3b)^2}=\frac{7}{2}(3b-a)$$
In other word... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3320802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to maximize area of a square inscribed in a equilateral triangle? We have an equilateral triangle and want to inscribe a square, in such way that maximizes the area of the square.
I sketched two possible ways, not to scale and not perfect.
Note I am not sure if the second way will really have all square corners t... |
Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below.
With the angles in the diagram,
$$d_1=\frac{1}{2\tan 30}a_1=\frac{\sqrt{3}}{2}a_1$$
$$d_2=\frac{\sin 15}{\sin 30}a_2=\frac{1}{2\cos 15}a_2$$
Assume both equilateral triangles have unit hei... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
} |
Prove inequality for a sum to $2^{n+1}$ Problem:I realize I made a mistake on one of the last questions I posted.Here is the problem:
Prove for all $n \in \mathbb{N}$
$\sum\limits_{k=1}^{2^{n+1}} \frac{1}{k} > \frac{n}{2}$
Attempt:
I'm having trouble recognizing the pattern for the general term of the inequality, can s... | Note that\begin{align}\left(\sum_{k=1}^{2^{n+2}}\frac1k\right)-\left(\sum_{k=1}^{2^{n+1}}\frac1k\right)&=\sum_{k=2^{n+1}+1}^{2^{n+2}}\frac1k\\&>\overbrace{\frac1{2^{n+2}}+\frac1{2^{n+2}}+\cdots+\frac1{2^{n+2}}}^{2^{n+1}\text{ times}}\\&=\frac{2^{n+1}}{2^{n+2}}\\&=\frac12.\end{align}Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3323536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Least value of $x+y+z$ where $ax=by=cz$ The following question is a generalization of the case $a=3$, $b=4$, $c=5$ from a MindYourDecisions YouTube video (which I am not going to actually link here).
Given positive integers $a$, $b$, and $c$, what is the smallest possible value of $x+y+z$ where $x$, $y$, and $z$ are p... | Let $N = ax = by = cz$. Since $N$ is a multiple of $a, b, c$ it must be a multiple of $\mathrm{lcm}(a, b, c)$. Thus, we should take
$$
\begin{align*}
x &= N/a \\
y &= N/b \\
z &= N/c
\end{align*}
$$
The sum will be minimal when $N = \mathrm{lcm}(a, b, c)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3324467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that the general value of $\theta$ satisfying $\sin\theta=\sin\alpha$ and $\cos\theta = \cos\alpha$ is given by $\theta = 2n\pi + \alpha$
The general value of $\theta$ simultaneously satisfying equations, $$\sin\theta = \sin\alpha \quad\text{and}\quad \cos\theta = \cos\alpha$$
is given by
$$\theta = 2n\pi + \... | One problem is you divided by $\sin\left(\frac{\theta - \alpha}{2}\right)$ when going from the third to fourth lines. However, with $\theta = 2n\pi + \alpha$, then $\frac{\theta - \alpha}{2} = n\pi$, but $\sin(n\pi) = 0$. When you divide by $0$, basically anything can occur from it, including incorrect results such as ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Why cant I prove this trigonometric equation straight down? The question is as follows:
Given that $\tan^2a - 2 \tan^2b = 1$. Show that $\cos2a + \sin^2b = 0$.
After a few attempts, I successfully came up with a solution as follows:
$$\tan^2a - 2 \tan^2b = 1$$
$$(\sec^2a - 1) - 2(\sec^2b - 1) = 1$$
$$\sec^2a - 2sec^2b ... | Starting from this line continue as follows (instead of dividing by $\cos ^2b$, express it as $1-\sin^2b$):
$$(\sin^2a - \cos^2a) \cos^2b - 2\sin^2b\cos^2a = 0 \\
(\sin^2a - \cos^2a) (1-\sin^2b) - 2\sin^2b\cos^2a = 0\\
-\cos 2a-\sin^2a\sin^2b+\sin^2b\cos^2a-2\sin^2b\cos^2a=0 \\
-\cos2a-\sin^2a\sin^2b-\sin^2b\cos^2a=0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3329683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!
| Why did things get compilicated?
If you replace $x$ with $y^2$ and $\sqrt x$ with $y$ you should get
$y^2 - 8y +7 =0$ and that should not be complicated.
========
Factor or use the quadratic formula so
$(y - 7)(y-1) = 0$ so $y = 7$ or $y=1$.
Or $y = \frac {-(-8) \pm \sqrt {64-4*7*1}}2 = \frac {8\pm {36}}2 = \frac {8\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 4
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If $x>4$, what is the minimum value of $\frac{x^4}{(x-4)^2}$. If $x>4$, what is the minimum value of $\frac {x^4}{(x-4)^2}$ ?
I have tried using AM-GM Inequality here by letting $y=x-4$ and ended up getting $224$ but that does not seem to be the correct answer. I find out by trial and error that the minimum value is wh... | Using AM-GM:
$$\frac {x^4}{(x-4)^2}=\left(\frac{x^2-16+16}{x-4}\right)^2=\left(\color{red}{x-4+\frac{16}{x-4}}+8\right)^2\ge (\color{red}8+8)^2=256,$$
equality occurs for $x=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Given two points endpoints of circle chord find the locus of midpoint
$A(2 \cos \theta_1 , 2 \sin \theta_1)$ and $B(2 \cos \theta_2 , 2 \sin \theta_2)$ are two end points of a variable chord $AB$ of the circle, and $M$ is the midpoint of the chord. Suppose that the slope of the chord $AB$ is $\textbf{always}$ equal to... | The slope of the line of midpoints must be $-\dfrac 1m$ and the line must contain the origin. So the equation is $y = -\dfrac 1mx$ or $x+my=0$
You know that
\begin{align}
M_{1,2}
&= \left(\cos\theta_1 + \cos\theta_2 , \
\sin\theta_1 + \sin\theta_2 \right) \\
&= \left(2\cos\dfrac{\theta_1+\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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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 ... | You've observed that $x\in[-1,2]$, but your mistake is that the maximum need not be at one of the endpoints. Once you know that
$$4(x+1)(x-2) + (y+1)^2=0,$$
you see that any solution $(x,y)$ must have $-1\leq x\leq 2$. Then
$$y=-1\pm2\sqrt{-(x+1)(x-2)},$$
and so you want to find the maximum of
$$5x+6y=5x-1\pm2\sqrt{-(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Integer $N$, in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b<16$, then what is the successor of $b$?
An integer $N$, expressed in base $b$, is $6789$. If $N$ is a multiple of $b-1$, and $b$ is less than $16$, then what is the successor of $b$?
I couldn't develop anything more than
$$N_b = \left(\;n\cdo... | If $N$ is a multiple of $b-1$ then by the rule of nines[1] the sum of the digits is a multiple of $b-1$ as well.
So $b-1| 6+ 7 + 8 + 9 = 30$. So $ b-1 = 1,2,3,5,6,10,15,30$
Bu $9$ is a digit so $b\ge 10$. And $b < 16$ so $b-1 < 15$.
... so $b-1 < 15$ and so $b-1 =10$ and $b =11$ and then successor of $b$ is $12$.
[1]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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The maximum of real function with 4 prime parameters and $\lfloor \ \rfloor$ Let $a$,$b$,$c$ and $d$ be prime numbers such that $a>b>c>d$. Let $x$ be an integer greater than $a$.
Let $f(x) = \left(\dfrac{x}{a}\right) – \left(\left(\dfrac{x}{ab}\right) + \left(\dfrac{x}{ac}\right) + \left(\dfrac{x}{ad}\right)\right) + ... | Note: Most of the work was already done by OP. This answer is a kind of summary.
In order to find the maximum difference $\max_{x\in\mathbb{N}}\{f(x)-g(x)\}$ the constant term $-1$ of $g(x)$ is not relevant and will be neglected. We denote with $\{x\}$ the fractional part of a real number $x\geq 0$.
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the solution to $y' = \frac{1+y}{x^2+x}$ that satisfies $y(1) = -1$ I'm having trouble when conditions like domain on variables are present in ODE's.
I am asked to find the solution to $y' = \dfrac{1+y}{x^2+x}$ that satisfies
$(a) \quad y(-2) = 1$
$(b) \quad y(1) = -1$
$(c) \quad y(1) = -2$
$(d) \quad y(-\frac{1... | If you consider $x < 0$ too, there will not be a unique solution to the IVP. Indeed, any solution to the IVP on the three intervals $(-\infty, -1), (-1, 0), (0, \infty)$ can be spliced together to form a solution to the differential equation. So long as the function is constantly $-1$ on $(0, \infty)$ (and whatever oth... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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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... | To detremine the $94$-th term, I can use the rules of the aritmethic progression, in particular: $$S_n=\frac{n}{2}(2a_0+(n-1)d)$$ Substituing the numbers, I have: $$S_n=\frac{n}{2}(2+2n)$$ Imposig $S_n=94$, I obtain: $n^2+n-94=0$ and so $n=10$.
| {
"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": 2
} |
If $\{e_1,e_2,e_3\}$ and $\{v_1,v_2,v_3\}$ are two different basis, $[v]_B\neq [v]_V$. So what is wrong in my argument What's wrong here ? Let $B=\{e_1,e_2,e_3\}$ the canonical basis of $\mathbb R^3$ and let $V\{v_1,v_2,v_3\}$ an other basis of $\mathbb R^3$. I denote $$\left[\begin{pmatrix}x\\ y\\ z\end{pmatrix}\right... | I see two problematic parts: the "red" and "blue" parts. I highlighted them from your argument. I hope you are not colorblind.
$\color{red}{\text{Let }v=\begin{pmatrix}1\\ 2\\ 3\end{pmatrix}\in\mathbb R^3}$.
I know that $$[e_1]={\color{red}{[v_1]_V}}=\begin{pmatrix}1\\0\\0\end{pmatrix},\quad [e_2]_B={\color{red}{[v_2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3341920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Integrating $\int\ln x \arccos\left( 7x^2-\sqrt{49x^4-50x^2+1}\right) dx$ This is a problem that my calculus professor gave to his students many years ago.
$$\int\ln x \arccos\left( 7x^2-\sqrt{49x^4-50x^2+1}\right) dx$$
Wolfram doesn't find any solution in terms of standard mathematical functions. I'm sure that thi... | Cross-posted on AoPS here. Answered by ysharifi. I will just put it here in length.
That is much neater than your other integral. Assuming you want to stay in real numbers, first we need to have $x > 0$ because of the presence of $\ln x.$ Secondly we also need to have
$$49x^4-50x^2+1=(1-x^2)(1-49x^2) \ge 0,$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3342115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Differentiate $\sqrt{\frac{1 +\sin x}{1 -\sin x}}$ I have tried a lot of ways to solve this question but I am unable to get the answer as same as my textbook.
The text book answer is as follow: $$\frac{1}{2}\sec^2\left(\frac{\pi}{4}+\frac{{x}}{2}\right)$$
The steps which I took is as follows:
$$\sqrt{\frac{1+ \sin x}{... | Alternatively, using the product rule:
$$\begin{align}\left(\sqrt{\dfrac{1 +\sin x}{1 -\sin x}}\right)'
&=(\sqrt{1+\sin x})'\cdot (1-\sin x)^{-1/2}+\sqrt{1+\sin x}\cdot ((1-\sin x)^{-1/2})'=\\
&=\frac{\cos x}{2\sqrt{1+\sin x}}\cdot \frac1{\sqrt{1-\sin x}}+\sqrt{1+\sin x}\cdot \frac{\cos x}{2(1-\sin x)\sqrt{1-\sin x}}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Prove that $a^2+u^2+d^2-b^2-c^2-v^2>-4w^2$ Let $ABCD$ be cyclic quadrilateral of the circle $O$ with: $$R=w\text{ is radius };AB=a;BC=b;CD=c;DA=d;AC=u;BD=v$$. Prove that $$a^2+u^2+d^2-b^2-c^2-v^2>-4w^2$$
We have $$u=\sqrt{\frac{\left(ac+bd\right)\left(ad+bc\right)}{ab+cd}};v=\sqrt{\frac{\left(ac+bd\right)\left(ab+cd... | Let us denote by $x,y$ the two angles in $A$ delimited by the diagonal $AC$ and the sides $AD$, respectively $AB$. Let us also denote by $s,t$ the two angles in $C$ delimited by the same diagonal $CA$ and the sides $CD$, respectively $CB$. We have
$$
x+y+s+t=\pi\ .
$$
Then we can express all the data in the inequality... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3348443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f... | As the degree of the divisor is $3$, so the degree of the remainder is at most $2$. Let the remainder $R\left(x\right)$ be $ax^2+bx+c$. As $x^3-x=x\left(x-1\right)\left(x+1\right)$, we'll find that: $$\begin{cases} R\left(-1\right)=a-b+c=\left(-1\right)^{81}+\left(-1\right)^{49}+\left(-1\right)^{25}+\left(-1\right)^9+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3352608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
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Closed form for series of multiple sums I have a series that goes like that:
\begin{align}
(n=1)\qquad &\sum_{i=0}^x\bigg(\sum_{j=1}^i 1\bigg)\\
(n=2)\qquad &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j 2\bigg)\bigg)\\
(n=3)\qquad &\sum_{i=0}^x\bigg(\sum_{j=0}^i \bigg(1+\sum_{k=1}^j \bigg(2+\sum_{l=2}^k 3\bigg... | Here we give a somewhat more compact expression. We look at the case $n=3$ again and derive from it a representation for the general case.
We have
\begin{align*}
\color{blue}{\sum_{j_0=0}^x}&\color{blue}{\left(\sum_{j_1=0}^{j_0}\left(1+\sum_{j_2=1}^{j_1}\left(2+\sum_{j_3=2}^{j_2}3\right)\right)\right)}\\
&=\sum_{j_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove that sum of the $k$ numbers in the $k$th group = ${\frac{1}{2}\left(k(k^2+1)\right)}$. Consider an arrangement of the positive integers, grouped as shown, so that the $k$th group has $k$ elements: $(1),(2,3),(4,5,6),(7,8,9,10), \ldots$.
The expression for the sum of the $k$ numbers in the $k$th group turns out to... | Let $a_{i,j}$ be the $j$'th term of the $i$'th group. The first element of the $k$'th group is the sum of the previous $1$ to $k - 1$ groups, plus $1$. Since each $i$'th group has $i$ elements, from Arithmetic progression, the first term of the $k$'th group is
$$a_{k,1} = 1 + \frac{(k-1)(1 + (k - 1))}{2} = 1 + \frac{k(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3355550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solving critical points and classifying the maxima, minima and saddle points I need help to find critical points of the function:
$$f(x,y,z)={(x^2+2y^2+3z^2) e ^{-(x^2+y^2+z^2)}}$$
Then I have to classify these critical points as local maxima/minima or saddle points.
What I would do is to take the partial derivatives $... | Computing the partial derivatives, we then have
$$\frac{\partial f(x, y, z)}{\partial x}=-2 x \left(x^2+2 y^2+3 z^2-1\right)e^{-(x^2+y^2+z^2)}$$
$$\frac{\partial f(x, y, z)}{\partial y}=-2 y \left(x^2+2 y^2+3 z^2-2\right)e^{-(x^2+y^2+z^2)}$$
$$\frac{\partial f(x, y, z)}{\partial z}=-2 z \left(x^2+2 y^2+3 z^2-3\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3357362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Discussion about $\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)}$ May I have some further discussion on
$$\sum_{n=0}^{\infty} {\frac{(-1)^{n+1}H_{2n}}{2n+1} \big(\psi(n+1)-\psi(n+\tfrac1{2})\big)} = \frac{\pi^3}{96} - \frac{\pi}{8}\ln^2\!2 \tag1$$
where $H_{n}$ is harmonic-nu... | Solution to question 1:
Since you managed to write your sum in (1) as $\int_0^1\frac{\arctan x\ln(1+x^2)}{x(1+x)}\ dx$, so lets evaluate the integral.
From here we have
$$\int_0^1\arctan x\left(\frac{3\ln(1+x^2)}{1+x}-\frac{2\ln(1+x)}{x}\right)\ dx=\frac{3\pi}{8}\ln^22-\frac{3\pi^3}{96}\tag{1}$$
and from here we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3361723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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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... | Since $x=\sec\theta$, so $\cos\theta=\frac1x$ and hence $\sin\theta=\frac{\sqrt{x^2-1}}{x}$. Thus
$$\sin(\sec^{-1}(x))=\frac{(x^2-1)^{3/2}}{x^3}. $$
| {
"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": 2
} |
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... | Let $y=\frac1{x}$. Then,
\begin{align}
(x+\frac1{x})^4(x-\frac1{x})^2&=(x+y)^4(x-y)^2\\
&=(x+y)^2(x+y)^2(x-y)^2\\
&=(x+y)^2(x^2-y^2)^2\\
&=\big((x+y)(x^2-y^2)\big)^2\\
&=(x^3+x^2y-xy^2-y^3)^2\\
&=(x^3+x-\frac1{x}-\frac1{x^3})^2\\
&=\bigg(\frac{x^6+x^4-x^2-1}{x^3}\bigg)^2\\
&=\frac{x^{12} + 2x^{10} - x^8 - 4 x^6 - x^4 ... | {
"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": 2
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A box with 3 red balls and 2 white balls. 3 balls taken randomly with replacement, A box containing 3 red balls and 2 white balls. A ball is taken randomly and with replacement. This is done 3 times.
*
*What is the probability that exactly one red ball is obtained?
*Probability at least 1 red ball is obtained?
A... | Both parts are correct. As regards the second one, it is easier to evaluate the complement: the probability that at least one red ball is obtained is $1$ minus the probability that all the three balls are white, namely
$$1-\left(\frac{2}{5}\right)^3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3366514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$ $\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5}$
So, getting rid of the denominators I got to:
$(x+1)(x+5) + x(x+5) < 2x(x+1)$
$\rightarrow x < \frac{-5}{9}$
And also keeping in mind we can't have $-1,-5$ in the solution set, I got that the solution set was:
$\left(-\i... | When multiply we need to take into account the sign; in this case your step are not correct since you are assuming positive terms.
As an alternative to avoid multiplication, we have
$$\frac{1}{x} + \frac{1}{x+1} < \frac{2}{x+5} \iff\frac{1}{x} + \frac{1}{x+1} - \frac{2}{x+5}<0 \iff \frac{9x+5}{x(x+1)(x+5)}<0$$
then stu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3372139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
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\\... | $$\iint_R x^2+y^2 dA=\int_0^{2\pi}\int_0^a r^3drd\theta=2\pi\int_0^a r^3 dr$$
| {
"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": 3
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Solve Integration
$$I = \int\frac{2\sin x+\sin 2x}{(\cos x-1)\sqrt{\cos x+\cos^2x+\cos^3x}}dx$$
I have tried for this question to solve it first I tried to separate the numerator but that doesn't work then i tried for substituting the value in the denominator but still I am not able to convert the whole function in t... | First, use the double angle identity $\sin2x = 2\sin x \cos x$ to get that
$$I = \int\frac{2\sin x + \sin 2x}{(\cos x - 1)\sqrt{\cos x + \cos^2 x + \cos^3 x}}dx$$ $$ = \int \frac{(1+\cos x)(-2\sin x)dx}{(1-\cos x)\sqrt{\cos x + \cos^2 x + \cos^3 x}}$$
Now the tricky part is deciding what substitution to choose. Using t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3379035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
How many solutions $x_1+\dots+x_8=30$ has?
Enumerate the solutions of
$$x_1+\dots+x_8=30$$
where $2\leq x_i\leq 5$ for $i=1,\dots,6$,
and $x_7$ and $x_8$ must be or 5 or 10.
I am not sure with my process.
and I would like to ask you if you have a better way to solve this?
and if I am right.
$S_0 - S_1 + S_2 - ... | A possible way is using generating functions:
*
*First set $y_i = x_i-2$ for $i=1,\ldots , 6$ and $y_i = x_i -5$ for $i=7,8$. So, you look for the number of solutions of
$$\mbox{}y_1 + \cdots + y_6 + y_7 + y_8 = 8 \mbox{ with } 0\leq y_i \leq 3 \;(i=1,\ldots ,6) \mbox{ and } y_i \in \{0,5\}\;(i=7,8)$$
*So you look... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the probability of the event that exactly two of the three events occur? My professor posted some practice problems with answers (without explanation) for our midterm and this was one of them:
Three independent events have respective probabilities, $\frac{1}{3},\frac{2}{5},\frac{1}{4}$
a.) Find the probabi... | In a) he's simply adding up the probabilities. The first term is the probability that the first two events happen and the third one does not, for instance.
In b) he's getting the complement of the event that no more than one event occurs. The first three terms are the probabilities that exactly one event occurs (simi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $z=-\frac12(1\pm i\cot(k\pi/8))$; and follow-up questions
*
*(a) Show that the equation $(z+1)^8-z^8=0$ has roots $z=-\frac12$, $-\frac12\left(1\pm i\cot\frac{k\pi}{8}\right)$, where $k=1,2,3$.
*(b) Hence show that
$$(z+1)^8-z^8=\tfrac18(2z+1)(2z^2+2z+1)\left(4z^2+4z... | By your method
$$(z+1)^8-z^8=0 \iff \left(1+\frac1z\right)^8=1$$
then let
$$w=1+\frac1z$$
and solve
$$w^8=1 \implies w=e^{\frac{2i\pi k}8}\quad k=0,1,2,3,4,5,6,7$$
but only $7$ solutions are valid since $k=0$ is not acceptable, indeed the original equation is of degree $7$.
The solutions are given by
$$z=\frac1{e^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3384259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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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:
$$\{a^3+(1-\sqrt{2})a^2-(3+\sqrt{2})a+3\sqrt{2}\}x^2+... | Well I got the answer, thanks to @Robert Z
I) First lets see what should be the condition when $a>0 \text{\{a here means coefficient of $x^2$\}}$
So $$\text{parameter }a\in \left(\frac{-1-\sqrt{13}}{2},\frac{-1+\sqrt{13}}{2}\right) \cup \left(\sqrt{2},\infty\right) $$
Now there can be two sub-cases:
1) When $D<0$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3395226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Ramanujan, sum of two cubes - how was it discovered? I'm looking for motivation for, and hopefully a derivation of, Ramanujan's sum of cubes formula
$$\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$$
I can't see where one could start to justify this with... | "OP" inquired about finding parameterization of two cubes in two different way's.
Consider the below equation;
$s^3+t^3=u^3+v^3$ ----$(1)$
Tomita has given a method to parameterize equation $(1)$.
In his solution, take $(p,q,r,s)=(-1,9,10,12)$
then the below mentioned solution is arrived at:
$s=36x^2+22xy+y^2$
$t=40... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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Is the function $f(x)=\int_0^{\infty} \left[ 1 - a^x \sin \left( \frac{1}{a^x} \right) \right] da$ writable in a nicer way? I was idly exploring properties of this strange function exploiting wolframalpha and I can't understand it. Some calculus seems to show that this function is defined only for $x>\frac{1}{2}$ (if $... | The substitution $u = a^{-x}$ yields
$$ f(x) = \int \limits_0^\infty [1 - a^x \sin(a^{-x})] \, \mathrm{d} a = \frac{1}{x} \int \limits_0^\infty \frac{u - \sin(u)}{u^{2 + \frac{1}{x}}} \, \mathrm{d} u \, .$$
The integrand behaves like $u^{1 - \frac{1}{x}}$ near $u = 0$, so $x > \frac{1}{2}$ is indeed necessary (and suff... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3396664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Using a line integral to find work For the life of me I can't see where I'm going wrong with this.
Given the vector field
$$ F(x, y) = \left[ xy, \frac{2y}{x} \right] $$
a particle travels on the hyperbola
$$ \frac{x^2}{4} - y^2 = 1 $$
from $ \left(2\sqrt{10}, -3 \right) $ to $ \left(2, 0 \right) $ how much work is d... | Since the $y$-coordinate of the trajectory is negative, you should parametrize the path as
$$ x = 2\sqrt{t}, \>\>\> y = -\sqrt{t-1},\>\>\> t\in\left[10,1\right]$$
to allow negative values for the $y$-variable. Then, the work integral becomes
$$ \int_C F(r(t)) \cdot r'(t) dt = \int_{10}^1 \left[2\sqrt{t}(-\sqrt{t-1})\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3400779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Uniform convergence of $\sum\limits_{k = 1}^{\infty} \frac{\sin (\sqrt{x}/k)}{\sqrt{x^2 + k^2}}$ I need to check uniform convergence of $\sum\limits_{k = 1}^{\infty} \frac{\sin (\sqrt{x}/k)}{\sqrt{x^2 + k^2}}, x \in [0, +\infty)$. I've tried Weierstrass Test, but it didn't work.
| The convergence is not uniform for $x \in [0,\infty)$.
With $x_n = \frac{\pi^2n^2}{4}$ and $n < k \leqslant 2n$ we have
$$\frac{\pi}{4} = \frac{\pi n}{2} \frac{1}{2n} \leqslant \frac{\sqrt{x_n}}{k} < \frac{\pi n}{2} \frac{1}{n}=\frac{\pi}{2}$$
Thus, for all $n \in \mathbb{N}$
$$\left|\sum_{k=n+1}^{2n}\frac{\sin \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3401857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$
$\sqrt{4^0+\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\cdots}}}}=?$
I found this problem in a book. I tried to solve this but couldn't. Using calculator, I found the value close to $2$. But how can this problem be solved with proper procedure?
| Take $f(x,n)=x+2^n$. We can see that;
$$\begin{aligned}
f(x,n) &= \sqrt{2^{2n}+x\left(x+2^{n+1}\right)} \\
&= \sqrt{2^{2n}+xf(x,n+1)} \\
&= \sqrt{2^{2n}+x\sqrt{2^{2\left(n+1\right)}+x\sqrt{2^{2\left(n+2\right)}+x\sqrt{...}}}}\\
&=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3402479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ \int_0^\frac{\pi}{2}\frac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta. $
Evaluate
$$
\int_0^\frac{\pi}{2}\frac{\cos\theta}{\cos\theta+\sin\theta}\,d\theta.\qquad\text{(1)}
$$
By letting $t=\tan\theta$, $(1)$ equals
$$
\int_0^\infty\frac{1}{(1+t)(1+t^2)}
\,dt,$$
and then?
By letting $t=\tan\frac{\theta}{2... | Use symmetry instead. Notice under the change of variable $\theta \mapsto \frac{\pi}{2}-\theta$
$$I = \int_0^{\frac{\pi}{2}} \frac{\sin\theta}{\cos\theta + \sin\theta}d\theta$$
by the trig identity $\cos\left(\frac{\pi}{2}-\theta\right) = \sin\theta$ and vice-versa. Then add the two integrals:
$$2I = \int_0^{\frac{\pi}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $x^3+x+57\equiv 0$ mod $125$
Solve $x^3+x+57\equiv 0$ mod $125$
So first I try to solve $x^3+x+2\equiv 0$ mod $5$
I have $f^\prime(x)=3x^2+1$
I found $f(4)$ to be the only solution.
And since $f^\prime(4)\equiv 4$ there is a unique solution mod $25$
So I then need to solve $f(4+t5)=f(4)+t\cdot 5\cdot f^\prime(4... | $x^3+x+57\equiv 0\pmod {125}\Rightarrow x^3+x-68\equiv 0\pmod {125}\Rightarrow (x-4)(x^2+4x+17)\equiv 0\pmod {125}.$
How did I get $x-4?$ I simply used the fact that $x-4$ is a factor of the polynomial $x^3+x-68.$
Since the quadratic $x^2+4x+17$ has no solutions $\pmod {125},$ the only solution is $x=4\pmod {125}.$ Y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3406462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Where is the error in this apparent contradiction arising from the function $f(\pm n)=\left(\frac{n}{n+1}\right)^{\pm 1}$? We define a function $ f \left( \pm n \right) =\left( \frac{ n }{ n + 1 } \right) ^ { \pm 1 } $.
That is, $$ f \left( m \right) = \begin{cases}\frac{ m }{ m + 1 } &\text{ if } m \ge 0 \\ \frac{ -m ... | The following is wrong: $f\left(-\frac{\gamma}{\delta}\right)=\frac{\gamma}{\gamma+\delta}$. In fact:
$$\frac{\gamma}{\gamma+\delta}=f\left(\frac{\gamma}{\delta}\right)=\frac{1}{f\left(-\frac{\gamma}{\delta}\right)}$$
hence
$$f\left(-\frac{\gamma}{\delta}\right)=\frac{\gamma+\delta}{\gamma}.$$
So $$f(-2)=f\left(-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3408268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all matrices $A\in \mathbb{R}^{2\times2}$ such that $A^2=\bf{0}$ Attempt:
Let's consider $A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}$.
$$\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... | If $A^2=0$, then $A$ is singular. Hence $Au=0$ for some nonzero vector $u$. Pick any vector $v$ that is linearly independent of $u$. Then $Av=au+bv$ for some scalars $a$ and $b$. It follows from $A^2=0$ that
$$
0=A^2v=A(Av)=A(au+bv)=aAu+bAv=bAv=b(au+bv)=abu+b^2v.
$$
Thus $b$ must be zero. Hence $Av=au$ and $Au=0$. Conv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3409265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
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+... | Write down $1+1+\cdots+1=X$. Now group runs of adjacent ones by going from left to right and for each $+$ sign you encounter deciding whether or not a group boundary occurs there. Parenthesize each group to get a partition of $X$ into a sum of nonnegative integers. Each possible choice of group boundaries corresponds t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3411080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
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... | $$\begin{aligned}\frac{\tan^3 x-3\tan x}{\cos\left(x+\frac{\pi}{6}\right)}
&=\tan(x)\cdot\frac{\tan^2 x-3}{\cos x\cos\left(\frac{\pi}{6}\right)-\sin x\sin\left(\frac {\pi}{6}\right)},\quad\quad x\neq \frac{\pi}{3}
\\
&=\frac{2\sin x}{\cos^{3} x}\cdot\frac{\sin x^{2}-3\cos x^{2}}{\sqrt{3}\cos x-\sin x}\quad\quad\text{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3412797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove $3|z-1|^2 = |z+1|^2 \iff |z-2|^2 = 3$ I'm struggling with this prove question. I tried starting with let $z = x + iy$, and substituting the three in the first eqn with $|z-2|^2$, with the intention of eventually equating the LHS with the RHS, and I ended up with something most likely incorrect.
Any help would be... | Generalization:
Let $$|z-a|=b,z=a+b\cos t+ib\sin t$$ where $a,b,t$ are real
$$\left|\dfrac{z-c}{z+c}\right|=\sqrt{\dfrac{(a-c+b\cos t)^2+b^2\sin^2t}{(a+c+b\cos t)^2+b^2\sin^2t}}=\sqrt{\dfrac{(a-c)^2+b^2+2b(a-c)\cos t}{(a+c)^2+b^2+2b(a+c)\cos t}}$$
which will be $$\left|\dfrac{a-c}{a+c}\right|$$ if
$$\dfrac{2b(a-c)}{2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2 t} dt$ \begin{align}
\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt
\end{align}
I am trying to evaluate above integral. The results is $2\pi$ according to Mathematica. I want to obtain this result by integrating properly
Can this integral be e... | Rewrite,
\begin{align}
I=\int_0^{2\pi} \frac{2-\cos t}{(2\cos t-1)^2 + \sin^2t} dt =
2\int_0^{\pi} \frac{2-\cos t}{3\cos^2 t -4\cos t +2} dt
\end{align}
and then substitute $\cos t = \frac{1-x^2}{1+x^2}$, along with $dt = \frac{2dx}{1+x^2}$,
$$I= 4\int_0^{\infty} \frac{1+3x^2}{9x^4 -2x^2 +1} dx $$
$$= 4\int_0^{\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3416545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove $\frac{1}{1^{5}\cosh(\frac{\pi}{2})}-\frac{1}{3^{5}\cosh(\frac{3\pi}{2})}+\frac{1}{5^{5}\cosh(\frac{5\pi}{2})}+\cdots=\frac{\pi^{5}}{768}$ This is an identity from Ramanujan's letter, I am just curious. How do you prove this. My math level knowledge is still very basic so a simplified proof is preferred:
$$\frac{... | This sum can be found using the partial fraction expansion of secant, with a similar approach to the one found here.
Beginning with the formula
$$ \frac{\pi}{4} \frac{1}{\cosh \left(\tfrac{\pi}{2}x \right)} = \sum_{k=0}^{\infty} \frac{(-1)^k (2k+1)}{(2k+1)^2+x^2} $$
valid for all real $x$ (see this post)
set $x=2n+1$, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3417312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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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... | Assuming $a,b,c,d\geq 0$ Maclaurin's inequality gives
$$\sqrt[3]{\frac{abc+abd+acd+bcd}{4}}\leq \frac{a+b+c+d}{4} $$
and by the AM-QM (i.e. Cauchy-Schwarz) inequality we have
$$ \frac{a+b+c+d}{4}\leq\sqrt{\frac{a^2+b^2+c^2+d^2}{4}}.$$
| {
"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": 1
} |
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{... | Define $f(y) = \ln(1+y)$. We know that, $f'(0) = 1$. Thus $$\lim_{y\to 0} \frac{\ln(1+y)}{y} = \lim_{y\to 0} \frac{f(y)-f(0)}{y-0} = f'(0) = 1 \hspace{1cm}(1)$$
Now, consider the limit $$\lim_{x\to 0^{-}}\frac{\ln(1+x+x^{2})-x}{x^{2}+x} \hspace{1cm} (2)$$ Note that $x^{2}+x = (x+\frac{1}{2})^{2}-\frac{1}{4}$. Take $u =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3424937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Parabola equation in form of quadratic $ax^2+bx+c=y$ where $a+b+c$ is an integer Suppose a parabola has vertex $(\frac{1}{4},\frac{-9}{8})$ and equation $ax^2+bx+c=y$ where $a>0$ and $a+b+c $ is an integer. Find the minimum possible value of $a$ under the given condition.
My approach
$(x-\frac{1}{4})^2=4a'(y+\frac{9}{8... | The equation of the parabola is:
$$a \left(x-\frac{1}{4}\right)^2 - \frac{9}{8}$$
$$=a \left(x^2-\frac{1}{2}x+\frac{1}{16} \right)- \frac{9}{8}$$
$$=ax^2-\frac{a}{2}x+ \left(\frac{a}{16}-\frac{9}{8} \right)$$
Then we find that: $$a+b+c = \frac{9}{16}(a-2).$$
When $a=0$, $a+b+c = -\frac{9}{8}$. Since the gradient of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3425766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Solve $x_{n+1}=Ax_n+\frac{B}{x_n^5}+\frac{C}{x_n^9}$
For $$x_{n+1}=Ax_n+\frac{B}{x_n^5}+\frac{C}{x_n^9}$$
a. Find $A,B,C$ that will give an optimal approximation (Highest order) for $\sqrt{2}$
b. Solve the system using LU decomposition, what is the given order?
So we are looking to approximate the function $f(x)=x... | Your answer is just fine... Your linear system is
$$
\begin{cases}
g(\sqrt 2) = \sqrt 2\\
g'(\sqrt 2)=0\\
g''(\sqrt 2)=0
\end{cases}
$$
or, in terms of $A,B,C$:
$$
\begin{bmatrix}
\sqrt 2 & \frac{\sqrt 2}{8} & \frac{\sqrt 2}{32}\\
1 & -\frac 58 & -\frac{9}{32}\\
0 & \frac{15 \sqrt 2}{8} &\frac{45 \sqrt 2}{32}
\end{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3430389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the values of a, b, c, and d from a matrix equation Here is the equation I'm working on:
\begin{align}
\begin{bmatrix}2a+b&a-2b\\5c-d&4c+3d\end{bmatrix} = \begin{bmatrix}4&-3\\11&24\end{bmatrix}\\
\end{align}
My solutions:
Row 1 column 1
\begin{align}
2a + b = 4\\
2a = 4 - b\\
a = 2 - \frac 12b\\
\end{align}
Row 1... | Yes your approach is correct. As an alternative you could consider the $4$ by $5$ augmented matrix for the system in the unknowns $a$,$b$,$c$ and $d$
$$\left[\begin{array}{cccc|c}
2&1&0&0&4 \\
1&0&0&-2&-3 \\
0&0&5&-1&11 \\
0&0&4&3&24
\end{array}\right]$$
and solve by gaussian elimination.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Problem regarding inconsistent solution
My attempt :
For First answer i try to used by elementary row operation $ A= \begin{bmatrix} 1&2&3\\2&4&8 \\3&6&7 \end{bmatrix}$ that give $ A= \begin{bmatrix} 1&2&3\\0&0&2 \\0&0&0 \end{bmatrix} \begin{bmatrix} x_1 \\x_2\\x_3 \end{bmatrix}=\begin{bmatrix} 1 \\0\\5\end{b... | That way to proceed is not correct, indeed we need to apply the RREF on the augmented matrix
$$\left[\begin{array}{ccc|c} 1&2&3&1\\2&4&8&0 \\3&6&7&5 \end{array}\right]$$
For the second question let consider
$$\left[\begin{array}{ccc|c} 1&2&3&b_1\\2&4&8&b_2 \\3&6&7&b_3 \end{array}\right]$$
and apply RREF again to obtain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432756",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For all real $x$ the expression$ \frac{x^2-2x+4}{x^2+2x+4}$ lies between $\frac 13$ and $3$.
The values between which the expression $$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$ is
A solution I saw says that the expression can be written as
$$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2.(3.3^x)+4}$$ which is analogous t... | $$\frac{9.3^{2x}+6.3^x+4}{9.3^{2x}-6.3^x+4}$$
$$\frac{(3)^2(3^x)^2+2(3)(3^x)+4}{(3)^2(3^x)^2-2(3)(3^x)+4}$$
$$\frac{(3.3^x)^2+2(3.3^x)+4}{(3.3^x)^2-2(3.3^x)+4}$$
$$\frac{(z)^2+2(z)+4}{(z)^2-2(z)+4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3434166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve this differential equation $(x^2+1)y' = xy$ $$\begin{align}
(x^2+1)\frac{dy}{dx} &= xy \\
\frac{1}{y}\frac{dy}{dx} &= \frac{x}{(x^2+1)} \\
\int \frac{1}{y} dy &= \int \frac{x}{(x^2+1)} dx \\
\ln(y) &= \frac{1}{2\ln(x^2+1)} + C \\
y &= e^{\frac{1}{2\ln(x^2+1)} + C}
\end{align}$$
When I integrated $\frac{x}{x^2+1}$... | Your hunch is right. Indeed there is a mistake at integration step:
$$\int \frac{x}{(x^2+1)} dx =\dfrac{1}{2}\ln|x^2+1|+C$$
Might help memorizing:
$$\int \dfrac{f'(x)}{f(x)}dx = \ln |f(x)|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Problem with equation in complex numbers
I am supposed to calculate:
$x^{2}=5+i$
I used formula:
$\left | \cos \frac{\alpha }{2} \right |=\sqrt{\frac{1+\cos \alpha }{2}}$
$\left | \sin \frac{\alpha }{2} \right |=\sqrt{\frac{1-\cos \alpha }{2}}$
and I came to the point where:
x= $\sqrt[4]{26}\left ( \frac{\sqrt{\sqr... | You can also write $$5+i=a^2-b^2+2abi$$ so we will get $$5=a^2-b^2$$ and $$1=2ab$$
Plugging $$b=\frac{1}{2a}$$ in the first equation you will get
$$20a^2=4a^4-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3438046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Sums and Sequences I want some help, please, to solve this problem.
Let $$\forall n\in\mathbb{N}^*,\ S_n=\sum_{k=1}^n\frac{1}{\sqrt{k}}$$
I want to prove that $\forall n\in\mathbb{N}^*,\ S_n\leqslant\sqrt{n-1}+\sqrt{n}$
and that $\forall n\in\mathbb{N}^*,\ S_n\geqslant2\sqrt{n+1}-2$
I used the Mathematical Induction m... | We have
$$S_{k+1}=S_k+\frac1{\sqrt{k+1}}\stackrel{?}\le \sqrt{k}+\sqrt{k+1}$$
which is true indeed
$$S_k+\frac1{\sqrt{k+1}}\le \sqrt{k-1}+\sqrt{k}+\frac1{\sqrt{k+1}}\le \sqrt{k}+\sqrt{k+1}$$
since
$$\sqrt{k-1}+\sqrt{k}+\frac1{\sqrt{k+1}}\le \sqrt{k}+\sqrt{k+1}\iff \sqrt{k-1}+\frac1{\sqrt{k+1}}\le \sqrt{k+1}$$
$$\sqrt{k... | {
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"url": "https://math.stackexchange.com/questions/3440662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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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^... | You can show divisiblity by $30$ by showing divisibilty by $2,3$ and $5$
So show $21\times5^{2k+3} - 5\times3^{2k+3} + 4\times30M$ is divisible by $2,3$ and by $5$
So
$21\times5^{2k+3} - 5\times3^{2k+3} + 4\times30M=$
$15(7\times 5^{2k+2} - 3^{2k+2} + 4\times 2M)$.
Obviously $15$ is divisible by $3$ and by $5$.
So j... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3441310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 5
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Problem of Apollonius: Find the radius of the circle tangent to $3$ other circles $O_1$, $O_2$ and $O_3$ have radius of $a$, $b$ and $c$ Find the radius of the circle tangent to $3$ other circles $O_1$, $O_2$ and $O_3$ have radius of $a$, $b$ and $c$
The Wikipedia page about the Problem of Apollonius can be found here
... | If you can endure the algebraic slog (a computer algebra system really helps), you can find some structure that simplifies the solution.
Let the given circles have radii $r_A$, $r_B$, $r_C$, and (non-collinear) centers $A:=(x_A,y_A)$, $B:=(x_B,y_B)$, $C:=(x_C,y_C)$; I specifically used
$$A = (0,0) \qquad B = (c,0) \qq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3443137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$ \int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx$ I am tring to obtain following formula [actually I obtained this resut via mathematica]
\begin{align}
\int \frac{\cosh(x)}{ (a^2 \sinh^2(x) + b^2 \cosh^2(x))^{\frac{3}{2}} } dx
= \frac{\sqrt{2} \sinh(x)}{a^2 \sqrt{a^2-b^2 + (a^2+b^2) \cosh... | Ket us use $$\int \frac{dt}{(m^2+t^2)^{1]2}}=\ln (t+\sqrt{m^2+t^2}]+C_1~~~)1)$$
D. w.r.t $a$ we get $$\int \frac{dt}{(m^2+t^2)^{3/2}}=\frac{t}{m^2\sqrt{m^2+t^2}}+C_2~~~~(2) $$
Then $$I= \int \frac{\cosh x}{[(a^2+b^2) \sinh^2 x+b^2]^{3/2}} dx= \frac{1}{(a^2+b^2)^{3/2}}\int \frac{dt} {(m^2+t^2)^{3/2}}, m=b/\sqrt{a^2+b^2}... | {
"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": 2
} |
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?
| $2a^2+b+2=2a^2+3-a=(\frac{a^2+1}{2}-a)+(\frac{3a^2+5}{2})\geq\frac{3a^2+5}{2}$.
On the other hand by QM-AM $\sqrt{\frac{a^2+a^2+a^2+1+1+1+1+1}{8}}\geq\frac{3a+5}{8}$, therefore $\sqrt{3a^2+5}\geq\frac{3a+5}{\sqrt{8}}$. Now we return back: $\sqrt{2a^2+b+2}\geq\sqrt{\frac{3a^2+5}{2}}\geq\frac{3a+5}{\sqrt{2\times8}}=\frac... | {
"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": 5
} |
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... | Examine the expression you have,
$$\underset{x\rightarrow \infty}\lim{\frac{\sqrt[6]{1-\cos{\frac{1}{x^3}}}\Big(\frac{\sqrt[x]{3}-\sqrt[x]{2}}{\sqrt[x]{6}}\Big)}{\frac{1}{x}\ln{\Big(1-\frac{1}{x}\Big)}}}$$
Note that in the limit $x\rightarrow \infty$,
$$ \sqrt[6]{1-\cos\frac 1{x^3}} \rightarrow
\sqrt[6]{1-(1-\frac12 \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How to find the limit of this function using L'Hospital's rule?
$$\lim_{x\rightarrow 0} \,\,\left( \sqrt[3]{1+2x+x^3} - \frac{2x}{2x+3} \right) ^ {\frac1{x^3}} $$
I have already tried several options, but the only answer I have gotten so far is
$e^{\infty}$, which is incorrect. The correct answer is $e^\frac{43}{81... | The standard approach here is to take logarithm and then evaluate the limit. If $f(x) $ denotes the expression under limit and it tends to a limit $L$ then
\begin{align}
\log L&=\lim_{x\to 0}\frac{1}{x^3}\log\left(\sqrt[3]{1+2x+x^3}-\frac{2x}{2x+3}\right)\notag\\
&=\lim_{x\to 0}\frac{g(x)}{x^3}\cdot\frac{\log(1+g(x))}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3448772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Manipulation of Taylor expansion of $e^x$ I have an expression:
$$ f(x) = \sum\limits_{N=0}^\infty (N-x)^2 \frac{x^N}{N!}$$
I want to figure out whose Taylor Expansion this is.
I've found that I can separate out the above expression, and get the following:
$$f(x) = \sum\limits_{N=0}^\infty N^2 \frac{x^N}{N!} - 2 \sum\l... | Hint:
$$(x-n)^2=x^2-2nx+n^2=x^2-2nx+n(n-1)+n$$
so that
$$\sum_{n=0}^\infty(x-n)^2\frac{x^n}{n!}\\
=\sum_{n=0}^\infty x^2\frac{x^n}{n!}-2\sum_{n=0}^\infty nx\frac{x^n}{n!}+\sum_{n=0}^\infty n(n-1)\frac{x^n}{n!}+\sum_{n=0}^\infty n\frac{x^n}{n!}\\
=x^2\sum_{n=0}^\infty \frac{x^{n}}{n!}-2x\sum_{n=1}^\infty \frac{x^{n}}{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3449582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
In a triangle, if $\tan(A/2)$, $\tan(B/2)$, $\tan(C/2)$ are in arithmetic progression, then so are $\cos A$, $\cos B$, $\cos C$
In a triangle, if $\tan\frac{A}{2}$, $\tan\frac{B}{2}$, $\tan\frac{C}{2}$ are in arithmetic progression, then show that $\cos A$, $\cos B$, $\cos C$ are in arithmetic progression.
$$2\tan\le... | Rewrite the equation as
$$2\frac{\sin\frac B2}{\cos\frac B2}
=\frac{\sin\frac A2}{\cos\frac A2}+\frac{\sin\frac C2}{\cos\frac C2}
=\frac{\sin\frac {A+C}2}{\cos\frac A2\cos\frac C2}$$
Then, with $A+C = \pi - B$,
$$2\sin\frac B2 \cos\frac A2 \cos\frac C2 = \cos\frac B2\sin\frac {\pi-B}2=\cos^2\frac B2$$
$$2\sin\frac B2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Proving $\sum_{cyc} \sqrt{a^2+ab+b^2}\geq\sqrt 3$ when $a+b+c=3$ Good evening everyone, I want to prove the following:
Let $a,b,c>0$ be real numbers such that $a+b+c=3$. Then $$\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{c^2+ca+a^2}\geq 3\sqrt 3.$$
My attempt: I try noting that $$\sum_{cyc} \sqrt{a^2+ab+b^2}=\sum_{cyc}... | I’m not 100% sure but this is what I got.
$$
a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2>(a+b/2)^2
$$
From this we get
$$
a+b/2+b+c/2+c+a/2=9/2>\sqrt3
$$
Or
$$
a^2+ab+b^2=(a+b/2)^2+\frac{3}{4}b^2> \frac{3}{4}b^2
$$
From this we get
$$
\frac{\sqrt{3}}{2}(a+b+c)= \frac{3\sqrt{3}}{2}>\sqrt{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3450882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Minimize $\frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, $x,y,z>0$ Minimize $\;\;\displaystyle \frac{(x^2+1)(y^2+1)(z^2+1)}{ (x+y+z)^2}$, if $x,y,z>0$.
By setting gradient to zero I found $x=y=z=\frac{1}{\displaystyle\sqrt{2}}$, which could minimize the function.
Question from Jalil Hajimir
| If you want some calculus/analysis argument:
After establishing there must exist a global minimum, let $p$ be the global minimum. Then we must have that
$$f(x) = x^2\left((y^2+1)(z^2+1) - p\right) - 2xp(y+z) + (y^2+1)(z^2+1) - p(y+z)^2\geq 0$$
as a quadratic in $x.$ So the discriminant is non-positive:
$$D =4\left[p^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3454095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\int \frac{dx}{x^3+x^2\sqrt{x^2-1}-x}$ solve:
$$\int \frac{\mathrm dx}{x^3+x^2\sqrt{x^2-1}-x}$$
I tried:
$$\begin{align}\int \frac{\mathrm dx}{x(x^2-1)+x^2\sqrt{x^2-1}}&=\int \frac{\mathrm dx}{x(\sqrt{x^2-1}\sqrt{x^2-1})+x^2\sqrt{x^2-1}}\\&=\int \frac{\mathrm dx}{x\sqrt{x^2-1}(\sqrt{x^2-1}+x)}\end{align}$$
$x=\sin t$
... | Hint: Put $x＝\dfrac{1}{\cos t}$, then
$$\int \frac{\cos^3 t}{\sin t(\sin t+1)}\frac{\sin t}{\cos^2 t}dt＝\int \frac{d\sin t}{\sin t+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Limit ${\lim\limits_{x \to \frac{\pi}{2}}\Big(\tan^2{x}\Big(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}}\Big)\Big)$ $\displaystyle\lim_{x \to \frac{\pi}{2}}\left(\tan^2{x}\left(\sqrt{2\sin^2{x}+3\sin{x}+4}-\sqrt{\sin^2{x}+6\sin{x}+2}\right)\right)$
I wanted to use L' Hopital's rule so I wrote the term as:
... | Hint:
Set $\sin x=s$
Rationalize the numerator
$$s^2\cdot\dfrac{\sqrt{2s^2+3s+4}-\sqrt{s^2+6s+2}}{1-s^2}$$
$$=s^2\cdot\dfrac{(s^2-3s+2)}{1-s^2}\cdot\dfrac1{\cdots}$$
As $s\to1,s\ne1$ so cancel out $s-1$ and set $s\to1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3457123",
"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.