Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Q regarding finding sum of first 2002 terms Q : A sequence of integers $a_{1}+a_{2}+\cdots+a_{n}$ satisfies $a_{n+2}=a_{n+1}$ $-a_{n}$ for $n \geq 1$. Suppose the sum of first 999 terms is 1003 and the sum of the first 1003 terms is - 999. Find the sum of the first 2002 terms.
My questions regarding this problem are:
... | M:1 = Telescoping
I can see we can’t do anything after this. So , I use telescoping method w/ the help of user dxiv.
$\begin{aligned} a_{n+2} &=a_{n+1}-a_{n} \\ \therefore \quad a_{n+3} &=a_{n+2}-a_{n+1} \\ &=a_{n+1}-a_{n}-a_{n+1} \\ &=-a_{n} \end{aligned}$ $$ \begin{array}{ll}a_{n+4} & =a_{n+3}-a_{n+2} \\ \therefore \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Are there any other "involutive" (a la the orthocenter) points on the Euler line? Throughout, given a triangle $T$ let $G(T)$ and $H(T)$ be the centroid and orthocenter of $T$, respectively. For $r\in\mathbb{R}$ and $p\in\mathbb{R}^2$, let $h_{p:r}$ be the homothety with focus $p$ and factor $r$ (with the understanding... | A point $P$ on the Euler line has barycentric coordinates we can parameterize as
$$(a^2 + b^2 - c^2) (a^2 - b^2 + c^2)-p\,(2 a^4 - a^2 b^2 - b^4 - a^2 c^2 + 2 b^2 c^2 - c^4)\;:\;\cdots\;:\;\cdots$$
(with the second and third coordinates derived cyclically from the first); here, $p$ is the dilation factor of the circumc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I expand $(1+x)^{1/x}$ for small $x$? The binomial expansion
$$(1+x)^{n} = 1 + nx + \frac{n(n-1)}{2}x^{2}+...$$
didn't work because of the $n$ term being undefined at $x=0$.
Taylor expansion doesn't work either since it too would depends on an undefined $1/x$ term.
How does one do it?
| Let's consider the function
$$
f(x)=\begin{cases}
\dfrac{\log(1+x)}{x} & x>-1, x\ne0 \\[6px]
1 & x=0
\end{cases}
$$
Then $f$ is everywhere differentiable and its Taylor expansion at 0 is
$$
f(x)=1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\dotsb
$$
Now we have $(1+x)^{1/x}=e^{f(x)}$ (with continuous extension at $x=0$), ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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2015 Cambridge Entrance Examination Q6
*
*Show : $\sec^2\left(\frac{\pi}{4}-\frac{1}{2}x\right)=\frac{2}{1+\sin(x)}\\$. Then evaluate$\int\frac{dx}{1+\sin(x)}$
*Show : $\int_{0}^{\pi}x \space f(\sin(x)) \space dx=\frac{\pi}{2}\int_{0}^{\pi}f(\sin(x)) \space dx$. Then evaluate $\int_{0}^{\pi}\frac{x}{1+\sin(x)}$
*Ev... | The beauty of this problem is that it used an allied function $\sin(x)$ with phase $\pi$
If you could see the graph of $y = E(x)\times A(x)$ it can help you understand the complexity of the problems
Basically if $E(x)$ is an odd function then this can be solved:
*
*Solution
$\int_0^\pi x^nf(\sin x)dx$ can be solved i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using De Moivre's theorem to solve $(z−3+2i)^4 = z^4$ What are all the solutions to: $(z−3+2i)^4 = z^4$?
I know I have to use De Moivre's theorem which states:
$$(\cos\theta + i\sin\theta)^n=\cos\theta n + i\sin\theta n$$
| To solve
$$
\left(\frac{z-3+2i}{z}\right)^4=1\tag1
$$
Using De Moivre, we want to find $\theta$ so that
$$
\begin{align}
1
&=(\cos(\theta)+i\sin(\theta))^4\tag{2a}\\
&=\cos(4\theta)+i\sin(4\theta)\tag{2b}
\end{align}
$$
which is $\theta\in\left\{0,\frac\pi2,\pi,\frac{3\pi}2\right\}$
$$
\frac{z-3+2i}z=\cos\left(\tfrac{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4363612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Solving the system $\tan x + \tan y = 1$ and $\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$ How can I solve this system of trigonometric equations:
$$\tan x + \tan y = 1$$
$$\cos x \cdot \sin y = \frac{\sqrt{2}}{2}$$
I tried to write tangent as $\sin/\cos$ and then multiply the first equation with the second one but it is ... | Let $ t = \tan \left(y\right)$ and $ u = \tan \left(x\right)$. One has $\sin \left(y\right) = \pm \frac{t}{\sqrt{1+{t}^{2}}}$ and $\cos \left(x\right) = \pm \frac{1}{\sqrt{1+{u}^{2}}}$. The second equation then implies
\begin{equation}2 {t}^{2} = \left(1+{u}^{2}\right) \left(1+{t}^{2}\right)\end{equation}
but acc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the solution to $a_n =a_{n-1}^2+a_{n-2}^2, a_0, a_1>1 $? I saw this on Quora
and have made
very little progress.
What is the solution
(exact or asymptotic) to
$a_n
=a_{n-1}^2+a_{n-2}^2,
a_0, a_1>1
$?
I would be satisfied
with a good asymptotic analysis.
Heck,
I would be happy with an
asymptotic form
of the log ... | For any $n \ge 1$, we have:
\begin{align*}
a_{n+1} &= a_n^2+a_{n-1}^2
\\
a_{n+1} &= a_n^2\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right)
\\
\log a_{n+1} &= 2\log a_n + \log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\right)
\\
\dfrac{1}{2^{n+1}}\log a_{n+1} &= \dfrac{1}{2^n}\log a_n + \dfrac{1}{2^n}\log\left(1+\dfrac{a_{n-1}^2}{a_n^2}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How do I evaluate $\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$ How do I evaluate the following integral when where $C$ is the square with vertices at $\pm2, \pm2+4i$,
$$\frac{1}{2 \pi i} \oint_{C} \frac{z^{2} d z}{z^{2}+4}$$
Using Cauchy integral:
$\frac{z^{2}}{z^{2}+4}=\frac{z^2}{(z+2i)(z-2i)}=\frac12\frac{... | I will perform the first computation in a slightly different manner:
$$
\begin{aligned}
\frac{1}{2 \pi i} \oint_{C} \frac{z^2 }{z^2+4}\; dz
&=
\frac{1}{2 \pi i} \oint_{C} \frac{(z^2 + 4) - 4}{z^2+4}\; dz
\\
&=
\frac{1}{2 \pi i} \oint_{C} dz
+
\frac{1}{2 \pi i} (-4)\oint_{C} \frac1{z^2+4}\; dz
\\
&=
0
+
\frac{1}{2 \pi i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding bounds for a function I would like to show that
$$\frac{1}{\pi^2}<\int_{\pi/2}^\pi\frac{\sin x}{x^3}<\frac{3}{2\pi^2}.$$
We know that $\frac{\sin x}{x^3}\leq\frac{1}{x^3}$ and integrating gives $\int_{\pi/2}^\pi\sin x/x^3\leq3/(2\pi^2)$. I don't know how to make $\leq$ into $<$. On the other hand, $\frac{1}{\pi... |
I don't know how to make $≤$ into $<$
Notice that $\frac{\sin(x)}{x^3}$ and $\frac{1}{x^3}$ are only equal at $\frac{\pi}{2}$. So if we split $\left[\frac{\pi}{2} , \pi \right]$ as $\left[\frac{\pi}{2} , \frac{3\pi}{4} \right] \cup \left[\frac{3\pi}{4} , \pi \right]$ for example, we can thus assert that
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Let $a,b,c,d,e$ be five numbers satisfying the following conditions...
Let $a,b,c,d,e$ be five numbers satisfying the following conditions: $$a+b+c+d+e =0$$ and $$abc+abd+abe+acd+ace+ade+bcd+bce+bde+cde=33$$ Find the value of $$\frac{a^3+b^3+c^3+d^3+e^3}{502}$$
My Approach:
$$(a+b+c+d+e)^3 = \sum_{a,b,c,d,e}{a^3} + 3... | OP's result is correct. For verification, Newton's identity $\,p_3=e_1^3-3e_1e_2+3e_3\,$ for the sum of cubes gives the same result directly (where the sums are the symmetric sums over $a,b,c,d,e$):
$$
\sum a^3 = \left(\sum a\right)^3 - 3\,\left(\sum a\right)\left(\sum ab\right) + 3 \left(\sum abc\right) = 0 - 3 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4372807",
"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{(x,y)\to(0,0)}\dfrac{x^2+y^2}{x^4+y^4}$
Evaluate the limit: $\displaystyle\lim_{(x,y)\to(0,0)}\dfrac{x^2+y^2}{x^4+y^4}$
To solve this, I converted it to polar coordinate and got: $\displaystyle\lim _{r\to0}\left(\frac{1}{r^2(\sin^4\theta+cos^4\theta)}\right)=\infty$
But after putting this on Wolfram... | We can also see that :
$$\dfrac{x^2 + y^2}{x^4 + y^4} \geq \dfrac{x^2 + y^2}{(x^2 + y^2)^2} = \dfrac{1}{x^2 + y^2} \underset{(x, y) \to (0, 0)}{\to} +\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4374717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})$
Question:
Show that, $$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$
My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead ... | You have$$3\arccos\left(\frac5{\sqrt{28}}\right)\in[0,3],$$since $\frac{\sqrt3}2<\frac5{\sqrt{28}}<1$, and therefore$$3>3\frac\pi6>3\arccos\left(\frac5{\sqrt{28}}\right)>0.$$You also have$$0\leqslant3\arctan\left(\frac{\sqrt3}2\right)<3\arctan\left(\sqrt3\right)=\pi,$$and therefore$$\arccos\left(\frac5{\sqrt{28}}\right... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Given real numbers $a,b,c >0$ and $a+b+c=3$. Prove that $\frac{a^2}{b}+\frac{b^2}{c} +\frac{c^2}{a} \ge3 + \frac{4}{3}\max\{(a-b)^2;(b-c)^2;(c-a)^2\}$ My first way is to put the inequality under the same degree, so I multipled both sides to $(a+b+c)$, however it leaded to a hard to solve result. Can anyone help me with... | For the current inequality of $ \sum \frac{a^2}{b} \geq 3 + \frac{4}{3} \max ( a-b)^2 $:
*
*Replace 3 with $ a+b+c$. The reason for this choice is the "well-known" inequality: $ \sum \frac{ a^2}{b} \geq \sum a $. We're then asking how much more leeway there is in the difference of these terms.
*We can write the dif... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorizing a quartic expression to show that it is a perfect square.
Show that $\frac{a^4+b^4+(a+b)^4}{2}$ is a perfect square.
I tried this,
$$\frac{a^4+b^4+(a+b)^4}{2}$$
$$\frac{a^4+b^4+(a^2+b^2+2ab)^2}{2}$$
$$\frac{2a^4+2b^4+4a^2b^2+2(a^2b^2+2a^3b+2ab^3)}{2}$$
$$a^4+b^4+2a^2b^2+ab(ab+2a^2+2b^2)$$
$$(a^2+b^2)^2+ab... | If $a^4+2a^3b+3a^2b^2+2ab^3+b^4$ is a square, it has to be of the form $(xa^2+yab+zb^2)^2$ for some coefficients $x,y,z$.
Then $a^4+2a^3b+3a^2b^2+2ab^3+b^4=x^2a^4+2xya^3b+(y^2+2xz)a^2b^2+2yzab^3+z^2b^4$.
Comparing coefficients we get $x^2=1, 2xy=2, y^2+2xz=3,2yz=2,z^2=1$.
Solving this system we get $x=y=z=1$ and $x=y=z... | {
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Prove that $\sum_{cyc}\frac{(a+b-c)^2}{(a+b)^2+c^2}\ge \frac{3}{5}$ $a,b,c$ are reals $ >0$ prove that $$\sum_{cyc}\frac{(a+b-c)^2}{(a+b)^2+c^2}\ge \frac{3}{5}$$
The inequality is homogeneous so assum WLOG $a+b+c=3$, the inequality is equivalent to $$\sum_{cyc}\frac{(3-2c)^2}{(3-c)^2+c^2} \ge 3/5$$
Set $$f(x)= \frac{(3... | Hint: Show $f(x)+\frac{18}{25}(x-1)\geqslant \frac15$ for all $x\in (0,3)$.
| {
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"url": "https://math.stackexchange.com/questions/4383599",
"timestamp": "2023-03-29T00:00:00",
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Evaluate $I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$ I am trying to calculate this integral:$$I=\int_{0}^{1}\frac{x^2-x}{(x+1)\ln{x}}dx$$.
I tried to find the antiderivative but it didn't exist.
So i changed variable by set $t=\frac{x-1}{x+1}$ due to factor numerator is $x(x-1)$ and it led to:$$I=2\int_{-1}^{0}\frac{t^... | Another approach can be as below:
First, by this link https://en.wikipedia.org/wiki/Frullani_integral one can easily obtain: $$\int_0^1 \frac{x^{a-1}-x^{b-1}}{\ln x}dx=\int_0^\infty \frac{e^{-bt}-e^{-at}}{t}dt=\ln\frac{a}{b},$$
where $a,b\gt0.$
Second, we have a well-known relation that states: $$\frac{\pi}{2}=(\frac{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4389061",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Which solution to $\int \frac{x^3}{(x^2+1)^2}dx$ is correct? I tried solving the following integral using integral by parts :
$$\int \frac{x^3}{(x^2+1)^2}dx$$
but I got a different answer from Wolfram Calculator This is the answer that I got :
$$\int \frac{x^3}{(x^2+1)^2}dx=\frac{1}{2(x^2+1)}+\frac{1}{2}\ln(x^2+1)+C$$
... | BOTH are correct since the two answers differ by a constant.
$$
\begin{aligned}
\int \frac{x^{3}}{\left(x^{2}+1\right)^{2}} d x &=-\frac{1}{2} \int x^{2} d\left(\frac{1}{x^{2}+1}\right) \\
&=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{2} \int \frac{2 x}{x^{2}+1} d x \\
&=-\frac{x^{2}}{2\left(x^{2}+1\right)}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4392187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Elementary inequalities exercise - how to 'spot' the right sum of squares? I have been working through CJ Bradley's Introduction to Inequalities with a high school student and have been at loss to see how one could stumble upon the solution given for Q3 in Exercise 2c.
Question:
If $ad-bc=1$ prove that $a^2+b^2+c^2+d^2... | Rather than $\sqrt 3 $ here is the matrix algorithm for coefficient $1.$ I will try $\sqrt 3$ in a few minutes
Positivity is shown in matrix $D.$ It is then matrix $Q$ that fills in the linear terms, as in: double your form (coefficient $1$) is
$$ 2 \left( a + \frac{c}{2} - \frac{d}{2} \right)^2 + 2 \left( ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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An exercise on the implicit function theorem I am trying to learn the implicit function theorem and this is one exercise about it; I have solved it and would be grateful for any feedback on my solution, thanks.
Let $f\begin{pmatrix}x\\ y\\ z\end{pmatrix}=x y^2+\sin(xz)+e^z$ and $\textbf{a}=\begin{bmatrix}1\\ -1\\ 0 \en... | (a) and (b) are fine. As regards (c), there is a minor error in your work: the tangent plane should be
$$\begin{bmatrix}1 & -2 & 2\end{bmatrix} \begin{bmatrix}x-1\\ y+1\\ z-0\end{bmatrix}=x-1-2(y+1)+2z=0\Leftrightarrow x-2y+2z=3.$$
Here it is another way to find the tangent plane. Since $\phi$ is differentiable at $\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Probability for Ben, Amos and Carl Three men Amos, Ben and Carl share an office at work with a single telephone. Calls call in at random with the proportions of $\dfrac{1}{2}$ for Amos, $\dfrac{1}{3}$ for Ben and $\dfrac{1}{6}$ for Carl. For any incoming, calls, the probabilities that it will be picked up by Amos, Ben,... | You got one number wrong. Your idea is correct, we have
\begin{align*}
\def\P{\mathbf P}\P(\text{call picked up by wrong person})
&=\underbrace{ \frac 12}_{\text{call for A}}\cdot \underbrace{\frac 12}_{\text{call not picked by A}} +
\underbrace{ \frac 13}_{\text{call for B}}\cdot \underbrace{\frac 7{10}}_{\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4399243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Calculating the spherical harmonic of θ=π/2 This is a very simple question, yet I'm not sure how to approach it. I want to calculate the spherical harmonic:
$$ Y_{l m}^{*}(\theta = \pi/2, \phi) $$
I know the general formula:
$$ Y_{l m}^{*}(\theta, \phi)=\sqrt{\frac{2 l+1}{4 \pi} \frac{(l-m) !}{(l+m) !}} P_{l}^{m}(\cos ... | For each choice of $l$ and $m$, you can find a closed form solution for $P_l^m(x)$ as a function of $x$. Typically, I would start there and only then evaluate it at a particular value of $x$ such as $x = 0$.
That said, if you prefer we can find a formula for $P_l^m(0)$ in terms of $l$ and $m$
Instead of starting from ... | {
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The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon? The solutions of the equation $z^4+4z^3i-6z^2-4zi-i=0$ are the vertices of a convex polygon in the complex plane. What is the area of the polygon?
Solution:
Looking at the c... | If we translate (preserves area) to the new variable $w=z+i$ we get $w^4=1+i$. Solutions are vertices of a square with side length $2^{5/8}$. Hence the area would be $2^{5/4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4404581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove : $\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$ Let $a,b,c>0$ satisfy $abc=1$, prove that:
$$\sqrt{\dfrac{ab}{bc^2+1}}+\sqrt{\dfrac{bc}{ca^2+1}}+\sqrt{\dfrac{ca}{ab^2+1}}\le\dfrac{a+b+c}{\sqrt{2}}$$
My attempt:
Let $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac... | Remark: As Calvin Lin pointed out, we can just deal with $a, b, c$, without the substitutions.
We have
\begin{align*}
&\sum_{\mathrm{cyc}}
\sqrt{\frac{ab}{bc^2 + 1}} \\
=\,& \sum_{\mathrm{cyc}}
\sqrt{\frac{ab ab}{(bc^2 + 1)ab}}\\
=\,& \sum_{\mathrm{cyc}}\frac{ab}{\sqrt{ab + bc}}\\
\le\,& \sqrt{(ab + bc + ca)\left... | {
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How do you evaluate: $\int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$ I want to find the value of
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}+1} \ \mathrm dx$
At first, I solved this elementary integral:
$\displaystyle \tag*{} \int _{0}^{\infty} \frac{\log x}{e^x+e^{-x}} \ \mathrm dx... | The similar problem was posted on AoPS the other day. There are nice answers posted. For the sake of completeness, I would like to add the solution, based on approach developed by Yaroslav Blagouchine; it is convenient for solving the problems with a specific symmetry by means of the integration along a rectangular con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Finding the slope of line intersecting the parabola A line $y=mx+c$ intersects the parabola $y=x^2$ at points $A$ and $B$. The line $AB$ intersects the $y$-axis at point $P$. If $AP−BP=1$, then find $m^2$. where $m > 0$.
so far I know $x^2−mx−c=0,$ and $P=(0,c)$.
$x = \frac{m \pm \sqrt{m^2 + 4c}}{2}$
$A_x = \frac{m + \... | Since you assume that $m > 0,$ this result of your calculations is good:
$$ AP = \frac{m + \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag1$$
Here's where you get in a bit of trouble:
$$ BP \stackrel?= \frac{m - \sqrt{m^2 + 4c}}{2} \sqrt{m^2 + 1}. \tag2$$
You want $AP - BP = 1,$ and I think the best interpretation of the prob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Rewriting $\cos^4 x \sin^2 x $ with exponent no higher than $1$ I'm having some trouble finishing this one off.
Rewrite with exponent no higher than $1$:
$$\cos^4 x \sin^2 x$$
The answer is:
$$\frac{2 + \cos(2x) - 2\cos(4x) - \cos(6x)}{32}$$
So I started like this:
$$\cos^4 x \sin^2 x = \frac{1+\cos(2x)}{2}\frac{1+... | $\cos x = \frac {e^{ix} + e^{-ix}}{2}\\
\sin x = \frac {e^{ix} - e^{-ix}}{2i}\\
\cos^4 x\sin^2 x = \frac {(e^{ix} + e^{-ix})^4(e^{ix} - e^{-ix})^2}{-64}\\
\frac {(e^{4ix} + 4e^{2ix} + 6 + 4e^{-2ix} + e^{-4ix})(e^{2ix} - 2 + e^{-2ix})}{-64}\\
\frac {e^{6ix} + 2e^{4ix} - e^{2ix} -4 - e^{-2ix}+2e^{-4ix} + e^{-6ix}}{-64}\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to? If $S_n=\Sigma_{k=1}^n \frac{1}{k(k+1)}= \frac{7}{8}$ then what is $n$ equal to?
So, the most obvious course of action in my mind is to find a closed form for the partial summations, but alas, this task eludes me. I started doing this by hand.... | There is a certain name for this type of sum...
First we can use partial fractions method:
$$
\frac{1}{k(k+1)} = \frac{A}k+\frac{B}{k+1} \quad \implies
$$
$$
1 = Ak+B(k-1) \quad \forall k
$$
So if we set $k=1$ we obtain that $A=1$.
If we set $k=0$ then $B = -1$ this means that
$$
\frac{1}{k(k+1)} = \frac{1}{k}-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Height of an irregular tetrahedron with an equilateral base and lateral faces making angles $60^\circ$, $60^\circ$, $80^\circ$ with that base
An irregular tetrahedron has a base that is an equilateral triangle of side length $10$. The lateral faces make angles of $60^\circ, 60^\circ$ and $80^\circ$ with the base. Fi... | Given a tetrahedron with base $ABC$ and apex $D$, let
*
*$E$ be the orthogonal projection of $D$ onto the plane holding $ABC$.
*$h = |DE|$ will be the height of tetrahedron.
*$\theta_A / \theta_B / \theta_C$ be the angle between faces $DBC$ / $DCA$ / $DAB$ and base $ABC$.
*$\ell_A / \ell_B / \ell_C$ be the distanc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Generating function for $a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$ Let $a_n$ be a sequence following the recurrence relation $$a_n = \sum_{k=0}^{n-2} a_ka_{n-k-2}$$ with initial conditions $a_0 = a_1 = 1$.
We have to find the generating function for $a_n$ that does not contain an infinite series.
Let $f(x) = \sum_{k=0}^{n}... | The expansion of $\sqrt{1+x}$ is $$\sum_{k=0}^{\infty}\binom{1/2}{k}x^k = 1 + \sum_{k=1}^\infty\frac{(1/2)(1/2 - 1) \cdots (1/2 - k + 1)}{k!}x^k = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 - \frac{5}{128}x^4 + \cdots.$$ Then the expansion of $\sqrt{1 - 4x^2(x + 1)}$ is \begin{align*}1 + \frac{1}{2}(-4x^2(x+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Has the sum of 4 cubes problem been proven? Today in class, my professor was lecturing on the sum of 3 cubes and sum of 4 cubes problems. Namely, can every number be written as the sum of 3 (or 4) cubes? He discussed their origins and showed a few examples, and showed how difficult they could be to find for certain num... | First:You can use this identity for finding numbers which are the sum of three cubes:
$$(x-y)^3+(y-z)^3+(z-x)^3=3(x-y)(y-z)(z-x)$$
For example:
$(3-5)^3+(5-7)^3+(7-3)^3=3(3-5)(5-7)(7-3)= 48$
Second : we solve this problem to find a number which it's cube is the sum of three cubes:
$$x^3+y^3+z^3=u^3$$
Let $u=-t$ we have... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4420427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer. Find all integers $n$ such that $\frac{3n^2+4n+5}{2n+1}$ is an integer.
Attempt:
We have
\begin{equation*}
\frac{3n^2+4n+5}{2n+1} = \frac{4n^2+4n+1 - (n^2-4)}{2n+1} = 2n+1 - \frac{n^2-4}{2n+1}.
\end{equation*}
So, we must have $(2n+1) \mid (n^2-4)$... | Write $k=2n+1$ then $n=(k-1)/2$ so $$3n^2+4n+5= {3(k^2-2k+1) + 8(k-1)+20\over 4} ={3k^2+2k+15\over 4}$$ and thus $$4k\mid 3k^2+2k+15\implies k\mid 15$$
Now you have only few values of $k$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to evaluate the definite integral $\int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$? I am struggling with this integral:
$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x$
What I tried so far:
$\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln... | Since
$$
\displaystyle \int _{0}^{\frac{\pi }{2}}\frac{\ln(\tan x)}{1-\tan x+\tan^{2} x}\mathrm{d} x=\int_0^{\infty} \frac{\ln t}{(1-t+t^2)(1+t^2)}dt,
$$
consider
$$
\begin{aligned}
\mathscr{I}(s)&=\int_0^{\infty} \frac{t^s}{(1-t+t^2)(1+t^2)}dt
\\&=\int_0^{\infty} \frac{t^{s-1}}{1-t+t^2}dt-\int_0^{\infty} \frac{t^{s-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4424853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A pattern of periodic continued fraction I am interested in the continued fractions which $1$s are consecutive appears.
For example, it is the following values.
$$
\sqrt{7} = [2;\overline{1,1,1,4}] \\
\sqrt{13} = [3;\overline{1,1,1,1,6}]
$$
In this article, let us denote n consecutive $1$s as $1_n$.
Applying this, the ... | I think your claims are correct but quite needlessly complicated. The theorem at the end of this answer shows a result which is both simpler to write out and more general.
Let $(F_n)_{n\geq 0}$ be the standard Fibonacci sequence, defined by $F_0=0,F_1=1$ and $F_{n+2}=F_n+F_{n+1}$ for $n\geq 1$.
Let $f(x)=\frac{1}{1+x}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Quadratic equation: understanding how the absolute values in the derivation correspond to the $\pm$ symbol in the classic quadratic formula expression I'd like if someone could help me understand the typical form of the quadratic formula, which, for the equation $ax^2+bx+c=0$, reads as $x=\frac{-b \pm \sqrt{b^2-4ac}}{2... | An appealing way of understanding why $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be written as $x=(\pm)\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}$ is to take a set perspective.
The statement $\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$ can be thought of as saying:
$x \in S$ where $S:=\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Integration of a piecewise defined discontinuous function In a proof i posted recently on this site (link) i made the mistake of thinking that a function $f$, which is bounded on a closed interval $[a,b]$, would assume its infimum at a certain $x$ in the domain of $f$. I was presented with a counter example which, for ... | In retrospection i think i found a mistake in my proof. I used $\sum_{j=1}^{n-1} j+1 = \frac{n(n+1)}{2}$ which is false, since $\sum_{j=1}^{n-1} j+1 = \frac{n^2 + n -2}{2}$.
With this discovery and the hint of @Andrea S. i found
$$ U(f, P) = \sum_{i=2}^n M_i \cdot \frac{b}{n} + M_1 \cdot \frac{b}{n} = \sum_{i=2}^n \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I find the eigenvalues of this $2n \times 2n$ matrix? The matrix I am dealing with is of the form below.
$$\begin{pmatrix} 1-\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & -\frac{1}{\sqrt{2^n}} & \cdots & -\frac{1}{\sqrt{2^n}}\\ -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqrt{2^n}}& -\frac{1}{\sqr... | Here's a similar approach to what KBS suggests. Begin with the observation that we can write $M = e_1e_1^T - \alpha 11^T$ for some $\alpha \in \Bbb R$. From there, we can write $M = AB$, with
$$
A = \pmatrix{e_1 & \alpha \mathbf 1}, \quad B = \pmatrix{e_1 & - \mathbf 1}^T.
$$
From the fact that $AB$ and $BA$ have the s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that $(a^2 - b^2)^2$ $ \ge $ $4ab(a-$ $b)^2$ An inequality problem from Beckenbach and Bellman:
Show that $(a^2 - b^2)^2 \ge 4ab(a-b)^2$
The given answer is simply
Equivalent to $(a - b)^4 \ge 0$
I have tried two approaches, one which agrees with the given answer, and the other which does not.
Approach one.... | Note that
$$
a^2 - b^2 = (a + b) (a - b)
$$
Thus,
$$
(a^2 - b^2)^2 = (a + b)^2 (a - b)^2
$$
Thus, the given inequality
$$
(a^2 - b^2)^2 \geq 4 a b (a - b)^2
$$
is equivalent to
$$
(a + b)^2 (a - b)^2 \geq 4 a b (a - b)^2
$$
or
$$
(a - b)^2 \left[ (a + b)^2 - 4 a b \right] \geq 0
$$
or
$$
(a - b)^2 \left[ a^2 + b^2 - 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4430076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Proving that $f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$ is strictly positive for odd integers $n\geq 5$
Let $n\geq 5$ be an odd integer. I'd like to prove that the function $$f_n(x)=\frac{(x+1)^n-x^n-1}{x(x+1)}$$ is strictly positive on $\mathbb{R}$.
(I already figured out that the numerator is divisible by the denominato... | First, we have
$$f_5(x) = 5(x^2 + x + 1) > 0, \, \forall x \in \mathbb{R}.$$
Second, for $k\ge 2$, let
\begin{align*}
g_k(x) &:= f_{2k+3}(x) - f_{2k+1}(x)\\[5pt]
&= \frac{(x + 1)^{2k + 3} - (x + 1)^{2k + 1} - x^{2k + 3} + x^{2k + 1}}{x(x + 1)}\\
&= \frac{(x + 1)^{2k}[(x + 1)^3 - (x + 1)] - x^{2k}(x^3 - x)}{x(x + 1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$ Prove that if the sum $S=\sum_{n=1}^{\infty}\frac{1}{n^2}$ then, $1+\frac{1}{3^2}+\frac{1}{5^2}+\frac{1}{7^2}+\dots=\frac{3}{4}S$
Attempt:
I did figure out that $1+\frac{1}{3^2}+\frac{1}{5^2... | Hint: Note that you get to this series by subtracting the even $n$ terms from the original.
Further, note that for any $n$,
$$
\frac{1}{(2n)^2}=\frac{1}{4}\cdot\frac{1}{n^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4443051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove that $\sum_{1\le iLet $n \ge 2$ be a an integer and $x_1,...,x_n$ are positive reals such that $$\sum_{i=1}^nx_i=\frac{1}{2}$$
Prove that
$$\sum_{1\le i<j\le n}\frac{x_ix_j}{(1-x_i)(1-x_j)}\le \frac{n(n-1)}{2(2n-1)^2}$$
Here is the source of the problem (in french) here
Edit:
I'll present my best bound yet on $$\... | Hint:
Indicating as $S_n , \, T_n$
$$
\begin{array}{l}
T_n = \sum\limits_{1 \le i < j \le n} {a_i a_j } \\
S_n = \sum\limits_{1 \le i,j \le n} {a_i a_j }
= \sum\limits_{1 \le i \le n} {\sum\limits_{1 \le j \le n} {a_i a_j } }
= \sum\limits_{1 \le i \le n} {a_i } \sum\limits_{1 \le j \le n} {a_j }
= \left( {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 3
} |
Geodesics on a pseudosphere I am trying to show the following:
Let
\begin{equation}
\gamma(t) = \begin{pmatrix}
\frac{1}{t} \\
\sqrt{1 - \frac{1}{t^2}} + \cosh^{-1}(t)
\end{pmatrix}
\end{equation}
for $t \geq 1$ be the unit-speed parametrization of the tractrix and
\begin{equation}
\sigma(u,v) = \begin{pmatrix}
\frac{1... | Comment
If $ (f, v) $ are polar coordinates it is in hyperbolic geodesic representation that have circles passing through origin and centered on the x-axis... or any radial line but not the lines on 2-D surface in 3-space.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4445778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Computing limits using Taylor expansions and $o$ notation on both sides of a fraction Let's define $o(g(x))$ as usually:
$$
\forall x \ne a.g(x) \ne 0 \\
f(x) = o(g(x)) \space \text{when} \space x \to a \implies \lim_{x \to a} \frac{f(x)}{g(x)}=0
$$
In theorem $7.8$, Tom Apostol in his Calculus Vol. $1$ gave and proved... | A half is the correct answer. For instance, something that is smaller than $ x^5 $ or $ x^6 $ is neglected in comparison to something that grows (or decreases) at $x^4$.
Remember that $o(x)$ is just a notation for something else, it may be a large polynomial approximation, so as it happens to be in this case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4446443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proof: $\sin \mathrm{A}+\sin \mathrm{B}+\sin \mathrm{C} \leq \cos \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{B}}{2}+\cos \frac{\mathrm{C}}{2}$ Let $\mathrm{I} \subseteq \mathbf{R}$ be an interval and
$\mathrm{f}: \mathrm{I} \rightarrow \mathbf{R}$.
We have this inequality for any $x, y, z \in I$,
$$
f\left(\frac{x+y}{2}\... | Following up on my comment, let $\,f(t) = \sin(t)\,$ and $\,x = B+C\,$, $\,y = C+A\,$, $\,z = A+B\,$.
*
*$f(x) = \sin \left(B+C\right) = \sin \left(\pi - A\right) = \sin \left(A\right)$
*$f\left(\frac{y+z}{2}\right) = f\left(\frac{(C+A)+(A+B)}{2}\right) = f\left(\frac{(A+B+C) + A}{2}\right) = f\left(\frac{\pi}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How can I prove that the perimeter is at most 60? Problem: Let $\Delta$ be a triangle in the plane. Let $P$ be the perimeter of the triangle and $A$ be the area. Let $a,b,c$ be the length of the sides and suppose they are positive integers. Suppose finally that $A=P$. How can I prove that $P \leq 60$?
My attempt: I tri... | $$hnk=4(h+n+k), h,n,k\in\mathbb{N}$$
$$(hk-4)n=4(h+k)$$
WLOG $h\geq n \geq k > 0$. Then
$$(hk-4)k \leq 4(h+k) \Rightarrow hk^2-8k \leq 4h \Rightarrow hk^2-8h\leq 4h\Rightarrow hk^2\leq12h\Rightarrow k\leq 3$$
$$n=\frac{4(h+k)}{hk-4}$$
At $k=1$:
$$n=\frac{4h+4}{h-4}=4+\frac{20}{h-4}$$
$$h+n+k=h+5+\frac{20}{h-4}=9+(h-4)+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\sum_{k=0}^\infty\frac{1}{k+1}\binom{3k+1}{k}\left(\frac{1}{2}\right)^{3k+2}$ converges to $\frac{3-\sqrt{5}}{2}$? I stumble upon the expression
$$
\sum_{k=0}^\infty
\frac{1}{k+1}
\binom{3k+1}{k}
\left( \frac{1}{2} \right)^{3k+2}
$$
and it seems to converge to
$$
\frac{3-\sqrt{5}}{2}
$$
Do they equate ? How to prove ... | Some thoughts:
We use the integral representation
$$\binom{3k + 1}{k} = \frac{1}{2\pi}\int_{-\pi}^\pi (1 + 2^{-1}\mathrm{e}^{\mathrm{i}t})^{3k + 1}(2^{-1}\mathrm{e}^{\mathrm{i}t})^{-k}\mathrm{d} t. \tag{1}$$
(Note: Similar to https://functions.wolfram.com/GammaBetaErf/Binomial/07/02/)
We have
\begin{align*}
&\sum_{k=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
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Other approaches to evaluate $\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}$
$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=?$$
I evaluated the limit by using the Hopital rule,$$\lim_{h\to 0} \frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}=4^x\lim_{h\to0}\frac{4^h+4^{-h}-2}{h^2}=4^x\lim_{h\to0}\frac{\l... | A possible way is using the Taylor expansion
$$2\cosh t = e^t+e^{-t} = 2+t^2 + o(t^2)$$
Hence,
\begin{eqnarray*}\frac{4^{x+h}+4^{x-h}-4^{x+\frac12}}{h^2}
& = & 4^x\frac{e^{h\ln 4}+e^{-h\ln 4}-2}{h^2}\\
& = & 4^x\cdot \frac{2+h^2\ln^2 4 + o(h^2)-2}{h^2} \\
& = & 4^x\cdot\ln^2 4 + o(1) \\
& \stackrel{h\to 0}{\longrightar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Possible positions of the knight after moving $n$ steps in Chessboard. Problem
There is a knight on an infinite chessboard. After moving one step, there are $8$ possible positions, and after moving two steps, there are $33$ possible positions. The possible position after moving n steps is $a_n$, find the formula for $a... | Mordechai Katzman demonstrates in section $3$ of his paper Counting monomials (pages $5$ - $8$) that
$$a_n = \begin{cases}
1 \quad \quad \quad \quad \quad \; \, n = 0 \\
8 \quad \quad \quad \quad \quad \; \, n = 1 \\
33 \quad \quad \quad \quad \quad n = 2 \\
1 + 4n + 7n^2 \quad \; \, n \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to find the probability of drawing a certain two cards from a pack Could I please ask for help on the last part of this question:
Two cards are drawn without replacement form a pack of playing cards. Calculate the probability:
a) That both cards are aces
b) that one (and only one) card is an ace
c) That the two ca... |
Two cards are drawn without replacement from a standard deck of playing cards. Given that at least one ace is drawn, find the probability that the two cards are of different suits.
Method 1: We correct your attempt.
As you observed, the probability that at least one ace is drawn is
$$\Pr(\text{at least one ace is dr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4457545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to evaluate $\int_{0}^{\infty} \frac{1}{x^{n}+1} d x?$ I first investigate the integral
$$\int_{0}^{\infty} \frac{1}{x^{6}+1} d x$$
using contour integration along the semicircle $\gamma=\gamma_{1} \cup \gamma_{2},$
$\textrm{ where }$ $$ \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} ... | This can be calculated explicitly, by the fomula $B(t,1-t)＝π/\sin πt$, where $B$ is the Beta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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$\frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001} = ?$ I found this problem in a high school text book.
Let $ \displaystyle s = \frac{1^2}{1\cdot3} + \frac{2^2}{3\cdot5} + \frac{3^2}{5\cdot7}+\cdots+\frac{500^2}{999\cdot1001}$. Find $s$.
How I tried:
Observe that $T_n... | $$T_n = \frac{n^2}{4n^2-1} = \frac{1}{4}\left( \frac{4n^2-1+1}{4n^2-1} \right) = \frac{1}{4}\left( 1 + \frac{1}{4n^2-1} \right)=\frac{1}{4}\left( 1 + \frac{1}{(2n-1)(2n+1)} \right). $$
Decomposing with partial fractions, we have:
$$ \frac{1}{(2n-1)(2n+1)} = \frac{\frac{1}{2}}{2n-1} - \frac{\frac{1}{2}}{2n+1} = \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4461891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is the triangle inequality equivalent to $a^4+b^4+c^4\leq 2(a^2b^2+b^2c^2+c^2a^2)$? Consider the existential problem of a triangle with side lengths $a,b,c\geq0$. Such a triangle exists if and only if the three triangle inequalities
$$a+b\geq c,\quad b+c\geq a\quad\text{and}\quad c+a\geq b\tag{0}$$
are all satisfie... | The quartic polynomial in (2) may be shown by algebra to equal the product
$(a+b+c)(-a+b+c)(a-b+c)(a+b-c),$
so for nonnegative $a,b,c$ the triangle inequality implies that the quartic expression is nonnegative.
But the quartic as defined above is also nonnegative if two or all four of the factors above are negative. So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4462950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove or disprove $\sum\limits_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac18$ for $\sum\limits_{i=1}^n x_i = \frac12$($x_i\ge 0, \forall i$)
Problem 1: Let $x_i \ge 0, \, i=1, 2, \cdots, n$ with $\sum_{i=1}^n x_i = \frac12$. Prove or disprove that
$$\sum_{1\le i < j \le n} \frac{x_ix_j}{1-x_i-x_j} \le \frac1... | Write $p_i = 2x_i$ and note that $\sum_i p_i = 1$. Then
\begin{align*}
1 + \sum_i \frac{p_i^2}{1 - p_i}
&= \sum_i \frac{p_i}{1 - p_i} \\
&= \sum_{i,j} \frac{1}{2} \left( \frac{1}{1 - p_i} + \frac{1}{1 - p_j} \right) p_i p_j \\
&\geq \sum_{i,j} \left( \frac{2}{2-p_i-p_j} \right) p_i p_j. \tag{by AM–HM}
\end{align*}
Rear... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{... | Maybe all has been said already.
Polynomial with real coefficients:
1)Complex roots occur in pairs, one is the complex conjugate of the other.
(Complex root theorem)
$x_1=1+i,$ $x_2=1-i;$
2)By inspection the real root is $x_3=-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Solving equation involving roots and powers . I'm trying to solve this equation :
$\sqrt{3}\sqrt{237x^2 + \frac{224}{x^2}}x^7 + \frac{35293}{222}x^8 + \frac{2}{999}\sqrt{3}{(\sqrt{237x^2 + \frac{224}{x^2}})}^3x^5 - \frac{44968}{111}x^4 + \frac{12544}{333}=0$.
It looks pretty complicated to me, but the computer gives me... |
how to obtain the exact solutions to the above equation?
Not something you'd want to do by hand, but the equation can be reduced to a quartic in $\,x^4\,$, which just "happens" to factor nicely.
Let $\,y = \sqrt{237x^2 + \dfrac{224}{x^2}}\,$ then, after eliminating the denominators, the original equation can be writt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving contour integral equal to zero Let $G$ be the path traversed once as shown:
Show that $\displaystyle{\int_{G}{\dfrac{1}{v^4-1} \text{d}v} = 0}$.
By partial fraction decomposition,
$\dfrac{1}{v^4 -1} = \dfrac{1}{4} \left( \dfrac{1}{v-1} - \dfrac{1}{v+1} + \dfrac{i}{v-i} - \dfrac{i}{v+i} \right)$
The singular p... | Since OP's first solution works just fine, I will provide yet another solution:
We "inflate" the contour $G$ so it becomes a CCW-oriented circle of radius $r > 1$.
Then
$$ \int_{G} \frac{\mathrm{d}z}{z^4 - 1}
= \int_{|z| = r} \frac{\mathrm{d}z}{z^4 - 1}
\stackrel{(w=1/z)}{=} \int_{|w|=\frac{1}{r}} \frac{-\mathrm{d}w/w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Proof $ \lim_{x \to \frac{\pi}{2}} \frac{1}{\cos^{2}x} - 4 = \infty $ I want to prove that
$$ \lim_{x \to \frac{\pi}{2}} \left(\frac{1}{\cos^{2}x} - 4\right) = \infty ,$$
is my proof correct?
Proof:
Given $ M \ge 1$, choose $ \delta = \arccos\left(\sqrt{\frac{1}{M}}\right) - \frac{\pi}{2}$.
Suppose $ 0 \lt \left|x - \f... | Given $M > 0$ we solve
$\vert\frac{1}{cos^2x} - 4\vert \geqslant M$ with $x \in [0, 2\pi]$
$$\frac{1}{cos^2x} \geqslant M + 4 \quad\vee\quad \frac{1}{cos^2x} \leqslant 4 - M$$
WLOG we assume $M > 4$ $$cos^2x \leqslant \frac{1}{M + 4} \quad\wedge\quad x \neq \frac{\pi}{2}$$
$$-\frac{1}{\sqrt[]{M + 4}} \leqslant cosx \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(1+\frac{a^2+b^2+c^2}{ab+bc+ca})^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq (1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})$
Assuming $a,b,c>0$, show that
$$\Big(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Big)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq \Big(1+\frac{a}{b}\Big)\Big(1+\frac{b}{c}\Big)\Big(1+\frac{c}{a}\Big).$$
I know... | I use $abc$'s method .
Let $a+b+c=3u,ab+bc+ca=3v^2,abc=w^3$ then the problem is :
$$\frac{3u3v^2}{w^3}-1\geq \left(1+\frac{9u^2-6v^2}{3v^2}\right)^{\frac{9u^2}{9u^2-6v^2}}$$
A bit of algebra and the inequality is linear in $w$ and as it's homogeneous we can assume $a=b=1$ so we need to show :
$$\Big(1+\frac{2+c^2}{1+2c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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A curve that intersects $2x^3y^3-30x^2y^2+11xy^3+2x^3-38x^2y+20xy^2-13y^3+16x^2+94xy+10y^2+301x-668y+662$ with multiplicity $3$ or more let $C=2x^3y^3-30x^2y^2+11xy^3+2x^3-38x^2y+20xy^2-13y^3+16x^2+94xy+10y^2+301x-668y+662$ be a curve in $\mathbb{C^2}$. Find a conic that intersects $C$ in $(2,2)$ with intersection mult... | If it is not required the conic should be smooth, you could take the union of the tangent to $C$ at $(2,2)$ and any other line that passes through $(2,2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $x$ and $y$ are positive integers such that $5x+3y=100$, what is the greatest possible value of $xy$? I first wrote $y$ in terms of $x$. in $5x + 3y = 100$, I subtracted $5x$ from both sides to get $3y = 100 - 5x$. Therefore, $y = \frac{100 - 5x}{3}$. I substituted this into $xy$ to get $x(\frac{100 - 5x}{3}$) – You... | Sometimes while finding maxima for positive integers you will have to implement some other techniques along with the traditional ones.In your problem you get $xy=\frac{-5(x-10)^2+500}{3}$.Note that $500\equiv 2$(mod 3) and $-5\equiv 1$(mod 3).So we must have $(x-10)^2\equiv 1$(mod 3).So x-10 can't be a multiple of 3.So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
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How to prove this summation of floor function I am supposed to show that
$$
\left\lfloor \frac{n+1}{2} \right\rfloor + \left\lfloor \frac{n+2}{2^2} \right\rfloor + \left\lfloor \frac{n+2^2}{2^3} \right\rfloor + \cdots + \left\lfloor \frac{n+2^k}{2^{k+1}} \right\rfloor + \cdots = n
$$
For all postive integers $n \geq 1 ... | So the first idea is to consider the base 2 expansion of $n$ (this is natural because you have the power of 2 everywhere). We then write
$$
n=\sum_{k=\geq 0} a_k 2^k, \quad a_k\in\{0,1\}
$$
Then we can actually compute each of the terms in your sum:
$$
\lfloor \frac{n+2^s}{2^{s+1}}\rfloor=\lfloor \sum_k (a_k +\delta_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Show that $0
Show that $0<e-\sum\limits_{k=0}^n\frac{1}{k!}<\frac{1}{n!n}$, where $n>0$.
Hint:
Show that $y_m:=\sum\limits_{k=n+1}^{m+n}\frac{1}{k!}$ has the limit $\lim\limits_{m\to\infty}y_m=e-\sum\limits_{k=0}^n\frac{1}{k!}$ and use that $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$.
If I use the hint then th... | I think I got it right this time!!
$e-\sum_{0}^{n}\frac{1}{k!}<\frac{1}{n!n}$ is equivalent to $ e-e+\sum_{n+1}^{\infty }\frac{1}{k!}<\frac{1}{n!n}$ equivalent to
$\frac{1}{(n+1)!}+\frac{1}{(n+1)!(n+2)}+.....<\frac{1}{n!n}$ and hence to
$\frac{1}{n!}({\frac{1}{(n+1)}}+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+1)(n+2)(n+3)}+....... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Minimum value of $ab+bc+ca$ Given that $a,b,c \in \mathbb{R^+}$ and $(a+b)(b+c)(c+a)=1$
Find the Minimum value of $ab+bc+ca$
My try: Letting $x=a+b, y=b+c, z=c+a$ we get
$$a=\frac{x+z-y}{2}, b=\frac{x+y-z}{2},c=\frac{y+z-x}{2}$$
$$xyz=1$$
Now the problem is to minimize $$ab+bc+ca=\frac{xy+yz+zx}{2}-\frac{x^2+y^2+z^2}... | With this separation you can't find a minimum ; since $x^2+y^2+z^2$ is not bounded. Just take the sequence $(x_n,y_n,z_n)=(n,n,1/n^2)$, $n\in\mathbb{N}^\ast$, so that $x_ny_nz_n=1$, then $x_n^2+y_n^2+z_n^2=2n^2+1/n^4$ is not upper bounded. You may maximize the 'whole' function $x^2+y^2+z^2-2(xy+yz+zx)$ by Lagrange mult... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluate $\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$. $$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx$$
Edit : $D> 0$.
My work:
Let $x = D\tan \theta$
$$\int_{-\infty}^\infty \frac{1}{(x^2 + D^2)^2}dx=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2}dx$$
$$=\int_{-\infty}^\infty \frac{1}{(D^2\sec^2 \theta)^2... | Do you know about complex analysis? Integrals like this became trivial:
We consider the contour integral
$$\oint_C \frac{1}{(z^2+D^2)^2}dz=\int_{\Gamma_R}\frac{1}{(z^2+D^2)^2}dz +\int_{[-R,R]}\frac{1}{(z^2+D^2)^2}dz $$
where $C$ is a semicircle in the upper half plane. By the residue theorem, we need only consider t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4477389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Complex Analysis to solve this integral? $\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$ Complex Analysis time! I need some help in figuring out how to proceed to calculate this integral:
$$\int_0^{\pi/2} \frac{\ln(\sin(x))}{\sqrt{1 + \cos^2(x)}}\text{d}x$$
I tried to apply what I have been studyi... | @Hans-André-Marie-Stamm, I hope you don't mind that I was unable to solve this problem using Complex Analysis, but here's a method that relies on the Beta Function and some algebric work.
$$\begin{align}I&=\int_{0}^{\pi/2}\frac{\log\left(\sin(x)\right)}{\sqrt{1+\cos^2(x)}}dx;\ \cos(x)\rightarrow y\\&=\frac{1}{2}\int_{0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Root in $(1,2]$ of Equation $x^n-x-n=0$ Consider the equation $x^n-x-n=0$ with $n\in\mathbb{N},n\geq2.$
$a)$ Show that this equation has exactly one solution $u_n\in(1,2].$
$b)$ Show that the sequence $\left\{u_n\right\}$ is decreasing.
$c)$ Determine $L=\lim_{n\rightarrow\infty}u_n.$
Here is what I've done so far on t... | First, I shall prove a stronger claim.
Claim: $u_n \geq 1 + \frac{2}{n}$ for $n \geq 2$.
Proof: Applying IVT on $\left(1 + \frac{2}{n}, 2\right)$, we simply have to evaluate $f_n\left(1 + \frac{2}{n}\right)$. Now,
$$
f_n\left(1 + \frac{2}{n}\right) = \left(1 + \frac{2}{n}\right)^n - \left(1 + \frac{2}{n}\right) - n
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Which one is the larger : $20!$ or $2^{60}$? Which one is the larger : $20!$ or $2^{60}$ ?
I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.
My solution:
Write $20!$ in prime factors and $2^{n}$:
$$... | Here is a way to go for the solution using only elementary computations.
(And building partial products that get bigger than and closer to powers of two. I wanted to use first very close approximations like $3\cdot 18\cdot 19=1026>1024$, but there is no need to be so economical at the beginning, and very generous at th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4482212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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How do I solve $\int \frac{20}{(x-1)(x^2+9)}dx$ I've been trying to solve the following integral:
$\int \frac{20}{(x-1)(x^2+9)}dx$
Sadly I'm kinda new to resolving fractional integrals and I'm not sure which method(s) I should use to solve it.
I've tried using partial fractions but I'm doing something incorrectly or ma... | We should resolve the integrand as below: $$\frac{20}{(x-1)\left(x^{2}+9\right)} \equiv \frac{A}{x-1}+\frac{B x+C}{x^{2}+9}$$
Then $$20 \equiv A\left(x^{2}+9\right)+(B x+C)(x-1)$$
Putting $x=1$ yields
$$
\begin{aligned}
20=A(10) & \Rightarrow A=2 \\
(B x+C)(x-1) &=20-2\left(x^{2}+9\right) \\
&=2-2 x^{2} \\
&=-2(x+1)(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
The question states the following: Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
My attempt
In order to solve this question, the first thing I think about is parametrize the surface so I c... | The surface is
$ x^2 + y^2 - z^2 = 1 $
Its standard parameterization is
$ P = (x, y, z) = ( \sec t \cos s , \sec t \sin s , \tan t ) $
So the surface area is
$ \text{A} = \displaystyle \int_{t = - \frac{\pi}{3} }^{ \frac{\pi}{3} } \int_{s = 0}^{2 \pi} \| P_t \times P_s \| \ d s \ d t $
And we have
$ P_t = (\sec t \tan ... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ge4$
Let $a,b,c>0$. Prove that
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{ab+bc+ca}{a^2+b^2+c^2}\ge4$$
I know $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge 3$ but $\dfrac{ab+bc+ca}{a^2+b^2+c^2}\le1$. And then I try $\dfrac{a}{b}+\dfrac{b}{c}+\d... | Alternative approach with hint :
We have the inequality for $a,b,c>0$:
$$\frac{\left(ab+bc+ca\right)}{a^{2}+b^{2}+c^{2}}+\left(\frac{a}{b}+\frac{c}{a}+\frac{b}{c}\right)\geq \left(\frac{a}{b}+\frac{c}{a}+\frac{b}{c}\right)+\frac{9}{\left(\frac{a}{b}+\frac{c}{a}+\frac{b}{c}\right)^{2}}$$
Remains to show the inequality f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Let $x = (0, 1)$ and $y = (−2, a)$ be two vectors in $\Bbb R^2$, where $a$ is a real number.
Problem:Let $x = (0, 1)$ and $y = (−2, a)$ be two vectors in $\Bbb R^2$, where $a$ is a real number.
Attempt: Please be nice.
(a) Compute the quantity $\frac{x · y}{||x||||y||}$, in terms of $a$.
$$
x·y=0(-2) + 1(a)=a
$$
$$
... | The angle and the dot product are related by $\cos\theta=(x\cdot y)/(\lVert x\rVert\lVert y\rVert)$. We're given $\theta=\pi/3$, and we know $\cos(\pi/3)=1/2$. And you've calculated $(x\cdot y)/(\lVert x\rVert\lVert y\rVert)=a/\sqrt{4+a^2}$. So we have this equation:
$$\frac12=\cos(\pi/3)=\cos\theta=\frac{x\cdot y}{\lV... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I show that $x$ and $y $ must have the same length where $ x+y $ and $x-y $ are non-zero vectors and perpendicular?
Problem: Let $x$ and $y$ be non-zero vectors in $\mathbb{R}^n$.
(a) Suppose that $\|x+y\|=\|x−y\|$. Show that $x$ and $y$ must be perpendicular.
(b) Suppose that $x+y$ and $x−y$ are non-zero and ... | $a)$ Using polarization identity
$4\langle x, y\rangle =\|x+y\|^2-\|x-y\|^2=0$
$\langle x, y\rangle =x\cdot y=0$
$b)$ Again using polarization identity
$\begin{align}4\langle x+y, x-y\rangle &=\|(x+y)+(x-y)\|^2-\|(x+y)-(x-y)\|^2\\&=4(\|x\|^2 -\|y\|^2)\end{align}$
Given $\langle x+y, x-y\rangle=0$ implies $\|x\|=\|y\|$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Is the $(3,4,5)$ triangle the only rectangular triangle with this property? While solving a loosely related exercise, by luck I found out that the $(3,4,5)$ triangle has the following property:
The product of the lengths ($\sqrt{2}$ and $\sqrt{5}$) of the two shorter line segments from a corner to the center of the ins... | In a Pythagorean right triangle $\triangle ABC$, we know that $a^2 + b^2 = c^2$ where $a, b, c$ are positive integers. We also know that $|\triangle ABC| = rs$, where $r$ is the inradius and $s = (a+b+c)/2$ is the semiperimeter. Thus we have $$\begin{align}
r &= \frac{ab}{a+b+c} \\
&= \frac{ab}{a+b+\sqrt{a^2+b^2}} \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find angle $\alpha$ in equation containing weighted sum of $\sin(\alpha)$ and $\cos(\alpha)$. The context of this question comes from calculating the Fidelity drop of quantum systems after applying a rotation with some arbitrary angle $\alpha$ on the state vector of the system, see equation below:
$$\begin{eqnarray}
\D... | Let us have the general (and prettier) equation $$B=m\sin\alpha -n\cos \alpha+K.$$
Transpose $K$ and divide the whole equation by $\sqrt{m^2+n^2}$:$$\frac{B-K}{\sqrt{m^2+n^2}}=\frac{m}{\sqrt{m^2+n^2} }\sin\alpha-\frac{n}{\sqrt{m^2+n^2} }\cos\alpha$$
Now, $\Bigg |\dfrac {m}{\sqrt{m^2+n^2}}\Bigg |<1$ and so there must ex... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $\int_0^1 x^4(1-x)^5dx$ quickly? This question came in the Rajshahi University admission exam 2018-19
Q) $\int_0^1 x^4(1-x)^5dx$=?
(a) $\frac{1}{1260}$
(b) $\frac{1}{280}$
(c)$\frac{1}{315}$
(d) None
This is a big integral (click on show steps):
$$\left[-\dfrac{\left(x-1\right)^6\left(126x^4+56x^3+21x^2+6x+... | I wouldn't say this method helps you in solving it in 30 seconds, but I think it can help you in simplifying the calculations so that the integral can be computed faster
First consider by usage of the Binomial Theorem
$$(1-x)^5=\sum_{k=0}^{5}\binom{5}{k}(1)^k(-x)^{5-k}=\sum_{k=0}^{5}\binom{5}{k}(-1)^{5-k}(x)^{5-k}$$
fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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Which interval is correct here$?$ The equation $$2\textrm{sin}^2\theta x^2-3\textrm{sin}\theta x+1=0$$ where $\theta \in \left(\frac{\pi}{4},\frac{\pi}{2}\right)$ has one root lying in the interval
$(0,1)$
$(1,2)$
$(2,3)$
$(-1,0)$
I know that if $f(a)$ and $f(b)$ are of opposite signs then at least $1$ or in ge... | Use the quadratic formula to solve for $x$:
$$x_\pm = \frac{3\sin\theta\pm\sqrt{9\sin^2\theta-8\sin^2\theta}}{4\sin^2\theta} = \begin{cases}\frac{1}{\sin\theta} & \text{ or } \\ \frac{1}{2\sin\theta} & \end{cases}$$
because $\sin\theta\in\left( \frac{1}{\sqrt 2},1\right)$ for $\theta\in\left( \frac{\pi}{4},\frac{\pi}{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inverse in Modular Exponent Properties I have a question about modular exponentiation that I would be very grateful to get some help with.
Assuming we have the values $x, a, r$ and the inverse of $a$ as $-a$ all under $mod \:N$. I know that the following property holds:
$(({x^a} \:mod \: N)^r \: mod \: N)^{-a} \: mod \... | You are confusing your self.
Short answer we do NOT have $a^{m} \equiv a^{m\pmod N} \pmod N$. Mod equivalences don't work on exponents as remainders are not preserved over exponents. And simple example we try will fail. Take for instance $2\pmod 5$. then $2^2 \equiv 4\pmod 5$ and $2^3\equiv 8\equiv 3\pmod 5$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the measure of angle $x$ in the triangle below?
Any point $D$, which is interior to the triangle $\triangle ABC$, determines vertex angles
having the following measures:
$m(BAD) = x,$
$m(ABD) = 2x,$
$m(BCD) = 3x,$
$m(ACD)= 4x,$
$m(DBC) = 5x.\\$
Find the measure of $x$.
(Answer:$10^\circ$)
My progress:
$\dfrac... | $\angle(CDB)=5x<\pi\implies x<\dfrac{\pi}{5} $
$\therefore 0<x<\dfrac{\pi}{5}\\
\dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}\\
\dfrac{\sin(3x)}{ {\sin(2x)}}=\dfrac{\sin(11x)}{2{\sin(2x)}\cos(2x)}\\
2\sin(3x)\cos(2x)=\sin(11x)\\
\sin(5x)+\sin(x)=\sin(11x)\\
\sin(5x)=\sin(11x)-\sin(x)\\
{\sin(5x)}=2\cos(6x){\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Computing the value of $(x+y)^4$ if $x^4+y^4=5$ and $x^2+xy+y^2=10$
Let $x^4+y^4=5$ and $x^2+xy+y^2=10.$ Find $(x+y)^4.$
First, I tried expanding $(x+y)^4$ using the binomial theorem to get $5+4x^3y+6x^2y^2+4xy^3,$ so simplifying I got $5+4xy(x^2+y^2)+6(xy)^2.$ Then I rearranged the given equation to get $x^2+y^2=10-... | We have $95 = (x^2+xy+y^2)^2 - (x^4+y^4) = 2x^3 y + 2xy^3 + 3x^2 y^2,$ which you may recognize. Any further and I'd just give away the entire solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solve the inequality $3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}$ I tried to group the summands so that I could decompose them into multipliers, but nothing worked...
$$3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}\Leftrightarrow 3^{x^2+6x+9}+\frac{1}{9}\leq 3^{x^2-2}+3^{6x+9}$$
$$3^{x^2+6x+7}+1\leq 3^{x^2-4}+3^... | Hint: You can rewrite the inequality as follows: $(3^{6x+11}-1)(3^{x^2}-1) \le 0$. Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is... | What about this : We have for example $$x^{10}+x^{12}=x^{10}(x^2+1)\equiv 0\mod x(x^2+1)$$
This way we can also cancel $11-13,14-16,15-17,18-20$. It remains $x^{19}$ for which you can use $x^6\equiv x^2$ giving $x^{18}\equiv x^6$ hence $x^{18}\equiv x^2$ hence $x^{19}\equiv x^3\equiv -x$ $\mod (x^3+x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Finding a horizontal line that divides a modulus function's area into 2 equal halves
How to find the horizontal line $y=k$ that divides the area of $y=2-\lvert x-2\rvert $ above the $x$-axis into two equal halves using Calculus?
I know of methods other than Calculus to solve this, but my question requires to solve it ... | The area above the $x$ axis is
$$
A =\int_0^4 2 - \lvert x-2 \rvert \, \mathrm{d}x = \int_0^2 2 -(2-x)\, \mathrm{d}x + \int_2^{4}2 -(x-2) \, \mathrm{d}x = 4 \tag{1}
$$
Now, if we draw a horizontal line $\color{green}{y=k}$ on top of the function $y= 2 - \lvert x-2 \rvert$ it splits the latter area in $2$ halves:
In... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\sum^n_{k=0} k^2 {n\choose k} = (n+n^2)2^{n-2}$ We can start with the expression of the binomial expansion, $\sum^n_{k=0} {n\choose k} x^ky^{n-k}= (x+y)^n$. Setting $x=y=1$ gives $\sum^n_{k=0} {n\choose k} = 2^n$
Differentiating both sides with respect to $x$ gives $\sum^n_{k=0} k{n\choose k} x^{k-1}y^{n-k} = ... | You have written the second derivative of $x^{k}$ as $k^{2}x^{k-2}$. It is $k(k-1)x^{k-2}$.
If you add $n2^{n-1}$ to $(n^{2}-n)2^{n-2}$ you get $(n^{2}+n)2^{n-2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$
Find $f^{(n)}(1)$ on $f(x)=(1+\sqrt{x})^{2n+2}$ .
Here is a solution by someone:
\begin{align*} f(x)&=(1+\sqrt{x})^{2n+2}=\sum_{k=0}^{2n+2}\binom{2n+2}{k}x^{\frac{k}{2}}\\ &=\sum_{k=0}^{2n+2}\binom{2n+2}{k}\sum_{j=0}^{\infty}\binom{\frac{k}{2}}{j}(x-1)^j\\ &=\sum_{j=0}^... | An approach using the Lagrange inversion theorem: let $f(z)$ be analytic around $z=z_0$ with $f'(z_0)\neq0$; then $w=f(z)$ has an inverse $z=g(w)$ analytic around $w=f(z_0)$ with $$g^{(n)}\big(f(z_0)\big)=\lim_{z\to z_0}\frac{d^{n-1}}{dz^{n-1}}\left(\frac{z-z_0}{f(z)-f(z_0)}\right)^n.\qquad(n>0)$$
We apply this to $f(z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4507583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2... | The solutions you got are (primitive) eighth roots of unity, $\zeta_8=e^{2\pi i/8},\dfrac 1{\zeta_8}=\bar {\zeta _8}$.
Now $$\zeta _8^{2022}=(\zeta _8^8)^{252}\cdot \zeta _8^6=1\cdot \zeta _8^6=\zeta_8^6$$. And $\dfrac 1{\zeta_8^{2022}}=\zeta_8^{-6}$.
So $$\zeta_8^6+\zeta_8^{-6}=-i+i=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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Other approaches to simplify $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$ I want to simplify the trigonometric expression $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$.
My approach,
Here I used the abbreviation $s,c,t$ for $\sin x$ and $\cos x$ and $\tan x$ respectively,
Numerator is, $$\frac{s^2}{c^2}-s^2=\frac{s^2-s^2c^2}{c... | $\begin{align}\tan^2 x-\sin^2 x&=\sin^2 x(\sec^2 x-1) \\&=\sin^2 x\tan^2x\space \tag1\end{align}$
$$\tan 2x=\frac{2 \tan x}{1-\tan ^2x}$$
$$\tan 2x-2\tan x=\tan 2x\tan^2 x\space \tag2$$
From (1) and (2),
$$\begin{align}\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}&=\frac{\sin^2 x\tan^2x}{\tan 2x\tan^2 x}\\&=\frac{\sin^2x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove if $\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}=\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}=\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}$, then $x=y=z$. Let $x$, $y$, $z$ be real numbers satisfying $$
\begin{align}
&\sqrt{x+1}+\sqrt{y+2}+\sqrt{z+3}\\
=&\sqrt{y+1}+\sqrt{z+2}+\sqrt{x+3}\\
=&\sqrt{z+1}+\sqrt{x+2}+\sqrt{y+3}.
\end{align}$$
Prove that... | The solution of tehtmi is wonderful, and I have a similar approach.
For each parameter $t \in \{x, y, z\}$ and each $1 \leq i \leq 3$, let $t_i = \sqrt{t + i}$. For example $x_2 = \sqrt{x + 2}$. So we have:
\begin{align*}
&x_1 + y_2 + z_3\\
=\ &y_1 + z_2 + x_3 \label{1}\tag{$*$}\\
=\ &z_1 + x_2 + y_3
\end{align*}
Suppo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 2,
"answer_id": 0
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Why this transformation matrix $A$ has $\begin{pmatrix}0 \\ 1\end{pmatrix}$ as Eigenvector? I have the following transformation matrix:
$$
A=\begin{pmatrix}
1 & 0 \\
-1 & 4
\end{pmatrix}
$$
If I resolve to find the eigenvalues I get:
$$
\begin{vmatrix}
A-\lambda I
\end{vmatrix} = 0
$$
which leads to:
$$
\lambda_1 = 1;... | For $\lambda_2$:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}-3 x_1 \\ -x_1 \end{pmatrix}=\begin{pmatrix}0 \\ 0 \end{pmatrix}\Rightarrow x_1=0\Rightarrow v_2=\begin{pmatrix}0 \\ x_2 \end{pmatrix}=x_2 \begin{pmatrix}0 \\ 1 \end{pmatrix}
$$
So your second eigenvector is $\begin{pmatrix}0 \\ 1 \end{... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}$
If $0 < x,y < \frac {\pi}{2}$, prove that:
$$
\frac {x \csc x + y \csc y}{2} < \sec \frac {x + y}{2}
$$
My attempt. First, I tried to change this inequality:
$$
\frac {\frac{x}{\sin x} + \frac{y}{\sin y}}{2} < \frac{1}{\cos \frac{x+y}{2}}
$$
Then,it'... | Lemma 1. $f(x)=\frac{x}{\sin x}$ is a positive, increasing and convex function on $I=\left(0,\frac{\pi}{2}\right)$.
Let us assume that $\mu = \frac{x+y}{2}\in I$ is fixed and $x\leq y$. Let us set $\delta=\frac{y-x}{2}$. By Lemma 1,
$$ \sup_{\substack{0\leq \delta < \min\left(\mu,\frac{\pi}{2}-\mu\right)\\ }}\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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$x^{x^2} + x^{x^8} =?$ Given $x^{x^4} = 4$ If x is any complex number such that $x^{x^4} = 4$ , then find all the possible values of :
$x^{x^2} + x^{x^8}$
First, I used laws of exponents to give $18$ as answer. However , I realised that I've misused it. Further I used logarithms which yielded
$$ x^{x^2} + x^{x^8} = 4^... | This answer only works (based on the original question) on the set of real numbers.
Hint:
\begin{align}
x^{x^4}=4&\implies \left(x^4\right)^{x^4}=4^4\\
&\implies x^4=4\\
&\implies x=\pm\sqrt 2=\pm 2^{\frac 12}.\end{align}
Justification about the step $$\left(x^4\right)^{x^4}=4^4\implies x^4=4$$
We know that if $x\ge0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find range of values of $a,b,c$ such that $ax^2+bx+c$ satisfies given conditions Let $y = f(x) = ax^2+bx+c$ where $a\neq0$. Find the range of values $a,b$ and $c$ such that it satisfies the following:
*
*$0 \le f(x) \le 1 \quad \forall \space x \in [0,1]$
What I have found so far?
*
*For $x=0$, we get $0 \le c \le... | We may obtain the necessary and sufficient conditions for
$$a, b, c \in \mathbb{R}, ~ 0 \le ax^2 + bx + c \le 1, \forall x \in [0, 1].$$
Fact 1: Let $A, B, C \in \mathbb{R}$. Then
$$Ax^2 + Bx + C \ge 0, ~ x \ge 0
\iff A \ge 0, ~ C \ge 0, ~ B \ge -\sqrt{4AC}.$$
(The proof is given at the end.)
Fact 2: Let $A, B, C \in \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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Evaluating $\sum_{k=0}^n(-1)^{\frac{k(k+1)}2}k$ I asked this question a few days ago, where I noticed $$\left\lfloor\frac{n}4\right\rfloor+\left\lfloor\frac{n+1}4\right\rfloor-\left\lfloor\frac{n+2}4\right\rfloor-\left\lfloor\frac{n+3}4\right\rfloor=\cos\left(\frac{n\pi}{2}\right)-1,\quad n\in\mathbb N$$
We can also wr... | We can write the series as $$\sum_{k=1}^n(-1)^{\frac{k(k+1)} 2}k=\sum_{4j\le n}4j+\sum_{4j-1\le n}(4j-1)-\sum_{4j-2\le n}(4j-2)-\sum_{4j-3\le n}(4j-3)$$
Since $$\sum_{4j\le n}4j=\left(4+8+\dots+4\left\lfloor\frac n4\right\rfloor\right)=\frac12\left\lfloor\frac n4\right\rfloor\left(2\cdot4+\left(\left\lfloor\frac n4\rig... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Rewrite this equation using cartesian coordiantes $x$ and $y$ In this question the $x$ and $y$ coordinates are given, however I do not know what to substitute in
$x=5t-2\;\quad y=-5t+7$
So far I have rearranged for $t$
$x+2=5t$
$\dfrac{x+2}5=t$
therefore would the next step by $y=-5\left(\dfrac{x+2}5\right)+7$ ?
| $x=5t-2$
$x+2=5t$
$\dfrac{x+2}5=t$
$y=-5t+7$
$y=-5\left(\dfrac{x+2}5\right)+7$
$y=-(x+2)+7$
$y=-x-2+7$
$y=-x+5$
| {
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"url": "https://math.stackexchange.com/questions/4518027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| HINT:
We have
$$x^2+7x+(4-2^y)=0\\
\Delta_x=2^{y+2}+33=z^2\\
$$
Let, $y=2k$, then
$$\left(z-2^{k+1}\right)\left(z+2^{k+1}\right)=33$$
and so
$$
\begin{cases}z- 2^{k+1}=\pm 1,3,11,33\\
z+ 2^{k+1}=\pm 33,11,3,1\end{cases}
$$
Then, let $y=2k-1$ we have
$$2^{2k+1}+1=(z^2+1)-33$$
Since, $2^{2k+1}+1 \mod 3=0$, putting $z=3n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Solving the equation $\overline z-z^2=i(\overline z+z^2)$ in $\mathbb{C}$ Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\overline z-z^2=i(\overline z+z^2)$ is _____________.
My approach is as fol... | We have
$$\bar{z} - z^2 = i(\bar{z} + z^2)$$
and multiplying by $i$
$$-\bar{z} - z^2 = i(\bar{z} -z^2)$$
then summing
$$z^2=-i\bar z\iff r^2e^{i2\theta}=re^{i\left(-\theta+\frac 3 2\pi\right)}$$
from which we can conclude that $r=0$ or $r=1$ with
$$2\theta=-\theta+\frac 3 2\pi+2k\pi \iff \theta =\frac \pi 2 +\frac 2 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solve $xdx-ydy=y^2(x^2-y^2)dy$ Question:
$$xdx-ydy=y^2(x^2-y^2)dy$$
I'm having trouble matching the solution in the book, which is $\frac{1}{2}\ln(x^2-y^2)=\frac{1}{3}y^3+C$.
I'm getting an integral that requires the incomplete gamma function.
My attempt:
Rewrite the equation:
$x + (-(x^2 - y^2) y^2 - y)\frac{dy}{dx} =... | Your way is correct.
https://www.wolframalpha.com/input?i=Integral+e%5E%7B-2%2F3y%5E3%7D%28y%5E4-y%29dy
I saw this trick:
$$\frac{d(x^2-y^2)}{x^2-y^2}=2y^2dy$$
$$\ln(x^2-y^2)=\frac{2}{3}y^3+2c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Given $3x+4y=15$, $\min(\sqrt{x^2+y^2})=?$ (looking for other approaches)
Given, $(x,y)$ follow $3x+4y=15$. Minimize $\sqrt{x^2+y^2}$.
I solved this problem as follows,
We have $y=\dfrac{15-3x}{4}$,
$$\sqrt{x^2+y^2}=\sqrt{x^2+\frac{(3x-15)^2}{16}}=\frac{\sqrt{25x^2-90x+225}}4=\frac{\sqrt{(5x-9)^2+144}}{4}$$Hence $\mi... |
$3x+4y=15$ represents a straight line and $\sqrt{x^2+y^2}$ represents the distance of the point $(x,y)$ from the origin. So the question is basically telling us to find the minimum distance of any point lying on the line $3x+4y=15$, from the origin.
This shortest distance must be the perpendicular distance from the o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1... | $$\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)=\dfrac{(\sin(\dfrac{3\pi}{4}+\alpha))^2}{(\cos(\dfrac{3\pi}{4}+\alpha))^2}=\dfrac{(\sin\dfrac{3\pi}{4}\cos\alpha+\cos\dfrac{3\pi}{4}\sin\alpha)^2}{(\cos\dfrac{3\pi}{4}\cos\alpha-\sin\dfrac{3\pi}{4}\sin\alpha)^2}=\dfrac{(\dfrac{1}{\sqrt2}\cos\alpha-\dfrac{1}{\sqrt2}\sin\alpha)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
Find all values of a so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2 My goal is to find all values of "a" so that the circle $x^2 - ax + y^2 + 2y = a$ has the radius 2
The correct answer is: $a = -6$ and $a = 2$
I tried solving it by doing this:
$x^2 - ax + y^2 +2y=a$
$x^2 - ax + (y+1)^2-1=a$
$(x - \frac ... | $\frac{a^2+4a+4}{4}$ is not a radius.
Actually, it is the square of radius.
So, you should solve $\frac{a^2+4a+4}{4}=2^2$
And its solution is a=-6, a=2
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | Suggestion
In case$-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$ simply means $a \ge 8$.
After re-writing the function in vertex form and plugging in some valid values of a (like 8, 10 etc) will give a graph on quadrant I for x in (0, 2). This means the function is greater than 0 for all x in that domain under that restriction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 6
} |
Prove that $\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2}$ Prove that $$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2}$$ for any number $x$.
My attempt:
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \\ \iff \frac{x}{1+x^2} \geq -\frac{1}{2} \land \frac{x}{1+x^2} \leq \frac{1}{2}$$ $$\iff (x+1)^2 \geq 0 \land (x-1)^2 \... | Your solution looks fine, as an alternative, by a single inequality, we have
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \iff \left|1+x^2\right|\ge 2|x|
\iff x^4-2x^2+1\ge 0\iff (x^2-1)^2 \ge 0$$
or also
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \iff 1+x^2\ge 2|x|
\iff x^2-2|x|^2+1\ge 0\iff (|x|-1)^2 \ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
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