Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What's wrong in my calculation of $\int \frac{\sin x}{1 + \sin x} dx$? I have the following function:
$$f: \bigg ( - \dfrac{\pi}{2}, \dfrac{\pi}{2} \bigg ) \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{\sin x}{1 + \sin x}$$
And I have to find $\displaystyle\int f(x) dx $. This is what I did:
$$\int \dfrac{\sin x}{1... | Your answer is correct and it matches with choice (E). You have to use half angle formulas:
$$\sin A = \frac{2 \tan \frac{A}{2}}{1+\tan^2 \frac{A}{2}} \quad \cos A = \frac{1-\tan^2 \frac{A}{2}}{1+\tan^2 \frac{A}{2}}$$
Observe that
\begin{align*}
\frac{1}{\cos x}-\tan x&=\frac{1-\sin x}{\cos x}\\
& = \frac{\left(1-\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$. Find the smallest integer $n$ greater than $1$ such that the last $3$ digits of $n^2$ are the same as the last $3$ digits of $n$.
So far I've got $n^2 = 1000k + n$ which means $n^2 ≡ n \mod ... | $$n(n-1)\equiv0\pmod{2^35^3}$$
As $(n,n-1)=1,$ we can have following four cases
$$n\equiv0\pmod{2^35^3}$$
$$n-1\equiv0\pmod{2^35^3}$$
$$n-1\equiv0\pmod{2^3}\text{ and } n\equiv0\pmod{5^3}$$
$$n-1\equiv0\pmod{5^3}\text{ and } n\equiv0\pmod{2^3}$$
For the last two cases use Chinese Remainder Theorem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluate $\int\frac{dx}{(1+\sqrt{x})(x-x^2)}$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}
$$
Set $\sqrt{x}=\cos2a\implies\dfrac{dx}{2\sqrt{x}}=-2\sin2a.da$
$$
\int\frac{dx}{(1+\sqrt{x})(x-x^2)}=\int\frac{dx}{(1+\sqrt{x})x(1-x)}=\int\frac{-4\sin2a\cos2a.da}{2\cos^2a.\cos^22a.\sin^22a}\\
=\int\frac{-2.\sec^2a.da}{\sin2a\cos2... | What's wrong with partial fractions? They are easy if you use the Heaviside method. If $\sqrt x=y$ then
$$\int\frac{dx}{\left(1+\sqrt x\right)\left(x-x^2\right)}=\int\frac{2dy}{y(1+y)^2(1-y)}$$
And if
$$\frac2{y(1+y)^2(1-y)}=\frac Ay+\frac B{1-y}+\frac C{1+y}+\frac D{(1+y)^2}$$
Then
$$\begin{align}A&=\left.\frac2{(1+y)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3487520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Calculating real integrals using complex functions and residue theory. $\int_0^{\pi}\frac{\cos^2x}{1-a\sin^2x}dx$, where $0<a<1$.
My professor thought us some ways of calculating the following type of real integrals using residue theorem in complex analysis:
$\int_0^{2\pi}P(\sin x,\cos x)dx=\int_Cf(z)dz$, $C$ is the ... | Use the half-angle formulas to simplify the integrand:
$$
\int_0^\pi \frac{\cos^2 x}{1-a\sin^2 x}dx = \int_0^\pi \frac{1+\cos(2x)}{2-a+a\cos(2x)}dx = \frac{1}{2a}\int_0^{2\pi} \frac{1+\cos(t)}{\frac{2}{a}-1+\cos(t)}dt.
$$
Then apply $z = e^{it}$:
$$
\frac{1}{2a}\int_0^{2\pi} \frac{1+\cos(t)}{b+\cos(t)}dt = \frac{1}{2a}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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let $f(x)=x+\frac{1}{x} \ \ x \geq 1$ and $g(x)=x^2+4x-6$ the find Min of $g(f(x))=?$
Let $f(x)=x+\frac{1}{x}$ for all $x \geq 1$ and $g(x)=x^2+4x-6$. Find minimum of $g\circ f$.
My try:
The domain of $g\circ f$ is $[1,+\infty)$ and we have
$$g(f(x))=\left(x+\frac{1}{x}\right)^2+4\left(x+\frac{1}{x}\right)-6.$$
At $x... | Since for positive numbers $a,b$ we have $a+b\geq 2\sqrt{ab}$ we have also $$f(x) \geq 2$$ (put $a=x$ and $b=1/x$) with equality iff $x=1$.
Now $g(x)= (x+2)^2-10$, so for $x\geq -2$ it is increasing and thus the minimum of $g$ is at $x=2$.
So the minimum of $g(f(x))$ is at $f(x)=2$ i.e. $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3490211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $\lim\limits_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \,dx$. I have the following limit to find:
$$\lim_{n \to \infty} \int_0^1 \sqrt[n]{x^n+(1-x)^n} \, dx$$
And I have to choose between the following options:
A. $0$
B. $1$
C. $\dfrac{3}{4}$
D. $\dfrac{1}{2}$
E. $\dfrac{1}{4}$
This is what I did:
$$\int_0^1 \... | The present problem is proposed by Prof. Ovidiu Furdui and appeared in Teme de analiza matematica. The solution presented is straightforward (using Squeeze theorem)
$$\small\frac{3}{4}=\int_0^{1/2}(1-x)\textrm{d}x+\int_{1/2}^1 x \textrm{d}x\le I_n \le \int_0^{1/2} \sqrt[n]{(1-x)^n+(1-x)^n}\textrm{d}x +\int_{1/2}^1\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3492509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to derive this series
The function $\dfrac1{1-x}$, equal to $$1 + x + x^2 + x^3 + \cdots,$$ can also be developed according to the series $$1 + \frac{x}{1 + x} + \frac{1\cdot2\cdot x^2}{(1 + x)(1 + 2x)} + \frac{1\cdot2\cdot3\cdot x^3}{(1 + x)(1 + 2x)(1 + 3x)} + \cdots $$ when $x$ is positive and smaller than $1$.
... | We can show the identity
\begin{align*}
\sum_{n=0}^\infty \frac{n!x^n}{\prod_{j=1}^n(1+jx)}=\frac{1}{1-x}\qquad\qquad0<x<1\tag{1}
\end{align*}
with the help of Gauss' summation formula.
We obtain
\begin{align*}
\color{blue}{\sum_{n=0}^\infty \frac{n!x^n}{\prod_{j=1}^n(1+jx)}}
&=\sum_{n=0}^{\infty}\frac{n!}{\left(1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 4,
"answer_id": 0
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Prove that there no $k$ such that : $3n^{2}+3n+7=k^{3}$ Problem :
Let $n,k$ be natural numbers
Prove that there no $k$ such that :
$$3n^{2}+3n+7=k^{3}$$
I see this short solution from a book :
We use $\pmod{9}$
$k^{3}\equiv 0,1,8\pmod{9}$
And $3n^{2}+3n+7\equiv 4,7\pmod{9}$
$\implies $ no solution ?
But I do... | The reason for choosing 9 is that n is not divisible by 3 , let $n=3m+r$ , we can write:
$n^2=9m^2+6mr+r^2$
⇒ $$3n^2+3n+7 ≡(R=3r^2+3r+7) \ mod (9)$$
$r= 0, 1, 2$
$r=0$ ⇒ $R=7$ ⇒ $3n^2+3n+7 ≡7 \mod(9)$
$r=1$ ⇒ $R=13$ ⇒ $$3n^2+3n+7 ≡13 \mod(9)≡4 \mod(9)$$
$r=2$ ⇒ $R=7 \mod (9)$
⇒ $$3n^2+3n+7 ≡(4, 7) \mod(9)$$
$k≡(0, 1, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$ without l'Hopital's rule or Taylor series Can anyone please help me find this limit without l'Hopital's rule, I already used it to evaluate the limit, but I didn't know how to calculate it without l'Hopital's rule.
$$\lim_{x\to0}{\frac{x^2+2\ln(\cos x)}{x^4}}$$
An... | Hint:
Use Taylor expansion at order $4$: as
$$\cos x=1-\frac{x^2}2+\frac{x^4}{24}+o(x^4),$$
setting $u=-\dfrac{x^2}2+\dfrac{x^4}{24}+o(x^4)$, we have to expand $\ln (1+u)$ at order $2$ in $u$ and truncate the result at order $4$ (in $x$):
\begin{align}
\ln(\cos x)&=\ln(1+u)=u-\frac{u^2}2+o(u^2)=-\dfrac{x^2}2+\dfrac{x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to evaluate $\int_{0}^{\pi}\frac{2x^3-3\pi x^2}{(1+\sin{x})^2}\,\textrm{d}x\,$? Problem:
Let $f$ be a bounded continuous function on the interval [0,1].
(a) show that $\int_{0}^{\pi} xf\,(\sin{x})\,\textrm{d}x = \frac{\pi}{2}\int_{0}^{\pi}f\,(\sin{x})\,\textrm{d}x\,$.
(b) Hence evaluate $\int_{0}^{\pi} \dfrac{x}{1+... | I guess it is really just one of approach, with your approach being perfectly fine.
Alternatively, note that
$$(x - \pi)^3 = x^3 - 3\pi x^2 + 3 \pi^2 x - \pi^3.$$
Thus
\begin{align}
I &= \int_0^\pi \frac{2x^3 - 3\pi x^2}{(1 + \sin x)^2} \, dx\\
&= 2\int_0^\pi \frac{(x - \pi)^3}{(1 + \sin x)^2} \, dx + 3\pi \int_0^\pi ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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sum of coefficient of all even power of $x$
The sum of all Coefficient of even power of $x$ in
$(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n}),n\in \mathbb{N}$
what i try
for $n=1,$ we have
$(1-x+x^2)(1+x+x^2)=(1+1)-(1)+(1+1)=3$
for $n=2,$ we have
$(1-x+x^2-x^3+x^4)(1+x+x^2+x^3+x^4)$
$=(1+1+1)-(1+1)+(1+1+1)... | $$f(x)=(1-x+x^2-x^3+\cdots +x^{2n})(1+x+x^2+x^3+\cdots +x^{2n})$$
$$f(x)=\Bigr[(1+x^2+\cdots +x^{2n})-(x+x^3+\cdots +x^{2n-1})\Bigr]\times \Bigr[(1+x^2+\cdots +x^{2n})+(x+x^3+\cdots +x^{2n-1})\Bigr]$$
$$f(x)=(1+x^2+\cdots +x^{2n})^2-(x+x^3+\cdots +x^{2n-1})^2$$
$$f(x)=\biggr(\frac{1-x^{2n}}{1-x^2}\biggr)^2-\biggr(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\begin{cases} x^3-y^3=19(x-y) \\ x^3+y^3=7(x+y) \end{cases}$
I should solve the following system: $$\begin{cases} x^3-y^3=19(x-y) \\
x^3+y^3=7(x+y) \end{cases}$$
by reducing the system to a system of second degree.
We can factor:
$$\begin{cases} (x-y)(x^2+xy+y^2)=19(x-y) \\
(x+y)(x^2-xy+y^2)=7(x+y) \end{cases}... | The thing that catches my eye is that $7,19$ are primes congruent to $1 \pmod 3,$ therefore integrally represented by both $x^2 + xy+y^2, \; x^2 -xy + y^2,$ meaning there are integer points on the two ellipses. It is worth drawing both, by hand (a valuable skill), to see whether that gives a simplified answer when $y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Solutions to $\prod_{n=1}^\infty \left ( 1+ \frac1{f(n)} \right ) = \varphi$ The constant $\varphi = \frac{1 + \sqrt5}{2}$ seems to show up everywhere. I am wondering what non-trivial function(s) $f$ satisfy
$$\prod_{n=1}^\infty \left ( 1+ \frac1{f(n)} \right ) = \varphi$$
I was suprised when I could not find any solut... | This is a little too long for a comment and may not be a complete answer, but here is one such $f(n)$.
We know that $2\cos\left(\frac{\pi}{5}\right)=\phi$, hence $\frac{\sin\left(2\pi/5\right)}{\sin\left(\pi/5\right)}=\phi.$ Then by Euler's product for the sine function
$$\begin{align*}
\frac{\sin\left(\frac{2\pi}{5} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Not sure what to do after trying to simplify the inequality and getting nowhere?
Here is the question:
If $x, y, z \in {\displaystyle \mathbb {R} }$, then prove
$$x^2+y^2+z^2+4\ge2(x+y+z)$$
Here is what I tried doing:
I tried simplifying the inequality and getting all the terms on one side, like so:
$$x^2+y^2+z^2+4... | \begin{align}
x^2 - 2x + y^2 - 2y + z^2 - 2z + 4
= (x-1)^2 + (y-1)^2 + (z-1)^2 + 1
\ge 1 > 0.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3508204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Verify that the equation of a plane is this determinant. Verify that the equation of a plane that passes through three points $(a_{1},a_{2},a_{3}),(b_{1},b_{2},b_{3}),(c_{1},c_{2},c_{13})$ of the space is:
$$\begin{vmatrix}
1 & x_{1} & x_{2} & x_{3} \\
1 & a_{1} & a_{2} & a_{3} \\
1 & b_{1} & b_{2} & b_{3} \\
1 & c_... | One approach is as follows. Let $a$ denote the vector $(a_1,a_2,a_3)$ and so-forth. It suffices to show that $x$ lies in the plane through $a,b,c$ if and only if the system of equations
$$
\begin{cases}
k_1 a + k_2 b + k_3 c = x\\
k_1 + k_2 + k_3 = 1
\end{cases}
$$
has a solution for $k_1,k_2,k_3 \in \Bbb R$. Indeed,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3509907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation. A circle is drawn with its centre on the line $x + y = 2$ to touch the line $4x – 3y + 4 = 0$ and pass through the point $(0, 1)$. Find its equation.
My attempt is as fol... | Let $(a,2-a)$ be a center of the circle.
Thus, $$\frac{|4a-3(2-a)+4|}{\sqrt{4^2+(-3)^2}}=\sqrt{(a-0)^2+(2-a-1)^2}.$$
It's easier than your attempt.
After squaring of the both sides we obtain:
$$a^2-22a+21=0$$ and the rest is smooth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$? How to solve $\sin 2x \sin x+(\cos x)^2 = \sin 5x \sin 4x+(\cos 4x)^2$?
\begin{align*}
2\cos x(\sin x)^2+(\cos x)^2 & = \frac{1}{2}(\cos x- \cos 9x) +(\cos 4x)^2\\
4\cos x(1-(\cos x)^2)+2(\cos x)^2 & = \cos x + \cos 9x +(2(\cos 4x)^2-1)+1\\
\cos x(... | As $5-4=2-1$
$$2(\sin x\sin2x-\sin5x\sin4x)=2(\cos^24x-\cos^2x)$$
Using http://mathworld.wolfram.com/WernerFormulas.html
$$\cos x-\cos3x-(\cos x-\cos9x)=-2(\sin^4x-\sin^2x)$$
As $\dfrac{9-3}2=4-1,$
using http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html and
Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $
$$2\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Prove that $\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$ I need to prove that: $$\left|\frac{x^3y^3}{9x^4+y^4}\right| \le \frac{x^2+y^2}{6}$$
I am new to inequalities so I only tried C-S and AM-GM but none of those work.
Any hints on how to proceed here?.
| For $xy=0$ it's obvious.
But for $xy\neq0$ by AM-GM we obtain:
$$\left|\frac{x^3y^3}{9x^4+y^4}\right|\leq\frac{|x^3y^3|}{2\sqrt{9x^4y^4}}=\frac{|xy|}{6}\leq\frac{\frac{x^2+y^2}{2}}{6}\leq\frac{x^2+y^2}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3512104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Logarithm Subtraction and Division with Same Bases I'm rusty on logarithms. What is the approach to a problem like this? Any hints would be appreciated.
I'm thinking the subtraction on the numerator and denominator can become division since the bases are the same?
| Often better to factor the arguments of the logarithms to get to much simpler logarithms more quickly... \begin{align*}
&\frac{\log_2 24 - \frac{1}{2} \log_2 72}{\log_3 18 - \frac{1}{3} \log_3 72} \\
&\quad{}= \frac{\log_2 (2^3 \cdot 3) - \frac{1}{2} \log_2 (2^3 \cdot 3^2)}{\log_3(2\cdot 3^2) - \frac{1}{3} \log_3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Angles in triangle with inscribed circle that touches one side in a ratio A triangle $\mathrm{ABC}$ has $\angle B=3\times\angle C$. The side $BC$ is devided by the touch point of the inscribed circle in the ratio 1:5. Can the angles be calculated?
With the side $a=\mathrm{BC}=6y$ I tried the "touch point" formula
$$d(B... |
Let ∠C = $x$, $r$ the radius of the incircle, and D the touch point. Then
$$\tan\frac{\angle B}2 = \tan \frac{3x}2 = \frac r{BD},\>\>\>\>\>
\tan\frac{\angle C}2=\tan \frac{x}2 = \frac r{CD}$$
Take the ratio of above equations,
$$\frac{\tan\frac{3x}2 }{\tan\frac{x}2} = \frac{CD}{BD}=5$$
Apply the identity $\tan 3a = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Conditional expectation of a geometric variable
$X, Y$ are two independent random variables which are both
Geometry($p$). We also define random variables $Z = |X-Y|, W = \min
\{X,Y\}$. Calculate $E [W | Z = 1]$.
I tried to use $\min\{X,Y\}=(|X+Y|-|X-Y|)/2$. At first I thought $Z,W$ are independent, because $Z$ is ... | For $Z$, first note that $\{|X-Y|\geqslant 0\}$ has probability one, and $|X-Y|= 0$ if and only if $X=Y$. So first we compute
\begin{align}
\mathbb P(Z=0) &= \mathbb P(X=Y)\\
&= \sum_{n=1}^\infty \mathbb P(X=Y\mid X=n)\mathbb P(X=n)\\
&= \sum_{n=1}^\infty \mathbb P(Y=n)\mathbb P(X=n)\\
&= \sum_{n=1}^\infty \mathbb P(X=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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A pastry shop sells 4 kind of pastries. How many distinct sets of 7 pastries can one buy? I HAVE SEEN stars and bars reference from this sum.
But still I dont get it .
I dont get it whether it is permutation or combination.
I have done the sum in this way.
All arrangements
C1 C2 C3 C4
7 0 0 0... | What matters here is how many of each type of pastry are selected. Selecting three pecan pies, two apple strudel, one croissant, and one baklava is different from selecting four baklava, two apple strudels, and one pecan pie.
Let's label the types of pastries 1, 2, 3, and 4. Let $x_i$, $1 \leq i \leq 4$, be the numbe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Minimum value of $p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$ where $x,y,z\in \mathbb{R}^+$.
Find the minimum value of $$p=3x+\frac{1}{15x}+5y+\frac{25}{y}+z+\frac{1}{36z},$$ where $x,y,z\in \mathbb{R}^+.$
Applying the AM-GM inequality,
$$
\begin{aligned}\frac{p}{6} & \geqslant\left(3x\cdot\frac{1}{15x}\cdo... | The calculus solution is correct.
Equality holds for AM-GM when all the values are the same, but this cannot be the case. (If 3x=5y, then it is not true that 1/(15x)=25/y.)
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x^2 - y^2 = 1995$, prove that there are no integers x and y divisible by 3
If $x^2 - y^2 = 1995$, prove that there are no integers $x$ and $y$ divisible by $3$.
I'm quite new to this. As in the title I would like to ask if my reasoning here is correct and if my answer would be considered acceptable.
*
*If $x$ ... | The question is a bit ambiguous. Anyway, you can indeed answer both interpretations.
If both $x$ and $y$ are divisible by $3$, then the left-hand side is divisible by $9$, but the right-hand side isn't (which is your proof).
Suppose $x$ is divisible by $3$. Then $-y^2\equiv0\pmod{3}$, because $1995\equiv0\pmod{3}$. Thu... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Largest integer such that $n+5$ divides $n^3+89$ I received a calendar with daily math problems as a Christmas gift, and so far most of the problems have been fun and challenging but not too difficult. However, I am completely stumped on the one for today, January 31. The problem reads:
What is the largest positive int... | Synthetic division.
$n^3 + 89 = n^3 + 5n^2 - 5n^2 + 89=$
$(n+5)n^2 - 5n^2 + 89=$
$(n+5)n^2 - 5n^2 - 25n + 25n +89=$
$(n+5)n^2 - (n+5)5n + 25n + 89 =$
$(n+5)n^2 - (n+5)5n + 25n + 125 -125+89=$
$(n+5)n^2 - (n+5)5n + (n+5)25 - 36=$
$(n+5)(n^2-5n + 25) -36$.
So if $n+5$ divides that, then we must have $n+5| 36$
If $n=31$ t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using Ratios for proof making If
$$a(y+z)=x,b(z+x)=y,c(x+y)=z$$
prove that
$$\frac{x^2}{a(1-bc)}=\frac{y^2}{b(1-ac)}=\frac{z^2}{c(1-ba)}$$
I haven't been able to get to the form in the proof, any suggestions on how to do it are welcome.
| If $y+z=0,$ we have $x=0,$ which leads to $y=z=0.$ Assuming that the fractions in the final relation are well-defined, we have the desired equality.
We arrive to the same conclusion if $z+x=0$ or $x+y=0.$
Assume that $(x+y)(y+z)(z+x)\neq 0.$
Rewrite $$a(y+z)=x,\;b(z+x)=y,\;c(x+y)=z$$ as $$a=\frac{x}{y+z},\;b=\frac{y}{z... | {
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$y'(1)$ and $z'(1)$ of $x^2+y^2-2z^2 = 0, x^2+2y^2+z^2 = 4$ Given this equation system
$$x^2+y^2-2z^2 = 0$$
$$x^2+2y^2+z^2 = 4$$
locally at $(1,1,1)$
It can be solved by:
$$ x^2+y^2-2z^2 = 0 \tag 1$$
$$x^2+2y^2+z^2 = 4 \tag 2$$
$2 \times (1) - (2)$ gives
$$x^2-5z^2 = -4 $$
$2 \times (2) + (1)$ gives
$$3x^2+5y^2 = 8 $... | From $x^2-5z(x)^2 = -4$ we derive $2x-10z(x)z'(x)=0.$ With $x=1$ and $z(1)=1$ we get
$$2-10z'(1)=0.$$
Hence $z'(1)=1/5.$
It is now your turn to compute $y'(1)$ with the aid of $3x^2+5y^2 = 8.$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove $a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$ for the recursive sequence $a_{n+1}=\frac{3a_n-1}{3-a_n}$
Prove the statement:
$$a_n=\frac{(2^{n-1}+1)a_1-2^{n-1}+1}{2^{n-1}+1-(2^{n-1}-1)a_1}$$
for the given recursive sequence:
$$a_{n+1}=\frac{3a_n-1}{3-a_n}. $$
My attempt:
Proof by induct... | Embed $a_n$ in a projective coordinate $[a_n,1]$ for each $n$. Then the given recursive relation may be written as
$$\left[\begin{array}{c}a_{n+1}\\ 1\end{array}\right]=\left[\begin{array}{cc}3&-1\\ -1&3\end{array}\right]\left[\begin{array}{c}a_n\\ 1\end{array}\right].$$
It follows by recursion that $$\left[\begin{arra... | {
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"timestamp": "2023-03-29T00:00:00",
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limit of $\lim_{x\to 7}(\frac{x}{7})^{(\frac{x^2-18x+80}{x-7})}$ $$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{x^2-18x+80}{x-7}\right)}$$
It is $1^{\infty}$
$$\lim_{x\to 7}\left(\frac{x}{7}\right)^{\large\left(\frac{(x-8)(x-10)}{x-7}\right)}$$
I tried to take
$$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^... | $$\lim_{x\to 7}e^{\ln\left(\frac{x}{7}\right)^{\left(\frac{(x-8)(x-10)}{x-7}\right)}}=\lim_{x\to 7}e^{\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)}$$
From your work we have:
$$\lim_{x\to 7}\left(\frac{(x-8)(x-10)}{x-7}\right)\ln\left(\frac{x}{7}\right)=\lim_{x\to 7}\left(\frac{(x-8)(x-10)}{1}\right)\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence of $u_{k+1} = \frac{1}{2-u_k u_{k-1}}$ Let $u_0, u_1 \in \mathbb{R}$ be arbitrary. I am interested in the sequence defined by:
$$u_{k+1} = \frac{1}{2-u_k u_{k-1}}.$$
In particular I would like to find $\lim_{k\rightarrow \infty} u_k$ for this sequence.
For a value $\bar u$ to be a potential limiting point, ... | We are lucky : one can find a closed-form expression for $u_n$ (see (1) below). This will allow us to show that the sequence always converges and to determine the limit.
Let us compute a first few terms. Putting $x=u_1,y=u_0$ we have $u_2=\frac{1}{2-xy}$, $u_3=\frac{2-xy}{4-x-2xy}$, $u_4=\frac{4-x-2xy}{7-2x-4xy}$, $u_... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Integral $\int{\frac{1}{(x^{3} \pm 1)^2}}$ I need to solve the following integrals:
$$\int{\frac{1}{(x^3+1)^2}}dx$$
and
$$\int{\frac{1}{(x^3-1)^2}}dx$$
My first thought was to use a trigonometric substitution but the $x^3$ messed it up. Can you guys suggest another method?
| Hint:
Simplify
$$\int{\frac{1}{(x^3+1)^2}}dx=\frac x{3(x^3+1)}+\frac13\int\frac2{x^3+1}dx$$
Then, decompose
$$\frac2{x^3+1}=\frac1{x+1}+\frac{1}{x^2-x+1}- \frac{x^2}{x^3+1}$$
| {
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"source": "stackexchange",
"question_score": "2",
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Asymptotic behaviour of an integral whose integrand cannot be expanded How does one evaluate how a limit is approached in the following case?
Consider the function
$$
f(z) = \int_{0}^1 d x \int_{0}^1 d y \left( \sqrt{1 - \frac{x y}{\sqrt{x^2 + z^2}\sqrt{y^2 + z^2}}} + \sqrt{1 + \frac{x y}{\sqrt{x^2 + z^2}\sqrt{y^2 + z... | Let $g(z) = f(z) - \sqrt{2}$, and consider the substitution
$$1-s = \frac{x}{\sqrt{x^2+z^2}}, \qquad 1-t = \frac{y}{\sqrt{y^2+z^2}}, \qquad w = 1-\frac{1}{\sqrt{z^2+1}}.$$
Then from the computation
$$\mathrm{d}x=-\frac{z}{s^{3/2}(2-s)^{3/2}} \mathrm{d}s, \qquad \mathrm{d}y=-\frac{z}{t^{3/2}(2-t)^{3/2}} \mathrm{d}t, $$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}$$
An approach I can think about is to expand $\cos$ using taylor series, is there another approach?
| This is another way to get the limit. The idea is the same as Quanto's. In fact.
\begin{eqnarray}
&&\lim _{x\to 0}\frac{1-\frac{x^2}{2}-\cos(\frac{x}{1-x^2})}{x^4}\\
&=&\lim _{x\to 0}\frac{2\sin^2(\frac{x}{2(1-x^2)})-\frac{x^2}{2}}{x^4}\\
&=&\frac12\lim _{x\to 0}\frac{4\sin^2(\frac{x}{2(1-x^2)})-x^2}{x^4}\\
&=&\frac12\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3542674",
"timestamp": "2023-03-29T00:00:00",
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Question: Using the Cauchy-Schwarz Inequality to compare between 2 expressions Use the Cauchy-Schwarz Inequality to determine whether $a^2+b^2+c^2$ is bigger than/smaller than/equal to $ab+bc+ac$, where $a,b,c$ are integers and $a<b<c$.
Cauchy-Schwarz Inequality:
$$(\sum_{i=1}^{n}a_ib_i)^2 \leq {\left(\sum_{i=1}^{n}a... | Scalar product:
$|(u,v)|\le ||u||$ $||v||$.
$|(a,b,c)\cdot (c,a,b)|\le$
$ ||(a,b,c)||$ $||(c,a,b)||$;
$ac+ba +bc \le |ac+ba+bc|\le$
$ a^2+b^2+c^2.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral of modified Bessel function of second kind first order multiply by incomplete gamma function? Is there any possible solution or approximation for that given integral? $$\int_0^\infty {\big(v^{\frac{m}{2}-\frac{1}{4}}\big)}K_1\Bigg[\frac{2\sqrt[4]{v}}{l}\Bigg]\Gamma\left[m,-\frac{a+b v}{c}\right]\text{d}v.$$
| I suppose that the approach could be the same as for your previous post.
If $a=0$, we have
$$\frac{1}{4 \pi }\left(-\frac{b}{c}\right)^{-\frac{2m+3}{4}}\,\,G_{2,5}^{4,2}\left(-\frac{c}{16\, b\, l^4}|
\begin{array}{c}
\frac{1-6 m}{4} ,\frac{1-2m}{4} \\
-\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{3}{4},-\frac{2m+3}{4} ... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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On the alternating quadratic Euler sum $\sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2}$ My question is:
Can a closed-form expression for the following alternating quadratic Euler sum be found? Here $H_n$ denotes the $n$th harmonic number $\sum_{k = 1}^n 1/k$.
$$S = \sum_{n = 1}^\infty \frac{(-1)^n H_n H_{2n}}{n^2... | Remark: I have noticed too late that this integral has already been solved (in the update of omegadot).
Nevertheless I don't delete the contribution because, together with this information, it shows that the hypergeometric functions appearing here can be simplified appreciably which gives hope for other cases.
Original... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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I can't understand a how the podium is built in my problem So we are going to build a podium with blocks with the measurements $1 \times 1 × 1$ inch the question is as follows.
The top layer should measure $2 \times 2$ inches and then each layer should project $1/2$ inch on all four sides relative to the layer above (... | Since the podium projects $\frac{1}{2}$ inch on each of the four sides, going from the top layer of $2 \times 2$ inches to the next layer would increase the size by $\frac{1}{2} + \frac{1}{2} = 1$ inch in its width, and the same in its height, to give that each dimension is now $2 + 1 = 3$ inches, i.e., the second laye... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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prove a sequence of functions that are partial sums of a series converges let
$f_n:[-\frac{\pi}{2},\frac{\pi}{2}]\to \mathbb{R}$
be defined by
$\displaystyle f_n(x):=\sum_{k=1}^n \left(\frac{x}{2}\right)^k\!\sin(kx)$.
Prove that the sequence $\{f_n\}_{n=1}^\infty$ is uniformly convergent.
My attempt:
$\sum_{k=1}^n|(... | $|(\frac x 2)^{k} \sin (kx)| \leq (\frac {\pi} 4)^{k}$ and $\frac {\pi} 4 <1$ so M-test applies.
| {
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Decomposition of $2+2^2+2^{2^2}+2^{2^{2^2}}+2^{2^{2^{2^2}}}$ It's a little question about number theory :
Remark that we have :
$$2+2^2=2*3$$
$$2+2^2+2^{2^2}=2*11$$
$$2+2^2+2^{2^2}+2^{2^{2^2}}=2*32779$$
Do we have :
$$2+2^2+2^{2^2}+2^{2^{2^2}}+2^{2^{2^{2^2}}}=2*p\quad?$$
Where $p$ is a prime number .
Thanks a lot for y... | No.$$2+2^2+2^{2^2}+2^{2^{2^2}}+2^{2^{2^{2^2}}} \equiv 2 + 1 + 1 +1 + 1 \equiv 0\pmod{3}$$ Because $2^{2n}\equiv (-1)^{2n}\equiv 1\pmod{3}$
| {
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Solving $\frac{5}{8} \cot36^\circ = \cos^3x$ without substituting the trig values for $36^\circ$
Find the value of $x$ such that
$$\frac{5}{8} \cot36^\circ = \cos^3x$$
The answer is $x=18^\circ$.
It's really messy to plug in the standard values of $\cos36^\circ$, $\sin36^\circ$ and miraculously guess a suitable ... | $$\dfrac58\cdot\cot36^\circ=\dfrac{5\cos36^\circ}{8\sin36^\circ}=\dfrac{5\cos^236^\circ}{4\cos18^\circ}$$
Using Proving trigonometric equation $\cos(36^\circ) - \cos(72^\circ) = 1/2$
$$\cos36^\circ-(2\cos^236^\circ-1)=\frac12\iff5\cos^2 36^\circ=(1+\cos36^\circ)^2=(2\cos^218^\circ)^2$$
Can you take it from here?
| {
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Which expression about power series is correct? I found something weird about power series.
In my textbook,
$$ \frac{1}{1-x} = 1+x+x^2+x^3+\cdots=\sum_{n=0}^\infty x^n \quad |x|<1 $$
and $ \frac{1}{2+x} $ could be expressed using above equation.
$$ \frac{1}{2+x} = \frac{1}{2 \left(1 + \frac{x}{2} \right)} = \frac{1}{2... | Notice that $\frac{1}{x+2}$ is undefined at $x = -2$. Any series you produce will not converge at this point.
We can write the series $\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}x^n$ as $\sum_{n=0}^\infty \frac{(-1)^n}{2^{n+1}}(x-0)^n$. We say this series is centered at $0$. We are obstructed at $x = -2$ and $|(-2)-0| ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For $n>1,$ prove that, for $\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$ For $n>1$ prove that for $$\sum_{n}^{3n-1}\frac{1}{x^2} < \frac{5}{4n}$$
I know that $\dfrac{1}{x}>\dfrac{1}{(x+1)}$ and I'm trying to break the summation into smaller sums to work with, but I'm just not making that final bridge to the $\dfrac{5}... | For $n=2$, we have:
$$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2} < \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{3^2}+\frac{1}{3^2}=\frac{1}{4}+\frac{1}{3}=\frac{7}{12}<\frac{5}{8}$$
Now assume $n\geq 3$. Since:
$$\frac{1}{x^2} < \frac{1}{x(x-1)}=\frac{1}{x-1}-\frac{1}{x}$$
we have
$$\sum_{n}^{3n-1}\frac{1}{x^2} < ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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how many five-digit integers can one make from the digits 2,2,6,0,9,4? I want to ask two questions:
(i) How many five-digit integers can one make from the digits $2, 2, 6, 0, 9, 4$?
(ii) How many digits in (i) satisfy that there is no $6$ followed by $2$?
My Approach
(i) I divided it into cases
*
*if we have $2$ in... | There seem to be some issues in your reasoning. The first question can be solved by considering three scenarios:
*
*The integer contains one $2$. There are four possibilities for the first digit (it cannot be $0$), after which the remaining four digits need to be assigned. The number of possibilities equals: $$4 \cd... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is the "solving for cubic equation roots general rule" sometimes not applicable while the equation obviously has roots? the general rule:
we have $ax^3+bx^2+cx+d=0$
$\Delta_0=b^2-3ac$
$\Delta_1=2b^3-9abc+27a^2d$
$C=\sqrt{\Delta_1^2-4\Delta_0^3}$
$D=(\frac{\Delta_1+C}{2})^\frac{1}{3}$
$x=-\frac{1}{3a}(b+D+\frac{\Del... | You use the Cardano formula to solve a cubic equation. Your formula for $x$ contains real numbers $C$ and $D$ if $\Delta_1^2-4\Delta_0^3 \ge 0$ and thus obviously gives you a real root. If $\Delta_1^2-4\Delta_0^3 > 0$, there are two additional non-real complex roots which are complex conjugate. If $\Delta_1^2-4\Delta_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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A closed form for $\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}$ Mathematica does give an analytic form for $$\sum_{k=0}^n \frac{(-1)^k {n \choose k}^2}{(k+1)^2}.$$
The question here is: How to find a simpler closed form for this alternating summation by hand. The summation of the absolute seies has been discuss... | Use Binomial identity:
$$
(1+t)^n=\sum_{k=0}^{n} {n \choose k}t^n.
\tag{1}
$$
Integration of $(1)$ from $t=0$ to $t=x$ gives
$$
\frac{(1+x)^{n+1}-1}{n+1}= \sum_{k=0}^n {n \choose k}\frac{x^{k+1}}{k+1}.\tag{2}
$$
We can change $x$ to $-1/x$ in $(2)$ to get
$$
\frac{(1-1/x)^{n+1}-1}{n+1}= \sum_{k=0}^n (-1)^k {n \choose k... | {
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Distance from $P(4,3)$ to tengent point on curve $x^2 + y^2 -2x-4=0$
Tangent line to curve $$x^2 + y^2 -2x-4=0$$ at A passes $P(4,3)$.
Find distance A and P.
Tangent of the line is $\frac{2-2x}{\sqrt{2x+4-x^2}}$
A(k,l) means
$l = \sqrt{2k+4-k^2}$
And
$l = \frac{2-2k}{\sqrt{2k+4-k^2}}$
So, I get $k=\frac{3\pm\sqrt{21}... |
Let $O(1,0)$ be the center of the given circle $(x-1)^2 + y^2 =5$. Then,
$$PO^2 = (4-1)^2+(3-0)^2 = 18, \>\>\>\>\>OA^2 = 5$$
Use the fact that PAO is a right triangle to calculate,
$$PA = \sqrt{PO^2-OA^2} = \sqrt{18-5}=\sqrt{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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For non-negative reals such that $a+b+c\geq x+y+z$, $ab+bc+ca\geq xy+yz+zx$, and $abc\geq xyz$, show $a^k+b^k+c^k\geq x^k+y^k+z^k$ for $0
Let $a$, $b$, $c$, $x$, $y$, $z$ be non-negative real numbers such that
$$a+b+c \geq x+y+z,$$
$$ab+bc+ca \geq xy+yz+zx,$$
$$ abc \geq xyz$$
Show that
$$a^k+b^k+c^k \geq x^k+y^k... | Thank @Michael Rozenberg, I have a proof when $k = \frac12,$ for weaker conditons
$$a+b+c = x+y+z,$$
$$\min(x, y, z) \leqslant \min(a, b, c),$$
$$\max(a, b, c) \leqslant \max(x, y, z).$$
Indeed, if $u, v > 0$ it's easy check
$$\sqrt{u} - \sqrt{v} \leqslant \frac{u-v}{2\sqrt{v}}.$$
Assume $x \geqslant y \geqslant z$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Roots of quadratic with $a$ and $8a+3b+c$ of the same sign cannot lie in (2,4) I have the second degree equation $ax^2+bx+c=0$ with $a$ non-zero such that $a$ and $8a+3b+c$ have the same sign and I need to prove that both the roots cannot lie in $(2,4)$.
My try:
I assume two roots are in $(2,4)$
$2 < \frac{-b+\sqrt{b^2... | Hint: Let's say the roots are $x_1$ and $x_2$. If the roots lie in $(2,4)$ simultaneously, then:
$$(x_1-2)(x_2-4)+(x_1-4)(x_2-2)<0$$
and then Vieta's. Can you end it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Evaluate $\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdot \ldots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$
Evaluate $$\sum_{r=1}^{\infty} \frac{1 \cdot 3 \cdots (2r-1)}{r!}\left(\frac{2}{5} \right)^{r}$$
Let $$y=x + \frac{1 \cdot 3 \cdot}{2!} x^2 + \frac{1 \cdot 3 \cdot 5}{3!} x^3+\ldots$$ be the given expression.(replaci... | To carry out the integration of
$y+1=(1-2x)\frac{dy}{dx}
$
we have
$\dfrac1{1-2x}
=\dfrac{y+1}{(y+1)'}
=(\ln(y+1))'
$
so
$\ln(y+1)
=\int \dfrac{dx}{1-2x}
=-\frac12\ln(1-2x)+c
$
so
$y+1
=\dfrac{C}{\sqrt{1-2x}}
$
so
$y
=\dfrac{C}{\sqrt{1-2x}}-1
$.
At $x=0, y=0$ so
$C=1$.
Then,
for $x = \frac25$,
$y
= \dfrac{1}{\sqrt{1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3581695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
simple Application of Gram-Schmidt Orthogonalization I wanted to apply the Gram-Schmidt orthogonalization to the following simple example of polynomials $1,x,x^2,x^3$ in $L^2[-1,1]$, is it correct?
$e_1 = 1$
$e'_2 = x - \int_{-1}^{1}xdx = x \implies e_2 = \frac{x}{\|x\|}$
$e'_3 = x^2 - \int_{-1}^{1}x^2dx -\int_{-1}^{1}... | For $e'_3$ the first integral is wrong. $\int_{-1}^1x^2dx=\frac23$. But you forgot to apply the norm at every step. For example
$$e'_3=x^2-\frac{\int_{-1}^1x^2\cdot 1dx}{\int_{-1}^11\cdot1dx}-x\frac{\int_{-1}^1x^2\cdot xdx}{\int_{-1}^1x\cdot xdx}=x^2-\frac 13$$
And so on. These are the Legendre polynomials.
$$e'_4=x^3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find the value of $\int_\frac{-\pi}{2}^\frac{\pi}{2} 2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx$
Find the value of $$\int_\frac{-\pi}{2}^\frac{\pi}{2}2^{\sin x} \text dx +\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \text dx.$$
I tried to solve it using the concept: $$\int_{a}^b f(x)\text dx... | It seems this problem is designed to use the formula
$$\int_a^bf(x)\;dx + \int_{f(a)}^{f(b)}f^{-1}(x)\;dx = bf(b)-af(a)$$
(See here)
You only need to rewrite the second integral using $x\mapsto x-2$ as
$$\int_\frac{5}{2}^4 \sin^{-1}(\log_2(x-2)) \;dx= \int_{\frac 12}^{2}\sin^{-1}(\log_2(x))\;dx$$
So, you get
$$\int_\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Verify the integral $\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$ I'm stuck solving the integral
$$\int_1^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}}-u} \,du$$
This is what I got so far
\begin{align}
\int_{1}^z \frac{1}{\frac{u}{1 + \sqrt{u^2 + 1}} - u} \,du &= \int_{1}^z \frac{1 + \sqrt{u^2 + 1}}{u - u\cdot(1 +... | HINT: (comment)
Substitution $ u= \sinh x $ makes it easy as it gives three log terms.
$$ \int \dfrac{-(1+\cosh x)dx}{\sinh x} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}$ Prove the following inequality
$$\frac{a}{b}+ \frac{b}{c} + \frac{c}{a} \geq \frac{9(a^2+b^2+c^2)}{(a+b+c)^2}, \enspace \forall a,b,c \in (0,\infty)$$
I tried by multying both sides by the denominator $(a+b+c)^2$ and then applying Holder for... | This seems to be deceptively cumbersome, needing a tight bound on the LHS while symmetrisation. The following is from my old notes, unfortunately no way to attribute it correctly....
First we need a well known inequality $4(x+y+z)^3 \geqslant 27(x^2y+y^2z+z^2x+xyz)$, which follows from AM-GM, as WLOG we may assume ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta \leq \frac{C}{(1+\vert v \vert)^2}$
$\textbf{Problem}$
\begin{equation*}
\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta
\leq \frac{C}{(1+\vert v \vert)^2}
\end{equation*}
For o... | We can compute the integral by rotating the $\sigma$-coordinate system in such a way that $v$ points towards the north pole of the sphere (i.e. in $e_3$-direction) and introducing polar coordinates $(\theta,\phi)$. Then the norm becomes $\lvert v + \rho \sigma \rvert = \sqrt{\lvert v \rvert^2 + \rho^2 + 2 \lvert v \rve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Use Taylor series to compute $\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $ Use Taylor series to solve $$\lim\limits_{x \to 0} \left( \frac{1}{\sin(x)}- \frac{1}{x} \right) $$
This equals $$\lim\limits_{x \to 0} \left( \sum_{n=0}^{\infty}\frac{(2n+1)!}{x^{2n+1}(-1)^{n+1}} - \sum_{n=0}^{\infty}\f... | Let $ x $ be a real from $ \mathbb{R}^{*} \cdot $
Since $ \sin{x}=\sum\limits_{n=0}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n+1}}{\left(2n+1\right)!}} $, we get that $ \frac{x-\sin{x}}{x^{3}}=\sum\limits_{n=0}^{+\infty}{\left(-1\right)^{n}\frac{x^{2n}}{\left(2n+3\right)!}}=\frac{1}{6}+x^{2}\sum\limits_{n=1}^{+\infty}{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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For which value of x $\sum_{n=1}^\infty \frac{n^{nx}}{n!}$ converges I want to know for which values of $x$ this series
$$\sum_{n=1}^\infty \frac{n^{nx}}{n!}$$ converges.
This series is defined $ \forall x \in R$.
$a_n=\frac{n^{nx}}{n!}= \frac{(n^x)^{n}}{n!}= \frac{(n^x)}{n} \frac{(n^x)}{n-1}
\frac{(n^x)}{n-2}...\fr... | For $ n\in\mathbb{N}^{*} $, denoting $ f_{n}:x\mapsto\frac{n^{nx}}{n!} \cdot $
If $ x\geq 1 $, observe that $ f_{n}\left(x\right)\geq\frac{n^{n}}{n!}=\prod\limits_{k=1}^{n-1}{\frac{n}{k}}\geq\prod\limits_{k=1}^{n-1}{\frac{k+1}{k}}=n $, meaning $ \lim\limits_{n\to +\infty}{f_{n}\left(x\right)}=+\infty\neq 0 $, which mea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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$y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes? $y=\dfrac{2x^2+3x-4}{-4x^2+3x+2}$. Find its horizontal asymptotes?
My attempt is as follows:-
Horizontal asymptotes are the horizontal lines which signify the values of $y$ which graph cannot ever attain.
There are two ways to find horizontal asymptote... | Another way of finding the horizontal asymptotes is by multiplying by $\frac{\frac1{x^2}}{\frac1{x^2}}$
$$
y = \dfrac{2x^2+3x-4}{-4x^2+3x+2} \sim \dfrac{2x^2+3x-4}{-4x^2+3x+2}\frac{\frac1{x^2}}{\frac1{x^2}} = \dfrac{2+\frac3x-\frac4{x^2}}{-4+\frac3x+\frac2{x^2}}
$$
and find the limit as $x\to\infty$
$$
\lim_{x\to\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Elementary proof of tangent half angle formula
Hi all, I am interested to find elementary proof of tangent half angle formula.
My solutions are the following:
Triangle $AOB$ is such that $|AB|=1$ and $\angle AOB=\theta$. We then extend $OB$ to $P$ and $Q$ such that $|OP|=|OQ|=1$. Thus we will have two isosceles triang... | Here's another proof:
$\tan(\frac{A}{2}) = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} = \frac{\sin\frac{A}{2}}{\cos\frac{A}{2}} \cdot \left(\frac{2\sin\frac{A}{2}}{2\sin\frac{A}{2}}\right) = \frac{2\sin^2(\frac{A}{2})}{2\sin(\frac{A}{2})\cos(\frac{A}{2})} = \frac{2 \cdot \frac{1-\cos A}{2}}{\sin \left(2 \cdot \frac{A}{2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3595919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Number of real roots of $3x^4+6x^3+x^2+6x+3$ How many real roots does the following quartic polynomial have?
$$3x^4+6x^3+x^2+6x+3$$
After dividing both sides by $x^2$, we get
$$3x^2+6x+1+\dfrac6x+\dfrac3{x^2}=0$$
Or,$$3\left(x^2+\dfrac1{x^2}\right)+6\left(x+\dfrac1x\right)+1=0$$
Taking $x+\dfrac1x$ as $t$
$$3t^2-2+6... | You have committed a mistake. The correct solution after substitution is
$$3(t^2-2)+6t+1=0\\
\implies3t^2+6t-5=0$$
Note that it takes value $3\times4+12-5>0$ at $2$ and $-5<0$ at $-2$. Hence, there is precisely one root between $2$ and $-2$. But we have $2$ solutions of the original quartic equation for each value of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square I am tring to find "integer $x$ which its plus one becomes perfect square and its half+1 is also perfect square". Make this statement into equation I have
\begin{align}
x + 1 = a^2, \qquad \frac{x}{2} + 1 = b^2
\end{ali... | We have:
$$2b^2-1=(x+2)-1=x+1=a^2 \implies a^2-2b^2=-1$$
This is a Pell Equation. We can observe that:
$$(a_1^2-2b_1^2)(a_2^2-2b_2^2)=(a_1a_2+2b_1b_2)^2-2(a_1b_2+a_2b_1)^2=A^2-2B^2$$
This means that multiplying values of the form $x^2-2y^2$ will result in values of the same form. It is easy to see that the first soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3598693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Function and Floor Properties Call a function $f : x \mapsto \lfloor x / 2\rfloor$. Let $x_0 = n$, then $x_1 = \lfloor n / 2 \rfloor$ and $x_2 = \lfloor \lfloor n / 2 \rfloor \rfloor / 2 \rfloor$.... $x_k = f(x_{k-1}).$
How do we prove that $x_k = \left\lfloor \frac{n}{2^k} \right\rfloor$?
My best guess is induction bu... | Induction is a good idea.
Note if $x_k = [\frac n{2^k}]$ then
$x_k \le \frac n{2^k} < x_k + 1$ so
$\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12$
Now if $x_k$ is even than $\frac {x_k}2$ is an integer and
$\frac {x_k}2 \le \frac n{2^{k+1}} < \frac {x_k}2 + \frac 12<\frac {x_k}2+1$
And $x_{k+1} = [\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3600887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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equation with quadratic power I was thinking how to solve:
*
*If $x^{(x-1)^2}=2x+1$, find $x-\frac{1}{x}$
*Solve $x^{x-x^2+13} = x^2-12$
I noticed that in both problems, the linear part can be constructed in the quadratic exponent, I tried a few change of variable and nothing.
| Here's a possible approach to the second problem:
Observe that $x-x^2+13=-(x^2-12)+x+1$, so let $a=x^2-12$.
Now, we wish to solve the equation $x^{-a+x+1}=a$. Note that $x=0$ is not a solution; this is by inspection.
It is easy to see that $x=a$ actually satisfies the preceding equation. Now, we argue that all other va... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Find the value of $\big|\frac{\cos\theta_1\cos\theta_0}{\cos^2\theta_2}+\frac{\sin\theta_1\sin\theta_0}{\sin^2\theta_2}\big|$ If $$\dfrac{\cos\theta_1}{\cos\theta_2}+\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{\cos\theta_0}{\cos\theta_2}+\dfrac{\sin\theta_0}{\sin\theta_2}=1,$$ where $\theta_1$ and $\theta_0$ do not diffe... | We have $$\sin\dfrac{\theta_0-\theta_1}2\cos\dfrac{\theta_0+\theta_1-2\theta_2}2=0$$
If $\sin\dfrac{\theta_0-\theta_1}2=0,$ this will make both equations identical
$$\implies\cos\dfrac{\theta_0+\theta_1-2\theta_2}2=0\implies\dfrac{\theta_0+\theta_1-2\theta_2}2=\dfrac{(2n+1)\pi}2$$ for some integer $n$
$\implies\theta_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to
My attempt is as follows:-
Squaring both equations and adding
$$9+16+24\sin(P+Q)=37$$
$$\sin(P+Q)=\dfrac{1}{2}$$
either $... | If $R=\frac{5\pi}6$, we have
$$3\sin P + 4\cos(\frac\pi6-P)=6$$
Note that the RHS is an increasing function of $P$ for $P\in(0,\frac\pi6]$, whose maximal value is at $P=\frac\pi6$, i.e.
$$RHS_{max}=3\cdot \frac12+4\cdot 1 = 5.5 <6$$
Thus, $R\ne \frac{5\pi}6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3603670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Studying convergenve of an integral I need to study the convergence of an integral
$$
\int_{2}^{\infty} \tan{(\frac{\arctan(x^2)}{x^2})}\frac{\sqrt{x^4+1}}{1+\sin^2(x)}\,dx
$$
We are only interested in the case where $x\rightarrow\infty$.
$$
\frac{\arctan(x^2)}{x^2} \rightarrow 0 \Rightarrow \tan{(\frac{\arctan(x^2)}... | You can even estimate the integrand from below by only a constant, because for $x\geq 2$ you have
$$\frac{\arctan x^2}{1+\sin^2x}\geq \frac{\arctan 4}{2}$$
Starting from the beginning you can use $\tan x\geq x$ for $0<x<\frac{\pi}2$. So, for $x\geq 2$ you surely have
$$\tan \left(\frac{\arctan x^2}{x^2}\right) \geq \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Quadratic Diophantine equation $x^2+6y^2-xy=47$ has no solutions. I am trying to show that $x^2 + 6y^2 - xy = 47$ has no integer solutions. I know that the an efficient way is to look at this equation modulo $n$; other equations can be easily be solved this way. I tried this for $n = 2,3,4,5,6$ so far and I still canno... | Let $z=x/y$. Since the roots of $z^2-z+6$ are $\frac{1\pm i\sqrt{23}}2$, we get
$$
z^2-z+6=\left(z-\frac{1+i\sqrt{23}}2\right)\left(z-\frac{1-i\sqrt{23}}2\right)
$$
Multiplying by $y^2$ gives
$$
\begin{align}
47
&=x^2-xy+6y^2\\[9pt]
&=\left(x-\frac{1+i\sqrt{23}}2\,y\right)\left(x-\frac{1-i\sqrt{23}}2\,y\right)\\
&=\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3608714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Question: If $a,b,c\in(0,1)$ satisfy $a+b+c=2$, prove that $\frac{abc}{(1-a)(1-b)(1-c)}\ge 8.$
Source: ISI BMath UGB 2010
My approach: We have $0<a,b,c<1$. Therefore we have $$0>-a,-b.-c>-1\implies 1>1-a,1-b,1-c>0\implies 0<1-a,1-b,1... | You may set
$$u=1-a,\; v= 1-b,\; w = 1-c$$
$$\Rightarrow u+v+w= 1 \text{ and } u,v,w\in (0,1)$$
To show is now
$$(1-u)(1-v)(1-w)\geq 8uvw$$
or, after expanding and using $u+v+w=1$
$$uv+vw+wu \geq 9uvw$$
or, because auf AM-HM
$$\frac 1 3 = \frac{u+v+w}{3}\geq \frac 3{\frac 1w + \frac 1u + \frac 1v}$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3609331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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System of 2 equations Solve the system of equations:
$$
\left\{\begin{matrix}2\sqrt{x+y} = y^2+y-x & \quad(1) \\ x(y^2+y)=(y^4-y^2)^2-2 & \quad(2) \end{matrix}\right.
$$
My attempt:
From $(1)$ I get:
$$(1) \implies (\sqrt{x+y}+y+2)(\sqrt{x+y}-y) = 0$$
So either $\sqrt{x+y}+y+2 = 0 \implies x = y^2+3y+4$ or $\sqrt{x+... | If $x=y^2-y$ so
$$y^4-y^2=(y^4-y^2)^2-2,$$
which gives $$y^4-y^2=-1,$$ which is impossible, or
$$y^4-y^2=2.$$
Can you end it now?
I got that the second case: $$\sqrt{x+y}+y+2=0$$ is impossible.
Indeed, $y\leq-2$ and $$x+y=(y+2)^2$$ or
$$x=y^2+3y+4,$$ which gives
$$(y^2+3y+4)(y^2+y)=(y^4-y^2)^2-2$$ or
$$y^8-2y^6-4y^3-7y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For all real $x, y$ that satisfy $x^3+y^3=7$ and $x^2+y^2+x+y+xy=4$ I started with $x+y=a$ and $xy=b$ and I rewrote the equations with a and b.
I got
$$b=a^2+a-4$$
$$x^3+y^3=2a^3+3a^2-12a^2=7$$
$$f(a)=-2a^3+3a^2-12a^2-7=0$$
I factorised it to get
$$f(a)=-(a-1)(a-1)(2a+7)=0$$
So $a=1,b=-2$ or $a=\frac{-7}{2},b=\frac{19}... | Let $t^2+u t+v$ be a quadratic polynomial with roots $a$ and $b$. Then
$$t^2+u t+v=(t-a)(t-b)=t^2-(a+b)t+ab,$$
and hence $v=ab$ and $a=-(a+b)$. Note the $-$-sign. In you particular case with $a=1$ and $b=-2$
$$u=-(a+b)=1\qquad\text{ and }\qquad v=ab=-2,$$
corresponding to the polynomial $t^2+t-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3610995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
$\alpha,\beta,\gamma$ are the roots of the equation $x^3 − 9x + 9 = 0$. Find the value of $ \alpha^{-5}+\beta^{-5}+\gamma^{-5}$ I've simplified the expression to get
$$\frac{(\alpha\beta)^5+(\beta\gamma)^5+(\gamma\alpha)^5}{(\alpha\beta\gamma)^5}.$$
Now all I need to find is $\sum (\alpha\beta)^5$ given that $\sum \alp... | Let $a=\alpha^{-1}, b=\beta^{-1},c=\gamma^{-1}.$ Then, we have $a+b+c=\alpha^{-1}+\beta^{-1}+\gamma^{-1}=(\alpha\beta+\beta\gamma+\gamma\alpha)/(\alpha\beta\gamma)=1,$ $abc=(\alpha\beta\gamma)^{-1}=-1/9,$ and $ab+bc+ca=(\alpha+\beta+\gamma)/(\alpha\beta\gamma)=0.$ Then, we have
\begin{equation}
\begin{split}
\alpha^{-5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3612482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$ Evaluate $\sum_{n = 1}^{99} \frac{2}{\sqrt{n} + \sqrt{n + 2}}.$
Listing out a few terms and simplifying the denominators gets me $$-1+\sqrt{3}+2-\sqrt{2}-\sqrt{3}+\sqrt{5}\cdots-\sqrt{97}+3\sqrt{11}-7\sqrt{2}+10-3\sqrt{11}+\sqrt{101}.$$
A lot of these term... | Let $ m \in\mathbb{N}^{*} : $
\begin{aligned}\sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n}\right)}\\ &=\sum_{n=1}^{m}{\left(\sqrt{n+2}-\sqrt{n+1}\right)}+\sum_{n=1}^{m}{\left(\sqrt{n+1}-\sqrt{n}\right)}\\ \sum_{n=1}^{m}{\frac{2}{\sqrt{n}+\sqrt{n+2}}}&=\sqrt{m+2}-\sqrt{2}+\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given $4$ variables and $5$ pairwise products, find the $6$th pairwise product?
Consider four positive numbers (not necessarily integers). The pairwise products are $2$, $3$, $4$, $5$, $6$, plus one more number.
What is the 6th product? What are the numbers?
I found this from Quora and I would be interested in a nice... | You can separate the six products into three pairs with each pair having different factors
$$x_1\cdot x_2\quad \ x_3\cdot x_4\\
x_1\cdot x_3\quad \ x_2 \cdot x_4\\
x_1 \cdot x_4\quad \ x_2 \cdot x_3$$
When we multiply the partial products on each line, we should get the same result. The only two pairs that have the sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3613961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
Four roots of the polynomial $x^4+px^3+qx^2+rx+1$ Let $a, b, c, d$ be four roots of the polynomial $x^4+px^3+qx^2+rx+1$ prove that $(a^4+1)(b^4+1)(c^4+1)(d^4+1)=(p^2+r^2)^2+q^4-4pq^2r$
Please provide hint.
| We seek a quartic whose roots are $a^4, b^4, c^4 , d^4$.
Naively, this is $ x + px^{3/4} + qx^{1/2} + rx^{1/4} + 1 = 0 $, but the powers need to be integers.
Shifting terms, $x + qx^{1/2} + 1 = - px^{3/4} - rx^{1/4}$.
Squaring both sides, $x^2 + q^2x + 1 + 2x + 2qx^{3/2} + 2qx^{1/2} = p^2 x^{3/2} + 2pr x + r^2 x^{1/2}$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Prove $\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$
Prove
$$\left(\frac{1+\cos2\theta+i\sin2\theta}{1+\cos2\theta-i\sin2\theta}\right)^n = \cos2n\theta+i\sin2n\theta$$
Not sure how to go about this proof, was thinking of putting numerator and denominator... | $$\begin{align}
e^{i2\theta n} \cdot (1+e^{-i2\theta})^n
&=e^{i2\theta n}\cdot \left(\binom{n}{0} e^{-i2\theta \cdot 0} + \binom{n}{1} e^{-i2\theta \cdot 1}+\binom{n}{2} e^{-i2\theta \cdot 2} + \dots + \binom{n}{n-1} e^{-i2\theta \cdot (n-1)}+ \binom{n}{n} e^{-i2\theta \cdot n}\right) \\
&= \binom{n}{0} e^{i2\theta \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3619860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$
Find all non-negative integers $a, b, c ,d$ such that $a! + b! + c! = 2^d$.
By trial I found $a= 2 , b= 3 , c= 5$ and $d= 7$ which is one solution. How to find all the solutions of it ?
| Without loss of generality $a\le b\le c$, so $a!|2^d\implies a!\in\{1,\,2\}$.
If $a!=1$, $b!+c!$ is odd so $b!=1$ and $c!=2^d-2$, so $c!\nmid4$ and $c\le3$. This gives the solutions $c=2$ and $c=3$.
If $a!=2$, $b!+c!$ isn't a multiple of $4$, so $b\le3$. In particular, if $a=b-2$ then $c!=2^d-4$ is a multiple of $4$ bu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3621564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove: $\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N}\setminus\{0\}, x\in{\mathbb{R}, x\neq k\pi}} $
Prove :
$\frac{\sin{nx}}{\sin{x}}\geqslant{n-\frac{n(n^2-1)x^2}{6}},n\in{\mathbb{N\setminus\{0\}}}, x\in{\mathbb{R}} $
I proved this relationship by incident. I tried to directly prove thi... | Alternative solution:
Remark: $n - \frac{n(n^2-1)x^2}{6}$ is the second order Taylor approximation of $f(x) = \frac{\sin nx}{\sin x}$ around $x = 0$.
First, we give the following result. The proof is given later.
Fact 1: Let $n\ge 3$ be a positive integer and $x\in (0, \frac{\pi}{2})$. Then
$\frac{\sin nx}{\sin x} \ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Evaluation of $\int^{1}_{0}\frac{x^4}{1+x^8}dx$
How can I Integrate $\displaystyle \int^{1}_{0}\frac{x^4}{1+x^8}dx$?
I have searched in that forum and get the result for $\displaystyle \int^{\infty}_{-\infty}\frac{x^4}{1+x^8}dx$ or for $\displaystyle \int^{\infty}_{0}\frac{x^4}{1+x^8}dx$
Although i know the formula $... | Knowing that the roots of the denominator are the primitives $x_k=e^{i\pi/8 + i2\pi k/8}$ with $k=0,1,2,...,7$ the integral is quickly calculated by partial fractions decomposition i.e.
$$\int_0^1 \frac{x^4}{1+x^8} \, {\rm d}x = \int_0^1 {\rm d}x \sum_{k=0}^7 \frac{A_k}{x-x_k} = \sum_{k=0}^7 A_k \log\left(1-\frac{1}{x_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Showing that $\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}$ via Fourier Transform If $a,b>0$, how can I prove this using Fourier Series
$$\int_{-\infty}^{\infty}\frac{x^2}{(x^2+a^2)(x^2+b^2)}dx=\frac{\pi}{a+b}.$$
I tried to split the product and calculate the integral using Parceval's Theorem... | HINT: Unless you’re dead-set on using the Fourier transform, I would try using that
$$\frac{x^2}{(x^2+a^2)(x^2+b^2)}=\frac{1}{a^2-b^2}\bigg(\frac{a^2}{x^2+a^2}-\frac{b^2}{x^2+b^2}\bigg)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3625913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
} |
Is there a nice way to reconcile the antiderivatives of $\frac{1}{x^2+a^2}$, $\frac{1}{x^2-a^2}$, and $\frac{1}{x^2}$ as $a\to0$? If $a$ is a (WLOG) positive real number, ignoring all constants of integration we have
$$
\int \frac{1}{x^2+a^2}\,dx = \frac{1}{a}\arctan\left(\frac{x}{a}\right)
$$
$$
\int \frac{1}{x^2-a^2}... | Remember that antiderivatives are only defined up to constant. One can select a sequence of constants $C_a$ so that $F_a(x)+C_a\to-1/x$ when $a\to0$ and $x\neq0$ for both those functions. It is easier to see with definite integrals like $\int_1^x \frac{1}{t^2+a^2}\,dt$, where it is justified to move the limit inside th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3626033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How can I get the $\frac{AP}{PD}+\frac{BP}{PE}+\frac{CP}{PF}=3$?
Let the triangle $ABC$ be inscribed in a circle, let $P$ denote the centroid of the triangle and let $O$ denote the circumcenter. Suppose that $A,B,C$ have coordinates $(0,0),(a,0)$ and $(b,c)$ respectively.
a) Express the coordinates of $P$ and $O$ in ... | I am certain that the theorem below has been proven on this site. Unfortunately, I could not find it.
Theorem. Let $ABC$ be a triangle with circumcenter $O$ and centroid $G$. If $R$ is the circumradius and $P$ is an arbitrary point on the plane (or in the space), then
$$\begin{align}PA^2+PB^2+PC^2&=GA^2+GB^2+GC^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3628664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How are the factors obtained at *this point* in the solution of proving $2^{2^n}+ 1 \mid 2^{2^m} -1$ I understood everything up until "BUT". My question is how are the factors on the right side for the bolded line of equation under the word "BUT" obtained? I tried to get these factors using property (5), but I was un... | More generally,
$x^{2^m}-1
=(x-1)\prod_{n=0}^{m-1}(x^{2^n}+1)
$.
For this,
put $x = 2$.
Proof.
For $m=1$ this is
$x^{2^1}-1
=(x-1)\prod_{n=0}^{0}(x^{2^n}+1)
$
or
$x^2-1
=(x-1)(x+1)
$.
If it is true for $m$ then
$\begin{array}\\
(x-1)\prod_{n=0}^{(m+1)-1}(x^{2^n}+1)
&=(x-1)\prod_{n=0}^{m}(x^{2^n}+1)\\
&=(x-1)(x^{2^m}+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle? If you know the Diagonal and Area of a Rectangle, can you find the sides of the rectangle?
I was doing strange math yesterday, and I can across the realization that two different rectangles can’t have the same area and same diag... | Sure:
You have $xy = A$ a known constant.
And $\sqrt{x^2 +y^2} = d$ a known constant.
So just substitute.
$x = \frac Ay$ (assuming $y\ne 0$ which if the area is positive must be so... or we could do $y = \frac Ax$.... it doesn't matter.
And $\sqrt{(\frac Ay)^2 + y^2} = d$.
So $(\frac Ay)^2 + y^2 = d^2$
So $A^2 + y^4 =d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3631995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Restrictions on a Function If $f(x) =$
\begin{cases}
x^2-4 &\quad \text{if } x \ge -4, \\
x + 3 &\quad \text{otherwise},
\end{cases}
then for how many values of $x$ is $f(f(x)) = 5$?
I'm not sure how to really start from this question other than bashing values. I need some help on a start, thanks!
| First find the values such that $f(x) = 5$:
$x^2-4=5\iff x^2=9 \iff x=\pm 3$ (both are greater than or equal to $-4$)
$x+3=5\iff x = 2$ (but this value is not less than $-4$).
Hence we must have $f(x) = \pm 3$
So now we find the values such that $f(x) = 3$
$x^2-4=3\iff x^2=7 \iff x=\pm \sqrt{7}$ (both are greater than ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3635132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$ Find the general solution and singular solution of $(y-px)^2 (1+p^2)=a^2p^2$ where $p=\frac {dy}{dx}$
My Attempt :
$$(y-px)^2 (1=p^2)=a^2p^2$$
$$y-px=\frac {ap}{\sqrt {1+p^2}}$$
$$y=px+\frac {ap}{\sqrt {1+p^2}} ... | The clairaut Eq.:
$$y=xy'+f(y')$$
D.w.r.t. $x$ to get
$$y''(x+f'(y')=0$$
Here $f'(z)$ is d.w.r. t to $z$ (the argument)
the general sdolution is gicen by $y''=0 \implies y/=C$
For the present case
$$y=xy'+\frac{ay'}{\sqrt{1+y'^2}}~~~~(1)$$
$$f'(y')=\frac{a}{(1+y'^2)^{3/2}}$$
$$\frac{x}{a}+\frac{1}{(1+y'^2)^{3/2}}=0, x<... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
y=x^3sinx,what's the 6th derivative of y at x = pi/6? question 1： $y=x^3\sin x$, what's $y^{\left(6\right)}\left(0\right)$ ?
I have solved it use Taylor's Formula in the follow way.
step1:
$y=\displaystyle \sum _{n=0}^{\infty }\frac{y^{\left(n\right)}\left(x_0\right)}{n!}\left(x-x_0\right)^n$
and x_0 = 0 for the questi... | Equivalently, we want $6^3(\sqrt{3}\sin y+\cos y)/2$, where $[y^k]f(y)$ is the $y^k$ coefficient in $f(y)$. So the result is$$\begin{align}&360[y^6]\left(y^3+\frac{\pi}{2}y^2+\frac{\pi^2}{12}y+\frac{\pi^3}{216}\right)\left(1+y\sqrt{3}-\frac12y^2-\frac{\sqrt{3}}{6}y^3+\frac{1}{24}y^4+\frac{\sqrt{3}}{120}y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3639956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Vieta's formulas for quadratic equation problem I'm using one hack, which I never though of why it works. But now I'm curious why it's works and how I can prove it.
Here's the deal: we have quadratic equation $ax^2 + bx + c = 0$, to find roots I just multiply $c$ by $a$ and solving $y^2 + by + ca = 0$, and then I divid... | Take the original equation $ax^2+bx+c=0$ then multiply by $a$ to obtain $$a^2x^2+abx+ac=0$$Now set $y=ax$ so that $$y^2+by+ac=0$$ and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$ Solve the equation $(D^2+4)y=x\sin^2 x$. It is given that $D=\frac {d}{dx}$
My Attempt:
The given equation is
$$(D^2+4)y=x\sin^2 x$$
It's auxiliary equation is
$$m^2+4=0$$
$$m^2=-4$$
$$m=\pm 2i$$
$$\textrm {Complementary Function (C.F)}=c_1 \c... | Starting from the 3rd step
$$P.I =\frac {x}{2(D^2+4)} - \frac {x\cdot \cos (2x)}{2(D^2+4)}
$$
$$=\frac{x}{4} -\frac {x\cdot \cos (2x)}{2(D^2+4)}...(1)$$
First we evaluate
$$\frac{cos {2x}}{D^2+4} $$
$$=\frac{1}{D^2+4}\frac{e^{2ix}+e^{-2ix}}{2}$$
$\text{ You can do the easy calculation. it will give. }$
$$=\frac{x\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3645134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show that $\int_{0}^{\infty} \cos(x^4+x+1) dx$ converges I need to show that $\int_{0}^{\infty} \cos(x^4+x+1) dx$ converges. I showed that $\int_{0}^{\infty} \cos(x^4) dx$ converges but I don't know how to continue. I can't say that $\cos(x^4) \sim \cos(x^4+x+1)$ to conclude as $\cos(x^4)$ keeps changing sign.
If I wri... | I've often used the technique of introducing a factor that allows me to do integration by parts. Setting $dv=\cos(x^4+x+1)\,dx$ is intractable, but if we introduce $4x^3+1$ then $dv=(4x^3+1)\cos(x^4+x+1)\,dx$ works great:
$$
\int _0^{\infty} \underbrace{\frac{1}{4x^3+1}}_{u}\cdot\underbrace{(4x^3+1)\cos(x^4+x+1)\,dx}_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3646223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$ABCD$ is a convex quadrilateral with $\angle CAB=60^\circ$, $\angle CAD=20^\circ$, $\angle ABD=50^\circ$, $\angle DBC=10^\circ$. Find $\angle ACD$. Can you help me solve this Olympiad angles problem?
Let $ABCD$ be a convex quadrilateral such that: $\widehat{CAB}=60^\circ$; $\widehat{CAD}=20^\circ$; $\widehat{ABD}=50^... | Let $E$ be the intersection point of lines $AC$ and $BD$.
Then:
$\angle BEC=60^\circ+50^\circ=110^\circ$
$\begin{align} \angle BCE&= 180^\circ-110^\circ - 10^\circ \\ &=60^\circ \\ &= \angle{BAC} \end{align}$
So that means $\Delta ABC$ is equilateral and hence:
$AB=AC \tag 1$
You can also deduce that:
$\begin{align} \a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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MOP 2011 inequality If $a,b,c$ are positive integers prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +\sqrt{(a^2+c^2-ac)} +9(abc)^{1/3} \le 4(a+b+c)$
My attempt:
I tried to split inequality and prove it bit by bit
$9(abc)^{1/3} \le 3(a+b+c)$
It suffices to prove that
$\sqrt{(a^2-ab+b^2)} +\sqrt{(b^2+c^2-bc)} +... | By Jensen $$\sum_{cyc}\sqrt{a^2-ab+b^2}\leq3\sqrt{\frac{\sum\limits_{cyc}(a^2-ab+b^2)}{3}}=\sqrt{3\sum_{cyc}(2a^2-ab)}.$$
Thus, it's enough to prove that:
$$\sqrt{3\sum_{cyc}(2a^2-ab)}+9\sqrt[3]{abc}\leq4(a+b+c).$$
Now, by $uvw$ (see here: https://artofproblemsolving.com/community/c6h278791 )
it's enough to prove the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Proof of 3-perfect codes I am reading a the proof of a theorem that says that $3$-perfect codes can only have length $7$ or $23$. I do not understand the following:
It follows from the definition that if $n$ is the length of a $3$-perfect then we must have that $(n+1)((n+1)^3-3(n+1)+8) = 3 \cdot 2^k$ for some $k \in \m... | Suppose the highest power of $2$ which divides $n+1$ be $m \ge 4$, i.e., $2^4 = 16$ divides $n + 1$. Thus, you have that
$$n + 1 = 2^{m}j \tag{1}\label{eq1A}$$
where $j$ is odd. Since $n + 1$ divides evenly into $3 \cdot 2^k$, this means $j = 1$ or $j = 3$. You also have the equation
$$\begin{equation}\begin{aligned}
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3652407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
Given $3x^2+x=1$, find $6x^3-x^2-3x+2010$.
I substituted $3x^2 = 1-x$ in $6x^3-x^2-3x+2010$ and it simplified to
$$\frac{-8x}{3}+2010$$
I know this is a simple problem but I can't solve it. I think there's some method I'm not trying. Please help me with this.
| $6x^3 = 2x-2x^2$
so
$6x^3-x^2 -3x+2010 = 2x-2x^2-x^2-3x+2010$
And finally $-x-3x^2+2010 = -x-(1-x)+2010=2009$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Direct product and inverse element of Group Let $(P_2[\mathbb{R}],+)$ and $(\text{GL}(2,\mathbb R), \cdot)$ be groups. How can I write direct product G of those groups. Let
$a = \left (2x^2+4x-3,\begin{pmatrix}3 & 1\\ 2 & 1 \end{pmatrix} \right)$ and
$b= \left (-x^2+x-1, \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix} ... | You have exactly the right idea!
Given two groups $(G,\star)$ and $(H, \cdot)$ we can form their Direct Product $G \times H$ whose elements are pairs $(g,h)$ with $g \in G$ and $h \in H$.
Now this object is a group "componentwise" in the following sense:
*
*$\text{Id}_{G \times H} = (\text{Id}_G, \text{Id}_H)$
*$(g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3658949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Homework Problem, Power Series Limit I am looking to find the solution for:
$$\lim_{x\rightarrow 0} {5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$$
A hint was provided:
transform: ${5^{tan^2(x) + 1} - 5 \over 1 - cos^2(x)}$ to $5y{5^{y-1} - 1 \over y-1}, y = {1\over cos^s(x)} \rightarrow 1$
The transformation is straigt fo... | Hint:
If you are using L'Hopital and differentiating with respect to $y$, then
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}y\,}5^{y-1} &= \frac{\mathrm{d}}{\mathrm{d}y\,}e^{(y-1) \ln5}\\
&=\ln 5 \cdot e^{(y-1) \ln5} \\
&=\ln 5 \cdot 5^{y-1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3659907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Can every odd integer greater than $1$ be written as a product of fractions $\frac{4m+1}{2m+1}$? Can every odd integer greater than $1$ be written as a product of fractions of the form $\frac{4m+1}{2m+1}$, $m$ a positive integer?
Here is a proof if the fractions were instead $\frac{4m-1}{2m+1}$. I tried to mimic it, bu... | $\def\N{\mathbb{N}}$This solution proves by induction on $n \geqslant 0$ that all odd integers $2n + 1$ can be expressed as products of numbers of the form $f(m) = \dfrac{4m + 1}{2m + 1}$.
For $n = 0$, there is $1 = f(0)$. Assume that the proposition holds for all odd integers less than $2n + 1$. Suppose $2n + 2 = 2^k ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3660897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to
Let an angle between the vectors $a$ and $b$ be $\frac{2\pi}{3}$. If $|b|=2|a|$, and the vectors $a+xb$ and $a-b$ are at right angles, then the value of $x$ is equal to $\frac23/\frac25/\frac13/\frac15$?
My attempt:... | There might be a problem with your dot product.
We have
$$(a+xb)\cdot(a-b) = a^2 -a\cdot b + x \ a \cdot b -x \ b^2=0 \\ \implies a^2 +(x-1)\left(2a^2 \cdot \cos \frac{2\pi}{3}\right) -4xa^2=0$$
$$\implies 1-(x-1)-4x=0\implies x=\frac 25$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$ with complex numbers (roots of unity) I want to prove that
Using $z^9=1$ and the fact that $1+w+w^2+w^3+w^4+w^5+w^6+w^7+w^8=0$ where $w=cis(\frac{2π}{9})$, that $$\cos\frac{π}{9} \cos\frac{2π}{9} \cos\frac{4π}{9}=\frac{1}{8}$$
I am able to do this b... | Note
$$\frac{1-z^9}{1-z}= \prod_{k=1}^8(w^k-z)
$$
Set $z=-1$ and use $\prod_{k=1}^8w^{k}=1$
\begin{align}
1&=\prod_{k=1}^8(w^k+1)= (\prod_{k=1}^8w^{k})^{1/2}\prod_{k=1}^8(w^{k/2}+ w^{-k/2})
= 2^8 \prod_{k=1}^8\cos\frac{k\pi}9
=\left( 2^4 \prod_{k=1}^4\cos\frac{k\pi}9 \right)^2
\end{align}
Then, substitute $\cos\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Integral $\int\frac{4x^4}{{x^8+1}}\;dx$ I'm attempting this integral but I'm unsure how to proceed. I've begun to suspect it's actually non-elementary. Can anyone do this? I could always use Taylor series after a small bit of u-sub and integration by parts, but I wanted to know if there were a somewhat more direct way ... | First of all, let me precise that, for this kind of integrals, Taylor expansions could me more thatn dangerous.
Considering the integrand, you can first write (by analogy with $(x^4+1)$
$$\frac {4x^4}{x^8+1}=\frac{\sqrt{2} x^2}{x^4-\sqrt{2} x^2+1}-\frac{\sqrt{2} x^2}{x^4+\sqrt{2} x^2+1}$$
Then
$$x^4-\sqrt{2} x^2+1=\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3665818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
if the polynomial $x^4+ax^3+2x^2+bx+1=0$ has four real roots ,then $a^2+b^2\ge 32?$ if such that the polynomial
$$P(x)=x^4+ax^3+2x^2+bx+1=0$$
has four real roots. prove or disprove $$a^2+b^2\ge 32?$$
I have solve this problem: if the polynomial $P(x)$ at least have one real root,then $a^2+b^2=8$,see a^2+b^2\ge 8,and ... | We will show that with some symmetry consideration we can either reduce to the case where all the roots have the same sign and then the result follows since if say $x_k >0$, $\sum {x_k} =|a| \ge 4(x_1..x_4)^{\frac{1}{4}} =4, a^2 \ge 16$ and then $\sum {1/x_k}=|b|$ and same applies so $b^2 \ge 16$ and we are done, or we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3666790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove $\sum_{n=1}^{\infty}((n+\frac{1}{2})\ln(1+\frac{1}{n})-1)=1-\ln(\sqrt{2\pi})$ I am looking for a derivation of the following sum:
$$\sum_{n=1}^{\infty}\bigg(\left(n+\frac{1}{2}\right)\ln\left(1+\frac{1}{n}\right)-1\bigg)=1-\ln(\sqrt{2\pi})$$
My current derivation(s) uses the zeta function at negative integers (an... | New Answer. Let $S_N$ denote the partial sum for the first $N$ terms. Then $S_N$ is related to the Stirling's Formula by the following computation:
\begin{align*}
S_N
&= \sum_{n=1}^{N} \left(n+\frac{1}{2}\right)\log(n+1) - \sum_{n=1}^{N} \left(n+\frac{1}{2}\right)\log n - N \\
&= \left(N+\frac{1}{2}\right)\log (N+1) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3667934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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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.