Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ where $x, y, z$ are real numbers satisfying the condition $x + y + z = 1 $.
My Attempt
$$
\begin{align}
\frac{x}{1+x^2}+\frac{y}{1+y^2}... | Ok rugi I will solve it using derivatives for your question consider x=y=z because for any condition like x+y+z=a we need to consider x=y=z to get maxima .So answer is 9÷10=0.9.This is for a <3.And when the degree in the numerator is less than the degree in denominator in which the value is to be found out
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1975480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Conjecture regarding integrals of the form $\int_0^\infty \frac{(\log{x})^n}{1+x^2}\,\mathrm{d}x$. I have been playing around a bit with integrals of the form $$I(n)=\int_0^\infty \frac{(\log{x})^n}{1+x^2}\,\mathrm{d}x,\,\,n\in\mathbb{Z}^+,$$ and I am trying to obtain a closed form solution for $I(n).$ I believe the sp... | \begin{align*}J_{n}&=\int_0^\infty \frac{\ln^{2n} x}{1+x^2}dx\\
J_0&=\int_0^1 \frac{1}{1+x^2}dx=\frac{\pi}{2}\\
K_{n}&=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}dxdy\\
&=\sum_{k=0}^{n}\binom{2n}{2k}J_kJ_{n-k}\\
J_n=&\overset{u(x)=yx}=\int_0^\infty\int_0^\infty\frac{y\ln^{2n}u}{(y^2+u^2)(1+y^2)}dudy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 5
} |
Show that if $x + \frac1x = 1$, then $x^5 + \frac1{x^5} = 1$. Suppose $x + \frac{1}{x} = 1$.
Without first working out what $x$ is, show that
$x^5 + \frac{1}{x^5} = 1$ as well.
| Notice that: $$x + \frac{1}{x} = 1 \;\implies\; x^2 - x + 1 = 0 \;\implies\; x^3 + 1 = (x+1)(x^2-x+1) = 0 \;\implies\; x^3=-1$$
Then $x^6=1$ so: $$x^5 + \frac{1}{x^5} = \frac{x^6}{x} + \frac{x}{x^6} = \frac{1}{x} + x = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1976629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
What is the largest perfect square that divides $2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$ I've tried this but didn't get the answer :
Let $S=2014^3-2013^3+2012^3-2011^3+\ldots+2^3-1^3$
Using $n^3-(n-1)^3 = 3n^2-3n+1$,
\begin{align}
S &= 3(2014^2)-3(2014)+1+3(2012^2)-3(2012)+1+\ldots+3(2^2)-3(2)+1
\\&= 3\left ( 2014(... | $$\begin{align}
\sum_{n=1}^{2m}(-1)^n n^3&=\sum_{n=1}^m (2n)^3-(2n-1)^3\\
&=\sum_{n=1}^m 12n^2-6n+1\\
&=\sum_{n=1}^m 24 \binom n2+6\binom n1+1\\
&=24\binom {m+1}3+6\binom {m+1}2+m\\
&=m\; \big[\;4(m+1)(m-1)+3(m+1)+1\;\big]\\
&=m^2(4m+3)\\
\end{align}$$
which is divisible by $m^2$.
Putting $m=1007$ gives the answer req... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Olympic problem - Probability A jar contains 3 cookies filled with chocolate, 4 cookies filled with strawberry and 3 cookies filled with vainilla. What is the probability of choosing 3 cookies with the same filling from the jar?
EDIT: I've tried this:
What I want to find is:
$$\frac{Favorable\ outcomes}{Possible\ outco... | Look at the problem as a choice tree and ask yourself: how many branches do I have that satisfy the conditions above?
In this case you have a tree with depth three (3). How many branch ends are there to satisfy your condition, and what is the probability of getting to each of them?
P(3 vanilla) = $\frac{3}{10} * \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How would you find the trigonometric roots of a cubic?
Question: How would you find the roots of the cubic$$x^3+x^2-10x-8=0\tag{1}$$
I'm not too sure where to begin. I'm thinking of somehow, implementing $\cos 3\theta=4\cos^3\theta-3\cos\theta$.
I've tried substituting $x$ with $t+t^{-1}$, but didn't get anywhere, a... | If you wish to express as roots of cubics the sums of trigonometric functions with arguments that use a prime of form $p=6\color{blue}n+1$, then the number of addends is $\color{blue}n$. Thus, the reason you couldn't find $p=31$ was that you should have used $\color{blue}5$ addends.
$p=31$
A root of
$$x^3+x^2-(2\tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\sin(x) + 2\sqrt{2}\cos x =3$ How to solve $\sin(x) + 2\sqrt{2}\cos x=3$ ?
What is general method for doing these kind of questions?
Thanks
| $$\sin x+2\sqrt { 2 } \cos { x } =3\\ 2\sin { \frac { x }{ 2 } \cos { \frac { x }{ 2 } +2\sqrt { 2 } \left( \cos ^{ 2 }{ \frac { x }{ 2 } -\sin ^{ 2 }{ \frac { x }{ 2 } } } \right) =3\sin ^{ 2 }{ \frac { x }{ 2 } +3\cos ^{ 2 }{ \frac { x }{ 2 } } } } } \\ \left( 3+2\sqrt { 2 } \right) \sin ^{ 2 }{ \frac { x }{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Mathemathic induction proof I need to prove that $$\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}=\frac{n}{n+1}$$
Here is what I tried:
\begin{align}
& \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4} + \cdots + \frac{1}{n(n+1)}+\frac{1}{n+1(n+2)} \\[10pt]
= {} & \frac{n}{n+1}+\f... | Hint: write
$$\frac1{k(k+1)}=\frac1k-\frac1{k+1}$$
and now do the telescopic sum carefully.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How would you go about find the values of a,b and c such that they f(0) = f'(0) = f''(0) = 1 ? The function is:
\begin{align}
f( x) &= \frac{a+bx}{1+cx} \\\\
\end{align}
The question also tells us that the function and both it's first and second derivatives are equal to the corresponding values of:
$$\ e^{x}$$
Woul... | Given $f(0) = f'(0) = f''(0) = 1$ with $$f(x) = \frac{a+bx}{1+cx}$$ quickly leads to $a=1$ and $$f(x) = \frac{1+bx}{1+cx} = \frac{b}{c} - \frac{b-c}{c} \, \frac{1}{1+cx}.$$
Taking derivatives provides:
\begin{align}
f'(x) &= \frac{b-c}{(1+cx)^{2}} \\
f''(x) &= - \frac{2c (b-c)}{(1+cx)^{3}}.
\end{align}
Using the remain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1983084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Are two sequences equal if the sums and sums of squares are equal? Are two sequences $(x_i)_{i=1,\ldots,n}$ and $(y_i)_{i=1,\ldots,n}$ equal if $\sum_{i=1}^nx_i=\sum_{i=1}^ny_i$ and $\sum_{i=1}^nx_i^2=\sum_{i=1}^ny_i^2$?
| Something interesting. On all $3 \times 3$ magic squares, the sums of the top and bottom rows is the same (of course) but, also, the sums of the squares of the top and bottom rows is the same. For example
\begin{array}{|c|c|c|}
\hline
8 & 3 & 4 \\
\hline
1 & 5 & 9 \\
\hline
6 & 7 & 2 \\
\hline
\end{array}
$8 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 3
} |
Finding an unknown coefficient of a polynomial given a factor Q:Find the value of $a$ given that $x^2+1$ is a factor of $x^4-3x^3+3x^2+ax+2$
No idea where to start, I was going to use the factor theorem but it didn't work out.
Question from year 10 Cambridge maths textbook
| Divide your polynomial by $x^{2} + 1$ and set the remainder equal to zero.
Or, since you are only interested in the remainder, set $x^{2} = -1$ in the polynomial to get that the remainder is
$$
(x^2)^{2}-3 x x^2+3 x^2+ax+2
=
(-1)^{2} + 3 x - 3 + a x + 2
=
(a + 3) x.
$$
So $x^{2} + 1$ divides your polynomial iff $a = -3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove the following Inequality. Given that $\alpha,\beta$ and $\gamma\in (0,\pi)$ and $\alpha+\gamma+\beta=\pi$, show that $$\cos\alpha+\cos\gamma+\sin\beta\leq\frac{3\sqrt3}{2}.$$
Now I am aware of the following Inequality: $\cos\alpha+\cos\gamma+\cos\beta\leq\frac{3}{2}$, but notice that the term $\cos\beta$ has been... | Because if $\cos\frac{\beta}{2}=x$ by AM-GM we obtain:
$$\cos\alpha+\cos\gamma+\sin\beta=2\sin\frac{\beta}{2}\cos\frac{\alpha-\gamma}{2}+\sin\beta\leq2\sin\frac{\beta}{2}+\sin\beta=$$
$$=2\sqrt{1-x^2}(1+x)=2\sqrt{(1-x)(1+x)^3}=2\sqrt{27(1-x)\left(\frac{1}{3}+\frac{x}{3}\right)^3}\leq$$
$$\leq2\sqrt{27\left(\frac{1-x+3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that Let $\alpha$, $\beta$, and $\gamma$ be acute angles such that $\alpha$ + $\beta$ = $\gamma$. Show that
cos$\alpha$ + cos$\beta$ + cos$\gamma$ - 1 $\geq$ $2\sqrt{\cos\alpha\times\cos\beta\times\cos\gamma}$.
Here's what... | Using that
$$
\cos\alpha\cos\beta-\cos\gamma = \sin\alpha\sin\beta
= 4 \sin\frac\alpha2 \cos\frac\alpha2 \sin\frac\beta2 \cos\frac\beta2 \ge 0
$$
and
$$
(1-\cos\alpha)(1-\cos\beta) = 4 \sin^2\frac\alpha2 \sin^2\frac\beta2,
$$
a possible proof:
$$
\cos\gamma - 2\sqrt{\cos\alpha\cos\beta}\cdot\sqrt{\cos\gamma}+
\cos\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\lim ( \sqrt{n^2+n}-n) = \frac{1}{2}$ Here's what I have so far:
Given $\epsilon > 0$, we want to find N such that $\sqrt{n^2+n}-n < \epsilon$ for all $n>N$. And so:
$( \sqrt{n^2+n}-n-\frac{1}{2}) \cdot \frac{\sqrt{n^2+n}-(n+\frac{1}{2})}{\sqrt{n^2+n}-(n+\frac{1}{2})}$
$= \frac{(n^2+n)-(n+\frac{1}{2})}{\sq... | Hint:
\begin{align}
\sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}
\end{align}
and
\begin{align}
\left|\frac{n}{\sqrt{n^2+n}+n} -\frac{1}{2}\right| = \left|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+2n}\right| = \frac{n}{2(\sqrt{n^2+n}+n)^2} \leq \frac{n}{8n^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1988705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the power series of $f(x)=\frac{1}{x^2+x+1}$ I want to find the power series of $$f(x)=\frac{1}{x^2+x+1}$$
How can I prove the following?
$$f(x)=\frac{2}{\sqrt{3}} \sum_{n=0}^{\infty} \mathrm{sin}\frac{2\pi(n+1)}{3} x^n \,\,\,\, |x|<1$$
In particular I would like to know how to proceed in this case. The polinomia... | The polynomial $x^2+x+1=\Phi_3(x)$ has no real roots, but it vanishes at $x=e^{\pm\frac{2\pi i}{3}}$. In particular, by setting $\omega=\exp\left(\frac{2\pi i}{3}\right)$ and $\overline{\omega}=\omega^2=\exp\left(\frac{4\pi i}{3}\right)$,
$$ \frac{1}{x^2+x+1} = \frac{1}{(x-\omega)(x-\omega^2)} = \frac{i\omega^2}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Finding maximum value from a statement/equation The statement has values of $x$ and $y$ as positive integers: $$\sqrt{x} - \sqrt{11} = \sqrt{y}$$
I have to find the maximum possible values of $\frac{x}{y}$, this what I have done so far:
$$x = (\sqrt{y} + \sqrt{11})^2$$
$$y = (\sqrt{x} - \sqrt{11})^2$$
therefore: $$\fra... | Dividing by $\sqrt{y}$ we get
$$\sqrt{\frac{x}{y}}=\sqrt{\frac{11}{y}}+1$$
and by squaring
$$\frac{x}{y}=\frac{11}{y}+2\sqrt{\frac{11}{y}}+1$$
So $\sqrt{\frac{11}{y}}$ should be rational and y is integer then for maximum we take y=11. so you get
$\frac{x}{y}=4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove $\int_{0}^{1} \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2$ I stumbled upon the interesting definite integral
\begin{equation}
\int\limits_0^1 \frac{\sin^{-1}(x)}{x} dx = \frac{\pi}{2}\ln2
\end{equation}
Here is my proof of this result.
Let $u=\sin^{-1}(x)$ then integrate by parts,
\begin{align}
\int \frac{\sin... | Integration by parts reduces the integral to,
$$\int_{0}^{1} \frac{\ln x}{\sqrt{1-x^2}} dx$$
And the substitution $x=\sin u$ reduces the integral to,
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\sin u) du$$
And the substitution $v=\frac{\pi}{2}-x$ reduces the integral to,
$$I=\int_{0}^{\frac{\pi}{2}} \ln (\cos v) dv$$
$$I=\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 1
} |
Show a chain is a composition series for a matrix ring
Consider the ring
$$
R =
\begin{pmatrix}
\mathbb{Q} & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix}
$$
as a left $R$-module over itself. I want to show that the following is a composition series
$$
0=\begin{pmatrix}
0 & 0 \\
0 & 0
\end{pmatrix}\sub... |
$ \begin{pmatrix}
\mathbb{Q} & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix} / \begin{pmatrix}
0 & 0 \\
\mathbb{R} & \mathbb{R}
\end{pmatrix}$. I see that this is isomorphic to $\mathbb{Q}$. Can I use this to show that the quotient is simple?
That alone is probably not a transparent enough explanation of why ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Stuck on showing $\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$ converges I am trying to show if this series converges or diverges and I know it converges since for very large values of n,
$$\sum_{n=0}^\infty \frac{(-1)^n(n^2+3n-7)}{n^3+1}$$ becomes $$\sum_{n=0}^\infty \frac{(-1)^n1}{n}$$
which is convergent from ... | We can write
$$\frac{n^2+3n-7}{n^3+1}= \frac{n^2-n +1}{n^3+1} + \frac{4(n +1)}{n^3+1} - \frac{12}{n^3+1}= \\
= \frac{1}{n+1} + \frac{4}{n^2-n +1} - \frac{12}{n^3+1}.
$$
Now, $\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n+1}}$ converges conditionally, however, $4\sum\limits_{n=0}^{\infty}{ \frac{(-1)^n}{n^2-n +1}}$ and $1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1993724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Prove: $\log_{2}{3} < \log_{3}{6}$ How should I prove that $\log_{2}{3} < \log_{3}{6}$?
I tried something like this: $2^{\log_{2}{3}}< 2^{\log_{3}{6}}$, $3<6^{\log_{3}{2}}$, $\log_{6}{3}<\log_{6}{6^{\log_{3}{2}}}=\log_{3}{2}$, $\frac{1}{\log_{3}{6}}< \log_{3}{2}$. $\frac{1}{1+\log_{3}{2}}<\log_{3}{2}$, but still nothi... | Use:
*
*$$\log_2(3)=\frac{\ln(3)}{\ln(2)}$$
*$$\log_3(6)=\frac{\ln(6)}{\ln(3)}=\frac{\ln(2\cdot3)}{\ln(3)}=\frac{\ln(2)+\ln(3)}{\ln(3)}=1+\frac{\ln(2)}{\ln(3)}=1+\frac{1}{\log_2(3)}$$
So, you need to prove that:
$$\log_2(3)<\log_3(6)\space\space\space\Longleftrightarrow\space\space\space\log_2(3)<1+\frac{1}{\log_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1994320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
How to determine if a function is increasing. How would I determine whether a function is increasing, decreasing or neither without using calculus? Like whether x^0.5 is increasing in interval [0, infinity)
Just curious
Thanks
| There are many non-calculus techniques that can be applied to show a function is increasing. One approach is to show for $k > 0$ that $f(x + k)$ is larger than $f(x)$.
Here’s a simple example.
Suppose that $f(x) = mx + b$
Now take $k > 0$ and compare $f(x)$ to $f(x+k)$.
$$f(x+k) - f(x) = mk$$
Since $k > 0$ we know tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove $(1+a+b)(1+b+c)(1+c+a)\ge 9(ab+bc+ca)$ How one can prove the following.
Let $a$, $b$ and $c$ be non-negative real numbers. Then the inequality holds:
$(1+a+b)(1+b+c)(1+c+a)\ge9(ab+bc+ca).$
WLOG one can assume that $0\le a\le b\le c$.
It is not difficult to prove the statement when $0\le a\le b\le c\le 1$ or $1\le... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $$(1+a+b)(1+a+c)(1+b+c)\geq9(ab+ac+bc)$$ or
$$1+2(a+b+c)+\sum_{cyc}(a^2+3ab)+\prod\limits_{cyc}(a+b)\geq9(ab+ac+bc)$$ or
$$1+6u+9u^2+3v^2+9uv^2-w^3\geq27v^2,$$ which is a linear inequality of $v^2$, which says that it remains to prove our inequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1996771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Calculating limit of $\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$ As the title says we want to calculate:
$$\lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}}$$
By multiplying nominator and denominator in their conjugates
$=\lim_{x\t... | Divide top and bottom by $\sqrt{x}$
$$ \lim_{x\to\infty}\dfrac{\sqrt{x+1}-2\sqrt{x+2}+\sqrt{x}}{\sqrt{x+2}-2\sqrt{x}+\sqrt{x-4}} = \lim_{x\to\infty}\dfrac{\sqrt{1+\frac1{x}}-2\sqrt{1+\frac{2}{x}}+\sqrt{1}}{\sqrt{1+\frac{2}{x}}-2\sqrt{1}+\sqrt{1-\frac{4}{x}}}$$
$$ \to \frac{0}{0}$$
so use L'Hopitals rule differentiate t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1998813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Probability: $a$ chosen randomly from $\{1,2,\dots,n\}$, find $P(a^2=1 \mod 10)$.
A number $a$ is chosen randomly from the set $\{1, 2,\dots , n\}$. Find
*
*The probability $p_n$ that $a^2=1 \mod 10 $.
*$\displaystyle \lim_{n\to\infty} p_n$.
I found that in $n$ numbers $\{1,2,3,\dots, n\}$ there are ... | Note that if $a^2 \equiv 1 \pmod{10}$ then we must have that $a^2 \equiv 1 \pmod{5}$ and $a^2 \equiv 1 \pmod{2}$. As both $5$ and $2$ are primes we must have that $a \equiv \pm 1 \pmod 5$ and $a \equiv 1 \pmod 2$. Combining these two with the Chinese Remainder Theorem we have that:
$$a \equiv \pm 1 \pmod{10}$$
So this ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
find the range of the $b$, if $a^2+b^2+c^2=21~~~~2b=a+c$ Let $\Delta ABC$ such $|AC|=b,|BC|=a,|AB|=c$,and such $$\begin{cases}
2b=a+c\\
a^2+b^2+c^2=21
\end{cases}$$
find the range of the $b$
since
$$(a+c)^2-2ac+b^2=21$$
so we have
$$2ac=5b^2-21$$
then $a,c$ is a equation
$$x^2-2bx+\dfrac{5b^2-21}{2}=0$$ roots.so we ha... | Upper bound
Since $a+c=2b$ we have that $a+b+c=3b.$ Now, squaring we get
$$9b^2=(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc.$$ Using that $a^2+b^2+c^2=21$ and $ab+ac+bc\le a^2+b^2+c^2=21$ we get $b^2\le 7.$ So, we have that upper bound $b\le \sqrt{7}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is this integral $\int {\sqrt{\frac{x+1}{x}}}\:dx$? I tried a lot of different approaches but found this problem very hard.
So can you help me with this integral?
$$\int {\sqrt{\frac{x+1}{x}}}\:dx$$
Thanks.
| I have finally understood what Surb meant in his answer and I'm posting it here for the benefit of future readers. I can't post it as a comment due to space restrictions.
$\int{\sqrt{\frac{x+1}{x}}}$
Let u = $\sqrt{\frac{x+1}{x}}$ and v = $\frac{x+1}{x} = 1 + \frac{1}{x}$
$\therefore u = \sqrt{v}$
$\frac{dv}{dx} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2001709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that exactly half of $1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}$ are congruent to 1 modulo $p$
Let $p$ be an odd prime number. Look at the numbers in the set
\begin{align*}
S \in \{1^{\frac{p-1}{2}}, 2^{\frac{p-1}{2}}, \dots, (p-1)^{\frac{p-1}{2}}\}
\end{align*}
Show that exactly half... | $x^p-1=0$ has $p-1$ solutions.
$$x^p-1=(x^{\frac{p-1}{2}}-1)(x^{\frac{p-1}{2}}+1)$$ and since each of
$x^{\frac{p-1}{2}}-1$ and $x^{\frac{p-1}{2}}+1$ has at most $\frac{p-1}{2}$ many solutions both must have exactly $\frac{p-1}{2}$ many solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Jensen's inequality application I tried finding the maximum value of the expression
$$\frac{2a}{3a+b}{\sqrt{3a+b}}+\frac{2b}{3b+c}{\sqrt{3b+c}}+\frac{2c}{3c+a}{\sqrt{3c+a}}$$
where a,b,c are positive real numbers.
I used Jensen's inequality for concave function: $$f(x)=\sqrt{x}$$
The expression is homogeneous, so WLOG ... | The maximum does not exist. Try $a=b=c\rightarrow+\infty$.
Your reasoning is true and gives
$\sum\limits_{cyc}\frac{2a}{3a+b}\sqrt{3a+b}\leq\sqrt{2a+2b+2c}=\sqrt{(2a+2b+2c)\sum\limits_{cyc}\frac{2a}{3a+b}}$,
which gives again that the maximum does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2002297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the coefficient of $x^{16}$ in expansion of $(x^2+5x+6)^4(x^3+6x^2+11x+6)^3$.
Find the coefficient of $x^{16}$ in expansion of $(x^2+5x+6)^4(x^3+6x^2+11x+6)^3$.
I simplified the expression and it turned out to be $(x+3)^7(x+2)^7(x+1)^3$. Now I'm stuck. I only know how to deal with 2 terms but there are 3 over ... | Considering $f(x) = (x+3)^7(x+2)^7(x+1)^3$
In expanded form, the sum of roots of this polynomial would be
$$S = \frac{-b}{a} $$
Here , $b$ would be the coefficient of $x^{16}$, and $a$ the coefficient of $x^{17}$[This is a standard result for any polynomial, that sum of roots is negative of coefficient of second highe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Limit square roots of polynomials I am trying to find $\lim \limits_{n \to \infty} {\sqrt{n^3+1}-n\sqrt{n} \over \sqrt{n^2+1}-n}$. I rewrite the fraction as
$${(\sqrt{n^3+1}-n\sqrt{n})(\sqrt{n^3+1}+n\sqrt{n}) \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})} = {1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}$$
I notice... | One may proceed the following way, as $n \to \infty$,
$$
\begin{align}
{1 \over (\sqrt{n^2+1}-n)(\sqrt{n^3+1}+n\sqrt{n})}&={\sqrt{n^2+1}+n \over ((n^2+1)-n^2)\cdot(\sqrt{n^3+1}+n\sqrt{n})}
\\\\&={\sqrt{n^2+1}+n \over 1\cdot(\sqrt{n^3+1}+n\sqrt{n})}
\\\\&=\frac1{n^{1/2}}{\sqrt{1+1/n^2}+1 \over (\sqrt{1+1/n^3}+1)}
\\\\&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find minimum of the $a+c+e$ if $a+b+c+d+e=100$.
Let $a\ge b\ge c \ge d \ge e \ge0$ and $a+b+c+d+e=100$. Find minimum of
$$a+c+e$$
My work so far:
$100=a+b+c+d+e\le 5a \Rightarrow a\ge20$
I think that $a+c+e \ge 50$ (if $a=40,b=40,c=10,d=10,e=0$ or $a=b=49, c=d=e=1$).
| $$100=a+b+c+d+e\le a+a+c+c+e=2a+2c+2e-e.$$ Thus $$50\le 50+\frac e2\le a+c+e.$$ So, the minimum of $a+c+e$ is $50$ and is achieved if $e=0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving simultaneous equations involving a quadratic I have the question
Solve the simultaneous equation pair
$$x^2 + y^2 = 25\tag1$$
$$2x - y = 5\tag2$$
I have found the value of $y$ from the second equation which is $2x-5$ and substituted this into the first equations $y$ value.
I get $x^2 + (2x -5)^2 = 25$
When I... | $$x^2 + y^2 = 25 $$
$$2x - y = 5 \Leftrightarrow y=2x-5$$
$$ x^2 + (2x-5)^2 = 25 $$
$$ x^2 +4x^2 -20x +25 = 25 $$
$$ 5x^2 -20x=0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2007864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Use mathematical induction to prove an assertion The assertion: $n^3 + 5n$ is divisible by $6$
I have completed the basis step $(n=1)$ and the first part of the induction step $(n=k)$, but I am stuck on the second part $(n=k+1)$.
This is what I have so far:
For $n=k$: $k^3 + 5k = 6t$
For $n= k+1$:
$(k+1)^3 + 5(k+1)$
$=... | You can complete your work this way:
$$
(k+1)^3 + 5(k+1)=(k^3 + 5k) + 3k + 3k^2 + 6=6t+3k+3k^2+6=6 \left(t+1+\frac{k(k+1)}{2} \right)
$$
and note that $k$ or $k+1$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2008111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Inequality on Quadratic Equation Coefficients
Let $a \in \mathbb{Z}_{>0}$ and $b, c \in \mathbb{Z}$, such that $$ax^2 + bx + c = 0$$ has two different solutions in the interval $(0, \frac{1}{2}]$. Prove that $a \geq 6$.
My work consists of just some observations:
*
*if the roots are $\alpha$ and $\beta$, then $\al... | \begin{align}
&f(x):=ax^2+bx+c\\\\
&f(0)>0\quad\rightarrow\quad c>0\tag1\\
&f\left(\frac12\right)>0\quad\rightarrow\quad a+2b+4c>0\tag2\\
&0<-\frac{b}{2a}<\frac12\quad\rightarrow\quad b^2<a^2\tag3\\
&b^2-4ac>0\quad\rightarrow\quad a<\frac{b^2}{4c}\tag4\\\\
&\text{As $a$ is maximuzed when $c$ is 1, let's set $c=1$. Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2010015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
factoring $(x^6 - y^6)$: what is going on here? I apologize if this question already exists, but it was quite difficult to word.
In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:
factor $x^6 - 64$
almost everyone other than me (including my teacher, a... | A complete factorization of $x^6 - y^6$ can be performed as follows: $$\begin{align*}
x^6 - y^6 &= (x^3)^2 - (y^3)^2 \\
&= (x^3 - y^3)(x^3 + y^3) \\
&= (x-y)(x^2 + xy + y^2)(x+y)(x^2 - xy + y^2). \end{align*}$$
The quadratic factors are irreducible in the ring of polynomials with rational coefficients $\mathbb Q[x]$ (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Show that a given equation has no solutions in integers Prove that there are no integers $x$ and $y$ such that
$$x^5-x^4y-13x^3y^2+13x^2y^3+36xy^4-36y^5=77.$$
Firstly setting $x=y$ gives LHS=0. This suggests that $x-y$ is a factor of it, factoring gives
\begin{align*}
(x-y)(x^4-13x^2y^2+36y^4) &=(x-y)(x^2-9y^2)(x^2-4... | By my comment above, we have that $(x-y)=\pm 1, \pm 7, \pm 11$ or $\pm 77$. Similarly, $(x-2y)=\pm 1, \pm 7, \pm 11$ or $\pm 77$.
Case I: Assume $x-y=1=x-2y$, then $y=0$ and $x=1$, but then $(x-3y)(x-2y)(x-y)(x+2y)(x+3y)=1\neq 77$.
Case II: Assume $x-y=-1$ and $x-2y=1$, then $y=-2$ and $x=-3$, but then $x+3y=-9$ which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\lim_\limits{n\to \infty}\ (\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + ... + \frac{1}{n+1})$? what is the value given to this limit?
$$\lim_{n\to \infty}\ \bigg(\frac{1}{2n} + \frac{1}{2n-1} + \frac{1}{2n-2} + \cdots + \frac{1}{n+1}\bigg)$$
Is it simply 0 because each term tends to 0 and you are just summing up ... | Alternatively,
\begin{align*}
1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n} &=
\ln n+\gamma+\frac{1}{2n}+O\left( \frac{1}{n^2} \right) \\
\frac{1}{n+1}+\frac{1}{n+2}+\ldots+\frac{1}{2n} &=
\left[ \ln (2n)+\gamma+\frac{1}{2(2n)} \right]-
\left( \ln n+\gamma+\frac{1}{2n} \right)+
O\left( \frac{1}{n^2} \right) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2018856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Question about solution to Putnam 1995 A2. To provide context Putnam 1995 A2 states:
For what pairs of integers $(a,b)$ of positive real numbers does the improper integral
$$ \int_{b}^{\infty} \left( \sqrt{ \sqrt{x+a \vphantom{b}} -\sqrt{x}} - \sqrt{\sqrt{x} - \sqrt{x-b}} \right) $$
converge?
The solution then proc... | I do not understand your change in notation; I will stick by the original. In any case, it is
$$\int_b^{\infty} dx \, \left (\sqrt{\sqrt{x+a}-\sqrt{x}} -\sqrt{\sqrt{x}-\sqrt{x-b}}\right ) $$
Note that we should have $a$ and $b \gt 0$ so that we have real quantities throughout the integration region.
We need to worry a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2020837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve a congruence $1978^{20}\equiv x\pmod{125}$ I have checked the solution ($x=26$).
Solving modulo $5$ gives
$$1978^{20}\equiv 1978^{2\cdot 10}\equiv 1\pmod{5}$$
Solving modulo $25$ also gives
$$1978^{20}\equiv 1\pmod{5}$$
How to evaluate the remainder $x$?
| Note that, modulo $125$, $11^2=121\equiv-4$ and $2^7=128\equiv3$.
Therefore, $1978^{20}\equiv(-22)^{20}\equiv 2^{20}11^{20}\equiv 2^{20}(-4)^{10}\equiv 2^{40}\equiv2^{35}2^5\equiv3^52^5\equiv6^5\equiv26$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the $n$ derivative of $y= e^{2x}\sin^2 x$ We have
\begin{align*}
y&= e^{2x}\sin^2 x\\
&= e^{2x}\left(\frac{1-\cos 2x}{2}\right)\\
&= \frac{e^{2x}}{2} - \frac{e^{2x}\cos 2x}{2}
\end{align*}
Then
\begin{align*}
y^{(n)} &= \left(\frac{e^{2x}}{2}\right)^{(n)} - \left(\frac{e^{2x}\cos 2x}{2}\right)^{(n)}\\
&= 2^{n-1}e... | Taking into consideration just the second part, let's call
$$p(x)=e^{2x}\cos 2x$$
then
\begin{align}
&p^{(1)}(x)=2e^{2x}\cos 2x-2e^{2x}\sin 2x\\
&p^{(1)}(x)=2e^{2x}(\cos 2x-\sin 2x)=2e^{2x}\sqrt{2}\left(\frac{\sqrt{2}}{2}\cos 2x-\frac{\sqrt{2}}{2}\sin 2x\right)\\
&p^{(1)}(x)=2^{3/2}e^{2x}\left(\cos(\pi/4)\cos 2x-\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2024756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Taylor expansion for $\frac{1}{x^2 + 2x + 2}$ I'm trying to find the Taylor expansion of the function:
$$ f(x) = \frac{1}{x^2 + 2x + 2} $$
about the point $ x = 0 $. I have worked out the terms up to the fourth derivative, which was very tedious. I found:
$$ f(x) = \frac{1}{2} - \frac{1}{2} x + \frac{1}{4} x^2 + 0 x^3 ... | $$
x^2 + 2x + 2 = (x+1)^2 + 1
$$
and, with $\ s = x+1 \ $ you have
$$
\frac 1{x^2 + 2x + 2} = \frac 1{(x+1)^2 + 1} = \frac 1{s^2 + 1} = \text{D} ( \arctan s ).
$$
Now, just derive the Taylor expansion of $\ \arctan s \ $ and subsitute $\ s = x +1 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2025920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 3
} |
Showing $\det(aM+bN)-\det(aM)-\det(bN)=ab\left[\det(M+N)-\det(M)-\det(N)\right]$
Why is $\det(aM+bN)-\det(aM)-\det(bN)=ab\left[\det(M+N)-\det(M)-\det(N)\right]$
$M,N$ are $2\times 2$ matrices, $a,b\in K$ for some field $K$. If I write $M=(m_{ij})_{i,j}$ for $\ 1\le i,j\le2$ then it is true, but a a long computation. ... | Just applying properties of determinants, we have:
$$\begin{vmatrix}aa_{11}+bb_{11} & aa_{12}+bb_{12}\\ aa_{21}+bb_{21}& aa_{22}+bb_{22}\end{vmatrix}- \begin{vmatrix}aa_{11} & aa_{12}\\ aa_{21}& aa_{22}\end{vmatrix}- \begin{vmatrix}bb_{11} & bb_{12}\\ bb_{21}& bb_{22}\end{vmatrix}$$
$$=\begin{vmatrix}aa_{11} & aa_{12}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How many positive integers less than 1,000,000 have the sum of their digits equal to 19? How many positive integers less than $1,000,000$ have the sum of their digits equal to $19$ ?
I tried to answer it by using stars and bars combinatorics method.
The question says that sum of $6$ digit numbers { considering the num... | If we treat each positive integer with fewer than six digits as a six-digit string with leading zeros, we seek the number of solutions of the equation
$$a + b + c + d + e + f = 19 \tag{1}$$
in the non-negative integers subject to the restrictions that $a, b, c, d, e, f \leq 9$. A particular solution of equation 1 cor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2028849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
point of symmetry of the graph of a function the graph of the function $$f(x)=\frac{1}{1+x}+\frac{2x}{1+x^2}+\cdots+\frac{2^{n-1}x^{2^{n-1}-1}}{1+x^{2^{n-1}}}+\cdots$$ where $x \in (-1,1)$ is symmetric about the point---
I tried to integrate the function and could not proceed to find centre of symmetry.I tried it in d... | Surprisingly $f(x)$ is a very simple function as shown below. To get it, we start from :
$F_0(x)=(1+x)$
$F_1(x)=(1+x)(1+x^2)=(1+x)+x^2(1+x)=1+x+x^2+x^3$
$\begin{cases}
F_2(x)=(1+x)(1+x^2)(1+x^4)=(1+x^2+x^3)+x^4(1+x^2+x^3) \\
F_2(x)=1+x^2+x^3+x^4+x^5+x^6+x^7
\end{cases}$
and so on ...
$\begin{cases}
F_n(x)=(1+x)(1+x^2).... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2029266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find $\int_{|z|=3} {1 \over P(z)}dz$ where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$
Evaluate
$$\int_{|z|=3} {1 \over P(z)}dz$$
where $P(z)=2z^5-16z^4-17z^3-11z^2+20z-18$.
I proved that $P(z)\neq 0$ for all $z$ outside of the ball of radius 3, except for z=9, so all the poles of ${1 \over P(z)}$ except $z=9$ are i... | The roots of $P(z)= 2z^4 + 2z^3+ z^2-2z+2 $ are $a_1 = -1-i$, $a_2 = \frac{-a_1}{2}$ , $\overline{a_1}$ and $\overline{a_2}$. Now by the residue Theorem, you get,
\begin{equation*}
\int_{|z|=3} \frac{dz}{P(z)} = 2i\pi \left( \frac{1}{(a_1 -9)P'(a_1)} + \frac{1}{(\overline{a_1}-9)P'(\overline{a_1})} + \frac{1}{(a_2-9)P'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find an $v,w \in \mathbb{C}$ such that $(1+2i)v=1-i$ and $w^{2}-i=0$ Task from an old exam:
Find an $v,w \in \mathbb{C}$ such that $(1+2i)v=1-i$ and $w^{2}-i=0$
and write the solutions in this form: $a+bi$
I'm currently trying to solve this, probably written several pages and always failed calculating it with succe... |
We obtain from $(1+2i)v=1-i$
\begin{align*}
v&=\frac{1-i}{1+2i}\\
&=\frac{(1-i)(1-2i)}{(1+2i)(1-2i)}\\
&=\frac{-1-3i}{5}\\
&=-\frac{1}{5}-\frac{3}{5}i
\end{align*}
We recall
\begin{align*}
w^2&=a+ib\\
&=r\left(\cos\phi+i\sin \phi\right)\\
w_{1,2}&=\pm\sqrt{r}\left(\cos\frac{\phi}{2}+i\sin \frac{\phi}{2}\right)\\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2033386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given $a^2+b^2=2$ prove $a+b\le2$
*
*Given $a^2+b^2=2$ prove $a+b\le2$
*Given $a+b=2$ prove $a^4+b^4\ge2$
I was trying to prove these using the fact that we know $a^2+b^2\ge2ab$ but not sure where to start.
| Courtsey : $\rightarrow$ Cauchy Schwarz
$$(a+b)^2\le(1+1)(a^2+b^2)\\\implies a+b\le |a+b|\le 2$$
$$$$Now by the fact that $a+b=2$ and $a^2+b^2\ge 2ab$, we deduce that $2(a^2+b^2)\ge 4\implies a^2+b^2\ge 2$ $$(a^2+b^2)^2\leq2(a^4+b^4)\\ \implies a^4+b^4\ge 2$$
BINGO!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2034351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
total number of positive and negative roots of the equation $ax^3+bx^2+cx+d=0$ Suppose $a,b,c,d$ are non zero real numbers and $ab>0,$ and
$\displaystyle \int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx = \int^{2}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0$
Then total number of positive and negative roots of the equation $ax^3+bx^2+... | Observe that all the functions involved in the integrand are continuous, furthermore the function $(1+e^{x^2})>0$ for all $x$.
From the first condition
$$\int^{1}_{0}(1+e^{x^2})(ax^3+bx^2+cx+d)dx=0.$$
This means the polynomial $ax^3+bx^2+cx+d$ will have at least one root in the interval $[0,1]$. So at least one positiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2035617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What's wrong with my approach - reciprocal integral I want to integrate $\int \frac{1}{2x}dx$
The way I would solve it: $$\frac{1}{a}f(ax + b)$$ where $$f(x) = \ln x$$
My answer $\frac{1}{2}\ln(2x) + k$
The correct way seems to be to factor out $\frac{1}{2}$ from the initial integral, leaving $\int \frac{1}{x}$. The co... | You are not wrong; however, your answer is just not simplified.
To check that your answer is correct, differentiate both your answer and the "correct" answer:
$$\frac{d}{dx} (\frac{1}{2} ln(2x) + k) = \frac{1}{2x}$$
$$\frac{d}{dx} (\frac{1}{2} ln(x) + k) = \frac{1}{2x}$$
To simplify:
$\frac{1}{2} ln(2x) + k = \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2037504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$ I am checking for convergence of
$$\sum_{n=2}^{\infty}{1 \over \sqrt{n}}\ln{n+1 \over n-1}$$
First, we notice
$${n+1 \over n-1}= {1+{2 \over n-1}}$$
Then, we use $\ln(1+x) \le x$ and get
$$\ln\left({1 + {2 \over n-1}}\right) \le {2 \over n-1}$$
$$... | You are right. Alternatively you can use the asymptotic comparison test by noting that
$${1 \over \sqrt{n}}\ln\left({n+1 \over n-1}\right)=
{1 \over \sqrt{n}}\ln\left({1+{2 \over n-1}} \right)\sim \frac{2}{n\sqrt{n}}=\frac{2}{n^{3/2}}$$
where we used that $\ln(1+x)\sim x$ when $x\to 0^+$.
It is similar to your approach... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2038591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the integral $\int_0^1 e^{-x^{4}}(1-x^{4}) dx $ How to find the definite integral of $$\int_0^1 e^{-x^{4}}(1-x^{4})dx $$ I tried solving this by using integration by parts and then by substitution but want able to solve this by either of those methods .
| Let $y=x^{4}$
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{-3/4} dy
= \frac{1}{4} \gamma\left(\frac{1}{4},1 \right)
\end{equation}
Using the same substitution, we also have
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} x^{4} dx = \frac{1}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2043049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to show $\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+...\right)^2 = 1 +\frac{2}{3}x+\frac{7}{36}x^2+\frac{1}{30}x^3+...$? I tried to find the right handside of the equation by manipulating the series but I failed at getting the right handside of it.
$$\left(1+\frac{1}{3}x+\frac{1}{24}x^2+\frac{1}{300}x^3+... | We first calculate the Cauchy product/product of power series
\begin{align*}
\left(\sum_{n=0}^\infty \frac{2}{n!(n+2)!}\cdot x^n\right)^2 &= \sum_{n=0}^\infty \left(\sum_{k=0}^n \frac{2}{k!(k+2)!}\cdot \frac{2}{(n-k)!(n-k+2)!} \right)\cdot x^n\\
&= \sum_{n=0}^\infty \left(\frac{4}{(n+2)!^2}\cdot \sum_{k=0}^n \binom {n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2044953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
What is wrong with the "proof" for $\ln(2) =\frac{1}{2}\ln(2)$? I have got a question which is as follows:
Is $\ln(2)=\frac{1}{2}\ln(2)$??
The following argument seems suggesting that the answer is yes:
We have the series $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$
which has a mathematically determined value $... | You may not rearrange the series if it's not absolutely convergent. Otherwise you can get results like these, and like $\sum_{n=0}^\infty (-1)^k = \frac{1}{2}$, which both are false (the last one being the famous Grandi's series, https://en.wikipedia.org/wiki/Grandi%27s_series). All of this is in the Riemann series the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2051691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
} |
Condition on a,b and c satisfying an equation(TIFR GS 2017) Let $a,b,c$ be positive real numbers satisfying $$(1+a+b+c)\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=16,$$ then $a+b+c=3$.
I thought about the application of the AM-GM-HM inequality, but in vain. I also thought about splitting 16 into factors and comp... | By the AM-HM inequality $$\frac{1 + a + b + c}{4} \geq \frac{4}{\frac{1}{1} + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$$ and equality holds if and only if $1 = a = b = c$. Since the problem statement tells us that equality holds we find $a + b + c = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2053861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the solution for $x$ with $0 \le x \lt 13$ so $13 \mid 3x^2+9x+7 $. Find the solution for $x$ with $0 \le x \lt 13$ so $13 \mid 3x^2+9x+7 $.
I got this question in my discrete mathematics class.
I don't really get the idea how 13 can divide $3x^2+9x+7 $
All of my friend told me to try all the number between 0 and... | Let's work in $\mod 13$. We want $3x^2+9x+7 \equiv 0 \pmod {13}$. Now $7 \equiv -6$ so you have $3x^2+9x-6 \equiv 0 \pmod {13}$. Because $\gcd(3, 13) = 1$, we can "divide both sides" by $3$ and get $x^2+3x-2 \equiv 0 \pmod {13}$
Notice that $-2 \equiv -28$, so $x^2+3x-28 \equiv 0 \pmod {13}$ which you can factor to $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2056673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Is $g(u)= \frac{E [ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} ] }{E [ \frac{1}{\sqrt{X}} e^{-\frac{u^2}{2X}} ]}$ decreasing in $u$ Let $X$ be a positive random variable, let us define a function
\begin{align}
g(u,a)= \frac{E \left[ \frac{1}{\sqrt{X}} e^{-\frac{a^2u^2}{2X}} \right] }{E \left[ \frac{1}{\sqrt{X}} e^{-\fra... | If by "monotonically decreasing" you mean decreasing for $u \in (-\infty, \infty)$, then the answer is trivially no: $g$ is an even function in $u$ on the reals, thus it is either constant or it has a local extremum. And we can already tell intuitively that $g$ is not constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
2-form integration example I encountered the following example and I'm not getting $8\pi$ as the answer. It seems to me that
$\int_{D} (x^2+y^2) \mathrm{dx}\wedge \mathrm{dy} = \int\int r^3 \mathrm{dr}\mathrm{d\theta}$
with $0<r<1$ and $0 < \theta < 2\pi$. This gives $\frac{\pi}{2}$. Am I missing something?
| You are right, the correct value is $\frac{\pi}{2}$, not $8\pi$. The wrong value is however the only error, the argument for transforming the integral into
$$\int_0^{2\pi} \int_0^1 r^3\,dr\,d\theta$$
is correct.
Let's go into "just stick to the definitions and plug things in" mode to verify the value of $\frac{\pi}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Inverse laplace 2/((s-1)^2+1)^2
Find inverse laplace transform of $$\frac{2}{((s-1)^2+1)^2}$$
I tried to decompose the fraction using
$$\because (s-1)^2+1=s^2-2s+2$$
$$\rightarrow \frac{2}{((s-1)^2+1)^2}=\frac{As+B}{s^2-2s+2}+\frac{Cs+D}{(s^2-2s+2)^2}$$
yet I get D=2, which leads me back to the same exact equation, a... | While there is a defined inverse Laplace transform, the integral is often difficult. Usually, it is easier to take the transforms we know and noodle with them until we find a fit.
$L\{e^t\cos t\} = \frac {s-1}{(s-1)^2+1}\\
L\{e^t\sin t\} = \frac 1{(s-1)^2+1} = \frac {s^2-2s +2}{((s-1)^2+1)^2}\\
L\{te^t\cos t\} = -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2057321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
what is the value of $ x$? If $\log_{2}3^4\cdot\log_{3}4^5\cdot\log_{4}5^6\cdot....\log_{63} {64}^{65}=x!$, what is the value of $ x$?
I've tried
$$\log_2 3^4\cdot\log_3 4^5\cdot\log_4 5^6\cdot.... \log_{63} 64^{65}$$
$$=\dfrac{4\log 3}{\log 2}\cdot\dfrac{5\log 4}{\log 3}\cdot\dfrac{6\log 5}{\log 4}\cdots\dfrac{65\log ... | You get $$4\cdot 5\cdot 6\cdots 65\cdot {\log 64\over \log 2}\\=1\cdot2\cdot3\cdot4\cdot5\cdots65=x!$$
Hence $x=65$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to find the value of an unknown exponent? E.g. I have the question:
$$2^{4x+1} = 128$$
I solved this by knowing that $128 = 2^7$ and therefore $x$ must equal $1.5$.
However, is there a way of solving this without knowing that $128 = 2^7$?
| ... and without logarithms or knowing any powers of $2$ other than the most trivial one ... \begin{align*}
2^{4x+1} &= 128 \\
2^{4x+1-1} = 2^{4x} &= 128/2 = 64 \\
2^{4x-1} &= 32 \\
2^{4x-2} &= 16 \\
2^{4x-3} &= 8 \\
2^{4x-4} &= 4 \\
2^{4x-5} &= 2 \\
2^{4x-6} &= 1 = 2^0 \text{,} \\
\end{align*}
so $4x-6 = 0$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2058765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 8,
"answer_id": 1
} |
Solve $\sqrt{3}^n=3^3$ I got answer but i could not understand, can anyone explain me ?
$\sqrt{3}^n=3^3$
$\implies{3}^{n/2}=3^3$
$\implies \frac{n}{2}=3\implies n=6$
How we get ${3}^{n/2}$ in the second line and how we get $\frac{n}{2}$ in third line.
| $ {\sqrt{3}}^{n} = 3^3$
$\Rightarrow$ $({\sqrt{3}}) ^{n} = 3^3$
$\Rightarrow$ $({3}^{1/2}) ^{n} = 3^3$
$\Rightarrow$ $({3}) ^{{(1/2)}.{n}} = 3^3$
$\Rightarrow$ $({3}) ^{n/2} = 3^3$
$\Rightarrow$ $n/2 = 3$
$\Rightarrow$ $n = 6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2060892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Evaluating $\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$ The following is a problem from an older exam which the instructor didn't provide solutions to.
Evaluate $$\int_0^{\infty} \frac{x^2}{x^5 + 1} \ dx$$
using only real-analytic techniques.
My work: this boils down to factoring $x^4 - x^3 + x^2 - x+ 1$ which I am... | Hint
$$x^4 - x^3 + x^2 - x+ 1=\left(x^2-\frac{\sqrt{5}+1}{2}x+1\right)\left(x^2+\frac{\sqrt{5}-1}{2}x+1\right)$$
Now apply the partial fraction decomposition method.
Edit
\begin{align*}x^4 - x^3 + x^2 - x+ 1&=x^2\left(x^2-x+1-\frac{1}{x}+\frac{1}{x^2}\right)\\
& =x^2\left((x^2+\frac{1}{x^2})-(x+\frac{1}{x})+1\right)\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Finding the linear combination of a vector that is in a span So say we have
Span S =
$
\begin{Bmatrix}
\begin{bmatrix}
1 \\
0 \\
1 \\
\end{bmatrix}
\begin{bmatrix}
-1 \\
1 \\
1 \\
\end{bmatrix}
\begin{bmatrix}
1 \\
... | $$\begin{pmatrix}-1\\4\\7\end{pmatrix}=3\begin{pmatrix}1\\0\\1\end{pmatrix}+4\begin{pmatrix}-1\\1\\1\end{pmatrix}.$$
$$\begin{pmatrix}-1\\4\\7\end{pmatrix}=-5\begin{pmatrix}1\\0\\1\end{pmatrix}+4\begin{pmatrix}1\\1\\3\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2061691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Asymptotic of integral I have this definite integral:
$$ \int_{0}^{\frac \pi 2} \left( 1 + t^2 \cot^2\theta \right )^{-\frac{n-1}{2}} d \theta , t\in (0,1)$$
where $n$ is an integer. I need to find the asymptotic as a function on $n$.
I suspect it should be $O(\frac{1}{\sqrt{n}})$ but wasn't able to complete the calcu... | *
*A behaviour as $n \to \infty$. Laplace's method ($3.$, p. $2$)
may be applied here, $$ \begin{align} \int_a^bf(x)e^{-\lambda
g(x)}dx\sim f(a)e^{-\lambda g(a)}\sqrt{\frac{\pi}{2\lambda
g''(a)}},\qquad \lambda \to \infty, \tag1 \end{align} $$ with
$g'(a)=0$, $g''(a)>0$, $f(a)\neq 0$.
One may write $$ \begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2062149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluating $\int_{-\infty}^{\infty}\frac{x^2 \cos(x)}{(x^2+1)^2}dx$ by method of residues. Im trying to solve $$\int_{-\infty}^{\infty}\frac{x^2 \cos(x)}{(x^2+1)^2}dx$$ using the method of residues. This function has two simple poles at $x=\pm i$ and so $$\int_{-\infty}^{\infty}\frac{x^2 \cos(x)}{(x^2+1)^2}dx=2\pi i\te... | Let $C$ be a closed contour comprised of $(i)$ the real-line segment from $-R$ to $R$ and $(ii)$ the semi-circular arc of radius $R$, centered at the origin, and in the upper-half plane. Then, we have for $R>1$
$$\begin{align}
\oint_C \frac{z^2e^{iz}}{(z^2+1)^2}\,dz&=2\pi i \text{Res}\left(\frac{z^2e^{iz}}{(z^2+1)^2},... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
An arctan exponential integral Prove the following for $\Re(z) >0$
$$ \log(\Gamma(z)) = 2\int_{0}^{\infty} \tan^{-1} \left(\dfrac{t}{z}\right)\dfrac{\mathrm{d}t}{e^{2\pi t} - 1} + \dfrac{\log(2z)}{2} + \left( z - \dfrac{1}{2} \right)\log (z) -z $$
I found this identity on Wolfram Functions here. I can't see where to s... |
Proposition : $$\int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right]$$
Proof : Let $ \displaystyle \text{I} (a) = \int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x$
$\displaystyle = -\sum_{r=1}^{\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2064719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integral $ \int_{0}^{\infty} \ln x\left[\ln \left( \frac{x+1}{2} \right) - \frac{1}{x+1} - \psi \left( \frac{x+1}{2} \right) \right] \mathrm{d}x $
Prove That : $$ \int_{0}^{\infty} \ln x\left[\ln \left( \dfrac{x+1}{2} \right) - \dfrac{1}{x+1} - \psi \left( \dfrac{x+1}{2} \right) \right] \mathrm{d}x = \dfrac{\ln^2 2}{2... | Two Auxiliary Identities
First, we shall establish two simple identities.
$\textbf{Identity }(*)$
$$\int^1_0\left[\frac{1}{\log{x}}+\frac{1}{1-x}\right]x^{s-1}\ dx=\log{s}-\psi_0(s)\tag{*}$$
$\text{Proof Outline: }$ Differentiate with respect to $s$, recognise the integral representation of the trigamma function, t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2065291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 1,
"answer_id": 0
} |
Prob. 4, Chap. 3 in Baby Rudin: How to show that these are the limit superior and the limit inferior? Here's Prob. 4, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:
Find the upper and lower limit of the sequence $\left\{ s_n \right\}$ defined by $$s_1 = 0; \ s_{2m} = \frac{s_{2m-... | Suppose there exists a sub sequence of $\{s_n\}$ converges to a point $s$. That is $\{s_{n_k} \}\to s$. Define $S_1$ as the set of terms with odd indices. That is, $n_k$ is odd and similarly $S_2$ for even terms. Note that either $S_1$ or $S_2$ has infinite elements, but not both. Hence, $s=1$ or $s=1/2$. Thus for any ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2067748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Equilateral triangle trisected In an equilateral triangle $ABC$ the side $BC$ is trisected by points $D$ and $E$. Prove $9|AD|^2 = 7|AB|^2$.
|
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = $\frac{BC}{2}$ = $\frac{a}{2}$
And using phythagorous theorem,
AE = $\sqrt{AB^2 - BE^2}$
AE = $\frac{\sqrt{3}a}{2}$
Given that, BD = $\frac{1}{3}$ BC
∴ BD = $\frac{a}{3}$
DE = BE - BD = $\frac a{2} - \frac a{3} = \frac a{6}$
A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
Let $a$ and $b$ be integers. Prove that if $10 \mid (a^2+ab+b^2)$, then $1000 \mid (a^3-b^3)$.
I saw that $a^3-b^3 = (a-b)(a^2+ab+b^2)$ and $(a^2+ab+b^2) = (a+b)^2-ab$. How can we use the fact that $10 \mid (a^2+ab+b^2)$ to solve this question?
| Clearly, $a^2+ab+b^2$ has to be even $\implies a,b$ both must be even(why?)
WLOG $a=2A,b=2B$
Now as $10|(a^2+ab+b^2),5|(a^2+ab+b^2)$
$$a^2+ab+b^2=(A+2B)^2+3A^2\implies(A+2B)^2\equiv-3A^2\pmod5$$
Again, for any integer $r,r\equiv0,\pm1,\pm2\pmod5\implies r^2\equiv0,\pm1$
$\implies(A+2B)^2\equiv0,\pm3\pmod5$ which is onl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Prove A New Method For Finding Primes. I was recently messing around with harmonic numbers - that is
$H_x = \sum^x_{k=1}\frac{1}{k}$ - and I was thinking what would happen if you applied Gauss' trick to the sum, which is to add up the first and last term, then the second and second-to-last term and so on. This is usual... | If you consider the $n$-th harmonic number, you add the terms $\frac{1}{k}$ and $\frac{1}{n - k + 1}$. Their sum is
$$\frac{1}{k} + \frac{1}{n - k + 1} = \frac{n - k + 1}{k(n - k + 1)} + \frac{k}{k(n - k + 1)} = \frac{n + 1}{k(n - k + 1)}$$
For $k = 1$ we get $\frac{n + 1}{n}$, which is a fraction in reduced form, sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum\limits_{cyc}\left(\frac{a+b}{a+b+c}\right)^2\geq\frac{16}{9}$ Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\left(\frac{a+b}{a+b+c}\right)^2+\left(\frac{b+c}{b+c+d}\right)^2+\left(\frac{c+d}{c+d+a}\right)^2+\left(\frac{d+a}{d+a+b}\right)^2\geq\frac{16}{9}$$
I tried C-S and more, but with... | HINT
Using substitutions $x =a+b+c, y = b+c+d, z = c+d+a, t = a+b+d$ we get $$(\frac{a+b}{a+b+c})^2=\frac 1 9(2 + \frac t x - \frac y x -\frac z x)^2$$
Similar for the other terms. Eliminate the squares then use the fact that $r + \frac 1 r \ge 2, r \gt 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2072190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Simplifying $9^{3/4}$, I get $3\sqrt[4]{9}$, but that's not the answer. Why? I am trying to simplify:
$9^\frac{3}{4}$
So this is what I did:
$9^\frac{3}{4} = \sqrt[4]{9^3}$
$\sqrt[4]{3*3*3*3*3*3}$
$3\sqrt[4]{3*3}$
$3\sqrt[4]{9}$
$3\sqrt[4]{3^2}$
I don't see how I can simplify this even more, however the answer I provid... | Simplify it this way:
$
9^\frac{3}{4} \\
= (3^2)^\frac{3}{4}\\
= 3^\frac{2\times 3}{4}\\
= 3^\frac{3}{2}\\
= 3^{\left(1 + \frac{1}{2}\right)}\\
= 3 \times 3^\frac{1}{2}\\
= 3 \sqrt{3}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Volume of sphere-cube intersection I am looking for a formula for the volume $V_i(r,L)$ of the intersection between a sphere of radius $r$ and a cube of edge $L$ with the same center.
For $r<\frac L 2$, $V_i$ is simply equal to the volume of the sphere:
$$V_i(r,L) = \frac 4 3 \pi r^3 \ \ \ \ r<\frac L 2$$
For $r>\frac{... | As John Hughes notes, you can fix the side of the cube or radius of the sphere and work in terms of the ratio. The strategy here is to fix the cube's edge length to be $2$, calculate the area of plane slices parallel to one face of the cube, then apply Cavalieri's theorem.
Theorem: The intersection of the square $[-1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Finding necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$ My question is that:
Find the necessary and sufficient condition for $x^3+y^3+z^3+kxyz$ to be divisible by $x+y+z$.
I know that answer is $k=-3$, as I know that $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
But I can't pr... | A low key answer is: We have $x^3+y^3+z^3+kxyz = (3+k)xyz + (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. Thus this means $(3+k)xyz$ must be divisible to $x+y+z$ for all $x,y,z$ integers. Put $x = y = 1$ we have $(3+k)z$ is divisible to $z+2$, thus you can write $(3+k)z = n(z+2)$, for all $z$ integer. Thus $n = \dfrac{(k+3)z}{z+2} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2077194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Let $P(x)=4x^2+6x+4$ and $Q(x) =4y^2-12y+25$. If $x,y$ satisfy the equation $P(x)Q(y)=28$, then find the value of $11y-26x$.
Let $P(x)=4x^2+6x+4$ and $Q(y) =4y^2-12y+25$. If $x,y$ satisfy the equation $P(x)Q(y)=28$, then find the value of $11y-26x$.
I know this question has been asked earlier here but the answer g... | It can be done by completing the square method,
$P(x)=4x^2+6x+4$
Let P(x) = 0.
Then,
$4x^2+6x+4 = 0$
First divide equation by 4,
So we have,
$x^2+\frac64x+1 = 0$
$x^2+\frac32x = -1$
The term of x is positive so compare it with,
$a^2 + 2ab + b^2 = (a + b)^2$
So, a = x,
And 2ab = $\frac32x$
2b = $\frac32$
b = $\frac34$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
finding value of $ \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$ finding value of $\displaystyle \int\frac{1}{x^3\sqrt{(1+x)^3}}dx$
Substituting $x=\tan^2 \theta\;, dx = 2\tan \theta \sec^2 \theta$
integral is $=\displaystyle \int \frac{2\tan \theta \sec^2 \theta }{\tan^6 \theta \cdot \sec^3 \theta}d\theta= 2\int\frac{\cos^6 \the... | A big partial fraction decomposition!
We set $u =\cos x$ to get $$I = -\int \frac{\cos ^6 x}{(\cos ^2x-1)^3} \sin x dx = \int \frac{u^6}{(u^2-1)^3} du =\int (\frac{3u^4-3u^2+1}{(u^2-1)^3} + 1) du = I_1 + I_2$$
For $I_1$, perform the partial fraction decomposition $$I_1 = \int \frac{3u^4-3u^2+1}{(u-1)^3(u+1)^3} du =-\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $A = \dfrac{f^2(1)+f^2(-1)}{2}$ is a composite number
Given $f(x) = a^{2016}x^2+bx+a^{2016}c-1$ where $a,b,c \in \mathbb{Z}$, suppose that the equation $f(x) = -2$ has two positive integer solutions. Prove that $A = \dfrac{f(1)^2+f(-1)^2}{2}$ is a composite number.
Let $g(x) = f(x)-2 = a^{2016}x^2+bx+a^{20... | The claim is not true: taking $a=1, \ b=2, \ c=3$, the equation $f(x) - 2 = 0$ is $x^2 + 2x = 0$, having both roots integers ($0$ and $-2$), but $A = \dfrac {5^2 + 1^2} 2 = 13$, clearly not a composite number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2078675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$? For which $n$ this inequality is always true $a^5+1\ge a^3+a^n$ for every $a>0$?
1.$1$
2.$2$
3.$3$
4.$4$
5.$5$
My attempt:It's clear that it is true for $n=2$.Because:
$(a^3-1)(a^2-1) \ge 0$
Is true because $a^3-1$ and $a^2-1$ are both n... | Clearly $\;n=5\;$ is out of question, as $\;a^3\le1\;$ isn't true for all $\;a>0\;$.
Something similar with $\;n=4\;$ , as then
$$a^5+1\ge a^3+a^4\iff (a+1)(a^4-a^3+a^2-a+1)\ge a^3(a+1)$$
which is false for values close to $\;1\;$ from the right, for example $\;a=1.1\;$ (check this).
With $\;n=3\;$ we'd get
$$a^5+1\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Is probability that $x<2$ equal to the probability that $x^2<4$ given $1Assuming $x$ is a real number uniformly distributed over the interval $(1,3).$
so $x^2$ is also uniformly distributed over the interval $(1,9)${As for every $x=a\in (1,3) $ there exists $x^2=a^2\in (1,9)$}.
Probability that $x<2$ would be $\frac{1}... | If $X$ is uniformly distributed on $[1,3]$, then $P(X^2\le a^2)=P(X\le a)=\frac{a-1}2$. Therefore, $P(X^2\le a)=\frac{\sqrt{a}-1}2$. Thus, the PDF of $X^2$ is $\frac1{4\sqrt{a}}$.
That is, $X^2$ is not uniformly distributed on $[1,9]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$
Find all natural numbers $k$ for which there exist natural numbers $n,m$ such that $m(m+k) = n(n+1)$.
Rearranging the given equation gives $k = \dfrac{n(n+1)}{m}-m$. Thus, $m \mid n(n+1)$ and so we have cases.
Case 1:... | For each positive integer $N$, let $\tau(N)$ denote the number of positive integers $d$ that divides $N$. For a given positive integer $u$, we also denote by $\bar{\tau}_u(N)$ the number of odd positive integers $d<u$ that divide $N$. If $N=2^r\tilde{N}$, where $r\in\mathbb{Z}_{\geq 0}$ and $\tilde{N}\in\mathbb{Z}$ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$. Find the number of three elements sets of positive integers $\{a,b,c\}$ such that $a\times b\times c=2310$.
My attempt:
Now, $2310=2\times 3\times 5\times 7\times 11=a\times b\times c$
$a=2^{x_{1}}3^{y_{1}}5^{z_{... | Since we are looking for a set, $\{2310,1,1\}$ is not an answer - that is, not all the prime factors are gathered in the same factor.
Then, to avoid double-counting, we can say, for example, that $11$ always divides $a$ and the largest prime factor $q$ that does not divide $a$, divides $b$.
So for $q=7$, we can distrib... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
When is the fraction $\frac{n+7}{2n+7}$ the square of a rational number?
Find all positive integers $n$ such that the fraction $\dfrac{n+7}{2n+7}$ is the square of a rational number.
The answer says $n = 9,57,477$, but how do we prove this? I wrote $\dfrac{n+7}{2n+7} = \dfrac{a^2}{b^2}$ for some $a,b \in \mathbb{Z^+}... | There are more solutions (n<1000000):
$$
\begin{array}{c}
9 \\
57 \\
168 \\
477 \\
2109 \\
5880 \\
16377 \\
71817 \\
199920 \\
556509 \\
\end{array}
$$
so you just plug these values of $n$ and check otherwise the question does not make sense. Also One solution in your list is missing for n<500.
Added after ob... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2083696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Integer solutions to nonlinear system of equations $(x+1)^2+y^2 = (x+2)^2+z^2$ and $(x+2)^2+z^2 = (x+3)^2+w^2$
Do there exist integers $x,y,z,w$ that satisfy \begin{align*}(x+1)^2+y^2 &= (x+2)^2+z^2\\(x+2)^2+z^2 &= (x+3)^2+w^2?\end{align*}
I was thinking about trying to show by contradiction that no such integers exi... | The differences between consecutive squares are consecutive odd numbers. So we have:
$y^2-z^2=$ odd number
$z^2-w^2=$ next odd number
Of course if these were reversed (if the word "next" were moved one equation up) then the solution would be simple; any three consecutive numbers would work. As it is, it's trickier, b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2085253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$ I need help understanding this:
Prove that $$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$$
$$(\sqrt n)^\frac{1}{n}\geq
(\sqrt1)^\frac{1}{n}=1, \forall n\in\Bbb N$$
By binomial theorem we get for $n\geq 2$
$$n=((\sqrt n)^\frac{1}{n})^n=[... | Since $n^{1/n}\ge 1$, then all of the terms in the binomial expansion
$$\left(1+(\sqrt[n]{n}-1)\right)^n=\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k$$
are non-negative. Therefore, we have for any $m\le n$
$$\sum_{k=0}^n\binom{n}{k}(\sqrt[n]{n}-1)^k\ge \sum_{k=0}^m\binom{n}{k}(\sqrt[n]{n}-1)^k$$
Taking $m=2$ and $n\ge 2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2086600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
roots of quadratic polynomials are positive integer numbers Find positive integer numbers $a,b,c$ such that all the roots of these quadratic polynomials are positive integer numbers
1) $x^2−2ax+b=0$
2) $x^2−2bx+c=0$
3) $x^2−2cx+a=0$
Sorry but I don't know what to do :(( please help me :((
| The only solution in positive integers is $a=b=c=1$.
For the equations to have integer roots, their discriminants must be perfect squares. Taking the first equation for example, this means $a^2-b$ must be a perfect square. But $a^2-b \lt a^2$ thus it must be less than or equal the next smaller perfect square below $a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Function is small $o$ of $x^2$ I have to solve the following exercise:
Give $a$, $b$, $c \in \mathbb R$ such that
$$\frac{1}{1-\cos x} = \frac{a}{x^2} + b +cx^2 + o(x^2)$$
for $x\to 0$.
Here's my attempt:
I know that $$\cos x = \frac{1}{2}\left( e^{ix} + e^{-ix}\right) = \frac{1}{2} \sum_{n=0}^\infty \frac{(ix)^... | Your expansion of $\cos(x)$ at $0$ is obviously wrong, since it implies $\lim_{x\to 0}\cos(x)=\frac 12$.
You needn't use the power series for $e^{ix}$. I'd suggest that you rather use Taylor expansion at order $6$ for $\cos$, since the derivatives at $0$ are very simple. This method yields $$\cos(x)=1-\frac{x^2}{2}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
About an integration technique My professor wrote this :
$\displaystyle \int x^2\log(\sqrt{1-x^2}) \,\mathrm{d}x = \frac{1}{3}\int \log(\sqrt{1-x^2}) \, \mathrm dx^3$
Can someone explain me what is going on with the integration variable ?
| $
\int x^2 \log \sqrt{1 - x^2} \ dx \\
= \int x^2 \log \sqrt{1 - x^2} \ \frac{dx}{dx^3} \ dx^3 \\
= \int x^2 \log \sqrt{1 - x^2} \ \frac{1}{3x^2} \ dx^3 \\
= \frac13 \int \log \sqrt{1 - x^2} \ dx^3
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proof of binomial formula to extract coefficients of a generating function A generating function ends up rearranged into a form:
\begin{align}
(1+x+x^2+\dotsb)^n&=\frac{1}{(1-x)^n}\\[6px]
&= 1+\binom{1+n-1}{1}x+\binom{2+n-1}{2}x^2+\binom{3+n-1}{3}x^3\\
&\phantom{=\;1}+\dots+\binom{r+n-1}{r}x^r+\dotsb
\end{align}
used t... | The statement is obviously true for $n=1$; suppose it is for $n$; then we can see that
\begin{align}
(1+x+\dotsb)^{n+1}=\frac{1}{n}D\frac{1}{(1-x)^n}=
\frac{1}{n}\sum_{k\ge1}k\binom{k+n-1}{k}x^{k-1}=
\sum_{k\ge0}\frac{k+1}{n}\binom{k+n}{k+1}x^k
\end{align}
and it's just a matter of proving that
$$
\frac{k+1}{n}\binom{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Differentiating $y=x^6e^{-4x^3}$ Can some one explain, how to solve this derivative.
I'm total beginner.
Would be preferable if some one explain step-by-step.
$$y=x^6e^{-4x^3}$$
| Breaking everything, let me start by remembering a couple of rules, if $f, g$ are functions:
$$(fg)' = f'g + fg' \\ [f(g)]' = f'(g)\cdot g$$
those are just the rules of the derivatives of the product and of the composition.
Let $f(x) = x^6$ and $g = e^{-4x^3}$. What you have is
$$y = fg$$
Therefore differentiating it g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that a limit of a sequence is zero Let $(a_n)$ positive sequence where $$\dfrac{n-1}{n} \leq \dfrac{a_{n+1}}{a_n} \leq \dfrac {n}{n+1}$$
Prove that: $\lim_{n\to\infty} a_n = 0$.
I was stuck here:
I: $$\lim_{n\to\infty} \dfrac{n-1}{n} = \lim_{n\to\infty}1-\dfrac{1}{n} = 1$$
II: $$\lim_{n\to\infty} \dfrac {n}{n+1} =... | Just observe that $$a_{n+1}\le \frac{n}{n+1}a_n\le \frac{n}{n+1}\cdot\frac{n-1}{n}a_{n-1}=\frac{n-1}{n+1}a_{n-1}\le\cdots\le \frac{1}{n}a_1\\a_{n+1}\ge \frac{n-1}{n}a_n\ge \frac{n-1}{n}\cdot\frac{n-2}{n-1}a_{n-1}=\frac{n-2}{n}a_{n-1}\ge\cdots\ge \frac{1}{n}a_2\\\implies \lim\sup_n a_n\le 0\le \lim\inf_n a_n\\\implies \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Problem with the area calculation I have to calculate the area of {$x^2+y^2+z^2\le1$}$\cap${$\frac{1}{2}\le z \le1$}.
So i try with the follow integral:
$\int_0^{2\pi}\int_0^{\sqrt{\frac{3}{4}}}\frac{2\rho}{\sqrt{4-\rho^2}}d\rho d\theta$
I got $z$ in function of $x$ and $y$
$z=\sqrt{1-x^2-y^2}$
Then i used the form... | Using spherical coordinates,
$$
x = \rho \cos \theta \sin \varphi
\\
y = \rho \sin \theta \sin \varphi
\\
z = \rho \cos \varphi\mbox{,}
$$
with $\rho \in (0 , \infty)$, $\theta \in (0 , 2 \pi)$ and $\varphi \in (0 , \pi)$, you can obtain, by the inequalities in
$$
A = \left\{(x , y , z) \in {\mathbb{R}}^3 : x^2 + y^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Write a product of two numbers in non decimal base If $A=(b-1) (b-1) (b-1)$ and $B=(b-1) (b-1)$ are written in base $b$, what is $A\times B$ in base $b$?
I've tried developing $A \times B$ in decimal base but couldn't get this product back in base $b$.
Beside this, is there a general procedure to compute product in n... | Let throw away the usual meanings of $8,9,10$ and replace them with $10$ means $b$, $9$ means simply $b -1$ and $8$ means $b-2$. We assume $b \ge 3$ and so if we were dealing with base $7$ or base $32$ then $9$ and $8$ means "six" and "five" in base $7$, and $9$ and $8$ means "thirty-one" and "thirty" in base $32$.
O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Decomposition of this partial function I came across this
$$\int \frac{dx}{x(x^2+1)^2}$$
in "Method of partial functions" in my Calculus I book.
The thing is that, however, decomposing $\frac{1}{x(x^2+1)^2}$ would take long in this way:
$$\frac{1}{x(x^2+1)^2}= \frac{A}{x}+\frac{Bx+C}{x^2+1}+\frac{Dx+E}{(x^2+1)^2}$$
... | Let $\displaystyle \mathcal{I} = \int\frac{1}{x(x^2+1)^2}dx$
substitute $\displaystyle x = \frac{1}{t}$ and $\displaystyle dx = -\frac{1}{t^2}$
So $\displaystyle \mathcal{I} = -\int\frac{t^5}{(1+t^2)^2}\cdot \frac{1}{t^2}dt = -\frac{1}{2}\int\frac{t^2\cdot 2t }{(1+t^2)^2}dt$
put $1+t^2=u\;,$ then $2tdt = du$
So $\displ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2096391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$ Differentiate by first principle,$f(x)=\frac{x^2}{\sin x}$
$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2}{\sin(x+h)}-\frac{x^2}{\sin x}}{h}$$
$$f'(x)=\lim_{h\to0}\frac{\frac{(x+h)^2\sin x-x^2\sin(x+h)}{\sin(x+h)\sin x}}{h}... | Note that $$\lim _{ h\to 0 } \frac { 2\sin ^{ 2 }{ \frac { h }{ 2 } } }{ h } =\lim _{ h\to 0 } \frac { \sin { \frac { h }{ 2 } \cdot \sin { \frac { h }{ 2 } } } }{ \frac { h }{ 2 } } =\lim _{ h\to 0 } \frac { \sin { \frac { h }{ 2 } } }{ \frac { h }{ 2 } } \sin { \frac { h }{ 2 } } =0$$
so
$$
\begin{align}
f'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Compute $5^{10,000}\equiv \mod 52$ I'm unsure how to solve questions like these, we are given an example below but I'm still confused about how they got from $5^{100\times10+0}$ to $5^{0}$.
ie) Compute $5^{1000} \bmod 77$
$$
5^{1000}=5^{(166⋅6+4)}=5^4=25^2=4^2=16=2 \pmod 7\\
5^{1000}=5^{(100⋅10+0)}=5^0=1 \pmod{11}\\
5^... | What you have to understand from the examples is that if $n$ is relatively prime to $m$, then there exists a exponent $k$ such that $n^k\equiv1\mod m$.
One exponent that works every time is $m-1$, but there may be others (divisors of $m-1$). For example, $5^6\equiv1\mod 7$, so
$$5{(166.6+4}\equiv {(5^6)}^{166}.5^4\equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to solve this equation? $a^2 \cdot \arcsin(\frac 4a)+4 \cdot\sqrt {a^2-16}=40$ Please, help me solve this equation:
$$a^2 \cdot \arcsin \left(\frac 4a \right)+4 \cdot\sqrt {a^2-16}=40$$
| There may not exist an explicit solution for $a$ in terms of elementary functions.
However, you can apply the Newton-Raphson Method (A numerical method) using a spreadsheet, or with more sophisticated software such as MATLAB:
$$a_{n+1}=a_n-\frac{f(a_n)}{f'(a_n)} \tag{1}$$
You have your function:
$$f(a)=a^2 \cdot \arcsi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$ So I am trtying to proof that $\frac{a}{a+3b+3c}+\frac{b}{b+3a+3c}+\frac{c}{c+3a+3b} \ge \frac{3}{7}$ for all $a,b,c > 0$. First I tried with Cauchy–Schwarz inequality but got nowhere.
Now I am trying to find a conve... | We need to prove that
$$\sum\limits_{cyc}\left(\frac{a}{a+3b+3c}-\frac{1}{7}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{2a-b-c}{a+3b+3c}\geq0$$ or
$$\sum\limits_{cyc}\frac{a-b-(c-a)}{a+3b+3c}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{1}{a+3b+3c}-\frac{1}{b+3c+3a}\right)\geq0$$ or
$$\sum\limits_{cyc}\frac{(a-b)^2}{(a+3b+3c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
What's the derivative of: $ \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$? let $ y=\displaystyle \sqrt{x+\sqrt{{x}+\sqrt{x+\cdots}}}$, i'm really interesting to know how do I find :
$\displaystyle \frac{dy}{dx}$ ?.
Note: I have used the definition of derivative of the square root function but i don't succed .
Thank you for any ... | Clarification to solve the question
The question posed is calcule Derivative $\sqrt{x+\sqrt{x+\sqrt{x+....}}}$
we put $y=\sqrt{x+\sqrt{x+\sqrt{x+....}}} $
Before calculating the derivative we simplify the expression $y$
$y^{2}=x+\sqrt{x+\sqrt{x+\sqrt{x+....}}}=x+y$
$y^{2}-y-x=0$
We solve the equation $y^{2}-y-x=0$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2101745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Two fair dice are rolled. What is the conditional probability that at least one lands on 6 given that the dice land on different numbers? $F =$ The probability that the dice land on different numbers
$F = {(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,1),(3,2),(3,4),(3,5),(3,6),(4,1),(4,2),(4,3),(4,5),... | You have made a mistake. In event E, $(6,6)$ is not an option, since it is given that dices are to land on different numbers. So you have only 10 such. So the probability is, going by your method, ${10\over{36}}\over{5\over6}$
This is $1\over3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2102523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
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.