Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integ... | A start: Rewrite as $2xy-101x-101y=0$ and then as $4xy-202x-202y=0$ and then as $(2x-101)(2y-101)=101^2$. There are not many ways to factor $101^2$.
Remark: The approach in the OP is fine, a little more complicated. If you go through the path outlined above, you will find that a couple of solutions were missed.
| {
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"url": "https://math.stackexchange.com/questions/1650875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How many $3$-tuples $(a, b, c) \in M^3$ are there such $a+b+c$ is even? The task is the following:
$M= \left \{ 1,2, ... 99,100 \right \}$
How many $3$-tuples $(a, b, c) \in M^3$ are there such $a+b+c$ is even?
I tried to solve it this way:
There are only two possibilities that $a+b+c$ is even:
First I look at this as ... | Your assumption that there are $3!$ of arranging each ordered triple $(a, b, c)$ is only valid if $a$, $b$, and $c$ are distinct. However, $(a, b, c) \in M^3$, so $a$, $b$, and $c$ need not be distinct.
Let's consider cases.
Three even numbers are selected:
*
*There are $50$ ordered triples of the form $(a, a, a)... | {
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"url": "https://math.stackexchange.com/questions/1651769",
"timestamp": "2023-03-29T00:00:00",
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Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
I understand that when factored out, you have $(x+3)^3 = x^3+9x^2+27x+27$ as opposed to the other side which is $x^3+27$, so even if you plug in $0$ the... | ${(x+3)^3 = x^3 + 3^3}$
Expand
${x^3 + 9x^2 + 27x + 27 = x^3 + 27}$
Cancel
${9x^2 + 27x = 0}$
Factor
${x(9x + 27) = 0}$
${9x(x + 3) = 0}$
Therefore ${x = 0, x = -3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0$,
If $$\sin A+\sin B+\sin C=\cos A+\cos B+\cos C=0,$$ prove that
$$\cos 3A+\cos 3B+\cos 3C=3\cos(A+B+C).$$
My solution:
From the given,
$$\cos^3A+\cos^3B+\cos^3C=3\cos A\cos B\cos C$$
Now,
L.H.S$$=\cos3A+\cos3B+\cos3C$$
$$=4\cos^3A-3\cos A+4\cos^3B-3\cos B+4\cos^3C... | $\cos (A+B+C)=\cos (A+B) \cos C- \sin (A+B) \sin C \\ = (\cos A \cos B -\sin A \sin B) \cos C -( \sin A \cos B +\cos A \sin B) \sin C \\ =\cos A \cos B \cos C- \sum_{cyc} \sin A \sin B \cos C $
$( \sin A+ \sin B)^2= (-\sin C)^2 , (\cos A +\cos B )^2=(-\cos C)^2 \\ \implies 2+ 2\sin A \sin B +2 \cos A \cos B =1 \\ \imp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What is wrong with this infinite sum We know that:
https://www.youtube.com/watch?v=w-I6XTVZXww
$$S=1+2+3+4+\cdots = -\frac{1}{12}$$
So multiplying each terms in the left hand side by $2$ gives:
$$2S =2+4+6+8+\cdots = -\frac{1}{6}$$
This is the sum of the even numbers
Furthermore, we can add it to itself but shifting th... | Lets try the same in a more general way: Instead of $-\frac{1}{12}$ I will use $C\in\mathbb{R}$.
So our infinite sum:
$$1+2+3+4+\cdots = C$$
Using the same technique you get these two equalities:
$$
2+4+6+8+\cdots = 2C
\\
1+3+5+7+\cdots = 2C
$$
And when you add them together:
\begin{align}
1+2+3+4+\cdots &= 4C
\\
C &= ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of ways to get 17 by rolling dice 4 times (with combinatorial argument)
A dice is rolled four times. In how many ways we can get total of 17?
We can solve this by finding coefficient of $x^{17}$ in the ordinary enumerator $(x+x^2+x^3+x^4+x^5+x^6)^4$. But I feel this requires little bit more effort.
We can dir... | Let $x_k$ be the outcome of the $k$th roll. Then we want the number of solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 17 \tag{1}$$
in the positive integers with the restriction that $x_k \leq 6$ for $1 \leq k \leq 4$.
Let $y_k = 6 - x_k$ for $1 \leq k \leq 4$. Then each $y_k$ is a non-negative integer satisfying... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve $ \sqrt{x^2 +\sqrt{4x^2 +\sqrt{16x^2+ \sqrt{64x^2+\dotsb} } } } =5\,$? How to find $x$ in:
$$
\sqrt{x^2 +\sqrt{4x^2 +\sqrt{16x^2+ \sqrt{64x^2+\dotsb} } } } =5
$$
| Hint: $~x+1=\sqrt{x^2+2x+1}=\sqrt{x^2+\sqrt{4x^2+4x+1}}=\ldots~$ Can you take it from here ? :-$)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to show $\lim\limits_{n\rightarrow \infty}\left(\raise{5pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $? How to show this is true?
$$\lim_{n\rightarrow \infty}\left(\raise{3pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $$
| Writing $I_n=\int_0^{\frac{\pi}{2}}\sin^n x \, dx$, one can establish:
*
*$I_0=\frac{\pi}{2}, I_1=1$
*$I_n=\frac{n-1}{n}I_{n-2}$
*$I_{2n}=\frac{\binom{2n}{n}}{2^{2n}}\frac{\pi}{2}, I_{2n-1}=\frac{1}{2n}\frac{2^{2n}}{\binom{2n}{n}},I_{2n+1}=\frac{1}{2n+1}\frac{2^{2n}}{\binom{2n}{n}}$
*$I_{n+1}\le I_n$ (as $0\le \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known. Is there a way to to find out what $a^3+b^3+c^3$ evaluates to, when the values of $abc$, $ab+bc+ca$ and $a+b+c$ are given?
Alternatively, is there a way to express $a^3+b^3+c^3$ in terms of the aforementioned expressions?
| Yes, indeed,
$$(a+b+c)^3$$
$$=a^3+b^3+c^3+3a^2b+3a^2c+3b^2a+3b^2c+3c^2a+3c^2b+6abc$$
$$=a^3+b^3+c^3+3a^2b+3a^2c+3abc+3b^2a+3b^2c+3abc+3c^2a+3c^2b+3abc-3abc$$
$$=a^3+b^3+c^3+3a(ac+ab+bc)+3b(ac+ab+bc)+3c(ac+ab+bc)-3abc$$
$$=a^3+b^3+c^3+3(a+b+c)(ac+ab+bc)-3abc$$
So
$$\color{red}{a^3+b^3+c^3=(a+b+c)^3+3abc-3(a+b+c)(ac+ab+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What's $\alpha+\beta$ if we have: $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$ ($\alpha$ and $\beta$ are Real) What's $\alpha+\beta$ if we have $\alpha^3-6\alpha^2+13\alpha=1$ and $\beta^3-6\beta^2+13\beta=19$? Here $\alpha$ and $\beta$ are real.
Firstly, I subtracted the two equations and got the... | In order to get rid of the quadratic terms we write $\alpha:=a+2$, $\beta:=b+2$. Then
$$1=\alpha^3-6\alpha^2+13\alpha=(a+2)^3-6(a+2)^2+13(a+2)=a^3+a+10\ ,$$
and similarly
$$19=\beta^3-6\beta^2+13\beta=b^3+b+10\ .$$
It follows that $a$ and $b$ satisfy the equations
$$a^3+a=-9,\quad b^3+b=9\ .$$
Now (an unexpected) symme... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Derivative of $4x^5 \tan(\frac{-1}{x})$ I can't seem to be able to get the correct signs for this derivative.
$\frac{d}{dx}(4x^5\tan(\frac{-1}{x}))$
Here's my work:
$= (4x^5)'\tan(\frac{-1}{x}) + 4x^5(\tan(\frac{-1}{x}))'$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\tan(\frac{-1}{x}))'$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\frac{-... | Hint $sec(-\theta)=sec(\theta)$ thats the only mistake.
| {
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"url": "https://math.stackexchange.com/questions/1661466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find a possible f(x) Do you think that it is possible to find a $f$ such that given a floating point constant $c \gt 0$ and an integer constant $n \gt 0$, then $\forall x_i \gt 1, i=1, 2,...,n$ :
$$f\left({1\over x_1}+{1\over x_2}+...+{1\over x_n}\right) = {1\over x_1 + c}+{1\over x_2 + c}+...+{1\over x_n + c}$$
| Update: Generalized for $n\geq3, c>0$, not just the case $n=3,c=1$.
For $n = 3, c>0, x_1 = x_2 = x_3 = 4c$:
\begin{gather}
\frac1{4c} + \frac1{4c} + \frac1{4c} = \frac3{4c}, \\
\frac1{4c+c} + \frac1{4c+c} + \frac1{4c+c} = \frac3{5c}.
\end{gather}
For $n = 3, c>0, x_1 = x_2 = 8c, x_3 = 2c$:
\begin{gather}
\frac1{8c} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there any formula for computing the product of $(n+1)(n-1)(n-3)(n-5)...(n+1-2k)$ I need to compute the product of $(n+1)(n-1)(n-3)(n-5)...(n+1-2n)$ in order to use the Stirling's formula to calculate, but I am stuck. Thanks!
I am a little bit confused now... should the last term of $w(n/2+1/2)$ be $(n+1-2n)$ or $(n... | While @alex.gordan beat me to a full proof that's simpler than mine functionally, I figured I would share the way I approached this problem!
First note that $(a)_k = a(a+1)(a+2)\cdots(a+k-1)$. Thus,
$$\begin{align}\left(\frac{1-n}{2}\right)_k & = \left(\frac{1-n}{2}\right)\left(\frac{1-n}{2}+1\right)\cdots\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac... | Before doing any substitutions are algebra, you can see that the value of this integral will be a finite positive number because you're integrating a positive function over a bounded interval.
Multiplying the numerator and denominator by $(1-v^2)(1+v^2)$, we get:
$$
\frac{\frac 2 {1+v^2}}{2+\frac{2v}{1-v^2}} = \frac{2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Althoug... | We'll prove that our inequality is true for all reals $x$, $y$ and $z$.
Indeed, let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Since $\prod\limits_{cyc}(x^2+xy+y^2)=27(3u^2v^4-u^3w^3-v^6)$,
we see that our inequality is equivalent to $f(w^3)\geq0$, where $f$ is a linear function.
Thus, $f$ get's a minimal value for an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Inequality $ab+bc+cd \geq 3abcd$ If $a,b,c,d$ are real numbers such that $a+b+c+d=6$ and $a^2+b^2+c^2+d^2=12$, then is this inequality true: $$ab+bc+cd\geq 3abcd$$
| Let $$a=\frac{5}{2},b=\frac{1}{2}\left(\frac{3}{2}-\frac{\sqrt{5}}{2}\right),c=2,,d=\frac{1}{2}\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)$$ we find $$a+b+c+d=6$$$$a^2+b^2+c^2+d^2=12$$ and $$ab+bc+cd=\frac{1}{8}\left(39-5\sqrt{5}\right)\approx 3.477$$ and $$3abcd=\frac{15}{4}=3.75$$ with $$ab+bc+cd<3abcd$$ so the inequ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$
This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.
| Integrate by parts
\begin{align}\int \sqrt{1 + \frac{1}{x^2}}\ dx
=x\sqrt{1 + \frac{1}{x^2}}+\int \frac{dx}{x^2 \sqrt{1 + \frac{1}{x^2}}}
=x\sqrt{1 + \frac{1}{x^2}}-\sinh^{-1}\frac1x
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
System of Equations , Finding value of coefficient so equations are inconsistent
Question; Calculate a value for the coefficient $'a'$ of $x$ so that the solutions to the three equations are inconsistent. Demonstrate the resulting system of equations are then inconsistent:
$$ x+2y+2z=6$$
$$ x-y+z=-4$$
$$ ... | Here is a computer doing basic algebra. It is tedious to do it by hand.
Leting $a= \frac{-28}{4}$ will reduce the system to $0x+0y+0za=1$ and $0 \neq 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$x^4 -10x^2 +1 $ is irreducible over $\mathbb Q$ I have seen the thread Show that $x^4-10x^2+1$ is irreducible over $\mathbb{Q}$ but this didn't really have a full solution.
Is it true that if it is reducible then it can be factored into a linear factor or quadratic factor in the form $x^2 - a$. Is $a$ in the rationals... | It's quite easy to find a factorization over $\mathbb{R}$:
\begin{align}
p(x)=x^4-10x^2+1
&=(x^2-1)^2-8x^2 \\[6px]
&=(x^2-2\sqrt{2}\,x-1)(x^2+2\sqrt{2}\,x-1) \\[6px]
&=(x-(\sqrt{2}+\sqrt{3}))(x-(\sqrt{2}-\sqrt{3}))\\
&\qquad \cdot (x-(-\sqrt{2}+\sqrt{3}))(x-(-\sqrt{2}-\sqrt{3}))\\
\quad
\end{align}
Our task is to show ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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Showing that the result of an equation may be a perfect square Consider an equation that, for a few select integers $x$, yields a value that is a perfect square. In the case of perfect square quadratics it is easy. For example $(x-2)^2$ will always yield a perfect square value. But what about for an equation like $(x-2... | We first show that $\gcd(x-2,x-3)=1$, which is to say that $x-2$ and $x-3$ have no common prime factor. Assume to the contrary that $x-2=pa$ and $x-3=pb$ for some prime $p \ge 2$ and integers $a$ and $b$. Now $$1 = (x-2)-(x-3) = pa-pb = p(a-b).$$ This implies $p \mid 1$, contradicting the hypothesis that $p$ is a prime... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| It seems you may not know l'Hôpital's rule. It's also possible without the need for differentiation. Multiply numerator and denominator with the so called conjugate expression of the numerator:
$$\frac{x^2 \sqrt{x+2} - 8}{4-x^2}
= \frac{\left( x^2 \sqrt{x+2} - 8 \right)\left( x^2 \sqrt{x+2} + 8 \right)}{\left(4-x^2\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) =... | By the mean value theorem, there exists $a<c<b$ such that
$$
\frac{b^{n+1}-a^{n+1}}{b-a}=(n+1)c^n.
$$
Hence,
$$
(n+1)a^n<\frac{b^{n+1}-a^{n+1}}{b-a}<(n+1)b^n.
$$
| {
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"url": "https://math.stackexchange.com/questions/1667450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
| Let $u=x^3+1$ then you have your $du=3x^2dx$, then from your original equation $x^2dx$ is just equal to $\frac{du}{3}$. Now solving for the whole equation that just equal to $\int\frac{du}{3u}$.
$\int\frac{du}{3u}=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}lnu+ C= \frac{1}{3}ln(x^3+1)+C.$
| {
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"url": "https://math.stackexchange.com/questions/1667729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make... | $$\frac{1+a+\ldots+a^{n-1}}n<\frac{1+a+\ldots+a^n}{n+1}\iff$$
$$\frac1n\,\frac{1-a^n}{1-a}<\frac1{n+1}\,\frac{1-a^{n+1}}{1-a}\iff$$
$$(n+1)(1-a^n)\stackrel{\text{since}\;1-a<0}>n(1-a^{n+1})\iff-na^n+1-a^n>-na^{n+1}\iff$$
$$n(a^{n+1}-a^n)>a^n-1\iff na^n(a-1)>(a-1)(a^{n-1}+a^{n-2}+\ldots+a+1)\iff$$
$$na^n>1+a+\ldots+a^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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How to evaluate this integral $\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$ How to evaluate this integral
$$\mathcal{I}=\int_{0}^{\infty }\frac{\ln\left ( 1+x^{3} \right )}{1+x^{2}}\mathrm{d}x$$
Mathematica gave me the answer below
$$\mathcal{I}=\frac{\pi }{4}\ln 2+\frac{2}{3}\pi \ln\left (... | We can attack this integral
$$I = \int_0^{\infty} dx \frac{\log{(1+x^3)}}{1+x^2}$$
by considering the complex contour integral
$$\oint_C dz \frac{\log{(1+z^3)} \log{z}}{1+z^2}$$
where $C$ is the following contour
This is a keyhole contour about the positive real axis, but with additional keyholes about the branch poin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
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Solve the following questions: Following are two questions and their respective answers(as given in textbook):
Q1. If $$ k=\sum_{r=0}^n \frac{1}{n \choose r}$$, then write $$\sum_{r=0}^n \frac{r}{n \choose r}$$ in terms of k.
Ans. $$\frac {nk}{2}$$
Q2. If $$x+y=1$$ find $$\sum_{r=0}^n r^{2} y^{n-r} x^r$$.
Ans. $$n(nx+y... | For the second one, start with:
$\begin{align}
\sum_{0 \le r \le n} x^r y^{n - r}
&= y^n \sum_{0 \le r \le n} \left( \frac{x}{y} \right)^r \\
&= y^n \frac{1 - (x/y)^{n + 1}}{1 - x/y} \\
&= \frac{y^{n + 1} - x^{n + 1}}{y - x} \\
x \frac{\mathrm{d}}{\mathrm{d} x}
\left(
x \frac{\mathrm{d}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is there another way to solve this equation? My problem is Solve the equation
$$2 \left(\sqrt{x^3-7 x^2+17x-14}+\sqrt{x^4-7 x^3+23x^2-37 x+28}\right)=4x^2-17 x+25.$$
And my solve.
We have
$$\sqrt{x^3-7 x^2+17x-14}= \sqrt{(x-2) \left(x^2-5 x+7\right)}\leqslant \dfrac{(x-2) + (x^2-5 x+7)}{2}= \dfrac{x^2 - 4x + 5}{2}.$$
A... | There is a very laborious way of doing it using successive squarings (which will create extra roots.
Starting with $$2(\sqrt A+\sqrt B)=C$$ a first squaring leads to $$4(A+B)-C^2=8\sqrt{AB}$$ and a second squaring leads to $$\Big(4(A+B)-C^2\Big)^2=64AB$$ Using $$A=x^3-7 x^2+17 x-14\qquad B=x^4-7 x^3+23 x^2-37 x+28\qqua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$a,b,c$ are real numbers $>0$. If $a+b+c=1$, show that $(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\ge\frac{100}{3}$ $a,b,c$ are real numbers $>0$. If $a+b+c=1$, show that
$$(a+\frac{1}{a})^2+(b+\frac{1}{b})^2+(c+\frac{1}{c})^2\geq\frac{100}{3}$$
| The function $f(x) =(x+x^{-1} )^2 $ is convex on $(0,\infty) $ hence $$ \frac{1}{3} (f(a) +f(b) +f(c))\geq f(\frac{a+b+c}{3}) =(3+3^{-1} )^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1671937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Hermite's identity for sum of floor function In Hermite's 1884 paper "Sur quelques conséquences arithmétiques des formules de la théorie des fonctions elliptiques", volume 5 of Acta Mathematica, pages 310-315, he proves what is often called "Hermite's identity" differently than the usual proof you'll find by Googling. ... | Assume that a function $f$ is given as a power series
$$ f(x) = A_0 + A_1x + A_2x^2 + \ldots; $$
then
$$ \frac{f(x)}{1-x} = A_0 + (A_0+A_1)x + (A_0 + A_1 + A_2)x^2 + \ldots $$
as can be verified easily by multiplying through by $1-x$.
Next
$$ f(x^a) = A_0 + A_1x^a + A_2x^{2a} + \ldots, $$
hence
$$ \frac{f(x^a)}{1-x} =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integration of $\frac{dx}{x\sqrt{\ln x^2 +6\ln x}}$ $\require{cancel}$
Hello everybody so this is my integration question. I have reached a point in my picture. I don't know how to continue. I appreciate the help.
Also , am I going in the right direction ??
Please note, square root is over $\ln x^2 +6\ln x$.
\begin{ali... | Set $t=\ln x$ and $dt=\frac{dx}{x}$
$$=\int\frac{dt}{\sqrt{t^2+6t}}\overset{\text{complete the square}}{=}\int \frac{dt}{\sqrt{(t+3)^2-9}}$$
Set $s=t+3$ and $ds=dt$
$$=\int\frac{ds}{\sqrt{s^2-9}}$$
Set $s=3\sec(p)$ and $ds=3\tan p \sec p dp$, then $\sqrt{s^2-9}=\sqrt{9\sec^2 p-9}=3\tan p$ and $p=\sec^{-1}\left(\frac s ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Integrating a function of tan inverse How would I carry out the following integration?
$$\int_0^12\arctan x^2$$
I tried to substitute $x^2 = \tan\theta$
From there I wrote the integral as:
$$\int_0^{\pi/4}\theta\cdot \frac{\sec^2\theta}{\sqrt{\tan\theta}}\ d\theta$$
Now, is my only option to use integration by parts? O... | Notice,you can use integration by parts as follows
$$\int_0^1 2\tan^{-1}(x^2)\ dx=2\int_0^1 \tan^{-1}(x^2)\cdot 1\ dx$$
$$=2\left(\tan^{-1}(x^2)\int 1\ dx-\int \left(\frac{d}{dx}(\tan^{-1}(x^2))\int 1\ dx \right)\ dx\right)_0^1$$
$$=2\left(x\tan^{-1}(x^2)-\int \frac{2x^2}{1+x^4}\ dx\right)_0^1$$
$$=2\left(x\tan^{-1}(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Isomorphism of a linear map of polynomials Let $V$ be the space of polynomials in one variable of degree $\leq3$.
Let $T$ be a map assigning to $p(x)$ the polynomial $ap(x)+(bx+c)p'(x)+(dx^2+ex+f)p''(x)$, for some real parameters $a, b, . . . , f.$
What are the set of parameters when $T$ is not an isomorphism?
I think ... | Let $T$ be $T(p(x))=a \cdot p(x)+(bx+c)\cdot p'(x)+(dx^{2}+ex+f)\cdot p''(x)$
Then, $T$ is a linear map as $T(k\cdot p(x))=k\cdot T(p(x))$ and $T(p(x)+q(x))=T(p(x))+T(q(x))$. So, $T$ can be expressed in the form of matrix(with respect to the basis $1, x, x^{2}, x^{3}$)
Thus, if we calculate $T$ with the fact that diffe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
| If you only need to know what $d$ is, then consider $z=\cos\theta+i\sin\theta$, in which case $$\left(z-\frac 1z\right)^7=(2i\sin\theta)^7=...+35z^4\left(-\frac 1z\right)^3+35z^3\left(-\frac 1z\right)^4+...$$
Hence
$$2^7i^7\sin^7\theta=...-35\left(z- \frac 1z\right)...=...-35.2i\sin\theta+...$$
therefore the required $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Factorize $(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$ I am unable to factorize this over $\mathbb{Z}:$
$$(x^2+y^2+z^2)(x+y+z)(x+y-z)(y+z-x)(z+x-y)-8x^2y^2z^2$$
Since, this from an exercise of a book (E. J. Barbeau, polynomials) it must have a neat factorization.
I tried to guess some factors by putting $x-y... | Note that
\begin{eqnarray}
&&(x^2+y^2+z^2)(x+y+z) (x+y-z) (y+z-x) (z+x-y)-8 x^2 y^2 z^2\\
&=&(x^2+y^2+z^2)(-x^4+2 x^2 y^2+2 x^2 z^2-y^4+2 y^2 z^2-z^4)-8x^2y^2z^2\\
&=&(x^2+y^2+z^2)[(-x^4+2 x^2 y^2-y^4)+(2 x^2 z^2+2 y^2 z^2-z^4)]-8 x^2 y^2 z^2\\
&=&(x^2+y^2+z^2)[z^4-(x^2-y^2)^2+2(x^2+y^2-z^2)z^2]-8 x^2 y^2 z^2\\
&=&[(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1678933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$
evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$
I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$
And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$
But how d... | $$\sum^{\infty}_{n=2} \frac{3}{10^n}=\frac{3}{100}+\frac{3}{1000}+...=0.03+0.003+...=0.0333...=\frac{0.333...}{10}=\frac{\frac{1}{3}}{10}=\frac{1}{30}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
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Value of an element of a symmetric matrix with given eigenvalue. Let \begin{bmatrix}
a & b & c \\
b & d & e \\
c & e & f
\end{bmatrix} be a real matrix with eigenvalues $1$,$0$ and $3$. If the eigenvectors corresponding to $1$ and $0$ are $(1,1,1)^{T}$ and$(1,-1,0)^{T}$ respectively, then what is the value... | We know trace $a+d+f=4$
Take help from equations you have got from $(A-\lambda I)X=0$
$a=b=d$ and $c=e$
Now $a+d+f=2a+f=4-----(1)$
Find $a$ which is not very hard
$a+b+c=2a+c=1------(2)$
$c+e+f=2c+f=1------(3)$
Solving $(2),(3)$ we have $4a-f=1----(4)$
Solving $1$ and $4$ we have $f=\dfrac{7}{3}$
$3f=7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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If $\sum_{n=-2}^{\infty}\cos^n x=8$, then find $x.$ Let $0<2x<\pi$. If $$\sum_{n=-2}^{\infty}\cos^n x=8,$$ then please find $x.$
I tried $\sum_{n=-2}^{\infty}\cos^n x=\frac{1}{1-\cos x}+\frac{1}{\cos x}+\frac{1}{\cos^2 x}=\frac{1+\cos x}{\cos^2 x \sin^2 x}=8.$ But I cant find $x.$
| Hint: If $-1 \lt \cos(x)\lt 1$ and $\cos(x)\ne 0$, we get that:
$$
\begin{align}
\sum_{n=-2}^{\infty} \cos(x)&=\frac{1}{\cos^2(x)}+\frac{1}{\cos(x)}+\frac{1}{1-\cos(x)}\\
&=\frac{1-\cos(x)+(1-\cos(x))\cos(x)+\cos^2(x)}{(1-\cos(x))\cos^2(x)}\\
&=\frac{1}{(1-\cos(x))\cos^2(x)}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1684045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Maximizing $\sqrt6xy+4yz$ Let $x, y, z$ be real numbers such that $x^2+y^2+z^2=1$.
Let $A$ be maximum value of $\sqrt6xy+4yz$. Find $2A^2-4$.
An initial approach was trying out the inequality $\text{RMS-AM-GM}$. Then I tried a parametric substitution: $x=\cos\alpha, y=\sin\alpha cos\beta, z=\sin\beta \sin\alpha$.
How w... | As $y$ is repeated in $\sqrt6xy+4yz,$
WLOG choose $y=\cos\alpha, x=\sin\alpha\cos\beta, z=\sin\beta\sin\alpha$
$\sqrt6xy+4yz=\cos\alpha\sin\alpha(\sqrt6\cos\beta+4\sin\beta)$
$=\dfrac{\sin2\alpha}2\cdot\sqrt{22}\cos\left(\beta-\arccos\sqrt{\dfrac6{22}}\right)$
$\le\dfrac{\sqrt{22}}2$
The equality occurs if $\sin2\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1685072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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$f(z)=\frac{1}{z}=u(x,y)+iv(x,y)$ - Circles of equations $u(x,y)=a$ and $v(x,y)=b$ are orthogonal curves. Let $f(z)=\frac{1}{z}=u(x,y)+iv(x,y)$. Show that the circles of equations $u(x,y)=a$ and $v(x,y)=b$ ($a\not=0$,$b \not=0$, constants) are orthogonal curves.
I am stuck for a while on this problem. How could do this... | One has $$\frac{1}{z} = \frac{1}{x+iy} = \frac{x}{x^2+y^2} +i \,\frac{-y}{x^2+y^2}.$$ Hence $u(x,y) = \frac{x}{x^2+y^2}$ and $v(x,y)=\frac{-y}{x^2+y^2}.$ The equation $u(x,y)=a$ gives $$ax^2+ay^2-x = 0.$$ Completing the square, you get $$\left(x-\frac{1}{2a}\right)^2+y^2 = \left(\frac{1}{2a}\right)^2.$$ Do the same wit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1686091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\int_{0} ^{1}{\sqrt{1-x^4}dx}=\frac{1}{12}\sqrt {\frac{2}{\pi}}(\Gamma\left(\frac{1}{4}\right))^2$? To prove ,
$$\int_{0} ^{1}{\sqrt{1-x^4}dx}=\frac{1}{12}\sqrt {\frac{2}{\pi}}(\Gamma\left(\frac{1}{4}\right))^2$$
When we substitute $x^4$ with t
we get the equation
$$\frac{1}{4}\int_{0}^1t^{\frac{-3}{4}}(1... | $$I=\frac { 1 }{ 4 } \int _{ 0 }^{ 1 } t^{ \frac { -3 }{ 4 } }(1-t)^{ \frac { 1 }{ 2 } }dt$$
Now, on using Beta function, we get $$I=\frac { 1 }{ 4 } B\left( \frac { 1 }{ 4 } ,\frac { 3 }{ 2 } \right) $$
Now, we use the relation of Beta and gamma function. $$I=\frac { 1 }{ 4 } \frac { \Gamma \left( \frac { 1 }{ 4 } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Palindromes on Keypad and divisibility by $111$ The integers 1 through 9 are arranged as follows on a rectangular keypad:
$\begin{array}{c c c}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{array}$
Consider the 6-digit palindromes formed by going back and forth across a diagonal, row, or column of the grid. (ex: 123321, 15995... | Here's the keypad :
$$\begin{array}{c c c}
1 & 2 & 3\\
4 & 5 & 6\\
7 & 8 & 9
\end{array}$$
Take a look at rows, columns and diagonals. What do they have in common? If you take any 3 numbers in a row, columns or diagonal, say $a\, b\, c$, then $b-a=c-b$. Let's call this difference $\epsilon:=b-a=c-b$.
Now take any pali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1689411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution,then all possible values of $a$ If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution, then all possible values of $a$ lie in the interval
$(A)(-1,0)\c... | First, let us see for which values of $a$ there will be solutions:
Since $\,0\leq t = \left\lfloor x \right\rfloor<1\,$ we have
\begin{align}
t &= \dfrac{1\pm \sqrt{1+3a^2}}{3} \in \left[0,1\right)
&\implies&& 0\le1\pm\sqrt{1+3a^2} < 3
\\
&\big(\text{subtract } 1 \text{ from each inequality}\big)
&\implies&& -1\le\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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If $tan^2 \theta = \frac{x}{y}$ how can we construct the angle $\theta$? If we are given the values of $x$ and $y$ and we know that $\tan^2 \theta = \dfrac{x}{y}$ is it possible for us to construct the angle $\theta$?
| Continuing from @egreg ‘s work
$\sin ^2 \theta = \dfrac {x}{x + y}$ and $\cos ^2 \theta = \dfrac {y}{x + y}$
Then, $\dfrac {y}{x + y} - \dfrac {x}{x + y} = \cos ^2 \theta - \sin ^2 \theta $
$∴\dfrac {y - x}{y + x} = \cos 2\theta$
Construct the above angle and then bisect it.
Note that this method can effectively avoid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find distance from point to line I am asked to find the distance between a point ( 5,1,1 ) and a line
$\displaystyle \left\{\begin{matrix} x + y + z = 0\\ x - 2y + z = 0 \end{matrix}\right.$
What ive done so far is to simplify by gauss elimination and I get to:
$\displaystyle \begin{bmatrix} 1 &1 &1 &0 \\ 0 &1 &0 &0 \... | A variational method :
Let $(x,y,z)$ the point on the line the closest from $(5,1,1)$. The distance is $D$.
$$D^2=(x-5)^2+(y-1)^2+(z-1)^2$$
For the smallest $D$, the differentiation leads to :
$$(x-5)dx+(y-1)dy+(z-1)dz=0$$
And on the line :
$$dx+dy+dz=0$$
$$dx-2dy+dz=0$$
This must be true any $dx,dy,dz$ so :
$\begin{V... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find the laurent series for $\frac{1}{z(z-2)^2}$ centered at z=2 and specify the region in which it converges. My attempt:
$$\frac{1}{z(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{A}{z}+\frac{B}{z-2}+\frac{C}{(z-2)^2}$$
$$\frac{1}{z(z-2)^2} = \frac{(1/4)}{z}+\frac{(-1/4)}{z-2}+\frac{(1/2)}{(z-2)^2}$$
This is where I get stu... |
The function
\begin{align*}
f(z)&=\frac{1}{z(z-2)^2} = \frac{1}{4z}-\frac{1}{4(z-2)}+\frac{1}{2(z-2)^2}\\
\end{align*}
has a simple pole at $z=0$ and a pole of order two at $z=2$.
Since we want to find a Laurent expansion with center $2$, we look at the other pole $0$ and see we have to distinguish two regions.
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Limit of $x(2\pi - 4\arctan(x))$ I am trying to decide the value of the limit stated below. But I am running out of ideas, since I can't figure out how to rewrite it in a beneficial way.
$$\lim_{x \to \infty} x(2\pi - 4\arctan(x))$$
My attempt
$\lim\limits_{x \to \infty} x(2\pi - 4\arctan(x))$
rewrite $2\pi$ as $4\arc... | L'Hopitals rule is more than sufficient here so this is overkill but you could also have arrived at the result by knowing the first term in the Laurent series for $\arctan(x)$ which says that
$\arctan(x) \approx \frac{\pi}{2} - \frac{1}{x} + \frac{1}{3x^3} + ...$ for large $x$. This is sort of like a Maclaurin series o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Insights needed for the following Lagrange Multipler problem Find the point of the paraboloid $z = \frac{x^2}{4} + \frac{y^2}{25}$ that is closest to the point $(3, 0, 0)$.
So this seems like a pretty standard question of Lagrange multiplier, except I ran into some problems but cant figure out where I went wrong.
Atte... | if $y=0$ then $z = \frac{x^2}{4}$ so $\lambda = -\frac{x^2}{2}$
so $$ 4(x-3) = -\frac{x^3}{2} \\ \implies x^3+8x-24 = 0 \\ \implies(x-2)(x^2+2x+12)=0$$
so your solution is the point $(2,0,1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the value of $(abc)-(ab+bc+ca)+3(a+b+c)$ is $0$ If the points $\big(\frac{a^3}{a-1}, \frac{a^2-3}{a-1}),(\frac{b^3}{b-1}, \frac{b^2-3}{b-1}) ,\big(\frac{c^3}{c-1}, \frac{c^2-3}{c-1}\big)$ are collinear for three distinct values of $a,b,c$ and $a,b,c\neq1$, then prove that the value of $(abc)-(ab+bc+ca)+3(a+b... | Let:
$$F=abc-(ab+bc+ca)+3(a+b+c) \ \ (0)$$
Let us consider the determinant
$$D=\begin{vmatrix}\frac{a^3}{a-1}&\frac{a^2-3}{a-1}&1\\\frac{b^3}{b-1}&\frac{b^2-3}{b-1}&1\\ \frac{c^3}{c-1}& \frac{c^2-3}{c-1}&1\end{vmatrix}=k\begin{vmatrix}a^3&a^2-3&a-1\\b^3&b^2-3&b-1\\ c^3&c^2-3&c-1\end{vmatrix} \ \ \ (1)$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int_a^{2a}\sqrt{2ax-x^2}\:dx$ where $a$ is a constant. Evaluate $$\int_a^{2a}\sqrt{2ax-x^2}\:dx$$ where $a$ is a constant.
I used substitution, $x=a(1-\sin k)$. This is my working
But I think I made some mistake since answer isn't correct. Please help.
|
Assume $a\ge0$:
$$\int_{a}^{2a}\sqrt{2ax-x^2}\space\text{d}x=\int_{a}^{2a}\sqrt{a^2-\left(x-a\right)^2}\space\text{d}x=$$
Substitute $u=x-a$ and $\text{d}u=\text{d}x$.
This gives a new lower bound $u=a-a=0$ and upper bound $u=2a-a=a$:
$$\int_{0}^{a}\sqrt{a^2-u^2}\space\text{d}u=$$
Substitute $u=a\sin(s)$ and $\tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1701333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is $P(|Y| > 1)$?
Suppose that $X_1,X_2,\ldots,X_{25}$ are independent random variables from $\mathcal{N}(1, 4)$. Let $X = \dfrac{1}{25} \sum\limits_{i=1}^{25} X_i$ and $Y =\dfrac{5}{2}X - \dfrac{2}{5}$. What is the probabili... | $Y$ is not a standard normal random variable!
If $X_i \sim N(1,4)$, then $X=\frac{1}{25}\sum_{i=1}^{25} X_i \sim N(1,\frac{4}{25}).$ So $Y$ is normal with mean
$$EY=\frac{5}{2} -\frac{2}{5}=\frac{21}{10},$$
and variance
$$\textrm{var}(Y)=\left(\frac{5}{2}\right)^2 \cdot \frac{4}{25} =1.$$
Can you finish it?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Find the sum of the distances of these four points The graphs of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points.Compute the sum of the distances of these four points from the point $(-3,2).$
$x^2+y^2+6x-24y... | We don't have to find the coordinates of the four points.
Since
$$(x^2+y^2+6x-24y+72)-(x^2-y^2+6x+16y-46)=0$$
$$\Rightarrow 2y^2-40y+118=0$$
we can set the four points as $(-3\pm\alpha,\beta),(-3\pm\gamma,\omega)$ where $$\beta+\omega=-(-40)/2=20,\qquad\beta\omega=118/2=59.$$
Now, noting that
$$(-3+\alpha)^2+\beta^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1702822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Series of squares of n integers - where is the mistake? Given the following two series:
$$1^3 + 2^3 + ... + n^3$$
$$0^3 + 1^3 + .... + (n-1)^3$$
I take the difference vertically of the two:
$$\left(1^3-0^3\right) + \left(2^3-1^3\right) + .... + \left(n^3-(n-1)^3\right)$$
This equals to $n^3$
If I now express this in su... | Your error is when you evaluate
$$\sum_{i=1}^n i = \frac{n(n-1)}{2}$$
The correct formula is
$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
plugging this, everything works.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1704944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Divisibility of $n^4 -n^2$ by 4 (induction proof) We have to show that $$ n^4 -n^2 $$ is divisible by 3 and 4 by mathematical induction
Proving the first case is easy however I do not know how what to do in the inductive step. Thank you.
| Since $a_n=n^4-n^2=n^2(n-1)(n+1)$, we can also write
$$
a_{n+1}=(n+1)^4-(n+1)^2=(n+1)^2n(n+2)
$$
Suppose $a_n=n^4-n^2$ is divisible by $3$; then $3$ divides one among $n$, $n-1$ and $n+1$. Since $n$ and $n+1$ appear in the decomposition of $a_{n+1}$. The only remaining case is when $3$ divides $n-1$; but in this case w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the limit as $n \to \infty $ of $n\ln\left(1+\frac{\ x}{n^2}\right)$ Find
$$\lim_{n\to \infty} n\ln\left(1+\frac{\ x}{n^2}\right)$$
My attempt: $\lim_{n\to \infty} n \left[\ln\left(\frac{\ n^2 +x}{n^2}\right)\right]$
= $\lim_{n\to \infty} n [\ln (n^2 +x) - \ln(n^2)]$
But I'm not sure how to get this out o... | $n\ln\left(1+\frac{\ x}{n^2}\right) = \frac{1}{n}n^2\ln\left(1+\frac{\ x}{n^2}\right) = \frac{1}{n} \ln \left( \left( 1+\frac{x}{n^2} \right)^{n^2} \right)$
If $n\to\infty$, then $ \frac{1}{n} \ln \left( \left( 1+\frac{x}{n^2} \right)^{n^2} \right) \equiv \frac{1}{n} \ln \left( e^x \right) \equiv \frac{x}{n} \equiv 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1705905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The graph of the equation $x+y=x^3+y^3$ is the union of The graph of the equation $x+y=x^3+y^3$ is the union of
$(A)$line and an ellipse$(B)$line and a parabola$(C)$line and hyperbola$(D)$line and a point
I tried to factorize the given equation.
$x^3-x+y^3-y=0$
$x(x^2-1)+y(y^2-1)=0$
The answer given is a line and an e... | $$(x+y)=x^3+y^3=(x+y)(x^2-xy+y^2)$$
It can be $\;y=-x\;$ , a straight line, or else, after cancelling:
$$x^2-xy+y^2=1\iff x^2-xy+y^2-1=0\iff$$
$$\left(x-\frac y2\right)^2+\frac34y^2=1\;\;(**)$$
Put now:
$$\begin{cases}u:=x-\frac y2\\{}\\v:=y\end{cases}\iff x=u+\frac v2\;,\;\;y=v$$
and you get the ellipse $\;u^2+\frac34... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reaso... | It is much shorter using equivalents:
$1+\sqrt n+\sqrt[3]n\sim_\infty\sqrt n$, hence $\;\ln(1+\sqrt n+\sqrt[3]n)\sim_\infty\frac12\ln(n)$.
Similarly, $\ln(1+\sqrt[3] n+\sqrt[4]n)\sim_\infty\frac13\ln(n)$, whence
$$\frac{\ln(1+\sqrt n+\sqrt[3]n)}{\ln(1+\sqrt[3] n+\sqrt[4]n)}\sim_\infty\frac{\frac12\ln(n)}{\frac13\ln(n)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1706508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Applying Euler's formula I try to solve the following task:
Show with Euler's formula (from complex analysis) that for $z\neq 2k\pi$ where $k\in\mathbb{Z}$ the following is true: $$\sum\limits_{v=0}^n \cos(vz) = \frac{1}{2}+\frac{1}{2}\frac{\sin(nz+\frac{z}{2})}{\sin \frac{z}{2}}$$
My attempt:
$$2\sum\limits_{v=0}^n \c... | $2\sum\limits_{v=0}^n \cos(vz) =\sum\limits_{v=0}^n (e^{ivz}+e^{-ivz}) = \frac{1-e^{(n+1)iz}}{1-e^{iz}}+\frac{1-e^{-(n+1)iz}}{1-e^{-iz}} = \frac{e^{\frac{n+1}{2}iz}}{e^{\frac{1}{2}iz}}(\frac{{e^{\frac{n+1}{2}iz}}-e^{\frac{-(n+1)}{2}iz}}{{e^{\frac{1}{2}iz}}-e^{\frac{-1}{2}iz}})+
\frac{e^{\frac{-(n+1)}{2}iz}}{e^{\frac{-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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find the limit of $\lim_{x \to 0} \frac{1}{8}\sin(x)^{1/x^2}$
$$\lim_{x \to 0} \frac{\sin(x)}{8}^{\frac{1}{x^2}}$$
$y=\lim_{x \to 0}\frac{1}{x^2}\cdot \lim_{x \to 0} \ln(\frac{\sin(x)}{8})=-\infty$
So we have $e^{-\infty}=0$?
| The limit is $0$.
We have for $0 < x < 1$ the inequality
$$-\log x = \int_x^1 \frac{dt}{t} \geqslant \frac{1-x}{1} \\ \implies \log x \leqslant x - 1$$
If $x > 0$, then as $x \to 0+$
$$\frac{\log \sin x}{x^2} \leqslant \frac{\sin x - 1}{x^2} \to -\infty.$$
If $x < 0$, then, extending into the complex plane, as $x \to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1707257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Limit of the sequence $\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right]$ How can we evaluate the following limit
$$
\lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\righ... | Using the first principle,
$$\begin{align} \\ & \lim_{n\to\infty}\frac{1}{n}\left[\log\left(\frac{n+1}{n}\right)+\log\left(\frac{n+2}{n}\right)+\dots+\log\left(\frac{n+n}{n}\right)\right]
\\ & =\lim_{n\to\infty}\frac{1}{n}\left[\log\left(1+\frac{1}{n}\right)+\log\left(1+\frac{2}{n}\right)+\dots+\log\left(1+\frac{n}{n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1708471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How one can show $(r!)^s$ divides $(rs)!$? I would like if anybody have suggestions to prove:
if $n = rs$
with: $r > 0$ and $s > 0$
then $(r!)^s \mid n!$
suggestions?
| Let illustrate it with an example. Say to prove that
$$
3!^4 | 12!
$$
We may analyze the products as follows:
$$
\eqalign
{
1 \cdot 2 \cdot 3 \cdot \ \ \ &\ \ \ \ \ 1 \cdot 2 \cdot 3 \cdot \cr
1 \cdot 2 \cdot 3 \cdot \ \ \ &\ \ \ \ \ 4 \cdot 5 \cdot 6 \cdot \cr
1 \cdot 2 \cdot 3 \cdot \ \ \ &| \ \ \ \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1710058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Let $BD$ be the internal angle bisector of $\Delta ABC$ with $D$ on $AC$. The incentre of $\Delta ABC$ is $(0,4)$ and $D$ is at $(1,3)$
Let $BD$ be the internal angle bisector of $\Delta ABC$ with $D$ on $AC$. The incentre of $\Delta ABC$ is $(0,4)$ and $D$ is at $(1,3)$. If $a,b,c$ are in arithmetic progression, find... | From the Bisector Theorem we have
\begin{align*}
\frac{|AD|}{|DC|}&=\frac{|AB|}{|BC|}=\frac{c}{a}\qquad\implies\qquad |AD|=\frac{c}a|DC|=\frac{c}a\left(b-|AD|\right)
\end{align*}
By solving for $|AD|$ the last equation we get
$$|AD|=\frac{bc}{a+c}$$
Since $a,b$ and $c$ are in arithmetic progression it follows that $a+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Relation between eigenvalues of symmetric $2\times2$ non-singular matrices The eigenvalues of a symmetric $2\times2$ non-singular matrix $$A=\left(\begin{matrix} a_{11} & a_{12}\\ a_{12} & a_{22}\end{matrix}\right)$$ are $\alpha_1$ and $\alpha_2$. How are the eigenvalues of
$$B=\left(\begin{matrix} b\cdot a_{11} & c\cd... | Let $\alpha_{1,2}(b,c)$ denote the eigenvalues of $B$. You have
$$
\alpha_1(b,c) + \alpha_2(b,c) = \operatorname{tr}(B) = b\operatorname{tr}(A) = b(\alpha_1+\alpha_2)
$$
and (which is the more messy part)
\begin{align*}
\alpha_1(b,c)\alpha_2(b,c)
&= \det(B) = b^2a_{11}a_{22} - c^2a_{12}^2 = b^2\left(\det(A)+a_{12}^2\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1711667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Product of its Prime Factors Given that $4095 = 8^4 - 1$ write $4095$ as a product of its prime factors.
I know how I could separate $4095$ into prime factors however I'm not sure how I could use $8^4 - 1$ to help me.
I could perhaps move the $1$ over to get $4096$ and then work from there...
| To be more systematic:
\begin{align*}
x^{12}-1 &= (x^6-1)(x^6+1)\\
&=(x^3-1)(x^3+1)(x^2+1)(x^4-x^2+1)\\
&=(x-1)(x^2+x+1)(x+1)(x^2-x+1)(x^2+1)(x^4-x^2+1)
\end{align*}
(As a polynomial it could be factored further).
Now set $x=2$. You get
$$4095=1\cdot 7\cdot3 \cdot 3 \cdot 5 \cdot13=3^2\cdot5\cdot 7\cdot 13.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Prove that $\lim_{x \to 5} \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}=2$ using $\delta-\epsilon$ method. I need to prove that Prove that $\lim_{x \to 5} \frac{2+\sqrt{x-3}-\sqrt{7-x}}{x-4}=2$ using $\delta-\epsilon$ method. So far, what I have is $\forall \epsilon >0$, there exists a $\delta>0$ such that if $0<|x-5|<\delta$,... | HINT:
$$\sqrt{x-3}-\sqrt{7-x}=\frac{2(x-5)}{\sqrt{x-3}+\sqrt{7-x}}$$
so that we have
$$\left|\frac{(-2x+10)+\sqrt{x-3}-\sqrt{7-x}}{x-4}\right|=2\frac{|x-5|}{|x-4|}\left|1-\frac{1}{\sqrt{x-3}+\sqrt{7-x}}\right|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Trigonometric substitution $\tan{\frac{x}{2}}=t$. What is $\cos{x}$ then? For example, the integral is:
$$\int \frac{\sin{x}}{3\sin{x}+4\cos{x}}dx$$
And we use the substitution: $\tan{\frac{x}{2}}=t$
Now, to get $\cos{x}$ in terms of $\tan\frac{x}{2}$, I first expressed $\cos^2\frac{x}{2}$ and $\sin^2\frac{x}{2}$ in te... | I hope it's better to write $$\sin x=A(3\sin x+4\cos x)+B\cdot\dfrac{d(3\sin x+4\cos x)}{dx}$$
and find $A,B$ by comparing the coefficients $\sin x,\cos x$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Proving that $\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$ I was wondering if anyone could give a hint on how to prove this expression, I have been stuck on it for hours. Thanks in advance!
Proving that $$\sum_{n=1}^{\infty}S(n,n-2)x^n = \frac{x^3(1+2x)}{(1-x)^5}$$ where $S(n,n-2)$ are the Stirling numb... | Given the known recurrence for Stirling No. of the $2$nd kind
$$
\left\{ \matrix{
n \cr
m \cr} \right\} = m\left\{ \matrix{
n - 1 \cr
m \cr} \right\} + \left\{ \matrix{
n - 1 \cr
m - 1 \cr} \right\}
$$
we obtain
$$
\eqalign{
& \left\{ \matrix{
n \cr
n - 1 \cr} \right\} = \left( {n - 1} \right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Average number of throws of 2 dice in order for the sum to be a prime number If I throw 2 dice, what's the average number of throws that I need to do in order for the sum of the 2 dice to be a prime number?
My attempt:
The probability that the sum will be a prime number is $\frac{4}{9}$. And, by applying the formula o... | Prime numbers between $2$ and $12$ are $2,3,5,7,11$
which can be obtained in $15$ ways as $\boxed{1-1}\boxed{ 1-2/2-1}\boxed{1-4/4-1,\; 2-3/3-2}\boxed{1-6/6-1,2-5/5-2,3-4/4-3}\boxed{ 5-6/6-5} $
Thus P(prime number) = $\dfrac{15}{36} = \dfrac5{12}$
and $E[X] = 1/p = \dfrac{12}5 = 2.4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1724987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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IVP solved with Laplace transformation - mistake? I want to solve following IVP with Laplace transformation.
\begin{align}
x''(t) + 2x'(t) + x(t) &=
\begin{cases}
0, & t < 0\\
1, & t \in (0;2)\\
3, & t > 2
\end{cases}\\
x(0) &= 0\\
x'(0)&=0\\
\end{align}
I've transformed left side of the equation like this:
\begin{alig... | $$
\frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2} + 2\frac{1}{s}e^{-2s} - 2\frac{1}{s+1}e^{-2s} - 2\frac{1}{(s+1)^2}e^{-2s}
$$
The inverse Laplace is :
$$
x(t) = u(t)- e^{-t}u(t) - te^{-t}u(t) + 2u(t-2) -2 e^{2-t}u(t-2) - 2e^{2-t}(t-2)u(t-2)
$$
Then
$$x(t) = 0 \forall t < 0 $$
$$
\int_{-\infty (or 0)}^{\infty} u(t) e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Find maximum $2\sin 5x-3\cos x$. Is it possible to find the maximum of $2\sin 5x-3\cos x $ without using calculus nor numerical methods?
I suspect there is a way to play around with trig identities until the expression is only in terms of either $\cos$ or $\sin$. I have done some trials but have only managed to make th... | We have
$$2\sin(5x)-3\cos(x)=\\2\sin(3x+2x)-3\cos(x)=\\2\sin(3x)\cos(2x)+2\cos(3x)\sin(2x)-3\cos(x)$$
We make use of the identities: $$\sin(3x)=3\sin(x)-4\sin^3(x)\\ \cos(3x)=4\cos^3(x)-3\cos(x)$$
and $$\sin(2x)=2\cos(x)\sin(x)\\\cos(2x)=1-2\sin^2(x)$$
Thus we obtain
$$2[(3\sin(x)-4\sin^3(x))(1-\sin^2(x))]+2[(4\cos^3(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Determine a matrix given two other matrices
Define the following matrices $C,P\in M_2(\mathbb{Z}_{26})$:
$$
C=\begin{pmatrix}22&13\\10&2\end{pmatrix},\quad P=\begin{pmatrix}6&21\\8&4\end{pmatrix}.
$$
Find an invertible matrix $A\in M_2(\mathbb{Z}_{26})$ such that $C=AP$.
My first attempt is to obtain $A$ as $CP^{... | I would rewrite this as $P^TA^T=C^T$ so it is more consistent with the way we normally view Gaussian elimination. Then
$$\begin{align}\begin{bmatrix}6&8&22&10\\21&4&13&2\end{bmatrix} & \xrightarrow{R_1\leftrightarrow R_2}\begin{bmatrix}21&4&13&2\\6&8&22&10\end{bmatrix}\\
& \xrightarrow{R_1\times5}\begin{bmatrix}1&20&1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Evaluate $\int_0^1 \sqrt{x^2+4x+1}\,dx$
Evaluate $\displaystyle\int_0^1 \sqrt{x^2+4x+1}\,dx$
I tried simplify by doing this:
$$\int_0^1 \sqrt{x^2+4x+1}\:dx=\displaystyle\int_0^1 \sqrt{(x+2)^2-3}\:dx$$
Then, by letting $t=x+2$, $dt=dx$
$$\int_0^1 \sqrt{t^2-3}\:dt=\displaystyle\int_0^1 \sqrt{(t-\sqrt 3)(t+\sqrt 3)}\:dt... | HINT:
Set $x+2=\sqrt3\sec u\implies\sqrt3\tan u=\sqrt{x^2+4x+1}$
$\int\sqrt{(x+2)^2-3}dx=\int\sqrt3\tan u(\sqrt3\sec u\tan u)du=3\int(\sec^3u-\sec u)\ du$
See How to integrate $\sec^3 x \, dx$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1731891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Trigonometric Indentities $\dfrac{1}{\cos x}+ \dfrac{2\cos x}{\cos 2x}$
If $x=\dfrac{\pi}{7}$ Prove that above expression is equal to $4$.
I worked a few steps and reached here:
$\dfrac{(4\cos^2 x -1)}{\cos x (2\cos^2 x-1)}$
Not able to proceed after this!!!
| For $x=\frac{\pi}{7}$ it follows
$$\cos (4x)=-\cos(3x)\tag{1}$$
Let $y=\cos x$, using two times the cosine double-angle formula we get
$$\cos (4x)=2\cos^2 (2x)-1=2(2y^2-1)^2-1=8y^4-8y^2+1\tag{2}$$
And, from the identity
$$\cos (3x)=4\cos^3x-3\cos x\qquad\text{i.e.}\qquad \cos(3x)=4y^3-3y\tag{3}$$
we get
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1737060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a,b,c>0\;,$ Then value of $\displaystyle \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\rfloor $,
If $a,b,c>0\;,$ Then value of $\displaystyle \bigg \lfloor \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\bigg\rfloor $, Where $\lfloor x \rfloor $ Rep. floor function of $x$.
$\bf{My\; Try::}$ Using $b+c>a$ and $c+a>b... | Hint a small trick is that if you take $3$ numbers greater than $0$ you will always have ywo such pairs whose sum is greater than $3$rd number so division of two pairs will give answer of division as $0<x<1$ thus the sum would always be less than $2$ implying iys sums floor function would be $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\int \frac{dx}{a+b \cos x}$ without Weierstrass substitution. I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique.
When $a,b=1$ we can just multiply the numerator and denominator... | If $a=b$ then you can modify the technique for $a=b=1$ slightly to obtain:
$\int \frac{dx}{b+b\cos x}=\int\frac{b-b\cos x}{(b+b\cos x)(b-b\cos x)}dx$
$=\int\frac{b-b\cos x}{b^2-b^2\cos^2 x}dx=\int\frac{b-b\cos x}{b^2(1-\cos^2 x)}dx=\frac{1}{b}\int\frac{1-\cos x}{\sin^2 x}dx$
Splitting the numerator, and further simplif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 4
} |
How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| A lot of these answers involve taking the limit as the term number approaches $\infty$, but this is pretty tedious sometimes, so I thought an alternative might help with the evaluation of the sum.
$$\sum_{n=1}^\infty \frac{1}{n(n+3)}=\sum_{n=1}^\infty \bigg(\frac{A}{n}+\frac{B}{n+3}\bigg)=\sum_{n=1}^\infty \bigg(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Convergence of $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ I need to prove $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges. The root test is inconclusive, so I check in W.A., $\sum \frac{1}{\sqrt{4n + 1}} - \frac{1}{\sqrt{4n + 3}}$ converges by comparison test, but I don't know what seri... | Asmptotics.
As $n \to +\infty$:
$$\begin{align}
(4n+1)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{1}{4n}\right)^{-1/2}
=\frac{1}{2\sqrt{n}}\left(1-\frac{1}{8n}+O(n^{-2})\right)
\\
(4n+3)^{-1/2} &= (4n)^{-1/2}\left(1+\frac{3}{4n}\right)^{-1/2}
=\frac{1}{2\sqrt{n}}\left(1-\frac{3}{8n}+O(n^{-2})\right)
\\
(4n+1)^{-1/2} - (4n+3)^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1743725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Closed form for the area of a convex cyclic n-gon, given the set of edge lengths Let's say we are given a set of positive reals, and we're told that these are the edges of a convex cyclic $n$-gon, and we must compute it's area.
For $n = 3$ there is the famous Heron's formula:
$$A = {1\over 4}\sqrt{4a^2b^2 - (a^2 + b^2... | In case the links go down (shout-out to Martin Buettner), at least for the pentagon we can triangulate it trigonometrically. But beware: it gets a bit complicated. What we see for the pentagon heralds the complexity of the problem for larger numbers of sides.
Let $ABCDE$ be the pentagon, with sides $AB=a, BC=b, CD=c,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1745620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
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How to compute the following series using taylor expansion manipulation? How to compute $\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}$ and $\sum^{\infty}_{n=0}(-1)^n \frac{(n+1)x^{2n+1}}{(2n+1)!}$ using taylor expansion manipulation?
$1.\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}=\frac{1}{x^2}\sum^{\infty}_{n=0} \frac{x^{n+2}}... | There were some flaws in the development in the OP. Following that development, we write instead
$$\begin{align}
\sum^{\infty}_{n=0} \frac{x^n}{(n+2)n!}&=\frac{1}{x^2}\sum^{\infty}_{n=0} \frac{x^{n+2}}{(n+2)n!}\\\\
&=\frac{1}{x^2} \sum^{\infty}_{n=0} \int_0^x\frac{t^{n+1}}{n!}\,dt\\\\
&= \frac{1}{x^2}\int_0^x \sum^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Lucas numbers proof I'm running through some example problems and encountered this one:
Define a sequence of integers $L_n$ by $L_1=1, L_2=3, L_{n+1}=L_n+L_{n-1}.$ Show that $L_n = a\cdot \left(\frac{1+\sqrt{5}}{2}\right)^n + b\cdot \left(\frac{1-\sqrt{5}}{2}\right)^n$ and find the values of $a$ and $b$.
I wasn't able... | We, as usual, assume that there is a solution $L_n=\alpha^n$ so we have $\alpha^{n+1}=\alpha^n+\alpha^{n-1}$ hence that $$\alpha^2=\alpha+1$$ It follows
$$\begin{cases} \alpha_1=\frac{1+\sqrt 5}{2}\\ \alpha_2=\frac{1-\sqrt 5}{2}\end{cases}$$
Hence $L_n=(\frac{1+\sqrt 5}{2})^n$ and $L_n=(\frac{1-\sqrt 5}{2})^n$ are solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit $\lim_{n\to\infty}\sum_{k=0}^n\frac{n}{n^2+k}$ I came across this problem that I'm supposed to be able to solve in under 5 minutes (for a competition).
$$\lim_{n\to \infty} \sum_{k=0}^n \frac {n}{n^2+k}$$
I tried solving this for small sums, $\sum_{k=0}^2 \frac {n}{n^2+k}$, $\sum_{k=0}^3 \frac {n}{n^2+k} $ and m... | Note: $\frac{n}{n^2+n}\leq\frac{n}{n^2+k}\leq\frac{n}{n^2+1}\leq\frac{1}{n}$ for all $k$ between $1$ and $n$. Hence, $\frac{1}{n+1}\leq\frac{n}{n^2+k}\leq\frac{1}{n}$. Note that $\sum_{k=0}^{n}\frac{1}{n}=\frac{n+1}{n}$ and $\sum_{k=0}^{n}\frac{1}{n+1}=\frac{n+1}{n+1}=1$. Therefore: $$1\leq\lim_{n\to\infty}\sum_{k=0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1748429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Calculate $\sin\frac{3\pi}{14}-\sin\frac{\pi}{14}-\sin\frac{5\pi}{14}$ I have interesting trigonometric expression for professionals in mathematical science. So, here it is:
$$\sin\dfrac{3\pi}{14}-\sin\dfrac{\pi}{14}-\sin\dfrac{5\pi}{14};$$
Okay! I attempt calculate it:
\begin{gather}
\sin\dfrac{3\pi}{14}-\left(\sin\df... | Let $14x=\pi$
$$S=\sin3x-\sin x-\sin5x=\sin3x+\sin(-x)+\sin(-5x)$$
Using How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? ,
$$\sin(-2x)\cdot S=\cos5x-\cos7x$$
$$-\sin2x\cdot S=\cos5x=\sin2x$$
As $\cos7x=0,\cos5x=\sin2x$ as $5x+2x=\dfrac\pi2$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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There does not exist any integer $m$ such that $3n^2+3n+7=m^3$ I have this really hard problem that I am working on and I just don't seem to get it. The question is: let $n$ be a positive integer; prove that there does not exist any integer $m$ such that $3n^2+3n+7=m^3$.
Please explain in detail otherwise I might not u... | I am making this community wiki because I'm really just unpacking the comments. With the right method, this problem may seem trivially easy, but it still requires a few tedious calculations and the answer is not instantaneously obvious, despite what some people may want you to believe.
We know that $m^3 \equiv 0, 1, 8 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
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Solving $\int_{0}^{1} \dfrac{\ln(x+1)}{x^2+1} dx$ The integral is:
$$\int_{0}^{1} \dfrac{\ln(x+1)}{x^2+1} dx$$
This is a Putnam question from my calculus book. It was in the section about trig substitution. However, I see no way of how mixing a trig function with a natural logarithm would make it any easier. I've tried... | A substitution other than a trigonometric substitution can be used to evaluate this integral.
Using a so-called self-similar substitution of $u = \dfrac{1 - x}{1 + x}$, one has $x = \dfrac{1 - u}{1 + u}$. Thus $dx = -\dfrac{2}{(1 + u)^2} \, du$ and for the limits of integration when $x = 0, u = 1$ and when $x = 1, u = ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Calculating integrals using method of Cauchy residues Calculate by the method of residues
$$\int_{0}^{\infty} \frac {x^2 dx}{x^4+6x^2+13}.$$
I am stuck in finding the residues and the contour, any hint?
| Let $I$ be given by
$$I=\int_0^\infty \frac{x^2}{x^4+6x^2+13}\,dx$$
Enforce the substitution $x\to 1/x$ to reveal
$$I=\int_0^\infty \frac{1}{13x^4+6x^2+1}\,dx$$
Then, noting that the integrand is an even function, we can write
$$I=\frac12 \int_{-\infty}^\infty \frac{1}{13x^4+6x^2+1}\,dx \tag 1$$
Now, we move to the com... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show convexity of a function via inequalities I am stuck with deriving the convexity of the function
$$
f(x) = \sqrt{1 + x^2}
$$
from first principles, that is I would like to show that for any $x,y \in \mathbb R$ and $\lambda \in (0,1)$ we have
$$
f(\lambda x + (1 - \lambda)y) \le \lambda f(x) + (1 - \lambda) f(y)
$... | We need to check
$$\sqrt{1 + (\lambda x + (1-\lambda)y)^2} \leq \lambda \sqrt{1+x^2} + (1-\lambda) \sqrt{1+y^2}.$$
This is equivalent to
$$1 + (\lambda x + (1-\lambda)y)^2 \leq \lambda ^2 (1+x^2) + (1-\lambda)^2(1+y^2) + 2\lambda (1-\lambda) \sqrt{1+x^2}\sqrt{1+y^2},$$
or equivalently
$$1 + xy \leq \sqrt{1+x^2}\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I simplify $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ using long division? The infinite series $\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}\dots}$ is supposed to simplify to $1-\frac{x^2}{2}+\frac{7x^4}{48}+\frac{19x^6}{576}\cdots$ but I don't know how this was calculated.
Di... | Assume that:
$$\frac{1}{1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}+\ldots} = 1+a_2 x^2+a_4 x^4 + a_6 x^6 + \ldots $$
and multiply both sides by $1+\frac{x^2}{2}+\frac{5x^4}{48}+\frac{7x^6}{576}+\ldots$. If, after that, we consider the coefficient of $x^2$ in both sides, we get $\frac{1}{2}+a_2=0$, so $a_2=-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find and solve simultaneous recurrence relations for determining n-digit ternary sequences whose sum of digits is a multiple of 3 I'm studying recurrence relations, and I ran into the following problem:
Find and solve simultaneous recurrence relations for determining $n$-digit ternary sequences whose sum of digits is a... | Here is a different method that can help you verify your answer once
you have found it. The generating function of these sequences is
$$f(z) = (1+z+z^2)^n.$$
With $\rho = \exp(2\pi i/3)$ we can extract the multiples of three
using
$$\left.\frac{1}{3}(f(z)+f(\rho z)+f(\rho^2 z))\right|_{z=1}.$$
We get
$$\frac{1}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1754787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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proving the inequality $ (\frac{n}{e})^n \leq n! \leq en ( \frac{n}{e})^n$ by induction I want to prove $ (\frac{n}{e})^n \leq n! \leq en ( \frac{n}{e})^n$ by induction. For this prove I want to use the inequality $(\frac{n+1}{n})^n < e <(\frac{n+1}{n})^{n+1}$.
for $n=1$ the inequality is obvious. I Assume it is true ... | I had originally written this up for another question but it seems fitting here as well. Maybe this can help someone.
Depending on how you introduced $e$, you might be able to use the fact that there are two sequences $(a_n)_{n \in \mathbb{N}}$, $(b_n)_{n \in \mathbb{N}}$ with
$$\begin{align}
a_n ~~~&:=~~~ \left ( 1 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ is equal to :
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}})^{\frac{1}{n}}$ is equal
to :
*
*$\frac{9}{e^2}$
*$3 \log3−2$
*$\frac{18}{e^4}$
*$\frac{27}{e^2}$
My attempt :
$\lim_{n \to \infty} (\frac{(n+1)(n+2)\dots(3n)}{n^{2n}}... | Similarly as in this question we can use the fact that:
If $a_n$ is a sequence of positive real numbers and the limit $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists, then $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$ also exists and
$$\lim\limits_{n\to\infty} \sqrt[n]{a_n} = \lim\limits_{n\to\infty} \frac{a_{n+1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
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Help with finding limit of function with 2 variables Find $$\lim_{(x,y)\to(0,0)} \frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}$$
I have tried taking limits when $y=x$ and when $y=-x$ but failing to get close to an answer.
| *
*Method 1. One may write, as $x^2+y^2 \to 0$, by the use of Taylor series expansions, $$ \begin{align}
\sin(x^2+y^2)&=x^2+y^2+O((x^2+y^2)^3)
\\\\e^{\sin(x^2+y^2)}&=1+x^2+y^2+O((x^2+y^2)^2) \end{align} $$
giving, as $x^2+y^2 \to 0$, $$
\frac{e^{\sin(x^2+y^2)}-1}{x^2+y^2}=1+O(x^2+y^2). $$
*Method 2. One may ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$. Prove that $(2\sqrt3+4)\sin x+4\cos x$ lies between $-2(2+\sqrt5)$ and $2(2+\sqrt5)$.
Since we know that the minimum and maximum values of $a\cos x+b\sin x$ is $-\sqrt{a^2+b^2}$ and $\sqrt{a^2+b^2}$
I applied this formula to get the ... | You want to show $\sqrt{11+4\sqrt3} < 2 + \sqrt5 \iff 11+4\sqrt3 < 9+4\sqrt5 \iff 1 < 2(\sqrt5-\sqrt3) \\ \iff 1 < 4 (8-2\sqrt{15}) \iff 8\sqrt{15} < 31 \iff 960 < 961$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Why can we use the following integration method? In order to integrate the following function, our professor taught us the following method:
$\dfrac{x^4+1}{x^3+x^2} = \dfrac{x^4+1}{x^2(x+1)}=\dfrac{A}{x^2}+\dfrac{B}{x}+\dfrac{C}{x+1}$
I don't understand why we can do that, in reality, $\dfrac{x^4+1}{x^3+x^2} = \dfrac{x... | You can see that your partial fraction decomposition cannot work simply noting that the numerator of the fraction: $A(x+1)+Bx(x+1)+Cx^2$ cannot be a polynomial of degree $4$. In other words, if the denominators are the factors of a polynomial of degree $n$, and the numerators of the partial fractions are numbers, that ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1770016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Explicit functions evaluated (a) Defined $f$ by $f(y):=\int_0^\infty\frac{xy}{(x^4+y^4)^{3/4}}dx$. Prove $f(y)$ is defined (i.e integral exists) for every $y\in\mathbb{R}$.
(b)Prove that actually $f(y)=c\operatorname{sign} y$ for some positive number $c$. *So in particular $f$ is not continuous at $0$, therefore a subs... | Part (a)
Clearly $f(0) = 0$.
For $x \geqslant 1,$
$$\left|\frac{xy}{(x^4+y^4)^{3/4}} \right| = \frac{x|y|}{(x^4+y^4)^{3/4} } \leqslant \frac{x|y|}{x^3 } = \frac{|y|}{x^2},$$
and, since,
$$\int_1^\infty \frac{dx}{x^2} = 1 < \infty,$$
it follows by the comparison test that
$$\int_1^\infty\frac{xy}{(x^4+y^4)^{3/4}} \, dx... | {
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"url": "https://math.stackexchange.com/questions/1770278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $a^2b+b^2c+c^2a \leqslant 3$ for $a,b,c >0$ with $a^ab^bc^c=1$
Let $a,b,c >0$ and $a^ab^bc^c=1$. Prove that
$$a^2b+b^2c+c^2a \leqslant 3.$$
I don't even know what to do with the condition $a^ab^bc^c=1$. At first I think $x^x>1$, but I was wrong. This inequality is true, following by the verification from M... | We employ of the rearrangement inequality.
First, since $x\mapsto x^2$ preserves the order for $x>0,$ we have $$a^2b+b^2c+c^2a\le a^3+b^3+c^3.$$
Next write the condition as $$a\ln a+b\ln b+c\ln c=0.$$ Again using the rearrangement inequality, as $x\mapsto\ln x$ preserves the order for $x>0,$ we have the inequalities: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.
Prove that if $2n+1$ and $3n+1$ are both perfect squares then $40|n$.
First, I took
$$2n+1 \equiv x^2 \equiv 0, 1 \pmod 4$$
which showed that $n$ was even.
Now,
$$3n + 1 \equiv y^2 \equiv 0, 1, 4 \pmod 8$$ But since $n$ is even, we get that $8|n$.... | Your proof is good.
I suppose you could be more direct by pointing out:
For any natural $m$, $m^2 \equiv 0,1,4 \mod 8$ and $m^2 \equiv 0,1,-1 \mod 5$.
So $2n+1$ and $3n+1$ being perfect squares implies:
a) $2n \equiv -1, 0 ,3 \mod 8$.
As $2n$ is even $2n \equiv -1, 3 \mod 8$ are impossible so $2n \equiv 0 \mod 8$ so $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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A rational function integration Evaluate $$\int \frac{3x^2+1}{(x^2-1)^3}dx$$ I tried breaking the numerator in terms of the denominator but it didn't help much. I also tried a few substitutions buy most of them were useless. Please give some hints. Thanks.
| Partial fractions will do the trick.
Try write $$\frac{(3x^2+1)}{(x^2-1)^3}=\frac{ax^2+bx+c}{(x-1)^3}+\frac{ex^2+fx+g}{(x+1)^3}.$$
Further note that $(x+1)^3-(x-1)^3$ leaves only the quadratic and constant terms as non-zero.
So $$\frac{1}{(x-1)^3}-\frac{1}{(x+1)^3}= \frac{-6x^2-2}{(x-1)^3(x+1)^3} =\frac {-1}{2} \cdot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776283",
"timestamp": "2023-03-29T00:00:00",
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When are we able to find a quadratic with roots that are a function of another quadratic? Motivation:
Given the roots of the quadratic $2x^2+6x+7=0$ find a quadratic with roots $\alpha^2-1$ and $\beta^2-1$
I was able to solve this problem in two ways:
Method 1:
Sum of the roots $\alpha+\beta=-\frac{b}{a}$
Product of ro... | This is my solution:
Note that $2x^2+6x+7=0$ is equivalent to $x^2= -3x-\dfrac 72$
So
*
*$\alpha + \beta = -3$
*$\alpha \, \beta = \dfrac 72$.
*$\alpha^2 - 1 = -3\alpha-\dfrac 92$
*$\beta^2 - 1 = -3\beta-\dfrac 92$
We compute
\begin{align}
\hline
(\alpha^2 - 1) + (\beta^2 - 1)
&= (-3\alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to differentiate this fraction $\frac{2}{x^2+3^3}$? $\frac{2}{(x^2+3)^3}$.
I have ${dy}/{dx}$ x 2 x ${x^2+3^3}$ - 2 x ${dy}/{dx}$ x ${x^2+3^3}$ over $({x^2+3)^6}$
And then simplifying to $-12x^5 + 36x^2$ over $({x^2+3)^6}$
I'm not sure if this is right.
| Classical approach:
$$\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$$ whereas $u$ and $v$ are functions in $x$.
So when we have
$$\frac{2}{x^2+3^3}$$
We have
$$u(x) = 2$$
$$v(x) = x^2+ 3^3$$
And from that it follows
$$u'(x) = 0$$
$$v'(x) = 2x$$
Using the above formula
$$\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1776669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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For real numbers $a,b,c$ calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$ if we have... For real numbers $a,b,c$ we have: $a+b+c=11$ and $\frac1{a+b}+\frac1{b+c}+\frac1{c+a}=\frac{13}{17}$, calculate the value of: $\frac{c}{a+b}+\frac{a}{b+c}+\frac{b}{c+a}$
I think we should use a trick to solve thi... | $$\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{a+b+c}{b+c}+\frac{a+b+c}{a+c}+\frac{a+b+c}{a+b}-3=$$
$$=(a+b+c)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c} \right)-3=$$
$$=11\cdot\frac{13}{17}-3=\frac{92}{17}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the Lagrange multipliers with one constraint: $f(x,y,z) = xyz$ and $g(x,y,z) = x^2+2y^2+3z^2 = 6$ Where
$f(x,y,z) = xyz$
and the constraint is $g(x,y,z) = x^2+2y^2+3z^2 = 6$
I have tried this problem like three or four times and not gotten the solution, I even asked this question once and got the wrong answer from... | You asked where exactly you have made a mistake:
Multiplying equations 1,2,3 by $x,y,z$ we obtain that $2x^2=4y^2=6z^2$.
This is correct. (And, since you know what the solution should be, you can check for yourself, that the solution fulfills this equation.)
From here I found the corresponding multiples that $x^2$ ... | {
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"url": "https://math.stackexchange.com/questions/1784599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fractions in Questions and Answers
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