Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Prove that $a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2$ In thinking about a base case in this problem, I came up with the following question.
Given real numbers $a \ge b \ge c \ge d \ge 0$, prove that the following holds:
$a^2 - b^2 + c^2 - d^2 \ge (a - b + c - d)^2 \tag{A}$
My attempt:
After simplification, this redu... | A standard trick for a problem like this is to make a variable substitution which simplifies the problem in some way; hopefully you hit on a substitution which makes the problem more amenable. One idea for such a substitution is to simplify the condition that $a\geq b\geq c\geq d\geq 0$, and we can do this by defining ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/734007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compositions - Fruit Salad I'm asked to find $s(n)$ which is the number of ways to make a fruit salad with $n$ pieces of fruit, given that we must use strawberries by the half-dozen, an odd number of apples, between 2 and 7 bananas and at most 1 pineapple.
I started evaluating the ordinary generating function $S(X)$. I... | You can write:
\begin{align}
S(z) &= \frac{1}{1 - z^6} \frac{z}{1 - z^2} \frac{z^2 - z^8}{1 - z} (1 + z) \\
&= \frac{1}{1 - z^6} \frac{z}{(1 - z) (1 + z)}
\frac{z^2 (1 - z^6)}{1 - z} (1 + z) \\
&= \frac{z^3}{(1 - z)^2} \\
&= z^3 \sum_{k \ge 0} \binom{-2}{k} (-1)^k z^k \\
&= z^3 \sum_{k \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/735324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Which performance enhancement is better? Suppose you are given the task of improving the performance of a program consisting of three parts. Part $A$ requires 20% of the overall run time, part $B$ requires 30%, and part $C$ requires 50%.
You determine that for \$1000 you could either speed up part $B$ by a factor of 3.... | Let $x$ be the runtime of your program.
Then
\begin{align}
t(A) :&= a = 0.2 x\\
t(B) :&= b = 0.3 x\\
t(C) :&= c = 0.5 x\\
\end{align}
Speed something up by a factor of 2 (double as fast) means to reduce the runtime to half of the time it needed before. The thought behind that is that you can do twice as much in the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/736170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generating functions for compositions Let $g(n)$ be the number of compositions of n where each part is an odd number. Let $h(n)$ number of compositions of $n$ where each part is either 1 or 2. Using the ordinary generating functions $G(x)$ and $H(x)$, show that $g(n) = h(n-1)$
| Let $\mathcal{O}$ denote the class of all odd numbers, so that it has generating function $O(z) = z + z^3 + z^5 + \dots = \dfrac{z}{1-z^2}$. Then the class of compositions into odd parts is
$$\mathcal{G} = \operatorname{S\scriptsize EQ}(\mathcal{O}) \implies G(z) = \frac{1}{1-O(z)} = \frac{1}{1-\frac{z}{1-z^2}} = \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/737996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that if $a,b > 1$, then $5^a - 3^b=16$ has only one solution with $a=2$ and $b=2$ This may be one of those problems that is easy to state but very hard to prove. I don't know.
I have tried to show that there is only one solution but I have not made much progress.
Here's what I have:
If there exists $a,b > 2$, ... | Consider everything modulo $3$. Then,
$$(-1)^a \equiv 1 \pmod 3$$
From this, $a$ must be even, i.e. $a = 2m$ for some integer $m \ge 1$.
Now, consider everything modulo $5$. Then,
$$-(-2)^b \equiv 1 \pmod 5$$
From this (why?), $b$ must even, i.e. $b = 2n$ for some $n \ge 1$.
So we rewrite:
$$5^{2m} - 3^{2n} = 16$$
$$(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/740449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
An almost impossible limit The following limit appeared in a qualification exam: Find the limit of
$$\lim_{x \to 0} \left( \frac{\tan (\sin (x))-\sin (\tan (x))}{x^7} \right).$$
I ended up doing it in Mathematica, is there any other way?
Thanks in advance!
| I have not been able to find an elementary solution which avoids Taylor's series but I think we can provide much simpler approach in terms of calculations while following the Taylor's series approach.
Let $\tan x = t$ so that $\sin x = t/\sqrt{1 + t^{2}} = s$ and then $$\tan(\sin x) - \sin (\tan x) = \tan s - \sin t$$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/741446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
question on surds i already asked this question but the answer I got did not match the one in the book $$\sqrt{ 3x }= x + \sqrt {3}$$
Give x in the form
$$A \sqrt {B} + C $$
Can you show me how this is done step by step.
The answer I have in the book is:
$$\frac {1}{2} \sqrt{3} + \frac {3}{2} $$
this is where I got s... | $\sqrt{3x}=x+\sqrt{3}$
$3x=x^2+2x\sqrt{3}+3$
$x^2+(2\sqrt{3}-3)x+3=0$
$D=(2\sqrt{3}-3)^2-4 \times 1 \times 3=12-12\sqrt{3}+9-12=9-12\sqrt{3}<0$
The equation nas no real solutions.
you can check that the answer you have is not the solution at all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/743508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Green balls and Red balls, probability problem I'm studying for my exam and I came across the following draw without replacement problem :
$N$ boxes filled with red and green balls.
The box $r$ contains $r-1$ red balls and $N-r$ green balls.
We pick a box at random and we take $2$ random balls inside it, without putt... | All of the boxes contain $N - 1$ balls. This is just a complicated conditional probability problem. Lets look at a single box with $r$ red balls and $g$ green balls. What would the probability be of getting green on the second? Well it depends on whether or not you draw a red or green first. If you draw a red firs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/743819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How many surjective functions are there from $A=${$1,2,3,4,5$} to $B=${$1,2,3$}? I want to find how many surjective functions there are from the set $A=${$1,2,3,4,5$} to the set $B=${$1,2,3$}? I think the best option is to count all the functions ($3^5$) and then to subtract the non-surjective functions. However, I'm n... | Pedestrian approach:
How many surjective functions from:
$A$ ={ $1, 2, 3, 4, 5$} to $B$= {$a, b, c$} ?
1) Let $3$ distinct elements of $A$ be mapped onto $a, b$, or $c$.
There are
$\binom{5}{3} = 10$,
ways to pick $3$ elements from $5$.
There are $3$ ways to map these elements onto $a,b$, or $c$.
Altogether $3×10 = 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/744404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
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Calculate for $(1+\tan 20^\circ)(1+\tan 25^\circ)$. Help me with my works I have no idea what I am doing here,
I started with $\tan 20^\circ=\tan(45^\circ-25^\circ)=(1-\tan 25^\circ)/(1+\tan 25^\circ)$
I am sure the work I have shown so far are ok, but how do you get $1+\tan 20^\circ=2/(1+\tan 25^\circ)$ from that?
| It's a consequence of the following trigonometric identity
\begin{equation*}
\color{blue}{\left( 1+\tan a\right) \left( 1+\tan b\right) =2+\tan a+\tan b-\dfrac{\tan
a+\tan b}{\tan \left( a+b\right) }.}\tag{1}
\end{equation*}
On the one hand we rewrite
\begin{equation*}
\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/745929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Roots of Taylor's series.
Show that there is exactly one value of x which satisfies the equation $$2\cos^2 (x^3+x)=2^x+2^{-x} $$
I solved this using Taylor's series:
$$2^x+2^{-x}=2\{1+\frac {x^2 \{\ln2\}^2}{2!}+\frac {x^4 \{\ln2\}^4}{4!}....\} $$
$$\cos^2 (x^3+x)=1+\cos (2x^3+2x)=2-\frac{{2x^3+2x}^2}{2!}.... $$
Equat... | As for real $x, 2^{\pm x}>0$
$$\frac{2^x+2^{-x}}2\ge \sqrt{2^x\cdot 2^{-x}}=1\ \ \ \ (1)$$
So, we have $$2\cos^2(x^3+x)=2^x+2^{-x}\ge2$$
$$\iff \sin^2(x^3+x)\le0$$
For real $x,\displaystyle\sin^2(x^3+x)\ge0\implies$ the relation will hold iff
$\displaystyle2^x+2^{-x}=2$ and $\displaystyle\sin^2(x^3+x)=0\iff\sin(x^3+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/751770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Need help with basic factoring equation I'm just trying to brush up on my factoring of quadratic equations.
$$\frac1{x+3} + \frac1{x^2 + 5x +6}$$
$$\frac1{x+3} + \frac1{(x+2)(x+3)}$$
$$\frac{(x+2)(x+3) + (x+3)}{(x+2)(x+2)(x+3)}$$
Then I'm stuck. I think you can factor out the $(x+2)(x+3)$ from top and bottom?
| $$\begin{align}
&{1\over{x+3}}+{1\over{x^2+5x+6}}\to\\
&{1\over{x+3}}+{1\over{(x+2)(x+3)}}\to\\
&{x+2+1\over{(x+2)(x+3)}}\to\\
&{x+3\over{(x+2)(x+3)}}\to\\
&1\over{x+2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/755869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find three numbers given their sum, product and sum of their squares Given three unknown positive integers. Is it possible to find the three numbers if we are given their
Sum->(a+b+c) = X
Product-> (abc) = Y
Sum of Squares-> (a^2 + b^2 + c^2) = Z
| Here is a method you can try. Let
$s_1=a+b+c$,
$s_2=a^2+b^2+c^2$,
$p_2=ab+bc+ac$,
$p_3=abc$
$a, b, c$ are the roots of the cubic $$0=(x-a)(x-b)(x-c)=x^3-s_1x^2+p_2x-p_3$$
We don't know $p_2$ but can calculate it using $s_1^2=s_2+2p_2$, and then solve the cubic to find $a,b,c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/756175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How find this $x^3-5x+10=2^y$ let $x,y$ is positive integer,and such
$$x^3-5x+10=2^y$$
find all $x,y$.
since $$x=1\Longrightarrow 1^3-5+10=6$$ can't
$$x=2,2^3-5\cdot 2+10=8=2^3$$
so $x=2,y=3$
$$x=3,LHS=27-15+10=22$$
$$x=4,LHS=64-20+10=54$$
$$x=5,LHS=125-25+10=110$$
$$x=6,LHS=216-30+10=236$$
$$\cdots$$
I find $$(x,y)=(... | Some experimenting reveals that $\pmod{7}$ is the way to go:
$$x^3-5x+10=2^y$$
\begin{align}
\begin{array}{|c|c|}
\hline
x \pmod{7} & x^3-5x+10 \pmod{7} \\
\hline
0 & 3 \\
\hline
1 & 6 \\
\hline
2 & 1 \\
\hline
3 & 1 \\
\hline
4 & 5 \\
\hline
5 & 5 \\
\hline
6 & 0 \\
\hline
\end{array}
&
\begin{array}{|c|c|}
\hline
y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/756355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$ If $a^2+b^2=1$ where $a,b>0$ then find the minimum value of $(a+b+{1\over{ab}})$
This can be easily done by calculas but is there any way to do do this by algebra
| Edit: The last line is incorrect.
In light of @Lemur's answer, it will suffice to show that
$$
a + b \geq \sqrt{2}
$$
where $a,b > 0, a^2 + b^2 = 1$ is enforced.
To see this, note that by AMGM, $ab \leq \frac{a^2 + b^2}{2} = \frac{1}{2}$, and so
$$
(a + b)^2 = a^2 + b^2 + 2 ab \geq 1 + 2 ab \geq 1 + 1 = 2
$$
Thus $a +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/756702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ Another problem that I already wasted hours on.
Given
$$4\sinθ +3\cosθ = 5$$
Find
$$4\cosθ -3\sinθ$$
Help me guys (PS:I'm not that good in maths)
| Its simple:
Given
$$4\sin\theta + 3\cos\theta=5$$
Squaring on both sides gives
$$16\sin^2\theta +24\sin\theta \cos\theta+9\cos^2\theta=25$$
$$16-16\cos^2\theta+24\sin\theta \cos\theta+9-9\sin^2\theta=25$$
$$16\cos^2\theta-24\sin\theta \cos\theta+9\sin^2\theta=0$$
$$(4\cos\theta-3\sin\theta)^2=0$$
$$4\cos\theta-3\sin\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/757497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 14,
"answer_id": 12
} |
Series $\sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{4n+1}$ Does anyone know the sums of the following two series?
$$\sum_{n=0}^\infty (-1)^n \frac{x^{4n+1}}{4n+1}$$
$$\sum_{n=1}^\infty (-1)^{n+1} \frac{x^{4n-1}}{4n-1}$$
I encounter such series in my work.
| Well, to attack the first sum, if we call it $f(x)$, then
$$f'(x) = \sum_{n=0}^{\infty} (-1)^n x^{4 n} = \frac1{1+x^4}$$
so that
$$f(x) = \int_0^x \frac{dt}{1+t^4}$$
You can use partial fractions to deduce that
$$\begin{align}f(x) &= \frac1{2 \sqrt{2}}\int_0^x dt \, \left [\frac{t+\sqrt{2}}{t^2+\sqrt{2} t+1}-\frac{t-\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/758616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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A different type binomial expansion problem Suppose we have $$(1+x+x^2)^n = a_0 + a_1 x + a_2 x^2 + \cdots + a_{2n} x^{2n}.$$
What will be the value of $a_0^2 - a_1^2 + a_2^2 - \cdots + a_{2n}^2$?
The answer is $a_n$, but I can't solve it.
See, what I've done is substitute $x$ as $-\frac{1}{x}$ and I've got:
${\frac{(x... | Let $(1+x+x^2)^n=\sum_ka_kx^k$. Then:
\begin{align}
(1+x^2+x^4)^n&=(1-x^{-1}+x^{-2})^n(1+x+x^2)^nx^{2n}\\
\sum_ja_jx^{2j}&=\sum_k(-1)^ka_kx^{-k}\sum_ja_jx^jx^{2n}\\
&=\sum_j\sum_k(-1)^ka_ka_jx^{2n+j-k}\\
&=\sum_j\sum_k(-1)^ka_ka_{k+j-2n}x^j\\
\end{align}
The $x^{2n}$ coefficient on the left side is $a_n$; the same coef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/762762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Any one help me solve Algebraically equation $\sqrt{x^2+x+1}+ x^{3}= \sqrt{2x+2}+x^{2}+x$ I can't solve this. Can any one show me detail solution
$$
\sqrt{x^2+x+1}+x^3=\sqrt{2x+2}+x^2+x
$$
| This can be solved easily by observing the aid equation once you rearrange the terms.
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} = -x^3 + x^2 + x
$$
$$
\sqrt{x^2 + x + 1} - \sqrt{2x+2} = -x(x^2 -x -1)
$$
Notice that on the LHS, the difference between squares of the terms in under the root is RHS/$x$
Let's multiply and divide... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/764508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Expression generating $\left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \frac{33}{100} + \frac{333}{1000}, \dots \right)$ I'm looking for a closed-form expression (in terms of $n$), that will give the sequence
$$
(s_n) = \left( \frac{3}{10}, \, \frac{3}{10} + \frac{33}{100}, \, \frac{3}{10} + \f... | Let
$$S=0.3+0.33+0.333+0.3333+\cdots \text{ to $n$ terms}$$
$$=\underbrace{\frac {3}{10}+\frac {33}{100}+\frac {333}{1000} + \frac {3333}{10000}+\cdots}_n$$
$$=3\left(\frac {1}{10}+\frac {11}{100}+\frac {111}{1000} + \frac {1111}{10000}+\cdots\right)$$
$$=3\left(\frac {1}{10^1}+\frac {11}{10^2}+\frac {111}{10^3} + \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/765660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the axis of rotation of a rotation matrix by inspection (NOT by solving $Kv=v$) $$K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$ by INSPECTION.
This is from my other thread click here to view it
Everything you see below is me finding... | I can offer a remark on how you could simplify in some sense the calculation, but I'm not sure whether you would call the final result "inspection" and not "calculation".
Sometimes it easier to do mental calculations with matrices by thinking of the result of matrix multiplication by a vector as a linear combination of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/766565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Solve the initial value problem x'=Ax $A =
\begin{array}{cc}
-2 & 1 \\
5 & -4 \\
\end{array}$
$x(0) =
\begin{array}{cc}
1\\
3\\
\end{array}$
I undertsand that this is an eigenvector problem, and I got the values of $ -3+ \sqrt{6}$ and $ -3- \sqrt{6}$. I am unable to calculate the eigenvectors from here.
| Using the eigenvalue $\lambda = -3 - \sqrt{6}$, we find $[A-\lambda I]v_1 = 0$ and arrive at a RREF of:
$$\left(
\begin{array}{cc}
1 & \frac{1}{5} \left(-1+\sqrt{6}\right) \\
0 & 0 \\
\end{array}
\right)v_1 = 0$$
We choose $b = 1, a = \dfrac{1}{5} \left(1-\sqrt{6}\right)$, so:
$$v_1 = \left(\dfrac{1}{5} \left(1-\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\... | Setting $x=\tan2y$ $$F=\frac{\tan2y+\sec2y-1}{\tan2y+\sec2y+1}=\frac{\sin2y+1-\cos2y}{\sin2y+1+\cos2y}$$
Using Double Angle formula, $$F=\frac{2\sin y\cos y+2\sin^2y}{2\sin y\cos y+2\cos^2y}=\frac{2\sin y(\cos y+\sin y)}{2\cos y(\sin y+\cos y)}=\tan y$$ assuming $\cos y+\sin y\ne0\iff\tan y\ne-1\implies x=\tan2y\ne\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Definite Integral of $1+\sqrt{9-x^2}$? Right, so I'm to find the definite integral (interpreting it as an area)...
$\int^0_{-3}(1+\sqrt{9-x^2})dx$
How do I go about doing this?
I am to specifically use the following theorem to work it out...
If $f$ is integrable on $[a,b]$ then...
$\int^b_a f(x)\,dx = \lim_{n\rightarro... | The expression you are having trouble with seems to be incorrect.
You have $\mathrm{f}(x) = 1+\sqrt{9-x^2}$. Since $a=-3$ and $b=0$, you have $\displaystyle{\Delta x = \frac{3}{n}}$ and $\displaystyle{x_k = -3+\frac{3k}{n}}$.
Applying your definition:
\begin{eqnarray*}
\mathrm{f}(x_k) &=& 1+\sqrt{9-\left(-3+\frac{3k}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/769390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Show that $\int_0^\pi \log^2\left(\tan\frac{ x}{4}\right)dx=\frac{\pi^3}{4}.$ Hi I am trying to prove the relation
$$
I:=\int_0^\pi \log^2\left(\tan\frac{ x}{4}\right)dx=\frac{\pi^3}{4}.
$$
I tried expanding the log argument by using $\sin x/ \cos x=\tan x,$ and than used $\log(a/b)=\log a-\log b$, I get
$$
I=\int_0^\p... | Letting $x\mapsto \frac{1}{x},$ we have
$$
I=\int_0^\pi \ln ^2\left(\tan \frac{x}{4}\right) d x=4 \int_0^{\frac{\pi}{4}} \ln ^2(\tan x) d x
$$
By my post,
$$\int_0^{\frac{\pi}{4}} \ln ^2(\tan x) d x= \frac{\pi^3}{16} ,$$
we can conclude that
$$
\boxed {I=\frac{\pi^3}{4}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/770654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 4,
"answer_id": 3
} |
How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$? How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$?
I tried by using integration by parts, but always got sucked. Should be very easy...
| $$\int \frac{ \sqrt{x}}{1 - \sqrt{x}} = - \int \frac{\sqrt{x}}{\sqrt{x}-1} = - \int \frac{ \sqrt{x}-1+1}{\sqrt{x}-1}= - \int dx - \int \frac{dx}{\sqrt{x}-1}$$
But, $ d( \sqrt{x} - 1) = \frac{1}{2 \sqrt{x} } dx $. Hence
$$\int \frac{dx}{\sqrt{x}-1} = 2 \int \frac{\sqrt{x} d( \sqrt{x}-1)}{\sqrt{x}-1}=2 \int \frac{(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Integration by parts with log function $$\int2x^3\ln(1-x^2)\;dx$$
How do I integrate this equation?
Thanks.
| Use substitution rule. Let:\begin{align}u&=1-x^2\\x^2&=u+1\\du&=2x \, dx\end{align}
Then \begin{align}\int 2x^3 \ln(1-x^2) \, dx &= \int x^2\ln(1-x^2) \,2x \, dx \\
&=\int(u+1)\ln (u) \, du\end{align}
Now use integration by parts:
\begin{align}a=u+1 &\implies da=du\\
db=\ln(u) \, du &\implies b = u \ln(u)-u \end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inequality problem about sides of a triangle and the semiperimeter Let $a,b,c$ the sides of a triangle and $s$ be the semi perimeter. Then show that
$$
a^2+b^2+c^2 > \frac{36}{35}(a^2+\frac{abc}{s})
$$
I tried it doing in many ways using some changes but cannot help my cause.
| The inequality seems really loose to me, and this approach shows that it is indeed loose. ( which makes me somewhat doubt the inequality is what we want to show)
We want to show that
$$ 35 b^2 +35 c^2 > a^2 + 36 abc/s .$$
We have $b+c > a$ and so $ 2s > 2a $ and so $abc/s < bc $. We will show that
$$ 35 b^2 +35 c^2 > a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/777876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find Polynomial coefficients How can I find the real coefficients a,b if the polynomial
$ P(X)=X^4-5X^3+8X^2+aX+b $ is divisible by $ Q(X)=(X-1)^2 $
So if 1 is a solution, I get a+b=-4; where should i look for other roots?
| Using synthetic division:
\begin{array}{rrrrrrr}
& | & 1 & -5 & 8 & a & b \\
1 & | & 0 & 1 & -4 & 4 & a+4 \\
& + & --- & --- & --- & --- & ----- \\
& & 1 & -4 & 4 & a+4 & a+b+4 \\
\end{array}
So $a+b+4 = 0$.
Dividing again by $(x-1)$ :
\begin{array}{rcrrrr}
& ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\frac{1}{2\pi }\int_{-\pi }^{\pi }\dfrac{\sin n\theta }{\sin\theta }d\theta $? How to evaluate the integral given below?
$$\dfrac{1}{2\pi }\int_{-\pi }^{\pi }\dfrac{\sin n\theta }{\sin\theta }d\theta $$
| Since the integrand is $ 2 \pi$-periodic,
$$ \begin{align} \frac{1}{2\pi } \int_{-\pi }^{\pi } \frac{\sin n\theta }{\sin\theta } \ d\theta &= \frac{1}{2 \pi} \int_{-\pi/2}^{3 \pi/2} \frac{\sin n \theta}{\sin \theta} \ d \theta \\ &=\frac{1}{2 \pi} \int_{-\pi/2}^{3 \pi/2} \text{Im} \frac{e^{in \theta}}{\sin \theta} \ d ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/780913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
$\int_{0}^{\infty}\frac{x}{x^3+1}dx$ =? So guys I have this improper integral $\int_{0}^{\infty}\frac{x}{x^3+1}dx$. I checked that it converges by $ \int_{0}^{1}\frac{x}{x^3+1}dx + \int_{1}^{\infty}\frac{x}{x^3+1}dx $ and using the $\frac{c}{(x-a)^\lambda} $ and $\frac{1}{x^\lambda} $ criteria. But for finding the val... | One thing that simplifies the computation is splitting the integral and substituting $y = x^{-1}$ in one part:
$$\begin{align}
\int_0^\infty \frac{x}{x^3+1}\,dx &= \int_0^1\frac{x}{x^3+1}\,dx + \int_1^\infty \frac{x}{x^3+1}\,dx\\
&= \int_0^1 \frac{x}{x^3+1}\,dx + \int_1^0 \frac{y^{-1}}{y^{-3}+1}\,d(y^{-1})\\
&= \int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/782531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$ Hi I am trying to show this astonishing result$$
\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.
$$
Note we can to use $\ln(a/b)=\ln a-\ln b$ b... | Let
$$$$
\begin{align}
I(a)&=\int_0^1 2\, \log\frac{\big(x+a\, \sqrt{1-x^2}\big)}{\big(x-a\, \sqrt{1-x^2}\big)} \frac{x\, dx}{1-x^2} \tag 1\\
\therefore \frac{\partial}{\partial a}I(a) &= 2\, \int_0^1 \left(\frac{\sqrt{1-x^2}}{\big(x+a\, \sqrt{1-x^2}\big)}+\frac{\sqrt{1-x^2}}{\big(x-a\, \sqrt{1-x^2}\big)}\right) \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/782964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 2
} |
Limit when x approaches a number $\ne \infty$ I was rushing through a analysis task book when I suddenly encountered
$$\text{Find }~\lim_{x \to 3}\frac{x^2-9}{x-3}~\text{ by using the definition of limits.}$$
I thought, as the term equals to x+3, the limit should be 6.
Here's my beginning: I have to show, that $$\fora... | Since you want to prove the limit, $L=6$, you need to show $\forall\epsilon>0,\ \exists\delta>0$ such that $|x-3|<\delta\implies \left|\dfrac{x^2-9}{x-3}-6\right|<\epsilon$. It would be a good idea to start with the right side of the $\implies$ and work backwards, to show you can actually find such a $\delta$ that mak... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $y =\sqrt{5+\sqrt{5-\sqrt{5+ \cdots}}}$, what is the value of $y^2-y$? If $$y =\sqrt{5+\sqrt{5-\sqrt{5 + \cdots}}},$$
what is the value of $y^2-y$ ?
I am unable to get the clue due to these alternative signs of plus and minus please help on this thanks...
| Hint:
$$\begin{align}
(y^2 -5)^2 - 5 &= -y \\
y^4 - 10y^2 + 25 - 5 + y &= 0 \\
y^4 -10y^2 + y + 20 &= 0.
\end{align}
$$
Edit: To elaborate a bit on the comments given below: you have produced a depressed quartic equation. As explained on the Wikipedia article you can try to factor this into quadratics
$$
y^4 -10y^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the indefinite integral of...? I have spent well over an hour trying to solve this equation:
$$\int\cos^4{x}\sin^3{x}\, dx$$
I have tried substituting u as $\cos{x}$, $\sin{x}$, $\cos^4{x}$, and $sin^3{x}$ to no avail. How can I solve this?
| Through "brute force", you can just write out the values...there are several identities you need to get there:
$$
\sin(a)\sin(b) = \frac{\cos(a - b)-\cos(a + b)}{2} = \frac{\cos(b - a)-\cos(a + b)}{2} \\
\cos(a)\cos(b) = \frac{\cos(a - b)+\cos(a + b)}{2} = \frac{\cos(b - a) + \cos(a + b)}{2} \\
\sin(a)\cos(b) = \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Points on $(x^2 + y^2)^2 = 2x^2 - 2y^2$ with slope of $1$ Let the curve in the plane defined by the equation: $(x^2 + y^2)^2 = 2x^2 - 2y^2$
How can i graph the curve in the plane and determine the points of the curve where $\frac{dy}{dx} = 1$.
My work:
First i found the roots of this equation with a change of variable ... | You've already got some good suggestions on the graphing part using polar coordinates (if you are familiar with that). For the other, implicitly differentiate the given relation to obtain ${dy\over dx}$:
\begin{align*}
(x^2 + y^2)^2 &= 2x^2 - 2y^2\\
2(x^2+y^2)(2x+2y{dy\over dx})&=4x-4y{dy\over dx}\\
2x^3+2x^2y{dy\over ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785900",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Inverse Laplace Transform of $ \left(\frac{1-s^{1/2}}{s^2}\right)^2$ I found this question in my N.P Bali's Engineering Mathematics 7th Edition.
I could not find any solved questions related to this.
How can I find the Inverse Laplace Transform of :
$\left({1-s^{1/2} \over s^{2}}\right)^2$
I know only to find the inv... | The partial fraction expansion yields;
$$\left(\dfrac{1-s^{1/2}}{s^{2}}\right)^2 = -\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3} $$
Using this table of Laplace Transforms (item $6$ and item $3$ (twice)) yields:
$$\mathscr{L}^{-1} \left(-\dfrac{2}{s^{7/2}}+\dfrac{1}{s^4}+\dfrac{1}{s^3}\right) = -\dfrac{2 \times 2^3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Summation of Fibonacci numbers $F_n$ with $n$ odd vs. even Compare the summation below:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i-1}}&=F_1+F_3+F_5+\cdots+F_{2n-1}\\
&=1+2+5+\cdots+F_{2n-1}\\
&=F_{2n}\\
\end{align}
$$
with this one:
$$\begin{align}
\smash[b]{\sum_{i=1}^n F_{2i}}&=F_2+F_4+F_6+\cdots+F_{2n}\\
&=1+3+8+\... | Here is a late answer using my favorite Fibonacci technique.
The matrix formulation for the Fibonacci sequence is well worth knowing and easily proved:
$$
\begin{pmatrix}1&1\\1&0\end{pmatrix}^n=
\begin{pmatrix}F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}
$$
Let
$
A=\begin{pmatrix}1&1\\1&0\end{pmatrix}
$.
Since the second colu... | {
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"url": "https://math.stackexchange.com/questions/787341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Calculating integral I need some help to solve this integral:
$$\int(x^2+y^2)^{-\frac32} \mathrm dx$$
Thank you.
| Put $x = y \tan t$ and $dx = y \sec^2 t\ dt$, then
$$
\begin{align}
\int\frac{dx}{\sqrt{(x^2+y^2)^3}}&=\int\frac{y \sec^2 t\ dt}{\sqrt{(y^2\tan^2 t+y^2)^3}}\\
&=\int\frac{y \sec^2 t\ dt}{y^3\sec^3 t}\\
&=\frac{1}{y^2}\int\cos t\ dt\\
&=\frac{\sin t}{y^2}+C\\
&=\frac{x}{y^2\sqrt{x^2+y^2}}+C.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/787879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that $n$ is divisible by $6$ Problem:
Let $x^2+mx+n$ and $x^2+mx-n$ give integer roots where $(m,n)$ are integers. Show that $n$ is divisible by $6$
My attempt:
Since the roots are integers then the discriminants of both the equations should be perfect squares.
Let $a=\sqrt{m^2-4n}$ and $b=\sqrt{m^2+4n}$, then $... | We have $a^2 + b^2 = 2 m^2$, whose solution in the integers is
$$a = x^2+2xy-y^2, b = y^2 + 2xy - x^2, m = x^2 + y^2, \text { for } x, y \in \mathbb{N}$$
Hence, $ 4n = m^2 - a^2 = 4 xy (x-y) ( x+y)$
It is clear that $ xy (x-y)(x+y)$ must be a multiple of 6.
Proof of classification: (I'm slightly surprised I can't find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/788993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sin(2\theta) -\tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi $ I want to use the fact that $$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$
to solve $\sin(2\theta) -tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi$
My solution:
$\frac{2tan(\theta)}{1 + tan^2(\theta)} - tan(\theta) = 0 $
so $\fra... | Another way to solve:
Since identity $\sin (2\theta)=\dfrac{2\tan\theta}{1+\tan^2\theta}$, equation $\sin (2\theta)-\tan\theta=0$ is equivalent to
\begin{align}
\left(\dfrac{2}{1+\tan^2\theta}-1\right)\tan\theta&=0\;\;\;\text{ or}\\
\left(\dfrac{1-\tan^2\theta}{1+\tan^2\theta}\right)\tan\theta&=0\;\;\;\text{ or}\\
\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Proving some trig identities. Stuck on two questions.
First one: $\tan^2x - \sin^2x = (\sin^2 x)(\tan^2 x)$
Tried solving that with the right side but wasn't able to.
Second one: $\csc x / \sec x = \cot x$
I tried solving this one with the left side doing,
L.S = $\csc x / \sec x $
L.S = $(1 / \sin x) / (1 / \cos x) $
L... | *
*$$
\begin{align}
\tan^2x - \sin^2x&=\frac{\sin^2x}{\cos^2x}-\sin^2x\quad;\ \text{where}\ \tan x=\frac{\sin x}{\cos x}\\
&=\sin^2x\left(\frac{1}{\cos^2x}-1\right)\\
&=\sin^2x\left(\frac{1}{\cos^2x}-\frac{\cos^2x}{\cos^2x}\right)\\
&=\frac{\sin^2x}{\cos^2x}\left(1-\cos^2x\right)\\
&=\color{blue}{\tan^2x \cdot \sin^2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solving a fractional quadratic equation problem by completing the square I have the following problem to solve using the method of completing the square.
$$2x^2-3x-1 = 0$$
Here is where I've gotten to so far on this problem.
$$2x^2-3x = 1$$
$$x^2-\frac{3}{2}x = \frac{1}{2}$$
$$x^2-\frac{3}{2}x +\frac{9}{16} = \frac{1}{... | You made a mistake in the last line. It should read
$(x-\frac{3}{4})^2=\frac{17}{16}$...
This gives $(x-\frac{3}{4})^2 - \frac{17}{16} = 0$. Using the third binomial formula, we obtain $(x-\frac{3}{4} + \sqrt \frac{17}{16})(x-\frac{3}{4} - \sqrt \frac{17}{16}) = 0$.
Remembering that a product is zero iff one of its co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/790892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How prove this $\frac{r}{R}\ge\frac{AM}{AX}$ Let acute $\triangle ABC$ have largest angle $\angle A$, and let $R,r$ be the circumradius and inradius respectively. Let $M$ be the midpoint of $BC$, and $X$ be the intersection of the tangents at $B,C$ to the circumcircle of $\triangle ABC$
Show that:
$$\frac{r}{R}\ge\frac... | Let us identify the plane with the set of complex numbers $\Bbb{C}$, with $\Gamma$ being the unit circle. (So that $R=1$). The complex number representing a point will be denoted by the same, but small, corresponding letter.
Now, $X$ is the image of $M$ under the inversion with respect to $\Gamma$, hence
$$
m=\frac{b+c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/791263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Length of Curve $6xy=3+x^4$ Question :
Find the length of the curve $6xy=3+x^4$ between $x=1$ and $x=2$.
Answer = 17/12
I have tried this:
I obtain a different answer. Where did I do wrongly?
Thank you for your help.
| Your integration is not correct. You should simplify the integrand before proceeding. We have $$\mathscr{L}=\int_{1}^{2}\sqrt{1+\left(\frac{x^{4}-1}{2x^2}\right)^{2}}dx=\int_{1}^{2}\sqrt{\frac{(x^{4}+1)^2}{4x^4}}dx\\=\int_{1}^{2}\frac{x^{4}+1}{2x^2}dx=\frac{1}{2}\int_{1}^{2} x^2\,dx+\frac{1}{2}\int_{1}^{2} \frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/791329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find limit of this recursive sequence $$
a_0=0,\ a_1=2,\ a_{n+1}=\sqrt{2 - \frac{a_{n-1}}{a_n}} \\
\lim_{n\to\infty}2^na_n\ =\ ?
$$
| Let's prove by induction that
$$
a_n=2\sin\left(\frac{\pi}{2^n}\right)
$$
Let $\theta_n=\pi/2^n$. Clearly, $2\sin(\pi)=0=a_0$ and $2\sin(\frac{\pi}{2})=2=a_1$. Furthermore,
$$
a_{n+1}=\sqrt{2-\frac{a_{n-1}}{a_n}}=2\sqrt{\frac{1-\frac{a_{n-1}}{2a_n}}{2}}
$$
But
$$
\frac{a_{n-1}}{2a_n}=\frac{\sin \theta_{n-1}}{2\sin\thet... | {
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"url": "https://math.stackexchange.com/questions/795306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \lo... | View $I$ as a function of $a$, differentiate under integral sign and let
$x = \sin\theta$, we have
$$\begin{align}
I'(a) &= \int_0^1 \left( \frac{x}{1+ax} - \frac{-x}{1-ax}\right) \frac{dx}{x\sqrt{1-x^2}}
= \int_{-1}^1 \frac{dx}{(1+ax)\sqrt{1-x^2}}\\
&= \int_{-\pi/2}^{\pi/2} \frac{d\theta}{1+a\sin\theta}
= \frac12 \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/795493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 2
} |
Integral comparison test $$ \int_a^b \frac{x+1}{\sqrt[4]{x^5+x^2}} \,dx $$
I want to know this integral converges or not when $(a, b) = (0, 1), (1, \infty)$. I was thinking of using the comparison test, but I can't think of anything to compare it to.
Thanks in advance.
| For $(a,b)=(0,1)$ notice that
$$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\le\frac{x+1}{\sqrt[4]{x^{2}}}=\frac{x+1}{\sqrt{x}}=\sqrt{x}+\frac{1}{\sqrt{x}}$$
and for $(a,b)=(1,\infty)$ notice that
$$\frac{x+1}{\sqrt[4]{x^{5}+x^{2}}}\ge\frac{x}{\sqrt[4]{x^{5}+x^{2}}}\ge\frac{x}{\sqrt[4]{2x^{5}}}=\left(\frac{1}{2}\right)^{\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/799851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Simple Integral $\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}$. Hellow I am trying to prove this result.
$$
I:=\int_0^\infty (1-x\cot^{-1} x)dx=\frac{\pi}{4}.
$$
The indefinite integral exists for this integral.
The function $\cot^{-1} x$ is the arc-cotangent function, not the multiplicative inverse. Note, we can no... | Let $y=\cot^{-1}x$, then $x=\cot y\;\Rightarrow\;dx=d(\cot y)$. The integral turns out to be
$$
\int(1-y\cot y)\ d(\cot y)=\int\ d(\cot y)-\int y\cot y\ d(\cot y).$$
The second integral in the RHS can be solved using IBP by taking $u=y$ and $dv=\cot y\ d(\cot y)$, we obtain $du=dy$ and $v=\dfrac12\cot^2 y$. Hence
$$
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
The $n$'th derivative of $x^x$ I want to know the $n$'th derivative of $f(x)=x^x$.
Then, I'll calculate $f(0)$ with Taylor expansion of $f(x)$ on $a=1$.
Here is my answer, but it is unfinished.
The derivative of $f(x)=x^x$
$$\begin{align}
f'(x)&=x^x(\log x+1)\\
f''(x)&=x^x(\log x+1)^2+x^{x-1}\\
f'''(x)&=x^x(\log x+1)^3... | I tried the simplest approach to test myself. Let $u=x^{x-1}=e^{(x-1)\ln x}$, then $u=1+\sum_{n=1}^{\infty}\frac{1}{n!}(x-1)^n(\sum_{s=1}^{\infty}\frac{(-1)^{s-1}}{s}(x-1)^s)^n$ and expanding the binomial we have
$$u=1+\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n!}C_{n,k}(x-1)^{n+k}$$
where
$$C_{n,k}=\sum_{r_1+r_2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
How to integrate $\int_{0}^{1} \int_{0}^{\pi} \int_{0}^{\pi} r^2 \sin\theta \sqrt{1 - r^2\cos^2\theta - r^2\sin^2\theta} \,d\phi\, d\theta \,dr$
Find the center of mass of the hemispherical region $W$ defined by the inequalities $x^2 + y^2 + z^2 \leq 1$ and $z \geq 0$ with unity density.
By symmetry we know the $x$ a... | The right way is to use cylindrical coordinates:
$$
Z=\frac{\int_0^1z (\pi(1-z^2))dz}{\int_0^1 (\pi(1-z^2))dz}=\frac{3}{2}\int_0^1(z-z^3)dz=\frac{3}{8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How prove $a_{n}=[\sqrt{2}n]+[\sqrt{5}n]$ Contains infinitely even numbers.
let sequence $$a_{n}=[\sqrt{2}n]+[\sqrt{5}n]$$
where $[x]$ is the largest integer not greater than $x$
show that $\{a_{n}\}$ Contains infinitely even numbers.
also I guess contains infinitely odd numbers.
before I have ask this How prov... | When you increase $n$ by one, $\lfloor \sqrt 2 n \rfloor$ increases by either $1$ (with frequency $2-\sqrt 2$ ) or $2$ (with frequency $\sqrt 2 - 1$), while $\lfloor\sqrt 5n \rfloor$ does by either $2$ (with frequency $3-\sqrt 5$) or $3$ (with frequency $\sqrt 5 - 2$).
If your sequence only contains finitely many odd (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx$ $$\int_0^1\frac{x\ln x}{(1+x^2)^2}\ dx
$$
Help me please. I don't know any ways of solution. Thank you.
| Rewrite:
$$
\int\frac{x\ln x}{(1+x^2)^2}\ dx=\frac14\int\frac{2x}{1+x^2}\cdot\ln x\cdot\frac{2\ dx}{1+x^2}.
$$
Now, consider Weierstrass substitution: $x=\tan\dfrac t2$, $\sin t=\dfrac{2x}{1+x^2}$, $\cos t=\dfrac{1-x^2}{1+x^2}$, and $dt=\dfrac{2\ dx}{1+x^2}$. The integral turns out to be
$$
\frac14\int\sin t\cdot \ln \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 0
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What is a simple way of computing the following fraction? Compute the value of the expression:
$$\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$$
| Note that
$$x^4+324=(x^2-6x+18)(x^2+6x+18).$$
Hence
$$
\begin{align*}
\frac{(n+6)^4+324}{n^4+324}
&= \frac{((n+6)^2-6(n+6)+18)((n+6)^2+6(n+6)+18)}{(n^2-6n+18)(n^2+6n+18)} \\
&= \frac{(n^2+6n+18)((n+6)^2+6(n+6)+18)}{(n^2-6n+18)(n^2+6n+18)} \\
&=\dfrac{(n+6)^2+6(n+6)+18}{n^2-6n+18}
\end{align*}$$
which means that most of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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$\int_{0}^{\infty}\int_{1}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$ diverges? I want someone to review my proof that $$\int_{1}^{\infty}\int_{0}^{\infty}\frac{x^2-y^2}{(x^2+y^2)^2}dxdy$$ does not converge.
To make things easier, I said let's look at the entire first quadrant and then subtract the integral over the smal... | As an iterated integral, this integral converges. Indeed
$$ \int_{0}^{\infty} \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} \, dx = \left[ - \frac{x}{x^{2} + y^{2}} \right]_{0}^{\infty} = 0. $$
So the whole integral also converges to 0. But it does not converge absolutely. Indeed,
$$ \int_{0}^{\infty} \left|\frac{x^{2} - y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve the following equation in positive integers $x$ and $y$ What are the solutions in positive integers of the equation:
$${1+2^x+2^{2x+1}=y^2}$$
I tried to factorize the equation but it didn't help much.
Clearly $y $ is an odd integer. Substituting $y =2n+1$, we get
$2^x+2^{2x+1}=(2n)\cdot{(2n+2)}$
$\Rightarrow (2... | $1+2^x+2^{2x+1}=y^2$ or $1+2^{(x-1)+1}+2^{2(x-1)+3}=y^2$ or $1+2*2^{(x-1)}+8*2^{2(x-1)}=y^2$. Now, let $p=2^{(x-1)}$, then, $1+2p+p^2+7p^2=y^2$ or $(p+1)^2=y^2-7p^2$ , this a Rational Pell equation, now, it can be shown that any such Pell equation has solutions; $(r^2-d)^2=(r^2+d)^2-d(2r)^2$. Hence, by comparison, $d=7... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Application of Definite Integral I been solving this problem for an hour, but I cant get the correct answer. Please help me with this.
| If you split the desired violet area into $2$ equals parts by drawing the line $x=\frac32$, you can evaluate it. In fact you only need to solve the system below
\begin{cases}
\int_0 ^\frac32 \sqrt{4-x^2} = A+S\\
\int_0 ^2 \sqrt{4-x^2} = A+2S\\
\end{cases}
letting $S=$ half of the violet area, and $A =$ the sector of th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $f\in \mathcal{R}[0,1]$ Let $f$ be defined by
$$f(x) =\left\{\frac{1}{n}, \frac{1}{n+1} \lt x \le \frac{1}{n} \right\}$$ and $f(x) = 0$ when $x=0$ and $n \in N$. Show that $f$ is integrable and $$\int_{0}^{1}f(x)dx=\frac{\pi^2}{6}-1.$$
| Let $\epsilon > 0$ be given, and choose $n$ such that $n > \dfrac{1}{\epsilon}$, consider the partition $P = \{0, \frac{1}{n}, \frac{1}{n-1},..., \frac{1}{2}, 1\}$, then:
On $[0,\frac{1}{n}]$, $M - m = \dfrac{1}{n} - 0 = \dfrac{1}{n}$,
On $(\frac{1}{n}, \frac{1}{n-1}]$, $M - m = \dfrac{1}{n-1} - \dfrac{1}{n-1} = 0$,
On... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/809842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Evaluating the integral with trigonometric integrand While solving another problem I have come across this integral which I am unable to evaluate. Can someone please evaluate the following integral? Thank you.
$$\int_0^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta.$$
| Here's the contour integration approach.
$$\int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta = \int_{0}^{2 \pi} \frac{1}{(2+\frac{e^{i \theta} + e^{- i \theta}}{2})^{2}} \ d \theta$$
Let $z=e^{i \theta}$.
Then
$$ \begin{align}\int_{0}^{2 \pi} \frac{1}{(2+ \cos \theta)^{2}} \ d \theta &= \int_{|z|=1} \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Laurent series for $\frac1{z^2+1}$ I have this problem:
Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.
I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$,
then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem.
D... | The answer by Santosh is not correct.
The partial fraction decomposition of $f(z)$ is
$$
f(z) = \frac{2}{z+2} - \frac{1-2i}{z+i} - \frac{1+2i}{z-i}
$$
Since we are looking for Laurent series in the annulus, $1<\lvert z\rvert < 2$, we have that $\frac{1}{\lvert z\rvert} < 1$ and $\frac{\lvert z\rvert}{2}<1$.
\begin{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Simplifying $(1/x-1/5 )/( 1/x^2-1/25)$ How do I get $\frac{5x}{x+5}$ from simplifying the following?
$$\frac{(\frac{1}{x}-\frac{1}{5} )}{( \frac{1}{x^2}-\frac{1}{25})}$$
My work:
I multiplied the top and bottom by the LCD: $25x^2$ (to get the same denominators).
Then I got: $\frac{25x-5x}{25-x^2}$
Then I got this for... | Easier: $\frac{1}{x^2} - \frac{1}{25} = \big( \frac{1}{x} -\frac{1}{5} \big)\big(\frac{1}{x} + \frac{1}{5} \big)$ and the first term happily cancels out, now sum two fractions and you are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Logarithm Fraction Contest Math Question The question is as follows:
If $\dfrac{\log_ba}{\log_ca}=\dfrac{19}{99}$ then $\dfrac{b}{c}=c^k$. Compute $k$.
| Rewrite
$$
\begin{align}
\dfrac{\log_ba}{\log_ca}&=\dfrac{19}{99}\\
\dfrac{\frac{\log a}{\log b}}{\frac{\log a}{\log c}}&=\dfrac{19}{99}\\
\dfrac{\log a}{\log b}\cdot\dfrac{\log c}{\log a}&=\dfrac{19}{99}\\
\dfrac{\log c}{\log b}&=\dfrac{19}{99}\\
99\log c&=19\log b\\
\log c^{99}&=\log b^{19}\\
c^{99}&=b^{19}\tag1
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/811884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Congruency by completing the square $x^2+x+1\equiv 0\mod 49$
We have the ring isomorphism $\mathbb{Z}/49\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$.
Consider $x^2+x+1\equiv 0\mod 7$
I usually solve these polynomial congruencies using the 'complete the square' method. How would I go about doing that... | Mod 7, we have $$x^2+x+1\equiv x^2-6x+8=0$$ or $$(x-3)^2=1$$
Since $7$ is prime, each coprime number has either two square roots, or none. Hence $x-3=1$ or $x-3=-1$, i.e. $x=4$ and $x=2$ are the two solutions mod $7$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$4 \sin 72^\circ \sin 36^\circ = \sqrt 5$ How do I establish this and similar values of trigonometric functions?
$$
4 \sin 72^\circ \sin 36^\circ = \sqrt 5
$$
| A key observation is $36 + 72 + 72 = 180$. This means $$\sin 72 = \sin (36 + 72) = \sin 36 \cos 72 + \cos 36 \sin 72$$
Using the identity $\sin 72 = 2\sin 36 \cos 36$, we get
$$2\sin 36 \cos 36 = \sin 36 \cos 72 + 2 \sin 36 \cos^2 36$$
Now notice $\sin 36$ appears everywhere, so we can divide by it...
$$2\cos 36 = \cos... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Applications of the identity $ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$ I am reading Euclid's elements
I found the algebraic identity
$ab + \left(\frac{a+b}{2} - b\right)^2 = \left(\frac{a+b}{2}\right)^2$
I ponder on usage of this identity for $2$ hours.
but I can't click anything.
$a^2 + b^... | If I understand properly $$ab + \left(\frac{a+b}{2} - b\right)^2 =ab+\left(\frac{a-b}{2} \right)^2=\frac{1}{4} \Big(4ab+(a-b)^2\Big)=\frac{1}{4} (a+b)^2=\left(\frac{a+b}{2}\right)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 1
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Another inequality to prove involving arbitrary reals Let $a,b$ be non-zero reals such that $ab\ge \frac{1}{a}+\frac{1}{b}+3$ then prove the following inequality :
$$ \sqrt[3]{ab}\ge \frac{1}{\sqrt[3]{a}}+\frac{1}{\sqrt[3]{b}}$$
This one stumped me completely as usual. A solution would be welcome.
| [Clearly we can substitute all the variables with their cubes, which I will do so to avoid writing so many cube-roots.]
For any $a,b < 0$:
$a b > 0 > \frac{1}{a} + \frac{1}{b}$
For any $a > 0$ and $b \ne 0$:
If $a b < \frac{1}{a} + \frac{1}{b}$:
Let $b' > b$ such that $a b' = \frac{1}{a} + \frac{1}{b'}$ and ( $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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With how many ways can we choose $9$ balls of a box? With how many ways can we choose $9$ balls of a box that contains $12$ balls, of which $3$ are green, $3$ are white, $3$ are blue and $3$ are red?
$$$$
I have done the following:
$x_1=\# \text{ green balls that we choose }$
$x_2=\# \text{ white balls that we choose }... | Generating function solution: Each color is represented by:
$$
1 + z + z^2 + z^3 = \frac{1 - z^4}{1 - z}
$$
All four colors are then:
$$
\left( \frac{1 - z^4}{1 - z} \right)^4
$$
and you want 9 balls, i.e., the coefficient of $z^9$, ellipses are terms that don't affect the result:
\begin{align}
[z^9] \left( \frac{1 - z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Trying to evaluate integral$\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx$ I am trying to integrating
$$
\int_0^\infty x \sqrt{1-e^{-x}}\,e^{-x}dx\equiv I
$$
but cannot get the answer, I would like a proof not a numerical answer. My attempt at proof:
$$
y=\sqrt{1-e^{-x}}\\
y(0)=0, \ y(\infty)=1\\
y^2=1-e^{-x}\\
2ydy=e^{... | Using the well-Known Series formula for the Digammafunction http://en.m.wikipedia.org/wiki/Digamma_function we have $$2\sum_{k=1}^{\infty}\frac{1}{n(2n+3)}=\frac{2}{3}\sum_{n=0}^{\infty}\frac{\frac{5}{2}-1}{(n+1)(n+\frac{5}{2})}=\frac{2}{3}(\gamma +\psi(\frac{5}{2}))$$
Recall the recurrence relation formula for Digamma... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 3
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Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Sorry to add a second answer, dont upvote if you dont like it, but this answer is different from my first and is essentially the same as some of the other answers, but I wanted to point a different formulation that is a little to long for a comment.
One can use the limit
$$\lim\limits_{y \to 0} \frac{(y+k)^{\alpha}-k^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| You need to Taylor expand the square root part as well:
Recall that $\sqrt{1+x} = 1 + \frac{1}{2}x + o(x)$ for $x \to 0$
$\sqrt{2/x^2+2/x+1} = (1+\frac{1}{2}2/x+o(1/x)) = 1+1/x + o(1/x)$ as $x \to \infty$
Now you get:
$1-x+x\sqrt{2/x^2+2/x+1} = 1-x+x[1+1/x + o(1/x)] = 2 + o(1) \to 2$ as $x \to \infty$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument:
$P_{k+1} = \frac{... | Here is an alternate solution that gives a bit more insight.
$\frac{1 \cdot 3 \cdots (2n-1)}{n!} = \frac{(2n)!}{2^n(n!)^2}$ so we have to prove $2^{2n} \ge \frac{(2n)!}{(n!)^2} = \dbinom{2n}{n}$.
However, $2^{2n} = \sum_{k = 0}^{2n} \dbinom{2n}k > \dbinom{2n}n$ and the conclusion follows. Obviously, the inequality is v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int x... | You are right till $$I = x\ln(2x+1) - \int\dfrac{2xdx}{2x+1}.$$
The last integral can be calculated like this:
$$\int\dfrac{2xdx}{2x+1} =\int \dfrac{2x+1 - 1}{2x+1}$$
$$\int dx - \int \dfrac{1}{(2x+1)}$$
Substituting this you get the desired integral:
$$ I = x\ln(2x+1) -x +\frac{ \ln(2x+1)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/823162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.
| Consider the polynomial
$$\sum_{k=0}^n(z+1)^k=\frac{(z+1)^{n+1}-1}{z}=(n+1)+\frac{(n+1)n}2z+\frac{(n+1)n(n-1)}6z^2+\frac{(n+1)n(n-1)(n-2)}{24}z^3...$$
when $z=0$, it takes the value $$\color{blue}{\sum_{k=0}^n1=n+1}.$$
Derive on $z$:
$$\sum_{k=0}^nk(z+1)^{k-1}=\frac{(n+1)n}2+\frac{2(n+1)n(n-1)}6z+\frac{3(n+1)n(n-1)(n-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Probability of getting the sum of two numbers selected from two ranges as odd. Consider two integers N and M.
One number is selected from a range of 1 to N.
Another number is selected from a range of 1 to M.
What is the probability that the sum of the two numbers is odd?
Note:
I figured out that, whenever either N or ... | $A \equiv \text{Number selected from the range $1\ldots N$ is odd}\\
B \equiv \text{Number selected from the range $1\ldots M$ is odd}
$
Now $P(A) = \frac 1 N\left\lceil\frac N 2\right\rceil,\ P(\bar{A}) = \frac{1}{N}\left\lfloor \frac N 2 \right\rfloor$
and $P(B) = \frac 1 M\left\lceil\frac M 2\right\rceil,\ P(\bar{B}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/825911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
What is the value of the following? $3^{302} \mod 5.$ I have to choose from
a. 0
b. 1
c. 2
d. 3
e. 4
I think its e. 4 because
$$3^{302} = 3^{300} \cdot 3^2 = 3^{4\cdot 75} \cdot 3^2 = (3^4)^{75} \cdot 3^2.$$
Applying Fermat's Little Theorem to turn $3^4$ into $1$, then evaluate and solve:
$$(3^4)^{75} \cdot 3^2 \e... | Hint $\ {\rm mod}\ 5\!:\,\ \color{#c00}{3^2\equiv\, -1}\ \Rightarrow\ 3^{2k}\equiv (\color{#c00}{3^2})^k\equiv (\color{#c00}{-1})^k\ $ [$\,\equiv -1\ $ if $\,k\,$ is odd, as in your case]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Help w/ Partial Fraction Decomposition I need some help figuring out how to decompose $\displaystyle\frac{1}{x^4+1}$ into partial fractions.
This is what I have done so far: $$\frac{1}{x^4+1} = \frac{1}{(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}$$
From there, I do not know how to decompose it into partial fractions, o... | My favorate partial fraction decomposition.
$$\frac{1}{x^4+1}=\frac{1}{2\sqrt{2}}\left(\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1}
-\frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}
\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/828598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How is $2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$ calculated? $$2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$$
I cant figure out why this is $\frac64$. I try to use telescopic series without success.
| Since OP is apparently aware this is a telescoping series, here's the computation.
$$
2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)} =
\sum_{n=2}^{\infty}\frac{2}{(n-1)(n+1)}=
\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n+1}\right) =\\
\left[
\left(1 - \frac 13\right) + \left(\frac 12 - \frac 14\right)
+ \left(\frac 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Inf and Sup for ratio and root test of convergent series I would like to use the ratio and root test on the following series:
$$s = 1/2 + 1/3 + (1/2)^2 + (1/3)^2 + \ldots = a1 + a2 + a3 + \ldots$$
where $a_2$ is $\left(\frac{1}{2} \right)^2 + \left(\frac{1}{3} \right)^2 $ for example.
I know we have a sum of two geomet... | Hint: Since $a_n=2^{-n}+3^{-n}$, we get
$$
\begin{align}
\frac{a_{n+1}}{a_n}
&=\frac{2^{-n-1}+3^{-n-1}}{2^{-n}+3^{-n}}\\
&=\frac{1/2+1/3(2/3)^n}{1+(2/3)^n}
\end{align}
$$
and
$$
\begin{align}
a_n^{1/n}
&=\left(2^{-n}+3^{-n}\right)^{1/n}\\
&=\frac12\left(1+(2/3)^n\right)^{1/n}
\end{align}
$$
To get the limits listed in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find a generating function for $\sum_{k=0}^{n} k^2$
Find a generating function for $\sum_{k=0}^{n} k^2$
I know my solution is wrong, but why?
My solution:
If $F(x)$ generates $\sum_{k=0}^{n} k^2$ then $F(x)(1-x)$ generates $k^2$.
$\frac{x}{(1-x)^4}: \left\{ 0,1,4,9,16,25... \right\}$
$\frac{x}{(1-x)^3}: \left\{ ... | Just consider:
\begin{align}
\sum_{n \ge 0} z^n
&= \frac{1}{1 - z} \\
\sum_{n \ge 0} n^2 z^n
&= z \frac{\mathrm{d}}{\mathrm{d} z}
\left( z \frac{\mathrm{d}}{\mathrm{d} z} \frac{1}{1 - z} \right) \\
&= \frac{z + z^2}{(1 - z)^3} \\
\sum_{n \ge 0} \left( \sum_{0 \le k \le n} k^2 \right) z^n
&= \frac{z + z^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Partial fraction (doubt) I have this partial fraction
$$\displaystyle\frac{1}{(2+x)^2(4+x)^2}$$
I tried to resolve using this method:
$$\displaystyle\frac{A}{2+x}+\displaystyle\frac{B}{(2+x)^2}+\displaystyle\frac{C}{4+x}+\displaystyle\frac{D}{(4+x)^2}$$
$$1=A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2$$
When x=-2
$$1=... | If you clear fractions you get $$A(2+x)(4+x)^2+B(4+x)^2+C(4+x)(2+x)^2+D(2+x)^2=1$$
The easy ones are $x=-2$ which gives $B=\frac 14$ and $x=-4$ which gives $D=\frac 14$
Now equate the coefficients of $x^3$ on each side to give $$A+C=0$$ and set $x=0$ (or equivalently equate constant coefficients), to give $$32A+16B+16C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
How Find the sum $\sum_{n=2}^{\infty}\left(\sum_{j=1}^{n}\frac{1}{(2j-1)^2} \binom{2n}{n}\frac{1}{2^{2n}(2n+1)}\right)$
Prove or disprove
$$I=\sum_{n=2}^{\infty}\left(1+\dfrac{1}{3^2}+\dfrac{1}{5^2}+\cdots+\dfrac{1}{(2n-1)^2}\right)
\binom{2n}{n}\dfrac{1}{2^{2n}(2n+1)}=\dfrac{\pi^3}{48}-\dfrac{1}{6}$$
My try: since... |
Lemma. For $x\in[-1,1]$, we have
$$
\frac{1}{6}(\arcsin x)^3=\sum_{n=1}^\infty a_nx^{2n+1}
$$
with
$$a_n=\left(\sum_{j=1}^n\frac{1}{(2j-1)^2}\right)\binom{2n}{n}\frac{1}{2^{2n}(2n+1)}.$$
Proof. First, we will suppose that $x \in(-1,1)$. Recall that,
$$\arcsin x= \sum_{n=0}^\infty \binom{2n}{n}\frac{x^{2n+1}}{2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find the intersections of the functions I have $f(x)=-x^2+4$ a parabola and $g(x)=\sqrt{(4-x^2})$ a semi circle with a raduis of $2$
if I say $g(x)=f(x)$ and solve for $x$. I should find the points at which $x$ intercepts
$\sqrt{4-x^2}=-x^2+4$
then
$4-x^2=x^4-8x^2+16\Rightarrow
-x^4+7x^2-12=0$
this is as far as i go... | I would use the quadratic formula.
if you set $f(x) = g(x)$,
you get $x^4 - 7x^2 + 12 = 0$
factoring, we get $(x^2-4)*(x^2-3) = 0$
So $x^2 -4 = 0$,
$x = 2, -2$
Then we have $x^2 - 3$ $= 0$,
$x$ = $\sqrt{3},$ $-\sqrt{3}$
But now we must check for extraneous solutions.
Seeing as there are none, our answers are $x$ = $\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Consider convergence of series: $\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$ Consider convergence of series:
$$\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$$
My tried:
We have
$$\sum_{n=1}^{\infty }\sin(\pi (2+\sqrt{3})^{n})=\sum_{n=1}^{\infty}\sin\left(\pi[(2+\sqrt{3})^{n}+(2-... | As you noticed,
$$\sin(\pi(2+\sqrt{3})^n+\pi(2-\sqrt{3})^n)=0$$
so$$\sin(\pi(2+\sqrt{3})^n)=-\dfrac{\sin(\pi(2-\sqrt{3})^n)\cos(\pi(2+\sqrt{3})^n)}{\cos(\pi(2-\sqrt{3})^n)}$$
or for $n$ large enough
$$\begin{cases} \sin(\pi(2-\sqrt{3})^n)=\pi(2-\sqrt{3})^n+o(\pi(2-\sqrt{3})^n)\\ \cos(\pi(2-\sqrt{3})^n)\ge \frac12 \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
If $I_n =\int \cot^nx\ dx$ then $I_0 +I_1 +2(I_2+I_3+ \cdots I_8) +I_9+I_{10}= $? If $\displaystyle I_n =\int \cot^nx\ dx$ then find :
$I_0 +I_1 +2(I_2+I_3+ \cdots I_8) +I_9+I_{10} $ = ?
My approach :
$I_n = \displaystyle\int \cot^{n-2} \cot^2x dx$
$\Rightarrow I_n = \displaystyle\int \cot^{n-2} (\csc^2x -1)dx$
$\R... | Consider using the integral
\begin{align}
I_{n} = \int \cot^{n}(x) \ dx
\end{align}
to evaluate the integral $J = I_{0} + I_{1} + 2(I_{3} + \cdots + I_{8}) + I_{9} + I_{10}$. Let $P$ be the integrand of $J$ for which
\begin{align}
P &= 1 + \cot^{1}(x) + 2( \cot^{2}(x) + \cdots + \cot^{8}(x)) + \cot^{9}(x) + \cot^{10}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta... | Yes, you can differentiate both sides and check whether they are equal: $$\eqalign{
\dfrac{\mathrm d}{\mathrm d\theta}\big[\sin3\theta+\sin\theta\big]&\overset?=\dfrac{\mathrm d}{\mathrm d\theta}\big[2\sin2\theta\cos\theta\big] \\
3\cos3\theta+\cos\theta&\overset?= 2\big[\tfrac12( \cos\theta+3\cos3\theta)\big]\tag{$\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Extrema of $f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$. (a) I need help in finding maxima and minima of the following funcion:
$$f(x)=\frac{\sin (5x)} 5 - \frac{2\sin(3x)} {3} + \sin (x)$$
therefore I need to find the roots of $f'(x)=\cos(5x)-2\cos(3x)+\cos(x)$.
(b) I need to find the minima of $$f(x)=4... | $$\begin{align}
\cos5x-2\cos3x+\cos x&=\cos5x-\cos 3x + \cos x-\cos 3x\\
&=-2\sin4x\sin x+2\sin2x\sin x\\
&=2\sin x(-\sin 4x+\sin2x)\\
&=-4\sin x\sin3x\cos x
\end{align}$$
Now you only have to equal $0$ to each factor.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/840793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute the indefinite integral $I=\int y^{-a}(1−y)^{b-1} dy$ or $I=\int_{d}^1 y^{-a}(1−y)^{b-1} dy$ I need to calculate the indefinite integral $I=\int y^{-a}(1−y)^{b-1} dy$, where $a$, $b$ are REAL NUMBERS and $b>0$.
(my goal is to determine the definite integral $I=\int_{d}^1 y^{-a}(1−y)^{b-1} dy$, where $d<1$).
Tha... | \begin{align}
\int_d^1 {y^{ - a} \left( {1 - y} \right)^{b - 1} dy} = \left\{ \begin{array}{l}
\int_d^1 {y^{r - 1} \left( {1 - y} \right)^{b - 1} dy} ,\,\,\,\,\,\,a < 0,\,\,\text{setting}\,\,a = 1 - r,r > 1 \\
\frac{1}{b}\left( {1 - d} \right)^b ,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,a = 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to calculate $\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}\ dx$? How to calculate $$\int\frac{1}{x + 1 + \sqrt{x^2 + 4x + 5}}dx?$$ I really don't know how to attack this integral. I tried $u=x^2 + 4x + 5$ but failed miserably. Help please.
| $$\int\frac{1}{x+1+\sqrt{x^{2}+4x+5}}dx=\int\frac{(x+1)-\sqrt{x^{2}+4x+5}}{-2x-4}$$
$$=\frac{-1}{2}\int\frac{x+1}{x+2}dx-\frac{1}{2}\int\frac{\sqrt{x^{2}+4x+5}}{x+2}dx$$
The first integral can be dealt with but noticing:
$$\int\frac{x+1}{x+2}dx=\int1dx-\int\frac{1}{x+2}dx$$
The second integral is dealt with as follows:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/850500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Upper bound for $\frac{x^{2k+2}}{1-x^2} $ My book says that if $x \in (-0.5, 0.5)$, then
$$\frac{x^{2k+2}}{1-x^2} \leq \frac{1}{3\cdot 4^k}$$
I don't understand how to get such upper bound?
Take $k=0 \Rightarrow \frac{x^{2}}{1-x^2} \leq \frac{0.25}{0.75} = \frac{1}{3}$
But how to figure out the general form?
| Note that if $|x| < .5 = \frac 1 2$, then
$$\frac{x^{2k + 2}}{1 - x^2} = \frac{x^2}{1 - x^2} \cdot (x^2)^k \le \frac 1 3 (x^2)^k \le \frac 1 3 \cdot \left(\frac 1 4\right)^k$$
using your previous estimate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $ I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real number... | HINT:
Let $x,y,z$ be the roots of $\displaystyle t^3+bt+c=0\ \ \ \ (1)$
$\displaystyle\implies xy+yz+zx=b, xyz=-c$
Multiplying $(1)$ by $t^n\ne0$
$\displaystyle \implies t^{n+3}+bt^{n+1}+ct^n=0$
$\displaystyle \implies\sum x^{n+3}=-b\sum x^{n+1}-c\sum x^n$
$\displaystyle n=1\implies \sum x^4=-b\sum x^2-c\sum x =-b[(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
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Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.... | Here is one approach:
Let $z=r e^{i\theta}$, then we want to compute the $\min,\max$ of
$I=\{ r \mid \, |r e^{i2\theta} + {2 \over r}| = 2 \}$. We see that $r \in I$ iff $|r \cos ( 2 \theta)+ {2 \over r} +i r \sin ( 2 \theta)|^2 = r^2 +{4 \over r^2} + 4 \cos(2 \theta)= 4$ for some $\theta$.
If we let $f(r) = 1-{1 \ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/853992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Extrema and inflection points of the function $y = \cos^2(x) - \cos(x)$ Please help me find the extrema and inflection points of the function $y = \cos^2(x) - \cos(x)$.
So far: $$y'=-2 \cos(x)\sin(x)+\sin(x)$$ $$y'' = 2\sin^2(x) - 2\cos^2(x) + \cos(x)$$
$y' = 0$ when $x = 0, \pi, \pi/3, 5\pi/3$
I plugged these values i... | You changed from addition to subtraction of the first two terms in $y''$:
Recall that $$y'' = 2sin^2(x) - 2cos^2(x) + cos(x)$$
$$y'' = 2(1-\cos^2 x) - 2 \cos^2(x) + \cos x \\
= -4\cos^2 x + \cos x + 2 \iff 4\cos^2 x - \cos x - 2$$
Try solving for $\cos x$ now, since we have a quadratic in $\cos x$. I suggest using the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/855641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\frac{2t-t^2}{t+2} \cdot (\frac{5t}{t-2} - \frac{2t}{t-2} )$ Simplify:
$$\frac{2t-t^2}{t+2} \cdot \left(\frac{5t}{t-2} - \frac{2t}{t-2} \right)$$
*
*I first subtracted the parenthesis because the denominator is equal. I then got:
$$\frac{2t-t^2}{t+2} \cdot \frac{3t}{t-2}$$
*
*Then I was lost. I tried multiplyi... | $$\begin{align}
\frac{2t-t^2}{t+2}\cdot\left(\frac{5t}{t-2}-\frac{2t}{t-2}\right) &= \frac{2t-t^2}{t+2}\cdot \frac{3t}{t-2} \\
&= \frac{3t\left(2t-t^2\right)}{(t+2)(t-2)} \\
&= \frac{6t^2-3t^3}{(t+2)(t-2)} \\
&= \frac{-3t^2(t-2)}{(t+2)(t-2)} \\
&= -\frac{3t^2}{t+2} \\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$
How do I find the sum of this series?
$$\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$$
Hints in the right direction would be appreciated.
| Hint: Use this
$\sin^3x=\frac{3}{4}\sin x - \frac{1}{4}\sin3x. $
Solution:
Let $a_n=\frac{\sin^33^n}{3^n}$ and $b_n=\frac{\sin3^n}{3^n}$. So from the identity mentioned above, one simply has that
$
a_n = \frac{1}{4}(3b_n -3b_{n+1}).
$
Therefore,
$$
\sum_{n\ge 0} \frac{\sin^33^n}{3^n} = \sum_{n\ge 0} a_n = \sum_{n\ge 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
} |
The sum of three consecutive cubes numbers produces 9 multiple I want to prove that $n^3 + (n+1)^3 + (n+2)^3$ is always a $9$ multiple
I used induction by the way.
I reach this expression: $(n+1)^3 + (n+2)^3 + (n+3)^3$
But is a lot of time to calculate each three terms, so could you help me to achieve the induction fo... | There is one really elegant and simple way of proving this by means of induction.
It goes as following:
Define the problem as: $n^3 + (n+1)^3 + (n+2)^3 = 9\lambda$
Test for $n=1$; $1^3+2^3+3^3=9\lambda$; $36=4*9$
Assume that for some $n \in N$ -> $n^3 + (n+1)^3 + (n+2)^3 = 9\lambda$
Test for $n+1$
$
(n+1)^3+(n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/859572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Find the Value of Integral Find the Value of $$\begin{align}I=\int_{0}^{1}\frac{\ln(x)\,dx}{1-x^2}\end{align}$$
I have tried like this: We have $$\begin{align}2I=\int_{0}^{1}\frac{\ln(x^2)\,dx}{1-x^2}=\int_{0}^{1}\frac{\ln(1-(1-x^2))\,dx}{1-x^2}\end{align}$$ So
$$2I=\begin{align}\int_{0}^{1}\frac{-(1-x^2)-\frac{(1-x^2)... | Integrating by parts,
$$ \begin{align} \int \frac{\ln (x)}{1-x^{2}} \ dx &= \ln(x) \ \text{arctanh} (x) - \int \frac{\text{arctanh}(x)}{x} \ dx \\ &= \ln (x) \ \text{arctanh} (x) - \frac{1}{2} \int \frac{\ln (1+x)}{x} \ dx + \frac{1}{2} \int \frac{\ln (1-x)}{x} \ dx \\ &= \ln(x) \ \text{arctanh}(x) + \frac{1}{2} \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/860091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find the maximum value of $ \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1} $ If $x\in\mathbb{R}$ find the maximum value of
$$ \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1} $$
I tried this:
Let $$y= \sqrt{x^4-3x^2-6x+13} - \sqrt{x^4-x^2+1}$$
For maxima $\frac{dy}{dx}=0$ and $\frac{d^2y}{dx^2} < 0$. However, the equation... | since
$$\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$
let $$P(x,x^2),A(3,2),B(0,1)$$
so
$$|PA|-|PB|\le |AB|=\sqrt{10}$$
if and only is $A,P,B$ on a line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/860305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
} |
Find expression for : $ S_n =\sum_{i=1}^{n} \frac{i}{i^4+i^2+1} $ I want to find a formula for the sum of this series using its general term.
How to do it?
Series
$$
S_n = \underbrace{1/3 + 2/21 + 3/91 + 4/273 + \cdots}_{n \text{ terms}}
$$
General Term
$$
S_n = \sum_{i=1}^{n} \frac{i}{i^4+i^2+1}
$$
| Write it out:
$$
\begin{eqnarray}
\sum_{i=1}^\infty \frac{i}{i^4 + i^2 + 1}
&=& \sum_{i=1}^\infty \left( \frac{2}{4i^2 - 4i + 4}
- \frac{2}{4i^2 + 4i + 4}\right)\\
&=& \sum_{i=1}^\infty \left( \frac{2}{\Big(2i-1\Big)^2 + 3}
- \frac{2}{\Big( 2i + 1\Big)^2 + 3}\right)\\
&=& \frac{2}{\Big( 2 - 1\Big)^2 + 3} + \sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Simplify expression $(x\sqrt{y}- y\sqrt{x})/(x\sqrt{y} + y\sqrt{x})$ I'm stuck at the expression: $\displaystyle \frac{x\sqrt{y} -y\sqrt{x}}{x\sqrt{y} + y\sqrt{x}}$.
I need to simplify the expression (by making the denominators rational) and this is what I did:
$$(x\sqrt{y} - y\sqrt{x}) \times (x\sqrt{y} - y\sqrt{x}) ... | I am assuming your ambiguous notation begins with the task of simplifying:
$$\frac{x\sqrt y - y\sqrt x}{x\sqrt y + y\sqrt x}.$$
Assuming I'm correct, then we can rationalize the denominator (get rid of the factors with square roots), as follows:
Multiply the numerator and denominator by $(x\sqrt{y}-y\sqrt{x})$ to get a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/869037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
find a polynomial whose roots are inverse of squares of roots of $x^3+px+q$ Question is :
Given a polynomial $f(x)=x^3+px+q\in \mathbb{Q}[x]$ find a polynomial whose roots are inverse of sqares of roots of $f(x)$
Supposing $a,b,c$ as roots of $f(x)$ we have :
*
*$a+b+c=0$
*$ab+bc+ca=p$
*$abc=-q$
Now i need to kn... | *
*$x^3+px-q$ has roots $-a,-b,-c$.
*$(x^3+px+q)(x^3+px-q)$ has roots $a,b,c,-a,-b,-c$.
*$x^3+2px^2+p^2x-q^2$ has roots $a^2,b^2,c^2$.
*$1+2px+p^2x^2-q^2x^3$ has roots $1/a^2,1/b^2,1/c^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
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.