Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^... | If you don't know the binomial theorem, you might rearrange the terms:
\begin{align*}
r^6-3r^4+3r^2-1 &= r^6-1-3(r^4-r^2) \\
&= (r^2-1)(r^4+r^2+1)-3r^2(r^2-1) \\
&= (r^2-1)(r^4+r^2+1-3r^2) \\
&= (r^2-1)(r^4-2r^2+1) \\
&= (r^2-1)(r^2-1)^2 \\
&= (r^2-1)^3 \\
&= (r+1)^3(r-1)^3
\end{align*}
Useful identities... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Finding $\int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$ Finding $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx$
Attempt: Assume $\displaystyle \int\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}dx =\int \frac{2x-x^{-2}}{\bigg[x^4-x^2+2x-\frac{2}{x}-\frac{1}{x}+\frac{1}{x^2}\bigg]}dx$
could some help me how to solve,thank... | As Grant B. commented
$$x^6-x^4+2x^3-2x^2-x+1=(x-1)(x^2+x-1)(x^3+x+1)$$ Then, Using partial fraction decomposition,
$$\frac{2x^3-1}{x^6-x^4+2x^3-2x^2-x+1}=\frac 13 \left(\frac{2 x+1}{ x^2+x-1}-\frac{3 x^2+1}{
x^3+x+1}+\frac{1}{ x-1}\right)$$ where you can notice that each numerator is the derivative of each denomi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Computing Matrix Representation of a Linear Transformation I'm having some trouble figuring out this question:
Let $V=P_3(\mathbb{R})$ and $W=P_4(\mathbb{R})$.
Let $D:W\to V$ be the derivative mapping $D(p)=p'$.
Let $Int:V\to W$ be the integration mapping $Int(p)=\int_0^x p(t)dt$.
Let $\alpha=\{1,x,x^2,x^3\}$ and $\be... | Let's apply the transformation $D$ to each element of the given basis $\beta=\{1,x,x^2,x^3,x^4\}$ of $W=P_4(\mathbb{R})$ in the order they are listed, and write each result with respect to the given basis $\alpha=\{1,x,x^2,x^3\}$ of $V=P_3(\mathbb{R})$.
$D(1)=(1)'=0=0\cdot1+0\cdot x+0\cdot x^2+0\cdot x^3$, therefore th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$3\mid a^3+b^3+c^3$ if only if $3\mid a+b+c $
Prove the following equivalence: $3\mid (a^3 + b^3 + c^3) $ if and only if $3\mid (a + b + c) $.
My try:
I know $a^3 + b^3 + c^3 = (a + b + c) (a^2 + b^2 + c^2 – ab – bc – ca) + 3abc$, but I can't seem to proceed from here.
Thanks all!
| *
*$1^3 = 1 \mod 3$
*$2^3 = 2 \mod 3$
*$3^3 = 3 \mod 3$
This means $x = x^3 \mod 3$ for any integer $x$. This also means that if $a^3 + b^3 + c^3 = 0 \mod 3$ then $a + b + c = 0 \mod 3$.
Thus it is divisible by three.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Last three digits of $\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$ Find the last three digits of $$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}$$
I don't know how to continue my work:
$$\sqrt{4^{2016}+2\cdot 6^{2016}+9^{2016}}=\sqrt{2^{2(2016)}+2\cdot (3\cdot 2)^{2016}+3^{2(2016)}}=\sqrt{2^{4032}+2\cdot 3^{2016}\cdot 2^{2016}... | HINT:
This can be factorised like this
$$(2^{2016})^2+2\cdot2^{2016}\cdot3^{2016}+({3^{2016}})^{2}=(2^{2016}+3^{2016})^2$$
So the question is to find last three digits of
$$2^{2016}+3^{2016}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the Laurent series expansion for: $\frac{1}{z^2-3z-4}$ Find the Laurent series expansion for:
$$\frac{1}{z^2-3z-4}$$
which converges for$\:$ $1 < |z| < 3.$
I understand the concept of Laurents expansion but I cant seem to make progress in this question so any help will be appreciated.
| $$\frac{1}{z^2-3z-4} = \frac{1}{(z+1)(z-4)}= \frac{1}{5}\left(\frac{1}{z-4}-\frac{1}{z+1}\right)$$
We look at the geometric series. For the first term we have
$$\frac{1}{z-4}=\frac{-1/4}{1-z/4}=\sum_{n=0}^{\infty }\left( -\frac{1}{4} \right) \left( \frac{z}{4}\right) ^{n}\qquad \left\vert z\right\vert <4\implies|z|<3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2200682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Constructive proof for $p|x^2+1$ It's easy to prove $x^2+1$ is never divisible by $4k+3$ primes. I know a non-constructive proof for existing $x$ so that $p|x^2+1$ for $4k+1$ primes. is there any constructive one?
| First note that
$(p-1)! = -1 \mod p$
by Wilson's Theorem. Since $p = 4k+ 1$, we can write this product as
$1\cdot 2 \cdots (\frac{p-1}{2}) (\frac{p+1}{2}) \cdots (p-1) \equiv -1 \mod p$
Note that this can be written as
$1 \cdot 2 \cdots (\frac{p-1}{2}) \cdot (p-\frac{p-1}{2}) \cdots (p-1) \equiv -1 \mod p$
so
$1 \cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2202339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $f(x) = -3x^2 + 6x + 2$, prove that $f(x)$ is $O(x^2)$ I had a question on Big-O.
If $f(x) = -3x^2 + 6x + 2$, prove that $f(x)$ is $O(x^2)$
Generally my teacher gave us these strategies to follow:
1. Eliminate Negative Terms
2. Multiply to match highest order term.
So in this case Would I be getting rid of $-3x^2... | Consider that by the triangle inequality we have,
$$|f(x)| = |-3x^2+6x+2| \leq |-3x^2|+|6x|+2$$
then for $x>0$ we have that
$$|f(x)| \leq 3x^2+6x+2$$
and for any $x>1$ (because $x^2>x>1$), we clearly than have that
$$|f(x)|\leq 3x^2+6x^2+2x^2 = 11x^2 = 11\cdot|x^2|.$$
Since this holds for all $x>1$, we have that $f(x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2204746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Integration with floor function I can't find the exact value of the following integral:
$$\int\limits_0^1 {[ax][bx]} dx$$
where $a$ and $b$ have as greatest common divisor 1.
(gcd(a,b)=1)
Thanks.
| As mentioned in the comments, we have
$$\int_0^1 \lfloor ax \rfloor \lfloor bx \rfloor \, dx = \frac{1}{ab} \sum_{n=0}^{ab-1} \lfloor n/a \rfloor \lfloor n/b \rfloor $$
We'll work with residues mod $a$ and $b$ instead of the floor function. We can write $\lfloor n/a \rfloor = n/a - \{n/a\}$, where $\{x\}$ denotes the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2210246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove that $3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be sides-lengths of a triangle. Prove that:
$$3(a^5b+b^5c+c^5a)\geq(a^2c+b^2a+c^2b)^2$$
I tried to apply the way like the proof of the following inequality, but without success.
Let $\sqrt{a}$, $\sqrt{b}$ and $\sqrt{c}$ be... | $$3(a^5b + b^5c + c^5a) \ge (a^2b + b^2c + c^2a)^2$$
$$\iff 3(abc)^2\left[\left(\frac{a}{b}\right)^3 \cdot \left(\frac{b}{c}\right)^2 + \left(\frac{b}{c}\right)^3 \cdot \left(\frac{c}{a}\right)^2 + \left(\frac{c}{a}\right)^3 \cdot \left(\frac{a}{b}\right)^2\right] \ge (abc)^2\left(\frac{a}{b} + \frac{b}{c} + \frac{c}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How can I show that $\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$? I want to show this
$\sum_{k=0}^n [\frac{(n-2k)}{n}{n \choose k}]^2 = \frac{2}{n}{2n - 2 \choose n - 1}$
I tried to solve it using this identities:
I) $\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$
II) ${n \choose... | Solution
$\sum_{k=0}^{n}[\frac{n-2k}{n}\binom{n}{k}]^{2}$
$= \frac{1}{n^{2}}\sum_{k=0}^{n}(n-2k)^2\binom{n}{k}^2 $
$=\frac{1}{n^{2}}[\sum_{k=0}^{n}n^2\binom{n}{k}^2 + 4\sum_{k=0}^{n}k^2\binom{n}{k}^2 - 4\sum_{k=0}^{n}nk\binom{n}{k}^2] $
$= \sum_{k=0}^{n}\binom{n}{k}^2 + \frac{4}{n^{2}}\sum_{k=0}^{n}n^2\binom{n-1}{k-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $f(x)$ in this question? I have a quation regarding composite functions.
if $f\left( \dfrac{x^2+1} x \right)=\dfrac{x^4+1}{x^2}$, then what is $f(x)$?
| HINT:
$$\left(\frac{x^2+1} x\right)^2=\frac{x^4+2x^2+1}{x^2}=\frac{x^4+1+2x^2}{x^2}$$
Alternatively you have $$\frac{x^2+1} x = x + \frac 1 x \\\frac{x^4+1}{x^2} = x^2 + \frac 1 {x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Area inside outer curve of limacon $r=1+2\cos\theta$ I need to find the area inside the outer curve of limacon $r=1+2\cos\theta$
Here is what I tried:
$$
0=1+2\cos\theta
$$
$$
\theta=\cos^{-1}(-\frac{1}{2})
$$
$$
\theta=\frac{2}{3}\pi , \theta=\frac{8}{3}\pi
$$
$$
A = 2\int^{\frac{8}{2}\pi}_{\frac{2}{3}\pi}r^2d\theta... | $\int\limits_{\theta = 0}^{2 \pi} (1 + 2 \cos (\theta )\ d \theta = 2 \pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find the positive $n$ such $2^{n-2i}|\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$ Question:
if such $\forall i\in\{1,2,\cdots,\lfloor\dfrac{n-1}{2}\rfloor\}$,have
$$2^{n-2i}|\sum_{k=i}^{\lfloor\frac{n-1}{2}\rfloor}\binom{k}{i}\binom{n}{2k+1}$$ Find all the positive intger $n$
I want to find th... | We have the sum
$$S_{n,q} = \sum_{k=q}^{\lfloor (n-1)/2\rfloor}
{k\choose q} {n\choose 2k+1}.$$
Introducing
$${n\choose 2k+1} = {n\choose n-2k-1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k}} (1+z)^n
\; dz$$
this controls the range and we may extend $k$ to infinity, getting for
the sum
$$\frac{1}{2\pi i}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2220704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find the degree and a basis for $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$ Find the degree and a basis for $\mathbb{Q}(\sqrt{3} + \sqrt{5})$ over $\mathbb{Q}(\sqrt{15})$.
My attempt:
I first proved that $\mathbb{Q}(\sqrt{3} + \sqrt{5}) = \mathbb{Q}(\sqrt{3}, \sqrt{5})$,
Then using tower law its fair... | One way to prove this is :
$a=(\sqrt{3}+\sqrt{5})$
$a^2=(\sqrt{3}+\sqrt{5})^2$
Thus $a^2-8+2\sqrt{15}=0$ and the polynomial $f(X)=X^2-8+2\sqrt{15}$ is irreducible over $\mathbb{Q}(\sqrt{15})$ because its roots do not lie in $\mathbb{Q}(\sqrt{15})$.
Thus a basis is the set $\{1,\sqrt{3}+\sqrt{5}\}$
Also $\{1,\sqrt{5}\}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Singular value decomposition works only for certain orthonormal eigenvectors, not all? I'm trying to find the SVD of the following matrix:
$$A=
\begin{pmatrix}
1 & 1 \\
2 & -2 \\
2 & 2 \\
\end{pmatrix}
$$
I found the eigenvalues and vectors for $A'A$:
$$
\begin{array}{cc}
\lambda_1=10 & \lambda_2=8 \\
e_1'=(1,1) ... | I'm posting an answer just to complete the previous answers.
Let $A=UDV'$.
Then, if we have dim of $U$ greater than $V$ we can use the formula $v_i=\frac{1}{\sigma_i} A'u_i$, for given $U$.
However, if we have dim of $U$ smaller than $V$, we can use the formula $u_i=\frac{1}{\sigma_i}Av_i$, for given $V$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
If $f(x+1)+f(x-1)=4x^2-2x+10$ then what is $f(x)$
If $$f(x+1) +f(x-1)= 4x^2 -2x +10$$ then what is $f(x)$
What is strategy of solving this kind of problems ?
Thank you for help
| $$ 4x^2 -2x +10 = 2(x+1)^2-(x+1)+3+2(x-1)^2-(x-1)+3$$
which means $f(x)=2x^2-x+3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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How can I simplify $\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$? This problem got me thinking hard:
$$\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$$
Considering the fact that the common base is '$3$', I decided to do this:
$$\log_3\left(\frac{(n+1)(n-1)}{(n+2)(n)}\right)$$
$$\log_3\left(\frac{n^2-n+n+1}{n^2+2n}\right)$$
$$\lo... | Factor out by $3^n $ to get
$$\frac{3^n (3+3^{-1})}{3^n (3^2+3^0)} $$
$$\frac {3+3^{-1}}{10}=\frac {1}{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2227857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Prove $8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$
Let $a$, $b$ and $c$ be angles of an acute triangle. Prove that:
$$8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$$
I tried am-gm, rm-gm rm-am and a couple of other inequalities but I didn't get anywhere.
| Let $u$, $v$ and $w$ be lengths-sides of the triangle such that
$\cos{a}=\frac{v^2+w^2-u^2}{2vw}$, $\cos{b}=\frac{u^2+w^2-v^2}{2uw}$, $\cos{c}=\frac{u^2+v^2-w^2}{2uv}$ and
let $u^2+v^2-w^2=r$, $u^2+w^2-v^2=q$, $v^2+w^2-u^2=p$.
Hence, we need to prove that
$$8\prod_{cyc}\left(1-\frac{r^2}{(p+r)(q+r)}\right)\geq\frac{27p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2228636",
"timestamp": "2023-03-29T00:00:00",
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Find the value of $3+7+12+18+25+\ldots=$ Now, this may be a very easy problem but I came across this in an examination and I could not solve it.
Find the value of
$$3+7+12+18+25+\ldots=$$
Now here is my try
$$3+7+12+18+25+\ldots=\\3+(3+4)+(3+4+5)+(3+4+5+6)+(3+4+5+6+7)+\ldots=\\3n+4(n-1)+5(n-2)+\ldots$$
After that, I co... | This answer is unique in that it does not use trial and error to find out the expression for the $n$-th term.
Our sum is given by :
$$
S=3+7+12+18+...+T_n
\\
S=\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;3+7+12+18+...+T_n-1+T_n
$$
Subtracting these two sums :
$$
0=3+4+...+(n+2)-T_n
\\
T_n=3+4+...+(n+2)
\\
T_n=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2230870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
definite integral- how to solve. I know the substitution. I know the substitution but how should I continue?
$$\int_0^1 x\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}dx$$
| An alternative way - by setting $x=u^2$, then $u=\frac{1-v}{1+v}$ and $v=z^2$, we are left with:
$$ \int_{0}^{1}2u^3\sqrt{\frac{1-u}{1+u}}\,du = 4\int_{0}^{1}\frac{(1-v)^3}{(1+v)^5}\sqrt{v}\,dv =8\int_{0}^{1}\frac{z^2(1-z^2)^3}{(1+z^2)^5}\,dz$$
Then by applying integration by parts we get
$$ \int_{0}^{1}\frac{1-9 z^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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QUESTION on Logarithm This question was easy to solve, but when it came to matching options I failed. Hope you can help. Let,
*
*$a= \log_{24}(12)$
*$b= \log_{48}(36)$
*$c= \log_{36}(24)$
then $abc +1 = ?$
*
*a) $2ab$
*b) $2bc$
*c) $2ca$
*d) $ba + bc$
FYI :- I tried to solve and found $abc= log_{48}(1... | It looks like you already know that $\log_a(b) = \frac{\log a}{\log b}$.
Since all of the numbers have $2$ and $3$ as their only prime factors, we can start by setting $x=\log 2$ and $y=\log 3$. We then have
$$ a = \log_{24}{12} = \frac{2x+y}{3x+y} \qquad
b = \log_{48}{36} = \frac{2x+2y}{4x+y} \qquad
c = \log_{36}{24} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2232472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $p\mid (3^n+1)$, then $p\equiv 1\pmod{3}$
Show that if $p> 2$ is a prime, $n > 1$ is odd and $p\mid (3^n+1)$,
then $p\equiv 1\pmod{3}$.
Since $n$ is odd, we have $3^{n+1} \equiv -3 \pmod{p}$ is a quadratic residue. Then I thought about using Quadratic Reciprocity but didn't see how to apply it.
We have $x^2 \eq... | Lemma: for an odd prime $q,$ we always have
$$ (-3|q) = (q |3). $$
Proof: if $q \equiv 1 \pmod 4,$ then
$$ (-3|q) = (-1|q)(3|q) = (3|q) = (q|3). $$
If $q \equiv 3 \pmod 4,$ then
$$ (-3|q) = (-1|q)(3|q) = -(3|q) = (q|3). $$
In the lemma, $3$ can be replaced by any prime $r \equiv 3 \pmod 4.$
Let $n = 2k+1.$ We have $1 ... | {
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"url": "https://math.stackexchange.com/questions/2232818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of $\sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$ Find the following sum
$$S= \sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$$
I am not able to make it telescopic series. Could someone help me with this?
| "Recall" this trigonometric identity:
$$
\arctan x + \arctan y = \arctan \frac{x+y}{1-xy}
$$
Now think about applying that:
\begin{align}
& \arctan \frac{an+b}{cn+d} + \arctan \frac{en+f}{gn+h} \\[10pt]
= {} & \arctan \frac{\dfrac{an+b}{cn+d} + \dfrac{en+f}{gn+h}}{1 - \dfrac{an+b}{cn+d} \cdot \dfrac{en+f}{gn+h}} \\[10p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 0
} |
The adjoint of a linear operator with respect to an inner product I have an operator $T (x_1, x_2) := (x_2, x_1)$ which is self-adjoint with respect to standard inner product, now I have a new inner product on $\Bbb R^2$
$$\langle x,y\rangle = x_1y_1 + \frac{1}{2} (x_1y_2 + x_2y_1) + \frac{1}{3} x_2y_2.$$
I find the m... | By definition, the adjoint operator of $T$ corresponding to the inner $\langle \cdot, \cdot\rangle_A$ is given by
\begin{align}
\langle v, Tv\rangle_A = \langle T^\dagger v, v\rangle_A.
\end{align}
where $T^\dagger$ will be the symbol of the adjoint with respect to $A$. However, since
\begin{align}
\langle v, Tv\rangle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Can the Laurent series be found in region where there is singularity? In class my teacher showed that we can use the expansion of $\frac{1}{1-z}$ to find an expansion for $\frac{1}{(z+1)(z+2)}$ in the regions $|z|<1$, $1<|z|<2$ and $|z|>2$
My question is can this method be used to find the expansion in the regions $|z|... |
The function
\begin{align*}
f(z)&=\frac{1}{(z+1)(z+2)}\\
&=\frac{1}{z+1}-\frac{1}{z+2}
\end{align*}
has two simple poles at $-1$ and $-2$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $-1$ and $-2$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1<|z|<2,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Inequality by telescope It can be shown through less elementary methods that:
$\displaystyle \large \frac{2}{3}n\sqrt{n} < \sum_{k=1}^n \sqrt{k} < \frac{2}{3}n\sqrt{n} + \frac{\sqrt{n}}{2}$
It is easy, but tedious, through squaring, to show that:
Lemma:
$\displaystyle \large \frac{3}{2}\sqrt{k} > \left(k+\frac{1}{2}\ri... | Let $\frac{3}{2}\sqrt{k}=a$, $\sqrt{\left(k+\frac{1}{2}\right)^3}=b$ and $\sqrt{\left(k-\frac{1}{2}\right)^3}=c$.
Since $b>c$, we see that $a+b-c>0$ and $$b+c=\left(k+\frac{1}{2}\right)\sqrt{k+\frac{1}{2}}+\left(k-\frac{1}{2}\right)\sqrt{k-\frac{1}{2}}\geq\sqrt{k}+1\cdot\frac{1}{2}\sqrt{k}=\frac{3}{2}\sqrt{k}=a,$$
whic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$ If $a^2-6a-1=0$, find the value of $a^2 +\dfrac {1}{a^2}$
My Attempt:
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2a.\dfrac {1}{a}$$
$$a^2+\dfrac {1}{a^2}=(a+\dfrac {1}{a})^2 - 2$$
How do I proceed further?
| This can be done by using Sridharacharya Formula.
The roots of a quadratic equation $ax^2+bx+c$ are $$\boxed{\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$
Thus,
$\Large a=\frac{6 \pm \sqrt{6^2+4}}{2}$
$\large a=3 \pm \sqrt{10}$
Now, $$a^2+a^{-2}=\bigg(a+\frac1a\bigg)^2-2$$
Now, one can put both the values of $a$, and then by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2235337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Transformation of $|z-1|=1$ and $\mathrm{Re}(z)^2 = \mathrm{Im }(z)^2-1$ under $f(z) = \sqrt{z}$ I have two regions in the complex plane, defined by
$|z-1|=1$ and $(\,\mathrm{Im}(z))^2 = (\,\mathrm{Re}(z))^2-1$, $\mathrm{Re}(z)>0$.
I am being asked to find and sketch the image of those regions under the mapping $f(z)... | Possible suggestion:
Let $z = x + yi$. We will emulate this problem in the Cartesian plane.
By transforming $Re(z)^2 = Im(z)^2 - 1$ under $f$, we are taking the root of each point. Conversely, $z^2 = (x^2 - y^2) + (2xy)i$ satisfies $Re(z^2)^2 = Im(z^2)^2 - 1$ instead for every point $z$ on the transformed curve.
This i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Show this $|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\frac85$ Let $x\in R$ show that
$$f(x)=|\sin{x}|+|\sin{(x+1)}|+|\sin{(x+2)}|>\dfrac{8}{5}$$
since
$$f(x)=f(x+\pi),$$it sufficient to show $x\in (0,\pi]$
| Drafting behind Michael Rozenberg's clever answer, appealing to the concavity of $\sin$ on $[0, \pi]$ quickly reduces the problem to showing the inequality
$$2 \sin 1 > \frac{8}{5} .$$
Then, from Maclaurin expansion, we have
$$ \sin 1 = 1 - \frac{1}{3!} + \frac{1}{5!} - \frac{1}{7!} + \ldots $$
Observe that the absolut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
For $\Bbb K^{n \times n}$ matrices $A$ and $B$, $\exists p[p(A)=B] \implies \exists p[p(B)=A]$? For $\Bbb K^{n \times n}$ matrices $A$ and $B$, is it true that if there exists a matrix polynomial $p$ such that $p(A)=B$, then there exists another matrix polynomial $p$ (not necessarily the same) such that $p(B) = A$?
My ... | In a word, no. Let
$A=\pmatrix{1&0\\0&-1}$ and $p(X)=X^2$.
Then
$$B=p(A)=\pmatrix{1&0\\0&1}.$$
How can there be a polynomial with $q(B)=A$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2239681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $x$ and $y$ such that $15\sin(x+y)+7\sin x+7\sin y$ is maximized Question is: Find $x$ and $y$ such that
$$15\sin(x+y)+7\sin x+7\sin y$$
is maximized.
What I tried was
$$f(x,y)=15\sin(x+y)+7\sin x+7\sin y=15(\sin x\cos y+\cos x\sin y)+7(\sin x +\sin y)$$
$$\frac{\partial f}{\partial x}=15\cos(x+y)+7\cos x=0$$
$$\f... | Since $f$ is periodic, we don't need to consider the other regions.
Hence, we can consider $x = \pm y$ without the $2n\pi$.
For the case $x=y$, we obtain:
$$\begin{array}{rcrcl}
\dfrac{\partial f}{\partial x} &=& 15\cos(x+y)+7\cos x &=& 0 \\
&& 15\cos(2x)+7\cos x &=& 0 \\
&& 15(2\cos^2x-1)+7\cos x &=& 0 \\
&& 30\cos^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2241291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$2 + 2 = 5$ (Fake proof, or ?) Basically, there is no error in the following steps(as it seems), but there is some error due to which 2 + 2 = 5. What is it?
-20 = -20
16-36 = 25-45
16-36+(81/4) = 25-45+(81/4)
(4^2)-(2*4*9/2)+((9/2)^2) = (5^2)-(2*5*(9/2))+((9/2)^2)
... | The problem is when you go from a quadratic to a linear.
$$(4-\frac{9}{2})^2 = (5-\frac{9}{2})^2$$
$$|4-\frac{9}{2}| = |5-\frac{9}{2}|$$
$$|\frac{-1}{2}| = |\frac{1}{2}|$$
$$\frac{1}{2} = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Perturbation series for $x^5+\varepsilon x-1=0$ I want to find a closed form for the perturbation coefficients $a_n$ defined by the perturbative solution
$$
x(\varepsilon)=1+\sum_{n=1}^\infty a_n \varepsilon^n
$$
to the quintic equation
$$
x^5+\varepsilon x-1=0.
$$
By computing the determinant, as was suggested for th... | For convenience let $\varepsilon = 5s$. Maple gives me the following:
$$\sum _{k=0}^\infty \frac{(-16 s^5)^k}{k!} \left(\frac{\left(\frac 2 5\right)_{2k} \left(-\frac 1 {10}\right)_{2k} }{\left(\frac 4 5\right)_k \left(\frac 3 5\right)_k \left(\frac 2 5\right)_k}
- \frac{\left(\frac 4 5\right)_{2k} \left(\frac 3 {10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2242657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Sum of series = S
If $S=\frac35+\frac{10}{5^2}+\frac{21}{5^3}+\frac{36}{5^4}+\frac{55}{5^5}+\cdots$
In this we have to find value of S
Intianlly I thought it is geometric series .
After that I try to makes terms which can cancel other terms.
| We have
$$\frac{S}{5} = \frac{3}{5^2}+\frac{10}{5^3}+\frac{21}{5^4}+\frac{36}{5^5}+\frac{55}{5^6}+\cdots$$
Hence
$$S_1 = \frac{4S}{5} = S - \frac{S}{5} = \frac{3}{5}+\frac{7}{5^2}+\frac{11}{5^3}+\frac{15}{5^4}+\frac{19}{5^6}+\cdots$$
$$\frac{1}{5}S_1 = \frac{3}{5^2}+\frac{7}{5^3}+\frac{11}{5^4}+\frac{15}{5^5}+\frac{19}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve a simple equation: $x+x\sqrt{(2x+2)}=3$
$x+x\sqrt{(2x+2)}=3$
I must solve this, but I always get to a point where I don't know what to do. The answer is 1.
Here is what I did:
$$\begin{align}
3&=x(1+\sqrt{2(x+1)}) \\
\frac{3}{x}&=1+\sqrt{2(x+1)} \\
\frac{3}{x}-1&=\sqrt{2(x+1)} \\
\frac{(3-x)^{2}}{x^{2}}&=2(x+1... | I usually check if the integer divisors of the independent term of the polynomial is a root in order to decompose it in factors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Find Laurent Series for $f(z) = \frac{1}{z} \cdot \frac{1}{1+z^2}$ when $1 < |z| < \infty$ Q: Find the Laurent series that represents the following function when $1 < |z| < \infty$
\begin{align*}
f(z) &= \frac{1}{z} \cdot \frac{1}{1+z^2} \\
\end{align*}
The answer is given:
\begin{align*}
f(z) &= \sum\limits_{n=1}^... | When $1 < |z| < \infty$ then $0 < \dfrac{1}{|z|} < 1$
$$\dfrac{1}{z}.\dfrac{1}{1+z^2}=\dfrac{1}{z^3}.\dfrac{1}{1+(\dfrac{1}{z})^2}=\\
\dfrac{1}{z^3}.\dfrac{1}{1-(-\dfrac{1}{z^2})}=\\
\dfrac{1}{z^3}(1+(-\dfrac{1}{z^2})+(-\dfrac{1}{z^2})^2+(-\dfrac{1}{z^2})^3+...)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2245910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$ If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$
I could not get any idea to solve. However I tried by using $\theta =0^\circ $. Bu... | Since $$\tan(\pi\cos\theta)=\cot(\pi\sin\theta),$$
we can write $$\tan(\pi\cos\theta)=\cot(90-(\pi\cos\theta)).$$
Thus $$\cot(90-(\pi\cos\theta))=\cot(\pi\sin\theta)$$
$$\therefore\quad 90-(\pi\cos\theta)=\pi\sin\theta$$
$$90=\pi(\sin\theta+\cos\theta)$$
$$\frac{90}{\pi}=\sin\theta+\cos\theta$$
$$\frac{1}{2}=\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\int_0^{\pi/2}x e^x \sin(x)dx$. $$I=\int_0^{\pi/2}x e^x \sin(x)dx$$
I gave a lot of efforts in this question and replaced x with $\pi/2 -x $ but to no avail, this even became more complicated.
Kindly give me an idea on how to proceed in this very question.
| Define $$S(x) = \int e^x \sin x \, dx, \quad C(x) = \int e^x \cos x \, dx.$$ Then with the choice $$u = \sin x, \quad du = \cos x \, dx, \\ dv = e^x \, dx, \quad v = e^x,$$ we find that $$S(x) = e^x \sin x - \int e^x \cos x \, dx = e^x \sin x - C(x).$$ Similarly, we find $$C(x) = e^x \cos x + \int e^x \sin x \, dx = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \int \sqrt{(x^2 + y^2)}$ $dx dy$ over a triangle with corners (0, 0), (4, 4), and (4, 0) using $x = u$, $y = uv$. Help?
So far I have:
$$a = x = 4\\ b = y = 0\\ c = y = x$$
$$y=x \rightarrow uv = u\rightarrow v=u/u \rightarrow v=1$$
$$x = 4 \rightarrow u = 4$$
$$y = 0\rightarrow uv = 0 \rightarrow u = 0... | The Jacobian is not correct. It should be $J=u$. Then, we have
$$\begin{align}
\int_0^4\int_0^y \sqrt{x^2+y^2}\,dx\,dy&=\int_0^1\int_0^4 u^2\sqrt{v^2+1}\,du\,dv\\\\
&=\int_0^4 u^2 \,du\,\int_0^1\sqrt{v^2+1}\,dv\\\\
&=\frac{64}{3}\,\left(\frac12\left(\sqrt 2+\log(1+\sqrt 2)\right)\right)\\\\
&=\bbox[5px,border:2px sol... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this... | Before cubing, separate your roots:
$$\sqrt[3]{5x+7}=\sqrt[3]{5x-12}+1$$
Now cube:
$$5x + 7 = 5x - 12 + 3\sqrt[3]{5x-12}^2 + 3\sqrt[3]{5x-12} + 1$$
Simplify:
$$ \sqrt[3]{5x-12}^2 + \sqrt[3]{5x-12} - 6 = 0$$
Solve for $\sqrt[3]{5x-12}$ (it is a quadratic equation). Finally solve to $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Series $(1+b^n/n)^n$ at infinity Is there a series expansion for
$$\left ( 1 + \frac{b^n}{n} \right )^n$$
at $n\rightarrow\infty$, assuming $0<b<1$?
Similar to
$$ \left ( 1 + \frac{b}{n} \right )^n \rightarrow e^b $$
I want to want to write this as
$$\left ( 1 + \frac{b^n}{n} \right )^n \rightarrow e^{b^n} $$
But I am ... | For $0 < b < 1$, we have $b^n \to 0$ as $n \to \infty$ and it is true that
$$\lim_{n \to \infty} \left(1 + \frac{b^n}{n} \right)^n = e^0 = 1.$$
This follows because
$$\tag{*}\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^n = e^x,$$
converging uniformly for $x$ in a compact set.
Justification for all of this requires... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to evaluate this improper integral $\int_{0}^{\infty}\frac{1-x}{1-x^{n}}\,\mathrm dx$?
$\def\d{\mathrm{d}}$How to evaluate this improper integral?
$$\int_{0}^{\infty}\frac{1-x}{1-x^{n}}\,\d x.$$
What I tried is a substitution i.e $x^{n}=t$, but then things got complicated, and I'm stuck.
| We are going to evaluate the integral
$I:=\displaystyle \int_{0}^{\infty} \frac{1-x}{1-x^{n}} d x \tag*{} $
by the theorem
$ \displaystyle \sum_{k=-\infty}^{\infty} \frac{1}{k+z}=\pi \cot (\pi z), \textrm{ where } z\notin Z.\tag*{} $
We first split the integral into $2$ integrals
$ \displaystyle \int_{0}^{\infty} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Q6 polynomial complex Find the equation whose roots are the fourth powers of the roots of the equation $x^3 + x + 1 = 0$. Hence find the sum of the fourth powers of the roots of the equation $x^3 + x + 1 = 0$
I am really stuck. The answer at the back says cubic is $u^3 - 2u^2 + 5u -1 = 0$ and sum of roots is 2
| So we seek a cubic in $y = x^4, $ where $x^3+x+1=0$.
Hence we have $y = x^4 = x(x^3) = -x(x+1) = -x^2-x$, using which, $y^2 = (x^4)^2 = x^2(x+1)^2 = x^4+2x^3+x^2 = -3x-2$.
Similarly, $y^3 = (x^4)^3 = -x^3(x+1)^3 = 5x^2-x-3$.
Now it is easy to eliminate (the powers of) $x$ and get $y^3-2y^2 +5y = 1\dots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the angle between 2 intersection lines of four planes?
Given $L_1=\begin{cases}P_1: x-y+z=4\\ P_2: 2x+y-z=6 \end{cases}$
and $L_2=\begin{cases}P_1:x+y+z=4\\ P_2:2x+3y-z=6 \end{cases}$
$L_1$ and $L_2$ are the intersection lines of the given planes. Find the angle between the lines $L_1,L_2$
First in order... | Let me answer the general case.
First finding intersection of planes:
In general planes in $\mathbb{R}^3$ are of the form:
$$ax+by+cz+d=0.$$
The intersection of two planes $ax+by+cz+d=0$ and $Ax+By+Cz+D=0$ exists if and only if $|aA+bB+cC|\neq ||(a,b,c)||.||(A,B,C)||$ or if $|aA+bB+cC|=||(a,b,c)||.||(A,B,C)||$, then $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2253072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Step-by-step derivative of $\left ( \frac{c_1 x}{c_2 x + c_3 + c_4 \sqrt{c_5 x}} \right)^{c_6x + c_7 + c_8 \sqrt{c_9 x}}$ Can someone please walk step by step on how to calculate the derivative
$\left ( \frac{c_1 x}{c_2 x + c_3 + c_4 \sqrt{c_5 x}} \right)^{c_6x + c_7 + c_8 \sqrt{c_9 x}}$
Where the $c_i$ are constants, ... | Define
$$
f(x) = g(x)^{h(x)} = e^{h(x)\ln g(x)}
$$
Compute
$$
f'(x) = e^{h(x)\ln g(x)} \left(
\frac{h(x)g'(x)}{g(x)} + h'(x) \ln g(x)
\right)
$$
Define
$$
\begin{align}
\alpha(x) &= c_{2} + c_{3} x + c_{4} \sqrt{c_{5}x} \\[2pt]
g(x) &= \frac{c_{1}x}{\alpha(x)} \\[2pt]
h(x) &= c_{6} + c_{7} x + c_{8} \sqrt{c_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Chain rule differentiation
Given that $\overline{r} = \frac{r}{r_0}$ and $u = \frac{1}{\overline{r}}$, show that
$$r \frac{d^2r}{d\theta^2} - 2\left(\frac{dr}{d\theta} \right)^2 = r^2 - 3\frac{Gm}{c^2}r$$
is equivalent to the following;
$$\frac{d^2u}{d\theta^2} + u = \epsilon u^2 : \epsilon = \frac{3Gm}{r_0c^2}... | \begin{equation}
r=\frac{r_{0}}{u}
\end{equation}
\begin{equation}
\frac{dr}{d\theta}=\frac{d}{d\theta}(\frac{r_{0}}{u})
\end{equation}
Applying chain rule of differentiation you'll get
\begin{equation}
\frac{dr}{d\theta}=\frac{-r_{0}}{u^{2}}\frac{du}{d\theta}
\end{equation}
\begin{equation}
\frac{d^{2}r}{d\theta^{2}}... | {
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Given $x+y+z=1$ Find Maximum value of $x^5y+y^5z+z^5x$ $x,y,z$ are Non negative reals such that $x+y+z=1$ Find Maximum value of $$E=x^5y+y^5z+z^5x$$
The only idea i have is $E$ can be written as $f(x,y)$ and using partial differentiation for maxima...but too lengthy
| Let $x\geq y\geq z$. Hence, by AM-GM
$$x^5y+y^5z+z^5x\leq(x+z)^5y=5^5\left(\frac{x+z}{5}\right)^5y\leq5^5\left(\frac{5\cdot\frac{x+z}{5}+y}{6}\right)^6=\frac{5^5}{6^6}.$$
In the case $x\geq z\geq y$ we can use the same idea.
$$x^5y+y^5z+z^5x\leq x^5z+y^5x+z^5y\leq(x+y)^5z\leq5^5\left(\frac{5\cdot\frac{x+y}{5}+z}{6}\r... | {
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Closed Form for an Alternating Sum Involving Binomial Coefficients
Question. For integer $n \geq 0$, find the closed form for
$$
S_n = \sum_{k \leq 2^n} \binom{2^n - k}{k}(-1)^k
$$
My Attempt: I tried some small $n$ and got $S_0 = 1$, $S_1 = 1$, $S_2 = 0$, $S_3 = -1$ and $S_4 = -1$, but failed to come up with some ... | The calculation is not correct in all aspects. We obtain
\begin{align*}
\color{blue}{S_0}&=\sum_{k=0}^1\binom{1-k}{k}(-1)^k=\binom{1}{0}=\color{blue}{1}\\
\color{blue}{S_1}&=\sum_{k=0}^2\binom{2-k}{k}(-1)^k=\binom{2}{0}-\binom{1}{1}
=1-1=\color{blue}{0}\\
\color{blue}{S_2}&=\sum_{k=0}^4\binom{4-k}{k}(-1)^k=\binom{4}{0}... | {
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How to solve $\frac{dx}{ds}=\frac{x}{x^2+y^2},\quad\frac{dy}{ds}=\frac{y}{x^2+y^2}$? I was wondering if my method to solve the following pair of equations:
$$\frac{dx}{ds}=\frac{x}{x^2+y^2},\quad\frac{dy}{ds}=\frac{y}{x^2+y^2}$$
is the right one.
My attempt: We have $\frac{x^2+y^2}{x}dx=ds$ and $\frac{x^2+y^2}{y}dy=ds$... | You have $$\frac{\text{d}}{\text{d}s}\,\left(x^2+y^2\right)=2\,.$$
So, $x^2+y^2=2(s+c)$ for some constant $c$. Plugging this in, you will see that $$x=\pm\sqrt{2A(s+c)}\text{ and }y=\pm\sqrt{2(1-A)(s+c)}$$ for some constant $A$.
| {
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Mysterious $w$ -- a function of triangle sides Suppose $a$, $b$, $c$ are the lengths of the three sides of a triangle. When studying rational Bézier curves inscribed in the triangle, a mysterious quantity $w$ emerges in the algebra:
$$
w = \frac{c}{\sqrt{2(a^2 + b^2)}}
$$
I'd like to get a better understanding of $w$. ... | It looks as if $w^2$ is more tractable. So
$$2w^2=\frac{c^2}{a^2+b^2}$$
and
$$1-2w^2=\frac{a^2+b^2-c^2}{a^2+b^2}=\frac{2ab\cos C}{a^2+b^2}.$$
For fixed $a$ and $b$ as $C$ varies, this varies between $\pm2ab/(a^2+b^2)$. So $2w^2$ varies between $|a-b|^2/(a^2+b^2)$
and $|a+b|^2/(a^2+b^2)$. Thus
$$2w^2\le\frac{a^2+2ab+b^2... | {
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Is this the correct way to find the Laurent Expansion of the complex function $f(z) = \frac{z}{z^2 - 1}$? The function
$$
f(z) = \frac{z}{z^2 - 1}
$$
has singularities at $z = \pm 1$. I will expand about the point $z = 1$. Then, with the substitution $w = z+1$,
\begin{align*}
f(x) =& \frac{z}{z^2 - 1} \\[3mm]
=& \frac{... | From where you left off:
\begin{align}
&= \sum_{n=0}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+
\sum_{n=1}^\infty \frac{(z+1)^{n-1}}{2^{n+1}} - \sum_{n=0}^\infty \frac{(z+1)^n}{2^{n+1}}\\[4px]
&=\frac{(z+1)^{-1}}{2}+
\sum_{n=0}^\infty \frac{(z+1)^{n}}{2... | {
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If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A$, $B$, and $C$.
If $\sin^4(x)=A+B\cos(2x)+C\cos(4x)$, then find $A, B$, and $C$.
Would I use half-angle identities for this problem?
| By the double-angle formula for $\cos{2x}$,
$$ \cos{2x} = \cos^2{x} - \sin^2{x} = 2\cos^2{x}-1 = 1-2\sin^2{x}. $$
Rearranging,
$$ \sin^2{x} = \frac{1}{2}(1-\cos{2x}). $$
Squaring both sides an expanding,
$$ \sin^4{x} = \frac{1}{4}(1-2\cos{2x}+\cos^2{2x}), $$
and then using the double-angle formula $\cos{4x} = 2\cos^2{x... | {
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Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$
The expression $\sin(3x)$ is equivalent to:
A. ...
My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x... | Almost there (from your 5th line):
$$
\sin x (4\cos^2 x - 1) = \sin x (3\cos^2 x + \cos^2 x - 1)
$$
$$
= \sin x (3\cos^2 x + (1-\sin^2 x) - 1) = 3\sin x\cos^2 x-\sin^3x.
$$
| {
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"answer_count": 4,
"answer_id": 2
} |
Using Taylor's theorem show that $1 - \frac{x^2}{2} < \cos x < 1- \frac{x^2}{2} + \frac{x^4}{24}$ Using Taylor's theorem show that
$1-\frac{x^2}{2} < \cos x < 1-\frac{x^2}{2} + \frac{x^4}{24} \, \text{ for all } x \in \mathbb{R}$
I know this is true because of the Alternating series truncation error that alternates in... | We keep integrating the inequality $\cos x\le x$, for $x\ge 0$, in the interval $[0,x]$, and we recursively obtain
$$
\cos x \le 1\quad\Longrightarrow\quad
\sin x \le x\quad\Longrightarrow\quad
1-\cos x \le \frac{x^2}{2!}\quad\Longrightarrow\quad
1- \frac{x^2}{2!}\le\cos x \quad\Longrightarrow\quad
x-\frac{x^3}{3!}\le ... | {
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"answer_count": 2,
"answer_id": 1
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Factoring $x^5 - 5x^4 + 1$ The original problem was stated like this:
Prove that the polynomial $x^5 - 5x^4 + 1$ does not have roots of multiplicity 4.
So, factoring the polynomial would answer the question, but I don't know how to do it.
Using the root test, ${1, -1}$ should be roots, but If I divide by $x-1$ or $x+1$... | extended gcd in $\mathbb Q[x].$ the original polynomial and its derivative are coprime; there are no multiple roots, not even double.
$$ \left( x^{5} - 5 x^{4} + 1 \right) $$
$$ \left( x^{4} - 4 x^{3} \right) $$
$$ \left( x^{5} - 5 x^{4} + 1 \right) = \left( x^{4} - 4 x^{3} \right) \cdot \colo... | {
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"answer_id": 5
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Differentiation - Finding derivatives of this function -
$h(y)= (y^{-3}+2y)(2+3y^3)^{-1/2} $
I tried and weren't sure on how to simplify it to get the answer
$\frac{8-27y^{-1} - 12y^4-6y^3}{2(2+3y^3)^{3/2}} $
This is my try -
$ (y^{-3} + 2y) (\frac{-1}{2} (2+3y^3)^{-3/2} (0+9y^2)) + (2+3y^3)(-3y^{-4} + 2) $
I sim... | There are two typos in your calculation
We obtain applying the product rule
\begin{align*}
&(y^{-3}+2y)(2+3y^3)^{-1/2}\\
&\qquad= (y^{-3} + 2y) \left(-\frac{1}{2} (2+3y^3)^{-3/2} (0+9y^2)\right) + (2+3y^3)^{\color{blue}{-1/2}}(-3y^{-4} + 2)\tag{1}\\
&\qquad= (2+3y^3)^{-3/2} \left(-\frac{9}{2}\color{blue}{y^2} (y^{-... | {
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How to find the sum of the series $(1) + (2 + 3) +(4 + 5 + 6) + \cdots$ to $n$ terms? How to find the sum of this series?
$$(1) + (2 + 3) +(4 + 5 + 6) + \cdots \text{ to }\, n \, \text{ terms }$$
Answer is given as:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$
| The $k$th term is composed of $k$ integers, so the first $n$ terms, cumulatively, will be the composed of $\displaystyle F(n) = \sum_{k=1}^{n} k = \dfrac{n(n+1)}{2}$ consecutive integers overall.
And to actually sum up those consecutive integers, we apply the same idea again.
$$S = F(F(n)) = F\left(\frac{n(n+1)}{2}\ri... | {
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Rational numbers inequality proof - is this proof valid?
Knowing that $x^2+y^2 = 2 $ prove that $x+y \le 2$
I rewrite this as
$$x^2+y^2 \ge x+y$$
Now, multiply both sides by 2
$$x^2 -2x + 1 + y^2-2y + 1 -2 +x^2 + y^2 \ge 0$$
I substitue $2$ for $x^2+y^2$
$$(x-1)^2+(y-1)^2\ge0$$
Which is true
| Using Cauchy-Schwarz:
$$(1+1)(x^2+y^2)\ge(x+y)^2 \Rightarrow (x+y)^2\le4 \Rightarrow |x+y|\le2.$$
| {
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Minimum value of $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$ Let $x,y,z>0$ and $x+y+z=xyz$. What is the minimum value of $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}?$$
In the case when $x=y=z$, the equation $x+y+z=xyz$ translates to $3x=x^3$, or $x=\sqrt{3}$. If $A$ denotes the quantity that we want to minimize, then $... | The constraint allows the substitution $x=\tan A, y = \tan B, z = \tan C$ for some acute triangle $\triangle ABC$. Further from rearrangement inequality we have:
$$\sum_{cyc} \frac{x}{y^2} \geqslant \sum_{cyc} \frac1x = \sum_{cyc} \cot A$$
Now $x \mapsto \cot x$ is convex for $x \in (0, \frac{\pi}2)$, so we may use Je... | {
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Substitution Integral: $\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx$ $$\int_0^3\frac {\sqrt{x}}{\sqrt{x}+\sqrt{3-x}} dx $$ I tried with substitution $t=3-x$ but I don't know what to do next.
Also how do I solve this limit: $$\lim_{x\to\infty}\frac{1}{x}\int_0^x\frac{dt}{2+\cos t} $$ I saw that most problems like... | Form the properties of definite integral we know that:
$$\int_b^af(x)\ dx=\int_b^af(a+b-x)\ dx\\
\implies \int_0^af(x)\ dx=\int_0^af(a-x)\ dx$$
Applying this property to your problem:
Let,
\begin{align*}I&=\int_0^3\dfrac{\sqrt x}{\sqrt{x}+\sqrt{3-x}}\ dx\\
&=\int_0^3\dfrac{\sqrt{3-x}}{\sqrt{3-x}+\sqrt{3-(3-x)}}\ dx\\
... | {
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Limit $\lim_{n\to\infty}\int _{2n}^{3n}\frac{x+3}{x^2+1}\,dx?$ How to evaluate
$$\lim_{n\to\infty}\int _{2n}^{3n}\frac{x+3}{x^2+1}\,dx?$$
The answer in the book is $\ln\frac a2$ where $a$ is a constant.
Attempt:
$\lim_{n\to\infty }\int_{2n}^{3n}\:\frac{x+3}{x^2+1}dx =\lim_{n\to\infty}[\frac12\ln(x^2+1)+3\arctan x]^{3n}... | Well, for the integral:
$$\mathscr{I}_\text{n}:=\int_{2\text{n}}^{3\text{n}}\frac{3+x}{1+x^2}\space\text{d}x=\int_{2\text{n}}^{3\text{n}}\frac{3}{1+x^2}\space\text{d}x+\int_{2\text{n}}^{3\text{n}}\frac{x}{1+x^2}\space\text{d}x\tag1$$
*
*Substitute $\text{u}=1+x^2$:
$$\int_{2\text{n}}^{3\text{n}}\frac{x}{1+x^2}\space... | {
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Prove that $\sum_{cyc}\frac{2a}{\sqrt{3a+b}}\le \sqrt{3(a+b+c)}$ Prove that if $a,b,c\in \mathbb{R}^+$ then $\sum_{cyc}\frac{2a}{\sqrt{3a+b}}\le \sqrt{3(a+b+c)}$
The equality case seems to hold if $a=b=c$ so I was thinking of using AM-GM but cant operate in any creative ways. Please help or provide alternate solution... | By C-S $$\left(\sum_{cyc}\frac{a}{\sqrt{3a+b}}\right)^2\leq(a+b+c)\sum_{cyc}\frac{a}{3a+b}.$$
Thus, it remains to prove that
$$\sum_{cyc}\frac{a}{3a+b}\leq\frac{3}{4}$$ or
$$\sum_{cyc}\left(\frac{a}{3a+b}-\frac{1}{3}\right)\leq\frac{3}{4}-1$$ or
$$\sum_{cyc}\frac{b}{3a+b}\geq\frac{3}{4},$$
which is true by C-S:
$$\sum_... | {
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Solve $\sqrt{3+\sqrt{3+x}}=x$
Solve
$$\sqrt{3+\sqrt{3+x}}=x$$
My try:
$$\sqrt{3+\sqrt{3+x}}=x \\
3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?
| There is no need to deal with any quartic equation in $x$. The key is the function
on LHS $x \mapsto \sqrt{ 3 + \sqrt{3+x}}$ is a composition of the map
$x \mapsto \sqrt{3+x}$ with itself.
The map $x \mapsto \sqrt{3+x}$ is strictly increasing whenever it is defined (i.e $x \ge -3$)
If $\sqrt{3+x} > x$, then $\sqrt{3 +... | {
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"answer_count": 4,
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Induction: $\frac1{1 \cdot 5} + \frac1{5 \cdot 9} + \cdots + \frac1{(4n-3)(4n+1)}$
For $n \ge 1$, let
$$a_n = \dfrac1{1 \cdot 5} + \dfrac1{5 \cdot 9} + \cdots + \dfrac1{(4n-3)(4n+1)}.$$
Guess a simple explicit formula for $a_n$ and prove it by induction.
Hi, I'm trying to answer this question. I was not provided ... | $$a_n=\frac{1}{4}\sum_{k=1}^n\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{4n-3}-\frac{1}{4n+1}\right)=$$
$$=\frac{1}{4}\left(1-\frac{1}{4n+1}\right)=\frac{n}{4n+1}$$
| {
"language": "en",
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What are the roots of the equation $8x^3-6x+1=0$? Let the roots be, $a,b$ and $c$
We have $$a+b+c=0\tag1$$ $$ab +bc +ca=-\frac68\tag2$$ $$abc = -\frac18\tag3$$
| Note that
$$\cos 3\theta=4\cos^3\theta-3\cos\theta$$
Let $x=\cos\theta$. Then
\begin{align*}
8\cos^3\theta-6\cos\theta+1&=0\\
2\cos3\theta+1&=0\\
\cos3\theta&=\frac{-1}{2}\\
3\theta&=2n\pi\pm\frac{2\pi}{3}\\
\theta&=\frac{2n\pi}{3}\pm\frac{2\pi}{9}
\end{align*}
So $x=\cos\frac{2\pi}{9}\approx 0.7660$, $\cos\frac{4\pi}... | {
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Prove an inequality involving powers of two Let $a,b,c$ be mutually distinct positive integers.
Prove
$$2^{a+b} + 2^{b+c} + 2^{a+c} < 2^{a+b+c} + 1$$
Any hints on how to start proving it?
| Note that the left-hand side is a sum of distinct powers of $2$, so (assuming WLOG that $a < b < c$)
$$2^{a+b}+2^{b+c}+2^{a+c}<\sum_{k=0}^{b+c}2^k=2^{b+c+1}-1<2^{a+b+c}+1.$$
| {
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"answer_id": 1
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How to find the "most upright" orthogonal vector to another vector? Given an arbitrary vector, I can find any orthogonal vector by solving $ax + by + cz = 0$.
I want to find the "most upright" orthogonal (unit) vector, where $z$ is maximized. There must be a straightforward closed form solution to this, right?
| Restating your problem (and adding a condition), you are given a vector $\vec{v}=\langle a,b,c\rangle$ and you want to find a vector $\vec{w}=\langle x,y,z\rangle$ so that
*
*$\vec{w}\cdot\vec{v}=0$,
*$\|\vec{w}\|=1$
*$\vec{w}\cdot\vec{e}_3$ is maximized
As you state, you want to maximize $z$ subject to
\begin{a... | {
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Finding coefficient of $x^{15}$ $$(x+x^2+x^3+x^4+x^5)\cdot (x^2+x^3+x^4+…)^5$$
I have done a little :
$$x(1 + x+x^2+x^3+x^4)\cdot x^{10}(1 + x^2+x^3+…)^5$$
By generating functions:
$$\begin{align}&x^{11}\cdot\frac{1 - x^5}{1-x}\cdot\frac{1}{(1-x)^5}\\[1ex]
\implies &x^{11}(1 - x^5)\cdot\frac{1}{(1-x)^6}\\[1.5ex]
\impl... | It is convenient to use the coefficient of operator $[x^k]$ to denote the coefficient of $x^k$ in a series.
We obtain
\begin{align*}
\color{blue}{[x^{15}]}&\color{blue}{x^{11}(1-x^5)\sum_{n=0}^\infty \binom{n+5}{5}x^n}\\
&=[x^4](1-x^5)\sum_{n=0}^\infty\binom{n+5}{5}x^n\tag{1}\\
&=[x^4]\sum_{n=0}^\infty\binom{n+5}{5}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which... |
I thought it might be instructive to present an approach that relies on elementary analysis only . To that end, we now proceed.
First, note that $\int_0^1 x^{2n}\,dx=\frac{1}{2n+1}$. Therefore, we can write
$$\begin{align}
\sum_{n=0}^N \frac{(-1)^n}{2n+1}&=\sum_{n=0}^N(-1)^n\int_0^1x^{2n}\,dx\\\\
&=\int_0^1 \sum_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +... | $$
\begin{aligned}
\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}
& = \lim _{t\to 0}\left(\frac{\frac{1}{\sqrt{\left(t+8\right)\:+1}}\:-\:\frac{1}{3}}{\left(t+8\right)-8}\right)
\\& = \lim _{t\to 0}\left(\frac{\left(3-\sqrt{t+9}\right)\sqrt{t+9}}{3t^2+27t}\right)
\\& = \lim _{t\to \:0}\left(-\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ then $x+\dfrac{1}{x}=?$
let :
$$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$$
then :
$$x+\dfrac{1}{x}=?$$
My try :
$$\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18}=\sqrt3+\s... | Suppose we say $u = \sqrt 2, v = \sqrt 3$
then
$\sqrt 2 + \sqrt 3 + \sqrt 4 = u+v + 2$
and $(\sqrt 3 +\sqrt 6 + \sqrt 16 + \sqrt 18) = (v+ uv + 3u + 4) = (uv + 2u+2)+(u+v+2)$
in the first term on the right hand side say $2 = u^2$
$(v + 2u+u^2)+(u+v+2) = u(u+v+2) + (u+v+2) = (u+1)(u+v+2)\\
x = \sqrt 2 +1$
$\frac 1x = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
I'm stuck at integrating an arc length over a circle + angle between curves Let $(\phi, \theta)$ be the usual spherical coordinates. And let $\gamma (t) = (\phi (t), \theta (t))$ be the curve given by
$$ \phi (t) = \ln\left(\cot\left(\frac{\pi}{4} - \frac{t}{2}\right)\right),$$ $$ \theta (t) = \frac{\pi}{2} - t.$$
$t \... | \begin{align}
& \int_0^1 \sqrt{ \left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt]
= {} & \int_0^1 \sqrt{\left(\frac{\sin\theta}{\cos t} \right)^2+1} \, dt = \int_0^1 \sqrt{ \left( \frac{\cos t}{\cos t} \right)^2 + 1} \, dt = \int_0^1 \sqrt 2 \, dt = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2296761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Point of Tangency between Two Circles I made sure to look around this site a couple times before asking my question. A lot of people have already asked how to find the point of tangency or intersection between two circles, but none of them could really help me solve the problem.
I'm being asked to find the point of tan... | The equations are:
\begin{gather}
x^2 + y^2 = 5 \tag{1} \\
x^2 -12x + y^2 + 6y + 25 = 0 \tag{2}
\end{gather}
If you substitute (1) into (2), you get
\begin{align}
-12 x + 6y + 30 &= 0 \implies y=2x-5\tag{3}
\end{align}
Substitute this back into (1):
\begin{align}
x^2 + (2x-5)^2 &= 5 \\\implies
x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Number of ways to roll a 6-sided die and a 12-sided die (alternating) to sum to n There are two dice, a six-sided die and a twelve-sided die.
You roll the dice, alternating between the two, starting with the six-sided die. You keep track of the running total and stop when the total reaches $n$.
Show that the number of ... | It's actually very simple:
*
*$x+x^2+\dots+x^6$ gives the number of ways to roll $n$ on the d6, and similarly $x+x^2+\dots+x^{12}$ for the d12
*Their product $(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})$ gives the number of ways to roll $n$ as the sum of a d6-d12 pair
*$\frac1{1-(x+x^2+\dots+x^6)(x+x^2+\dots+x^{12})}$ g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Maximum product when the sum is constant and the variables are positive integers Today I came across an olympiad problem which after some time translated into a simple problem. I was supposed to prove that given $x+y+z = 73$ and $x,y,z ∈ I^+$. Show that the maximum value of $xyz = 14400$.
Now using the AM-GM inequality... | Let
$$x + y + z = c \tag{A}$$
and
$$f(x,y,z) = xyz$$
then using (A) to sub we have
\begin{align}
f(x,y) &= xy(c - x - y) \\
&=xyc - x^2y - xy^2
\end{align}
Take partial derivatives
\begin{align}
f_x(x,y) &= yc - 2xy - y^2 \tag{1}\\
f_y(x,y) &= xc - 2xy - x^2 \tag{2}
\end{align}
to find critical points of the 3d surface... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$
My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lflo... | Note that
\begin{align}
\frac{2\cos x}{\cos 2x}-2&=\frac{2\cos x-2\cos 2x}{\cos 2x}\\
&=\frac{4\sin\frac{3x}{2}\sin\frac{x}{2}}{\cos 2x}\\
&>0
\end{align}
for $x\in(\frac{-\pi}{4},\frac{\pi}{4})$.
We also have
\begin{align}
3-\frac{2\cos x}{\cos 2x}&=\frac{3\cos 2x-2\cos x}{\cos 2x}\\
&=\frac{6\cos^2x-2\cos x-3}{\cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
$\int_{\alpha}z^2 \log\left(\frac{z+1}{z-1}\right)\,dz$ where $\alpha$ is $|z-1|=1$ Calculate
$$\int_{\alpha}z^2 \log\left(\frac{z+1}{z-1}\right)\,dz$$
where $\alpha$ is $|z-1|=1$ and the initial point of integration $z_1=1+i$.
The function isn't analytic at $1$ and $-1$. Is there a theorem to solve it?
Transformation... | $g(z)=z^2 (\log(1+z)-i\pi)$ is analytic on $|z-1| < 2$ so that $\int_{|z-1| = 1} g(z)dz = 0$.
Thus, it reduces to finding a primitive of $$h(z) = -z^2 \log(1-z) = \sum_{n=1}^\infty \frac{z^{n+2}}{n}$$ ie. $$H(z) = \sum_{n=1}^\infty \frac{z^{n+3}}{n(n+3)}=\frac13\sum_{n=1}^\infty z^{n+3} (\frac{1}{n+3}-\frac{1}{n}) \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Does $\{f_n\}$ converge uniformly on $(0,1) $ and $(0,\infty)?$
Let $g_n(x)=\sin^2(x+\frac{1}{n}),x\in (0,\infty)$
and $f_n(x)=\int _0^x g_n(t) dt$.
Then show that $\{f_n\}$ converges uniformly on $(0,1)$.
Does $\{f_n\}$ converge uniformly on $(0,\infty)?$.
$f(x)=\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty}\int _0... | Note that
$$\int_0^x \sin^2(t +1/n) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x+2/n) + \frac{1}{4} \sin (2/n),$$
and
$$\int_0^x \sin^2(t) \, dt = \frac{x}{2} - \frac{1}{4}\sin(2x) .$$
Hence,
$$|f_n(x) - f(x)| \leqslant \frac{1}{4}\left|\sin(2x+2/n) - \sin(2x) \right|+ \frac{1}{4} |\sin(2/n)| \leqslant \frac{1}{2n} + \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\int \sqrt{2x + 3} dx$ Integrate: $\int \sqrt{2x + 3} \ dx$
Doing some guesswork gives me: $\frac{1}{2}(2x + 3)^{3/2}$ to try. Differentiating this gives: $2\frac{3}{2}\frac{1}{2}(2x + 3)^{1/2}$ = $\frac{6}{4}(2x + 3)^{1/2}$
Too bad the answer should be: $\frac{1}{3}(2x + 3)^{1/2}$, so clearly I'm missing so... | $$\int\sqrt{2x+3}dx=\frac{1}{2}\int(2x+3)^{\frac{1}{2}}d(2x+3)=$$
$$=\frac{1}{2}\frac{(2x+3)^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}\sqrt{(2x+3)^3}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
minimizing vectors Could please check whether my solution is right?
Q. $$A=\begin{pmatrix}
1 & 1 \\
1 & 1 \\
0 & 0 \\
\end{pmatrix}$$Find the set of vectors $x$ that minimize the value $$||
A
x-
\begin{pmatrix}
1 \\
2 \\
3 \\
\end{pmatri... | Problem statment
Given the matrix
$$
\mathbf{A} =
\left[
\begin{array}{cc}
1 & 1 \\
1 & 1 \\
0 & 0 \\
\end{array}
\right]
\in \mathbb{C}^{m\times n}_{\rho}
$$
and the data vector
$$
b =
\left[
\begin{array}{cc}
1 \\
2 \\
3 \\
\end{array}
\right]
$$
find the least squares solution
$$
x_{LS} = \left\{
x\in\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Giving a closed form combinatorial solution to finding the larger of two balls drawn from an urn I understand how to solve the following problem by listing out the sample space; however, I would like to know how to derive a combinatorial solution to the problem (that is, I would like to give a closed form solution to n... | Case 0: $X=0$
The only way that the largest ball drawn is $0$ is if we pick $0$ both times, so
$$n(X=0) = \boxed{1}$$
Case 1: $X=1$
We can either have both draws be $1$, or one draw be $0$ and the other $1$. There are two ways to draw $0$ and $1$, one for each order, so we get
$$n(X=1) = 1+2 = \boxed{3}$$
Case 2: $X=2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
What are the different methods to determine the derivative of $f(x)=\sqrt{1+x}$? The methods that I can think of are:
1) Chain rule
2) Binomial series of $f(x)$
3) Through the formula $f\prime(g(x))=\frac{1}{g\prime (x)}$ at a particular point
What are the other methods to determine the derivative of $f(x)$?
| Using logarithmic differentiation:
\begin{align}
f'(x) &= f(x) \cdot \frac{d}{dx} \ln f(x)
= \sqrt{1+x} \cdot \frac{d}{dx} \ln \sqrt{1+x}\\
&= \sqrt{1+x} \cdot\frac{1}{2} \frac{d}{dx} \ln(1+x)
= \sqrt{1+x} \cdot \frac{1}{2} \frac{1}{1+x}\\
&= \frac{1}{2\sqrt{1+x}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate the definite integral $\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$
$$\int_0^{2016} x (x-1)(x-2)(x-3)... (x-2016)\,dx$$
I have tried integration by parts using calling $f (x)=(x-1)(x-2)... (x-2016)$ but it doesn't help. I can't think of anything else. Please help.
| Note that
$$\scriptsize\begin{align}
\int_0^2 x(x-1)(x-2)dx&=\int_{-1}^1 (x-1)x(x+1)dx&=0\\
\int_0^4 x(x-1)(x-2)(x-3)(x-4)dx&=\int_{-2}^2 (x-2)(x-1)x(x+1)(x+2)dx&=0\\
&\vdots\\
\int_0^{2016} x(x-1)(x-2)(x-3)\cdots (x-2016)dx&=\int_{-1008}^{1008} (x-1008)(x-1007)\cdots x\cdots (x+1007)(x+1008)dx&=\color{red}0\\
\end{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2310750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:
Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a... | Hint:$$\frac{a+b+c}{b+c}=1+\frac{a}{b+c}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 0
} |
Finding the geometric shape
Draw the geomatrix region of $|z^2-1|=1$
$$|z^2-1|=1$$
$$|(x+iy)^2-1|=1$$
$$|x^2-y^2-1+ixy|=1$$
$$\sqrt{(x^2-y^2-1)^2+(xy)^2}=1$$
$$\sqrt{(x^2-y^2-1)^2+(xy)^2}=1$$
Let $a=x^2, b=y^2$
$$\sqrt{a^2-ab+b^2-2a+2b+1}=1$$
$$a^2+b^2-ab-2(a+b)=0$$
How should I continue?
| Let us write $ z = r e^{i\varphi} $, then the equation says :
$$1=|z^2-1|^2=(z^2-1)\overline{ (z^2-1) } = \left(r^2e^{2i\varphi}-1\right)\cdot\left(r^2e^{-2i\varphi}-1\right)$$
Expanding the expression above rise in equation (using the Euler's formula in the form $e^{ix}+e^{-ix}=2\cos x$) :
$$1=1+r^4-2r^2 \cos(2\varph... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2313743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\sin(10^{\circ})$, if $\cos(40^{\circ}) = a$ I have this trig problem where it asks to find $ \sin(10^{\circ})$, given $\cos(40^{\circ})=a$.
I tried the write the $\sin(10^{\circ})$ as $\sin(40^{\circ}-30^{\circ})$ to use the sum formula but i didn't get anywhere near to the options given in the test:
a)$\sin(10... | You are on the right lines
\begin{align*}
\sin(10) &= \sin(40-30) \\
&= \sin(40)\cos(30)-\cos(40)\sin(30) \\
&= \frac{\sqrt{3}}{2}\sin(40) - \frac{1}{2}\cos(40) \\
&= \frac{\sqrt{3}}{2}\sqrt{1-\cos^2(40)} - \frac{1}{2}\cos(40) \\
&= \frac{\sqrt{3}}{2}\sqrt{1-a^2} - \frac{1}{2}a \\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2315299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $p, q, r$ are the roots of $x^3 -x+1=0$, what is $p^5 + q^5 + r^5$? Can someone please help us solve this equation. It was on our college entry exam, but no one managed to solve it. The problem is:
Let $p,q,r$ be the roots of $x^3 -x+1=0$. Then $p^5+q^5+r^5 = ?$
Correct answer was $-5$, but no one managed to solv... | Alternative approach: the characteristic polynomial of
$$ M = \begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$$
is exactly $p(x)=x^3-x+1$, since $M$ is the companion matrix of $p(x)$. It follows that the wanted sum is
$$ p^5+q^5+r^5=\text{Tr }\begin{pmatrix}0 & 0 & -1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving algebraic inequality I been struggling trying to solve this inequality by hand
$$x+\sqrt{x+3} < 0$$
how should I proceed?
| $$x+\sqrt { x+3 } <0\Rightarrow \begin{cases} x+3\ge 0 \\ x+3<{ x }^{ 2 } \end{cases}\Rightarrow \begin{cases} x\ge -3 \\ { x }^{ 2 }-x-3>0 \end{cases}\Rightarrow \cap \begin{cases} x\ge -3 \\ x\in \left( -\infty ;\frac { 1-\sqrt { 13 } }{ 2 } \right) \cup \left( \frac { 1+\sqrt { 13 } }{ 2 } ;+\infty \right) \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2320106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Elements needed to derive the Riemann-Siegel Z function On this post I got a comment to motivate the present question:
To prove a zero $\zeta(s_0)=0, s_0 \approx 1/2+iy$ is exactly on the critical line, it is enough to prove the real function $Z(t) = \zeta(1/2+it) e^{i \,\text{arg}(\Gamma(1/4+it/2)}\, \pi^{-it/2}$ cha... | The Riemann functional
$$\large \pi^{\frac{-s}{2}}\Gamma\left(\frac{s}{2}\right)\,\zeta(s)=\pi^{-\frac{1-s}{2}}\Gamma\left(\frac{1-s}{2}\right)\,\zeta(1-s)$$
relates $\zeta(s)$ to $\zeta(s-1)$, as a fraction expressed as
$$\chi(s) = \frac{\zeta(s)}{\zeta(s-1)}
=\frac{\Gamma\left(\frac{1}{2}(1-s)\right)}{\Gamma\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Spectral Decomposition calculating the Eigenvalue I currently have this problem i can't seem to figure out.
A symmetric matrix
.. 6 2/5 ..
S = [.. .. ..]
.. .. ..
It has the Eigenvalues 9, 18 and c is not equel to 9 and 18 with
eigenspace:
a
E9 = Span([ b ])
... | The eigenspaces of a symmetric matrix associated with distinct eigenvalues are always orthogonal. Since all the eigenvalues of your matrix are distinct, this means we must have
$$ \left< \begin{pmatrix} a \\ b \\ -2.5 \end{pmatrix}, \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} \right> = 2a + b - 5 = 0, \\
\left< \begin{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2321451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Alternate method for solving missing area question I recently saw a puzzle in an advert for the website Brilliant.org, which went as follows:
What is the blue area?
Hint: Think outside the box
My answer:
I set the area to be found to $x$, the side length of the square to be $y$, and the sections to be $a,b,c,d$ as ... | Here is an alternative solution. See the graph for labeling:
From the trapezium area formula we get:
$$\frac{a+x}{2}\cdot x=5+A \Rightarrow ax=10+2A-(9+A) \Rightarrow ax=A+1 \ \
(1)$$
$$\frac{b+x}{2}\cdot x=7+A \Rightarrow bx=14+2A-(9+A) \Rightarrow bx=A+5 \ \
(2)$$
$$ab=6 \ \ (3)$$
Now multiply $(1)$ and $(2)$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 12,
"answer_id": 1
} |
Can anyone explain this solution (contractive function, interval, first derivative related)
We are trying to find a positive null $x \geq 0$ of $$f(x) =
\sqrt{x+1}+\frac{\text{sin}(x)}{10}-2$$
Choose $X=\left[0,8\right]$. For $x \in X$ we have that
$$\frac{1}{15}= \frac{1}{6}-\frac{1}{10} \leq f'(x) =
\frac{1}{2\sqr... | Consider $-1\leq \cos(x)\leq 1$ for all $x\in\mathbb{R}$ yields
$$
-\frac1{10}\leq \frac{\cos(x)}{10}\leq \frac1{10}.
$$
And consider that $x\mapsto\sqrt{x+1}$ is monoton increasing. Therefore you get $$
2\leq2\sqrt{x+1}\leq 6
$$
for $x\in[0,8]$ using the boundary points $x=0$ and $x=8$. This yields
$$
\frac16\leq \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Area under the graph - integration
The region $P$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the line $x=a$ . The region $Q$ is bounded by the curve $y= 3x-x^2$ , the $x$-axis and the lines $x=2a$ and $x=a$. Given that the area of $Q$ is twice the area of $P$, find the value of $a$ .
Firstly , on the fir... | Since $a\neq 0$ (as this would be absurd), we can solve your equation as follows
\begin{align}9a^2−2a^3&=18a^2−4a^3\\
9-2a&=18-4a\tag{divide through by $a$}\\
2a&=9\\
a&=\frac 92\end{align}
However, as noted in other answers, this is incorrect.
Your error is somewhere in your calculation of the area of $Q$, you shoul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Sitting in chairs with empty space There are n chairs in a row. In how many ways can a teacher sit k students on these chairs so that no 2 students sit next to each other (and obviously no 2 students sit on 1 chair)?
I was thinking about inclusion-exclusion, that each student will sit next to each other, however it isn... | Consider strings over the alphabet $\{0,1\}$ where $0$ denotes a chair and $1$ denotes a student. The strings that do not contain $11$ are captured by the regular expression $0^* (10^+)^* \{\varepsilon, 1\}$. If we give $0$ a weight of $x$ and $1$ a weight of $xy$ then we are looking for the coefficient of $[x^ny^k]$.
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
What is wrong with my integral calculation? I want to solve the integral $\int_1^\infty\frac{a}{a^2+x^2}dx$ where $a>0$ is a constant
I first tried simplifying so
\begin{align}
\frac{a}{a^2+x^2}
&=\frac{1}{a+\frac{x^2}{a}} \\[6px]
&=\frac{1}{a}\cdot\frac{1}{1+(\frac{x}{a})^2}
\end{align}
which got me
\begin{align}
\int... | \begin{align}
\int_1^\infty\frac{a}{a^2+x^2}dx
&=\frac{1}{a}\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}dx\\[6px]
&=\int_1^\infty\frac{1}{1+(\frac{x}{a})^2}d\frac{x}{a}\\[6px]
&=\left[\arctan \frac{x}{a}\right]_1^\infty\\[6px]
&=\frac{\pi}{2}-\arctan\frac{1}{a}=\arctan a
\end{align}
If $a<0$, then the final step will give ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2334955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Laplace Transform on Beam (w/ Loads) (Sorry about my english I'm a South American student)
The deflection of a fixed beam is zero at $x=0$ and $x=L$ on its boundaries. The beam resists a distributed load of $Wo$ per unit length. The equation is:
$$\frac{d^{4}y}{dx^4}=\frac{W_{0}}{EI}$$
$$0 < x < L$$
Where $Wo$, $E$, a... | Take Laplace Transform of the equation, but perform this with respect to x (where normally, it is performed with respect to time, t):
$$ \mathcal{L} \left[ \frac{d^4 y}{dx^4} \right] = \mathcal{L} \left[ \frac{W_0}{EI} \right] $$
$$ s^4 Y(s) - s^3 y(0) - s^2 y^\prime (0) - s y^{\prime \prime}(0) - y^{\prime \prime \pri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Non-Symmetric Algebraic inequality
Let $a,b,c$ be positive real numbers. Prove that : $$\frac{ab}{3a+b}+\frac{bc}{2c+b}+\frac{ac}{c+2a}\le \frac{2a+20b+27c}{49}$$
| Let $a=6x$, $b=3y$ and $c=2z$.
Hence, we need to prove that
$$\frac{xy}{6x+y}+\frac{yz}{3y+4z}+\frac{zx}{z+6x}\leq\frac{2x+10y+9z}{49},$$
which is true by AM-GM.
Indeed,
$$\frac{xy}{6x+y}+\frac{yz}{3y+4z}+\frac{zx}{z+6x}\leq\frac{x+6y}{49}+\frac{4y+3z}{49}+\frac{6z+x}{49}=\frac{2x+10y+9z}{49}.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\frac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$, find the value of $P(n+1)$ Suppose that $P(x)$ is a polynomial of degree $n$ such that $P(k)=\dfrac{k^{2}}{k^{3}+1}$ for $k=0,1,\ldots,n$. Find the value of $P(n+1)$
I tried by making a $f(x) = (x^3+1) P(x) - ... | Using (Lagrange) polynomial interpolation:
\begin{align*}
P(x) &= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \prod_{j \neq k} \frac{x - j}{k-j} \right) \\
&= \sum_{k=0}^n \frac{k^2}{k^3+1} \left( \frac{x (x-1) \cdots (x - (k-1)) (x-(k+1)) \cdots (x-n) }{ k (k-1) \cdots (k-(k-1)) (k-(k+1)) \cdots (k-n)}\right) \\
&= \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
How to solve these functional equations $f(x) = 1 + f(x) + f(x^2) + \ldots $ and $f(x) = 1 + f(x) + \big(f(x^2)\big)^2 + \ldots $? The identity $\frac{1}{1-f(x)}=1+f(x)+\big(f(x)\big)^2+\big(f(x)\big)^3 + \ldots$ is well known.
Is there somewhere in the literature a method allowing us to solve the following functional ... | There is no solution for your first functional equation. Indeed :
$f\left( x\right) =1+f\left( x\right) +f\left( x^{2}\right) +f\left(
x^{3}\right) +f\left( x^{4}\right) +f\left( x^{5}\right) +f\left(
x^{6}\right) +\cdots $
so
$-1=f\left( x^{2}\right) +f\left( x^{3}\right) +f\left( x^{4}\right)
+f\left( x^{5}\right) +f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.