Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Find length BC of triangle with incircle and circumcircle
Some thoughts... some chord theorem to get the angle between PQ and BC... AB and AC are tangent to the circle, there has to be another theorem about that, perhaps ADE is isosceles and that helps looking at the angled PDE and AEQ, and with all this one should be... |
Let $r$ and $R$ be the radii of the incircle and circumcircle, respectively. Then
$$\frac r{4R} =\sin \frac A2 \sin \frac B2 \sin \frac C2\tag1$$
Note that $ab= PD\cdot DQ =6$ and $ac= PE\cdot EQ =4$
$$r\cos \frac A2 =a \sin\frac A2 =\frac{DE}2=1$$
$$2R\sin A = BC = b+ c=\frac{6+4}{a}=10\sin\frac A2\tag2
$$
which lead... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4065917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given $x^2 +px + q$ has roots -1 & 4, find the values for p & q.
Given $x^2 +px + q$ has roots $-1$ & $4$, find the values for $p$ & $q$.
Attempt:
$$
x = \frac{-p\pm\sqrt{p^2-4q}}{2}\\(p + 2x)^2 = p^2 - 4q\\
p^2 + 4px + 4x^2 = p^2 -4q
\\
q = -x^2 - px
\\
q = -(-1)^2 -p(-1)
\\
q = -1 - p
\\
q = -(4)^2-p(4)
\\
q = -16 ... | Instead of substituting the roots later, you could have substituted them right at the start:
$(-1)^2+p·(-1)+q = 0$.
$(4)^2+p·(4)+q = 0$.
And it is easy from there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4066478",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to evaluate $ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$? I was given the series:
$$ \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ...$$
Making some observations I realized that the $ a_{n} $ term would be the following:
$$ a_... | It is acceptable to split one convergent series in two as long as both are convergent.
$$ 0 < \frac{1}{3} + \frac{2}{3^{2} } + \frac{1}{3^{3}} + \frac{2}{3^{4}} + \frac{1}{3^{5}} + ... < \frac{2}{3} + \frac{2}{3^{2} } + \frac{2}{3^{3}} + \frac{2}{3^{4}} + \frac{2}{3^{5}} + ...$$
and since
$$\frac{1}{3} + \frac{2}{3^{2}... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Computing $\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$ I'm trying to compute
$$I=\int_0^\infty \frac{\ln x}{(x^2+1)^2}dx$$
The following is my effort,
$$I(a)=\int_0^\infty\frac{\ln x}{x^2+a^2}dx$$
Let $x=a^2/y$ so that $dx=-(a^2/y^2)dy$ which leads to
$$I(a)=\int_0^\infty \frac{\ln(a/y)}{a^2+y^2}dy=\int_0^\infty\frac{\ln... | I would like to find a more general integral using trigonometric substitution $x=a\tan \theta.$
$\displaystyle \begin{aligned} \int_0^{\infty} \frac{\ln x}{\left(a^2+x^2\right)^2} d x &=\frac{1}{a^3} \int_0^{\frac{\pi}{2}} \cos ^2 \theta \ln (a \tan \theta) d \theta\\ &=\frac{\ln a}{a^3}\int _0^\frac{\pi}{2} \cos ^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4069094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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Why can the Taylor formula be used to evaluate the limit,Why does its remainder converge to zero？ A simple example $\lim\limits_{x\to0}\frac{\cos(x)-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac{1}{2}$
If taken out alone, its limit is zero.How can I judge that the sum of infinite limits converges to zero.
Such ... | Apply the rules of asymptotic analysis:
$$\frac{\cos x-1}{x^2}=\frac{1-\frac{1}{2}x^2+o(x^2)-1}{x^2}=-\frac 12+o(1).$$
Added: an elementary way to determine the limit:
$$\frac{\cos x-1}{x^2}=\frac{\cos^2 x-1}{x^2(\cos x +1)}=-\underbrace{\frac{\sin^2x}{x^2}}_{\substack{\downarrow\\1^2=1}}\,\underbrace{\frac1{\cos x+1}}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the following limit using Taylor Evaluate the following limit:
\begin{equation*}
\lim_{x\to 0} \frac{e^{x^2} + 2\cos x -3}{\sin^2 x^2}.
\end{equation*}
I know the Taylor series of $e^x$ at $a=0$ is $\sum_{k=0}^{\infty} \frac{x^k}{k!}$. And if we substitute $x$ with $x^2$ we get $e^{x^2}=\sum_{k=0}^{\infty} \fr... | Neither the numerator nor the denominator has been correctly analysed.
The numerator is $O(x^4)$, but is asymptotic not to $x^4$ as your calculation incorrectly assumes, but $\tfrac12x^4+\tfrac{1}{12}x^4=\tfrac{7}{12}x^4$.
The denominator $\sin^2x^2=\tfrac12(1-\cos2x^2)$ has leading order term $\tfrac14(2x^2)^2=x^4$. Y... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $T=\mathop {\lim }\limits_{n \to \infty } {\left( {1+ \frac{{1+\frac{1}{2}+ \frac{1}{3}+ . +\frac{1}{n}}}{{{n^2}}}} \right)^n}$ I am trying to evaluate
$$T = \mathop {\lim }\limits_{n \to \infty } {\left( {1 + \frac{{1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}}}{{{n^2}}}} \right)^n}.$$
My sol... | Your doubt is right - it is not possible to split limit in variable amount of summands, as shows example $1=\lim 1=\lim\left(\frac{1}{n}+\cdots + \frac{1}{n}\right)=\lim \frac{1}{n} +\cdots + \lim\frac{1}{n} = 0$.
On another way you can use representation
$$\sum_{i=1}^{n}\frac{1}{i} = \ln n + \gamma + \varepsilon_{n}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4071409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Why do we multiply by $-1$ here? WolframAlpha solves $$\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2}$$ by multiplying by $-1$ as such:
$$\sin\left(-x+\fracπ4\right)=-\frac{1+\sqrt3}{2\sqrt2}$$ then arcsin both sides, etc. but why multiplying by $-1$ and not just directly find the answer? Mainly say this because I... | Apply trig addition identity,we can do this step by step as the following (there is no requirement to multiply -1 for both sides):
\begin{align}\sin\left(x-\fracπ4\right)=\frac{1+\sqrt3}{2\sqrt2} \end{align}
\begin{align} \sin x\cos\frac{\pi}{4}-\cos x\sin \frac{π}{4}=\frac{1+\sqrt3}{2\sqrt2} \end{align}\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4072398",
"timestamp": "2023-03-29T00:00:00",
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Prove for every $1 \leq n$, that $47\mid3^n4^{2n}-1$ I tried to solve with induction, base $n=1$ is correct such that, $\frac{3*16-1}{47}=1$.
Then assuming that for $n$ the statement is correct such that, $3^n4^{2n} \equiv 0 \pmod {47}$.
We need prove that for $n+1$ is true such that, $\frac{(3*16*48^n)-1}{47}$ .
Then... | Transform ($3^n4^{2n}-1) \to$ ($3^n16^n-1) \to (48^n-1)$
For sum of geometric series
$1+x+x^2+x^3+...+x^{n-1}=\frac{x^n-1}{x-1}$
or
$x^n-1=(x-1)(1+x+x^2+x^3+...+x^{n-1})$
Set x=48,
$48^n-1=(48-1)(1+48+48^2+48^3+...+48^{n-1})$
=
$47 (1+48+48^2+48^3+...+48^{n-1}) \to 47\mid3^n4^{2n}-1$
| {
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"url": "https://math.stackexchange.com/questions/4075029",
"timestamp": "2023-03-29T00:00:00",
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Finding $p$ and $t$ such that the system $px+y+z=1$, $x+2y+4z=t$, $x+4y+10z=t^2$ has one, infinitely-many, or no solutions
Consider the following equations:
$$\begin{align}
px+\phantom{2}y+\phantom{10}z &=1 \\
x+2y+\phantom{1}4z &=t \\
x+4y+10z &=t^2
\end{align}$$
Now find the values of $p$ and $t$ for which
(a) there... | System of equations give the below matrix:
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1 & 2 & 4 & t\\
1 & 4 & 10 & t^2
\end{array}\right)$
We set third row as $(R_3 - 2 \cdot R_2)$ and second row as $(R_2 - 2 \cdot R_1)$.
$\left(\begin{array}{rrr|r}
p & 1 & 1 & 1\\
1-2p & 0 & 2 & t-2\\
-1 & 0 & 2 & t^2-2t
\end{array}\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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In checking whether $\tan^{-1}(2\sin^2x-1)+C$ agrees with $\tan^{-1}(\tan^2x)+C$, why am I getting $\sin^{2} 2x = 2$? So this began with the integral-
$$\int \frac{\sin 2x}{\sin^4x + \cos^4x}dx$$
The integral is simple, I solved it as following-
$$\int \frac{\sin 2x}{\sin^4x + \cos^4x}dx=\int \frac{2\sin x\cos x}{\sin^... | Using
$$\tan(A-B) = \frac{\tan A - \tan B}{1+\tan A \tan B}$$
we have that
$$\tan\left(\tan^{-1}(\tan^2 x) - \tan^{-1}\left(2\sin^2 x - 1\right)\right) = 1$$
So the two solutions are off by a $\frac{\pi}{4}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given a triangle $\triangle ABC$, which fulfills the equations $6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$, find the value of $\angle C$ Given a triangle $\triangle ABC$, which fulfills the equations $6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$, find the value of $\angle C$.
I though of solving it ... | $6\cos{A}+2\cos{B}=7$ and $6\sin{A}-2\sin{B}=\sqrt{3}$
Squaring and adding these two equations, we get
$$36(\sin^2 A+\cos^2 A)+4(\cos^2 B+\sin^2 b)+24(\cos A \cos B-\sin A \sin B)=52$$
$$\implies 36+4+24\sin (A+B)=52 \implies \cos(A+B)=1/2=-\cos C \implies C= 2\pi/3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4079650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where does the plus-minus come from in the quadratic formula? In the formula
$$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
for solving quadratic equations, where does the $\pm$ come from?
The square root already results in both a positive and negative term, is the $\pm$ not therefore extraneous?
| What about completing squares?
\begin{align*}
ax^{2} + bx + c & = a\left(x^{2} + \frac{bx}{a}\right) + c\\\\
& = a\left(x^{2} + \frac{bx}{a} + \frac{b^{2}}{4a^{2}}\right) + c - \frac{b^{2}}{4a}\\\\
& = a\left(x + \frac{b}{2a}\right)^{2} + \frac{4ac - b^{2}}{4a}
\end{align*}
Hence we deduce that
\begin{align*}
ax^{2} + ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I prove this formula $\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x=\frac{1}{a}\int_0^\infty f(x^2+2ab)\mathrm{d}x$? While solving this trigonometric integral :
$$\int_0^\infty \sin\left(a^2x^2+\frac{b^2}{x^2}\right)\mathrm{d}x$$
I came across this formula :
$$\int_0^\infty f\left( a^2x^2+\frac... | With $a,\>b>0$
\begin{align}
\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right){d}x
\overset{x\to \frac b{ax}}= &
\int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \frac b{ax^2} dx\\
= & \frac12 \int_0^\infty f\left( a^2x^2+\frac{b^2}{x^2}\right) \left(1+\frac b{ax^2}\right) dx\\
= & \frac1{2a}\int_0^\infty f\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4085987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many five-digit numbers can be formed using digits $1,2,3,4,5,6$ which are divisible by $3$ and $5$? How many five-digit numbers can be formed using digits $1,2,3,4,5,6$ which are divisible by $3$ and $5$, without any of the digits repeating?
A number is divisible by $5$ if and only if last digit is $0$ or $5$ and;... | hint
The mostright digit should be $ 5$.
now, we have to choose $ 4 $ digits from $5 $. there are five possibilities
$$\{1,2,3,4\}\;;\;\{1,3,4,6\}\;;\;\{1,2,4,6\}\;;\;\{2,3,4,6\}\;;\;\{1,2,3,6\}$$
take those for which the (sum +5) is divisible by $ 3$.
we find
$$\{1,2,4,6,5\}\;;\;\{1,2,3,4,5\}$$
The final answer is $2\... | {
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"timestamp": "2023-03-29T00:00:00",
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Geometric Proof of $a^2-b^2=(a+b)(a-b )$ and its applications I am teaching a student on the subject of factoring. One commonly used formula is $a^2-b^2=(a+b)(a-b) $. It is easy to prove it from RHS to LHS algebraically, but how to prove it geometrically? I would also like to find some of its applications, such as this... | Hint:
$$a^2 - b^2 = (a-b)b + (a-b)b + (a-b)^2 $$
$$ = (a-b)[b + b + (a-b)] $$
or
$$a^2 - b^2 = (a-b)a + (a-b)b$$
$$ = (a-b)(a+ b) $$
Draw two squares inside each other ($a>b$) and look at the three or two leftover rectangular areas.
| {
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"url": "https://math.stackexchange.com/questions/4089139",
"timestamp": "2023-03-29T00:00:00",
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Find the value of $A=\frac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt{7}}}-\frac{\sqrt{2\left(3+\sqrt7\right)}}{\sqrt{3-\sqrt7}}$ Find the value of $$A=\dfrac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt{7}}}-\dfrac{\sqrt{2\left(3+\sqrt7\right)}}{\sqrt{3-\sqrt7}}.$$
Answer: $A=1$
Since $$8+2\sqrt7=8+2\cdot1\cdot\sqrt{7}=1^2+2\cdot1\cdo... | $\begin{align}&A = \frac{3\sqrt{8+2\sqrt7}}{\sqrt{8-2\sqrt7}}\frac{\sqrt{8+2\sqrt7}}{\sqrt{8+2\sqrt7}}- \frac{\sqrt2\sqrt{3+\sqrt7}}{\sqrt{3-\sqrt7}}\frac{\sqrt{3+\sqrt7}}{\sqrt{3+\sqrt7}}\\\Rightarrow&A=\frac{3(8+2\sqrt7)}{6}-\sqrt2\left(\frac{3 +\sqrt7}{\sqrt2}\right) = 4+\sqrt7-3-\sqrt7=1\end{align}$
| {
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"timestamp": "2023-03-29T00:00:00",
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If x, y, w, z >0 and $x^4$+$y^4$+$w^4$+$z^4$ <=4 prove 1/$x^4$+1/$y^4$+1/$w^4$+1/$z^4$>=4 I would appreciate suggestions to solve:
If x, y, w, z > 0 and $x^4$ + $y^4$ + $w^4$ + $z^4$ <=4 prove the following:
1/$x^4$ + 1/$y^4$ + 1/$w^4$ + 1/$z^4$ >= 4
From plugging in numbers into Excel, it looks like x, y, w, z must be... | Remember the Cauchy-Schwarz inequality:
$$(x_1y_1+ \ldots +x_ny_n)^2\leq (x_1^2+ \ldots + x_n^2)(y_1^2+\ldots +y_n^2)$$
In this case we have $n=4$. Set $x_1=x^2, x_2=y^2, x_3=z^2, x_4=w^2$ and $y_1=1/x^2, y_2=1/y^2, y_3=1/z^2, y_4=1/w^2$. Now, applying the inequality we have
$$4^2 \leq (x^4+y^4+z^4+w^4)\left( \frac{1}{... | {
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Calculate $\sum_{i=1}^{n-1} i\alpha^{2i}$ I try to calculate $\sum_{i=1}^{n-1} i\alpha^{2i}$
I think one week but still have no ideas to this sigma,thank everyone
| Call $S = \sum_{i=1}^{n-1} i \alpha^{2i}=\sum_{i=1}^{n-1} i (\alpha^2)^{i}$ and $\alpha^2 = \beta$. If $\beta \neq 1$, then \begin{align} \beta S = \sum_{i=1}^{n-1} i \beta^{i+1} &= \sum_{i=1}^{n-1}\left((i+1)-1\right) \beta^{i+1} \\ &=\sum_{i=1}^{n-1} (i+1)\beta^{i+1} - \sum_{i=1}^{n-1} \beta^{i+1} \\ &= \sum_{i=2}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4096461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I solve $x\left(y^2+z\right)z_x-y\left(x^2+z\right)z_y=\left(x^2-y^2\right)z$? I have this right now:
$$x\left(y^2+z\right)z_x-y\left(x^2+z\right)z_y=\left(x^2-y^2\right)z$$
$$\frac{dx}{x\left(y^2+z\right)}=\frac{dy}{-y\left(x^2+z\right)}=\frac{dz}{\left(x^2-y^2\right)z}$$
I get the first first integral like th... | In the same way you found the first identity you can also sum up
$$
ds=\frac{dx/x }{y^2+z}=-\frac{dy/y}{x^2+z}=\frac{dz/z}{x^2-y^2}
$$
to get
$$
ds=\frac{dx/x+dy/y+dz/z}{0}\implies xyz=C_2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4097001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Minimize $3\sqrt{5-2x}+\sqrt{13-6y}$ subject to $x^2+y^2=4$
If $x, y \in \mathbb{R}$ such that $x^2+y^2=4$, find the minimum value of $3\sqrt{5-2x}+\sqrt{13-6y}$.
I could observe that we can write
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{x^2+y^2+1-2x}+\sqrt{x^2+y^2+9-6y}$$
$\implies$
$$3\sqrt{5-2x}+\sqrt{13-6y}=3\sqrt{(x-1... | Comment: you may use this modification:
We rewrite final relation as:
$$A=\sqrt{[3(x-1)=a]^2+(3y=b)^2}+\sqrt{(x=a')^2+(y-3=b')^2}$$
and use this inequality:
$\sqrt{a^2+b^2}+\sqrt{a'^2+b'^2}\geq\sqrt{(a+a')^2+(b+b'^2)}$
we get:
$A\geq\sqrt{16(x^2+y^2)-24(x+y)+18}$
$x^2+y^2=4$
$\Rightarrow$
$A\geq \sqrt{82-24(x+y)}$
If $... | {
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"url": "https://math.stackexchange.com/questions/4098030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$ Integrate $\int\frac{x}{(x^4-1)\cdot\sqrt{x^2+1}} \, dx$
I tried substituting $x=\tan(t)$ in order to get away with square root. ($\:dx=\frac{1}{\cos^2(t)}dt\:$)
$\sqrt{x^2+1}=\frac{1}{\cos(t)}\:\:$ and $\:\:x^4-1=\frac{\sin^4(t)-\cos^4(t)}{\cos^4(t)}$
Now after ... | Note
\begin{align}
\int\frac{\sin t\cos^2t}{\sin^4t-\cos^4t} \, dt
= &\int\frac{\sin t\cos^2t}{(1-\cos^2t)^2-\cos^4t} \, dt
=\int\frac{\sin t\cos^2t}{1-2\cos^2t} \, dt\\
= & \frac12 \int\left( 1- \frac{1}{1-2\cos^2t} \right) \, d(\cos t)\\
=&\frac12\cos t -\frac1{4\sqrt2}\ln \frac{1+\sqrt2\cos t}{1-\sqrt2\cos t}
\end{... | {
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"url": "https://math.stackexchange.com/questions/4098156",
"timestamp": "2023-03-29T00:00:00",
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"answer_count": 4,
"answer_id": 2
} |
Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$ Integrate $\int\frac{3x^2-1}{\sqrt{x^2+x-1}}dx$
I solved this integral by euler substitution by replacing
$\sqrt{x^2+x-1}=x+t$
but it's not allowed by the problem.
p.s Is there any other method to solve with?
Thank you in advance :)
| Hint:
We observe that
$$p':=\left(\sqrt{x^2+x-1}\right)'=\frac{2x+1}{2\sqrt{x^2+x-1}},$$
$$q':=\left(x{\sqrt{x^2+x-1}}\right)'=\frac{4x^2+3x-2}{2\sqrt{x^2+x-1}}.$$
Also, by completing the square,
$$x^2+x-1=\left(x+\frac12\right)^2-\frac54=\frac54\left(\left(\frac{2x+1}{\sqrt5}\right)^2-1\right)$$
and
$$r':=\left(\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4100874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$ Evaluate $\int \frac{x^3+4x^2+x-1}{x^3+x^2}dx$.
Where do I start with this integral? I can easily see that it is possible to fator $x^{2}$ out on the denominator and use partial fractions. The numerator is also factorable but it does not have any integer roots. Can someone... | \begin{align}
\frac{x^3+4x^2+x-1}{x^3+x^2}
&= \frac{\color{red}{x^3 + x^2} + 3x^2+x-1}{x^3+x^2} \\
&= \color{red}{1} + \frac{3x^2+x-1}{x^3+x^2} \\
\end{align}
Now notice that the derivative of the denominator is almost the numerator (but not quite). But let's start with:
\begin{align}
\int \frac{3x^2+x-1}{x^3+x^2}dx
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve for z for this inverse matrix? How do I go about solving this for z?
$$
\begin{pmatrix}
a & b & c \\
1 & 2 & 3 \\
d & e & f \\
\end{pmatrix}
^{-1}= \begin{pmatrix}
1 & 2 & 3 \\
x & y & z \\
4 & 5 & 6 \\
\end{pmatrix}
$$
| As :
$$\underbrace{\begin{pmatrix}
a & b & c \\
\color{red}{1} & \color{red}{2} & \color{red}{3} \\
d & e & f \\
\end{pmatrix}}_A
\underbrace{\begin{pmatrix}
1 & 2 & \color{red}{3} \\
x & y & \color{red}{z} \\
4 & 5 & \color{red}{6} \\
\end{pmatrix}}_B=
\begin{pmatrix}
1 & 0 & 0 \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4110344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Two form surface integral over sphere I'm trying to compute $\int_M \omega$ with
\begin{align*}
\omega &= x^4 dy \wedge dz + y^4 dz \wedge dx + z^4 dx \wedge dy, \\
M &: x^2 + y^2 + z^2 = R^2.
\end{align*}
I have done this in two ways: Stokes' theorem and direct computation of the wedge product via a spherical coordin... | In case of sphere, the surface integral is indeed zero.
Please note in below working of mine, I use $\theta$ as azimuthal angle and $\phi$ as polar. I think you have the other way round. So want to call out to avoid confusion.
In case of hemisphere above $z = 0$,
Parametrization of sphere is $r(\phi, \theta) = (R \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4111353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Solve the equation $\frac{x-13}{x-14}-\frac{x-15}{x-16}=-\frac{1}{12}$ Solve the equation $$\dfrac{x-13}{x-14}-\dfrac{x-15}{x-16}=-\dfrac{1}{12}.$$
For $x\ne14$ and $x\ne 16$ by multiplying the whole equation by $$12(x-14)(x-16)$$ we get: $$12(x-16)(x-13)-12(x-14)(x-15)=-(x-14)(x-16).$$ This doesn't look very nice. Can... | $$\frac{x-13}{x-14} - \frac{x-15}{x-16} = -\frac{1}{12}$$
Let $x = t+ 15$
\begin{align}
\frac{t+2}{t+1} - \frac{t}{t-1} &= -\frac{1}{12} \\
\dfrac{(t^2+t-2) - (t^2+t)}{t^2-1} &= -\frac{1}{12} \\
\dfrac{-2}{t^2-1} &= -\frac{1}{12} \\
t^2-1 &= 24 \\
t^2 &= 25 \\
t &= \pm 5 \\
x &= 20, 10
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4117579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How the right term of the derivative is gained? This deduction is one of the typical ones I think.
What I want to deduce is the right term from the left term of the below equation.
$$\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)=\frac{-a}{x\sqrt{a^{2}+x^{2}}}$$
$\frac{d}{dx}\left(\log\left(\f... | $$f(x)=\frac{d}{dx}\left(\log\left(\frac{a+\sqrt{a^{2}+x^{2}}}{x}\right)\right)=\frac{d}{dx}\left(\log\left(a+\sqrt{a^{2}+x^{2}}\right)\right)-\frac 1x$$
$$\frac{d}{dx}\left(\log\left(a+\sqrt{a^{2}+x^{2}}\right)\right)=\frac{\Big[a+\sqrt{a^{2}+x^{2}}\Big]' }{ a+\sqrt{a^{2}+x^{2}}}$$
$$\Big[a+\sqrt{a^{2}+x^{2}}\Big]'=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4120409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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When ${u_n}$ is converge, find $u_0$ and solve $\lim(nu_n)$ with $ {u_n} : 2u_{n+1}-2u_n+u_n^2=0$ With ${u_n}: 2u_{n+1}-2u_n+u_n^2=0$. its show $${u_{n + 1}} - {u_n} = \dfrac{{ - u_n^2 + 2{u_n}}}{2} - {u_n} = \dfrac{- u_n^2}{2} \leqslant 0$$
${u_n}$ is decrease.
When $n \to \infty $ we have $\lim(u_n)=0$ (it's happen... | If the $\left\{ u_n \right\} $ is converge,let $\displaystyle\lim_{n\rightarrow \infty} u_n=A$,we can get that$$A=-\frac{A^2}{2}-A\Rightarrow A=0 \text{or} -4$$
i.e. $\displaystyle\lim_{n\rightarrow \infty} u_n=0$ or $\displaystyle\lim_{n\rightarrow \infty} u_n=-4$.
Due to $\displaystyle u_{n+1}-u_n=-\frac{u_{n}^{2}}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4122846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt79+\sqrt81}$ Find the sum of $\frac{1}{\sqrt1+\sqrt3} + \frac{1}{\sqrt3+\sqrt5} + \frac{1}{\sqrt5+\sqrt7} + ... \frac{1}{\sqrt{79}+\sqrt{81}}$
I've thought about multiplying every fraction by 1, but like thi... | Hint: $\frac 1{\sqrt{n} + \sqrt{n+2}} \cdot \frac {\sqrt{n+2} - \sqrt{n}}{\sqrt{n+2} - \sqrt{n}} = \frac{\sqrt{n+2} - \sqrt{n}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4126580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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$\sum\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ converges or diverges? The original question is to show that $\;\sum\left(\dfrac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2} $ either converges or diverges.
I know it diverges but I'm having difficulty arriving at something useful for $ a_n $.
Here's what I did:
$ a_n = \l... | $$a_n=\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)^{n^2}\implies \log(a_n)=n^2\log\left(\frac{ n^2 + 1}{n^2 +n + 1}\right)$$
For large $n$
$$\log(a_n)=-n+\frac{1}{2}+\frac{2}{3 n}-\frac{3}{4 n^2}-\frac{1}{5
n^3}+O\left(\frac{1}{n^4}\right)$$ Repeat it for $a_{n+1}$ to obtain
$$\log(a_{n+1})-\log(a_n)=-1-\frac{2}{3 n^2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4132689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Distribution of a function of a random variable and the steps/range that is needed to take. The Question that I need to solve is:
Let $X\sim~\text{Unif}[-1,2]$. Find the probability density function of the random variable $Y = X^2$.
Now I have solved it as follows, but the steps I have to take after that are unclear to... | We have that $X \in [-1, 2]$.
Consider the graph $Y = X^2$. This is not a one-to-one function. (Observe that for $X \in [-1, 1]$ that there are two $X$ values that yield the same $Y$ value.)
Now, by definition, for each $y \in \mathbb{R}$,
$$F_{Y}(y) = \mathbb{P}(Y \leq y) = \mathbb{P}(X^2 \leq y)\text{.}$$
Partition t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Convergence of $ \sum\limits_{n=1}^{\infty}\frac{1}{n(n+m)} $ Let $ m \in \mathbb{N} $. Find to what does the series $ \sum\limits_{n=1}^{\infty}\frac{1}{n(n+m)} $ converge to
My attempt: ( I did as discussed in Infinite Series $\sum 1/(n(n+1))$ )
Let $ N>m $.
$
\begin{align}
S_N & = \sum\limits_{n=1}^{N}\frac{1}{n(n... | You are almost done. We have $$\begin{align}
S_N &= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = N+1}^{N+m} \frac{1}{n} \\
&= \frac{1}{m} \sum_{n=1}^m \frac{1}{n} - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N} \\
&= \frac{1}{m} H_m - \frac{1}{m} \sum_{n = 1}^{m} \frac{1}{n+N},
\end{align}$$
where I have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4134403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding distance from Local maximum of $f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant
What is the distance of the local maximum of the function
$f(x)=x+\sqrt{4x-x^2}$ to bisector of first quadrant?
$1)1\qquad\qquad2)\sqrt2\qquad\qquad3)2\qquad\qquad3)2\sqrt2$
It is a question from timed exam so the fastest answ... | Let $u$ be the maximum value of $f(x)$. Then:
$$\sqrt{4x-x^2} = u-x \Rightarrow 4x-x^2 = u^2 - 2ux + x^2$$
$$\Rightarrow 2x^2 - x(2u + 4) + u^2 = 0 \tag{*}$$
$$\Delta = 0 \Rightarrow (2u+4)^2-4(2)(u^2) = 0$$
$$\Rightarrow -4u^2+16u+16=0 \Rightarrow u^2 - 4u - 4 = 0$$
$$\Rightarrow u = 2 ± 2\sqrt{2}$$
hence the maximum ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4144247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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How to use polar coordinate in ODE? I don't understand how to use polar coordinate.
\begin{cases} \frac{dx(t)}{dt}=2x-y \\ \frac{dy(t)}{dt}=5x-2y \\ \end{cases}
$$ \frac{d}{dt} \left( \begin{array}{c} x \\ y \end{array} \right) = \begin{pmatrix} 2 & -1\\ 5 & -2 \end{pmatrix} \left( \begin{array}{c} x \\ y \end{array} ... | You can tackle it, but it's not necessarily pretty, and as you've noticed - you can solve it perfectly fine in Cartesian coordinates so that's probably how you're expected to approach it.
However, if you're keen, then you can start like this (I'm using dots for time derivatives by convention and because $\theta'$ looks... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4151308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve the inequality $\frac{3-x}{x^2-2x-3}\le\frac{3-x}{x^2+2x-3}$ Solve the inequality $$\dfrac{3-x}{x^2-2x-3}\le\dfrac{3-x}{x^2+2x-3}$$
We have $D: \begin{cases}x^2-2x-3\ne0\Rightarrow x\ne -1;3 \\ x^2+2x-3\ne0\Rightarrow x\ne-3;1\end{cases}$
Is the given equality equivalent (in $D$) to $$x^2-2x-3\ge x^2+2x-3\\\iff x... | hint
For $ x\notin\{-1,1,-3,3\}$, the inequation is equivalent to
$$(3-x)\Bigl(\frac{1}{(x-1)(x+3)}-\frac{1}{(x+1)(x-3)}\Bigr)\ge 0$$
$$\iff x(x-1)(x+1)(x+3)\ge 0$$
You will find that the solution is
$$S=(-\infty,-3)\cup(-1,0]\cup(1,3)\cup(3,+\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4152338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Could someone check my work for this problem for vectors? Find the angle between the two planes. (Round your answer to two decimal places. If the planes do not intersect, enter DNE.)
I found this to be 65.28°
(b) Find parametric equations of their line of intersection. (Let $z = t$, then solve for $x(t)$ and $y(t)$ in... | $$\cos^{-1} \left( \frac{3+2}{\sqrt{(1^2+3^2+1)(3^2+2^2)}}\right)=\cos^{-1}\left( \frac{5}{\sqrt{143}}\right)\approx 65.28^\circ$$
To verify your solution, we just have to substitute them in:
$$x-3y+z=-\frac{2t}{3}-\frac53-\frac{t}3-\frac{10}3+t=-5$$
$$3x+2z+5=-2t-5+2t+5=0$$
It is fine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4153196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Flaw in my reasoning for the maximum of $ab$ if $a,b\ge0$ and $a+2b=3$? Problem statement: What is the maximum value of the product $ab$ if $a$,$b$ are non-negative numbers such that $a+2b=3$?
What is the flaw in my solution?
We know that $\sqrt{ab} ≤ \frac{ (a+b)}{2}$ and that $a=3-2b$. The product $ab$ will be maximu... | Building upon OP's attempt to use AM-GM, the following will work, instead, because it is arranged such that the right-hand side of the inequality matches the known constant sum.
$$
\sqrt{a \cdot 2b} \le \frac{a + 2b}{2}=\frac{3}{2} \;\implies\; 2ab \le \left(\frac{3}{2}\right)^2 = \frac{9}{4} \;\implies\; ab \le \frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find $x$ for $3^x = 5/3$ Why my answer is wrong? $3^x = 5/3$
$e^{\ln 3^x} = 5/3$
$e^{x \ln 3}=5/3$
$e^x \times e^{\ln 3} = 5/3$
$e^x \times 3 = 5/3$
$e^x = 5/9$
$x = \ln(5/9)$
| $3^x=\frac{5}{3}\Leftrightarrow e^{x ln 3}=\frac{5}{3}$; as you see you have a mistake right after the first equal sign.
Try instead $$3^x=\frac{5}{3}|\log_3 \Leftrightarrow \log_3(3^x)=\log_3\left( \frac{5}{3}\right )$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4154947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the equations of circles passing through $(1, -1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
Find the equations of circles passing through $(1,-1)$ touching the lines $4x+3y+5=0$ and $3x-4y-10=0$
The point of intersection of the lines is $(\frac25,-\frac{11}5)$
If we want this point of intersection to be $... | As pointed out in comments, the original lines are not parallel to coordinate axes. So if you want them to be coordinate axes, you will need rotation, in addition to shifting of origin. Instead you can find the equation of lines in new coordinate system after shifting the origin and complete the work using your approac... | {
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"url": "https://math.stackexchange.com/questions/4157083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Can there be a triangle ABC if $\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}$? Can there be a triangle ABC if $$\frac{\cos A}{1}=\frac{\cos B}{2}=\frac{\cos C}{3}\;?$$ Equating the ratios to $k$ we get $\cos A=k$, $\cos B=2k$, $\cos C=3k$.
Then the identity $$\cos^2A+\cos^2B+\cos^2C+2\cos A \cos B \cos C=1 \impl... | Well, not as sophisticated.
But for any angle $\frac \pi 2 > A > \frac \pi 3$ we can have $\cos A$ be any value from $0 < \cos A < \frac 1 2$ we can have $B_A= \cos^{-1} (2\cos A)$ so $\cos B_A = 2\cos A$ and $\frac \pi 2 > B_A > 0$ and $0 < \cos B < 1$.
These can be two angles of a triangle with the third angle being ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4158972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ Solve the equation: $ 3^{2x+1}+4 \cdot 3^x = 15 $ where $x$ is a real number.
Background: Doing Olympiad question and got one from the book.
Attempt:
Let $3^x$ be $u$.
\begin{align*}
3^{2x+1} + 4 \cdot 3x - 15 &= 0 \\
3^{2x} \cdot 3^1 + 4 \cdot 3^x - 15 &= 0 \\
3^{2x} \... | HINT:
You want to find value of $x$
It can be proceeded as
$x\log3=\log(\frac{5}{3})$
$x=\frac{\log(\frac{5}{3})}{\log3}$
$x=\log_3(\frac{5}{3})$ by converse of base change theorem
$x=\log_3 5-1$
To make your solution short by two or three steps take $\log_3()$ on both sides
$3^x=5/3$
$\log_3 3^x=\log_3 (5/3)$
$x=\log_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4164980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given that $\int \frac{1}{x\sqrt{x^2-1}}dx=\arccos(\frac{1}{x})+C$, what is $\int\frac{1}{x\sqrt{x^2-a^2}}dx$? The following is clear:
$x\sqrt{x^2-a^2}=x\sqrt{a^2}\sqrt{\frac{x^2}{a^2}-1}=ax\sqrt{\frac{x^2}{a^2}-1}= a^2\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}$.
So I get that $$\int\frac{1}{x\sqrt{x^2-a^2}}dx=\frac{1}{a^2}\i... | $$ \frac{1}{a^2}\int\frac{1}{\frac{x}{a}\sqrt{\frac{x^2}{a^2}-1}}dx $$
Substitute $u=\frac{x}{a} \Longrightarrow dx = a*du $
$$ \frac{1}{a}\int\frac{1}{u\sqrt{u^{2}-1}}du = \frac{1}{a} \arccos(\frac{1}{u})= \frac{1}{a} \arccos(\frac{a}{x})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4165124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Spot the mistake when computing the density function of $Y = 1/X$ when the density function of $X$ is known. The random variable $X$ has density function $f_{X}$ given by
\begin{equation}
f_{X}(x) =
\begin{cases}
\frac{1}{2} \ \ \ \ \ x \in (0,1] \\
\frac{1}{2x^2} \ \ x \in (1,\infty) \\
0 \ \ \ \ \ \ x \in (-\infty... | A simpler approach is possible via the formula $$f_Y(y) = f_X(g^{-1}(y)) \left|\frac{dg^{-1}}{dy}\right|,$$ where $Y = g(X) = 1/X$, since $g$ is an injective function. Then since the density of $X$ may be more compactly written as
$$f_X(x) = \frac{1}{2 \max(1, x^2)} \mathbb 1 (x > 0)$$
then for $g^{-1}(y) = 1/y$ we im... | {
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"timestamp": "2023-03-29T00:00:00",
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$P(p^2,p^3)$ and $Q(q^2,q^3)$ are two points on the curve $y^2=x^3$. Find the values of p and q. I am working through a pure maths book as a hobby. I am tackling this problem.
$P(p^2,p^3)$ and $Q(q^2,q^3)$ are two points on the curve $y^2=x^3$. Find the equation of the chord PQ and deduce the equation of the tangent at... | Given curve is $y^2 = x^3$. Slope of its tangent is given by,
$\dfrac{dy}{dx} = \dfrac{3x^2}{2y}$
Now as you found, the gradient of line $PQ$ is $\dfrac{p^2+pq+q^2}{p+q}$.
But we also know that $PQ$ is tangent to the curve at point $P$ and is normal to the curve at point $Q$.
Slope of tangent at point $P$ is $\dfrac{3 ... | {
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"timestamp": "2023-03-29T00:00:00",
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Solvability of a particular Negative Pell's Equation How to show that the equation $x^2-(k^2-4)y^2=-1$ is solvable only when k=3?
I can show that 3 must divide d otherwise 3 divides $k^2-4$ and the equation will not be solvable.Again k=3(2m+1) because other wise $x^2\equiv -1(mod8)$ which is not possible.Now how should... | Another approach:
Consider equation $x^2-Dy=-1$
Let $D=m^2+1$. we have:
$x^2-m^2y^2-y^2=-1$
$\Rightarrow (x-y)(x+y)=(my-1)(my+1)$
Suppose:
$\begin{cases} x-y=my-1\\x+y=my+1\end{cases}$
which gives :
($x=my$) and ($y=1$) and we have:
$D=m^2-1=k^2-4\Rightarrow (m-k)(m+k)=-5$
This is possible only if ($m=2$) and ($k=3$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4169872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$ For positive real numbers satisfying $a+b+c=2013$. Prove that
$$\frac{a}{a+\sqrt{2013a+bc}}+\frac{b}{b+\sqrt{2013b+ca}}+\frac{c}{c+\sqrt{2013c+ab}}\leq 1$$
This is my attempt.
We have
$$\frac{a}{a+\sqrt{2013a+bc}}+\fr... | $$\sum_\text{cyclic} \frac{a}{a+\sqrt{2013a+bc}}=\sum_\text{cyclic} \frac{a}{a+\sqrt{(a+b+c)a+bc}}=\sum_\text{cyclic}\frac{a}{a+\sqrt{(a+b)(a+c)}} $$
Using AM-GM inequlaity,
$$a^2+bc\ge 2a{\sqrt{bc}}\;\;\Longleftrightarrow \;\;\sqrt{(a+b)(a+c)}\ge \sqrt{ab}+\sqrt{ac} $$
Therefore,
$$\sum_\text{cyclic}\frac{a}{a+\sqrt{(... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Show that $(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$
Let $a,b,c\in\Bbb N$ such that $a>b>c$. Then $K:=(a^2-b^2)(a^2-c^2)(b^2-c^2)$ is divisible by $12$.
My attempt : Since each $a,b,c$ are either even or odd, WLOG we may assume $a,b$ are both even or odd. For both cases, $a+b$ and $a-b$ are divisible by $2$ s... | Note that $x^2\equiv 0 \quad\textrm{or}\quad 1 \mod 3$ for $x\in\mathbb{N}$.
By Pigeonhole Principle, at least two among $a^2$, $b^2$ and $c^2$ have the same remainder when divided by $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4172927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int\frac{1}{x\sqrt{1-x^2}}dx$ without using trigonometric substitution. The integral is
$$\int\frac{1}{x\sqrt{1-x^2}}dx\tag{1}$$
I tried solving it by parts, but that didn't work out. I couldn't integrate the result of substituting $t=1-x^2$ either.
The answer is
$$\ln\left|\dfrac{1-\sqrt{1-x^2}}{x}\right|$$... | With the substitution $t=\sqrt{1-x^2}$, the integral can be integrated as follows
$$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int \frac1{1-t^2}dt= \frac12 \ln\frac{1-t}{1+t}= \frac12\ln \frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}+C
$$
which is the same as $\ln\frac{1-\sqrt{1-x^2}}{x}$ after rationalizing the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4173559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
high order integer equation
Find all tuple $(x,y)$ such that $x,y$ are integers and $(x^2-y^2)^2=20y+1$.
First i see that $x^2-y^2$ is odd and from the fact that a difference between square of two odd is multiple of $8$ and thus $y$ is a multiple of $2$.
Moreover, we have $(x^2-y^2+1)(x^2-y^2-1)=20y$.
Somebody can gi... | Since $x^2-y^2$ is odd then $x\not=y$ and
$$20y+1=(x^2-y^2)^2\geq ((y-1)^2-y^2)^2=(-2y+1)^2=1-4y+4y^2.$$
Hence
$$0\geq 4y^2-24y\Leftrightarrow y(y-6)\leq 0 $$
which implies that $0\leq y\leq 6$.
Moreover $20y+1$ is a perfect square, and therefore $y\in\{0,4,6\}$:
*
*if $y=0$ then $(x^2-0)^2=1^2$ and $x=\pm 1$;
*if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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In this problem of a moving point in a graph, why is $aθ+θ/3=π/2$ being used to find the angle in which the points $O$,$P$, and $Q$ meet? In this problem the circles $C_1:x^2+y^2=1$ and $C_2:x^2+y^2=4$ are given along with the points $P(\cos(a\theta), \sin(a\theta))$ and $Q(2\cos(\frac{\pi}2-\frac{\theta}3), 2\sin(\fra... | Since the points P, Q, O are collinear, so the area of the triangle formed by them will be zero.
$\mathrm{P\equiv\left(cos(a\theta),sin(a\theta)\right)\,\,\,\,\&\,\,\,\,Q\equiv\left(2\,cos\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right),2\,sin\left(\dfrac{\pi}{2}-\dfrac{\theta}{3}\right)\right)}$
So,
$\left|\begin{array}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4178266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing$ \int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$ Showing $$\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$$
We can show this by re-writing $I$ as
$$
\implies I=6\int_{0}^{\infty}\frac{\frac{1-\cos(2x)}{2x}-\frac{1-\cos(3x)}{3x}}{x}\,\mathrm dx,
$$
which is Frullani Integral.
$$J=\int_{0}^{\infty... | Here is an approach that uses the following identity for Laplace transforms
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (s)\mathcal{L}^{-1} \{g(x)\} (s) \, ds.$$
Recently this identity has been referred as the Maz identity for Laplace transforms.
Setting $f(x) = 1 - 3\cos 2x + 3 \cos 3x$ and $g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4179582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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A proof in $\varepsilon$-language for $\lim \sqrt[n]{1^2+2^2+...+n^2} = 1$ I found a proof that $\lim \sqrt[n]{1^2+2^2+...+n^2}=1$ by $\varepsilon$-language, but I think it's quite complicated and not sure that it's correct.
My question is:
1- Is my proof correct?
2- Is there another simpler proof in the sense of $\var... | because
\begin{gather*}
1\leq \sqrt[n]{1^2+2^2+\cdots+n^2}\leq \sqrt[n]{n\cdot n^2}
=(\sqrt[n]{n})^3,\end{gather*}
and
\begin{gather*}
\lim_{n\to\infty}1=1=\lim_{n\to\infty}(\sqrt[n]{n})^3=\left(\lim_{n\to\infty}\sqrt[n]{n}\right)^3=1,
\end{gather*}
we have, by the squeeze test,
$\lim_{n\to\infty}\sqrt[n]{1^2+2^2+\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$? How many points are common to the graphs of the two equations $(x-y+2)(3x+y-4)=0$ and $(x+y-2)(2x-5y+7)=0$?
\begin{align*}
(x-y+2)(3x+y-4) &= 0\tag{1}\\
(x+y-2)(2x-5y+7) &= 0\tag{2}
\end{align*}
In equation $... | If $x-y+2=0,y=?$
Replace the value of$y$ in terms of $x,$ in the second equation to find
$$0=(x+x+2-2)(3x+x+2-4)=2x(4x-2)$$
$$x=?,?$$
Had there been $n$ factors, we would have $n$ degree equation in $x$
Similarly for $3x+y-4=0\implies y=4-3x$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question about the properties of radicals all. I was solving a simple problem when I realized I had something seemingly contradictory come up. I completed the square on a polynomial as such:
$$
\begin{eqnarray}
y^2-3y&=&-1\\
y^2-3y+\frac{9}{4}&=&-1+\frac{9}{4}\\
(y-\frac{3}{2})^2&=&\frac{5}{4}\\
y-\frac{3}{2}&=& \pm \s... | The way you are adding fractions is not correct. Note that$$\frac32+\frac{\sqrt5}{-2}=\frac32+\frac{-\sqrt5}2=\frac{3-\sqrt5}2$$and that$$\frac32-\frac{\sqrt5}{-2}=\frac32+\frac{\sqrt5}2=\frac{3+\sqrt5}2.$$So, you get$$\frac32\pm\frac{\sqrt5}{-2}=\frac{3\mp\sqrt5}2,$$which are the same numbers that you had got before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to solve $5^{x + 1} = 3^{x + 2}$
Solve $5^{x + 1} = 3^{x + 2}$.
I got this far, but I'm not sure how to continue:
\begin{align}
5^{x + 1} &= 3^{x + 2} \\
(5^x)(5^1) &= (3^x)(3^2) \\
5(5^x) &= 9(3^x)
\end{align}
Where do I go from here?
| Your equation is equivalent to
$$\left(\frac{5}{3}\right)^x = \frac{9}{5}$$
Therefore, $x = \text{log}_{\frac{5}{3}} \frac{9}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Determine the conditional extreme for $u = x^2 + y^2 + z^2$ for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$. Determine the conditional extreme for $u = x^2 + y^2 + z^2$ for $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1$.
Using Lagrange's multipliers we get:
$$<2x, 2y, 2z> = \lambda<\frac{2x}{a^... | $a,b,c$ are all constants so you cannot conclude that they are all equal. What you can conclude from your four equations is that either
*
*$\lambda = a^2, y = z = 0$, $x = \pm a$
*$\lambda = b^2, x = z = 0$, $y = \pm b$
*$\lambda = c^2, x = y = 0$, $z = \pm c$
Then, you can take the maximum of $a,b,c$ and get your ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does $\sin(\frac{\pi}{2}+h)$ become $\cos h$, and $\cos h-1$ become $-2 \sin^2 (\frac{h}{2})$? I have questions regarding trigonometry used in the solution of this problem:
Discuss the differentiability of $f(x)$ at the point $x=\frac{\pi}{2}$.
$$f(x) = \begin{cases}
1, & x<0 \\[4pt]
1+\sin x, &0\leq x<\frac\pi2 \... | Both of your questions are answered by formulas which can be derived from the angle sum formulas for sine and cosine given by
\begin{align}
\sin(\alpha \pm \beta) &= \sin \alpha \cos \beta \pm \cos \alpha \sin \beta \\
\cos(\alpha \pm \beta) &= \cos \alpha \cos \beta \mp \sin \alpha \sin \beta
\end{align}
For the firs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4190526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Orthogonally diagonalizable matrix If $A$ a $3\times3$ matrix, where
$A \begin{bmatrix} 1 \\ 2\\4\\ \end{bmatrix}=3\begin{bmatrix} 1 \\ 2\\4\\ \end{bmatrix}$, $A \begin{bmatrix} 2 \\ -5\\ 2\\ \end{bmatrix}=3\begin{bmatrix} 2 \\ -5\\ 2\\ \end{bmatrix}$, $A \begin{bmatrix} -6 \\ 1\\ 1\\ \end{bmatrix}=2\begin{bmatrix} -6 ... | Recall that a real square matrix is orthogonally diagonalizable if and only if $A$ is a symmetric matrix. (The forwards implication is easy, the reverse implication is the spectral theorem).
Note that $$A\begin{bmatrix} 1&2&-6 \\ 2&-5&1\\ 4&2&1\\ \end{bmatrix}=\begin{bmatrix} 1&2&-6 \\ 2&-5&1\\ 4&2&1\\ \end{bmatrix}\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4192374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int_{0}^{\frac{\pi}{2}} \ln(1+\sin^3 x)\text{d}x$ Here's the integral that I would like to solve. Purely for recreational purposes:
$$I=\int_{0}^{\frac{\pi}{2}} \ln(1+\sin^3x)\text{d}x$$
Here's my shot at it. I would like to stick to this method if possible. Let $I(\alpha)$ be defined as follows:
$$I(\alpha)... | Not a finished answer, just the beginning of a variation of the $I(\alpha)$ approach from the OP.
Using $$I(\alpha)=\int_{0}^{\pi/2} \log(1+\alpha^3\sin^3x)\,dx$$
Then use $$1+z^3=(1+z)\left(1+\omega z\right) \left(1+\omega^2z\right)$$
where $\omega=e^{2\pi i/3}.$
Then, for $k=0,1,2,$ let$$I_k(\alpha)=\int\log(1+\omega... | {
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"source": "stackexchange",
"question_score": "12",
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We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of $\frac{d^2y}{dx^2}$?
We have $(x^2+y^2)^2-3(x^2+y^2)+1=0$. What is the value of
$\frac{d^2y}{dx^2}$?
$1)-\frac{x^2+y^2}{y^2}\qquad\qquad2)-\frac{x^2+y^2}{y^3}\qquad\qquad3)\frac{x+y}{x^2+y^2}\qquad\qquad4)\frac{xy}{x^2+y^2}$
Here is my approach:
We have a qua... | You missed the $-\dfrac{1}{y} $ because you forgot the $2$ on the left-side of the equation.
$$ y'' = \dfrac{-2(y')^2 \color{red}{-2}}{2y} \ne \dfrac{-2(y')^2 }{2y} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4194890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Evaluate $\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$
Evaluate: $$\lim\limits_{n\to\infty}\frac{1\cdot3+2\cdot4+\dots+n(n+2)}{n^3}.$$
I'm learning limits for the first time and this is an exercise problem from my book. Here is my solution to the problem:
Let $S=1\cdot3+2\cdot4+\dots+n(n+2)\\ =(... | Yet another solution, using Stolz Cesaro theorem, which states that if $(b_n)_{n \in \mathbb N_+}$ is an increasing sequence diverging to $+\infty$ and if limit $\lim_{n \to \infty} \frac{a_n-a_{n-1}}{b_n - b_{n-1}}$ exists, then $\lim_n \frac{a_n}{b_n} = \lim_n \frac{a_n - a_{n-1}}{b_n-b_{n-1}}$.
Using this with $b_n ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
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Proving on the equation $(x^2+mx+n)(x^2+px+q)=0$ Find all real numbers k such that if $a,b,c,d \in \mathbb R$ and $a>b>c>d \geq k$ then there exist permutations $( m ,n ,p,q)$ of $(a,b,c,d)$ so that the following equation has 4 distinct real solutions :
$(x^2+mx+n)(x^2+px+q)=0$
Here all i did :
$(x^2+mx+n)(x^2+px+q)=0$... | The question needs to be answered in three steps: First, we need to place constraints on $m,n,p,q$ so that all four roots are real. Second, we need to add constraints so that all four roots are distinct. Lastly, we need to find $k$ such that for all $a>b>c>d>k$, at least one permutation exists where all the constrain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Base $11$ representation of a number must equal to another representation in base $8$. Credit: 2020 AIME I (Problem #$3$ on the test)
Question: A positive integer $N$ has base-eleven representation $\underline{a}\kern 0.1em\underline{b}\kern 0.1em\underline{c}$ and base-eight representation $\underline1\kern 0.1em\unde... | Yes your solution is correct. Another solution also suggested by AoPS is
The conditions of the problem imply that $121a + 11b + c = 512 + 64b + 8 c + a$, so $120 a = 512+ 53b+7c$. The maximum digit in base eight is $7,$ and because $120a \ge 512$, it must be that $a$ is $5, 6,$ or $7.$ When $a = 5$, it follows that $60... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Prove that $2^n + 5^n + 56$ is divisible by $9$, where $n$ is an odd integer
Prove that $9 \mid2^n + 5^n + 56$ where n is odd
I have proved this by using division into cases based on the value of $n\bmod3$ but it seems a little bit clumsy to me and I wonder if there are other ways to prove it, probably by using modul... | Mathematical induction approach
We know that $n$ is an odd integer, i.e. we can use $n= 2k - 1, k \in \mathbb{N}$. Hence, we can reformulate a starting statement:
$$9 \mid 2^{2k-1} + 5^{2k-1} + 56, \quad k \in \mathbb{N}.$$
$\text{1.}$ Basis of Induction:
For $\ k=1 \ $ we have:
$$\ 2^1 + 5^1 + 56 = 63 = 9 \cdot 7, \ $... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
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Any trick for evaluating $\left(\frac{\sqrt{3}}{2}\cos(\theta) + \frac{i}{2}\sin(\theta)\right)^7$? Expressions of the form $(a\cos(\theta) + bi\sin(\theta))^n$ come up from time to time in applications of complex analysis, but to my knowledge the De Moivre's formula can only be applied with $a = b$. Is there some tric... | Long Comment:
For your particular problem there is a small tiny tiny algebraic simplification as
$$T=\sin \left(\frac{\pi }{3}\right) \cos (\theta )-i \cos \left(\frac{\pi }{3}\right) \sin (\theta )=\frac{1}{2} \sqrt{3} \cos (\theta )-\frac{1}{2} i \sin (\theta )\tag1$$
or
$$T=\cos \left(\frac{\pi }{6}\right) \cos (\th... | {
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"url": "https://math.stackexchange.com/questions/4206791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Count the 4-digit integers that are multiple of 9 I want to count the total number of 4-digit integers that are multiple of 9, without any zero digit.
I was wondering if the best strategy is to treat it as a count problem or just apply some properties from number theory.
Naive way is to solve $a+b+c+d = 9, 18, 27, 36$ ... | Okay, so you have determined that $abcd = 1000a + 100b + 10c + d$ is a multiple of $9$ if and only if $a+b+c+d$ is a multiple of $9$.
But now the trick is to realize that $a+b+c+d$ is a multiple of $9$ if and only if
$a+b+c+d \equiv 0 \pmod 9$ if and only if
$d\equiv -(a+b+c) \pmod 9$.
Now for any possible value $a+b+c... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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What are the values of $x$, $y$ and $r$ if $\cot\theta=\frac{5}{6}$ and $\csc\theta <0$? I solved this problem, where $r$ is a radius of a circle. Is my solution right?
What are the values of $x$, $y$ and $r$ if $\cot\theta=\frac{5}{6}$ and $\csc\theta <0$?
Since $\cot\theta$ is positive and $\csc\theta$ is negative, $... | You can't conclude that $x=-5$ and $y=-6$.
You are right that it is in the third quadrant.
$$\cot \theta = \frac{r\cos \theta}{r \sin \theta}=\frac{x}y=\frac{5}{6}.$$
Hence, we have the half-line:
$$6x=5y, x < 0.$$
Any point on the half line satisfies the condition.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where am I going wrong with this proof for partial fraction decomposition? We have to prove that $ \frac{F}{G} $ can be written as $\frac{F_1}{G_1}+ \frac{F_2}{G_2}$ if $G=G_1G_2$ and $G_1,G_2$ are co-prime polynomials. This was the proof given in my textbook:
$$There \space exists \space polynomials \space C,D \space... | That is a very strange proof. It seems easier this way: From $CG_1 + DG_2 = 1$ we have
$$
FC\cdot G_1 + FD \cdot G_2 = F
$$
Now let $F_1 = FD$ and $F_2 = FC$. Then
\begin{align*}
\frac{F_1}{G_1} + \frac{F_2}{G_2} &= \frac{F_1 G_2}{G_1 G_2} + \frac{F_2 G_1}{G_2 G_1} \\&= \frac{F_1 G_2 + F_2 G_1}{G_1 G_2} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4212032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A JEE Exam problem on determinants and matrices I am first stating the question:
Let $A=\{a_{ij}\}$ be a $3\times 3$ matrix, where
$$a_{ij}=\begin{cases}
(-1)^{j-i}&\text{if $i<j$,}\\
2&\text{if $i=j$,}\\
(-1)^{i-j}&\text{if $i>j$,}
\end{cases}$$
then $\det(3\,\text{adj}(2A^{-1}))$ is equal to __________
I solved th... | Alternatively, use the properties:
$$\begin{align}\text{adj}(cA)&=c^{n-1}\text{adj}(A) \quad (1)\\
\det(cA)&=c^n\det(A) \quad (2)\\
\det(\text{adj}(A))&=(\det(A))^{n-1} \quad (3)\\
\det(A^{-1})&=(\det(A))^{-1} \quad (4)
\end{align}$$
to get:
$$\det(3\,\text{adj}(2A^{-1}))\stackrel{(1)}{=}\\
\det(3\cdot 2^2\,\text{adj}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion.
It is given that $(x^2 +y +2t +3k)^{10}$. What is the sum of the coefficients of the terms containing $x^2t^3$ in the given expansion. For example the terms are $x^2y^4t^3k^2$ etc.
I said that let $(x^2 +2t)$ be $"a"$ and $(... | Following OPs approach somewhat more formally we obtain
\begin{align*}
\color{blue}{[x^2t^3]}&\color{blue}{\left(\left(x^2+2t\right)+(y+3k)\right)^{10}}\\
&=[x^2t^3]\sum_{q=0}^{10}\binom{10}{q}\left(x^2+2t\right)^q\left(y+3k\right)^{10-q}\\
&=\sum_{q=0}^{10}\binom{10}{q}\left([x^2t^3]\sum_{r=0}^q\binom{q}{r}x^{2r}(2t)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
$a^a\cdot{b^b}\ge \bigl(\frac{a+b}{2}\bigl)^{a+b}\ge{a^b}\cdot{b^a}$ If $a$ and $b$ are positive rational numbers, prove that
$$a^a\cdot b^b\ge \left(\frac{a+b}{2}\right)^{a+b} \ge a^b \cdot{b^a}$$
My try:
consider $\frac{a}{b}$ and $\frac{b}{a}$ be two positive numbers with associated weights $b$ and $a$.
Then $\displ... | Part 1: by GM-HM:
$\left(\underbrace{(a.a.a\dots a)}_{\text{a times}}\underbrace{(b.b.b\dots b)}_{\text{b times}}\right)^{\frac{1}{a+b}} \ge \frac{a + b}{\left(\underbrace{\frac{1}{a}+ .... \frac{1}{a}}_{\text{a times}}\right)+\left(\underbrace{\frac{1}{b}+ .... \frac{1}{b}}_{\text{b times}}\right)}$
$\implies a^ab^b \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Show that $ \frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2+x^2}} \geq \frac{1}{\sqrt{2}}(x+y+z) $ Show that for positive reals $x,y,z$ the following inequality holds and that the constant cannot be improved
$$
\frac{x^2}{\sqrt{x^2+y^2}} + \frac{y^2}{\sqrt{y^2+z^2}} + \frac{z^2}{\sqrt{z^2... | For $x=y=z$ the inequality $\sum\limits_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}\geq k(x+y+z)$ gives $k\leq\frac{1}{\sqrt2}$.
The Peter Scholze's solution for $k=\frac{1}{\sqrt2}.$:
By Rearrangement $$\sum_{cyc}\frac{x^2}{\sqrt{x^2+y^2}}=\sqrt{\sum_{cyc}\left(\frac{x^4}{x^2+y^2}+\frac{2x^2y^2}{\sqrt{x^2+y^2}}\cdot\frac{1}{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4215933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Proof-verification: of numbers such that $(a^2+b^2)(c^2+d^2)=u^2+w^2$ I would like to ask if someone can verify that my solution to the following question is correct:
Question:
Prove that for any $a,b,c,d\in\mathbb{Z}$ there exists integers $u,v$ such that
$$(a^2+b^2)(c^2+d^2)=u^2+v^2$$
My solution:
Let $z_1 = a+ib$ an... | There are an infinite number of solutions where the right side is a hypotenuse and the left side is the product of hypotenuses of smaller Pythagorean triples. One example is
$$(3,4,5)\rightarrow 3^2+4^2=5^2\quad (5,12,13)\rightarrow 5^2+12^2=13^2\\
5\times13=65\quad \implies\quad a=2, b=1, c=3, d=2\\
\implies u=7, v=4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Determinant of a Pascal Matrix, sort of Let $A_{n}$ be the $(n+1) \times(n+1)$ matrix with coefficients
$$
a_{i j}={i+j \choose i}
$$
(binomial coefficients), where the rows and columns are indexed by the numbers from
0 to $n$ are indexed.
Now I want to determine the Determinant and with the first 5 matrices i found ou... | Let's consider the version where the columns are indexed from $0$ to $n$. Consider for example
$$ A_3 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ 1 & 3 & 6 & 10 \\ 1 & 4 & 10 & 20 \end{bmatrix}. $$
Note that each element in the matrix (except the elements in the $0$th row and column) is the sum of the element a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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Can the formula for the cubes $a^3 + b^3 + c^3 - 3abc$ be generalized for powers other than 3? I recently learnt out about this formula:
$$a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - ac - bc)$$
Is there a way of generalizing for powers other than $3$, i.e.
$$a^n + b^n + c^n + \mathop{???} = \mathop{???... | For cases of $n=5$ and $n=7$, let's have a look at the following identities
$$(x+y+z)^3 - (x^3+y^3+z^3) = 3(x+y)(x+z)(y+z)$$
$$(x+y+z)^5 - (x^5+y^5+z^5) = 5(x+y)(x+z)(y+z)(x^2+y^2+z^2+xy+xz+yz)$$
$$(x+y+z)^7 - (x^7+y^7+z^7) = 7(x+y)(x+z)(y+z)((x^2+y^2+z^2+xy+xz+yz)^2+xyz(x+y+z))$$
The three identities listed above are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$ To prove: $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18 = \cot^{-1}3$
My Attempt:
First Method: we know that $\cot^{-1}x = \tan^{-1}\frac{1}{x}$ for $x>0$ and $\tan^{-1}x+\tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}, xy<1$
Now $\cot^{-1}7+\cot^{-1}8+\cot^{-1}18$ = $\tan^{-1}\fr... | Obviously $\pi+\cot^{-1} 3 \neq \cot^{-1} 3$, since $\pi \neq 0$. Your second method is incorrect because $$\tan^{-1} a+\tan^{-1}b+\tan^{-1}c=\tan^{-1} \left(\frac {a+b+c-abc}{1-ab-bc-ac}\right)$$ is only true for $a,b,c>0$ if $ab+bc+ac<1$.
It is better for you to combine the $\arctan$s two at a time. We have:
$$\tan^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4220297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the second real root for cubic $x^3+1-x/b=0$ A cubic of the form $$x^3+1-x/b=0$$ has has three real roots
Using the Lagrange inversion theorem one of the roots is given by
$$x = \sum_{k=0}^\infty \binom{3k}{k} \frac{b^{3k+1} }{(2)k+1} $$
How do you find the second one? I cannot find any info online. I am looking... | The related OEIS sequences A206300 and
A224884 help to answer your question. Define
$$ a_n := \frac{2^{2n-1}}{n!}\frac{\Gamma(3n/2-1/2)}{\Gamma(n/2+1/2)}. \tag{1} $$
Now define
$$ y_2 := \sum_{n=0}^\infty -a_n z^{3n-1},\quad
y_3 := \sum_{n=0}^\infty\, (-1)^na_n z^{3n-1},\quad
y_1 := -y_2-y_3. \tag{2} $$
These are the t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4224731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
Find the derivative of $\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac x2\right)$.
I'm learning differentiation and this is an exercise problem from my book. I used chain rule and got the following:
$\begin{align}
\dfrac d{dx}\left[\tan^{-1}\... | Power of $t$-formula
Let $t=\tan \dfrac{x}{2}$, then
\begin{align}
\frac{dt}{dx} &= \frac{1}{2} \sec^2 \frac{x}{2} \\
&= \frac{1+t^2}{2} \\
\tan y &= t\sqrt{\frac{a-b}{a+b}} \\
\sec^2 y \times \frac{dy}{dx} &=
\frac{dt}{dx} \times \sqrt{\frac{a-b}{a+b}} \\
\left( 1+\frac{a-b}{a+b} t^2 \right) \frac{dy}{dx}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Approximate Trigonometric Integral Let the function $I(x)$ be defined as
$$I(x) = \int_{-\pi/2}^{\pi/2}{\rm d} \theta \ \frac{\cos^2{\theta}}{\sqrt{\cos^2\theta+x^2\sin^2\theta}} \ , $$
Then we have for instance $I(0)=2$. Is there a nice answer for $I(x)$ for $x$ small? The most obvious thing to do would be to binomial... | There is something interesting : starting from the result given by Wofram Alpha
$$I(x)=\frac{2 \left(x^2 K\left(1-x^2\right)-E\left(1-x^2\right)\right)}{x^2-1}$$ Expanding as a series around $x=0$
$$I(x)=2 +\sum_{n=1}^\infty \Big[a_n-b_n \log(2)+c_n \log(x^2)\Big]\, x^{2n}$$ and the coefficients are
$$\left(
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve for when this trigonometric function intersects the line $y=1$? How can I solve for $\alpha$ in
$$4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right)\left(t-r\right)+\sin\left(\frac{\alpha}{2}\right)=1$$
on the domain $0\leq\alpha\leq\pi$? Clearly, one solution is when $\alpha=\pi$, but t... | Excluding the trivial $a=\pi$, reset the problem as
$$(t-r)=\frac{1-\sin\left(\frac{\alpha}{2}\right)}{4\sin\left(\frac{\alpha}{2}\right)\cos^3\left(\frac{\alpha}{2}\right) }=\frac{1}{4} \left(1-\sin \left(\frac{a}{2}\right)\right) \csc \left(\frac{a}{2}\right)
\sec ^3\left(\frac{a}{2}\right)\tag 1$$
Let $a=2 \csc ^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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The sequence $(x_n)$ : $x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}{x_{n+1}x_n-(\sqrt{x_{n+1}^2+1}-1)(\sqrt{x_n^2+1}-1)}$ The sequence $(x_n)$ is defined by the formula:
$$\left\{\begin{array}{cc}
x_1=1, x_2=\sqrt{3}\\
x_{n+2}=\frac{x_{n+1}\sqrt{x_n^2+1}+x_{n}\sqrt{x_{n+1}^2+1}-x_n-x_{n+1}}... | By rationalising the denominator, $x_{n+2}$ simplifies to $$\frac{\sqrt{x_n^2+1}\sqrt{x_{n+1}^2+1}+x_nx_{n+1}-1}{x_n+x_{n+1}}.$$
Letting $$x_n=\frac12\left(p_n-\frac1{p_n}\right),$$ that is (say) $$p_n=x_n+\sqrt{x_n^2+1}$$ and simplifying further, you get $$x_{n+2}=\frac{p_np_{n+1}-1}{p_n+p_{n+1}}$$
So, if $p_n=\cot\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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If $T_1=7^7,T_2=7^{7^{7}},T_3=7^{7^{7^{7}}}$ and so on, what will be the tens digit of $T_{1000}$? $7^4$ ends with $2301$, so $7^{4k+r}$ ends with $7^r$ digits
$7^2\equiv1 \mod(4) $, and $7^7\equiv 3 \mod(4) $
Can we use the modulo function in exponent form? I think we will use these two properties, how can I proceed f... | If i correctly understand the question, we have to compute the value of $T_{1000}$ modulo $100$.
Long solution:
This is done in a few steps applying at each one the Euler indicator function $\varphi$, noting that basis $7$ is relatively prime to each of the numbers $\varphi(100)=40$, $\varphi(40)=16$, $\varphi(16)=8$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Remainder Theorem Technique
Determine the remainder when $(x^4-1)(x^2-1)$ is divided by $1 + x + x^2$ (HMMT 2000, Guts Round)
A. Write the division in the form:
$$(x^4-1)(x^2-1)= (1 + x + x^2)Q(x) + R(x)$$
B. Multiply both sides by $x-1$:
$$(x-1)(x^4-1)(x^2-1)= (x^3-1)Q(x) + R(x)(x-1)$$
C. Substitute $x^3=1,x\neq1$, ... | Yes, you can skip those steps.
The verbose equivalent is that $$\begin{align}(x^4-1)(x^2-1)&=(x^4-x)(x^2-1)+(x-1)(x^2-1)\\
&=(x^2+x+1)x(x-1)(x^2-1)+(x-1)(x^2-1)
\end{align} $$
So $(x-1)(x^2-1)$ has the same remainder as $(x^4-1)(x^2-1)$ when dividing by $x^2+x+1.$
Answering the question in comments, yes, you can repla... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Coefficient of $x^{10}$ in $f(f(x))$ Let $f\left( x \right) = x + {x^2} + {x^4} + {x^8} + {x^{16}} + {x^{32}}+ ..$, then the coefficient of $x^{10}$ in $f(f(x))$ is _____.
My approach is as follow
$f\left( {f\left( x \right)} \right) = f\left( x \right) + {\left( {f\left( x \right)} \right)^2} + {\left( {f\left( x \rig... | Let's replace $10$ with a lower number, $5$, and illustrate the role of partitions in the result.
Only $f+f^2+f^4$ contributes to the coefficient of $x^5$. The $f$ term has none, of course.
The $f^2=f\cdot f$ term contributes a coefficient of $2$ from $x^1\cdot x^4$ and $x^4\cdot x^1$. Note that we get two terms becaus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4238019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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How to calculate the residue of this function quickly (or by mathematica)? While looking at this post (The second answer), I find that it is to tedious to calculate the Residue of
$$ f(z)=-\left(\frac{z-1}{z+1}\right)^2\frac{2n/z}{z^{2n}-1} $$
at $-1$.
I do know that we can do this:
$$\operatorname*{Res}_{z=-1}f(z)=\f... | 1. Note that we have
\begin{align*}
-\left(\frac{z-1}{z+1}\right)^2\frac{1}{z(z^2-1)}
&= -\frac{z-1}{(z+1)^3 z} \\
&= -\frac{2}{(z+1)^3} - \frac{1}{(z+1)^2} - \frac{1}{z+1} + \frac{1}{z}.
\end{align*}
Also, if we write $g(z) = \frac{z^{2n}-1}{z^2-1} = 1 + z^2 + z^4 + \cdots + z^{2(n-1)}$, then the Taylor series for $\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4239900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to integrate $ \int e^{\cos( x)} \ dx$ I mainly did this for fun of it and am posting it here to have it reviewed and corrected if I made a mistake. This is a considerably simpler version of this solution I posted a couple months back.
Please comment with any corrections, questions, comments or concerns and hopeful... | $\displaystyle \begin{array}{{>{\displaystyle}l}}
Evaluating\ \int e^{cos( x)} \ dx:\\
Starting\ with\ the\ taylor\ series\ definition\ we\ have\ e^{cos( x)} =\ \sum _{n=0}^{\infty }\frac{cos^{n}( x)}{n!}\\
Now,\ using\ the\ complex\ exponential\ definition\ of\ cosine,\ we\ can\ obtain\ a\ generalized\ summation\\
fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4240328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Prove by induction $1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$ I am practicing mathematical induction and I got this question.
$$1^2 + 2^2 + ... + n^2 = \frac{n(n+1)(2n+1)}{6}$$
I am wondering, if this is correct way to do it? Would my answer get accepted if it was an exam?
I put n+1 in the places of n. So Im try... | It is quite well done, but you should check first that the statement is true when $n=1$, and then to say that if$$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}6,\tag1$$then$$1^2+2^2+\cdots+n^2+(n+1)^2=\frac{n(n+1)(2n+1)}6+(n+1)^2$$(due to $(1)$), followed by the computations that you have shown us at the end of your question.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving an analytic geometry problem with euclidean geometry
$\mathrm{OABC}$ is a tetrahedron with $\overline{\mathrm{OA}}=1$.
There is a point $P$ on $\triangle \mathrm{ABC}$ such that $\cos^2 \alpha+\cos^2 \beta + \cos^2 \gamma = \frac{11}{6}$ where $\alpha=\angle\mathrm{AOP}$, $\beta=\angle\mathrm{BOP}$ and $\gamma... | Here's a much shorter answer.
Assuming $OABC$ is a regular tetrahedron of edge length 1.
Place point O at the origin $(0,0,0)$, and let face $ABC$ be parallel to the $xy$ plane, then points $A, B, C$ can be taken as
$A = (\sin \theta , 0, \cos \theta )$
$B = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Applying binomial expansion and using algebra to find the value of $a+b$ Q) In the expansion of
$f(x)=(1 + ax)^4 (1 + bx)^5$
where $a$ and $b$ are positive integers,
the coefficient of $x^2$ is 66.
Evaluate $a+b$.
My working:
After expanding the expression I simplified it and got
$5b^2+10ab+3a^2=33$
After further simpl... | Your result is correct.
If $a,b\in\mathbb Z^{+}$ then we can write,
$$\begin{align}&5b^2+3a^2≥2\sqrt {15}ab\\
\implies &5b^2+3a^2+10ab≥ab \left(10+2\sqrt {15}\right)\\
\implies &33≥\left(10+2\sqrt {15}\right)ab\\
\implies &ab≤\frac{33}{10+2\sqrt {15}}<2\\
\implies &a=b=1.\end{align}$$
This means, the solution doesn't e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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All Unique Three Digit Combinations using four 8s, three 9s, seven 1's and three 5s So basically I have find the unique possible combinations of 3 digit number using four 8s, three 9s, seven 1's and three 5s, e.g.
"888"
"819"
"891"
"855", ..., etc
Because of having same number present multiple times I am not able to de... | These types of questions can be easily solved by exponential generating functions. It will give you power.
It is said that there are four $8's$ , three $9's$ , seven $1's$ , and three $5's$ .In other words , we have $8,8,8,8,9,9,9,1,1,1,1,1,1,1,5,5,5.$
We want to find the number of all possible $3$ digits numbers using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4243825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Intuition behind getting two straight lines as result Question:
Find the equation of the straight line that passes through $(6,7)$ and makes an angle $45^{\circ}$ with the straight line $3x+4y=11$.
My solution (if you want, you can skip to the bottom):
Manipulating the given equation to get it to the slope-intercept fo... | We have to be mindful about sign of slope
To the given inclination $ \tan^{-1} (-3/4) = - 36. 87 ^{\circ}$ add, resulting in $ 45 ^{\circ}- 36.87 ^{\circ} = 8.13^{\circ}$ positive ( counter clockwise) to x-axis and you have only one straight line solution AB. Other solution is spurious, needs to be discarded.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Prove that $\frac{1}{2(n+2)}<\int_0^1\frac{x^{n+1}}{x+1}dx$ $\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\left[\frac{x^{n+2}}{(n+2)(x+1)}\right]_0^1+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
$\displaystyle\int\limits_0^1\frac{x^{n+1}}{x+1}dx=\frac{1}{2(n+2)}+\int\limits_0^1\frac{x^{n+2}}{(x+1)^2(n+2)}dx$
If ... | Hint
You know that $x \in \left[0;1\right]$ so
$$
\frac{1}{x+1}>\frac{1}{2}
$$
What does this imply for $\displaystyle \int_{0}^{1}\frac{x^n}{1+x}$ ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Independent chances of $3$ events ${a\over{a+x}}$, ${b\over{b+x}}$, ${c\over{c+x}}$ Here's a problem from my probability textbook:
Of three independent events the chance that the first only should happen is a; the chance of the second only is $b$; the chance of the third only is $c$. Show that the independent chances ... | Let $\alpha$ be a root of $(a+x)(b+x)(c+x)=x^2$
$\implies (a+\alpha)(b+\alpha)(c+\alpha)=\alpha^2$
Now, $(p_1, p_2, p_3)$ satisfies the system of equations
$\begin{align}
(1) \quad p_1(1 - p_2)(1 - p_3) = a \\
(2) \quad (1-p_1)p_2(1-p_3) = b \\
(3) \quad (1-p_1)(1-p_2)p_3 = c
\end{align}$
It's sufficient to show that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Definite Integral $\int_{1-\sqrt{a}}^{1+\sqrt{a}} \ln(1-z t^2)\frac{\sqrt{4a-(t^2-1-a)^2}}{t}\, dt$ Suppose $z<\frac{1}{(1+\sqrt{a})^2}$, $0<a<1$. I need to compute the following integral as function of z:
$$
I(z)= \int_{1-\sqrt{a}}^{1+\sqrt{a}} \ln(1-z t^2)\frac{\sqrt{4a-(t^2-1-a)^2}}{t}\, dt
$$
So far, following the ... | With the substitution $x=\frac{ t^2-1-a}{2\sqrt a}$, along with the shorthands $p= \frac{2z\sqrt a}{1-z-az}$, $q= \frac{1+a}{2\sqrt a} $, the integral simplifies to
\begin{align}
I(z)=& \int_{1-\sqrt{a}}^{1+\sqrt{a}} \ln(1-z t^2)\frac{\sqrt{4a-(t^2-1-a)^2}}{t}\, dt\\
=& \sqrt a\int_{-1}^1 [\ln(1-z-az)+\ln (1-p x)]\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4246957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How can I find the limit of the following? What is the limit of this
$$\lim_{x\to+\infty}\left(1+\frac{4}{2x+3}\right)^x$$
I know that $$\lim_{x\to+\infty}\left(1+\frac{4}{2x}\right)^x$$ will give me $$e^2$$ but the I dont know what to do with the 3.
I have tried bringing them to a common denominator so I got
$$\lim_{x... | Let $f(x) = 1 + \frac{4}{ 2x+3}$
And $g(x) = x $
Your limit is of the form $(1)^{\infty}$ , whose value is equal to $$e^{ \lim_{ x \to {\infty}} {(f(x) - 1 )}{g(x)}} $$
Therefore , $$L = e^{ \lim_{ x \to {\infty}} {(\frac{4x}{2x+3} )}} $$
Now $$\frac{4x}{2x+3} = {\frac{4x+6}{2x+3} } - \frac{6}{2x+3} = 2 - \frac{6}{2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4249669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Prove that $ \int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\frac{1}{36}\Big(\psi^{(1)}(2/3)-\psi^{(1)}(1/3)\Big) $ I am having trouble with the following integral
Prove that $$ \int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\frac{1}{36}\Big(\psi^{(1)}(2/3)-\psi^{(1)}(1/3)\Big)$$
$$I=\int_0^1\frac{x\ln(x)}{1+x^2+x^4}dx=\int_0^1\frac{\ln(u... | Alternatively
\begin{align} \int_0^1\frac{x\ln x}{1+x^2+x^4}dx
= &\int_0^1\frac{x\ln x}{(e^{i\frac\pi3}+x^2)(e^{-i\frac\pi3}+x^2)}dx \\
=& \>\frac1{\sin\frac\pi3 }\>\Im \int_0^1 \frac{e^{i\frac\pi3}x\ln x}{1+ e^{i\frac\pi3}x^2 }dx
\overset{x^2\to x} =\frac{1}{2\sqrt3}\Im \text{Li}_2(- e^{i\frac\pi3})
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $n(x-y)(xy)^{(n-1)/2} \leq x^n - y^n$ for nonnegative $x,y$ and integers $n$ Show for $x,y\geq 0$ and $n \in \mathbb N$ that:
$$n(x-y)(xy)^{(n-1)/2} \leq x^n - y^n$$
What I've tried
Proving by induction seems natural since the case for $n=2$ and $n=3$ are straightforward to show.
Assuming the claim is true fo... | Yes, AM-GM. Let $x\ge y\ge0$, $n\in\mathbb N$. Then
\begin{align*}
x^n-y^n&=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k\\
&=\frac12(x-y)\sum_{k=0}^{n-1}\left(x^ky^{n-1-k}+x^{n-1-k}y^k\right)\\
&\ge\frac12(x-y)\sum_{k=0}^{n-1}2\sqrt{x^ky^{n-1-k}\cdot x^{n-1-k}y^k}\\
&=(x-y)\sum_{k=0}^{n-1}(xy)^{\frac{n-1}2}\\
&=n(x-y)(xy)^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the sum of series $ \sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$ Let it be known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}
{6}.$$ Given such—find $$\sum_{n=1}^{\infty}\frac{1}{n^3(n+1)^3}$$
Attempt:
I have tried using the fact that $\displaystyle \frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ and then expandi... | Hint:
$$
\frac{1}{n^3} - \frac{1}{{(n+1)}^3} \;=\;
\frac{3n(n+1) + 1}{n^3{(n+1)}^3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
If $ax+by = a^n + b^n$ then $\left[\frac{x}{b}\right]+\left[\frac{y}{a}\right]=\left[\frac{a^{n-1}}{b}\right]+\left[\frac{b^{n-1}}{a}\right]$
Let a,b,n be positive integers such that $(a,b) = 1$.
Prove that if $(x,y)$ is a solution of the equation $ax+by = a^n + b^n$ then
$$\left[\frac{x}{b}\right]+\left[\frac{y}{a}\r... | As I stated in my comment, if $(x, y)$ is allowed to be any real solution, then \eqref{eq6A} will not always hold, so I'll assume that $x$ and $y$ must be integers. Also, I believe the square brackets are meant to represent the floor function, as I've seen being used elsewhere on this site.
Given the above, we can proc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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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.