Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Mistake with Integration with Beta, Gamma, Digamma Fuctions Problem: Evaluate:
$$I=\int_0^{\pi/2} \ln(\sin(x))\tan(x)dx$$
I tried to attempt it by using the Beta, Gamma and Digamma Functions. My approach was as follows:
$$$$
Consider $$I(a,b)=\int_0^{\pi/2} \sin^a(x)\sin^b(x)\cos^{-b}(x)dx$$
$$$$
$$=\dfrac{1}{2}\beta... | $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\new... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Questionnaire probability When sent a questionnaire, the probability is .5 that any particular individual to whom it is sent will respond immediately to that questionnaire. For an individual who did not respond immediately, there is a probability of .4 that the individual will respond when sent a follow-up letter. If t... | Let $X$ be the number of those who respond immediately and Let $Y$ be the number of those who never respond. The probability that at least $3$ will never respond is
$$P(Y\ge 3)=P(Y=3)+P(Y=4).$$
So far, so good. Then
$$P(Y=3)=P(Y=3\mid X=0)P(X=0)+P(Y=3 \mid X=1)P(X=1).$$
and
$$P(Y=4)=P(Y=4\mid X=0)P(X=0).$$
Now
*
*$P... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find out the greater number from $15^{1/20}$ and $20^{1/15}$? I have two numbers $15^{\frac{1}{20}}$ & $20^{\frac{1}{15}}$.
How to find out the greater number out of above two?
I am in 12th grade. Thanks for help!
| This is just another way of looking at alkabary's proof.
\begin{align}
\left[ \dfrac{20^{\frac{1}{15}}}
{15^{\frac{1}{20}}} \right]^{20}
&=\dfrac{20^{\frac 43}}{15} \\
&= \dfrac 43 20^{\frac 13} \\
&> \dfrac 43 \cdot 2 \\
&> 1
\end{align}
$\left[ \dfrac{20^{\frac{1}{15}}}
{15^{\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Solution to Fibonacci Recursion Equations Let the sequence $(a_n)_{n\geq0}$ of the fibonacci numbers: $a_0 = a_1 = 1, a_{n+2} = a_{n+1} + a_n, n \geq 0$
Show that:
i) $$a^2_n - a_{n+1}a_{n-1} = (-1)^n \text{ for }n\geq1$$
I try to show this with induction:
$n=1:$
\begin{align*}
a^2_1 - a_{2}a_{0}&=1 ... | Another flavor on the inductive step argument:
$$
\begin{split}
a_n^2-a_{n+1}a_{n-1}
&= a_n^2-(a_n + a_{n-1})a_{n-1} \\
&= a_n^2 - a_n a_{n-1} - a_{n-1}^2 \\
&= a_n(a_n-a_{n-1}) - a_{n-1}^2 \\
&= a_n a_{n-2} - a_{n-1}^2 \\
&= -\left( a_{n-1}^2 - a_n a_{n-2}\right),
\end{split}
$$
reducing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solve the inequation: $5+3\sqrt{1-x^2}\geq x+4\sqrt{1-x}+3\sqrt{1+x}$ Solve the inequation: $5+3\sqrt{1-x^2}\geq x+4\sqrt{1-x}+3\sqrt{1+x}$
I tried to substitute $x=\cos t$ but don't get any result. Who can help me?
| First of all, we need to have $1-x^2\ge 0$, $1-x\ge 0$ and $1+x\ge 0$, i.e. $-1\le x\le 1$.
Now since the both sides of
$$5-x+3\sqrt{1-x^2}\ge 4\sqrt{1-x}+3\sqrt{1+x}\tag 1$$
are non-negative, this is equivalent to
$$\left(5-x+3\sqrt{1-x^2}\right)^2\ge \left(4\sqrt{1-x}+3\sqrt{1+x}\right)^2,$$
i.e.
$$6(1-x)\sqrt{1-x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(... | $$x^3=4+3\left ( \sqrt[3]{2-\sqrt{3}} \right )\left ( \sqrt[3]{2+\sqrt{3}} \right )\left ( \sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}} \right )$$
$$=4+3\sqrt[3]{2^2-3}(x)=4+3x$$
$$x^3=4+3x$$
$$x^3-3x-4=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
} |
Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:
If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$
I tried to substitute the value of x in the expression, but that comes out to be very big.
| Here is a slightly weird way of doing it. $x$ looks like the quadratic formula, so if we can cook up a quadratic equation that it satisfies, we won't actually have to square it. The solutions to
$$ y^2+2by+c = 0 $$
are
$$ y = -b \pm \sqrt{b^2-c} $$
(because $a=1$ and the $b$ has a $2$ multiplying it that cancels the $2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 1
} |
Improper integral and residues
Evaluate $\int_0^\infty \frac{dx}{x^4+1}$
By the residue theorem $$\int_{-R}^Rf(x)dx+\int_{C_R}dz=2\pi i\sum Res(f,z_i)$$
but I have problems to evaluate it because
$$z^4+1=0\Rightarrow z^4=-1=e^{i\pi},e^{i3\pi},e^{i5\pi},e^{i7\pi}$$
$$z=e^{\frac{i\pi}{4}},e^{\frac{i3\pi}{4}},e^{\frac{i... | Well, I'm learning the subject now, too, we might as well teach each other, no? :)
Note that: $$\int_0^{+\infty} \frac{1}{x^4+1}\,{\rm d}x = \frac{1}{2}\int_{-\infty}^{+\infty}\frac{1}{x^4+1}\,{\rm d}x,$$so we'll compute that second integral. Let $R > 1$, $\gamma_1$ be the line segment joining $-R$ to $R$, and $\gamma_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find the infinite sum of a sequence Define a sequence $a_n$ such that $$a_{n+1}=3a_n+1$$ and $a_1=3$ for $n=1,2,\ldots$. Find the sum $$\sum_{n=1} ^\infty \frac{a_n}{5^n}$$ I am unable to find a general expression for $a_n$. Thanks.
| Let $x=\sum_{n=1}^{\infty}\dfrac{a_{n}}{5^n}$,since
$$\dfrac{a_{n+1}}{5^{n+1}}=\dfrac{3}{5}\dfrac{a_{n}}{5^n}+\dfrac{1}{5^{n+1}}$$
so
$$\sum_{n=1}^{\infty}\dfrac{a_{n+1}}{5^{n+1}}=\dfrac{3}{5}\sum_{n=1}^{\infty}\dfrac{a_{n}}{5^n}+\sum_{n=1}^{\infty}\dfrac{1}{5^{n+1}}$$
then we have
$$x-\dfrac{a_{1}}{5}=\dfrac{3}{5}x+\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Combining sums and/or differences of squares I'd like to combine a sum of as many squares as possible into a sum of as few squares as possible. The signs of the squares doesn't matter.
For example, the Brahmagupta-Fibonacci Identity combines a sum of four squares into a sum of two squares. Thus the compression ratio ... | The "compression ratio" can be made as high as you want.
I. Euler-Aida Ammei
$$(-x_0^2+x_1^2+x_2^2+\dots+x_n^2)^2+(2x_0x_1)^2+(2x_0x_2)^2+\dots+(2x_0x_n)^2 = (x_0^2+x_1^2+x_2^2+\dots+x_n^2)^2$$
Thus, there can be arbitrarily many squares on the LHS.
II. Fauquembergue $n$-Squares Identity
Using the Brahmagupta-Fibon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
3 variable symmetric inequality Show that for positive reals $a,b,c$,
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{3a^3+3b^3+3c^3}{2a^2+2b^2+2c^2}$
What I did was WLOG $a+b+c=1$ (since the inequality is homogenous)
Then I substituted into the LHS to get $\sum_{\text{cyc}} \frac{a^2}{1-a}\geq \frac{3a^3+... | your inequalitiy is equivalent to
$$\left(a^3-2 a b c+b^3+c^3\right) \left(2 a^3-a^2 b-a^2 c-a b^2-a c^2+2 b^3-b^2
c-b c^2+2 c^3\right)\geq 0$$
since $$a^3+b^3+c^3\geq 3abc>2abc$$ is the first factor positive,
and
$$2(a^3+b^3+c^3)\geq ab(a+b)+ac(a+c)+bc(b+c)$$
since $a^3+b^3=(a+b)(a^2+b^2-ab)\geq ab(a+b)$ etc is the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral of $\int \frac{x+1}{(x^2+x+1)^2}dx$ I'm currently learning Calculus II and I have the following integral:
Integal of $\large{\int \frac{x+1}{(x^2+x+1)^2}dx}$
I've tried with partial fractions but it led nowhere, I've tried with substitution, but I failed again.
If it helps, I know the answer, but I don't know... | Hint:
\begin{align*}
\int\frac{x+1}{(x^2+x+1)^2}dx&=\frac{1}{2}\int{\frac{2x+1}{(x^2+x+1)^2}}dx+\frac{1}{2}\int\frac{1}{(x^2+x+1)^2}dx\\
&=\frac{1}{2}\int\frac{d(x^2+x+1)}{(x^2+x+1)^2}+\frac{1}{2}\int{\frac{1}{\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]^2}}dx
\end{align*}
The last of these integrals can be fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for $x$ from an equation containing inverse trigonometric functions
How to solve the following for $x$?
$$
\sin^{-1}\left(\frac{2a}{1+a^{2}}\right)+
\sin^{-1}\left(\frac{2b}{1+b^{2}}\right)=
2 \tan^{-1}(x )
$$
What conditions apply?
| Hint: Let $\theta = \sin^{-1}\left(\dfrac{2a}{1+a^2}\right) \Rightarrow \sin \theta = \dfrac{2a}{1+a^2}$, and similarly $\sin \beta = \dfrac{2b}{1+b^2}$. Thus:
$\tan^{-1}x = \dfrac{\theta}{2}+\dfrac{\beta}{2} \Rightarrow x = \tan\left(\dfrac{\theta}{2}+\dfrac{\beta}{2}\right)$. You can compute $\tan\left(\dfrac{\theta}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1339791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{... | For $a,b>0$, let
\begin{equation}
A(a,b)=\frac{a+b}{2}, \quad G(a,b)=\sqrt{ab}\,, \quad H(a,b)=\frac{2}{\dfrac1{a}+\dfrac1{b}}
\end{equation}
are respectively the arithmetic, geometric, and harmonic means of two positive numbers $a,b$.
It is commom knowledge that
\begin{equation}
A(a,b)\ge G(a,b) \quad\text{and}\quad H... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
| Let $x=\frac{1}{2}\sin t+\frac{1}{2}$, then we have
\begin{align*}
\int_0^1\frac{\sqrt{x-x^2}}{x+2}dx&=\int_0^1\frac{\sqrt{\left(\frac{1}{2}\right)^2-\left(x-\frac{1}{2}\right)^2}}{x+2}dx\\
&=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sqrt{\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\sin t\right)^2}}{\frac{1}{2}\sin t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
Minimization on compact region I need to solve the minimization problem
$$\begin{matrix} \min & x^2 + 2y^2 + 3z^2 \\ subject\;to & x^2 + y^2 + z^2 =1\\ \; & x+y+z=0
\end{matrix}$$
I was trying to verify the first order conditions using lagrange multipliers
$$\left\{\begin{matrix}
2x = 2\lambda_1x + \lambda_2 \\
4y = 2 ... | since you have
$$x=\dfrac{\lambda_{2}}{2(1-\lambda_{1})},y=\dfrac{\lambda_{2}}{4-2\lambda_{1}},z=\dfrac{\lambda_{2}}{6-2\lambda_{1}}$$
and $x+y+z=0$ so we have
$$\dfrac{\lambda_{2}}{2}\left(\dfrac{1}{1-\lambda_{1}}+\dfrac{1}{2-\lambda_{1}}+\dfrac{1}{3-\lambda_{1}}\right)=0$$
so we have$\lambda_{2}=0$ or
$$\dfrac{(2-\la... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
This general inequality maybe is true? $\sum_{i=1}^{n}\frac{i}{1+a_{1}+\cdots+a_{i}}<\frac{n}{2}\sqrt{\sum_{i=1}^{n}\frac{1}{a_{i}}}$
Let $a_{1},a_{2},\ldots,a_{n}>0$ and prove or disprove
$$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{n}{1+a_{1}+a_{2}+\cdots+a_{n}}\le\dfrac{n}{2}\sqrt{\dfrac{1}{a_{1}}+... | suppose $n=k, \\ \dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}} \\ \le\dfrac{k}{2}\sqrt{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{k}}}$
when $n=k+1 $
LHS=$\dfrac{1}{1+a_{1}}+\dfrac{2}{1+a_{1}+a_{2}}+\cdots+\dfrac{k}{1+a_{1}+a_{2}+\cdots+a_{k}}+\dfrac{k+1}{1+a_{1}+a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 0
} |
Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$ Convert the equation to rectangular form $r = \frac {6}{1-\sinθ}$
The answer should be: $y = \frac{1}{12} x^2 -3$
But how to arrive at the answer?
I tried replacing r with $\sqrt{x^2 + y^2}$, then $\sinθ$ as $\frac{y}{r}$ but to no avail.
I also ended up... | $$r = \frac {6}{1-\sinθ}$$
$${r-r\sinθ}=6$$
$$\sqrt{x^2 + y^2}-y=6$$
$$\sqrt{x^2 + y^2}=y+6$$ now squaring both side and simplyfying we get
$$y = \frac{1}{12} x^2 -3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solving a system of non-linear equations Let
$$(\star)\begin{cases}
\begin{vmatrix}
x&y\\
z&x\\
\end{vmatrix}=1, \\
\begin{vmatrix}
y&z\\
x&y\\
\end{vmatrix}=2, \\
\begin{vmatrix}
z&x\\
y&z\\
\end{vmatrix}=3.
\end{cases}$$
Solving the above system of three non-linear equations with three unknowns.
... | Let's rewrite the system as $$\mathbf r \times \mathbf P\mathbf{r} = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\x&y&z\\z&x&y\end{vmatrix} = 2\mathbf{i} + 3\mathbf{j} + 1\mathbf{k} \equiv \mathbf f$$
where $\mathbf{r} = (x,y,z)^\top$ and
$$
\mathbf P = \begin{pmatrix}
0 & 0 & 1\\
1 & 0 & 0\\
0 & 1 & 0
\end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
} |
value of an $\sum_3^\infty\frac{3n-4}{(n-2)(n-1)n}$ I ran into this sum $$\sum_{n=3}^{\infty} \frac{3n-4}{n(n-1)(n-2)}$$
I tried to derive it from a standard sequence using integration and derivatives, but couldn't find a proper function to describe it.
Any ideas?
| Using the Heaviside Method of Partial Fractions to solve
$$
\frac{3n-4}{n(n-1)(n-2)}=\frac{A}{n}+\frac{B}{n-1}+\frac{C}{n-2}
$$
we get
multiply by $n$ and set $n=0\implies\frac{3\cdot\color{#C00000}{0}-4}{(\color{#C00000}{0}-1)(\color{#C00000}{0}-2)}=A\implies A=-2$
multiply by $n-1$ and set $n=1\implies\frac{3\cdot\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the area of triangle, given an angle and the length of the segments cut by the projection of the incenter on the opposite side.
In a triangle $ABC$, one of the angles (say $\widehat{C}$) equals $60^\circ$.
Given that the incircle touches the opposite side ($AB$) in a point that splits it in two segments having ... |
Area $S_{ABC}$ in terms of $|AB|$ and altitude $h_C$ at the point $C$:
\begin{align}
h_C&=(b+r\sqrt{3})\sin(60)
=\frac{\sqrt{3}}{2}(b+r\sqrt{3})
\\
S&=
\frac{\sqrt{3}}{4}
(a+r\sqrt{3})
(b+r\sqrt{3})
\\
&=
\frac{3}{4}\left(
\sqrt{3}r^2+(a+b)r
\right)
+\frac{\sqrt{3}}{4}ab
\quad(1)
\end{align}
Also, $S_{ABC}=S_{OAB}+S_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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According to Stewart Calculus Early Transcendentals 5th Edition on page 140, in example 5, how does he simplify this problem? In Stewart's Calculus: Early Transcendentals 5th Edition on page 140, in example 5, how does
$$\lim\limits_{x \to \infty} \frac{\dfrac{1}{x}}{\dfrac{\sqrt{x^2 + 1} + x}{x}}$$
simplify to
$$\li... | $$
\frac{\sqrt{x^2 + 1}}{x} = \frac{\sqrt{x^2 + 1}}{\sqrt{x^2}} = \sqrt{\frac{x^2+1}{x^2}} = \sqrt{\frac{x^2}{x^2} + \frac 1 {x^2}} = \cdots
$$
$\frac{\left(\dfrac{1}{x}\right)}{\left(\dfrac{\sqrt{x^2 + 1} + x}{x}\right)}$ is exactly the same thing as $\dfrac 1 {\sqrt{x^2+1}+x}$. To get that just multiply the numerato... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Show that $\frac{n}{n^2-3}$ converges Hi I need help with this epsilon delta proof. The subtraction in the denominator as well as being left with $n$ in multiple places is causing problems.
| @Ian has provided a very efficient way to tackle a problem such as this. I thought it might be instructive to see how tight a bound we can achieve. To that end, let's write for $n\ge2$
$$\begin{align}
\left|\frac{n}{n^2-3}\right|<\epsilon \implies n&>\frac{1}{2\epsilon}+\frac{1}{2\epsilon}(1+12\epsilon^2)^{1/2} \tag ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the number of solutions of the trigonometric equation in $(0,\pi)$ Find the number of solutions of the equation $$\sec x+\csc x=\sqrt {15}$$ in $(0,\pi)$. The question is easy. But when you solve, you get would get $4$ as the answer. I am sure the method gives $4$ as answer, but the correct answer is $3$. I don't ... | If you set $X=\cos x$ and $Y=\sin X$, the equation is
$$
\frac{1}{Y}+\frac{1}{X}=\sqrt{15}
$$
to be considered together with $X^2+Y^2=1$.
Note that we must have $Y>0$, since $x\in(0,\pi)$.
The equation becomes $X+Y=\sqrt{15}\,XY$ and it's convenient to make another substitution, namely $S=X+Y$, $P=XY$, so we get
$$
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Closed form expression for products How can I find a closed form expression for products of the following form $$\prod_{k=1}^n (ak^2+bk+c)\space \text{?}$$
| As commented by r9m, the key idea is to factorize the polynomial. When this is done, you can either use Pochhammer functions which would give $$\prod_{k=1}^n (ak^2+bk+c)=a^n \left(\frac{2 a+b-\sqrt{b^2-4 a c}}{2 a}\right){}_n \left(\frac{2 a+b+\sqrt{b^2-4
a c}}{2 a}\right){}_n$$ or transform to Gamma functions $$\p... | {
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"url": "https://math.stackexchange.com/questions/1355975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How can I get the exact value of this infinite series? I want to compute the exact value of this infinite series
$$\sum_{n=2}^\infty\arcsin{\left(\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)}$$
By comparison test, we can get the series is convengence.
I tried to find some hints from the exact value of $\displ... | The sum to 1000 terms,
according to Wolfy,
is
1.55713
or $0.481892\pi$.
Would be amusing if the sum were
$\pi/2$.
$\sum_{n=2}^\infty\arcsin{\left(\dfrac{2}{\sqrt{n(n+1)}(\sqrt{n}+\sqrt{n-1})}\right)}
$
More seriously,
I will try
$\arcsin(x)
=\arctan(\frac{x}{\sqrt{1-x^2}})
$
to see if
$\arctan(u)+\arctan(v)
=\arctan(\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and... | You can actually get away with computing only one binomial coefficient; I'll leave out one detail so this qualifies as a hint.
As @GrahamKemp observes, both $X$ and $Y$ are distributed as $\mathcal{Bin}(5,1/2)$; we want the probability that $X\ge Y$, i.e. that $Z:=X-Y\ge 0$. But $Z+5$ is distributed as $\mathcal{Bin}(... | {
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"url": "https://math.stackexchange.com/questions/1358776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Find all postive integers $n$ such $(2n+7)\mid (n!-1)$ Find all postive integers $n$ such that
$$(2n+7)\mid(n!-1).$$
I have $n=1,5$, but can not find any other and can not prove whether there is any other solution or not.
| Building on nayrb's observation that $2n+7$ must be prime (for $n\ge 4$), let $p = 2n+7$. Then by assumption, $n! \equiv 1 \pmod p$. Symmetrically, we also have $$(2n+6)(2n+5)\cdots(n+7) \equiv (-1)(-2)\cdots(-n) \equiv (-1)^n n! \equiv \pm 1 \pmod p.$$
Now combine this with Wilson's theorem that $(2n+6)! \equiv -1 \... | {
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"url": "https://math.stackexchange.com/questions/1359021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$
let $0<a\le b\le c\le d$, and such $abcd=1$,show that
$$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$
it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
| it is trivial $d \ge 1$
there are two cases: $c\ge 1$ , or $c\le 1$
$c\ge 1$
let $x=\dfrac{1}{a},y=\dfrac{1}{b} \implies x\ge y\ge 1 ,xy=cd$
let $f(x)=x-\dfrac{1}{x}$, it is trivial $f(x)$ is mono increasing
function
now we need to prove :
$f(c^3)+f(d^4) \ge f(x)+f(y^2)$
1)if $ c^3< x \cap c^3<y^2 \implies yc^3 <xy=cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
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is there a general formula for cases like $\sqrt{2}$ = $\frac {2}{\sqrt{2}}$? I just noticed that $\sqrt{2}$ is equal to $\frac {2}{\sqrt{2}}$:
$\sqrt{2} = 1.414213562$
$\frac {2}{\sqrt{2}} = 1.414213562$
It is confirmed by a hand-calculator.
I tried to proof this as follows:
$\sqrt{2}$ = $\frac {2}{\sqrt{2}}$
$2^{\fra... | The steps are: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}.$$You do the same thing for any positive number $a>0$: $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a}.$$The trick is "multiplying by $1$ in a smart way".
P.s.: Congratulations... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
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Evaluating the infinite product $ \prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $ How does one evaluate
$ \displaystyle\prod\limits_{n=1}^{\infty} \cos(\frac{y}{2^n}) $?
Seems fairly straightforward, as I just plugged in some initial values $n = 1, 2, 3,\dotsc$
$n = 1$
$ \sin(y)= 2\sin(\frac{y}{2})\cos(\frac{y}{2})$... | Let's take a look at the finite product first:
$$\prod_{i=0}^k \cos\frac{x}{2^i} = \frac1{2^k}\frac{\sin 2 x}{\sin \frac{x}{2^k}}$$
which is the standard trick of multiplying the product by $\sin \frac{x}{2^k}$.
Now take limit as $k \rightarrow +\infty$ and you get:
$$\prod_{i=0}^\infty \cos\frac{x}{2^i} = \frac{\sin{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital... | Learn to love asymptotics;
$$\frac {\tan x - x}{x \tan x} \sim \frac{x + x^3/3 - x}{x^2} \sim x/3 \to 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 1
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Prove by induction that $4$ divides $n^3+(n+1)^3+(n+2)^3+(n+3)^3$ Just looking for someone to check my work and for feedback, thanks!
Base case: $n=0$
$0+1+8+27 = 36$
$4$ divides $36.$
Inductive step: Assume $4$ divides $k^3+(k+1)^3+(k+2)^3+(k+3)^3$ for some number where $k$ is a natural number including zero. So $k^... | Your answer seems right and looks like a short route, I would use modules to avoid the algebra but it wouldn't be induction.
For $K \equiv 0 \pmod 4$ the modules of $4$ for the elements of the equation would be $0^3 + 1^3 + 2^3 + (-1)^3 \equiv 0 + 1 + 0 + (-1) = 0 \pmod 4$.
And because there are always $4$ consecutive ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Repeated roots of biquadratic equation What is condition for repeated roots of the fourth order polynomial
$$ x^4 + a x^3 + b x^2 + c x + d = 0 ?$$
| A polynomial has roots with multiplicity greater than one iff its discriminant is zero.
The discriminant of your generic fourth-degree polynomial is:
$$a^2 b^2 c^2-4 b^3 c^2-4 a^3 c^3+18 a b c^3-27 c^4-4 a^2 b^3 d+16 b^4 d+18 a^3 b c d-80 a b^2 c d-6 a^2 c^2 d+144 b c^2 d-27 a^4 d^2+144 a^2 b d^2-128 b^2 d^2-192 a c d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the values of $\cos(\alpha+\beta) $ if the roots of an equation are given in terms of tan It is given that $ \tan\frac{\alpha}{2} $ and $ \tan\frac{\beta}{2} $ are the zeroes of the equation $ 8x^2-26x+15=0$ then find the value of $\cos(\alpha+\beta$).
I attempted to solve this but I don't know if my solution is r... | Since, $\tan\frac{\alpha}{2}$ & $\tan\frac{\beta}{2}$ are roots of the equation $8x^2-26x+15=0$ Hence, we have $$\tan\frac{\alpha}{2}+\tan\frac{\beta}{2}=\frac{-(-26)}{8}=\frac{13}{4}$$
$$\tan\frac{\alpha}{2}\tan\frac{\beta}{2}=\frac{15}{8}$$
$$\implies \tan \left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\tan\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Span of similar matrices Given $D=\begin{pmatrix} 3 & 0 \\ 0 & 2\\ \end{pmatrix}$
How can i find the span of all matrices A that are similar to D.
thus $\exists P$ $A=P^{-1}DP$
| Here are four matrices similar to $D$ which span all of $\mathbb{R}^{2\times 2}$:
$$ \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix},
\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix},
\begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix},
\begin{pmatrix} 2 & 0 \\ 1 & 3 \end{pmatrix} $$
So the span of matrices $A$ similar to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solution verification for $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$ I was required to find $y'$ when $y=\sin^{-1}(\frac{2x}{1+x^2})$
This is my solution.
Above when I put $\sqrt{x^4-2x+1}=\sqrt{(1-x^2)^2}$ then I get the correct answer but when I put $\sqrt{x^4-2x+1}=\sqrt{(x^2-1)^2}$ then I get something else.
Now my... | Hint:
Given
$$
y = \sin^{-1}\left( \frac{2x}{1+x^2}\right),
$$
then
$$
\sin(y) = \frac{2x}{1+x^2},
$$
so
$$
\cos(y) y' = \left( \frac{2x}{1+x^2} \right)'.
$$
Now
$$
\cos(y) = \sqrt{ 1 - \sin^2(y) } =
\sqrt{ 1 - \left( \frac{2x}{1+x^2} \right)^2 },
$$
so you get
$$
y' = \frac{ \displaystyle \left( \frac{2x}{1+x^2} \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1367747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the r... | Replace $(a,b,c) \to \left(\frac{a^2}{bc},\frac{b^2}{ca},\frac{c^2}{ab}\right)$ the inequality become
$$ \sum \frac{a}{\sqrt{a^2+8bc}} \geqslant 1.$$
We have
$$\left[5(a^2+b^2+c^2)+4(ab+bc+ca)\right]^2-(a^2+8bc)(5a+2b+2c)^2$$
$$=[5a(b+c)+17bc](2a-b-c)^2+[82a^2+25(b^2+c^2)+35a(b+c)+41bc](b-c)^2 \geqslant 0.$$
Therefore
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1369441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 3
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Finding the Jordan Form and basis $$A= \begin{pmatrix}
2&1&2\\ -1&0&2 \\ 0&0&1
\end{pmatrix}$$
I found that $$f_A(x)=m_A(x) = (x-1)^3.$$
So the Jordan form must be:
$$J= \begin{pmatrix}
1&0&0\\ 1&1&0 \\ 0&1&1
\end{pmatrix}$$
Now,
$$A-I= \begin{pmatrix}
1&1&2\\ -1&-1&2 \\ 0&0&0
\end{pmatrix} $$
$$(A-I)^2 = \begin{pmatri... | *
*If you choose a basis such that $Ab_2 = b_1 + \lambda b_2$ the ones will be above the diagonal. If you choose a basis such that $A b_1 = b_2 + \lambda b_1$ they will be below. It's a matter of whether the chains go up or down your indexing.
*No. These two matrices have different Jordan forms, but the same chara... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$ without finding the integration
I don't know how to start
| HINT:
For $n^2+1<x<n^2+n+1$, we hav
$$\arctan(n^2+1)<\arctan(x)<\arctan(n^2+n+1)$$
and
$$\frac{1}{n+1}<\frac{1}{\sqrt{n^2+n+1}}<\frac{1}{\sqrt{x}}<\frac{1}{\sqrt{n^2+1}}<\frac1n$$
SPOLIER ALERT
SCROLL OVER SHADED AREA
We have $$I(n)=\int_{n^2+1}^{n^2+n+1}\frac{\arctan(x)}{x^{1/2}}dx$$Then, since the arctangent is inc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
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Polynomial of $11^{th}$ degree Let $f(x)$ be a polynomial of degree $11$ such that $f(x)=\frac{1}{x+1}$,for $x=0,1,2,3.......,11$.Then what is the value of $f(12)?$
My attempt at this is:
Let $f(x)=a_0+a_1x+a_2x^2+a_3x^3+......+a_{11}x^{11}$
$f(0)=\frac{1}{0+1}=1=a_0$
$f(1)=\frac{1}{1+1}=\frac{1}{2}=a_0+a_1+a_2+a_3+...... | For convenience, let us shift the variable: $g(y)=\dfrac1y$ for $y=1,2,\cdots12$.
Then $$\frac{A(y-1)(y-2)\cdots(y-12)+1}y$$ is a polynomial of degree $11$ and coincides with $\dfrac1y$ at the given points, provided that the numerator has no independent term, i.e. $12!A+1=0$.
From this,
$$f(12)=g(13)=\frac{A(13-1)(13-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Find a polynomial from an equality Find all polynomials for which
What I have done so far:
for $x=8$ we get $p(8)=0$
for $x=1$ we get $p(2)=0$
So there exists a polynomial $p(x) = (x-2)(x-8)q(x)$
This is where I get stuck. How do I continue?
UPDATE
After substituting and simplifying I get
$(x-4)(2ax+b)=4(x-2)(ax+b)$
F... | The route you take is fruitful.
$p\left(x\right)=\left(x-2\right)\left(x-8\right)q\left(x\right)$
leads to:
$$\left(x-4\right)q\left(2x\right)=2\left(x-2\right)q\left(x\right)$$
Then $4$ must be a root of $q$, so $q\left(x\right)=\left(x-4\right)r\left(x\right)$ leading to:
$$r\left(2x\right)=r\left(x\right)$$
Then $r\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Find $x$ in the triangle
the triangle without point F is drawn on scale, while I made the point F is explained below
So, I have used $\sin, \cos, \tan$ to calculate it
Let $\angle ACB = \theta$, $\angle DFC = \angle BAC = 90^\circ$, and $DF$ is perpendicular to $BC$ (the reason for it is to have same $\sin, \cos, ... | Let $AD=y$ so that $\cos C=\frac{2y}{12}$
Then using the cosine rule, $$x^2=y^2+3^2-6y\cos C$$
So $x=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Not the toughest integral, not the easiest one Perhaps it's not amongst the toughest integrals, but it's interesting to try to find an elegant
approach for the integral
$$I_1=\int_0^1 \frac{\log (x)}{\sqrt{x (x+1)}} \, dx$$
$$=4 \text{Li}_2\left(-\sqrt{2}\right)-4 \text{Li}_2\left(-1-\sqrt{2}\right)+2 \log ^2\left(1+... | By making the Euler substitution $\sqrt{x^{2}+x} = x+t$, we find
$$ \begin{align} \int_{0}^{1} \frac{\log(x)}{\sqrt{x^{2}+x}} \, dx &= 2 \int_{0}^{\sqrt{2}-1} \frac{\log \left(\frac{t^{2}}{1-2t}\right)}{1-2t} \, dt \\ &= 4 \int_{0}^{\sqrt{2}-1} \frac{\log t}{1-2t} \, dt - 2\int_{0}^{\sqrt{2}-1} \frac{\log(1-2t)}{1-2t} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Calculate the limit $\lim_{x\to 0} \left(\frac 1{x^2}-\cot^2x\right)$ The answer of the given limit is $2/3$, but I cannot reach it. I have tried to use the L'Hospital rule, but I couldn't drive it to the end. Please give a detailed solution!
$$\lim_{x\to 0} \left(\dfrac 1{x^2}-\cot^2x\right)$$
|
Solution by using L-Hospital's rule (no series expansion)
$$\lim_{x\to 0} \left(\frac{1}{x^2}-\cot^2 x\right)$$ $$=\lim_{x\to 0} \left(\frac{1}{x^2}-\frac{\cos^2 x}{\sin^2x}\right)$$ $$=\lim_{x\to 0} \left(\frac{\sin^2x-x^2\cos^2 x}{x^2\sin^2x}\right)$$ $$=\lim_{x\to 0} \left(\frac{\sin^2x-x^2(1-\sin^2 x)}{x^2\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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how can i prove this trigonometry equation I need help on proving the following:
$$\frac{\cos {7x} - \cos {x} + \sin {3x}}{ \sin {7x} + \sin {x} - \cos {3x} }= -\tan {3x}$$
So far I've only gotten to this step:
$$\frac{-2 \sin {4x} \sin {3x} + \sin {x}}{ 2 \sin {4x} \cos {3x} - \cos {x}}$$
Any help would be appreciated... | Notice, the following formula $$\color{blue}{\cos A-\cos B=2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{B-A}{2}\right)}$$ & $$\color{blue}{\sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right)}$$ Now, we have
$$\frac{\cos 7x-\cos x+{\sin 3x}}{\sin 7x+\sin x-\cos 3x}=-\tan 3x$$
$$\implies LHS=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove using mathematical induction that $x^{2n} - y^{2n}$ is divisible by $x+y$ Prove using mathematical induction that
$(x^{2n} - y^{2n})$ is divisible by $(x+y)$.
Step 1: Proving that the equation is true for $n=1 $
$(x^{2\cdot 1} - y^{2\cdot 1})$ is divisible by $(x+y)$
Step 2: Taking $n=k$
$(x^{2k} - y^{2k})$ is... | For $n=k$, assume $P(k)$ is true, we have
$$x^{2k}-y^{2k}=A(x-y)$$, where A is a polynomial.
For $n=k+1$,
\begin{align}
x^{2k+2}-y^{k+2}&=x^2[A(x-y)+y^{2k}]-y^{2k+2}\\&=A(x-y)x^2+x^2y^{2k}-y^{2k+2}\\&=A(x-y)x^2+y^{2k}(x^2-y^2)\\&=A(x-y)x^2+y^{2k}(x-y)(x+y)\\&=(x-y)[Ax^2+y^{2k}(x+y)]\\&=B(x-y)\text{, where } B \text{ i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
| $$\int \frac{x^3}{(x^2+1)^{3/2}}dx=\int \frac{x(x^2+1)-x}{(x^2+1)^{3/2}}dx=\int \frac{x}{(x^2+1)^{1/2}}dx-\int \frac{x}{(x^2+1)^{3/2}}dx\\
=\sqrt{x^2+1}+\frac1{\sqrt{x^2+1}}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Is $\left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$ an integer? The problem is the following:
Prove that this number
$$x = \left(45+29\sqrt{2}\right)^{1/3} + \left(45-29\sqrt{2}\right)^{1/3}$$
is an integer. Show which integer it is.
I thought that it has some relations with something like c... | Let $a,b$ be the first and second term of the sum then $x = a+b$. Observe that: $ab = 7 \to x^3 = (a+b)^3 = a^3+b^3 + 3ab(a+b)=90+3\times 7\times x \to x^3-21x-90 = 0\to (x-6)(x^2+6x+15)=0\to (x-6)((x+3)^2+6)=0\to x = 6.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$ (B) $\dfrac{\pi \ln2}{3}$ (C) $\dfrac{\pi \ln2}{4}$ (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x... | In General $$\int_0^{\infty}\frac{\ln(x)}{x^2+b^2}dx=\frac{\pi\ln(b)}{2b}$$ for $b>0, b\neq 1$
$\bf{Solution::}$ Let we introduce a parameter, say $a > 0$ as follows:
Set $$F(a) = \int_{0}^{\infty} \frac{\ln ax}{x^2 + b^2} dx$$
So that we get $$\frac{dF}{da} = \frac{1}{a}\int_{0}^{\infty} \frac{1}{x^2 + b^2} dx = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
How to factor intricate polynomial $ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $ I would like to know how to factor the following polynomial.
$$ ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c $$
What is the method I should use to factor it? If anyone could help.. Thanks in advance.
| \begin{align*}
ab^3 - a^3b + a^3c -ac^3 +bc^3 - b^3c&=-a^3(b-c)+b^3(a-c)+c^3(b-a)\\
&=-a^3(b-c)+b^3(a-c)+c^3[(b-c)+(c-a)]\\
&=(c^3-a^3)(b-c)+(b^3-c^3)(a-c)\\
&=(c-a)(c^2+ac+a^2)(b-c)+(b-c)(b^2+bc+c^2)(a-c)\\
&=(a-c)(b-c)(-c^2-ac-a^2+b^2+bc+c^2)\\
&=(a-c)(b-c)(b^2+bc-ac-a^2)\\
&=(a-c)(b-c)[(b+a)(b-a)+c(b-a)]\\
&=(a-c)(b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Integration by Partial Fractions $\int\frac{1}{(x+1)^3(x+2)}dx$ I'm trying to do a problem regarding partial fractions and I'm not sure if I have gone about this right as my answer here doesn't compare to the answer provided by wolfram alpha. Is it that I can't Seperate things that are raised to powers on the denominat... | Let $$\frac{1}{(x+1)^3(x+2)}=\frac{A_1}{(x+1)^3}+f_1(x)$$
$$A_1=\frac{1}{(x+2)}|_{x=-1}=1,f_1(x)=-\frac{1}{(x+1)^2(x+2)}$$
Let $$f_1(x)=-\frac{1}{(x+1)^2(x+2)}=\frac{A_2}{(x+1)^2}+f_2(x)$$
$$A_2=-\frac{1}{(x+2)}|_{x=-1}=-1,f_2(x)=\frac{1}{(x+1)(x+2)}$$
$$f_2(x)=\frac{1}{(x+1)}-\frac{1}{(x+2)}$$
thus:
$$\frac{1}{(x+1)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Minimize the area of a wire divided into a circle and square.
A wire is divided into two parts. One part is shaped into a square, and the other part is shaped into a circle. Let r be the ratio of the circumference of the circle to the perimeter of the square when the sum of the areas of the square and circle is minimi... | For Minimization of $$\displaystyle \frac{x^2}{16}+\frac{(y-x)^2}{4\pi}\;,$$ We Also use $\bf{Cauchy-Schwartz}$ Modified Inequality.
Which is $$\displaystyle \frac{A^2}{X}+\frac{B^2}{Y}\geq\frac{(A+B)^2}{X+Y}$$ and equality hold When $\displaystyle \frac{A}{X} = \frac{B}{Y}$.
So $$\displaystyle \frac{x^2}{16}+\frac{(y-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Two circles touching a line and the axes If the circle $C_1$ touches x-axis and the line $y=x \tan\theta$,$\theta \in (0,\frac{\pi}{2})$ in the first quadrant and circle $C_2$ touches the line $y=x \tan\theta$,y-axis and circle $C_1$ in such a way that ratio of radius of $C_1$ to the radius of $C_2$ is 2:1,then value o... | Let the radius of circle $C_2$ be $R_2$ & center $O_2$ & the radius of the circle $C_1$ will be $R_1$ & center $O_2$ $\forall\ \frac{R_1}{R_2}=2$
Now, let the circles touch at any point say $P(d, d)$ on the line $y=x$ then in right $\triangle OPO_1$ $$\tan\frac{\theta}{2}=\frac{R_1}{d}$$ $$d=R_1\cot\frac{\theta}{2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Showing that $\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$ for all $n\geq 1$
Show that $$\frac{1}{2^n +1} + \frac{1}{2^n +2} + \cdots + \frac{1}{2^{n+1}}\geq \frac{1}{2}$$ for all $n\geq 1$
I need this in order to complete my proof that $1 + \frac{n}{2} \leq H_{2^n}$, but I don't ... | $$\sum_{k=1}^{2^n} \frac{1}{2^n+k} \geq \sum_{k=1}^{2^n} \frac{1}{2^n+2^n} =\sum_{k=1}^{2^n} \frac{1}{2^{n+1}}$$
the rightmost sum equals $\frac{1}{2}$, in fact we have a strict inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Big an open ball inside small open ball in metric space When i was reading the book "Elements of the Theory of Functions and Functional Analysis" (A. N. Kolmogorov, S. V. Fomin), I encountered a very interesting (for me) problem.
Problem:
Create a metric space with two open balls $B(x,\rho_1), B(y,\rho_2)$ such that $... | One example:
metric space $(X,d)$ such that $X(0,1)$ with metrica $d\left( x,y \right) =\begin{cases} 2-\left| x-y \right| \quad ,x\neq y \\ 0\quad \quad \quad \quad \quad \quad ,x=y \end{cases}$
$$d\left( \frac { 1 }{ 2 } ,\frac { 1 }{ 2 } +\frac { 1 }{ 5 } \right) =2-\frac { 1 }{ 5 } =\frac { 9 }{ 5 } >\frac { 7 }... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does $x^2+47y^2 = z^5$ involve solvable quintics? This is related to the post on $x^2+ny^2=z^k$. In response to my answer on,
$$x^2+47y^2 = z^3\tag1$$
where $z$ is not of form $p^2+nq^2$, Will Jagy provided one for,
$$x^2+47y^2 = z^5\tag2$$
$$ (14p^5 + 405p^4q + 3780p^3q^2 + 13410p^2q^3 + 11550pq^4 - 14647q^5)^2 + ... | (A partial answer.) Thanks to Jagy's two parameterizations for degs $5$ and $7$, a general identity has been found. The clue was to transform $(3)$ using $p=u-14v,\,q=3v$ to the form,
$$(14 u^5 + 235 u^4 v - 6580 u^3 v^2 - 22090 u^2 v^3 + 154630 u v^4 + 47^3 v^5)^2 + 47(u^5 - 70 u^4 v - 470 u^3 v^2 + 6580 u^2 v^3 + 110... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Solving simple mod equations Solve $3x^2 + 2x + 1 \equiv 0 \mod 11$
Additionally, I have an example problem, but a step in the middle has confused me:
$3x^2 + 5x - 7 \equiv 0 \mod 17$. Rearrange to get $3x^2 + 5x \equiv 7 \mod 17$.
$\implies 6\cdot 3x^2 + 6\cdot 5x \equiv 6*7 \mod 17$.
The next line reads $x^2 + 30x \e... | $3x^2 + 2x + 1 \equiv 0 \pmod{11}$
Since $3 \cdot 4 \equiv 12 \equiv 1 \pmod{11}, \quad \frac 13 \equiv 4 \pmod{11}$
So, multiplying both sides by $4$, we get
$x^2 + 8x + 4 \equiv 0 \pmod{11}$
Half of $8$ is $4$ and $4^2 = 16 \equiv 5 \pmod{11}$. So
$x^2 + 8x + 4
\equiv x^2 + 8x + 5 - 1
\equiv (x+4)^2 - 1 \pmod{11}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove the series has positive integer coefficients How can I show that the Maclaurin series for
$$
\mu(x) = (x^4+12x^3+14x^2-12x+1)^{-1/4}
\\
= 1+3\,x+19\,{x}^{2}+147\,{x}^{3}+1251\,{x}^{4}+11193\,{x}^{5}+103279\,
{x}^{6}+973167\,{x}^{7}+9311071\,{x}^{8}+\cdots
$$
has positive integer coefficients? (I have others to d... | Here is a proof for positivitiy. Use the recurrence
$$
(n+1)Q(n)+(12n+21)Q(n+1)+(14n+35)Q(n+2)+(-12n-39)Q(n+3)+(n+4)Q(n+4) = 0
$$
$Q(0)=1,Q(1)=3,Q(2)=19,Q(3)=147$. Prove by induction that
$Q(n) \ge 3 Q(n-1)$ for $n \ge 1$. This is true for the first few terms.
Assume true up to $Q(n+3)$, then prove it for $Q(n+4)$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1386819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
Find the value below
Let $a,b,c$ be the roots of the equation $$8x^{3}-4x^{2}-4x+1=0$$
Find $$\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}$$
It's just for sharing a new ideas, thanks:)
| It has been observed by others that the reciprocals $x_i$ $(1\leq i\leq 3)$ of $a$, $b$, $c$ are the solutions of the equation
$$x^3-4x^2-4x+8=0\ .\tag{1}$$
The power sum $$p_3:=x_1^3+x_2^3+x_3^3={1\over a^3}+{1\over b^3}+{1\over c^3}$$ is a symmetric function of the $x_i$. Therefore it can be expressed as a polynomial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Proving that $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2n-1}-\frac{1}{2n}=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$ I have written the left side of the equation as $$\left(1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\cdots+\frac{1}{2n}\right).$$ I... | $$\begin{align}
S&=\sum_{r=1}^n\frac 1{2r-1}-\sum_{r=1}^n\frac 1{2r}\color{orange}{+\sum_{r=1}^n\frac 1{2r}-\sum_{r=1}^n\frac 1{2r}}\\
&=\underbrace{\sum_{r=1}^n\frac 1{2r-1}+\sum_{r=1}^n\frac 1{2r}}_{}-\underbrace{2\sum_{r=1}^n\frac 1{2r}}\\
&=\qquad\quad\sum_{r=1}^{2n}\frac 1r\qquad \qquad-\sum_{r=1}^n\frac 1r\\
&=\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387744",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
How do I find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17? Recently I came across a question,
Find the remainder of $4^0+4^1+4^2+4^3+ \cdots + 4^{40}$ divided by 17?
At first I applied sum of G.P. formula but ended up with the expression $1\cdot \dfrac{4^{41}-1}{4-1}$. I couldn't figure out how ... | If you calculate $4^{k}\ (\operatorname{mod}\ 17)$ for some $k$'s, you might quickly notice that the values are periodically $1,4,-1,-4$. This is not coincidence, since $4^4 \equiv 1 \ (\operatorname{mod}\ 17)$, and thus $$4^{4k+l}\equiv 4^{4k}\cdot 4^l \equiv (4^4)^k\cdot 4^l \equiv 4^l \ (\operatorname{mod}\ 17)$$ So... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
} |
Find the sum of binomial coefficients
Calculate the value of the sum
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
... | $\bf{My\; Solution::}$ Using $$\displaystyle (1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+..........+\binom{n}{n}x^n$$
Now Diff. both side w.r to $x\;,$ We get
$$\displaystyle n(1+x)^{n-1} = \binom{n}{1}x+\binom{n}{2}\cdot 2x+\binom{n}{3}\cdot 3x^2+.........+\binom{n}{n}\cdot nx^{n-1}$$
Now Put $x=1$ and $n=10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find the general solution of $\cos(x)-\cos(2x)=\sin(3x)$ Problem:
Find the general solution of $$\cos(x)-\cos(2x)=\sin(3x)$$
I tried attempting this by using the formula$$\cos C-\cos D=-2\sin(\dfrac{C+D}{2})\sin(\dfrac{C-D}{2})$$
Thus, $$-2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)=\sin 3x$$
$$\Right... | \begin{align}
−2 \sin\left(\frac{x}{2}\right) \, \sin\left(\frac{3x}{2}\right)−\sin 3x &=0 \\
−2 \sin\left(\frac{x}{2}\right) \, \sin\left(\frac{3x}{2}\right)-2 \sin\left(\frac{3x}{2}\right) \, \cos\left(\frac{3x}{2}\right) &=0 \\
\sin\left(\frac{3x}{2}\right) =0 \quad \text{or} \quad \sin\left(\frac{x}{2}\right)-\cos\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
If $(y^2-5y+3)\cdot (x^2+x+1)<2x$ for all $x\in \mathbb{R}\;,$ Then Range of $y$ is
If $(y^2-5y+3)\cdot (x^2+x+1)<2x$ for all $x\in \mathbb{R}\;,$ Then Range of $y$ is
$\bf{My\; Try::}$ We can write the given inequality as $\displaystyle y^2-5y+2<\frac{2x}{x^2+x+1}.$
Now we will calculate range of $$\displaystyle z =... | If $x<0$, then by AM-GM, $|x|+\frac{1}{|x|}\geq 2$, whence
$$\frac{2x}{x^2+x+1}=\frac{2}{1-\left(|x|+\frac{1}{|x|}\right)}\geq \frac{2}{1-2}=-2\,.$$
The equality holds iff $x=-1$. If $x\geq 0$, then $\frac{2x}{x^2+x+1}\geq 0>-2$. That is, $y^2-5y+3<-2$, meaning that all possible values of $y$ are $y\in\mathbb{R}$ suc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What substitution should I do to solve the following integral? Evaluate $$ \int{\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2+x^4}}}dx $$
I substituted $x- \frac{1}{x} $ with u so $$-\int{\frac{du}{u\sqrt{u^2+3}}} $$
Now I put $u=\frac1t$
so $$\int{\frac{dt}{\sqrt{1+3t^2}}} $$
| HINT:
$$\text{Set }\int\left(1+\dfrac1{x^2}\right)dx=u\text{ so that }\dfrac{du}{dx}=\cdots$$
$$\text{in }\frac{(1+x^2)}{(1-x^2)\sqrt{1+x^2+x^4}}=\dfrac{1+\dfrac1{x^2}}{\left(x-\dfrac1x\right)\sqrt{\dfrac1{x^2}+1+x^2}}$$
and use
$$\dfrac1{x^2}+x^2=\left(x-\dfrac1x\right)^2+2$$
For $\int\dfrac{du}{u\sqrt{u^2+3}}$
either... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Counting integral solutions Suppose $a + b + c = 15$
Using stars and bars method, number of non-negative integral solutions for the above equation can be found out as $15+3-1\choose15$ $ =$ $17\choose15$
How to extend this principle for finding number of positive integral solutions of
$a + b + 3c = 15$?
I tried to do i... | One way to solve this is using generating functions. If you multiply $A(z) = \sum_{n \ge 0} a_n z^n$ by $B(z) = \sum_{n \ge 0} b_n z^n$, you get:
$$
A(z) \cdot B(z)
= \sum_{n \ge 0} \left(\sum_{0 \le k \le n} a_k b_{n - k}\right) z^n
$$
If $a_n$ is the number of ways of picking $a$ to have value $n$, and similarly f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find Sum of n terms of the Series: $ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \cdots $ I want to find the sum to n terms of the following series.
$$ \frac{1}{1\times2\times3} + \frac{3}{2\times3\times4} + \frac{5}{3\times4\times5} + \frac{7}{4\times5\times6} + \cdots $$
I have... | $$\displaystyle \frac{2n-1}{n(n+1)(n+2)} = \frac{2}{(n+1)(n+2)}-\frac{1}{n(n+1)(n+2)} = A-B$$
Now $$\displaystyle A = \frac{2}{(n+1)(n+2)} = 2\left[\frac{(n+2)-(n+1)}{(n+1)(n+2)}\right] = 2\left[\frac{1}{n+1}-\frac{1}{n+2}\right]$$
And $$\displaystyle B = \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left[\frac{(n+2)-(n)}{n(n+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int e^x \sin^2 x \mathrm{d}x$ Is the following evaluation of correct?
\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 -... | Let $$\displaystyle I = \frac{1}{2}\int e^x 2\sin^2 xdx = \frac{1}{2}\int e^x-\frac{1}{2}\int e^x \cos 2x dx$$
Now Using $$\displaystyle \cos \phi +i\sin \phi = e^{i\phi}$$ and $$\cos \phi-i\sin \phi = e^{-i\phi}.$$
So $$\displaystyle \cos \phi = \frac{e^{i\phi}+e^{-i\phi}}{2}$$
So we get $$\displaystyle \cos 2x = \le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1398965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
How do I evaluate this improper integral $\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}$ Given integral is $$\int_{-1}^{1}\frac{dx}{(2-x)\sqrt{1-x^{2}}}.$$ I tried to split it up at $0$, but I donot know what to do ahead.
Thanks.
| Let $\displaystyle I =\int_{-1}^{1}\frac{1}{(2-x)\sqrt{1-x^2}}dx....(1)$
Noe Let $x = -t\;,$ Then $dx = -dt$ and changing Limit, we get
$\displaystyle I = \int_{1}^{-1}\frac{1}{(2+t)\sqrt{1-t^2}}\times -dt = \int_{-1}^{1}\frac{1}{(2+x)\sqrt{1-x^2}}......(2)$
Now Add $(1)$ and $(2)\;,$ we get $\displaystyle 2I = \int_{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to get to this equality $\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1$?
How to get to this equality $$\prod_{m=1}^{\infty} \frac{m+1}{m}\times\frac{m+x}{m+x+1}=x+1?$$
I was studying the Euler Gamma function as it gave at the beginning of its history, and need to solve the following product operat... | If you write out the first few terms of the product, you'll see that most of the factors cancel out:
$$
\require{cancel}
\begin{align}
\prod_{m=1}^{\infty} \frac{m+1}{m}\cdot\frac{m+x}{m+x+1} &= \left(\frac{\cancel2}{1}\cdot\frac{x+1}{\cancel{x+2}}\right)\left(\frac{\cancel3}{\cancel2}\cdot\frac{\cancel{x+2}}{\cancel{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove area of a quadrilateral is $\frac14[4m^2n^2-(b^2+d^2-a^2-c^2)^2]^{\frac12}$ Someone asked me this question which I am really stuck at, any help is appreciated.
If $a,b,c,d$ are the sides of a quadrilateral and $m,n$ are diagonals of the quadrilateral, then prove that area of the quadrilateral is $$\frac14[4m^2n... | For convex $\square ABCD$, we can proceed as follows:
Let the diagonals meet a $P$, which subdivides the diagonals into segments of length $m_1$, $m_2$, $n_2$, $n_2$; and let the diagonals make an angle of $\theta$ at $P$ as shown. Then we can write:
$$\begin{align}
|\square ABCD| &= |\triangle APB| + |\triangle BPC| ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Power series of z/sin(z)? So I need to compute the coefficient of the $z^4$ in the power series of $\frac{z}{\sin z}$.
I tried differentiating the function and obtaining coefficients like in Taylor's expansions but had a really hard time. In general I'm finding it extremely hard to obtain the power series of various co... | Here is a simple method that works well to find the first few terms in the power-series.
We first expand the denominator in a power-series around $z=0$: $$\frac{z}{\sin(z)} = \frac{1}{1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \ldots}$$
This has the form of a geometrical series $\frac{1}{1-x} = 1+x + x^2 + \ldots$ for $x = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
limit involving $e$, ending up without $e$. Compute the limit
$$ \lim_{n \rightarrow \infty} \sqrt n \cdot \left[\left(1+\dfrac 1 {n+1}\right)^{n+1}-\left(1+\dfrac 1 {n}\right)^{n}\right]$$
we have a bit complicated solution using Mean value theorem. Looking for others
| Taylor expansion always works:
\begin{align}
& \left(1 + \frac{1}{n + 1}\right)^{n + 1} - \left(1 + \frac{1}{n}\right)^{1/n}\\
= & \exp\left[(n + 1)\log\left(1 + \frac{1}{n + 1}\right)\right] - \exp\left[n\log\left(1 + \frac{1}{n}\right)\right] \\
= & \exp\left[(n + 1)\left(\frac{1}{n + 1} - \frac{1}{2(n + 1)^2} + o\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$ Determine the equation of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ such that it has the least area but contains the circle $(x-1)^2+y^2=1$
Since the area of ellipse i... | Firstly, define $f(x):=b\sqrt{1-\frac{x^2}{a^2}}$ and $g(x):=\sqrt{1-(x-1)^2}$ such that $f$ and $g$ represent the ellipse and the circle respectively in the upper half of the coordinate system.
In order to guarantee that the ellipse contains the circle, we need to have:
$$
f(x)≥g(x)\iff b\sqrt{1-\frac{x^2}{a^2}}≥\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1401256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A circle of radius $r$ is dropped into the parabola $y=x^{2}$. Find the largest $r$ so the circle will touch the vertex. If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle wil... | Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0\ \ \vee\ \ x^2+1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How to find the MacLaurin series of $\frac{1}{1+e^x}$ Mathcad software gives me the answer as: $$ \frac{1}{1+e^x} = \frac{1}{2} -\frac{x}{4} +\frac{x^3}{48} -\frac{x^5}{480} +\cdots$$
I have no idea how it found that and i don't understand. What i did is expand $(1+e^x)^{-1} $ as in the binomial MacLaurin expansion. I ... | Let $\dfrac1{1+e^x}=a_0+a_1x+a_2x^2+\cdots$
$$\implies(1+e^x)\left(a_0+a_1x+a_2x^2+\cdots\right)=1$$
As $e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$
$$\implies\left(2+\dfrac x1+\dfrac{x^2}2+\dfrac{x^3}{3!}+\cdots\right)\left(a_0+a_1x+a_2x^2+\cdots\right)=1$$
Comparing the coefficients of different powers of $x,$
$\implies1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Integrating $\frac{1}{(x^4 -1)^2}$ How to solve the the following integral? $$\int{\frac{1}{(x^4 -1)^2}}\, dx$$
| In general, the following method will always work. Solve for the roots of the denominator in the complex plane:
$$
\begin{split}
z^4 &= 1 \Longrightarrow\\
z&=z_{n}\equiv\exp\left(\frac{i\pi n}{2}\right)
\end{split}
$$
In this case things are simple, you have $z_0=1$, $z_1=i$, $z_2=-1$, $z_3 = -i$.
Then the trick is to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1406034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why is $\binom{2n}{n} \asymp \Theta \big(\frac{2^{2n}}{\sqrt{n}}\big)$? I saw this statement : $$\binom{2n}{n} \asymp \Theta \bigg(\frac{2^{2n}}{\sqrt{n}}\bigg) \asymp \Theta\bigg(\frac{4^n}{\sqrt{n}}\bigg)$$
How did we go from the first statement to the second? I tried Stirling's aproximation, but that didn't get me a... | Stirling's approximation gives
$$\binom{2n}{n} = \frac{(2n)!}{n!^2} \sim \frac{\sqrt{2\pi \cdot 2n} \left( \frac{2n}{e} \right)^{2n}}{\left[\sqrt{2\pi n} \left( \frac{n}{e} \right)^n \right]^2} = \frac{2\sqrt{\pi n} \cdot 2^{2n} \cdot \left( \frac{n}{e} \right)^{2n}}{2 \pi n \cdot \left( \frac{n}{e} \right)^{2n}} = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Local extremes of $f(x) = (x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}}$ The task is to find local extremes of $f: \mathbb R \to \mathbb R$, $f(x) = (x-2)^{\frac{1}{5}}(x-7)^{\frac{1}{9}}$
There is theorem that if $x_{0}$ is local extreme of $f(x)$ then $f'(x_0) = 0$
So I start from calculating derivative of $f(x)$
$$f'(x) = ... | for $$f'(x)$$ i have got $$f'(x)=1/45\,{\frac {14\,x-73}{ \left( x-2 \right) ^{4/5}} \left( x-7
\right) ^{-{\frac {8}{9}}}}
$$ and for the second derivative we have
$$f''(x)=-{\frac {434\,{x}^{2}-4526\,x+15416}{2025\, \left( x-2 \right) ^{9/5}}
\left( x-7 \right) ^{-{\frac {17}{9}}}}
$$ here you must plugg in $$x_0=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1407914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no i... | For any positive integer x this is true: $x \leqslant x^2$ (From $1 \leqslant x$ for any positive ineger x ). So $a + b + c \leqslant a^2 + b^2 + c^2$. But for 2 positive integers $x$, $y$ $x$ is divisible by $y$ only if $x \geqslant y$. So $a + b + c \geqslant a^2 + b^2 + c^2$ if $a + b + c \mid a^2 + b^2 + c^2$.
Fro... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 6,
"answer_id": 5
} |
Evaluating the ratio $ {{a_{n+1}}\over{a_n}}$ in calculating the radius of convergence for a power series In calculating the radius of convergence for the power series
$$ \sum_{n=1}^\infty {{(2n)!}\over(n!)^2}\ x^n $$
By the ratio test, we let
$$ a_n = \lvert {{(2n)!}\over(n!)^2}\ x^n \rvert \quad\quad a_{n+1} = \lvert... | Hint
\begin{equation}
\frac{(2n + 2)!}{(2n)!} \frac{(n!)^2}{(n + 1)!} = \frac{(2n + 2)(2n + 1)}{(n + 1)(n + 1)} = \frac{4n^2 + 6n + 2}{n^2 + 2n + 1} = \frac{4 + \frac{6}{n} + \frac{2}{n^2}}{1 + \frac{2}{n} + \frac{1}{n^2}}
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find min of $M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$
Find min of $$M=\frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}$$, where $A, B, C$ are three angle of triangle $ABC$
Using Cauchy-Schwarz, we obtain: \begin{align*}
M &= \frac{1}{2+\cos2A}+\frac{1}{2+\cos2B}+\frac{1}{2-\cos2C}\\
&\ge... | The change might not be correct, but before that the numerator is
$$
f(A,B)=6+\cos 2A+\cos 2B-\cos 2C=6+\cos 2A+\cos 2B-\cos 2(A+B)
$$
$$
\frac{\partial f}{\partial A}=2 \sin (2 A+2 B)-2 \sin (2 A)
$$
$$
\frac{\partial f}{\partial B}=2 \sin (2 A+2 B)-2 \sin (2 B)
$$
We can only have extrema at the border of the set, n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\cos \arctan 1/2 = 2/\sqrt{5}$ How can we prove the following? $$\cos \left( \arctan \left( \frac{1}{2}\right) \right) =\frac{2}{\sqrt{5}}$$
| Let $\displaystyle \tan^{-1}\left(\frac{1}{2}\right) = \theta\;,$ Then $\displaystyle \frac{1}{2} = \tan \theta$
So $\displaystyle \cos\left[\tan^{-1}\left(\frac{1}{2}\right)\right] = \cos \theta = \frac{1}{\sec \theta} = \frac{1}{\sqrt{1+\tan^2 \theta}}= \frac{2}{\sqrt{5}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If complex no. ($z$) satisfying $\frac{1}{2}\leq |z|\leq 4\;,$ Then Max. and Min. of $\left|z+\frac{1}{z}\right|$
Let $z$ be a complex no. satisfying $\displaystyle \frac{1}{2}\leq |z|\leq 4\;,$ then the Sum of greatest and
least value of $\displaystyle \left|z+\frac{1}{z}\right|$ is
$\bf{My\; Try::}$ Let $z=re^{i\th... | $g(r,\alpha)= r^2+\dfrac{1}{r^2}+2cos2\alpha$
for min:$r^2+\dfrac{1}{r^2} \ge 2,cos2\alpha \ge -1$
for max: $r^2+\dfrac{1}{r^2}$ will get max at two ends, $cos2\alpha \le 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1413252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Prove that $s(n-1)s(n)s(n+1)$ is always an even number Let $n$ be a natural number, and let $s(n)$ denote the sum of all positive divisors of $n$. Show that for any $n>1$ the product $s(n-1)s(n)s(n+1)$ is always an even number.
I calculated the sum of the divisors(in the form a geometric Progression) and then tried pro... | Express the numbers $n-1,n,n+1$ in their prime factorisations.
Let $x$ be any one of $n-1,n,n+1$ and express $$x=2^a\cdot3^b\cdot5^c\cdot\ldots \tag{1}$$
Then $$s(x)=(1+2+\ldots 2^a)\times(1+3+\ldots 3^b)\times(1+5+\ldots 5^c)\times\ldots \tag{2}$$
* Assume now that all of $s(n-1),s(n),s(n+1)$ are odd *
The first term... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve this system of nonlinear equations? How to solve these equations for $a$, $b$, $c$ and $x$?
I have the following:
\begin{align}
1 &= 2a+b+c\\
a &= (a+b)x + 0.25(a+c)\\
a&=(a+c)(1-x)\\
b&=a(1-x)+c(x-0.25)\\
c&=b(1-x)+a(x-0.25)
\end{align}
I tried, but ended circular at the point I started. Can someone help... | Using Grobner basis in maxima gives:
grobner_basis([2*a+b+c-1,(a+b)x+(a+c)/4-a,(a+c)(1-x)-a,a*(1-x)+c*(x-1/4)-b,b*(1-x)+a*(x-1/4)-c]);
(- 44) x + (- 144) a^2 + 213 a + (- 36),
(- 11) c + (- 48) a^2 + 60 a + (- 12),
(- 11) b + 48 a^2 + (- 82) a + 23,
(- 144) a^3 + 381 a^2 + (- 312) a + 64
The last equation is a c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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"Trig Substitutions", I tried half- angle and trig indentity in this one, but doesn't work I´m really lost in this one.
$\int \sin^3 (2x) \cos^2 (2x) dx$
I know that the answer is:
$\frac{1}{10}cos^5(2x)-\frac{1}{6}cos^3(2x) + c$
Please help
| HINT:
Write $\sin^3(2x)\cos^2(2x)=(1-\cos^2(2x))\cos^2(2x)\sin(2x)$
SPOLIER ALERT: SCROLL OVER THE SHADED AREA TO REVEAL SOLUTION
We have $$\begin{align}\sin^3(2x)\cos^2(2x)&=\left(1-\cos^2(2x)\right)\cos^2(2x)\sin(2x)\\\\&=\left(\cos^2(2x)-\cos^4(2x)\right)\sin(2x)\end{align}$$Then, $$\int \sin^3(2x)\cos^2(2x)\,dx=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find tangent to trigonometric function I want to find the tangent to the curve:
$x\sin{y} + y\sin{x} = \frac{\pi}{4}(1+\sqrt{2})$
through the point $(\frac{\pi}{2}, \frac{\pi}{4})$
Now I know I can fill certain information into this formula
$y-y_0 = a(x-x_0)$ to get a tangent, but I dont really know how to find
$a$ i... | We will use implicit differentiation in this problem to extract $\frac{dy}{dx}$:
$$
x\sin{y}+y\sin{x}=\frac{\pi}{4}(1+\sqrt{2})\\
\cos{y}+x\frac{dy}{dx}\cos{y}+\frac{dy}{dx}\sin{x}+y\cos{x}=0\\
\frac{dy}{dx}(x\cos{y}+\sin{x})=-(\sin{y}+y\cos{x})\\
\frac{dy}{dx}=-\frac{\sin{y}+y\cos{x}}{x\cos{y}+\sin{x}}
$$
And for $x=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1420212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to evaluate $45^\frac {1-a-b}{2-2a}$ where $90^a=2$ and $90^b=5$ without using logarithm? Let $90^a=2$ and $90^b=5$, Evaluate
$45^\frac {1-a-b}{2-2a}$
I know that the answer is 3 when I used logarithm, but I need to show to a student how to evaluate this without involving logarithm. Also, no calculators.
| Let me try.
$$10 = 90^{a+b} \Rightarrow 3^2 = 90^{1-a-b} \Rightarrow 3 = 90^{\frac{1-a-b}{2}}.$$
Then, $$45 = 90^{(1-a-b)+b} = 90^{1-a}.$$
So, $$45^{\frac{1}{1-a}} = 90 \Rightarrow 45^{\frac{1-a-b}{2(1-a)}} = 90^{\frac{1-a-b}{2}} = 3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find $z$ when $z^4=-i$? Consider $z^4=-i$, find $z$.
I'd recall the fact that $z^n=r^n(\cos(n\theta)+(i\sin(n\theta))$
$\implies z^4=|z^4|(\cos(4\theta)+(i\sin(4\theta))$
$|z^4|=\sqrt{(-1)^2}=1$
$\implies z^4=(\cos(4\theta)+(i\sin(4\theta))$
$\cos(4\theta)=Re(z^4)=0 \iff \arccos(0)=4\theta =\frac{\pi}{2} \iff \theta=\... | the last step, you need $$-i = \cos(3\pi/2) + \sin (3\pi/2) = \cos (4 \theta) + i \sin(4 \theta) $$ which will give you $$4\theta = 3\pi/2, 3\pi/2 + 2\pi, 3\pi/2 + 4\pi, 3\pi/2 + 6\pi$$ now you can solve for $\theta.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1421839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Solve using: complete quadratic equation? $$x^2 - x - 20 > 0$$
$$x^2 - x > 20$$
Step 3:
So to complete it, let's use: $1/2$ so it ends up with $x$ as the middle term.
$$x^2 - x + 1/2 > 20 + 1/2$$
Step 4:
But then:
It would be transform to:
$$(x - 1/2)^2 > 20 + 1/2$$
Step 5:
But resolving it, the initial inequation of... | Given $$\displaystyle x^2-x-20=0\Rightarrow \underbrace{x^2-x+\left(\frac{1}{2}\right)^2}-\underbrace{20-\left(\frac{1}{2}\right)^2} =0$$
So we get $$\displaystyle \left(x-\frac{1}{2}\right)^2-\left(\frac{9}{2}\right)^2>0$$
So we get $$\displaystyle \left[x-\frac{1}{2}-\frac{9}{2}\right]\cdot \left[x-\frac{1}{2}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Conjecture $\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)=\frac{7\pi^2}{48}-\frac13\arctan^22-\frac16\arctan^23-\frac18\ln^2(\tfrac{18}5)$ I numerically discovered the following conjecture:
$$\Re\,\operatorname{Li}_2\left(\frac12+\frac i6\right)\stackrel{\color{gray}?}=\frac{7\pi^2}{48}-\frac{\arctan^22}3-\frac... | I don't know how you got your conjecture but I checked and on further simplifications I found it to be correct.
We know that :
$$
{Li}_{2}(\bar{z})=\bar{{Li}_{2}(z)}
$$
So :
$$
\Re{{Li}_{2}(z)}=\frac{\bar{{Li}_{2}(z)}+{Li}_{2}(z)}{2}
$$
So:
$$
\Re{{Li}_{2}(\frac{1}{2}+\frac{i}{6})}=\frac{{Li}_{2}(\frac{1}{2}+\frac{i}{6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 0
} |
What is the integral for this expression? (I tried complete the square) But complete the square, doesn´t lead me to something coherent.
I got this:
$$\int\frac{x \, dx}{\sqrt{3-2x-x^2}}$$
| $$
\int\frac{x \, dx}{\sqrt{3-2x-x^2}}
$$
First let $u=3-2x-x^2$ so that $du = -2(1+x)\,dx$ and $\dfrac{du}{-2} = (1+x)\,dx$. Then we can say
\begin{align}
\int\frac{x \, dx}{\sqrt{3-2x-x^2}} = \int \frac{(1+x) \, dx}{\sqrt{3-2x-x^2}} - \int \frac{dx}{\sqrt{3-2x-x^2}}.
\end{align}
For the first integral, use the subst... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1427629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to get the RHS from the LHS $\sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1}$ How to get this required RHS from the given LHS.
$$\sum_{k=0}^n2^kx^{k+2}=\frac{x^2-2^{n+1}x^{n+3}}{1-2x}\tag{1}$$
This was used in a solution to the following question I asked. I couldn't understand the step and hence to un... | $$
\sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\sum_{k=0}^{n}{2^k{x}^{k}}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\sum_{k=0}^{n}{{(2x)}^{k}}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = x^2\frac{1-{(2x)}^{n+1}}{1-2x}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = \frac{x^2-x^2{2}^{(n+1)}{x}^{n+1}}{1-2x}\\
\sum_{k=0}^{n}{2^k{x}^{k+2}} = \frac{x^2-{2}^{n+1}{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1428052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculating the $n^\text{th}$ derivative How do we calculate the $n^{\text{th}}$ derivative for
$$
\frac{x^3}{(x-a)(x-b)(x-c)}?
$$
How can I obtain the partial fraction for the given term?
| If we assume that $a,b,c$ are distinct numbers, from:
$$ \text{Res}\left(\frac{z^3}{(z-a)(z-b)(z-c)},z=a\right)=\frac{a^3}{(a-b)(a-c)}\tag{1}$$
it follows that:
$$\begin{eqnarray*} f(z)&=&\frac{z^3}{(z-a)(z-b)(z-c)}\tag{2}\\&=&1+\frac{a^3}{(a-b)(a-c)(z-a)}+\frac{b^3}{(b-a)(b-c)(z-b)}+\frac{c^3}{(c-a)(c-b)(z-c)}\end{eqn... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Convergence/divergence of: $\sum_{n = 2}^\infty \ln\left( 1-\frac{1}{n^2}\right)$? How to prove the convergence/divergence of:
$$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right)$$ ?
I've tried:
$$\sum_{n = 2}^{+\infty} \ln\left( 1-\frac{1}{n^2}\right) = \sum_{n = 2}^{+\infty} \ln\left(\frac{n^2-1}{n^2}\right) $... | \begin{array} \\
\displaystyle\sum_{n=2}^N \ln\left(1-\frac{1}{n^2}\right) &=& \displaystyle\sum_{n=2}^N \ln\left(\frac{n^2-1}{n^2}\right) \\
&=& \displaystyle\sum_{n=2}^N \ln\left(n^2-1\right)-\ln\left(n^2\right) \\
&=& \displaystyle\sum_{n=2}^N \ln\left((n-1)(n+1)\right)-2\ln n \\
&=& \displaystyle\sum_{n=2}^N \ln(n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Simple limit of a sequence
Need to solve this very simple limit $$ \lim _{x\to \infty
\:}\left(\sqrt[3]{3x^2+4x+1}-\sqrt[3]{3x^2+9x+2}\right) $$
I know how to solve these limits: by using
$a−b= \frac{a^3−b^3}{a^2+ab+b^2}$. The problem is that the standard way (not by using L'Hospital's rule) to solve this limit - ver... | $$\lim _{ x\to \infty \: } \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) =$$
$\lim _{ x\to \infty \: } \frac { \left( \sqrt [ 3 ]{ 3x^{ 2 }+4x+1 } -\sqrt [ 3 ]{ 3x^{ 2 }+9x+2 } \right) \left( \sqrt [ 3 ]{ { \left( 3x^{ 2 }+4x+1 \right) }^{ 2 } } +\sqrt [ 3 ]{ \left( 3x^{ 2 }+4x+1 \right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1433216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Show that $e^{\sqrt 2}$ is irrational I'm trying to prove that $e^{\sqrt 2}$ is irrational. My approach:
$$
e^{\sqrt 2}+e^{-\sqrt 2}=2\sum_{k=0}^{\infty}\frac{2^k}{(2k)!}=:2s
$$
Define $s_n:=\sum_{k=0}^{n}\frac{2^k}{(2k)!}$, then:
$$
s-s_n=\sum_{k=n+1}^{\infty}\frac{2^k}{(2k)!}=\frac{2^{n+1}}{(2n+2)!}\sum_{k=0}^{\infty... | Since the sum of two rational numbers is rational, one or both of $e^{\sqrt{2}}$
and $e^{-\sqrt{2}}$ is irrational. But, $e^{-\sqrt{2}}=1/e^{\sqrt{2}}$,
and hence both are irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1441292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 3,
"answer_id": 0
} |
Minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ If the minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ when $x>0$ is $\frac{p}{q}$ where $p,q\in N$ then find the least value of $(p+q).$
I can find the minimum value of $\frac{4}{x}+2x$ using $AM- GM$ inequality but that too comes irrational and when i tr... | The derivative of $f(x) = \frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ is
$$
\begin{align}
f'(x) &= \frac{-4}{x^2}+2+\frac{(4x^2+1)(1) - (3+x)(8x)}{(4x^2+1)^2}\\
&= \frac{-4}{x^2}+2+\frac{-4x^2-24x+1}{(4x^2+1)^2}\\
\end{align}
$$
So when $f'(x) = 0$, we have
$$
\begin{align}
-4(4x^2+1)^2 + 2x^2(4x^2+1)^2 + (-4x^2-24x+1)x^2 &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"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.