Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Which is bigger: $\sqrt{1001} - \sqrt{1000}$, or $\frac{1}{10}$?
Which is bigger: $\sqrt{1001} - \sqrt{1000}$, or $\frac{1}{10}$?
I can calculate the answer using a calculator, however I suspect to do so may be missing the point of the question.
The problem appears in a book immediately after a section called 'Rules ... | We want to find the sign of $\sqrt{1001} - \sqrt{1000} - \dfrac{1}{10}$. Multiplying by the obviously positive $\sqrt{1001} + \sqrt{1000} + \dfrac{1}{10}$ we get $(\sqrt{1001})^2 - (\sqrt{1000} + \dfrac{1}{10})^2 = 1001 - 1000 - \dfrac{2\sqrt{1000}}{10} - \dfrac{1}{100} = \dfrac{99}{100} - \dfrac{2\sqrt{1000}}{10}$
Si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Find the value of $\sin(B-A)$. If $A$ is an acute angle whose tangent is $\frac{15}{8}$ and $B$ is and obtuse angle whose sine is $\frac{12}{13}$, find $\sin (B-A)$.
[Without calculators]
I suppose I gotta use this formula: $\sin B \cos A - \cos B \sin A$
Before that they asked me to find $\tan 2A$ which is $\frac{-240... | Label a right triangle with one acute angle $A$. Since $\tan A = \dfrac{15}{8}$ you can label the opposite side with length $15$ and the adjacent side with length $8$. Then the hypotenuse has length $17$ so that $\sin A = \dfrac{15}{17}$ and $\cos A = \dfrac{8}{17}$.
For $B$, use $\sin^2 B + \cos^2 B = 1$. Since $B$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/763523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Question about a sum Why is it that
$$\sum_{k=1}^n(n+k+1)(n+1)=\frac{3}{2}\sum_{k=1}^n3k^2+k.$$
I cannot understand it. This is not homework, I am just a little interested in this!
| The first sum can be written as$$\sum_{k=1}^n(n+k+1)(n+1)=(n+1)\sum_{k=1}^n(n+k+1)=(n+1)n(n+1)+(n+1)\sum_{k=1}^nk\\=n(n+1)^2+\frac{n(n+1)^2}{2}=\frac{3n(n+1)^2}{2}$$
instead the second one as
$$\frac{3}{2}\sum_{k=1}^n3k^2+k=\frac{3}{2}\left(\sum_{k=1}^n3k^2+\sum_{k=1}^nk\right)=\frac{3}{2}\left(3\sum_{k=1}^nk^2+\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/764579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove inequality $(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$ I am trying to prove the following inequality:
$$(\sqrt{a} - \sqrt{b})^2 \leq \frac{1}{4}(a-b)(\ln(a)-\ln(b))$$
for all $a>0, b>0$.
Does anyone know how to prove it?
Thanks a lot in advance!
| Inequality that is another form of a well-known inequalities:
For $0<y<x$
$$ \sqrt{xy}<\frac{x-y}{\ln x-\ln y} < \frac{x+y}{2}. (1)$$
For $a>b$ inequality of enunciation transform so:
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(a-b)(\ln a-\ln b)<=>$$
$$(\sqrt{a} - \sqrt{b})^2 < \frac{1}{4}(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/765738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
non-homogeneous recurrence relations The question is:
$a_{n}=12a_{n-2}+16a_{n-3}+9\cdot 4^{n}+81n$with $a_{0}=0 ,a_{1}=1 ,a_{2}=98$
I tried to deal with the particular solution first by:
$A4^{n}+Bn=12[A4^{n-2}+B(n-2)]+16[A4^{n-3}+B(n-3)]+9\cdot 4^{n}+81n$.
However, turns out the coefficient of $4^{n}$being 0 which mean... | Go for generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, rewrite the recurrence without index subtractions:
$$
a_{n + 3} = 12 a_{n + 1} + 16 a_n + 576 \cdot 4^n + 81 n + 243
$$
Multiply by $z^n$, sum over $n \ge 0$, and recognize a few sums to get:
$$
\frac{A(z) - a_0 - a_1 z - a_2 z^2}{z^3}
= 12 \frac{A(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/765894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the axis of rotation of a rotation matrix by inspection (NOT by solving $Kv=v$) $$K=\
\begin{pmatrix}
0 & 0 & 1\\
-1 & 0 & 0\\
0 & -1 & 0
\end{pmatrix}$$
Find the axis of rotation for the rotation matrix $K$ by INSPECTION.
This is from my other thread click here to view it
Everything you see below is me finding... | Edit. For any $3\times3$ rotation matrix $K$, if it is not symmetric, you can read off the rotation axis directly from its skew-symmetric part $W=K-K^T$. More specifically, the rotation axis is parallel to $(w_{32}, w_{13}, w_{21})^T=-(w_{23}, w_{31}, w_{12})^T$.
Reason: any anticlockwise rotation for an angle $\theta$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/766565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Definite integral problem I was solving a definite integral problem which was reduced to :
$$\int^{1}_{0} \frac{\ln(1+t)}{t} dt$$
I couldn't solve it and when I saw the solution, the answer was simply given as $\frac{\pi^2}{12}$, and claimed that this is an identity.
Can anybody give me a proof of this identity?
| \begin{align*}
\int_0^1 \frac{\log(x+1)}{x} \, dx &= \int_0^1 \sum_{n=1}^\infty (-1)^{n+1} \frac{x^{n-1}}{n}dx\\
&= \sum_{n=1}^\infty (-1)^{n+1}\int_0^1 \frac{x^{n-1}}{n}dx\\
&= \sum_{n=1}^\infty (-1)^{n+1} \frac{1}{n^2}\\
&= \frac{\pi^2}{12}\\
\end{align*}
To calculate that sum, let us assume that the value of... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/766786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Calculation of $\int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$ Calculation of $\displaystyle \int_{0}^{1}\frac{x-1+\sqrt{x^2+1}}{x+1+\sqrt{x^2+1}}dx$
$\bf{My\; Try::}$ Let $x=\tan \psi\;,$ Then $\displaystyle dx = \sec^2 \psi$
So Integral convert into $\displaystyle \int_{0}^{\frac{\pi}{4}}\frac{\tan \psi-1+\... | Solution:
Let $$t=x+1+\sqrt{1+x^2}\to 1+x^2=(t-1-x)^2\to x=\frac{t^2-2t}{2t-2}\to dx=\frac{t^2-2t+2}{2(t-1)^2}dt$$
We have $$x-1+\sqrt{1+x^2}=t-2$$
And $$\frac{x-1+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}=\frac{t-2}{t}$$
Hence $$I=\int_{0}^{1}\frac{x-1+\sqrt{1+x^2}}{x+1+\sqrt{1+x^2}}dx=\int_{2}^{2+\sqrt{2}}\frac{t-2}{t}\frac{t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
If $x\equiv2\pmod{3}$ prove that $3|4x^2+2x+1$ I've tried many different things to get a factor of $k-2$ but keep failing.
If $x\equiv2\pmod{3}$ prove that $3 \mid 4x^2+2x+1$
| Hint $\ {\rm mod}\ 3\!:\ x\equiv 2\equiv -1\,\Rightarrow\, 1+2x+4x^2\equiv 1-2+4\equiv 3\equiv 0.\,$
Generally for polynomial $\ p(x),\,\ \ p(-1) = p_0-p_1+p_2-p_3+\cdots = $ alternating coef sum.
Alternatively $\,\color{#c00}{4\equiv1}\,\Rightarrow\, \color{#c00}{4x^2}\!+2x+1\equiv \color{#c00}{x^2}\!+2x+1\equiv (x\!... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/769293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Find equation of a circle given a sector A sector of a semi-circle which is $60^\circ$ has an area of $\frac{3\pi}{2}$ units squared.
Show that the curved section is a function of the form $f(x)= \sqrt{9-x^2}$ with domain $[0,\frac{3\sqrt{3}}{2}]$
| Area of sector
\begin{align}
A&=\frac{60^\circ}{360^\circ}\pi r^2\\
\frac{3\pi}{2}&=\frac{1}{6}\pi r^2\\
r&=3
\end{align}
Equation of circle with center on the origin
\begin{align}
x^2+y^2&=r^2\\
x^2+y^2&=3^2\\
y&=\sqrt{9-x^2}
\end{align}
then use parametric equation (polar coordinate) $x=r\cos\theta$ and $y=r\sin\thet... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Factoring in terms of Irreducibles Factor the polynomial $x^5 + 2x^3 + 3x^2 + 1$ as a product of irreducible polynomials in $\mathbb{Z}_5[x]$.
My thoughts: I know what the definition of an irreducible function is but as far as methods to find the correct answer as easy as possible is difficult to understand.
| You found the linear factors already, leaving the quartic $x^4 + 3x^3 + x^2 + x + 3$. The only possibility is that this factors into two quadratics, thus:
$$
(x^2+ax+b)(x^2+cx+d)\equiv x^4 + 3x^3 + x^2 + x + 3\pmod5
$$
which you can expand as
$$
x^4+(a+c)x^3+(ac+b+d)x^2+(ad+bc)x+bd\equiv x^4 + 3x^3 + x^2 + x + 3\pmod5
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/771608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$? How to integrate $\frac{\sqrt{x}}{1-\sqrt{x}}$?
I tried by using integration by parts, but always got sucked. Should be very easy...
| Let $u=\sqrt{x}$ and $x=u^2$, $\frac{\sqrt{x}}{1-\sqrt{x}}=-1+\frac{1}{1-\sqrt{x}}=-1+\frac{1}{1-u}$, then
\begin{align}
\int{\frac{\sqrt{x}}{1-\sqrt{x}}dx}&=\int{\left(-1+\frac{1}{1-u}\right)(2udu)}\\
&=-\int{2udu}+2\int{\frac{u}{1-u}du} \\
&=-u^2+2\int{\left(-1+\frac{1}{1-u}\right)du} \\
&=-u^2-2u+2\ln|1-u|+C \\
&=-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Non homogeneous Recurrence relation problem So here i have this non homogeneous recurrence relation i need to solve:
$$a_{n}=12a_{n-2}+16a_{n-3}+9\cdot 4^{n}+81n,$$
where $a_{0}=0$, $a_{1}=1$ $a_{2}=98$. I'm confused at the homogeneous relation part of this relation. What's the characteristic of this type of homogene... | Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence with no index subtractions:
$$
a_{n + 3}
= 12 a_{n + 1} + 16 a_n
+ 576 \cdot 4^n + 81 n + 243
$$
The resulting recurrence is now valid for $n \ge 0$ ( no terms for negative index allowed originally).
Multiply by $z^n$, sum ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/774125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the value of one-sided limits and greatest integer function.
$$
\lim_{x \to 0} \frac{a}{x} \left\lfloor\frac{x}{b} \right\rfloor
$$
The $\lfloor \rfloor$ stands for the greatest integer function.
I have calculated and the left-hand limit is coming as (ab). But, I have doubt in the right-hand limit. I did this... | I assume that
$a > 0$
and $b > 0$.
If $0 < x < b$,
$\lfloor \frac{x}{b} \rfloor
= 0$.
Therefore,
for
$0 < x < b$,
$\frac{a}{x}\lfloor \frac{x}{b} \rfloor
= 0$.
Therefore,
$\lim_{x \to 0^{+}}\frac{a}{x}\lfloor \frac{x}{b} \rfloor
= 0$.
The expression is not defined at $x = 0$.
If $0 > x > -b$,
$\lfloor \frac{x}{b} \rflo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/775776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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finding square $k$ such that $k\mid 2^{k-1} -1$ Can we find a positive square integer $k>1$, which satisfies $k\mid 2^{k-1} -1$ ?
If yes, what are such $k>1$ values? Here $k = n^2$ and $n$ is some positive integer. If we cannot find such square integer $k>1$, then how to disprove the statement $k\mid 2^{k-1} -1$?
High ... | Given that $n^2|(2^{n^2-1}-1)$.
Now it is obvious that $(2^r-1)|(2^{n^2-1}-1)$, where $r|(n^2-1)$, hence,
$2^r-1=n^2$ for some positive integer $r$. Now $r$ cannot be even, (otherwise, $2^{2k}-n^2=1$ for $r=2k$, and factorising yield $2^k-n=1$ and $2^k+n=1$ for which $n=0$, or $2^k-n=-1$ and $2^k+n=-1$ for which $n=1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/775978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving simultaneous congruences with the Chinese remainder theorem Solve the following system of congruences using the Chinese remainder theorem:
$$\begin{align*}
2x &\equiv 3 \pmod{7} \\
x &\equiv 4 \pmod{6} \\
5x &\equiv 50 \pmod{55}
\end{align*}
$$
I was a little confused how to reduce the congruences into a form w... | The congruence $2x\equiv 3\pmod{7}$ holds if and only if $x\equiv 5\pmod{7}$. And $5x\equiv 50\pmod{55}$ if and only if $x\equiv 10\pmod{11}$. Now the problem is in "standard" form.
Remark: To get the first result, we can multiply both sides of $2x\equiv 3\pmod{7}$ by the inverse of $2$ modulo $7$. Note that $4$ is th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/778854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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How many ways can you build a football/soccer team? I'm reading a book on the history of soccer. There are several ways to position a team, for example, four defenders, four midfielders and two forwards is the common 4-4-2. How many ways can you build a team, given that there are 10 players that move because the keeper... | You are looking for the number of compositions of 10, this is $2^{10 - 1} = 512$.
The proof of this fact is similar to stars-and-bars: To divide $n$ into $k$ nonempty pieces is to lay down $n$ stars, and place separators (bars) in $k - 1$ of the $n - 1$ positions between stars, i.e, $\binom{n - 1}{k - 1}$ ways. Adding ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/780341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Pie Integral $\int_0^1 \log\frac{(x+\sqrt{1-x^2})^2}{(x-\sqrt{1-x^2})^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.$ Hi I am trying to show this astonishing result$$
\int_0^1 \log\frac{\big(x+\sqrt{1-x^2}\big)^2}{\big(x-\sqrt{1-x^2}\big)^2} \frac{x\, dx}{1-x^2}=\frac{\pi^2}{2}.
$$
Note we can to use $\ln(a/b)=\ln a-\ln b$ b... | So here is a solution which follows the logic of both @gar's and @xpaul's but avoids the issue of poles.
Let $I(a)$ be defined by
$$ I(a) = \int_{0}^{1} \frac{2x}{1-x^{2}} \log \left| \frac{x + \sqrt{1-x^{2}}}{x + a\sqrt{1-x^{2}}} \right| \, dx. $$
The value we want to calculate is $I(-1)$. With the substitution $t = x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/782964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
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Proof via induction $1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$ (b) Prove that for every integer $n \ge 1$, $$1\cdot3 + 2\cdot4 + 3\cdot5 + \cdots + n(n+2) = \frac{n(n+1)(2n+7)}{6}$$
This is the second part of a two part question. Part (a) was the following:
Write the sum: $1\cdot3 + 2\cd... | Note that
$$
k(k + 2) = k^2 + 2k = k^2 - k + 3k = 2{k \choose 2} + 3{k \choose 1}
$$
Definition 1: Let $\{x_n\}$ a sequence. $\Delta x_n = x_{n+1} - x_n$.
Definition 2: If $\Delta X_n = x_n$, then $X_n = \Delta^{-1}x_n + C$ with $C \in \mathbb{R}$.
Proposition 1: $\Delta^{-1}$ is linear and
$$
\Delta \biggl(\sum_{k =n_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Finding the image of a disk Find the image of the disk $D= \{ z \in \mathbb{C}: |z|<1 \}$ under the mapping $z \to w=\frac{1}{z-1}$.
My attempt:
First I graphed the boundary.
When $z=i, f(z)=-i/2-1/2$
When $z=-i, f(z)=i/2-1/2$
When $z=-1, f(z)=-1/2$
When $z=1, f(z)$ is undefined.
Then I graphed a few other points, $z=... | Say, $z=x+iy$, then $f(z)=f(x+iy)$ becomes $$\frac{x-1}{(x-1)^2+y^2}-\frac{i y}{(x-1)^2+y^2}$$
Note that in $D$, $x$ and $y$ are between $0$ and $1$.
Now considering the real part we have $(x-1)^2+y^2 = x^2 +1 -2x +y^2 < 2-2x$. It follows $$\frac{1}{x^2 +1 -2x +y^2} > \frac{1}{2-2x}$$ which implies $$\frac{x-1}{x^2 +1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How did my book simplify this? How did my book go from:
$\frac{4}{5}=\frac{x}{30}$
to
$\frac{4}{1}=\frac{x}{6}$
I understand that I could have cross multiplied it in the first place but what I don't understand is why my book changes the denominators without changing the numerators when simplifying the problem.
| You can rewrite 30 as $5\times 6$:
$$\frac{4}{5}=\frac{x}{5\times 6}\;\; \text{ multiply both sides by 5 } \;\;5\times \frac{4}{5}=5\times \frac{x}{5 \times 6}$$
if you then do some rearrangement you can get:
$$ \frac{5}{5}\times \frac{4}{1}=\frac{5}{5}\times \frac{x}{6}$$
Then replacing $5/5$ with 1 gives your result:... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solve $\sin(2\theta) -\tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi $ I want to use the fact that $$\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}$$
to solve $\sin(2\theta) -tan(\theta) = 0 \ $ for $ 0\leq \theta \leq 2\pi$
My solution:
$\frac{2tan(\theta)}{1 + tan^2(\theta)} - tan(\theta) = 0 $
so $\fra... | You're missing a solution
$$ \sin(2x) = 2 \cos x \sin x = \tan x = \frac{\sin x}{\cos x} \implies \cos^2x = \frac{1}{2} $$
$$\therefore \cos x = \pm \frac{1}{\sqrt{2}} \quad or \quad \sin x = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/789707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
calculating the sum of a series I was wondering if there is a way to use complex-analysis to solve the sum of the following series? (just like you can use it to solve integrals of some kinds (using fourier's transform for integrals).
$$\sum^{\infty}_{n=0}\frac{1}{n^2+25}$$
and
$$\sum^{\infty}_{n=0}\frac{(-1)^n}{n^2+... | Solution via Fourier's series:
Note that (see here)
$$
\int \cos {nx} ~ \cosh {bx} dx = \dfrac{1}{n^2+b^2}
\left(n \sin nx \cosh bx + b \cos nx \sinh bx \right) + C
$$
(form is very close to given $\pm\dfrac{1}{n^2+25}$).
Now consider function
$$
f(x)=\cosh {5x}, \qquad x\in [-\pi,\pi];
$$
and make it $2\pi$-periodic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to find the sum $\sum\limits_{n=1}^{\infty}\frac{1}{n!(n^4+n^2+1)}$ Find this sum
$$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}$$
My try: since
$$n^4+n^2+1=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$
then
$$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}=\sum_{n=1}^{\infty}\dfrac{1}{2n\cdot n!}\left(\dfrac{1}{n(n-1)+1}-\dfrac{1... | Let $S$ denote the sum to be evaluated. For every $n$,
$$
\frac2{n^4+n^2+1}=\frac{n+1}{n(n+1)+1}-\frac{n-1}{(n-1)n+1},
$$
hence
$$
2S=\sum_{n\geqslant1}\frac{n+1}{n(n+1)+1}\frac1{n!}-\sum_{k\geqslant1}\frac{k-1}{(k-1)k+1}\frac1{k!}.
$$
The RHS is almost a telescoping series. To wit, the change of variable $n=k-1$ in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/794272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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Integral $\int_0^1 \log \frac{1+ax}{1-ax}\frac{dx}{x\sqrt{1-x^2}}=\pi\arcsin a$ Hi I am trying to solve this integral $$
I:=\int_0^1 \log\left(\frac{1+ax}{1-ax}\right)\,\frac{{\rm d}x}{x\sqrt{1-x^2}}=\pi\arcsin\left(a\right),\qquad
\left\vert a\right\vert \leq 1.
$$
It gives beautiful result for $a = 1$
$$
\int_0^1 \lo... | It's easier to take the derivative of both sides according to $a$, than to perform the integration:
$$\int_0^1 \frac{2}{1-a^2x^2}\frac{dx}{\sqrt{1-x^2}}=\frac{\pi}{\sqrt{1-a^2}}$$
The left hand side giving:
$$2 \text{arctanh}\left(\frac{\sqrt{a^2-1}}{\sqrt{1 - x^2}}x\right)\frac{1}{\sqrt{a^2-1}}+C$$
Which, when applyin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/795493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 1
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How should I solve integrals of this type? The general form of the integral I want to solve is:
$$ \int e^{bx}\sin(ax) dx$$
Euler's formula has a nice connection, but the i makes it too complicated.
Doing it by parts doesn't seem to get me anywhere.
Do you have any tips for how to begin solving this?
| A third (or maybe fourth, depending on how you're counting) method (related to the one in David's answer) which uses a bit of linear algebra is as follows:
\begin{align*}
\left[\begin{array}{c}
e^{ax}\cos(bx)\\
e^{ax}\sin(bx)
\end{array}\right]' &= \left[
\begin{array}{c}
ae^{ax}\cos(bx) - be^{ax}\sin(bx)\\
ae^{ax}\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/796730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
Integral $\int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx=-\frac{\pi}{4\alpha}\text{arcsinh}\alpha$ Hi I am trying to prove this
$$
I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\,
\frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x
=-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\lef... | Integrate by parts; then you get that
$$I(\alpha) = \left [\frac1{2 \alpha} \arctan{(\alpha \sin{2 x})} \log{(\tan{x})} \right ]_0^{\pi/4} - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\alpha \sin{2 x}}$$
The first term on the RHS is zero. To evaluate the integral, expand the arctan into a Taylor series and g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/798227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
Please solve this recurrence relation question for $8a_na_{n+1}-16a_{n+1}+2a_n+5=0$ Suppose $a_1=1$ and $$8a_na_{n+1}-16a_{n+1}+2a_n+5=0,\forall n\geq1,$$Please help to sort out the general form of $a_n$.
Here are the first a few values of the series. Not sure if they are useful as they seem quite random to me.
$$1,{7\... | Let $p_n$, $q_n$ be two sequences to be determined such that $\displaystyle\;a_n = \frac{p_n}{q_n}\;$. In terms of $p_n$, $q_n$, the recurrence relation for $a_n$ can be rewritten as
$$a_{n+1} = \frac{2a_n+5}{-8a_n+16}
\quad\iff\quad\frac{p_{n+1}}{q_{n+1}} = \frac{2p_n+5q_n}{-8p_n+16q_n}\tag{*1}
$$
If we scale $p_n, q... | {
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"url": "https://math.stackexchange.com/questions/799006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Find all real solutions of $6^x+1=8^x-27^{x-1}$
Find all real solutions of $6^x+1=8^x-27^{x-1}$.
Things I tried:
We want solutions of
$$2^x3^x+1 = (2^x)^3-\frac{(3^x)^3}{27}.$$ Write $a=2^x$ and $b=3^x$. This gives
$$ab+1 = a^3-\frac{b^3}{27}$$
or
$$27ab+27=27a^3-b^3$$
How to continue?
| As you did, let $a=2^x, b=3^x$. We have $$ab+1=a^3-\frac{b^3}{27}$$
Note the standard identity $$u^3+v^3+w^3-3uvw=(u+v+w)\left(\frac{(u-v)^2+(v-w)^2+(w-u)^2}{2}\right)$$
Take $u=a, v=-\frac{b}{3}, w=-1$. Then $u^3+v^3+w^3-3uvw=a^3-\frac{b^3}{27}-1-ab=0$.
Thus $u+v+w=0$ or $(u-v)^2+(v-w)^2+(w-u)^2=0$. In the latter case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find the last column of a matrix. Find the matrix. $A\left[\begin{matrix} 1 & 0 \\ 0 & 0 \\ 1 & 1 \end{matrix} \right]$ = $\left[\begin{matrix} 2 & 3 \\ -1 & 0 \\ 5 & -7 \\ 0 & 6 \end{matrix} \right]$
(1)Find the last column of $A$.
(2)Find the matrix $A$.
I figured that the size of $A$ has to be $4 \times 3$. There... | Multiplying $A$ by $\begin{bmatrix} 1 \\ 0 \\ 1\end{bmatrix}$ on the right sums the first column of $A$ with its last column.
Multiplying $A$ by $\begin{bmatrix} 0 \\ 0 \\ 1\end{bmatrix}$ on the right maintains the last column.
Using the last column and the first line of this answer you're able to obtain the first colu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Cylindrical coordinates: $\iiint (x^2 + y^2 + z^2) dxdydz$ Show that
$$ I= \iiint_S (x^2 + y^2 + z^2) dxdydz = \frac{2^{10} a^5 k}{75} \left(1 + \frac{k^2}{3} \right), a>0, k>0$$
where $S$ is the region bounded by the cilinder $x^2 + y^2 = 2ax$ and the upper and lower sections of the cone $z^2 = k^2(x^2 + y^2)$.
My at... | Using Cylinderical coordinates, Mathematica evaluates the following integral as
$$\int_{-\pi/2}^{\pi/2} \int_0^{2 a \cos \theta } \int_{-kr}^{kr} (r^2 + z^2) r dz dr d\theta = \frac{1024}{225} a^5 k \left(k^2+3\right)$$
I suppose you made error in the bounds of $r$ since center of cylinder is $(a, 0)$ and polar equatio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/801587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Probability Question: Drawing balls from an urn Question:
An urn contains 12 balls where 3 balls colored red, 3 balls colored blue, 3 balls colored green, and 3 balls colored white. If we draw 4 balls in succession without replacement, what are the probabilities that
*
*the first is not red, the second is not blue,... | For the second problem, we first find the probability of the complement, that the first is red or the second is blue or the third is green or the fourth is white.
We use Inclusion/Exclusion.
The probability the first is red is $\frac{3}{12}$, with similar expressions for the others. If we add these $\frac{3}{12}$, we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $2^{15} - 2^3 $ divides $ a^{15} - a^3$ Prove that $$2^{15} - 2^3 $$ divides $$ a^{15} - a^3$$ for any integer $a$.
Hint: $$ 2^{15} - 2^3 = 5\cdot7\cdot8\cdot9\cdot13$$
| We can factor:
$$a^{15}-a^3=a^3(a-1)(a+1)(a^2+1)(a^2+a+1)(a^2-a+1)(a^4-a^2+1)$$
Now, we just need to prove that this is divisible by $2^3\cdot3^2\cdot5\cdot7\cdot13$.
If $a$ is odd, $(a-1)(a+1)(a^2+1)$ is the product of three even numbers, and is divisible by 8. If $a$ is even, $a^3$ is divisible by 8.
If $a$ is a mul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/803097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 5
} |
Show that $2 < e^{1/(n+1)} + e^{-1/n}$ I'm trying to show $2 < e^{1/(n+1)} + e^{-1/n}$.
I can show that $ 2 < e^{1/n} + e^{-1/(n+1)}$ since
$$2 \leq 2\cosh\left(\frac{1}{n}\right) = e^{1/n} + e^{-1/n} < e^{1/n} + e^{-1/(n+1)}$$
but I'm still having trouble with the other inequality. I though using $\cosh$ again might ... | well, since
$$ \left(\frac1{4! n^4}-\frac1{5!n^5}\right)+\cdots+\left(\frac1{(4+2k)!n^{4+2k}} -\frac1{(4+2k+1)!n^{4+2k+1}}\right)>0 $$
and
$$ \left(\frac1{4! n^4}-\frac1{5!n^5}\right)+\cdots+\left(\frac1{(4+2k)!n^{4+k}} -\frac1{(4+2k+1)!n^{4+2k}}\right) +\frac1{(4+2k+2)!n^{4+2k+2}} >0 $$
then,
$$\sum_{k\geq4}\frac{(-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Laurent series for $\frac1{z^2+1}$ I have this problem:
Find the Laurent series around $z=0$, for $\dfrac{10}{(z+2)(z^2+1)}$ in the region $1<|z|<2$.
I did partial fractions and found this: $\dfrac{2}{z+2}-\dfrac{2z-4}{z^2+1}$,
then I have to know what's the Laurent series for $\dfrac{1}{z^2+1}$ to solve the problem.
D... | Realize the last series you ask about as a geometric series...
$$
\frac{1}{z^2+1} = \frac{1}{z^2}\;\frac{1}{1-(-1/z^2)} =
\frac{1}{z^2} - \frac{1}{z^4} +\frac{1}{z^6} - \frac{1}{z^8}+\dots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Partial Fractions Integration Question $$\int\frac{x^5+x-1}{x^3 +1} dx$$
Have tried everything ... polynomial long division, partial fractions, trig substitution etc...
Not for an assignment, so if a complete solution could be provided that'd be much appreaciated
| First divide the denominator into the numerator to get
$$\frac{x^5+x-1}{x^3+1}=x^2 -\frac{x^2-x+1}{x^3+1}$$
Now we factor the denominator, $$x^3+1=(x+1)(x^2-x+1)$$
Note that we have cancellation so we get
$$\int x^2-\frac{1}{x+1} dx=\frac{x^3}{3}-\ln |x+1|+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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calculate the Integral $ \int \ln\left(1+2a\cos x+a^2\right)dx\;,$ where $a\in \mathbb{R}.$ How can we calculate the Integral $\displaystyle \int \ln\left(1+2a\cos x+a^2\right)dx\;,$ where $a\in \mathbb{R}.$
$\bf{My\; Try::}$ Using $\displaystyle \cos (x) = \frac{e^{ix}+e^{-ix}}{2},$ we get ...
\begin{align}\displaysty... | The integral
\begin{align}
I_{+} = \int \ln(1+a e^{ix}) \ dx
\end{align}
can be evaluated by making the substitution $t = e^{ix}$ which leads to
\begin{align}
I_{+} = - i \int \frac{\ln(1+at)}{t} \ dt = \left. i Li_{2}(-a t) \right|_{t = e^{ix}} = i \ Li_{2}(-a e^{ix}),
\end{align}
where $Li_{2}(x)$ is the dilogarith... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Simplifying fraction with square root as denominator I'm trying to find the integral of:
$$\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}$$
but I first need to simplify it so I tried dividing by the $\sqrt{x}$ for each of the numbers on the top like so:
$$\dfrac {2\sqrt{x}}{\sqrt{x}}$$
and did the same for the others. For the... | HINT :
Rewrite
$$
\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}=\dfrac {2\sqrt{x} - 3\sqrt{x}\sqrt{x} + x\cdot x}{\sqrt{x}}=\dfrac {2\sqrt{x} - 3\sqrt{x}\sqrt{x} + \sqrt{x}\sqrt{x}\cdot x}{\sqrt{x}}
$$
or
$$
\dfrac {2\sqrt{x} - 3x + x^2}{\sqrt{x}}=\dfrac {2x^{\large\frac12} - 3x^1 + x^2}{x^{\large\frac12}},
$$
where $x=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/812315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Congruency by completing the square $x^2+x+1\equiv 0\mod 49$
We have the ring isomorphism $\mathbb{Z}/49\mathbb{Z}\to\mathbb{Z}/7\mathbb{Z}\times\mathbb{Z}/7\mathbb{Z}$.
Consider $x^2+x+1\equiv 0\mod 7$
I usually solve these polynomial congruencies using the 'complete the square' method. How would I go about doing that... | A start: Equivalently, we solve $4x^2+4x+4\equiv 0\pmod{49}$, that is, $(2x+1)^2+3\equiv 0\pmod{49}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find $\lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$ and $\lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$ How can I determine these limits:
$$ a) \lim_{n\rightarrow 0} \cos(\frac{\pi}{n} \sin n \cos n)$$
$$ b) \lim_{n\rightarrow 0} \frac{n}{\sin(2n) - \cos(\frac{n}{2}) +1}$$
Note I ca... | For (a), you can use a very similar approach. Since $\cos(x)$ is continuous on $\mathbb{R}$, you can pull the limit inside, and you can rewrite the limit as:
\begin{align*}
\lim_{n\to 0} \cos\left(\frac{\pi}{n}\sin(n)\cos(n)\right) &= \lim_{n\to 0} \cos\left(\frac{\sin(n)}{n}\pi\cos(n) \right)\\
&= \cos\left(\lim_{n\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Math Olympiad Algebraic Question Comprising Square Roots
If $m$ and $n$ are positive real numbers satisfying the equation $$m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n=3$$
find the value of $$\frac{\sqrt{m}+2\sqrt{n}+2014}{4-\sqrt{m}-2\sqrt{n}}$$
I came across this question in a Math Olympiad Competition and had no idea h... | We have
$$
\begin{align*}
m+4\sqrt{mn}-2\sqrt{m}-4\sqrt{n}+4n&=3\\
m+4\sqrt{mn}+4n-2\sqrt{m}-4\sqrt{n}-3&=0\\
(\sqrt{m}+2\sqrt{n})^2-2(\sqrt{m}+2\sqrt{n})-3&=0\\
(\sqrt{m}+2\sqrt{n}-3)(\sqrt{m}+2\sqrt{n}+1)&=0
\end{align*}
$$
which gives
$$
\sqrt{m}+2\sqrt{n}=3\,\,\text{or}\,\,\sqrt{m}+2\sqrt{n}=-1.
$$
We can disregard... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Find the limit of $ \lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2} $ I need to evaluate the limit without using L'Hopital's rule.
$$\Large
\lim_{x \to 7} \frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}
$$
| Because $(a-b)(a+b)(a^2+b^2)=a^4-b^4$, and $\lim_{x\to 7}a=\lim_{x\to 7}b=2$, we rewrite as
$$\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{\sqrt[4]{x+9}-2}=\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{(x+9)-16}(2+2)(2^2+2^2)$$
$$=32\lim_{x\to 7}\frac{\sqrt{x+2}-\sqrt[3]{x+20}}{x-7}$$
Because $(a-b)(a+b)(a^2+ab+b^2)(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Limit $\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}$ $$\lim_{x\rightarrow\infty}1-x+\sqrt{2+2x+x^2}=\lim_{x\rightarrow\infty}1-x+x\sqrt{\frac{2}{x^2}+\frac 2x+1}=\lim_{x\rightarrow\infty}1=1\neq2$$ as Wolfram Alpha state. Where I miss something?
| You should note that you get
$$\sqrt{x^2 + 2x + 2} = \sqrt{(x+1)^2 + 1} \approx (x+1)$$
as you are squaring i.e. $(x+1)^2$ is not approximately equal to $x^2$ since $(x+1)^2 - x^2 \approx 2x$ you can only ignore the constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Evaluating e using limits What algebraic operations can I use on the $RHS$ to show $RHS = LHS$
$$e=\lim_{k\to\infty}\left(\frac{2+\sqrt{3+9k^2}}{3k-1}\right)^k$$
| Since$$\left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)^k=e^{k \log \left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)} $$
it's sufficient to show that
$$\lim_{k \to \infty} k \log \left( \frac{2+\sqrt{3+9k^2}}{3k-1} \right)=1 .$$
We can rewrite this expression as
$$\frac{\log \left(\frac{2+\sqrt{3+9k^2}}{3k-1} \right)}{\frac{1}{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/822469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving by induction: $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ WTS $ \frac{1\cdot3\cdot5\cdot \ldots \cdot (2n-1)}{1\cdot2\cdot3\cdot\ldots\cdot n} \leq 2^n $ for all natural $n$.
Have checked $P_1$, and assumed $P_k$.
Trying the following argument:
$P_{k+1} = \frac{... | All you need is to prove that: $\dfrac{2k+1}{k+1} \leq 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$ Let $n$ be a positive whole number. Given $2^7 \equiv 2 \mod n$ and $3^7 \equiv 3 \mod n$, prove for all $a$ in $\mathbb{Z}$: $a^7 \equiv a \mod n$, without using a computer.
$n$ must be greater than 3 and l... | Using Fermat's Little Theorem,
$$a^7\equiv a\pmod7$$
$$a^7-a=a(a^6-1)=a(a-1)\{a^5+a^4+\cdots+a+1\}$$ where by Little Theorem, $\displaystyle2|a(a-1)$
$$a^7-a=a(a^6-1)=a(a^2-1)(a^4+a^2+1)=(a^3-a)(a^4+a^2+1)$$ where by Little Theorem, $\displaystyle3|(a^3-a)$
So, $\displaystyle a^7-a$ will be divisible by lcm$(2,3,7)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/822796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Integration by parts - $\int \ln (2x+1) \text{dx}$ Use integration by parts to find
$\int \ln (2x+1) \text{dx}$.
So far I have:
$$x\ln(2x+1)-\int\dfrac{2x}{2x+1}dx+c$$
Using integration by substitution to find the integral
$$u=2x+1\Rightarrow\text{du}=2\text{dx}$$
$$\int\dfrac{2x}{2x+1}\cdot\dfrac{1}{2}\text{du}=\int x... | Your issue is in what follows this step:
\begin{align*}
x\ln(2x+1) - \int \frac{2x}{2x+1}\,\text{dx} &= x\ln(2x+1) - \int \frac{2x+1-1}{2x+1}\,\text{dx}\\
&= x\ln(2x+1) - \left(\int \frac{2x+1}{2x+1}\,\text{d}x - \int \frac{1}{2x+1}\,\text{d}x\right)\\
&= x\ln(2x+1) - \left(\int \,\text{d}x - \int \frac{1}{2x+1}\,\text... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Find a closed form for $\sum_{k=0}^{n} k^3$ Find a closed form for $\sum_{k=0}^{n} k^3$.
I would appreciate ideas for approaching questions like this in general as well.
Thanks.
| Notice $$
\begin{align}
k^3
&= \color{red}{(k^3 - k)} + \color{blue}{k}\\
&= \color{red}{(k-1)k(k+1)} + \color{blue}{k}\\
&= \frac{\color{red}{(k-1)k(k+1)}(k+2)-(k-2)\color{red}{(k-1)k(k+1)}}{4}
+ \frac{\color{blue}{k}(k+1) - (k-1)\color{blue}{k}}{2}
\end{align}$$
is telescoping. We have
$$\begin{align}
\sum_{k=0}^n k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/823897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Help w/ Partial Fraction Decomposition I need some help figuring out how to decompose $\displaystyle\frac{1}{x^4+1}$ into partial fractions.
This is what I have done so far: $$\frac{1}{x^4+1} = \frac{1}{(x^2 - \sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}$$
From there, I do not know how to decompose it into partial fractions, o... | I'm guessing that this problem is homework so I'll provide a step forward from where you're stuck to see if it'll help you.
Currently, you have this factored into two distinct irreducible quadratics, each of which gets a term in the form $\frac{Ax+B}{ax^2+bx+c}$. Therefore, you should set up the equation to be:
$$\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How is $2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$ calculated? $$2\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}=\frac{6}{4}$$
I cant figure out why this is $\frac64$. I try to use telescopic series without success.
| Ok well first of all $\frac{6}{4}=\frac{3}{2}$.
The series telescopes as has been pointed out, using that formula we have
$$\left(\frac{1}{1}-\frac{1}{3}\right)+
\left(\frac{1}{2}-\frac{1}{4}\right)+
\left(\frac{1}{3}-\frac{1}{5}\right)+
\left(\frac{1}{4}-\frac{1}{6}\right)+\cdots =1+\frac{1}{2}=\frac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/829220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Value of the given limit I need to calculate the value of :
$$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{2n}{\frac{r}{\sqrt{n^2+r^2}}}$$
I had been trying to use Cesàro summation but somehow, I might be messing up.
The options are : $$\sqrt{5}+1,\sqrt{5}-1,\sqrt{2}-1,\sqrt{2}+1$$
| Hint: As a Riemann Sum:
$$
\lim_{n\to\infty}\sum_{r=1}^{2n}\frac{r/n}{\sqrt{1+r^2/n^2}}\frac1n=\int_0^2\frac{x}{\sqrt{1+x^2}}\mathrm{d}x\tag{1}
$$
We can also use the fact that
$$
\sqrt{n^2+r^2}-\sqrt{n^2+(r-1)^2}=\frac{2r-1}{\sqrt{n^2+r^2}+\sqrt{n^2+(r-1)^2}}\tag{2}
$$
and
$$
\begin{align}
&\left|\,\frac{2r-1}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find a generating function for $\sum_{k=0}^{n} k^2$
Find a generating function for $\sum_{k=0}^{n} k^2$
I know my solution is wrong, but why?
My solution:
If $F(x)$ generates $\sum_{k=0}^{n} k^2$ then $F(x)(1-x)$ generates $k^2$.
$\frac{x}{(1-x)^4}: \left\{ 0,1,4,9,16,25... \right\}$
$\frac{x}{(1-x)^3}: \left\{ ... | The differences of $\{0,1,2,2,2,2,\dots\}$ are $\{0,1,1,0,0,\dotsc\}$, not $\{0,1,1,1,1,\dotsc\}$.
So you should get
$$\begin{align}
x+x^2 &: \{0,1,1,0,0,\dotsc\}\\
\frac{x+x^2}{1-x} &: \{ 0,1,2,2,2,\dotsc\}\\
\frac{x+x^2}{(1-x)^2} &: \{ 0,1,3,5,7,\dotsc\}\\
\frac{x(1+x)}{(1-x)^3} &: \{ 0,1,4,9,\dotsc\}
\end{align}$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/832858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that limit $\lim_{n\to\infty}\sqrt{4+\frac{1}{n^2}}+\sqrt{4+\frac{2}{n^2}}+\cdots+ \sqrt{4+\frac{n}{n^2}}-2n=\frac{1}{8}$
Let $$a_{n}=\sqrt{4+\dfrac{1}{n^2}}+\sqrt{4+\dfrac{2}{n^2}}+\cdots+
\sqrt{4+\dfrac{n}{n^2}}-2n,$$
show that
$$\lim_{n\to\infty}a_{n}=\dfrac{1}{8}$$
My attempt:
Since
$$\sqrt{4+\dfrac{i}... |
My solution: Note
$$\sqrt{4+\dfrac{k}{n^2}}<\sqrt{2^2+\dfrac{k}{n^2}+(\dfrac{k}{4n^2})^2}=2+\dfrac{k}{4n^2}$$
$$\sqrt{4+\dfrac{k}{n^2}}=\sqrt{4+\dfrac{k-1}{n^2}+\dfrac{1}{n^2}}>\sqrt{2^2+\dfrac{k-1}{n^2}
+(\dfrac{k-1}{4n^2})^2}=2+\dfrac{k-1}{4n^2}$$
and
$$\lim_{n\to\infty}\dfrac{k-1}{4n^2}=\lim_{n\to\infty}\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/833824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
Limit by diagonalisation I want to compute the limit of the sequence $x_0=a$, $x_1=b$ and $x_n=\frac{x_{n-1}+x_{n-2}}{2}$.
I did it by setting:
$\begin{pmatrix}
x_{n+1}\\ x_{n}
\end{pmatrix}=$
$\begin{pmatrix}
\frac{1}{2} &\frac{1}{2}\\
1 &0\\
\end{pmatrix}$
$\begin{pmatrix}
x_{n}\\ x_{n-1}
\end{pmatrix}$
and calculat... | $x_2 = \dfrac{a + b}{2}$
$x_3 = \dfrac{a+3b}{4}$
It is already not symmetric in $a$ and $b$. There is no error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/835317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Consider convergence of series: $\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$ Consider convergence of series:
$$\sum_{n=1}^{\infty}\sin\left[\pi\left(2+\sqrt{3}\right)^n\right]$$
My tried:
We have
$$\sum_{n=1}^{\infty }\sin(\pi (2+\sqrt{3})^{n})=\sum_{n=1}^{\infty}\sin\left(\pi[(2+\sqrt{3})^{n}+(2-... | A simple way to show convergence is to note that $$(2+\sqrt{3})^2=7+4\sqrt{3}=2*7-(2- \sqrt{3})^2$$
$$(2+\sqrt{3})^3=26+15\sqrt{3}=2*26-(2-\sqrt{3})^3$$
$$(2+\sqrt{3})^4=97+56\sqrt{3}=2*97-(2-\sqrt{3})^4$$
and in general
$$(2+\sqrt{3})^n=2k-(2-\sqrt{3})^n$$
with k integer.
The summation then reduces to $$-\sum_{n=1}^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/835554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Easier Proof of $\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$ I am curious to see whether anybody can give me a proof that takes less steps.
Here is how I did it:
$$\sin{3\theta} + \sin\theta = 2\sin{2\theta}\cos\theta$$
LHS $$\eqalign{\sin(2\theta + \theta) + \sin\theta &= \sin2\theta\cos\theta + \cos2\theta... | Use the sum of angles identities for the $\sin$ function:-
$$\sin(3\theta)=\sin(2\theta+\theta)=\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta)$$
$$\sin(\theta)=\sin(2\theta-\theta)=\sin(2\theta)\cos(\theta)-\cos(2\theta)\sin(\theta)$$
Adding both equations results in
$$\sin(3\theta)+\sin(\theta)=2\sin(2\theta)\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/837430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
If $T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$. Determine $||T||$. Define $T : \mathbb{R}^2 → \mathbb{R}$ ( $\mathbb{R}^2$ & $\mathbb{R}$ being
equipped with the Euclidean norm) by
$T ( x,y ) = 2x + y , ∀( x,y )∈ \mathbb{R}^2$.
Determine $||T||$.
My thoughts:-
We know that
$ \qquad \left\|{T}\right\| = \sup \left\{... | Yes, you are right. To do so, note that
$$ T'(x,y) = \begin{pmatrix} 2 & 1 \end{pmatrix} $$
and for $F(x,y) = x^2 + y^2 - 1$ we have
$$ F'(x,y) = \begin{pmatrix} 2x & 2y \end{pmatrix} $$
We hence have to solve
\begin{align*}
2 &= 2x \lambda\\
1 &= 2y\lambda\\
x^2 + y^2 &= 1
\end{align*}
So $x = 1/\lambda$, $y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/840418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Given a sequence $ (a_{n})_{n \in \mathbb{N}} $, find $ \sum_{n = 1}^{\infty} \frac{1}{a_{n} + 2} $. I would appreciate it if somebody could help me with the following problem:
If the sequence $ (a_{n})_{n \in \mathbb{N}} $ satisfies
$$
a_{1} = 1
\quad \text{and} \quad
\forall n \in \mathbb{N}, n \ge 2: \quad
a_{n} ... | For $n \ge 2$, we have: $a_{n+1} = \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)(a_n+2)}{2^{n+1}}$ $= \dfrac{(a_1+2)(a_2+2)\cdots(a_{n-1}+2)}{2^n}\cdot \dfrac{a_n+2}{2} = \dfrac{a_n(a_n+2)}{2} = \dfrac{1}{2}a_n^2+a_n$.
Thus, $\dfrac{1}{a_n} - \dfrac{1}{a_{n+1}} = \dfrac{a_{n+1}-a_n}{a_na_{n+1}} = \dfrac{\tfrac{1}{2}a_n^2}{a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trigonometric Triangle Equality $A, B, C$ are the angles of a triangle then $tan^2(A/2)+tan^2(B/2)+tan^2(C/2)$ is
always greater than what integral value.
| Here $A,B,C$ are the angles of a $\triangle.$ So Here $$\displaystyle 0<\frac{A}{2},\frac{B}{2},\frac{C}{2}<\frac{\pi}{2}$$.
Now Using $\bf{A.M\geq G.M}\;,$ We Get
So we get $$\displaystyle \frac{\tan^2 \left(\frac{A}{2}\right)+\tan^2 \left(\frac{B}{2}\right)}{2}\geq \sqrt{\tan^2 \left(\frac{A}{2}\right)\cdot \tan^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/842881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Inequality for harmonic means Prove that for real numbers $a_1 ,a_2 ,...,a_n >0$ the following inequality holds
$$\frac{1}{a_1 } +\frac{2}{a_1 +a_2 } +...+\frac{n}{a_1 +a_2 +...+a_n }\leq 4\cdot \left(\frac{1}{a_1} +\frac{1}{a_2 } +...+\frac{1}{a_n} \right).$$
| Upper Bound
By Cauchy-Schwarz, we have
$$
\begin{align}
\left(\sum_{j=1}^ka_j\right)\left(\sum_{j=1}^k\frac{j^2}{a_j}\right)
&\ge\left(\sum_{j=1}^kj\right)^2\\[3pt]
&=\frac{k^2(k+1)^2}{4}\tag{1}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{k=1}^n\frac{k}{\sum\limits_{j=1}^ka_j}
&\le\sum_{k=1}^n\frac4{k(k+1)^2}\sum_{j=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/844995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
limit $f(x)=((x-7)(x+4))^{1/2}$ I have to calculate the Asymptotes in the infinity(and minus infinity) for this function:
$f(x)=((x-7)(x+4))^{1/2}$
I know that
$\lim_{x \to\infty} f(x)/x= 1$
And I get into trouble with:
$\lim_{x \to\infty} f(x)-x$
which is
$\lim_{x \to\infty} ((x-7)(x+4))^{1/2}-x = \lim_{x \to\infty} \... | $$
\sqrt{x^2-3x-28} - x = \frac{(\sqrt{x^2-3x-28} - x)(\sqrt{x^2-3x-28} + x)}{(\sqrt{x^2-3x-28} + x)}\\
= \frac{x^2-3x-28-x^2}{(\sqrt{x^2-3x-28} + x)}
= \frac{-3x-28}{x(\sqrt{1-3/x-28/x^2} + 1)} \to \frac {-3}{2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/848317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Apostol (6.22.10): Finding $\int \frac{\arcsin x}{x^2} dx$ I'm trying to solve another integral from Apostol (Chapter 6, Section 6.22, Question 10) which says to show the following:
$$
\int \frac {\arcsin x}{x^2}dx = \log|{\frac {1-\sqrt{1-x^2}}{x}}| - \frac {\arcsin x}{x} +C
$$
I have tried the following:
With $\int \... | A much more computationally straightforward approach follows. The integrand suggests the substitution $$\theta = \sin^{-1} x, \quad x = \sin \theta, \quad dx = \cos \theta \, d\theta,$$ giving $$I = \int \frac{\sin^{-1} x}{x^2} \, dx = \int \theta \csc \theta \cot \theta \, d\theta.$$ Then integration by parts with t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/849163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Jordan Canonical Form of $A^{-1}$
Let $A$ be a $5 \times 5$ invertible matrix over $\mathbb{R}$. Suppose
$c_A(x)=(x+1)^3(x+2)^2$ and $m_A(x)=(x+1)(x+2)^2$, what is a Jordan
Canonical Form of $A^{-1}$?
Is it possible to obtain a JCF of $A^{-1}$ from that of $A$, or we have to explicitly solve for $c_{A^{-1}}(x)$ a... | Its easy to see that
$$A=\begin{pmatrix}
-2&1&0&0&0\\
0&-2&0&0&0\\
0&0&-1&0&0\\
0&0&0&-1&0\\
0&0&0&0&-1\\
\end{pmatrix}$$
so we see
$$A^{-1}=\begin{pmatrix}
-\frac{1}{2}&-\frac{1}{4}&0&0&0\\
0&-\frac{1}{2}&0&0&0\\
0&0&-1&0&0\\
0&0&0&-1&0\\
0&0&0&0&-1\\
\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/850395",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$ \Big(\dfrac{x^7+y^7+z^7}{7}\Big)^2=\Big(\dfrac{x^5+y^5+z^5}{5}\Big)^2\cdot\Big(\dfrac{x^4+y^4+z^4}{2}\Big) $ I have a question. I tried so to solve it, but there is a problem.
that is i don't have any idea to findout how can i work with degrees 4,5,7 ...
this is the problem :
let $ x , y $ and $ z $ three real number... | Let $x,y,z$ be the roots of $t^3-Qt - P=0$, $Q= -(xy+yz+zx), P = xyz$.
$0 = (x+y+z)^2 = -2Q + x^2+y^2+z^2 \implies x^2+y^2+z^2 = 2Q$.
You have $t^3 = P + Qt$ and when you replace with $x,y,z$ and sum:
$$
x^3+y^3+z^3 = 3P + Q(x+y+z) = 3P
$$
You also get the identities:
$$
x^4+y^4+z^4 = Q(x^2+y^2+z^2) = 2Q^2
\\
x^{n+3}+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/851985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
} |
How to get the Normal line? My book proposed to me to find the the Normal lines to the curve that pass through the origin.The answer must be the intersection points between them.
The curve: $\dfrac{2}{1+x^2}$
My first idea is to get the derivative: $f'(x) = $ $\frac{-4x}{x^4 + 2x^2 + 1}$, then get the perpendicular of ... | If $\displaystyle x=\tan\theta, y=\frac2{1+x^2}=2\cos^2\theta$
So, $\displaystyle\frac{dy}{dx}=-\frac{4x}{(1+x^2)^2}=-4\tan\theta\cos^4\theta=-4\sin\theta\cos^3\theta$
So, the equation of the normal at $(\tan\theta,2\cos^2\theta)$
$$\frac{y-2\cos^2\theta}{x-\tan\theta}=-\frac1{-4\sin\theta\cos^3\theta}$$
$$\implies(y-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/852490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Max. and Min. value of $|z|$ in $\left|z+\frac{2}{z}\right| = 2\,$ If $z$ is a complex no. such that $\displaystyle \left|z+\frac{2}{z}\right| = 2\,$ Then find max. and min. value of $\left|z\right|$.
$\bf{My\; Try:}$ Given $\displaystyle \left|z+\frac{2}{z}\right| = 2\Rightarrow \left|z+\frac{2}{z}\right|^2 = 2^2=4$.... | Let $z=\color{red}{\sqrt{2}}re^{i\theta}$, then $\left|z+\frac{2}{z}\right|=2$ reduces to $$r^2+\frac{1}{r^2}=2-2\cos{2\theta} \tag{1}$$
Observe that $(1)$(considering it as a quadratic in $r^2$) will have two real roots of the form $\alpha$ and $\frac{1}{\alpha}$ only if $\cos{2\theta} \le 0$. Thus if the maximum valu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/853992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$
How do I find the sum of this series?
$$\sum_{n=0}^{\infty}{\sin^{3}\left(3^{n}\right) \over 3^{n}}$$
Hints in the right direction would be appreciated.
| From the identity $ \displaystyle \left(\sin(x)\right)^3 = \left(\frac{e^{ix}-e^{-ix}}{2i}\right)^3$ you get $ \displaystyle \left(\sin(x)\right)^3 = \frac{3}{4}\sin(x)-\frac{1}{4}\sin(3x) $
In particular : $ \displaystyle \frac{\sin^3(3^n)}{3^n} = \frac{3}{4}\left(\frac{\sin(3^n)}{3^n}-\frac{\sin(3^{n+1})}{3^{n+1}}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/858821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Show that the following series is less than $4 \pi^2 / 3$. Show: For any $k = 0,1,2,...$,
$$ \sum_{i=0}^{i=k} \frac{(k+1)^2}{(i+1)^2 (k-i+1)^2} \leq \frac{4 \pi^2}{3}. $$
| Hint:
1) From $(\frac{a+b}{2})^2\le\frac{a^2+b^2}{2}$ verify that $(k+1)^2 < 2[(i+1)^2+(k-i+1)^2]$,
2) $\sum_{n\ge 1}\frac{1}{n^2} = \frac{\pi^2}{6}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/858885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find the limit of $\lim\limits_{x \to+\infty} \frac{4x-3}{2x+5}$ and $\lim \limits_{x \to -\infty} \frac{2x^2-x+5}{4x^3-1}$ Hello guys help me with these:
1.) $\lim \limits_{x \to +\infty} frac{4x-3}{2x+5}$
$=\lim \frac{4-\frac{3}{x}}{2+\frac{5}{x}}$
$= \lim \frac{4-3\frac{1}{x}}{2+5\frac{1}{x}}$
=$\frac{\lim 4- \lim ... | When you evaluate a limit approaching either positive or negative infinity, you could easily evaluate it by dividing top and bottom by $x$ which has the highest power.
$$
\lim_{x \rightarrow \infty} \frac{x}{x^2}
$$
Given / $x^2$:
$$
\lim_{x \rightarrow \infty} \frac{\frac{1}{x}}{1} = \lim_{x \rightarrow \infty} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/861236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving a trigonometric limit $\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3}$ First off, please excuse my n00bishness I have only just begun learning about algebraic manipulation of limits so this is probably a really dumb or obvious question.
I'm trying to solve the following limit:
$$
\lim_{x\to\pi/6}\fra... | Another way is to use change of variables. Let $y = x - \pi/6$.
$$
\lim_{x\to\pi/6}\frac{1 - 2\sin{x}}{2\sqrt{3}\cos{x} - 3} = \lim_{y \to 0}\frac{1 - 2\sin{(y + \pi/6)}}{2\sqrt{3}\cos{(y+\pi/6)} - 3} = L
$$
But,
$$
1 - 2\sin(y + \pi/6) = (1 - \cos y) - \sqrt{3}\sin y
$$
and
$$
2\sqrt{3}\cos{(y+\pi/6)} - 3 = 2\sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/865509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
A closed form of the series $\sum_{n=1}^{\infty} \frac{H_n^2-(\gamma + \ln n)^2}{n}$ I have found a closed form of the following new series involving non-linear harmonic numbers.
Proposition. $$\sum_{n=1}^{\infty} \dfrac{H_n^2-(\gamma + \ln n)^2}{n} = \dfrac{5}{3}\zeta(3)-\dfrac{2}{3}\gamma^3-2\gamma \gamma_{1}-\... |
Here is a nice solution from R. Tauraso, for those interested in this problem:
Solution proposed by Roberto Tauraso, Roma, Italy.
The Multiple Harmonic Sum is defined by $$H_{n}(s_{1}, \dots, s_{l}) == \sum_{0 < k_{1} < k_{2} < \dots < k_{l} \leq n}\prod_{i = 1}^{l}\frac{1}{k_{i}^{s_{i}}}$$ Then, by the known propert... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 1,
"answer_id": 0
} |
$12\frac{\sin 45^\circ}{\sin 60^\circ}$ Need help breaking this down. Otherwise known as $12\dfrac{\left(\frac{1}{\sqrt2}\right)}{\left(\frac{\sqrt3}{2}\right)}$
How do you simplify this multi level fractional radical expression into $4\sqrt{6}$.
| Here's yet another approach: $\frac1{\sqrt{2}}$ is often (equivalently) written as $\frac{\sqrt{2}}2$ (an equivalency that can be easily seen by multiplying both numerator and denominator by $\sqrt{2}$). Using this form turns the original formula into $12\cdot\dfrac{\frac{\sqrt{2}}2}{\frac{\sqrt{3}}2}$ $= 12\cdot\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/866442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Another integral for $\pi$ Here is a new integral for $\pi$.
$$\int_{0}^{1}\sqrt{\frac{\left\{1/x\right\}}{1-\left\{1/x\right\}}}\, \frac{\mathrm{d}x}{1-x} = \pi $$
where $\left\{x\right\}$ denotes the fractional part of $x$.
Do you have any proof?
| Let me suggest a general approach.
We start with a theorem.
Theorem. (O. Oloa) Let $f$ be a piecewise continuous function on $(0,1)$ verifying $ \displaystyle \int_{0}^{1} \frac{ \left|f (x) \right| }{x} \: \mathrm{d}x <\infty.$
Then $$ \int_{0}^{1} f \left(\left\{1/x\right\}\right) \frac{ \mathrm{d}x}{1-x} = ... | {
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"url": "https://math.stackexchange.com/questions/870680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
Denesting radicals like $\sqrt[3]{\sqrt[3]{2} - 1}$ The following result discussed by Ramanujan is very famous: $$\sqrt[3]{\sqrt[3]{2} - 1} = \sqrt[3]{\frac{1}{9}} - \sqrt[3]{\frac{2}{9}} + \sqrt[3]{\frac{4}{9}}\tag {1}$$ and can be easily proved by cubing both sides and using $x = \sqrt[3]{2}$ for simplified typing.
R... | We also have the following identity,
$$\sqrt[3]{m^3-n^3+6m^2n+3mn^2-3(m^2+mn+n^2)\sqrt[3]{mn(m+n)}}=\\ \sqrt[3]{m^2(m+n)}-\sqrt[3]{mn^2}-\sqrt[3]{(m+n)^2n}$$
For $m=n=1$ we get $(1)$.
For $m=4$ and $n=1$ we get $(5)$.
| {
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"url": "https://math.stackexchange.com/questions/871639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "86",
"answer_count": 3,
"answer_id": 2
} |
Find the $n$th term of $1, 2, 5, 10, 13, 26, 29, ...$ How would you find the $n$th term of a sequence like this?
$1, 2, 5, 10, 13, 26, 29, ...$
I see the sequence has a group of three terms it repeats: Double first term to get second term, add three to get third term, repeat.
What about: $2, 3, 6, 7, 14, 15, 30,... $?... | A more general approach to solving for the $n$th term of such sequences uses matrix multiplication. Suppose the even terms are nonzero constant $r\neq 1$ times the preceding odd terms, while the odd terms are constant $d$ plus the preceding even terms. We have:
$$ \begin{pmatrix} 1 & 0 & d \\ 1 & 0 & 0 \\ 0 & 0 & 1 \... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
} |
Show that {$a_n$} is convergent and find sup{$a_n| n \in Z_+ $} $a_1 = 1$ and $ a_{n+1} = \frac{4+3a_n}{3+2a_n} ; \forall n \in Z_+$
Show that {$a_n$} is convergent, find its limit and find sup{$a_n| n \in Z_+ $} if exists.
I found the limit as follows -
Let$$\lim_{n \to \infty} a_n = L$$
$$\lim_{n +1\to \infty} a_n+1 ... | $a_{n+1} - a_n = \dfrac{4+3a_n}{3+2a_n} - a_n = \dfrac{4-2a_n^2}{3+2a_n}$. So consider $f(x) = \dfrac{4+3x}{3+2x}$, we have: $f'(x) = \dfrac{1}{(3+2x)^2} > 0$, thus $f$ increases strictly. Next assume by induction that : $0 < a_n < \sqrt{2}$, then: $a_{n+1} = f(a_n) < f(\sqrt{2}) = \dfrac{4+3\sqrt{2}}{3+2\sqrt{2}} = \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/873270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the probability of each outcome when a biased die is rolled, if rolling a 2 or 4 is three times as likely as rolling each of the other$\dots$ Question:Find the probability of each outcome when a biased die is rolled, if rolling a $2$ or $4$ is three times as likely as rolling each of the other four numbers on the ... | Your confusion stems from the obscure description of the question at hand.
The actual meaning of the question is apparently $\displaystyle P(2,4) = 3 \cdot P(1,3,5,6)$.
Therefore, $\displaystyle P(2,4) = \frac{12}{16} = 3 \cdot \frac{4}{16} = 3 \cdot P(1,3,5,6)$ implies the following:
*
*$\displaystyle P(2) = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/875688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$ I'm an eight-grader and I need help to answer this math problem.
Problem:
Calculate $$\frac{1}{5^1}+\frac{3}{5^3}+\frac{5}{5^5}+\frac{7}{5^7}+\frac{9}{5^9}+\cdots$$
This one is very hard for me. It seems unsolvable. How to calcul... | Let
$$\begin{align}f(x)&=\sum_{n=0}^\infty(2n+1)x^{2n+1}\\&=x\sum_{n=0}^\infty(2n+1)x^{2n}\\&=x\frac{d}{dx}\left(\sum_{n=0}^\infty x^{2n+1}\right)\\&=x\frac{d}{dx}\left( \frac{x}{1-x^2}\right)\\&=x\frac{x^2+1}{(1-x^2)^2}\end{align}$$
and notice that the desired sum is $f\left(\frac15\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/876893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
} |
Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$ Please help me to evaluate this integral:
$$
I={\large\int}_{0}^{\infty}{\ln\left(x\right) \over 1 + x}\,
\,\sqrt{\,x + \sqrt{\,1 + x^{2}\,}\, \over 1 + x^{2}\,}\,\,{\rm d}x.\tag1
$$
Mathematica could not evaluate it in a c... | \begin{align}
-\frac{1}{4}\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x
&=-\frac{1}{4}\int^\infty_0\frac{\ln{\sinh{x}}}{1+\sinh{x}}e^{x/2}{\rm d}x\\
&=-\frac{1}{2}\int^\infty_1\frac{\ln\left(\frac{x^2-x^{-2}}{2}\right)}{1+\frac{x^2-x^{-2}}{2}}{\rm d}x\\
&=\int^1_0\frac{\ln(1-x^4)-\ln(2x^2)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 3,
"answer_id": 1
} |
prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$
Show that if $a$, $b$, $c$ and $d$ are non-negatives and $ab+bc+cd+da=1$ then:
$$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$
yet again it should be solved with Cauchy inequality.
thing i have done so far:
$(\sum\limits_{cyc} \frac {a... | To finish your idea, just add up all inequalities of type $a^2+b^2 \geq 2ab$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/877621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Antiderivative of $\frac{1}{1+\sin {x} +\cos {x}}$ How do we arrive at the following integral
$$\displaystyle\int\dfrac{dx}{1+\sin {x}+\cos {x}}=\log {\left(\sin {\frac{x}{2}}+\cos {\frac{x}{2}}\right)}-\log {\left(\cos {\frac{x}{2}}\right)}+C\ ?$$
| Another way :
\begin{align}
\int\frac{1}{1+\color{blue}{\sin x}+\color{red}{\cos x}}\ dx&=\int\frac{1}{\sin^2\frac x2+\cos^2\frac x2+\color{blue}{2\sin\frac x2\cos\frac x2}+\color{red}{\cos^2\frac x2-\sin^2\frac x2}}\ dx\\
&=\int\frac{1}{2\cos\frac x2\left(\sin\frac x2+\cos\frac x2\right)}\ dx\\
&=\int\frac{1}{2\cos^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/877681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
In a triangle ABC, prove that cot(A/2)+cot(B/2)+cot(C/2) =cot(A/2)cot(B/2)cot(C/2) In a triangle ABC, prove that $\cot \left ( \frac{A}{2} \right )+\cot \left ( \frac{B}{2} \right )+\cot \left ( \frac{C}{2} \right )=\cot \left ( \frac{A}{2} \right )\times \cot \left ( \frac{B}{2} \right )\times \cot \left ( \frac{C}{2}... | It shouldn't be that hard cause you have this two.
$\frac{\cos (\frac{A}{2})}{\sin (\frac{A}{2})}+\frac{\cos (\frac{B}{2})}{\sin (\frac{B}{2})}+\frac{\cos (\frac{C}{2})}{\sin (\frac{C}{2})}=\frac{\cos (\frac{A}{2})}{\sin (\frac{A}{2})}\times \frac{\cos (\frac{B}{2})}{\sin (\frac{B}{2})}\times \frac{\cos (\frac{C}{2})}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Find limit of a fraction without derivatives: $\lim\limits_{h\rightarrow0} \frac{\sqrt[3]{x+h} - \sqrt[3]{x}}h$ How to find limit of this?
\begin{equation}
\lim_{h \rightarrow 0} \frac {\sqrt[3]{x+h} - \sqrt[3]{x}} {h}.
\end{equation}
I see here the derivative definition, so I understand in "derivative mind" that answe... | Rationalise the numerator by the identity $a^3-b^3=(a-b)(a^2+ab+b^2)$
$\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}$
$=\lim _{h\rightarrow 0}{\frac {\sqrt [3]{x+h}-\sqrt [3]{x}}{h}}{\frac { \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt [3]{x}+{x}^{2/3
}}{ \left( x+h \right) ^{2/3}+\sqrt [3]{x+h}\sqrt ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/878680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Verification of the Stokes theorem for the surface that is a part of a cone
Let $S$ consist of the part of the cone $z=(x^2+y^2)^{1/2}$ for $x^2+y^2\leq9$ and suppose $${\bf A}=(-y,x,-xyz).$$ Verify that Stokes theorem is satisfied for this choice of $\bf A$ and $S$.
In the solution to this question in order to work ... | By Stoke's theorem,
$$\int_S (\nabla \times \mathbf{A}) \cdot \mathbf{n} \,d\sigma = \oint_C \mathbf{A} \cdot d\mathbf{r},$$
where $C = \{(x,y,z):x^2+y^2=9,z =3\}$ oriented clockwise.
With $\mathbf{A} = -y\mathbf{i}+x\mathbf{j}+-xyz\mathbf{k},$ the line integral is
$$\eqalign{
\oint_C \mathbf{A} \cdot d\mathbf{r}&=-\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/879456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find depth of a half-filled parabolic cross-section Given a cross-section of an object that is parabolic in shape, how do you find the depth of the object when it is "half full".
A full example given in an exam:
A long trough whose cross-section is parabolic is $1\frac{1}{2}$ metres wide at the top and $2$ metres deep... | Model the cross-section as a parabola centred at the origin. Then we know that:
$$
y = ax^2
$$
for some constant $a$. But from the given dimensions, we also know that the parabola contains the point $(3/4, 2)$. Substituting, we find that:
$$
2 = a(3/4)^2 \iff 2 = \frac{9a}{16} \iff a = \frac{32}{9}
$$
Now the area of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Prove $\int_{\mathbb{R^{+}}} \frac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\frac{\pi}{32} (3\pi-4)^2$ How do you arrive at the result $$I=\displaystyle\int_{\mathbb{R^{+}}} \dfrac{\sin^3 {(\pi x^2)} \cos {(4x^2)}}{x^5} dx=\dfrac{\pi}{32} (3\pi-4)^2\ ?$$
Wolfram Alpha agrees numerically.
I tried replacing $\sin p$ by ... | The following is an approach using contour integration.
$ $
$$ \begin{align} \int_{0}^{\infty} \frac{\sin^{3}(\pi x^{2}) \cos(4x^{2})}{x^{5}} \ dx &= \frac{1}{2}\int_0^{\infty} \frac{\sin^3(\pi t)\cos(4t)}{t^3} \ dt \\ &= \frac{1}{4} \int_{-\infty}^{\infty} \frac{\sin^{3} (\pi t) \cos(4t)}{t^{3}} \ dt \\ &= \frac{1}{4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/881289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How do you prove this integral involving the Glaisher–Kinkelin constant According to wikipedia on the page Glaisher–Kinkelin constant
$$\int_0^{1/2} \ln\Gamma(x) dx=\frac32\ln \text{A}+\frac5{24}\ln 2+\frac14\ln\pi$$
I got interested in how you possibly could prove something like that, but couldn't find any citations a... | In general, $$ \begin{align} \int_{0}^{z} \log \Gamma(x) \ dx &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} - \zeta^{'}(-1) + \zeta^{'}(-1,z) \\ &= \frac{z}{2} \log(2 \pi) + \frac{z(1-z)}{2} + \ln A - \frac{1}{12} + \zeta^{'}(-1,z) \end{align}$$
where $\zeta(a,z)$ is the Hurwitz zeta function and where I used the fac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Integrate a division of polynomials Hi I have the following integral:
$$\int \frac{2x}{x^2+6x+3}\, dx$$
I made some changes like:
$$\int \dfrac{2x+6-6}{x^2+6x+3}\, dx$$
then I have:
$$\int \dfrac{2x+6}{x^2+6x+3}\, dx -\int\dfrac{6}{x^2+6x+3}\, dx$$
and thus: $$\ln(x^2+6x+3)-\int\dfrac{6}{x^2+6x+3}\, dx$$
Ok, I have de... | Another idea (just reducing it to another form):
Let $$I=6\int \frac{1}{x^2+6x+3} dx=6\int \frac{1}{(x+3)^2-6} dx=\int \frac{1}{(\frac{1}{\sqrt{6}}(x+3))^2-1} dx$$.
Now let $$\frac{1}{\sqrt{6}}(x+3)=\cosh a$$, hence using $$\cosh^2 a - 1 = \sinh ^2 a$$ and $$\frac{1}{\sqrt{6}} = \sinh a \frac{da}{dx}\Leftrightarrow dx ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/882692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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prove $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$
If $a,b,c$ are positive real numbers and $a+b+c=1$,Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geq 3(a^2+b^2+c^2)$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have done so... | lemma:
$$\sum_{cyc}\dfrac{a^2}{b}\ge\dfrac{(a+b+c)(a^2+b^2+c^2)}{ab+bc+ac}$$
Proof:
$$\Longleftrightarrow (ab+bc+ac)(\sum_{cyc}\dfrac{a^2}{b})\ge (a+b+c)(a^2+b^2+c^2)$$
$$\Longleftrightarrow a^3+b^3+c^3+a^2c+c^2b+b^2a+\sum_{cyc}\dfrac{a^3c}{b}\ge a^3+b^3+c^3+\sum_{sym }a^2b$$
$$\Longleftrightarrow\dfrac{a^3c}{b}+\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
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Deciding whether a number is rational (2 examples) 1) Prove that number irrational $\sqrt{7-\sqrt{2}}$
I created a polynomial $x=\sqrt{7-\sqrt{2}}$ so
$P(x)=x^4-14x^2+47$ and since $47$ is prime we check $P(x)$ for $ {1,-1,47,-47}$ and since all of them are $P(x)\neq0$ it means our number is irrational.
Is my prof OK... | For 2).
If $p=\sqrt{\sqrt 5+3}+\sqrt{\sqrt 5-2}$ is a rational number, then
$$\begin{align}p-\sqrt{\sqrt5+3}=\sqrt{\sqrt5-2}&\Rightarrow p^2-2p\sqrt{\sqrt5+3}+\sqrt5+3=\sqrt5-2\\&\Rightarrow 2p\sqrt{\sqrt5+3}=p^2+5\\&\Rightarrow 4p^2(\sqrt 5+3)=(p^2+5)^2\\&\Rightarrow \sqrt5=\frac{(p^2+5)^2}{4p^2}-3\end{align}$$
implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/883710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
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Prove $\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$ Question:
Prove or disprove
$$I=\lim_{n\to\infty} \frac{(2^{2^n}+1)(2^{2^n}+3)(2^{2^n}+5)\cdots (2^{2^n+1}+1)}{(2^{2^n})(2^{2^n}+2)(2^{2^n}+4)\cdots (2^{2^n+1})}=\sqrt{2}$$... | The logarithm of the expression under the limit can be rewritten as
$$\sum_{k=0}^{2^{2^n-1}}\ln\left(1+\frac{1}{2^{2^n}+2k}\right)=\sum_{k=0}^{2^{2^n-1}}\frac{1}{2^{2^n}+2k}+O\left(2^{-2^n}\right).$$
Denoting $N=2^{2^n-1}$, it is easy to see that the limit of the logarithm can be computed as the limit of a Riemann sum:... | {
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"url": "https://math.stackexchange.com/questions/885448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Integral $\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)} dx$ =? After reading articles on differentiation under the integral sign, I hit this post from MIT OCW, where after introducing the power tool, it challenges reader to do
$$\int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)} dx$$
Obviously I have no clue where to start. Could... | This is a possible way to evaluate the integral. Partial fraction decomposition and the double angle formula yield
$$\int^\infty_0\frac{\sin^2{x}}{x^2(1+x^2)}dx=\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{x^2}dx-\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{1+x^2}dx$$
The first integral can be evaluated in many ways, differe... | {
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"url": "https://math.stackexchange.com/questions/886909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find $\int_{0}^{\pi/2} \log ({1+\cos x}) dx$ using real-variable methods? How do you find the value of this integral, using real methods?
$$I=\displaystyle\int_{0}^{\pi/2} \log ({1+\cos x}) dx$$
The answer is $2C-\dfrac{\pi}{2}\log {2}$ where $C$ is Catalan's constant.
| \begin{align}J&=\int_0^{\frac{\pi}{2}}\ln(1+\cos x)dx\\
K&=\int_0^{\frac{\pi}{2}}\ln(1-\cos x)dx\\
J+K&=2\int_0^{\frac{\pi}{2}}\ln(\sin x)dx\\
&=\int_0^{\frac{\pi}{2}}\ln(\sin x)dx+\underbrace{\int_0^{\frac{\pi}{2}}\ln(\sin x)dx}_{u=\frac{\pi}{2}-x}\\
&=\int_0^{\frac{\pi}{2}}\ln(\sin u\cos u)du\\
&=\int_0^{\frac{\pi}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 5
} |
limit of trigonometric series Given
$$a_n = \frac 1{n^2}\sum_{k=1}^n {(2k+1)\sin\left(\frac{k^2 \pi}{n^2} \right)}$$
find $\lim_{n \to \infty} a_n$.
My try: Because $k = 1,2,\dots, n$, $$0\le\frac {k^2}{n^2} \le 1$$
hence
$$0\le\frac {k^2}{n^2} \pi \le \pi$$
so
$$0 \le \sin \left( \frac {k^2}{n^2} \pi\right) \le 1.$$
T... | Consider that:
$$a_n = \frac{1}{n^2}\sum_{k=1}^{n}\sin\left(\frac{k^2\pi}{n^2}\right)+\frac{2}{n}\sum_{k=1}^{n}\frac{k}{n}\sin\left(\frac{k^2\pi}{n^2}\right),\tag{1}$$
where the first term on the right is $O(1/n)$ while the second one is a Riemann sum converging to the integral:
$$\int_{0}^{1}2x\sin(\pi x^2)\,dx=\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/887133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
log-trig integral with sin, cos, and tan Here is another log-trig integral you may find challenging/fun. Or not :)
$$\int_{0}^{\frac{\pi}{2}}\ln(1+\sin(x))\ln(1+\cos(x))\tan(x)dx=\frac{\pi^{2}}{8}\ln(2)-\frac{5}{16}\zeta(3)$$
| The integral equals
$$
\int_0^1 dx\, \ln \left( 1 + \sqrt{1-x^2} \right) \frac{\ln(1+x)}{x}
\\= \left. -\text{Li}_2(-x) \ln \left( 1 + \sqrt{1-x^2} \right) \right\lvert_0^1 + \int_0^1 dx\, \text{Li}_2(-x)\frac{1}{x} \left[1 - \frac{1}{\sqrt{1-x^2}} \right]
\\=\text{Li}_3(-1)-\int_0^1 dx\, \frac{\text{Li}_2(-x)}{x\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/890373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Simplifying a trigonometric identity Simplify $1 + \tan^2x$
My attempt:
$$\begin{align}1 + \tan^2x&\\
&= \frac{1}{1} + \frac{\sin^2x}{\cos^2x}\\
&= \frac{1(\cos^2x)}{1(\cos^2x)} +\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{\cos^2x + \sin^2x}{\cos^2x\cos^2x}\\
&= \frac{\sin^2x}{... | 4 th line of your solution is clearly wrong. In denominator their must be
lcm of [(cosx)^2 * (cosx)^2] which is (cosx)^2
and in 5th line you dont have used well known identity (cosx)^2 + (sinx)^2 = 1.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/892065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Find a real numbers $a,b$ such $a^n+b^n$ is rational Question:
prove or disprove :there exsit real numbers $a,b$ such follow two condition:
(1):$a+b$ is irrational
(2): for any postive integer $n\ge 2$, then $a^n+b^n$ is rational.
I have know if
$n=2k$ case is true,because I let $a=\sqrt{2}+1,b=\sqrt{2}-1$,so
$$a^{... | Suppose that $a,b$ are real numbers such that $a^n+b^n$ is rational for all integers $n \ge 2$.
Since $a^2+b^2$ and $a^4+b^4$ are rational, $\dfrac{(a^2+b^2)^2-(a^4+b^4)}{2} = a^2b^2$ is also rational.
Then, since $(a^5+b^5)-(a^2+b^2)(a^3+b^3)+a^2b^2(a+b) = 0$, we have that
$\dfrac{(a^2+b^2)(a^3+b^3)-(a^5+b^5)}{a^2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/895796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to solve the inequality $x^2>10$ using square roots? Solve the inequality: $$x^2>10$$
How am I supposed to do this? It doesn't make sense when I take into account that if $x^2=10$ then $x=+\sqrt{10}$ and $x=-\sqrt{10}$
But how am I supposed to apply this to an inequality, I would get $x>\sqrt{10}$ and $x>-\sqrt{10}... | Here is how I think about problems like this:
If $x^2 > 10$, then
$x^2 - 10 > 0$
$(x + \sqrt {10})(x - \sqrt {10}) > 0 $
$(x + \sqrt {10})$ and $(x - \sqrt {10})$ are either both positive or both negative.
Thus, when they are both positive and we divide both sides by one of the terms:
$x + \sqrt {10} > 0$ and $x - \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/898665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 9,
"answer_id": 1
} |
Where did the negative answer come from? The question is to evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$
$$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$
$$x^2=2+x$$
$$x^2-x-2=0$$
$$(x-2)(x+1)=0$$
$$x=2,-1$$
because $x$ is positive $x=2$ is the answer. bu... | $x=\sqrt{2+\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}_{\text{$x$}}}$, so we get $x=\sqrt{2+x}$.
Now there is only one solution. If we square both sides, we add the case $-x=\sqrt{2+x}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/898964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 1
} |
Finding the sum to infinity Question:
Find the sum to infinity for the following series
$$1, -\frac{1}{2}, \frac{1}{2^2}, -\frac{1}{2^3},\cdots$$
What would be the technique used to find such a sum?
| Write the sum of the $n$ and $(n+1)^{th}$ terms as
$$
\frac{1}{2^n} - \frac{1}{2^{n+1}}
= \frac{1}{2^{n+1}}
$$
then sum them up to get
$$
\sum_{n \in 2 \mathbb{N} }^\infty \frac{1}{2^{n+1}}
= \sum_{k=0}^\infty \frac{1}{2^{2k+1}}
= \frac{1}{2} \sum_{k=0}^\infty \frac{1}{4^k}
= 2/3.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/901672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
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.