Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solutions for $\frac{3}{x+1}\le\frac{2}{2x+5}$ Im in search of the solutions for:
$$\frac{3}{x+1}\le\frac{2}{2x+5}$$
So first i tried to combine the two sites:
$$\frac{6x + 15 - 2x + 2}{2x^2 +7x + 5}\le{0}$$
$$\frac{4x + 17}{2x^2 +7x +5}\le{0}$$
My problem is that now i have two solutions for the denominator and i dont know how to continue:
$2x^2+7x+5 = -1 \text{ and } -2.5$
The solution should be: $(-2.5;-1) \cup (-\infty;-3.25)$
| If you assume assume $2x+5>0$ and $x+1>0$ you end up with :$$3(2x+5)\leq 2(x+1)$$
$$3(2x+5)\leq 2(x+1)\Rightarrow 6x+15\leq2x+2\Rightarrow 4x\leq -13\Rightarrow x\leq??$$
Suppose we assume $x+1$ is positive, then we would end up with above conclusion (Why??)
Suppose we assume $2x+5$ is positive but $x+1 <0$ then???
Please do it by yourself :)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/618661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$. If $n\in\mathbb N$ and $4^n+2^n+1$ is prime, prove that there exists an $m\in\mathbb N\cup\{0\}$ such that $n=3^m$.
I.e. if $4^n+2^n+1$ is prime, prove that $n=3^m$, where $m\in\mathbb N\cup\{0\}$.
I don't know how I could solve this, I don't have any ideas. I know that $4^n+2^n+1$ is obviously odd. Assume $m=0$, i.e. $n=1$. Then $3$ is prime. So that doesn't disprove what we're trying to show.
So assume $m>0$. Then $4^n+2^n+1>3$, so $4^n+2^n+1$ is either $3k+1$ or $3k+2$, where $k\in\mathbb N$. So either $4^n+2^n$ or $4^n+2^n-1$ is divisible by 3. I think it's not something that could help out at solving this. Just a little observation. I don't have any ideas.
| Lemma: $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$
Proof: Several proofs can be found here
$4^n+2^n+1 = \frac{8^n-1}{2^n-1}$. Let $n=3^t \times k$, where $k$ is not divisible by $3$. The numerator becomes $8^{3^t \times k}-1 = \left( {8^{3^t}} \right) ^k-1$, which has a factor $\left(8^{3^t}-1 \right)$.
The denominator becomes $2^{3^t \times k}-1$. By our lemma, $\gcd(8^{3^t}-1, 2^{3^t \times k}-1)$ $= \gcd(8^{3^t}-1, 8^{3^{t-1} \times k}-1)$ $ = 8^{\gcd(3^t, {3^{t-1} \times k})} - 1 = 8^{3^{t-1}}-1 = 2^{3^{t}}-1$, since $3 \nmid k$.
We now rewrite the fraction $\frac{8^n-1}{2^n-1}$as follows:
$$\frac{ \left( 2^{3^{t}}-1 \right) \left( {\left(2^{3^{t}}\right)}^2 + 2^{3^{t}} +1\right) \left( {\left(8^{3^{t}}\right)}^{k-1} + {\left(8^{3^{t}}\right)}^{k-2} + ... +1\right)}{\left( 2^{3^{t}}-1 \right) \left( {\left(2^{3^{t}}\right)}^{k-1} + {\left(2^{3^{t}}\right)}^{k-2} + ... +1\right)}$$
$\gcd\left( \left( {\left(2^{3^{t}}\right)}^2 + 2^{3^{t}} +1\right),\left( {\left(2^{3^{t}}\right)}^{k-1} + {\left(2^{3^{t}}\right)}^{k-2} + ... +1\right) \right) = 1$, so the only way the fraction cancels to a prime is when $\left( {\left(8^{3^{t}}\right)}^{k-1} + {\left(8^{3^{t}}\right)}^{k-2} + ... +1\right)$ $=\left( {\left(2^{3^{t}}\right)}^{k-1} + {\left(2^{3^{t}}\right)}^{k-2} + ... +1\right)$. This only occurs when $k=1$. Hence, $n$ is of the form $3^t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
} |
Solving quadratic system If $a,b,c\in \mathbb{R}$ satisfy the system
$a^2+ab+b^2=9$;
$b^2+bc+c^2=16$;.
$c^2+ac+a^2=25$.
Find $ab+ac+bc$
| The following is a geometric solution. Let $P$ be a point, and consider three line segments $PA,PB$ and $PC$ making an angle of $120$ degrees with each other. Thus $\angle APB = \angle BPC = \angle CPA = 120^o.$ Here $|PA| = a, |PB| = b$ and $|PC| = c.$
Then by the law of cosines, triangle $ABC$ has sides $3,4$ and $5.$
Further the area of the triangle $ABC$ is $\frac{1}{2}\cdot 3 \cdot 4 = 6.$ This can be calculated another way, namely $\frac{1}{2}\left( ab + bc + ca\right) \sin 120,$ since area of $ABC = $ areas of $PAB + PBC + PCA.$
Comparing the two, we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How does mod multiplication work? For example, $10^{10} \equiv 4\pmod{6}$
If I used $\pmod{2}$ and $\pmod{3}$, how does the multiplication process work?
Since $10^{10} \equiv 0 \pmod{2}$ and $10^{10}\equiv 1\pmod{3}$,
$$
10^{10}\equiv (0,1) \pmod{(2,3)}
$$
how do we get the value $4$ at the end? do we list out the possible values of $0\pmod{2}$ and $1\pmod{3}$?
$$
1\pmod{3} = 1, 4, 7, 10
$$
so on.
Since only $4, 10$ and so on satisfy $\pmod{2}$, only values that satisfy both criteria can be used.
In general, can we do this for $\pmod{n}$, $n$ being any integer?
| For integer $n\ge0$
$$10^{n+1}-10=10(10^n-1)\equiv0\pmod{10\cdot9}$$ as $10^n-1$ is divisible by $10-1=9$
$$\implies 10^n\equiv10\pmod{90}\equiv10\pmod6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620994",
"timestamp": "2023-03-29T00:00:00",
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Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist? Why the integral $\int_0^3 \frac{1}{x^2-6x+5}dx$ doesn't exist?
$$
I=\frac{1}{2} \lim_{t\to 1^{-}} [\arctan(2-3/2) - \arctan(-3/2)] + \lim_{t\to 1^+}[\arctan(0/2) - \arctan(1-3/2)]
$$
I can find $\arctan(0)= \pi/2$
So why the answer in my book is doesn't exist?
| Note that $\frac{1}{(x-1)(x-5)}$ is undefined at $x = 1$ so
$$\int_0^3\frac{1}{(x - 1)(x-5)}dx = \lim_{s\to 1^-}\int_0^s\frac{1}{(x-1)(x-5)}dx + \lim_{t\to 1^+}\int_t^3\frac{1}{(x-1)(x-5)}dx.$$
To evaluate the two integrals on the right, we need to find an antiderivative. We do this by first applying partial fraction. Doing so we obtain
$$\frac{1}{(x-1)(x-5)} = \frac{-\frac{1}{4}}{x-1} +\frac{\frac{1}{4}}{x-5}.$$
Therefore $-\frac{1}{4}\ln|x-1| + \frac{1}{4}\ln|x-5| = \frac{1}{4}\ln\left|\frac{x-5}{x-1}\right|$ is an antiderivative of $\frac{1}{(x-1)(x-5)}$. Therefore
$$\int_0^s\frac{1}{(x-1)(x-5)}dx = \frac{1}{4}\ln\left|\frac{s-5}{s-1}\right| - \frac{1}{4}\ln\left|\frac{0-5}{0-1}\right| = \frac{1}{4}\ln\left|\frac{s-5}{s-1}\right| -\frac{1}{4}\ln 5.$$
As $\lim\limits_{s\to 1^-}\frac{1}{4}\ln\left|\frac{s-5}{s-1}\right|$ does not exist, $\lim\limits_{s\to 1^-}\int_0^s\frac{1}{(x-1)(x-5)}dx$ does not exist, so $\int_0^3\frac{1}{(x-1)(x-5)}dx$ does not exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Finite sum related to Stirling numbers I am wondering if there is a closed form solution for the following sum:
$$
\sum _{k =0}^{n-1} \frac{(-1)^{k} (n-k)^{n+1} }{(k+1)(k+2)}\binom{n}{k}.
$$
If the the factors $(k+1)(k+2)$ in the denominator weren't there, the sum would equal $n!S(n+1,n)$, where $S(n,m)$ denotes a Stirling number of the second kind. This made me wonder if 'Stirling polynomials' of the form
$$
\mathcal{S}(x)=\sum _{k =0}^{n-1} (n-k)^{n+1}\binom{n}{k}x^k
$$
admit a closed form solution. The value of my sum could then be obtained by integrating $\mathcal{S}$ twice.
Any other strategy to evaluate the sum are also very welcome.
| We can get a better closed form using a different technique.
Suppose we seek to evaluate
$$\sum_{k=0}^{n-1} \frac{(-1)^k(n-k)^{n+1}}{(k+1)(k+2)}
{n\choose k}.$$
This is
$$\frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n-1} (-1)^k (n-k)^{n+1}
{n+2\choose k+2}
\\ = \frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n-1} (-1)^{k+2} (n+2-(k+2))^{n+1}
{n+2\choose k+2}
\\ = \frac{1}{(n+2)(n+1)}
\sum_{k=2}^{n+1} (-1)^{k} (n+2-k)^{n+1}
{n+2\choose k}
\\ = -\frac{(n+2)^n}{n+1}
+ (n+1)^n +
\frac{1}{(n+2)(n+1)}
\sum_{k=0}^{n+2} (-1)^{k} (n+2-k)^{n+1}
{n+2\choose k}.$$
Restrict to the sum for a moment and introduce
$$(n+2-k)^{n+1}
= \frac{(n+1)!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} \exp((n+2-k)z) \; dz.$$
Observe that this is zero when $k=n+2$ which is the correct value.
We get for the sum
$$\frac{n!}{(n+2)\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} \exp((n+2)z)
\sum_{k=0}^{n+2} (-1)^{k}
{n+2\choose k} \exp(-kz) \; dz
\\ = \frac{n!}{(n+2)\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}} \exp((n+2)z)
(1-\exp(-z))^{n+2} \; dz
\\ = \frac{n!}{(n+2)\times 2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+2}}
(\exp(z)-1)^{n+2} \; dz$$
This is
$$\frac{n!}{(n+2)} [z^{n+1}] (\exp(z)-1)^{n+2} = 0$$
because $\exp(z)-1 = z +\frac{1}{2} z^2 + \frac{1}{6} z^3+\cdots.$
We conclude that the initial sum is equal to
$$-\frac{(n+2)^n}{n+1} + (n+1)^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Analytic hierarchy process (AHP), what is the significance of eigenvalues/eigenvectors? I was to a new years party today. Mathematics interested as I am I tried to discuss
eigenvalues with my friend there. He did not know what eigenvalues are but said
that he has heard about them in his research in something called
Analytic Hierarchy Process (AHP).
Analytic hierarchy process, wikipedia link
I found the following passage in wikipedia that mentions the word eigenvector, and the reference to Oskar Perron:
"Non-Monotony of some weight extraction methods
Within a comparison matrix one may replace a judgement with a less favourable judgement and then check to see if the indication of the new priority becomes less favourable then the original priority. In the context of tournament matrices, it has been proven by Oskar Perron in,[32] that the principle right eigenvector method is not monotonic. This behaviour can also be demonstrated for reciprocal n x n matrices, where n > 3. Alternative approaches are discussed in.[33][34][35]"
Reading about Oskar Perron I found that he worked in differential equations. Then reading about differential equations and eigenvalues in my book Advanced Engineering Mathematics, in the chapter Eigenvectors from the Start, page 299, I found:
"The general solution of system of differential equations $x' = Ax$ is known as soon as the eigenvalues and eigenvectors are known. In fact, if the eigenvalues are $\lambda_1$, $\lambda_2$
and the corresponding eigenvectors are $v_1, v_2$ then the general solution would be
$$x = a \cdot \exp(\lambda_1 \cdot t) \cdot v_1 + b \cdot \exp(\lambda_2 \cdot t) \cdot v_2$$
So my question is: What is the significance of eigenvalues in Analytic Hierarchy Process (AHP)?
| The second link by Amzoti in the comment above is very clear about how the algorithm goes.
Illustration of Analytic hierarchy process
I have begun to study this by looking at the more arithmetic table $n/k$ as the pairwise comparison:
$$\begin{array}{llllll}
1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} \\
2 & 1 & \frac{2}{3} & \frac{1}{2} & \frac{2}{5} & \frac{1}{3} \\
3 & \frac{3}{2} & 1 & \frac{3}{4} & \frac{3}{5} & \frac{1}{2} \\
4 & 2 & \frac{4}{3} & 1 & \frac{4}{5} & \frac{2}{3} \\
5 & \frac{5}{2} & \frac{5}{3} & \frac{5}{4} & 1 & \frac{5}{6} \\
6 & 3 & 2 & \frac{3}{2} & \frac{6}{5} & 1
\end{array}$$
Mathematica:
TableForm[Table[
Rationalize[
Total[Transpose[
MatrixPower[N[Table[Table[n/k, {k, 1, i}], {n, 1, i}]], i]]]/
Total[Total[
Transpose[
MatrixPower[N[Table[Table[n/k, {k, 1, i}], {n, 1, i}]],
i]]]]], {i, 1, 6}]]
The output is very orderly:
$$\begin{array}{llllll}
1 & \text{} & \text{} & \text{} & \text{} & \text{} \\
\frac{1}{3} & \frac{2}{3} & \text{} & \text{} & \text{} & \text{} \\
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} & \text{} & \text{} & \text{} \\
\frac{1}{10} & \frac{1}{5} & \frac{3}{10} & \frac{2}{5} & \text{} & \text{} \\
\frac{1}{15} & \frac{2}{15} & \frac{1}{5} & \frac{4}{15} & \frac{1}{3} & \text{} \\
\frac{1}{21} & \frac{2}{21} & \frac{1}{7} & \frac{4}{21} & \frac{5}{21} & \frac{2}{7}
\end{array}$$
which is $$\frac{k}{n(n+1)/2}$$
I will continue editing this later when I have the time.
nn = 5
Do[
A = N[Table[Table[n/k, {k, 1, nn}], {n, 1, nn}]];
A = MatrixPower[A, n];
MatrixForm[N[A]];
MatrixForm[Total[Transpose[N[A]]]];
Total[Total[Transpose[N[A]]]];
Print[Rationalize[
Total[Transpose[N[A]]]/Total[Total[Transpose[N[A]]]]]], {n, 1, 6}]
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $n$ such that $(n-6)$ is divisible by $6$, $(n-7)$ is divisible by $7$ and $(n-8)$ is divisible by $8$. If $(n-6)$ is divisible by $6$, $(n-7)$ is divisible by $7$ and $(n-8)$ is divisible by $8$, then what is the value of $n $?
| If $n - 6$ is divisible by $6$, then so is $n$. Likewise for $7$ and $8$; that is, $n$ is divisible by $6$, $7$ and $8$. There are infinitely many such values; the smallest positive value is the least common multiple of $6$, $7$ and $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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manipulation of subtraction I am trying to solve an induction problem and got stuck at this part.
$$ 1 - \frac{n+2}{(n+2)!} + \frac{n+1}{(n+2)!} = 1 - \frac{(n+2) - (n+1)}{(n+2)!} $$
Shouldn't it be
$$ 1 - \frac{n+2}{(n+2)!} + \frac{n+1}{(n+2)!} = 1 - \frac{(n+2) + (n+1)}{(n+2)!} $$
How do you get the left expression to the right expression?
| Note that
$$-\frac{A+B}{C}=(-1)\times \left(\frac{A+B}{C}\right)=(-1)\times\left(\frac{A}{C}+\frac{B}{C}\right)=-\frac{A}{C}-\frac{B}{C}=\frac{-A-B}{C}.$$
The minus sign in front of the first fraction does influence the plus sign in front of $B$ of the numerator of the first fraction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/626056",
"timestamp": "2023-03-29T00:00:00",
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prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$ prove by induction that $n(n+1)(n+2)(n+3)$ is an integer multiple of $24$
Let $P(n)$ be the proposition we want to prov, ie: $P(n):=24 \mid(n)(n+1)(n+2)(n+3)$
For $P(1)$ we have: $24 \mid(1)(1+1)(1+2)(1+3)\implies6 \mid(1)(2)(3)(4)\implies24 \mid24$, so $P(1)$is true.
For $P(2)$ we have: $24 \mid(2)(2+1)(2+2)(2+3)\implies6 \mid(2)(3)(4)(5)\implies24 \mid120$, so $P(1)$is true
Inductive Hypothesis: Let $n=k$ and we assume that $P(k):=24\mid k(k+1)(k+2)(k+3)$ is true.
Inductive Step:
$$(k+1)(k+2)(k+3)(k+4)$$
$$k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$
Using the assumption of $P(k) \implies \exists a\in \mathbb Z$, such that, $(k+1)(k+2)(k+3)=24\cdot a$
so: $$=24\cdot a +4(k+1)(k+2)(k+3)$$
| Continuing from where you left, we just need to prove that for $n=k+1$, $P(n)$ is an integer multiple of $24$.
$$P(k+1) = (k+1)(k+2)(k+3)(k+4)$$
$$P(k+1) = k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)$$
$1st$ term on right hand side is $P(k)$ which is an integer multiple of $24$ from your inductive hypothesis.
$2nd$ term on the right hand side has a product of $3$ consecutive integers and hence divisible by $6$. So on a whole divisible by $4*6=24$.
On a whole the right hand side is divisible by $24$. Hence $P(k+1)$ is an integral multiple of $24$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/632245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How find this $\sum_{i=0}^{5}\frac{1}{2+\cos{\left(x+\frac{i\pi}{3}\right)}}\cdot \frac{1}{2+\cos{\left(x+\frac{(i+1)\pi}{3}\right)}}$ Find this follow function $f(x)$ range ,where $x\in R$,
$$f(x)=\sum_{i=0}^{5}\dfrac{1}{2+\cos{\left(x+\dfrac{i\pi}{3}\right)}}\cdot \dfrac{1}{2+\cos{\left(x+\dfrac{(i+1)\pi}{3}\right)}}$$
or
$$f(x)=\dfrac{1}{2+\cos{x}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{6\pi}{3}\right)}}$$
I think $f(x)$ have simple form,becasue this is exam problem
I think maybe can use
$$f(2\pi-x)+f(x)=?$$
or maybe have use this
$$\cos{x}\cos{y}=\dfrac{1}{2}[\cos{(x-y)}+\cos{(x+y)}]$$
I use this two idea all solve this problem,
Thank you
| HINT
I should start simplifying the sum of the first and the sixth terms (result = A), then the sum of the second and fifth terms (result = B), then the sum of the third and fourth terms (result = C). Now, I should simplify A + B (result = D) and finally simplify C + D.
You will arrive to a surprizingly simple result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Searching for an explicit functional form Let $f:\mathbb N \to \mathbb R$ be a strictly decreasing function. Suppose $$\frac{f(x)}{f(x+1)}=2^x,\qquad \forall x\in\mathbb N.$$ Is it possible to find an explicit functional form of $f$? Any hint at finding a solution is appreciated.
| Hint: We compute a little to find out what's going on.
Let us rewrite the equation as $f(n+1)=\frac{f(n)}{2^{n}}$. Let $f(1)=a$.
We have $f(2)=\frac{a}{2^1}$, and therefore $f(3)=\frac{a}{2^{1}\cdot 2^2}$, and therefore $f(4)=\frac{a}{2^{1}\cdot 2^2\cdot 2^3}$, and therefore $f(5)=\frac{a}{2^{1}\cdot 2^2\cdot 2^3\cdot 2^4}$, and so on.
To get simpler-looking expressions, note that for example $2^1\cdot 2^2\cdot 2^3\cdot 2^4=2^{1+2+3+4}$, and use the formula for the sum of the first $k$ positive integers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining y(2) knowing that $y(1)=1+e^3$ and $ty'+(2t^2)y=2t^2$ How do I determine y(2) knowing that $y(1)=1+e^3$ and $ty'+(2t^2)y=2t^2$ ?
| By dividing both sides by $t$ (to obtain the standard form), express the ordinary differential equation as
$$\begin{aligned}
t\dfrac{dy}{dt} + 2t^2y &= 2t^2\\
\dfrac{dy}{dt} + 2ty &= 2t
\end{aligned}$$
Since $p(t) = 2t$ by the standard form of differential equation, the integrating factor of this equation is
$$\begin{aligned}
\mu(t) &= e^{\int 2t\,dt}\\
&= e^{t^2}
\end{aligned}$$
So we have
$$\begin{aligned}
\mu(t)\left(\dfrac{dy}{dt} + 2ty = 2t \right) \longrightarrow e^{t^2}\dfrac{dy}{dt} + 2te^{t^2}y = 2te^{t^2}
\end{aligned}$$
So the general solution is
$$\begin{aligned}
\left[e^{t^2}y\right]' &= 2te^{t^2}\\
e^{t^2}y &= \int 2te^{t^2}\,dt\\
e^{t^2}y &= e^{t^2} + \mbox{C}
\end{aligned}$$
Since $y(1) = 1 + e^3$, we have
$$\begin{aligned}
e(1 + e^3) &= e + \mbox{C}\\
e + e^4 &= e + \mbox{C}\\
\mbox{C} &= e^4
\end{aligned}$$
So this IVP has the solution
$$\begin{aligned}
e^{t^2}y &= e^{t^2} + e^4\\
y(t) &= 1 + e^{4 - t^2}
\end{aligned}$$
Thus,
$$\begin{aligned}
y(2) &= 1 + e^{4 - 2^2}\\
&= 1 + 1 = 2
\end{aligned}$$
| {
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Sum of a Hyper-geometric series. (NBHM 2011)
How to find the sum of the following series
$$\frac{1}{5} - \frac{1\cdot 4}{5\cdot 10} + \frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15} - \dots\,.?$$
I have no idea. I have written the general term and tested its convergence by Gauss' test for convergence, but they are neither the question nor the answer.
| From the Generalized Binomial Theorem (for $|x|<1$), $$\left(1+x\right)^n=1+nx+\frac{n(n-1)}{2!}x^2+\frac{n(n-1)(n-2)}{3!}x^3+\cdots$$
Method $1:$
$\displaystyle -\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}=\frac{-\frac13\left(-\frac13-1\right)\left(-\frac13-2\right)3^3}{3!\cdot 5^3}=\frac{-\frac13\left(-\frac13-1\right)\left(-\frac13-2\right)}{3!}\left(\frac35\right)^3 $
and $\displaystyle \frac{1\cdot 4}{5\cdot 10}=\frac{-\frac13\left(-\frac13-1\right)3^2}{2!\cdot 5^2}=\frac{-\frac13\left(-\frac13-1\right)}{2!}\left(\frac35\right)^2$
and $\displaystyle-\frac15=\left(-\frac13\right)\left(\frac35\right)$
So, the given series $\displaystyle=1-\left(1+\frac35\right)^{-\frac13}$
Method $2:$
If $\displaystyle nx=-\frac15 \ \ \ \ (1)$
and $\displaystyle\frac{n(n-1)}{2!}x^2=-\frac{1\cdot4}{5\cdot10}=\frac2{25} \ \ \ \ (2)$
Divide $(2)$ with the square of $(1)$ to get $n$ and then $x$ using $(1)$
and check that $x,n$ satisfies $\displaystyle\frac{n(n-1)(n-2)}{3!}x^3=-\frac{1\cdot 4\cdot 7}{5\cdot 10\cdot 15}$
| {
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Partial sum of a geometric series Consider the geometric series $\sum_{n=0}^\infty ar^n$ where $a=1$
and $r=-\frac{1}{2}$. Since $|r| < 1$, the series converges to
$S = \sum_{n=0}^\infty ar^n = \frac{1}{1+\frac{1}{2}} = \frac{2}{3}$.
I would like to arrive at the same sum by computing
$\lim_{N \to \infty} S_N$ where $S_N$ is the partial sum of $N$ terms of
the goemetric series.
First few terms of the geometric series are: $1, -\frac{1}{2}, \frac{1}{4},
-\frac{1}{8}, \frac{1}{16}, -\frac{1}{32}, \cdots$.
Here are my attempts to find $S_N$:
\begin{align*}
S_1 &= 1 = \frac{2^{(1-1)}}{2^{(1-1)}} \\
S_2 &= 1 - \frac{1}{2} = \frac{1}{2} = \frac{2^{(2-1)} - 1}{2^{(2-1)}} \\
S_3 &= \frac{1}{2} + \frac{1}{4} = \frac{3}{4} =
\frac{2^{(3-1)} - 1}{2^{(3-1)}} \\
S_4 &= \frac{3}{4} - \frac{1}{8} = \frac{5}{8} =
\frac{2^{(4-1)} - 3}{2^{(4-1)}} \\
\cdots \\
S_N &= ??
\end{align*}
\begin{align*}
S_1 &= 1 - \frac{0}{1} \\
S_2 &= 1 - \frac{1}{2} \\
S_3 &= 1 - \frac{1}{2} + \frac{1}{4} = 1 - \frac{1}{4} \\
S_4 &= 1 - \frac{1}{4} - \frac{1}{8} = 1 - \frac{3}{8} \\
\cdots \\
S_N &= ??
\end{align*}
I cannot find a pattern that will help me find $S_N$.
| Hint:
$$a + ar + ar^2 + \ldots + ar^{n-1} = \frac{ a( 1- r^n) } { (1-r) }. $$
| {
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If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$
By squaring both sides, we get,
$\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$
$\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
So, now, we have to prove that,
$-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$
I cannot do this. Please help!
| One's first inclination might be to reduce powers by invoking the Half-Angle Identities, which can be written as
$$\sin^2 x = \frac{1}{2}( 1 - k ) \qquad \cos^2 x = \frac{1}{2}(1 + k) \qquad\text{, where}\qquad k := \cos 2x$$
Then,
$$\begin{align}
\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b} = \frac{1}{a+b} \quad
&\implies\quad \frac{(1-k)^2}{4a}+\frac{(1+k)^2}{4b}=\frac{1}{a+b} \\[4pt]
&\implies\quad \big(\;k\;(a+b)+a-b\;\big)^2 = 0 \\[4pt]
&\implies\quad k = -\frac{a-b}{a+b} \\[4pt]
&\implies\quad \sin^2 x = \frac{a}{a+b} \quad\text{and}\quad \cos^2 x = \frac{b}{a+b}
\end{align}$$
Therefore,
$$\frac{\sin^6 x}{a^2} + \frac{\cos^6 x}{b^2} = \frac{1}{a^2}\left(\frac{a}{a+b}\right)^3 + \frac{1}{b^2}\left(\frac{b}{a+b}\right)^3 = \frac{a+b}{(a+b)^3} = \frac{1}{(a+b)^2}$$
| {
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Limit $\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2$ How can I find the following limit :$$\lim_{n\to\infty}\left(1-\frac{1}{3}\right)^2\left(1-\frac{1}{6}\right)^2\left(1-\frac{1}{10}\right)^2\cdots\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^2$$There are four options $(a)\frac1{3}(b)\frac1{9}(c)\frac1{81}(d)0$
Thanks!
| $$1-\frac{2}{n(n+1)}=\frac{(n-1)(n+2)}{n(n+1)}$$
In the following equation, you can simplify $(i+1)$ with the next $i$, except for the first (that is equals to $1$ anyway), and $(i+3)$ with the next $(i+2)$ except for the first one (that is equals to $3$)
$$\prod_{i=1}^{\infty}\left(1-\frac{2}{(i+1)(i+2)}\right)^2=\prod_{i=1}^{\infty}\left(\frac{i(i+3)}{(i+1)(i+2)}\right)^2=\prod_{i=1}^{\infty}\left(\frac{i+3}{i+2}\right)^2=\frac{1}{3^2}$$
Note that we proved in fact that
$$ \prod_{i=1}^{n}\left(1-\frac{2}{(i+1)(i+2)}\right)^2=\left(\frac{n+3}{3(n+1)}\right)^2$$
So the answer is $(b)$
| {
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Show that $\int_0^{\pi/3} \big((\sqrt{3}\cos x-\sin x)\sin x\big)^{1/2}\cos x \,dx =\frac{\pi\sqrt{3}}{8\sqrt{2}}. $ I have run a FORTRAN code and I have obtained strong evidence that
$$\int_0^{\pi/3} \!\!
\big((\sqrt{3}\cos\vartheta-\sin\vartheta)\sin\vartheta\big)^{\!1/2}\!\cos\vartheta \,d\vartheta
=\frac{\pi\sqrt{3}}{8\sqrt{2}}.
$$
In fact, it looks like my numerical integration method (the trapezoid rule) converges to the value $\dfrac{\pi\sqrt{3}}{8\sqrt{2}}$, with at least $12$ significant digits.
Any ideas how to prove this result analytically?
| OK, you can start by noting that
$$\sqrt{3} \cos{x} - \sin{x} = 2 \sin{\left ( \frac{\pi}{3}-x\right)}$$
which means that the integral is
$$ \sqrt{2} \int_0^{\pi/3} dx \, \cos{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} = \sqrt{2} I$$
You can make a substitution $x \mapsto \frac{\pi}{3}-x$ and see that
$$I = \sqrt{3} \int_0^{\pi/3} dx \, \sin{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}$$
EDIT
Integrate by parts:
$$\begin{align}I &= \left [\sin{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} \right ]_0^{\pi/3} - \int_0^{\pi/3} dx \, \sin{x} \frac{d}{dx} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}\\ &= -\frac12 \int_0^{\pi/3} dx \, \sin{x} \frac{\cos{x} \sin{\left ( \frac{\pi}{3}-x\right)} - \sin{x} \cos{\left ( \frac{\pi}{3}-x\right)}}{\sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}}\\ &= -\frac12 I + \frac12 \int_0^{\pi/3} dx \, \sin{x} \sqrt{\frac{\sin{x}}{\sin{\left ( \frac{\pi}{3}-x\right)}} } \cos{\left ( \frac{\pi}{3}-x\right)} \end{align}$$
This means that
$$\begin{align}3 I &= \int_0^{\pi/3} dx \, \frac{\cos{x}}{\sin{x}} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} \cos{\left ( \frac{\pi}{3}-x\right)}\\ &= \frac{\sqrt{3}}{2} \int_0^{\pi/3} dx \, \frac{\cos^2{x}}{\sin{x}} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}} - \frac12 I \end{align}$$
Combining again...
$$7 I = \sqrt{3} \int_0^{\pi/3} dx \sqrt{\frac{\sin{\left ( \frac{\pi}{3}-x\right)}}{\sin{x}}} - \underbrace{\sqrt{3} \int_0^{\pi/3} dx \, \sin{x} \sqrt{\sin{x} \sin{\left ( \frac{\pi}{3}-x\right)}}}_{\text{We know from above this equals } I}$$
Thus
END EDIT
$$I = \frac{\sqrt{3}}{8} \int_0^{\pi/3} dx \sqrt{\frac{\sin{x}}{\sin{\left ( \frac{\pi}{3}-x\right)}}}$$
Now, in a similar manipulation as in the evaluation of this integral, sub $u = \sin{x}/\sin{(\pi/3-x)}$ and find that the integral becomes
$$I = \frac{3}{16} \int_0^{\infty} du \frac{\sqrt{u}}{1+u+u^2}$$
This integral is very straightforward to evaluate via residues using, e.g., a keyhole contour about the positive real axis. By the residue theorem, the original integral is then
$$\sqrt{2} I = \frac{3 \sqrt{2}}{16} \frac12 i 2 \pi \left (\frac{e^{i \pi/3}}{i \sqrt{3}} - \frac{e^{i 2 \pi/3}}{i \sqrt{3}} \right ) = \frac{\pi}{8} \sqrt{\frac{3}{2}}$$
| {
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Calculus $T_1=\prod_{k=1}^{n-1} \cos\frac{k\pi}{2n}$ Calculus: $$T_1=\prod_{k=1}^{n-1} \cos\frac{k\pi}{2n}$$ and $$T_2=\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}$$
My tried:
I use Euler's formal: $$z_k=e^{i\frac{k\pi}{2n}}=\cos\frac{k\pi}{2n}+i\sin \frac{k\pi}{2n}$$
$$\to\left\{\begin{matrix}\cos\frac{k\pi}{2n}=\frac{1}{2}\left(z_k+\frac{1}{z_k}\right)\\\sin\frac{k\pi}{2n}=\frac{1}{2i}\left(z_k-\frac{1}{z_k}\right)\end{matrix}\right.$$
| First of all, $\displaystyle\cos\frac{k\pi}{2n}=\sin\left(\frac\pi2-\frac{k\pi}
{2n}\right)=\sin\left(\frac{(n-k)\pi}{2n}\right)$
So, $T_1=T_2$
Now, $\displaystyle\sin\frac{k\pi}{2n}=\sin\left(\pi-\frac{k\pi}{2n}\right)=\cos\frac{(2n-k)\pi}{2n}$
So, $\displaystyle T_2=\prod_{k=1}^{n-1}\sin\frac{k\pi}{2n}=\prod_{k=n+1}^{2n-1}\cos\frac{k\pi}{2n}$
$\displaystyle T_2^2=\prod_{k=1,k\ne n}^{2n-1}\sin\frac{k\pi}{2n}=\prod_{k=1}^{2n-1}\sin\frac{k\pi}{2n}$ as $\sin\frac{n\pi}{2n}=1$
Observe that $0<\frac{k\pi}{2n}<\pi\implies\sin\frac{k\pi}{2n}>0$
$\displaystyle\implies T_2=+\sqrt{\prod_{k=1}^{2n-1}\sin\frac{k\pi}{2n}}$
Now from this,
$$\prod_{k=1}^{m-1}\sin\frac{k\pi}m=\frac m{2^{m-1}}$$
Here $m=2n$
| {
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How to integrate fraction Hi can anybody give me some hint how to integrate this type of fractions
$$\int \frac{1}{(x^2 +a)^2} dx $$ where $ a \in \mathbb N $
Thanks
| Take the integral
$$\int \frac{dx}{x^2+a^2}$$
and we do an integration by parts:
$$\int 1\cdot\frac{dx}{x^2+a^2}=\frac{x}{x^2+a^2}+2\int\frac{x^2dx}{(x^2+a^2)^2}$$
for the last integral we add and subtract $1$ in the numerator and we find
$$\int\frac{dx}{(x^2+a^2)^2}=\frac 1 2\left(\int \frac{dx}{x^2+a^2}+\frac{x}{x^2+a^2}\right)$$
Notice that $$\int \frac{dx}{x^2+a^2}=\frac 1 a \arctan \left(\frac x a\right)+C$$
| {
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Need assistance on geometry problem
Having a really hard time solving this problem. Given:
*
*a circle of radius $a$
*an ellipse with minor axis $g$ and major axis $f$
*the ellipse is oriented so that the major axis is parallel with the vector between the circle and ellipse
*lines which are tangent to both the ellipse and the circle while crossing between them
*$d_1$ is the distance from the center of the circle to the crossing point
*$d_2$ is the distance from the center of the ellipse to the crossing point
*$d_3$ is the horizontal distance from the ellipse tangent to the crossing point
*$d_4$ is the distance from the center of the circle to the center of the ellipse
*$d_5$ is the horizontal distance from the center of the ellipse to the ellipse tangent
*$b$ is the vertical distance to the ellipse tangent
*$L$ is the distance from the crossing point to the ellipse tangent
*I have determined the following relationships
*
*$\sin(\theta_1) = \frac{a}{d_1}$
*$\sin(\theta_2) = \frac{b}{L}$
*$\theta_1$ = $\theta_2$
*$L = \sqrt{d_3^2 + b^2}$
*$d_4 = d_1 + d_2$
*$d_2 = d_3 + d_5$
*$\frac{b^2}{g^2} + \frac{d_5^2}{f^2} = 1$
*$\tan(\theta_1) = \frac{b}{d_3}$
To clarify, the known values are:
*
*$a$
*$g$
*$f$
*$d_4$
I would like to solve for $d_2$. Any assistance would be greatly appreciated.
| Here's a coordinate-based approach.
Set the center of the circle at the origin. Then the tangent line through $T(a \cos\phi, a \sin\phi)$ has $x$- and $y$-intercepts $a\sec\phi$ and $a\csc\phi$, respectively. (Note that the $x$-intercept is $d_1$ in your figure.) In intercept-intercept form, the line's equation is
$$\frac{x}{a\sec\phi} + \frac{y}{a\csc\phi}=1 \qquad \text{or, more simply,} \qquad x \cos\phi + y\sin\phi = a$$
We can determine the intersections of the line with the ellipse
$$\frac{(x-d_4)^2}{f^2}+\frac{y^2}{g^2}=1$$
by replacing $y$ with $(a - x \cos\phi)/\sin\phi$. This gives
$$\begin{align}0 &= x^2 ( f^2 \cos^2\phi + g^2 \sin^2\phi ) - 2 x ( a f^2 \cos\phi + d_4 g^2 \sin^2\phi ) &(\star)\\
&\quad+ a^2 f^2 + d_4^2 g^2 \sin^2\phi - f^2 g^2 \sin^2\phi
\end{align}$$
Because (for $\phi$ corresponding to a line that is tangent to the ellipse) the line should have just one intersection with the ellipse, we know that $(\star)$ must have only one root, $x$; thus, its discriminant must vanish:
$$-4 f^2 g^2 \sin^2\phi \; ( ( d_4^2 - f^2 + g^2 ) \cos^2\phi - 2 a d_4 \cos\phi + a^2 - g^2 ) = 0 \quad (\star\star)$$
Presumably, $f$ and $g$ are non-zero. If $\sin\phi = 0$, then our common tangent line is vertical; this requires that the ellipse be tangent to the circle, so that $d_2 = f$. In general, we must have that the final factor of $(\star\star)$ vanishes; we can solve this quadratic for $\cos\phi$:
$$\cos\phi = \frac{a d_4 \pm \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{d_4^2-f^2+g^2} \qquad (\star\star\star)$$
Recall that $d_1 = a\sec\phi = a/\cos\phi$. From $(\star\star\star)$, we have
$$\begin{align}
d_1 &= \frac{a(d_4^2-f^2+g^2)}{a d_4 \pm \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}} \\
&= \frac{a(d_4^2-f^2+g^2)(a d_4 \mp \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{a^2 d_4^2 - \left(a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)\right)} \\
&= \frac{a(a d_4 \mp \sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{a^2-g^2}
\end{align}$$
and then
$$\begin{align}
d_2 = d_4 - d_1 = \frac{-g^2 d_4 \pm a\sqrt{a^2 d_4^2 - (d_4^2-f^2+g^2)(a^2-g^2)}}{a^2-g^2}
\end{align}$$
Some notes:
*
*The "$\pm$" ambiguity accounts for the fact that there are four tangent lines (corresponding to two values of $\cos\phi$) common to the circle and ellipse. The "internal" pair cross between the centers; the "external" pair enclose the circle and ellipse. (The external pair may be parallel. Specifically, the external pair is parallel when $g=a$, which is why the formula for $d_2$ seems averse to this situation: if the lines are parallel, $d_2$ is infinite.)
*When $d_4 = a + f$ (the circle and ellipse are tangent, and the tangent line is vertical), the above formula (with $\pm=+$) reduces to $d_2 = f$. So, the formula incorporates the $\sin\phi = 0$ case from before.
*When the ellipse is a circle of radius $b$ (not the $b$ in your figure) ---that is, when $f = g = b$--- then the formula (with $\pm=+$) reduces to $d_2 = \frac{d_4 b}{a+b}$. (Correspondingly, $d_1 = \frac{d_4 a}{a+b}$.) It's left to the reader to verify that this is consistent with the figure.
*When the ellipse is a circle congruent to the given circle ---that is, when $f = g = a$--- the formula in the previous bullet reduces to $d_2 = d_4/2$. (It's problematic to directly substitute $f=a$ and $g=a$ into the $d_2$ formula.) This is clearly consistent with the figure.
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Determine if $\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$ converges or diverges. Another series I found I'm struggling with.
Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$
Ratio test and n-th root test are both inconclusive, Leibniz - criterion cannot be applied since the sequence given is not in the form of $(-1)^na_n$. I am sure the problem can be solved with the limit comparison test, though, the $n^{(...)}$ look pretty inviting after all. Let $a_n:=\frac{(-1)^nn^2+n}{n^3+1}$ then $|a_n|= \frac{n^2+n}{n^3+1}$. A try showing that the series diverges using the divergence of $\sum\frac{1}{n}$. For $n≥1$ it is clear that
$$ \frac{n^3+n^2}{n^3+1} ≥ 1 $$ dividing by $n$ yields:
$$ \frac{n^2+n}{n^3+1} = |a_n| ≥ \frac{1}{n}$$
Thus the series diverges. (?)
EDIT: Hints you gave me yield:
$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1} = \sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} +
\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ With the first series converging by Leibniz-theorem and the second by limit comparison test with $\frac{1}{n^2}$.
| The series
$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1}$$ converges by Leibniz criterion, in addition the series
$$\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ converges by the comparison test.
Hence
$$\sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} + \sum_{n=1}^{\infty}\frac{n}{n^3+1} = \sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$ also converges
| {
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Finding a least common multiple (LCM) My Algebra 2 book explains how to find a least common multiple:
Find the least common multiple of $4x^2 - 16$ and $6x^2 - 24x + 24$.
Solution
Step 1 Factor each polynomial. Write numerical factors as products of primes.
$4x^2 - 16 = 4(x^2 - 4) = (2^2)(x + 2)(x - 2)$
$6x^2 - 24x + 24 = 6(x^2 - 4x + 4) = (2)(3)(x-2)^2$
Step 2 Form the LCM by writing each factor to the highest power it occurs in either polynomial.
LCM = $(2^2)(3)(x + 2)(x - 2)^2 = 12(x + 2)(x - 2)^2$
I don't understand their wording, and I don't want to go onto the rest of my assignment that includes finding the least common denominators until I know how to do it correctly, instead of going back and doing it over when I find out I'm doing it wrong.
| It's kind of a weird way of saying it. $$\mathrm{lcm}(P(x),Q(x)) = \frac{P(x)\cdot Q(x)}{\gcd(P(x),Q(x))}$$
Where $$\gcd(P(x),Q(x))$$ is just the product of the terms that appear in both factorizations.
For instance $$\mathrm{lcm}(((x+1)(x+1)(x-1)),((x+1)(x-1)(x-1)))=\frac{(x+1)^3(x-1)^3}{(x+1)(x-1)}=(x+1)^2(x-1)^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/648118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Use the disc method to find the volume of solid of revolution. Use the disc method to find the volume of solid of revolution obtained by revolving the the given region around the x-axis.
$R\quad is\quad bounded\quad by\quad the\quad graphs\quad of\quad k(x)\quad =\quad { x }^{ 2 }+x+2\quad and\quad m(x)\quad =\quad 2x+4$
My try :
$ V\quad =\quad \int _{ a }^{ b }{ f(x)^{ 2 } } -g(x)^{ 2 }\quad dx\\ \\ k(x)^{ 2 }\quad =\quad { x }^{ 4 }+2{ x }^{ 3 }+4{ x }^{ 2 }+2x+4\\ m(x)^{ 2 }\quad =\quad 4{ x }^{ 2 }+24x+16\\ \\ \pi \int _{ -1 }^{ 2 } \left| { x }^{ 4 }+2{ x }^{ 3 }-22x-12 \right| \quad dx\\ \\ \left| { \pi \left[ \frac { 1 }{ 4 } { x }^{ 4 }+\frac { 1 }{ 2 } { x }^{ 4 }+11{ x }^{ 2 }-12x \right] }_{ x=-1 }^{ x=2 } \right| \\ \\ \left| \pi \left[ 4+8-44-24 \right] \right| -\left| \pi \left[ \frac { 1 }{ 4 } +\frac { 1 }{ 2 } -11+12 \right] \right| \\ \\ \frac { 217\pi }{ 4 } $
The answer in the book is $\frac{369\pi}{10}$
So what did i do wrong ?
--Correction
$ So\quad i\quad corrected\quad the\quad { k }^{ 2 }\quad and\quad m^{ 2 }\quad \\ \\ k^{ 2 }={ x }^{ 4 }+2{ x }^{ 3 }+5{ x }^{ 2 }+4x+4\\ { m }^{ 2 }=\quad { 4x }^{ 2 }+16x+16\\ \\ \\ \\ \pi \int _{ -1 }^{ 2 }{ \left| { x }^{ 4 }+2{ x }^{ 3 }+{ x }^{ 2 }-12x-12 \right| \quad dx } \\ \\ \left| { \pi \left[ \frac { 1 }{ 5 } { x }^{ 5 }+\frac { 1 }{ 2 } { x }^{ 4 }+\frac { 1 }{ 3 } { x }^{ 3 }-6{ x }^{ 2 }-12x \right] }_{ x=-1 }^{ x=2 } \right| \\ \\ \left| \pi \left[ \frac { 32 }{ 5 } +8+\frac { 8 }{ 3 } -24-24 \right] \right| -\left| \pi \left[ \frac { -1 }{ 5 } +\frac { 1 }{ 2 } -\frac { 1 }{ 3 } -6+12 \right] \right| \\ \\ \frac { 464\pi }{ 15 } -\frac { 179\pi }{ 30 } \\ \frac { 749\pi }{ 30 } $
still wrong answer.
| You may want to do these without the absolute value brackets, as I really think that is confusing the issue (and isn't even really correct...). The line is the "upper function", so the evaluation should be
$$ \pi \ \left[ \ ( - \frac{32}{5} - 8 - \frac{8}{3} + 24 + 24 ) \ - \ (\frac{1}{5} - \frac{1}{2} + \frac{1}{3} + 6 - 12 ) \right] $$
$$ = \ \pi \ ( - \frac{33}{5} + \frac{1}{2} - 8 - 3 + 18 + 36 ) \ = \ \pi \ (43 \ - \ \frac{61}{10} ) \ . $$
You need to be really careful with this evaluation, because the second term is evaluated at negative 1 , so you have to make sure you subtract the odd powers of $ \ x \ $ in the right way. (I had to re-do this three times myself because there are so many opportunities for sign errors...)
Sign Errors: the #1 cause of math grief, from beginners to professionals!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/648288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$AX=B$ solve for $X$ ....... in MATRIX $$ 2x - 3y + 4z = -19\\
6x + 4y - 2z =8 \\
x + 5y + 4z = 23
$$
what I have done so far is I put the nubmer and $x, y $ and $ z$ in matrix form:
$$
\begin{bmatrix}
2 & -3 & 4\\
6 & 4 &-2\\
1 & 5 & 4
\end{bmatrix}
\begin{bmatrix}
x\\
y\\
z
\end{bmatrix}=
\begin{bmatrix}
-19\\
8\\
23
\end{bmatrix}
$$
step 2: I don't know where to go from here
| You can also use Cramer's Rule:
$$
x=\frac{\left|\begin{array}{r}\color{#C00000}{-19}&-3&4\\\color{#C00000}{8}&4&-2\\\color{#C00000}{23}&5&4\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=-2
$$
$$
y=\frac{\left|\begin{array}{r}2&\color{#C00000}{-19}&4\\6&\color{#C00000}{8}&-2\\1&\color{#C00000}{23}&4\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=5
$$
$$
z=\frac{\left|\begin{array}{r}2&-3&\color{#C00000}{-19}\\6&4&\color{#C00000}{8}\\1&5&\color{#C00000}{23}\end{array}\right|}{\left|\begin{array}{r}2&-3&4\\6&4&-2\\1&5&4\end{array}\right|}=0
$$
In the numerator, replace the column in the matrix corresponding to the given variable by the column of results. Note that the bars denote the determinant of the matrix.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/649080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(\displaystyle\frac{1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}.
\end{align*}$$
Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But
$$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider
$$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$
So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$
It looks too easy, is there any crucial lack?
$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
| Here is an elementary way to compute the integral.
First, let us prove some initial results.
*
*Making the substitution $x \mapsto \pi - x$ yields
$I(a) = \int^\pi_0 \log \left (1 + 2a\cos x + a^2 \right ) \, dx = I(-a)$
so that
$$I(a) = I(-a). \tag{$\dagger$}$$
*Then, consider
$$\begin{align*}
I(a) + I(-a)
&= \int^{\pi}_{0}\log \! \Big ( \left (1 - 2a\cos x + a^2 \right ) \left (1 + 2a\cos x + a^2 \right ) \Big) \> dx\\
&= \int^{\pi}_{0}\log \! \Big ( \left (1 + a^2 \right )^2 - \left (2a\cos x \right )^2 \Big) \> dx.\\
\end{align*}$$
Using double angle formulae produces
$$\begin{align*}
I(a) + I(-a)&= \int^{\pi}_{0}\log \left ( 1 + 2a^2 + a^4 - 2a^2 \left ( 1 + \cos 2x \right ) \right) \, dx\\
&= \int^{\pi}_{0}\log \left ( 1 - 2a^2\cos 2x + a^4 \right) \, dx,\\
\end{align*}$$
so we may let $x \mapsto \frac{1}{2}x$ to give
$$\begin{align*}
I(a) + I(-a) &= \frac{1}{2}\int^{2\pi}_{0}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, dx.\\
\end{align*}$$
We can then split the integral at $\pi$ and set $x \mapsto 2\pi - x$ for the second integral:
$$\begin{align*}
I(a) + I(-a) &= \frac{1}{2} I(a^2) + \frac{1}{2}\int^{2\pi}_{\pi}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, dx\\
&= \frac{1}{2} I(a^2) + \frac{1}{2}\int^{\pi}_{0}\log \left ( 1 - 2a^2\cos x + a^4 \right) \, dx\\
&= I(a^2).
\end{align*}$$
We thus have (applying $(\dagger)$)
$$I(a)= \frac{1}{2}I(a^2). \tag{$\star$}$$
It follows from $(\star)$ that $I(0) = 0$ and $I(1) = 0$.
Consider the case when $0 \le a < 1$. We may use $(\star)$ iteratively $n$ times to write
$$I(a) = \frac{1}{2^n} I \left ( a^{2^{n}} \right ). $$
Setting $n \to \infty$ allows $\frac{1}{2^n} \to 0$ and $a^{2^{n}} \to 0$ so that $I \left ( a^{2^{n}} \right ) \to 0$ which gives the result
$$ I(a) = 0. $$
When $a > 1$, it follows that $0 < 1/a < 1$ and consequently $I(1/a) = 0$. We have
$$\begin{align*}
I(a) &= \int^\pi_0 \log \! \Big ( a^2 \left ((1/a)^2 + (1/a)\cos x + 1 \right ) \Big ) \> dx\\
&= 2\pi\log(a) + I(1/a)\\
&= 2\pi\log\left(a\right).
\end{align*}$$
We could use $(\dagger)$ to extend the result to negative $a$, obtaining the final solution valid for all real $a$,
$$I(a) =
\begin{cases}
0 &\text{if } |a| \le 1;\\
2\pi\log|a| &\text{otherwise}.
\end{cases}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 3
} |
Solve $2000x^6+100x^5+10x^3+x-2=0$ One of the roots of the equation $2000x^6+100x^5+10x^3+x-2=0$ is of the form $\frac{m+\sqrt{n}}r$, where $m$ is a non-zero integer and $n$ and $r$ are relatively prime integers.Then the value of $m+n+r$ is?
Tried to use the fact that another root will be $\frac{m-\sqrt{n}}r$ as coefficients are rational but there are six roots and using sum and product formulas would allow many variables in the equations.
| The usual trick is to divide through by $x^3.$ Then notice that taking $w = 10x - \frac{1}{x}$ seems to allow writing the thing, and we get
$$ 2 w^3 + w^2 + 60 w + 30 = 0. $$
This has a rational root, namely $-1/2,$ and factors as
$$ (2w+1)(w^2 + 30) $$
Setting
$$ 10 x - \frac{1}{x} = -\frac{1}{2} $$
leads to
$$ 20 x^2 + x - 2 = 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/651024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Evaluating $\int_{\pi/3}^{\pi/2}\frac{1}{\sin\left(x\right)-\cos\left(x\right)+1} \operatorname dx$ $$\int \limits_{\pi/3}^{\pi/2}\dfrac{1}{\sin\left(x\right)-\cos\left(x\right)+1} \operatorname dx$$
If I try solving this integral with universal substitutions: $$\tan\left(\frac{x}{2}\right)=t;\:\sin\left(x\right)=\frac{2t}{1+t^2};\:\cos\left(x\right)=\frac{1-t^2}{1+t^2};\:dx=\frac{2dt}{1+t^2}$$
$$\implies 2\int \limits_{\frac{\sqrt{3}}{3}}^1 \frac{\left(\dfrac{dt}{ 1+t^2}\right)}{ \left(\dfrac{2t-1+t^2+1+t^2}{1+t^2}\right)} $$
I get: $\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)}$
The final solution reached according to my calculations is: $-\ln\left(\frac{\sqrt{3}}{3}\right)-\left(\ln\left(2\right)+\ln\left(\frac{\sqrt{3}}{3}\right)+1\right)$
That is incorrect taking into account the answer given by Symbolab: $\:\ln \left(\frac{1+\sqrt{3}}{2}\right)$
Could the mistake be the approach or something with the calculations?
| What you have got
$$\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)}$$
is certainly correct. Now,
\begin{align}\displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:\dfrac{dt}{t\left(t+1\right)} & = \displaystyle \int \limits_{\frac{\sqrt{3}}{3}}^1\:dt\left(\frac{1}{t}-\frac{1}{t+1}\right)\\ & = \log t \biggl|_{\sqrt{3}/3}^1 - \log(1+t)\biggl|_{\sqrt{3}/3}^1 \\ & = \log 1 - \log\frac{\sqrt{3}}{3} -\left(\log 2 - \log\left(1+\frac{\sqrt{3}}{3}\right)\right) \end{align}
and this simplifies to the correct answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/652582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
is there a nicer way to $\int e^{2x} \sin x\, dx$? I have to solve this:
$\int e^{2x} \sin x\, dx$
I managed to do it like this:
let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$
$\therefore \dfrac{du}{dx}_1 = \cos x \space $ and $\space v_1 = \frac{1}{2}e^{2x}$
If I substitute these values into the general equation:
$\int u\dfrac{dv}{dx}dx = uv - \int v \dfrac{du}{dx}dx$
I get:
$\int e^{2x} \sin x dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{2}\int e^{2x}\cos x\, dx$
Now I once again do integration by parts and say:
let $u_2 = \cos x$ and let $\dfrac{dv}{dx}_2 = e^{2x}$
$\therefore \dfrac{du}{dx}_2 = -\sin x$ and $v_2 = \frac{1}{2}e^{2x}$
If I once again substitute these values into the general equation I get:
$\int e^{2x}\sin x dx =\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}\int e^{2x}\sin x dx$
$\therefore \int e^{2x}\sin x dx = \frac{4}{5}(\frac{1}{2}e^{2x}\sin x -
\frac{1}{4}e^{2x}\cos x) + C$
$\therefore \int e^{2x}\sin x\, dx = \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + C$
I was just wondering whether there was a nicer and more efficient way to solve this?
Thank you :)
| $$
\begin{align*}
\\ \int e^{2x}\sin x dx &= \Im \int e^{2x}(\cos x + i\sin x) dx
\\ &= \Im \int e^{(2 + i)x}dx
\\ &= \Im \frac{e^{(2 + i)x}}{2+i} + C
\\ &= \Im \frac15 e^{2x}(\cos x + i\sin x)(2-i) + C
\\ &= \frac15 e^{2x}(2\sin x - \cos x) +C
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/657389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Simplify the surd expression. Simplify the surd.
$(2\sqrt 3 + 3\sqrt 2)^2$
I know I should us this formula: $(a^2+2ab+b^2)$
But this gets complicated later. Please explain. :(
| We are given:
$$(2\sqrt 3 + 3\sqrt 2)^2$$
Expand:
$$(2\sqrt 3 + 3\sqrt 2)(2\sqrt 3 + 3\sqrt 2)$$
Multiply using techniques such as the FOIL method:
$$(2\sqrt 3 + 3\sqrt 2)(2\sqrt 3 + 3\sqrt 2)=4\sqrt9+6\sqrt6+6\sqrt6+9\sqrt4$$
Simplify:
$$4(3)+12\sqrt6+9(2)\\=12+18+12\sqrt6\\=30+12\sqrt6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/658736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Limit $\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$ The limit is
$$\lim\limits_{x\to0}\frac1{\ln(x+1)}-\frac1x$$
The problem is I don't know if I can calculate it normally like with a change of variables or not. Keep in mind that I'm not allowed to use L'Hôpital's rule nor the $\mathcal O$-notation.
| It is bit difficult to avoid LHR or series expansions here. I present here a technique which is almost like using series expansion, but a bit simpler conceptually. For this purpose I need to use the standard definition of $\log x$ as $\int_{1}^{x}(1/t)\,dt$.
Let us assume that $0 < t < 1$. Then it can be checked using algebra that $$1 - t < \frac{1}{1 + t} < 1 - t + t^{2}$$ If $0 < x < 1$ then upon integrating above inequality in the interval $[0, x]$ we get $$x - \frac{x^{2}}{2} < \log (1 + x) < x - \frac{x^{2}}{2} + \frac{x^{3}}{3}$$ or $$\dfrac{1}{x - \dfrac{x^{2}}{2} + \dfrac{x^{3}}{3}} < \dfrac{1}{\log(1 + x)} < \dfrac{1}{x - \dfrac{x^{2}}{2}}$$ or $$\frac{6}{6x - 3x^{2} + 2x^{3}} < \frac{1}{\log(1 + x)} < \frac{2}{2x - x^{2}}$$ Subtracting $(1/x)$ from each term in above inequality we get (after some simplification) $$\frac{3 - 2x}{6 - 3x + 2x^{2}} < \frac{1}{\log(1 + x)} - \frac{1}{x} < \frac{1}{2 - x}$$ Taking limits as $x \to 0^{+}$ and using Squeeze theorem we get $$\lim_{x \to 0^{+}}\frac{1}{\log(1 + x)} - \frac{1}{x} = \frac{1}{2}$$ To handle the case when $x \to 0^{-}$ we need to substitute $x = -y$ to get $$\frac{1}{\log(1 + x)} - \frac{1}{x} = \frac{1}{\log(1 - y)} + \frac{1}{y}$$ and $y \to 0^{+}$.
Next we can see that if $0 < y < 1$ then $$\log(1 - y) = \log(1 - y^{2}) - \log(1 + y)$$ Using $\log(1 - y^{2}) < -y^{2}$ and $\log(1 + y) > y - (y^{2}/2)$ we can see that $$\log(1 - y) < -y - \frac{y^{2}}{2}$$ or $$\log(1 - y) + y < -\frac{y^{2}}{2}\,\,\,\cdots (1)$$ Again we can see that $$\frac{y^{2}}{y^{2} - 1} < \log(1 - y^{2})$$ and $$\log (1 + y) < y - \frac{y^{2}}{2} + \frac{y^{3}}{3}$$ so that $$\log(1 - y) > \frac{y^{2}}{y^{2} - 1} - y + \frac{y^{2}}{2} - \frac{y^{3}}{3}$$ or $$\frac{y^{2}}{y^{2} - 1} + \frac{y^{2}}{2} - \frac{y^{3}}{3} < \log(1 - y) + y \,\,\,\cdots (2)$$ From the equations $(1)$ and $(2)$ we can see that $$\frac{1}{y^{2} - 1} + \frac{1}{2} - \frac{y}{3} < \frac{\log(1 - y) + y}{y^{2}} < -\frac{1}{2}$$ Taking limits as $y \to 0^{+}$ and using Squeeze theorem we get $$\lim_{y \to 0^{+}}\frac{\log(1 - y) + y}{y^{2}} = -\frac{1}{2}$$ It is now easy to observe that
$\displaystyle \begin{aligned}\lim_{y \to 0^{+}}\frac{1}{\log(1 - y)} + \frac{1}{y} &= \lim_{y \to 0^{+}}\frac{\log(1 - y) + y}{y\log(1 - y)}\\
&= \lim_{y \to 0^{+}}\dfrac{\log(1 - y) + y}{-y^{2}\cdot\dfrac{\log(1 - y)}{-y}}\\
&= \lim_{y \to 0^{+}}\dfrac{\log(1 - y) + y}{-y^{2}\cdot 1}\\
&= -\lim_{y \to 0^{+}}\dfrac{\log(1 - y) + y}{y^{2}}\\
&= \frac{1}{2}\end{aligned}$
The above derivation is bit lengthy because it establishes the inequalities satisfied by $\log (1 + x)$ function using integration and their extensions to negative values of $x = -y$ by further algebraic manipulation. This method is the conceptually simpler (but taking more space and calculations) equivalent of using the Taylor's expansion $\log(1 + x) = x - x^{2}/2 + x^{3}/3 - \cdots$ In my view it is better to use the Taylor's expansion or LHR for such problems. However even when we apply Taylor or LHR it is better to change the problem into a different form as follows:
$\displaystyle \begin{aligned}\lim_{x \to 0}\frac{1}{\log(1 + x)} - \frac{1}{x} &= \lim_{x \to 0}\frac{x - \log(1 + x)}{x\log(1 + x)}\\
&= \lim_{x \to 0}\dfrac{x - \log(1 + x)}{x^{2}\cdot\dfrac{\log(1 + x)}{x}}\\
&= \lim_{x \to 0}\dfrac{x - \log(1 + x)}{x^{2}\cdot 1}\\
&= \lim_{x \to 0}\dfrac{x - \log(1 + x)}{x^{2}}\\\end{aligned}$
Doing this above simplification avoids taking reciprocal of a series (if you use Taylor's expansion) and also avoids complicated differentiation (if you use LHR).
| {
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"url": "https://math.stackexchange.com/questions/659037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Prove that $\sum_{k=0}^{\infty} (k-1)/2^k = 0$ How to prove that this series converges, and that the limit is 0 ?
| $$
\sum_{k=0}^\infty \frac{k-1}{2^k} = -1 + \frac{1}{2}\sum_{k=1}^\infty \frac{k-1}{2^{k-1}}.
$$
If we call $n=k-1$, then
$$
\sum_{k=0}^\infty \frac{k-1}{2^k} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{n}{2^n} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n}+ \frac{1}{2}\sum_{n=0}^\infty \frac{{n-1}}{2^n}.
$$
Therefore
$$
\frac{1}{2}\sum_{k=0}^\infty \frac{{k-1}}{2^k} = -1 + \frac{1}{2}\sum_{n=0}^\infty \frac{1}{2^n} = 0.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/660852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplify the expression where $x>y>0$ Simplify $$\frac{\sqrt {x^2+y^2}+x}{y+\sqrt {x^2-y^2}}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}$$
Help please, I tried but the answer doesn't match. I did it by multiplying of course and then normally simplifying.
I tried and what came is $\dfrac{-2xy -y^2}{2xy + x^2}$ and the correct answer is $\dfrac{y^2}{2y^2-x^2}$
| $$\frac{\sqrt {x^2+y^2}+x}{y+\sqrt {x^2-y^2}}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}= \frac{x + \sqrt {x^2+y^2}}{\sqrt {x^2-y^2}+y}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}$$
Now you've got a function of the form $$\dfrac{(a+b)(a- b)}{(c+d)(c- d)} = \dfrac{a^2 - b^2}{c^2 - d^2}$$
The rest is then simplification.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Computing this limit $$\lim_{n\to\infty}{\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4}}$$
At first glance, we see that it's an indeterminate form ($\infty-\infty$). Here are my tries:
I.) I tried to form $a^2-b^2$ in the numerator, where $\cdots$ represents something that $\to 0$:
$$\frac{n^2(\sqrt[3]{1+\cdots}-1)-3n-4}{n(\sqrt[3]{1+\cdots}+\sqrt{1+\cdots})}$$
II.) I tried to form $a^3-b^3$ in the numerator, same as I.):
$$\frac{n^3(1-\sqrt[3]{1+\cdots})+5n^2+6}{n^2(\sqrt{1+\cdots}+\sqrt{1+\cdots}\sqrt{1+\cdots}+1)+3n+4}$$
What method can I apply here?
Thank you.
| Using series:
\begin{align*}
\sqrt[3]{n^3+5n^2+6}-\sqrt{n^2+3n+4} &= n\left( \left(1+\frac{5}{n}+\frac{6}{n^3}\right)^{1/3} - \left(1+\frac{3}{n}+\frac{4}{n^2}\right)^{1/2} \right) \\
&= n\left( 1+\frac{5}{3n} + o\left(\frac{1}{n}\right) - \left(1+\frac{3}{2n}+o\left(\frac{1}{n}\right)\right) \right) \\
&= n\cdot\frac{1}{6n} + o(1) \xrightarrow[n\to\infty]{}\frac{1}{6}
\end{align*}
| {
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Evaluate $\sum_{k=1}^{\infty} (2k+1)\,x^{2k}$,$\,$ for $|x| < 1$ Let $f(x)$ be $\sum_{k=1}^{\infty} x^{2k+1}$. This sum equals $$x + x^3 + x^5 + \dots - x= x(1 + x^2 + (x^2)^2 + \dots) - x = \frac{x}{1-x^2} - x = \frac{x^3}{1-x^2}$$
Now, okay, the original sum is the derivative of $f$, but I don't think it is the conclusion the author expected...?
| Yes you are allowed to differentiate, as $f(x)$ is differentiable for every $|x|<1$.
Then
$$
f'(x)=\frac{3x^2}{1-x^2}+\frac{2x^4}{(1-x^2)^2}.
$$
| {
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"source": "stackexchange",
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Is it true that $(2^n+n^2)(n^3+3^n)$ is $O(6^n)$? $(2^n+n^2)$ is $O(2^n)$ and $(n^3+3^n)$ is $O(3^n)$, therefore I conclude that $(2^n+n^2)(n^3+3^n)$ is $O(2^n*3^n)=O(6^n)$
| Your answer is correct, yet if you were not 100% sure you can always check using the definition of "big O":
$$\limsup_{n\rightarrow\infty}\frac{(2^n+n^2)(n^3+3^n)}{6^n}=\limsup_{n\rightarrow\infty}\frac{n^5+n^32^n+n^23^n+6^n}{6^n}=\limsup_{n\rightarrow\infty}(n^5(1/6)^n+n^3(1/3)^n+n^2(1/2)^n+1)=1$$
$$\Rightarrow (2^n+n^2)(n^3+3^n)=\mathcal{O
}(6^n)$$
| {
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Evaluate integral as a logarithm plus an arctangent. Evaluate the integral as a logarithm plus an arctangent.
$$ \int \frac{x}{3x^2-18x+45} \ dx $$
I just completed the square and couldn't continue.
$$ \int \frac{x}{3(x-3)^2+18} \ dx $$
Fixed a typo $18$ changed to $18 x$
| $$\begin{align}
\int \frac{x}{3(x-3)^2+18} \ dx & = \frac 1 3 \cdot \frac 1 2 \int\frac{2(x-3)}{(x-3)^2 + 6} + \frac 1 3 \int \frac{3}{(x-3)^2+6}\,dx.
\end{align}
$$
The first integral can be handled by the substitution $u=(x-3)^2+6$ and $du=2(x-3)\,dx$.
Then:
$$
\int \frac{dx}{(x-3)^2+6} = \frac 1 6 \int \frac{dx}{\frac{(x-3)^2}{6}+ 1}.
$$
Let $u = \dfrac{x-3}{\sqrt{6}}$ and then $dx = \sqrt{6} \, du$. You get an arctangent.
| {
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Solving $\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$ I try to solve this equation:
$$\sqrt{x+2}+\sqrt{x-3}=\sqrt{3x+4}$$
So what i did was:
$$x+2+2*\sqrt{x+2}*\sqrt{x-3}+x-3=3x+4$$
$$2*\sqrt{x+2}*\sqrt{x-3}=x+5$$
$$4*{(x+2)}*(x-3)=x^2+25+10x$$
$$4x^2-4x-24=x^2+25+10x$$
$$3x^2-14x-49$$
But this seems to be wrong! What did i wrong?
| first step on the right hand side should be $3x+5$ not $3x+4$
Then proceeding the same way:
$$
\begin{split}
4(x^2-x-6) &= x^2 + 12x + 36 \\
3x^2 - 16x - 60 &= 0
\end{split}
$$
Since the discriminant of that equation is $16^2 - 4 \cdot 3 \cdot 60 < 0$ there are no solutions.
UPDATE you changed the problem now, so you get
$4(x^2-x-6) = x^2 + 10x + 25$ which results in a quadratic $3x^2 - 14x - 49 = 0$ so $(x-7)(3x+7)=0$ and $x=7$ or $x = -7/3$. The second solution is impossible (drives roots to be complex) so $x=7$ is the only one.
| {
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For which $x$ does the series $∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n})\, x^n$ converge? Determine for what value of $x$ the series converges
$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}\right) x^n $$
Observe that
$∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}) x^n =∑_{n=1}^∞ (∑_{n=1}^∞\frac{1}{n}) x^n$
Can I consider $(\sum_{n=1}^\infty\frac{1}{n})$ as $a_n$ and use the ratio test here or I must use the Hadamard?
| Note that
$$
1\le 1+\frac{1}{2}+\cdots+\frac{1}{n}\le \log (n+1),
$$
and hence, using for example the comparison test, the radius of convergence of the series
$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}\right) x^n $$
is $r=1$, since the radius of convergence of both series
$$
\sum_{n=0}^\infty x^n, \quad \sum_{n=0}^\infty \log n\, x^n
$$
is equal to $1$.
| {
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Partial fractions: why does $\int dt \implies t + C$ I am working on a partial fraction problem here, I understand everything in the problem except $t+C$, so I'd like to know where did the $t+C$ come from ?
I want to solve this integral
$$
\int \frac{dy}{(y+2)(1-y)} = \int dt
$$
$$\begin{align}
1 &= \frac{A}{y+2} + \frac{B}{1-y} \\
1 &= A(1-y) + B(y+2)
\end{align}$$
Let $y=1$, then $1=B(3)$ and $B=1/3$.
Let $y=-2$, then $1=A(3)$ and $A=1/3$
$$\begin{align}
\int \frac{1/3}{y+2} + \frac{1/3}{1-y} \;dy &= \int dt\\
\frac{1}{3}(\ln|y+2| - \ln|1-y|) &= \color{red}{t+C} & \text{?}\\
\ln\left|\frac{y+2}{1-y}\right| &=3(t+C)\\
\frac{y+2}{1-y} = Ce^{3t}
\end{align}$$
| Your original equation is $$\int \frac{dy}{(y+2)(1-y)} = \int \,dt$$
$t + C$ comes from integrating the righthand side of the original equation: $$\int \,dt = t + C$$
The $C$ is the constant of integration.
| {
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How can I check the convergence of the sequence? Does it diverge? How can I check the convergence of the sequence $\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}$? I think that it diverges,because it is bounded below from $\frac{n(n+1)}{2\sqrt{n^2+n}} $ and above from $\frac{n(n+1)}{2\sqrt{n^2+1}}$..Is this correct?
| HINT: you can use Squeeze Theorem.
$\frac{1}{\sqrt{n^2+n}}+\frac{2}{\sqrt{n^2+n}}+\cdots+\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}\leq\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+1}}+\cdots+\frac{n}{\sqrt{n^2+1}}$.
| {
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Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.
| Function $f(x)=\frac{x}{1+x}$ is increasing (it is easy to check first derivative). Therefore $f(a)<f(b+c)$, for $a<b+c$. Because of that, we have $$\frac{a}{1+a}<\frac{b+c}{1+b+c}=\frac{b}{1+b+c}+\frac{c}{1+b+c},$$ from where we get (since $a,b,c>0$):
$$\frac{a}{1+a}<\frac{b}{1+b}+\frac{c}{1+c}.$$
| {
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Generalization of "Sum of cube of any 3 consecutive integers is divisible by 3" I have this question posted by professor in graduate Number Theory class. First he asked for proof that the sum of cube of 3 consecutive integers is divisible by 3, which is very easy to prove, but then he continued by asking to prove its generalization, ie., n | 1^n + 2^n + 3^n + ... + n^n.
Here you can easily find a counterexample that if n is even, the generalization fails. But if n is odd, looks like it works. I tried using mathematical induction but did not go anywhere. Then I tried using Binomial Expansion, Pascal Triangle, and using representation of consecutive numbers as ... (a-3), (a-2), (a-1), a, (a+1), (a+2), (a+3), ... in order to cancel out, but still did not go anywhere.
I would appreciate any help. Thank you for your time.
| Since this is a graduate level number theory class, I think it's safe to assume that you are familiar with modulo arithmetic?
Given any list of $n$ consecutive integers, $a, a+1, a+2, \dots, a+n-1$, modulo $n$ this list is equivalent to $0,1,2,3,\dots,n-1$ modulo $n$. (Note that I am not saying $a \equiv 0 \pmod{n}$). This list can be rewritten as:
$1 \equiv 1 \pmod{n}$
$2 \equiv 2 \pmod{n}$
$3 \equiv 3 \pmod{n}$
$\dots$
$\dfrac{(n-1)}{2} \equiv \dfrac{(n-1)}{2} \pmod{n}$
$n-1 \equiv -1 \pmod{n}$
$n-2 \equiv -2 \pmod{n}$
$n-3 \equiv -3 \pmod{n}$
$\dots$
$\dfrac{(n+1)}{2} \equiv -\dfrac{(n-1)}{2} \pmod{n}$
Since $n$ is odd, exponentiation preserves the sign. And so
$$0^n + 1^n + 2^n + \dots + \left(\dfrac{n-1}{2}\right)^n + \left(\dfrac{n+1}{2}\right)^n + \dots + (n-2)^n + (n-1)^n + n^n$$
is equivalent to
$$1^n + 2^n + \dots + \left(\dfrac{n-1}{2}\right)^n - \left(\dfrac{n-1}{2}\right)^n + \dots - 2^n - 1^n$$
modulo $n$, and so the sum becomes $0$ modulo $n$. Note that the exponent could be replaced by any odd integer and the statement will still hold.
EDIT: Here is the example you requested in the comments.
Let's say we have the list 5, 6, 7, 8, 9, 10, 11, with $n=7$.
Okay, so the first thing I will do is to find their representatives in $\mod 7$ between $0$ and $6$ inclusive.
So,
5, 6, 7, 8, 9, 10, 11 becomes 5, 6, 0, 1, 2, 3, 4.
Now, let's look at $(n-1)/2$. For $n=7$, this number is 3. That is the cut off for the positive terms. The rest of them I turn them into negatives:
$5, 6, 7, 8, 9, 10, 11$ becomes
$5, 6, 0, 1, 2, 3, 4,$ which is
$7-2, 7-1, 0, 1, 2, 3, 7-3$, which becomes
$-2, -1, 0, 1, 2, 3, -3$.
Now, take any odd power of these numbers, sum, you get 0 in modulo 7.
Technically, I could've gone directly from the original list to $-2, -1, 0, 1, 2, 3, -3$, without going through the intermediate step, but I wanted to illustrate how the proof applies to this particular example.
| {
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Find all the $a$ such $539|a3^{253}+5^{44}$ This is what i thought:
Given that $539|a3^{253}+5^{44}$ then $11|a3^{253}+5^{44}$ and $7^2|a3^{253}+5^{44}$
using congruences I get:
$$a3^{253}+5^{44} \equiv 0 \pmod{7^2}$$
and
$$a3^{253}+5^{44} \equiv 0 \pmod{11}$$
then as a condition, I get $(a , 11)\not=1$ and $(a, 49)\not= 49$
given that $(3 , 7^2) = 1$ and $(5 , 7^2) = 1$
$$a3+5^{2} \equiv 0 \pmod{7^2}$$
$$a3 \equiv -5^{2} \equiv 24 \pmod{7^2}$$
$$a \equiv 8 \pmod{7^2}$$
Doing the same for $11$
$$a3^{253}+5^{44} \equiv 0 \pmod{11}$$
$$a3^{3}+5^{4} \equiv 0 \pmod{11}$$
$$a3^{3} \equiv 2 \pmod{11}$$
$$a \equiv 7 \pmod{11}$$
then whith this two conditions, using the Chinese remainder theorem:
$$a \equiv 502 \pmod{539}$$
Is this correct? what happens if $(a, 49) = 7$?
| $539=7^2\cdot11$. Per Fermat's Little Theorem, we have $3^{10}\equiv1\text{ mod }11\iff3^{253}\equiv3^3\equiv5\text{ mod }11$ and $5^{10}\equiv1\text{ mod }11\iff5^{44}\equiv5^4\equiv9\equiv-2\text{ mod }11\iff5a-2\equiv0\text{ mod }11\iff5a\equiv2$ $\text{mod }11\iff a\equiv7\text{ mod }11$: The first condition.
By brute force, $5^{21}\equiv-1\text{ mod }49\iff5^{44}\equiv5^2=25\text{ mod }49$, and $3^{42}\equiv1\text{ mod }49\iff3^{253}$ $\equiv3^1=3\text{ mod }49\iff3a+25\equiv0\text{ mod }49\iff3a\equiv24\text{ mod }49\iff a\equiv8\text{ mod }49$: The second condition.
$11x+7=49y+8\iff11x=49y+1\iff11(x-4y)=5y+1\iff y=2$ and $x=9$, thus $a\equiv106\text{ mod }539$. Indeed, the computer can confirm the result.
| {
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Finding the real solutions to $16^{x^{2} + y } + 16^{y^{2}+ x} = 1$ We have , $16^{x^{2} + y } + 16^{y^{2}+ x} = 1$ , then we have to find all the real values of $x$ and $y$.I tried this question but i am not able to proceed because I am not able to simplify this expression to an extent that it could be solved.
| $x^2 + y^2 + x + y = (x + 1/2)^2 + (y + 1/2)^2 - 1/2 \geq -1/2$ and equality occurs only when $x = y = - 1/2$.
Using AM-GM inequality $16^{x^2 + y} + 16^{y^2 + x} \geq 2\cdot\sqrt{16^{x^2+y^2+x+y}} \geq 2\cdot16^{-1/4} = 1$ and equality occurs only when $x=y=-1/2$
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility test for $4$
Claim: A number is divisible by $4$ if and only if the number formed by the last two digits is divisible by $4$.
Here's where I've gotten so far.
Let $x$ be an $(n+1)$-digit number. So $x= a_na_{n-1} \dots a_2a_1a_0$. If $a_1 = 0$ and $a_0 =0$, then $x$ is a multiple of $100$ and therefore clearly divisible by $4$. So we must deal with the case when $(a_1 \neq 0 \lor a_0 \neq 0)$.
Then if $10a_1 + a_0 \equiv 0 \mod 4$ is true, then $x$ is divisible by $4$.
Do I need to do anything else or is this done? I feel like it's not quite complete, but I'm not sure how to proceed.
| All integers can be written in the form:
$$a_{0}10^n+a_{1}10^{n-1}+...+a_{n-1}10^1+a_{n}$$
That can be rewritten as:
$$100(a_{0}10^{n-2}+a_{1}10^{n-3}+...+a_{n-3}10^1+a_{n-2})+a_{n-1}10^1+a_{n}$$
$$=4\times 25(a_{0}10^{n-2}+a_{1}10^{n-3}+...+a_{n-3}10^1+a_{n-2})+a_{n-1}10^1+a_{n}$$
Since the term $(a_{0}10^{n-2}+a_{1}10^{n-3}+...+a_{n-3}10^1+a_{n-2})$ is divisible by $4$, then $a_{0}10^n+a_{1}10^{n-1}+...+a_{n-1}10^1+a_{n}$ is divisible by 4 if and only if the last two digits, $a_{n-1}10^1$ and $a_{n}$, can combine together to form a number divisible by $4$. This concludes the proof.
| {
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How find this maximum $f(x)=\sqrt{\frac{1+x^2}{2}}+\sqrt{x}-x$ let $x\ge 0$,Find the follow maximum
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x$$
I find
$$f'(x)=\dfrac{1}{2}\left(\dfrac{\sqrt{2}x}{\sqrt{x^2+1}}+\dfrac{1}{\sqrt{x}}-2\right)$$
so
$$f'(x)=0\Longrightarrow x=1$$
so I think
$$f(x)_{max}=f(1)$$
But I this methods is ugly, and I think this have AM-GM inequality or Cauchy-Schwarz inequality to solve this problem
$$f(x)=\sqrt{\dfrac{1+x^2}{2}}+\sqrt{x}-x\le\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1+x)-x=\sqrt{\dfrac{1+x^2}{2}}+\dfrac{1}{2}(1-x)$$
Then I can't,Thank you
| Let us show that $f(x)+x\leqslant x+1$ for every $x$. This holds true if and only if
$$
\sqrt{\frac{x+\frac1x}2}+1\leqslant\sqrt{x}+\frac1{\sqrt{x}}.
$$
Let $u=\sqrt{x}+\frac1{\sqrt{x}}$, then $u\geqslant2$ for every $x\geqslant0$ and $x+\frac1x=u^2-2$ hence it is enough to show that $g(u)\leqslant0$ for every $u\geqslant2$, where
$$
g(u)=\sqrt{\frac{u^2-2}2}+1-u.
$$
Now, $g(2)=0$ and
$$
g'(u)=\frac{u-\sqrt{2u^2-4}}{\sqrt{2u^2-4}}=\frac{4-u^2}{(u+\sqrt{2u^2-4})\sqrt{2u^2-4}}\leqslant0,
$$
hence $g(u)\leqslant0$ for every $u\geqslant2$. Finally, $f(x)\leqslant1$ for every $x$ and $f(1)=1$ hence $f(x)$ is maximum at $x=1$, and only there.
| {
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Flexibility of form Prove that for all real $x,y,a,b$, one has
$$(5ax+ay+bx+3by)^2\leq(5a^2+2ab+3b^2)(5x^2+2xy+3y^2)$$
Here's my try but I can't continue.
$$(5ax+ay+bx+3by)^2=(\sqrt5a\cdot\sqrt5x+a\cdot y+b\cdot x+\sqrt3b\cdot\sqrt3y)^2\leq(5a^2+a^2+b^2+3b^2)(5x^2+y^2+x^2+3y^2)$$
| By C-S we obtain:
$$(5a^2+2ab+3b^2)(5x^2+2xy+3y^2)=$$
$$=\left(5a^2+2ab+\frac{1}{5}b^2+\frac{14}{5}b^2\right)\left(5x^2+2xy+\frac{1}{5}y^2+\frac{14}{5}y^2\right)=$$
$$=\left(\left(\sqrt5a+\frac{1}{\sqrt5}b\right)^2+\frac{14}{5}b^2\right)\left(\left(\sqrt5x+\frac{1}{\sqrt5}y\right)^2+\frac{14}{5}y^2\right)\geq$$
$$\geq\left(\left(\sqrt5a+\frac{1}{\sqrt5}b\right)\left(\sqrt5x+\frac{1}{\sqrt5}y\right)+\frac{14}{5}by\right)^2=(5ax+ay+bx+3by)^2.$$
Done!
| {
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Ratio and proportion problem
Question:
If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.
My solution:
*
*$$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$
*$$\Rightarrow a+b=6x \space\space \text{and} \space \space b+c=7x$$$$\Rightarrow b=6x-a\space\space\space\text{and}\space\space\space b=7x-c$$$$\Rightarrow \text{solving we get}\space\space x = c-a$$$$\Rightarrow ????$$
My query:
I am totally stuck on this problem. Please help.
Thanks a lot!
| Let $$x=b+c,$$ $$y=a+c,$$ $$z=a+b.$$ Then, $x+y+z=2(a+b+c)=28$ and $z:x:y=6:7:8$.
By substituting $z=6k$, $x=7k$, $y=8k$ into $x+y+z=28$, we have $21 k=28$, $k=\frac{4}{3}$, from where we get: $$z=6k=8,$$ $$x=7k=\frac{28}{3},$$ $$y=8k=\frac{32}{3}.$$
Finally, $$a=a+b+c-x=\frac{14}{3},$$ $$b=a+b+c-y=\frac{10}{3},$$ $$c=a+b+c-z=6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/677516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| \begin{align}
x+\frac{1}{x}&=5\\
x^2+1&=5x\\
x^2-5x+1&=0\\
x &=\frac{1}{2} \left( 5 +\sqrt{21}\right)\\
x^5+\frac{1}{x^5}&=\cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 3
} |
how to form a quadratic equation If $\alpha$ and $\beta$ are the root of the equation $x^2 - 4x +6 =0$ , find the equation whose roots are $\alpha + 1/\beta$ and $\beta + 1/\alpha$.
| $x^2 - \frac{28}{6}x + \frac{49}{6}$.
$\alpha + \beta = 4$ and $\alpha\beta = 6$. Expand $(x - (\alpha + \frac{1}{\beta}))(x-(\beta + \frac{1}{\alpha}))$. This gives us $x^2 - (\alpha + \beta)(1 + \frac{1}{\alpha\beta})x + (\alpha\beta + 2 + \frac{1}{\alpha\beta})$ So, doing the necessary substitutions gives us: $x^2 - (4\times \frac{7}{6}) x + (6 + 2 + \frac{1}{6})$ which is equal to $x^2 - \frac{28}{6}x + \frac{49}{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Change of variables in Double integral - explanation of solved example. I'm exercising double integrals and here's an solved example that I'm not understand something.
Calculate $\iint_D xy(x^2 + y^2) dx\,dy$ where $D$ in the first quadrant bounded by:
$1 \le xy \le 2$ and $5 \le x^2 - y^2 \le 9.$
Here's the solution:
Let $u = xy$ and $v = x^2 - y^2$.
$1 \le u \le 2$ and $5 \le v \le 9$.
Calculate the Jacobian. $$
\begin{vmatrix}
u'_x & u'_y \\
v'_x & v'_y \\
\end{vmatrix} =
\begin{vmatrix}
y & x \\
2x & -2y \\
\end{vmatrix} = -2y^2 - 2x^2 = -2(x^2 + y^2).$$
$J = \frac{1}{-2(x^2 + y^2)}$ so need to plug $dx\, dy = -\frac{1}{2(x^2+y^2)}du\,dv$.
$$\iint_D xy(x^2 + y^2)dx\,dy = -\int^9_5\int^2_1 u(x^2 + y^2) \frac{1}{2(x^2+y^2)} = \color{red}{-\frac{1}{2}\int^9_5\int^2_1udu\,dv} = -\frac{1}{2}\int^9_5\bigg(\frac{y^2}{2}\bigg|^2_1\bigg)dv = -\frac{1}{2}\int^9_5(2-\frac{1}{2})dv = -\frac{3}{4}\bigg(v\bigg|^9_5\bigg) = -3.$$
My question:
In the part I marked in $\text{$\color{red}{red}$}$, how has $(x^2 + y^2)$ just gone away?
Thanks in advance!!
| It was cancelled. There is an $x^2+y^2$ in the numerator and in the denominator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/680953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For which $n \in \mathbb{Z^+}$ is $(4n+9)/(2n^2+7n+6)$ a terminating decimal? I saw this problem here.
My approach:
Let $A = (4n+9)/(2n^2+7n+6) = \frac{6}{2n + 3} - \frac{1}{n + 2}$
If $\frac{6}{2n + 3}$ and $\frac{1}{n + 2}$ are terminating decimals, then so is $A$.
A number terminates if and only if it can be written as a ratio of the form $\frac{k}{2^m\cdot5^n}$
We know that $2\nmid2n+3$
So $2n+3=3^x\cdot5^y$ where $0\le x\le1$, $1\le y$ and $n+2=(3^x\cdot5^y+1)/2$
So we have $A = \frac{6}{3^x\cdot5^y} - \frac{2}{3^x\cdot5^y+1}$
But I don't even know if this approach makes any sense.
| The only prime that can simultaneously divide numerator and denominator is $3$. For if an odd prime $p$ divides $4n+9$ and $2n+3$, it must divide $(4n+9)-2(2n+3)=3$. And it is easy to see that $4n+9$ and $n+2$ are relatively prime.
We first examine the case where $3$ does not simultaneously divide numerator and denominator. Then $2n+3$ must be a power of $5$. What about $n+2$? In principle it could be divisible by a power of $2$ greater than $1$. However, that would make $2n+3$ of the shape $4k+1$, which no power of $5$ can be.
So $2n+3$ is a power of $5$ and $n+2$ is a power of $2$. This is obviously impossible if $n$ is positive (it only happens if $n=-1$).
It remains to deal with the case where $3$ divides both $4n+9$ and $2n+3$. Then $3$ must divide $n$. Let $n=3m$. Then our expression becomes
$$\frac{4m+3}{(2m+1)(3m+2)}.$$
It is easy to see that $3$ cannot divide both numerator and denominator. So $2m+1$ is a power of $5$. Since $2m+1$ and $3m+2$ are relatively prime, that forces $3m+2$ to be a power of $2$. There are the obvious solutions $m=0$ (which the wording of the problem does not allow) and $m=2$. That gives the solution $n=4$.
Are there other solutions? Note that $2(3m+2)-3(2m+1)=1$. So we need to consider solutions of the exponential Diophantine equation $2^x-3\cdot 5^y=1$. We do not know whether this equation has solutions other than the ones mentioned.
Added: Thanks to Gerry Myerson for pointing out that the exponential Diophantine equation has no other solutions.
Added 2: Zafer Cesur has pointed out that the analysis of the equation $2^x-3\cdot 5^y=1$ does not require fancy machinery. By working modulo $3$, we can see that $x$ is even, say $x=2n$. Then the equation can be rewritten as $(2^n-1)(2^n+1)=3\cdot 5^y$. Because $2^n-1$ and $2^n+1$ are relatively prime, the only possibilities are $2^n-1=1$ and $2^n-1=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that, $\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Prove that, $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{d}+\dfrac{d}{a}\ge 4$ where we do not use AM-GM inequality on the given statement to prove it. Typically, I am actually looking for a little advanced and elegant solution.
EDIT: $a,b,c,d>0$
| $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}=\left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right)=$$
$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{\left(\frac{a}{b}+\frac{c}{d}\right)\cdot\left(\frac{b}{c}+\frac{d}{a}\right)}=$$
$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{d}+\frac{d}{b}\right)}=$$
$$=\left[\sqrt{\frac{a}{b}+\frac{c}{d}}-\sqrt{\frac{b}{c}+\frac{d}{a}}\right]^2+2\sqrt{4+\left(\sqrt{\frac{a}{c}}-\sqrt{\frac{c}{a}}\right)^2+\left(\sqrt{\frac{b}{d}}-\sqrt{\frac{d}{b}}\right)^2}\ge$$
$$\ge 2\sqrt{4}=4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/686791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Integral of $\cos(3x) \cos(4x) \cos(5x)$ The integral $$\int_0^{\pi/8}\cos(3x)\cos(4x)\cos(5x) \,dx$$ is equal to $k/24$. Find the constant $k$.
So far, I assume that the best way to solve this question is to solve the integral and compare the answer to find $k$.
I have thought about rewriting the integral by using the identity
$\cos(2x) = \cos^2(x) - \sin^2(x)$ or $\cos(2x) = 1 - 2\sin^2(x)$,
but that results in a very long and confusing function.
I don't see any possible substitutions.
Please help!
| Using
\begin{align}
\cos(x) \cos(y) &= \frac{1}{2} \, (\cos(x+y) + \cos(x-y) ) \\
\sin\left(\frac{\pi}{4} \right) &= \sin\left(\frac{3 \pi}{4}\right) = \frac{1}{\sqrt{2}} \\
\sin\left(\frac{3 \pi}{2}\right) &= - \sin\left(\frac{\pi}{2}\right) = -1
\end{align}
then
\begin{align}
\cos(3 x) \cos(4 x) \cos(5 x) &= \frac{1}{2} \, (\cos(2 x) \cos(4 x) + \cos(4 x) \cos(8 x) ) \\
&= \frac{1}{4} \, (\cos(2 x) + \cos(4 x) + \cos(6 x) + \cos(12 x)).
\end{align}
The integral in question becomes:
\begin{align}
I &= \int_{0}^{\pi/8} \cos(3 x) \cos(4 x) \cos(5 x) \, dx \\
&= \frac{1}{4} \, \int_{0}^{\pi/8} (\cos(2 x) + \cos(4 x) + \cos(6 x) + \cos(12 x)) \, dx \\
&= \frac{1}{48} \, [ 6 \, \sin(2 x) + 3 \sin(4 x) + 2 \sin(6 x) + \sin(12 x) ]_{0}^{\pi/8} \\
&= \frac{1}{48} \, \left( 6 \sin\left(\frac{\pi}{4} \right) + 3 \sin\left(\frac{\pi}{2} \right) + 2 \sin\left(\frac{3 \pi}{4} \right) + \sin\left(\frac{3 \pi}{2} \right) \right) \\
&= \frac{1}{24} \, \left( 1 + \frac{4}{\sqrt{2}} \right) \\
&= \frac{1 + 2 \sqrt{2}}{24}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/687630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
} |
Looking for the recurrence relation for certain trigonometric integrals By assuming that:
$$ \int_{\pi/4}^{\pi/2} \frac{\cos^4(x)}{\sin^5(x)}\,dx = k,$$
what does the integral $$ \int_{\pi/4}^{\pi/2} \frac{\cos^6(x)}{\sin^7(x)}\,dx$$ equal in terms of k?
I have manipulated both integrals to get $\int_{\pi/4}^{\pi/2} (\csc^2(x)-1)^2(\csc(x))\,dx$ and $\int_{\pi/4}^{\pi/2}\csc^2(x)-1)^3(\csc(x))\,dx$.
I'm not sure where to go from here.
I multiplied $(\csc^2(x)-1)^2$ and $(\csc^2(x)-1)^3$ out, but I still couldn't find a solution.
Please help!
| By setting
$$ I_n = \int_{\pi/4}^{\pi/2}\frac{\cos^{2n}x}{\sin^{2n+1}x}dx$$
we have
$$ I_n = \int_{0}^{1}\frac{u^{2n}}{\sqrt{1+u^2}}du.\tag{1}$$
This follows from the variable changes $x=\arcsin t, t=\sqrt{s}, s=1/r, r=u+1$, or just $x=\arcsin\left(\frac{1}{\sqrt{1+u}}\right)$, for short.
Since:
$$\frac{d}{du}\operatorname{arcsinh} u = \frac{1}{\sqrt{1+u^2}},\qquad \frac{d}{du}\sqrt{1+u^2}=\frac{u}{\sqrt{1+u^2}},$$
integration by parts gives that $I_n$ is always a linear combination of $\sqrt{2}$ and $\operatorname{arcsinh}(1)=-\log\tan\frac{\pi}{8}=\log(1+\sqrt{2})$ with rational coefficients. In facts:
$$I_n + I_{n+1} = \int_{0}^{1}u^{2n}\sqrt{1+u^2}du=\frac{\sqrt{2}}{2n+1}-\frac{1}{2n+1}\cdot I_{n+1},$$
or:
$$I_{n+1} = \frac{\sqrt{2}-(2n+1)\cdot I_n}{2n+2},$$
$$ (2n+1)\cdot I_n + (2n+2)\cdot I_{n+1} = \sqrt{2},$$
$$ I_0 = \log(1+\sqrt{2}).\tag{2}$$
We can also write $I_n$ as a value of the incomplete beta function:
$$I_n = \frac{1}{2}\cdot B\left(\frac{1}{2},n+\frac{1}{2},-n\right),$$
or recover $I_n$ from the coefficients of a Taylor series, since $(1)$ gives:
$$\sum_{n=0}^{+\infty}I_n\, x^{2n}=\frac{1}{\sqrt{1+x^2}}\cdot\operatorname{arctanh}\left(\sqrt{\frac{1+x^2}{2}}\right)=\frac{1}{2\sqrt{1+x^2}}\cdot\log\left(\frac{\sqrt{2}+\sqrt{1+x^2}}{\sqrt{2}-\sqrt{1+x^2}}\right).\tag{3}$$
Exploiting convexity we have:
$$I_n\geq \int_{1-\frac{2}{4n-1}}^{1}\frac{(4n-1)u-(4n-3)}{2\sqrt{2}}\,du = \frac{1}{\sqrt{2}(4n-1)}>\frac{1}{4\sqrt{2}\,n},$$
and by plugging this inequality into $(2)$ we get:
$$ I_n < \frac{3}{4\sqrt{2}\,n}.$$
Since $I_n = \frac{\sqrt{2}}{2n+1}-\frac{I_{n+1}}{2n+1}$, we get the asymptotic:
$$ I_n = \frac{1}{2\sqrt{2}\,n}+\Theta\left(\frac{1}{n^2}\right).\tag{4}$$
Integration by parts gives other terms of the asymptotics, in virtue of:
$$I_n -\frac{1}{(2n+1)\sqrt{2}}= \int_{0}^{1}\left(\frac{u^{2n}}{\sqrt{1+u^2}}-\frac{u^{2n}}{\sqrt{2}}\right)du=\int_{0}^{1}\frac{u^{2n}\left(\sqrt{2}+\sqrt{1+u^2}\right)}{(1-u^2)\sqrt{2(1+u^2)}}\,du.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/689240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Solve polynomial congruence using Hensel's lemma Solve $x^4+2x+46 \equiv 0$ $(\mod 4375 )$ for x.
.
My attempt:
$x^4+2x+46 \equiv 0$ $(\mod 5^47 )$
breaks down to
a Chinese Remainder Problem with the 2 following congruence's':
(1) $x^4+2x+46 \equiv 0$ $(\mod 7 )$ which by inspection is x $\equiv 2,6 (\mod 7)$
and
(2) $x^4+2x+46 \equiv 0$ $(\mod 5^4)$
It's (2) that's really difficult for me and I know I am supposed to use Hensel's Lemma, can any one help with this?
| The Hensel lemma states that if there exist integer $x$ such that:
$$f(x) \equiv 0 \pmod {p^k} \quad \quad \text{and} \quad \quad f'(x) \equiv 0 \pmod p$$
then there exist integer $s$ such that:
$$f(s) \equiv 0 \pmod {p^{k+m}} \quad \quad \text{and} \quad \quad s \equiv r \pmod {p^k}$$
Now it's obvious that $s = r + tp^k$ for some $t \in \mathbb{Z}$.
Now let's move to the problem. Find a solution for the equation:
$$x^4 + 2x + 46 \equiv 0 \pmod 5$$
Since $x\equiv 0 \pmod 5$ obviously isn't solution, Fermat's Little Theorem come in use so we have:
$$x^4 + 2x + 46 \equiv 1 + 2x + 46 \equiv 2x + 47 \equiv 2x + 52 \equiv 0 \pmod 5$$
From this obviously $x \equiv 4 \pmod 5$ is a solution. So take $x_1 = 4$
Now find the derivative of the function. This is easy and we have:
$$f'(x) = 4x^3 + 2$$
Now we need to find a solution: $f(x_2) \equiv 0 \pmod {5^2}$ i.e. $f(x_1 + 5t) \equiv 0 \pmod {5^2}$. The last expression is equivalent to:
$$f(x_1) + 5tf'(x_1) \equiv \pmod {5^2}$$
Now substitute $x_1 = 4$ and we have:
$$f(4) + 5tf'(4) \equiv 0 \pmod {5^2}$$
$$310 + 5t \cdot 258 \equiv 0 \pmod {5^2}$$
Divide with $5$.
$$62 + 258t \equiv 2 + 3t \equiv \pmod 5$$
From this we get: $t \equiv 1 \pmod 5$. Choose $t=1$ and we have: $x_2 = x_1 + 5t = 9$
So now we have $f(9) \equiv 0 \pmod {5^2}$...
Did you get the concept. Can you continue on your own and lift the exponent first to $3$ and then to $4$?
| {
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"url": "https://math.stackexchange.com/questions/689362",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Find the maximum or minimum value of the quadratic function. Find the maximum or minimum value of the quadratic function by completing the squares. Also, state the value of $x$ at which the function is maximum or minimum.
$y=2x^2-4x+7$
$x^2$ has a coefficient of $2$, how should I complete the squares?
| $$y=2x^2-4x+7=2\left(x^2-2\cdot x\cdot1+1^2\right)+7-2\cdot1=2(x-1)^2+5$$
or $$2y=4x^2-8x+14=(2x)^2-2\cdot2x\cdot2+2^2+14-4=(2x-2)^2+10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Making sure I'm not doing something wrong I'm solving a matrix equation: $$2(A-B+X) = 3(X-A)$$
Where $A = \left( \begin{matrix} 1&2 \\ 3&4 \end{matrix} \right)$ and $B = \left( \begin{matrix} -1&0 \\ 1&1 \end{matrix} \right) $
So far I've done: $$2(A-B+x) = 3(X - A)$$ $$\Rightarrow 2 \left( \begin{matrix} 1&2 \\ 3&4 \end{matrix} - \begin{matrix} -1&0 \\ 1&1 \end{matrix} + X \right) = 3 \left( X - \begin{matrix} 1&2 \\ 3&4 \end{matrix} \right)$$
$$ \Rightarrow 2 \left( \begin{matrix} 2&2 \\ 2&3 \end{matrix} + X \right) = 3X - \begin{matrix} 3&6 \\ 9&12 \end{matrix}$$
$$\Rightarrow \begin{matrix} 4&4 \\ 4&6 \end{matrix} + 2X = 3X - \begin{matrix} 3&6 \\ 9&12 \end{matrix}$$
But I'm not sure if I can put it on this form, asssuming the 2 matrices form the matrix $C$ when added:
$$C + 2X = 3X$$
Because I remember something about watching out for positioning when subtracting two matrices..
Some help needed with the next step.
| It might be more clear if you don't write out the elements until necessary.
$$2(A-B+X) = 3(X-A) \\
2A-2B+2X = 3X-3A \\
5A-2B = X$$
Matrices add and subtract normally; that is, $A+A = 2A, 10Q-8Q = 2Q, 3X-2X = X,$ etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$ How can you evaluate $$\int\limits_0^{\pi/2}\log\cos(x)\,\mathrm{d}x\;?$$
| I first treat the integral as a derivative of a beta function
$$
I(a)=\int_0^{\frac{\pi}{2}} \cos ^a x d x=\frac{1}{2} B\left(\frac{a+1}{2}, \frac{1}{2}\right)
$$
$$
\begin{aligned}
I&\left.=\frac{\partial}{\partial a} I(a)\right|_{a=0} \left.=\left.\frac{1}{4} B\left(\frac{a+1}{2}, \frac{1}{2}\right)\left(\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a}{2}+1\right)\right)\right) \right|_{a=0}\\
&=\frac{1}{4} B\left(\frac{1}{2}, \frac{1}{2}\right)\left[\psi\left(\frac{1}{2}\right)-\psi(1)\right]=\frac{1}{4} \pi(-\ln 4) =-\frac{\pi}{2} \ln 2
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/690644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 9,
"answer_id": 8
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Proof by mathematical induction - Fibonacci numbers and matrices Using mathematical induction I am to prove:
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n $ =
$ \left( \begin{array}{ccc}
F_{n+1} & F_n \\
F_n & F_{n-1} \end{array} \right) $
where $F_k$ represents the $k^{th}$ Fibonacci number.
my base case is $n =2$
LHS: $ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right) \times$
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right) $$=$
$ \left( \begin{array}{ccc}
2 & 1 \\
1 & 1 \end{array} \right) $
RHS: $ \left( \begin{array}{ccc}
F_3 & F_2 \\
F_2 & F_1 \end{array} \right) $$=$
$ \left( \begin{array}{ccc}
2 & 1 \\
1 & 1 \end{array} \right) $
So $n = k + 1$
$ \left( \begin{array}{ccc}
F_{k+2} & F_{k+1} \\
F_{k+1} & F_k \end{array} \right) $
So for my inductive step I did:
$ \left( \begin{array}{ccc}
F_{k+1} & F_k \\
F_k & F_{k-1} \end{array} \right) $ $+$
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^{k+1} $
And now I'm not sure where to proceed from here. Can anyone point me in the right direction? Assuming my previous work is correct.
| To prove it for $n=1$ you just need to verify that
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^1 $ =
$ \left( \begin{array}{ccc}
F_2 & F_1 \\
F_1 & F_0 \end{array} \right) $
which is trivial.
After you established the base case, you only need to show that assuming it holds for $n$ it also holds for $n+1$.
So assume
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n $ =
$ \left( \begin{array}{ccc}
F_{n+1} & F_n \\
F_n & F_{n-1} \end{array} \right) $
and try to prove
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^{n+1} $ =
$ \left( \begin{array}{ccc}
F_{n+2} & F_{n+1} \\
F_{n+1} & F_n \end{array} \right) $
Hint: Write $ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^{n+1} $ as $ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n $$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right) $.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/693905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving a curve integral around part of an elipse I'm having trouble calculating a curve integral in a vector field:
$\int_C y (18x + 1)\ dx + 2y^2\ dy$
where $C$ is the curve along the ellipse
$9x^2 + y^2 = 64$
going counterclockwise from the point ( $-\frac{4}{3}\sqrt{3} $ , $4$) to the point (-$\frac{4}{3}$ , $4\sqrt{3} $)
Thats almost one "lap" around the ellipse..
When making a parametrisation I come up with:
x = $\sqrt{ \frac{64}{9}\ } \cos t $
y = $\sqrt{ 64 } \sin t $
$- \frac{\Pi}{3} \le t < \arctan(3 \sqrt{3}) $
But the integral created from this parametrization give an answer involving arctan. Any ideas to get a rational answer?
I had an idea to split the curve into multiple curves and integrating them piecewise, but that gets really messy aswell.
This is my calculations when making a variable change:
First the variable change:
$ u= 3x $
$ du = 3dx $
$ \ u^2 + y^2 =64 $
$ \int y(18x + 1)dx + 2y^2dy = \int y(6u + 1)\frac{du}{3} + 2y^2dy $
$ u = \sqrt{64}\cos t $
$ y = \sqrt{64}\sin t $
$ du = -\sqrt{64}\sin tdt $
$ dy = \sqrt{64}\cos tdt $
The startpoint after making variable change:
$ ( -4\sqrt{3},4)$
$ arctan( \frac{4}{-4\sqrt{3}}) = -\Pi/6$
Startangle: $ \frac{5\Pi}{6} $
The endpoint after making variable change: $ ( -4,4\sqrt{3})$
$ arctan( \frac{-4\sqrt{3}}{4}) = -\Pi/3$
End angle: $\frac{2\Pi}{3} $
$\int y(6u + 1)\frac{du}{3} + 2y^2dy = \int -\frac{1}{3}\sqrt{64}\sin t(6\sqrt{64}\cos t + 1)\sqrt{64}\sin tdt + 128\sin^2t\sqrt{64}\cos tdt = $
$\int -\frac{64}{3}\sin^2tdt $
$ \frac{5\Pi}{6} \le t \le \frac{2\Pi}{3} $
The answer becomes: $\frac{16}{9}\Pi $
| As @dfan suggested, substitute $u=3x$ leading to a circular curve $C'$ defined by $$u^2+y^2=r^2$$ of radius $r=8$. Your integral becomes
$$J:=\int_C dx\,y(18x+1)+\int_C dy\,2y^2=\frac{1}{3}\int_{C'}du\,y(6u+1)+2\int_{C'}dy\,y^2.$$ Taking the total differential of the circle equation, we have $udu=-ydy$, so the second integral cancels against the $6u$-term from the first one and we are left with
$$J=\frac{1}{3}\int_{C'}du\,y.$$
Next, introduce polar coordinates
$$u=r\cos\phi,\\y=r\sin\phi,$$parametrizing the curve with $\phi$. Your integral limits transform to
$$(x,y)_0=(-4/\sqrt{3},4)\to u_0=-4\sqrt{3}\quad \to\phi_0=\arccos\frac{-\sqrt{3}}{2}=\frac{5\pi}{6},$$
$$(x,y)_1=(-4/3,4\sqrt{3})\to u_1=-4\quad \to\phi_1=\arccos\frac{-1}{2}+2\pi=\frac{8\pi}{3}.$$
Here it is essential to add $2\pi$ to the upper limit $\phi_1$, such that $\phi_0<\phi_1$ and the integral passes the curve in the correct rotation sense. Finally, with the line element
$$du=-d\phi\,r\sin\phi,$$ the integral becomes
$$J=-\frac{r^2}{3}\int_{\phi_0}^{\phi_1} d\phi\,\sin^2\phi=-\frac{r^2}{6}\left( \phi-\frac{1}{2}\sin2\phi \right)_{\phi_0}^{\phi_1}$$ and using the given values for $r,\phi_0,\phi_1$ we obtain
$$J=-\frac{176\pi}{9}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots?
$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}-\frac{3}{\sqrt{n+3}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}=\sum _{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\frac{4}{\sqrt{n+4}}\right)$$
$$\cdots$$
And is there some easy cancellation that I have missed on the right hand side?
Mathematica:
Clear[s, i, n, j]
s = 1/2;
i = 1;
j = 0;
Sum[1/(n + 0)^s - 1/(n + 1)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s - 2/(n + 2)^s + 1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s - 3/(n + 3)^s + 1/(n + 0)^s,
{n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s - 4/(n + 4)^s +
1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
N[Accumulate[Sqrt[Range[4]]], 20]
| Let's see how we find the first sum which's known as telescoping sum and the other sums are almost the same: the idea is to change the index and then cancel most of the terms
$$\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=\lim_{N\to\infty}\sum\limits_{n=1}^{N} \frac{1}{\sqrt{n}}-\sum\limits_{n=2}^{N+1}\frac{1}{\sqrt{n}}\\=\lim_{N\to\infty}\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{N+1}}=\frac{1}{\sqrt{1}}$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 0
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Do not understand formula... How does;
$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$
work on $x^{2} - y^{2}$
When I attempt to apply the formula on $x^{2} - y^{2}$
I get the following
$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$
$x^{2} - y^{2} =(x-y)(2x+2y)$
which is obviously false. What is the correct way to use the formula?
| $$x^2 - y^2 = (x -y)(x^1y^0 + x^0y^1) =(x-y)(x+y)$$
| {
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"answer_id": 0
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How to evaluate $\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx$? Let $\operatorname{erfc}x$ be the complementary error function.
I successfully evaluated these integrals:
$$\int_0^\infty\operatorname{erfc}x\ \mathrm dx=\frac1{\sqrt\pi}\tag1$$
$$\int_0^\infty\operatorname{erfc}^2x\ \mathrm dx=\frac{2-\sqrt2}{\sqrt\pi}\tag2$$
(Both $\operatorname{erfc}x$ and $\operatorname{erfc}^{2}x$ have primitive functions in terms of the error function.)
But I have problems with
$$\int_0^\infty\operatorname{erfc}^3x\ \mathrm dx\tag3$$
and a general case
$$\int_0^\infty\operatorname{erfc}^n x\ \mathrm dx.\tag4$$
Could you suggest an approach to evaluate them as well?
| Integrating by parts twice, we get
$$ \begin{align} \int_{0}^{\infty} \text{erfc}^{3}(x) \, dx &= x \, \text{erfc}^{3}(x) \Bigg|^{\infty}_{0} + \frac{6} {\sqrt{\pi}} \int_{0}^{\infty}x \, \text{erfc}^{2}(x) e^{-x^{2}} \, dx \\ &= \frac{6} {\sqrt{\pi}} \int_{0}^{\infty}x \, \text{erfc}^{2}(x) e^{-x^{2}} \, dx \\ &= - \frac{3 \, \text{erfc}(x) e^{-x^{2}}}{\sqrt{\pi}} \Bigg|_{0}^{\infty} - \frac{12}{\pi} \int_{0}^{\infty}\text{erfc}(x) e^{-2x^{2}} \, dx \\ &= \frac{3}{\sqrt{\pi}} - \frac{12}{\pi} \int_{0}^{\infty} \text{erfc}(x) e^{-2x^{2}} \, dx. \end{align}$$
And using the integral definition of the complementary error function, we get
$$ \begin{align} \int_{0}^{\infty} \text{erfc}(x) e^{-2x^{2}} \, dx &= \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \int_{1}^{\infty} x e^{-x^{2}t^{2}} e^{-2x^{2}} \, dt \, dx \\ &= \frac{2}{\sqrt{\pi}} \int_{1}^{\infty} \int_{0}^{\infty} x e^{-(2+t^{2})x^{2}} \, dx \, dt \\ &= \frac{1}{\sqrt{\pi}} \int_{1}^{\infty} \frac{1}{2+t^{2}} \, dt \\ &= \frac{\sqrt{2}}{2 \sqrt{\pi}} \int_{1 / \sqrt{2}}^{\infty} \frac{1}{1+u^{2}} \, du \\ &= \frac{\sqrt{2}}{2 \sqrt{\pi}} \left[ \frac{\pi}{2} - \arctan \left(\frac{1}{\sqrt{2}} \right) \right] \\ &= \frac{\sqrt{2}}{2 \sqrt{\pi}} \, \arctan (\sqrt{2}). \end{align}$$
Therefore,
$$ \begin{align} \int_{0}^{\infty} \text{erfc}^{3}(x) \ dx &= \frac{3}{\sqrt{\pi}} - \frac{12}{\pi} \left( \frac{\sqrt{2}}{2 \sqrt{\pi}} \arctan \sqrt{2}\right) \\ &= \frac{3}{\sqrt{\pi}} - \frac{6 \sqrt{2}}{\pi^{3/2}}\, \arctan (\sqrt{2}). \end{align}$$
EDIT:
Again integrating by parts twice, we get
$$\int_{0}^{\infty} \text{erfc}^{4}(x) \ dx = \frac{4}{\sqrt{\pi}} + \frac{24}{\pi} \int_{0}^{\infty} \text{erfc}^{2}(x) e^{-2x^{2}} \, dx.$$
In general, for positive parameters $a,b$, and $p$,
$$ \begin{align} I(a,b,p) &= \int_{0}^{\infty} \text{erfc}(ax) \, \text{erfc}(bx) e^{-px^{2}} \, dx \\ &= \frac{4}{\pi} \int_{0}^{\infty} \int_{a}^{\infty} \int_{b}^{\infty} x^{2} e^{-x^{2}y^{2}} e^{-x^{2} z^{2}} e^{-px^{2}} \, dy \, dz \, dx \\ &= \frac{4}{\pi} \int_{a}^{\infty} \int_{b}^{\infty} \int_{0}^{\infty} x^{2} e^{-(p+y^{2}+z^{2})x^{2}} \, dx \, dy \, dz \\ &= \frac{1}{\sqrt{\pi}} \int_{a}^{\infty} \int_{b}^{\infty} \frac{1}{(p+y^{2}+z^{2})^{3/2}} \, dy \, dz. \end{align}$$
Let $y = \sqrt{p+z^{2}} \tan \theta$.
$$ \begin{align} I(a,b,p) &= \frac{1}{\sqrt{\pi}} \int_{a}^{\infty} \int_{\arctan(b / \sqrt{p+z^{2}})}^{\pi /2} \frac{\cos \theta}{p+z^{2}} \, d \theta \, dz \\ &= \frac{1}{\sqrt{\pi}} \int_{a}^{\infty} \frac{1}{p+z^{2}} \left(1- \frac{b}{\sqrt{p+b^{2}+z^{2}}} \right) \, dz \\ &= \frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{b}{\sqrt{\pi}} \int_{a}^{\infty} \frac{1}{(p+z^{2})\sqrt{p+b^{2}+z^{2}}} \, dz \end{align}$$
Now let $ \displaystyle t = \frac{1}{z}$.
$$ I(a,b,p) =\frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{b}{\sqrt{\pi}} \int_{0}^{1/a} \frac{t}{(1+pt^{2})\sqrt{1+b^{2}t^{2}+pt^{2}}} \, dt$$
Finally let $u^{2} = 1+b^{2}t^{2} + p t^{2}$.
$$ \begin{align} I(a,b,p) &= \frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{b}{\sqrt{\pi}} \int_{1}^{\sqrt{p+a^{2}+b^{2}}/a} \frac{1}{b^{2}+pu^{2}} \ du \\ &=\frac{1}{\sqrt{\pi p}} \arctan \left(\frac{\sqrt{p}}{a} \right) - \frac{1}{\sqrt{\pi p}}\int_{\sqrt{p} / b}^{\sqrt{p(p+a^{2}+b^{2)}}/(ab)} \frac{1}{1+w^{2}} \ dw \\ &= \frac{1}{\sqrt{\pi p}} \left[\arctan \left(\frac{\sqrt{p}}{a} \right) +\arctan \left(\frac{\sqrt{p}}{b} \right) - \arctan \left( \frac{\sqrt{p(p+a^{2}+b^{2})}}{ab}\right) \right] \end{align}$$
Therefore,
$$ I(1,1,2) = \int_{0}^{\infty} \text{erfc}^{2}(x) e^{-2x^{2}} \ dx = \frac{1}{\sqrt{2 \pi}} \left( 2 \arctan (\sqrt{2})- \arctan (2\sqrt{2}) \right) ,$$
and
$$ \begin{align} \int_{0}^{\infty} \text{erfc}^{4}(x) \ dx &= \frac{4}{\sqrt{\pi}} - \frac{12 \sqrt{2}}{\pi^{3/2}} \Big(2 \arctan (\sqrt{2})- \arctan (2 \sqrt{2}) \Big) \\ &= \frac{4}{\sqrt{\pi}} - \frac{24 \sqrt{2}}{\pi^{3/2}} \arctan \left( \frac{1}{2\sqrt{2}} \right). \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 2,
"answer_id": 0
} |
Determine the value of the integral $I=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$ Determine the value of the integral $$I(a)=\int_{0}^{\infty} \frac{\ln\left(a^2+x^2\right)}{b^2+x^2}dx$$
My try:
$\to I'(a)=\int_{0}^{\infty}\frac{2a}{(a^2+x^2)(b^2+x^2)}dx=\frac{\pi}{b(a+b)}$
Hence $I(a)=\frac{\pi}{b}\ln(a+b)+C$
Question: Find C???
Thank you!
| Firstly, we introduce a lemma:
\begin{equation}
\int_0^{\frac{\pi}{2}}\ln(\sin(x))dx=\int_0^{\frac{\pi}{2}}\ln(\cos(x))dx\\
\mbox{Then}\\
\int_0^{\frac{\pi}{2}}\ln(\tan(x))dx=0
\end{equation}
The proof see here.
Set $a=0$ and use the substitution that $x=b\tan(u)$, we have:
\begin{equation}
I(0)=2\int_0^\infty\frac{\ln(x)}{b^2+x^2}dx=\frac{2}{b^2}\int_0^\frac{\pi}{2}\frac{\ln(b\tan(u))}{(1+\tan(u)^2)\cos(u)^2}du\\
=\frac{2}{b^2}\int_0^\frac{\pi}{2}\ln(b\tan(u))du\\
=\frac{\pi\ln(b)}{b}
\end{equation}
Then, you can solve the constant $C$.
| {
"language": "en",
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"source": "stackexchange",
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Prove that $n^2+n+41$ is prime for $n<40$ Here's a problem that showed up on an exam I took, I'm interested in seeing if there are other ways to approach it.
Let $n\in\{0,1,...,39\}$. Prove that $n^2+n+41$ is prime.
I shall provide my own solution, though I am curious does anyone know how to do this without using PIDs?
| Below is problem $\rm 6\ (B3)$ from IMO $1987$ and its solution (from AoPS, or Scholes)
Theorem. If $k^2+k+n$ is prime for all integers $k$ such that $0\le k\le \sqrt\frac{n}{3}$, then $k^2+k+n$ is prime for all integers $0\le k\le n-2$.
Proof. First observe that if $m$ is relatively prime to $b+1$, $b+2$, $\cdots$, $2b$, then $m$ is relatively prime to any number less than $2b$. Indeed if $c\leq b$, then we can choose some $i$ to make $2^ic$ lies in range $b+1,b+2,\cdots,2b$, so $2^ic$ is relatively prime to $m$. Hence $c$ is also a prime. If we also have $(2b+1)^2>m$, then we can conclude that $m$ is a prime, otherwise there must be a factor of $m$ less than $\sqrt{m}$.
Let $n=3r^2+h$ where $0\leq h<6r+3$, so $r$ is the greatest integer less than or equal to $\sqrt{n/3}$. (to see this, just let $r=\lfloor\sqrt{n/3}\rfloor$, then we can write $n=3(r+\epsilon)^2(0\leq\epsilon< 1)$, so $h=6r\epsilon+3\epsilon^2\leq 6r+3$).
Assume that $n+k(k+1)$ is prime for $k=1,2,3\cdots,r$. We show that $N=n+(r+s)(r+s+1)$ is prime for $s=0,1,2,\cdots,n-r-2$. By our observation above, it is sufficient to show that $(2s+2r+t)^2>N$ and $N$ is relatively prime to all of $r+s+1,r+s+2,\cdots,2r+2s$. We have $(2r+2s+1)^2=4r^2+4s^2+8rs+4r+4s+1$. Since $s,r\ge1$, we have $4s+1>s+2$, $4s^2>s^2$ and $6rs>3r$. Hence $$(2s+2r+1)^2>4r^2+2rs+s^2+7r+s+2=3r^2+6r+2+(r+s)(r+s+1)\ge N.$$ Now if $N$ has a factor which divides $2r-i$ int the range $-2s$ to $r-s-1$, then so does $N-(i+2s+1)(2r-i)=n+(r-i-s-1)(r-i-s)$ which have the form $n+s'(s'+1)$ with $s'$ in range $0$ to $r$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Can $\frac {100-100}{100-100}=2$? \begin{align*}
\frac{0}{0} &= \frac{100-100}{100-100} \\
&= \frac{10^2-10^2}{10(10-10)} \\
&= \frac{(10+10)(10-10)}{10(10-10)} \\
&= \frac{10+10}{10} \\
&= \frac{20}{10} \\
&= 2
\end{align*}
I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after searching it on Google, any help would be appreciated.
| You do some thing like this
$$\frac{0\cdot 5}{0\cdot 9}=\frac{5}{9}$$you haven't to divide over zero
| {
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Arithmetic Problem Prime numbers Let $p\equiv 2\mod 3$ an odd prime number.
Prove that:
$p \mid (x^3+y^3) \implies p \mid (x+y)$ , for any integers $x,y$
$p\mid (x+y)(x^2-xy+y^2) \implies p\mid (x+y)$ or $p\mid (x^2-xy+y^2) $
if $p\mid (x+y)$ so problem solved
if $p\mid (x^2-xy+y^2)$ I can't find a contradiction.
| We give two proofs, one by a Legendre symbol calculation, the second much simpler.
First Proof: If $p$ divides one of $x$ or $y$, then it must divide the other, and swe are finished. So suppose $p$ divides neither $x$ nor $y$. We show that $p$ cannot divide $x^2-xy+y^2$.
Suppose to the contrary that it does. Then $p$ divides $4x^2-4xy+4y^2$, that is, $p$ divides $(2x-y)^2+3y^2$.
Then $(2x-y)^2\equiv -3y^2\pmod{p}$. But $y$ is invertible modulo $p$. Multiplying through by the inverse of $y^2$, we find that the congruence
$$w^2\equiv -3\pmod{p}$$
has a solution.
We reach a contradiction by showing that $-3$ is not a quadratic residue of $p$.
There are two cases, (i) $p\equiv 1\pmod{4}$ and (ii) $p\equiv -1\pmod{4}$.
For each case, we do a Legendre symbol calculation.
Case (i): We have $(-3/p)=(-1/p)(3/p)$. But $(-1/p)=1$, and by Quadratic Reciprocity $(3/p)=(p/3)=(2/3)=-1$. So $-3$ is not a quadratic residue of $p$.
The calculation for Case (ii) is very similar. Again, it turns out that $-3$ is not a quadratic residue of $p$.
Second proof: It is convenient to look at the equivalent problem of showing that if $p$ divides $x^3-y^3$, then $p$ divides $x-y$.
As in the first argument, we can assume that $y$ is not divisible by $p$. Then from the congruence $x^3\equiv y^3\pmod{p}$, multiplying through by $y^{-1}$, we get that $(xy^{-1})^3\equiv 1\pmod{p}$. We want to conclude that $x\equiv y\pmod{p}$.
Since $3$ and $\varphi(p)$ are relatively prime, the congruence $t^3\equiv 1\pmod{p}$ has precisely one solution, and we are finished.
If detail is needed, there exist integers $u$ and $v$ such that $3u+(p-1)v=1$. If $t^3\equiv 1\pmod{p}$, then we get the result from $t=t^1=t^{3u+(p-1)v}$.
| {
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"source": "stackexchange",
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On last digit of 4 consecutive primes less than 10 apart $\def\mod{\mathrm{\;mod\;}}\def\pq{\{p_1,p_2,p_3,p_4\}}$$\def\dq{\{d_1,d_2,d_3,d_4\}}$
This question concerns quartets of consecutive primes $p_1 < p_2 < p_3 < p_4$ such that $p_4 - p_1 < 10$, and the associated quartets $\dq$ consisting of the last decimal digit of each of the $p_i$.
For example, if $\pq = \{11,13,17,19\}$, then $\dq = \{1,3,7,9\}$.
Assuming my code is correct (i.e. no bugs), then among the primes $< 10^8$ there are only three such quartets of primes whose associated quartets of last digits is not $\{1,3,7,9\}$. These three quartets are:
$$
\begin{array}{l}
\{2, 3, 5, 7\} \\
\{3, 5, 7, 11\} \\
\{5, 7, 11, 13\}
\end{array}
$$
Are these really the only ones? If so, how does one prove it?
To state the question differently, let $x = 10\,n$ for some integer $n \geq 0$. Consider the following (overlapping) quartets of odd positive integers:
$$
\begin{array}{llllllll}
\{& x+3, & x+7, & x+9, & x+11 & & &\} \\
\{& & x+7, & x+9, & x+11, & x+13 & &\} \\
\{& & & x+9, & x+11, & x+13, & x+17 &\}
\end{array}
$$
Can any one such quartet consist solely of prime numbers?
I have not been able to find one, but it's not immediately obvious to me why.
FWIW, at least in the range $[0, 10^8]$, there seems to be no shortage of positive integers $x = 10\,n$ such that the quartet $\{x+1, x+3, x+7, x+9\}$ consists solely of primes. I found $4767$ such $x$, the smallest and largest ones being being $x = 10$ and $x = 99982240$.
| If you consider the additions to $x$ modulo $3$ within
$\begin{array}{llllllll}
\{& x+3, & x+7, & x+9, & x+11 & & &\} \\
\{& & x+7, & x+9, & x+11, & x+13 & &\} \\
\{& & & x+9, & x+11, & x+13, & x+17 &\}
\end{array}$
You get
$\begin{array}{llllllll}
\{& 0, & 1, & 0, & 2 & & &\} \\
\{& & 1, & 0, & 2, & 1 & &\} \\
\{& & & 0, & 2, & 1, & 2 &\}
\end{array}$
Since each row contains a $0$, a $1$ and a $2$, whatever $x$ is modulo $3$ each row will have a number divisible by $3$. So it either contains $3$ itself, or contains a composite number. The only way it can be $3$ is the first row, with $x = 0$ but that would make $x + 9 = 9$, so there's always a composite number.
Note however ${1,3,7,9} \equiv {1,0,1,0} \mod 3$, so the argument doesn't apply to that case.
| {
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if $f'''(x)$ is continuous everywhere and $\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$ Compute $f''(0)$
if $f'''(x)$ is continuous everywhere and $$\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$$ Compute $f''(0)$
The limit equals to $$\begin{align} \lim_{x \to 0} \frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=0. \end{align}$$
From $$\frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=o(1)$$ as $x \to 0$, I get $$1+x+\frac{f(x)}{x} = e^{3x+o(x)},$$ and $$f(x)=x(e^{3x+o(x)}-x-1),\frac{f(x)}{x}=e^{3x+o(x)}-x-1$$ as $x \to 0$. So both $f(0)$ and $f'(0)$ are $0$. Approximating $e^{3x+o(x)}=1+3x+o(x)$ I get $$\begin{align} f(x) &= x(1+3x+o(x)-x-1) =2x^2+o(x^2). \end{align}$$
Now I try to use the definition of derivative to calculate the $f''(x)$
$$f''(x)=\lim_{x \to 0}\frac{f'(x)-f'(0)}{x}=\lim_{x \to 0} \frac{f'(x)}{x}$$
I'm not sure whether I can get $f'(x)$ by differentiating the approximation $2x^2+o(x^2)$ and how to differentiate $o(x^2)$.
| I will use the following formula for $f''(0)$. We have
$$
f(x)=f(0)+f'(0)x+f''(0)\,\frac{x^2}2+o(x^3),
$$
$$
f(-x)=f(0)-f'(0)x+f''(0)\,\frac{x^2}2-o(x^3).
$$
Adding and solving for $f''(0)$, we get
$$\tag{1}
f''(0)=\frac{f(x)+f(-x)-2f(0)}{x^2}+o(x).
$$
Starting from your
$$
1+x+\frac{f(x)}{x} = e^{3x+o(x)},
$$
we have
$$\tag{2}
x+x^2+f(x)=xe^{3x+o(x)}.
$$
Then, taking $x=0$, we get $f(0)=0$; and, using $(1)$,
\begin{align}
f''(0)&=\lim_{x\to0}\frac{f(x)+f(-x)-2f(0)}{x^2}=\lim_{x\to0}\frac{f(x)+f(-x)}{x^2}\\ \ \\
&=\lim_{x\to0}\frac{xe^{3x+o(x)}-x-x^2-xe^{-3x+o(-x)}+x-x^2}{x^2}\\ \ \\
&=-2+\lim_{x\to0}\frac{e^{3x+o(x)}-e^{-3x+o(-x)}}{x}\\ \ \\
&=-2+\lim_{x\to0}\frac{3x+3x+o(x^2)}{x}\\ \ \\ &=-2+6=4.
\end{align}
In the last limit I'm cancelling $o(x)$ with $o(-x)$; this is ok since both terms are coming from the same expression $(2)$.
| {
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} |
Infinite Geometric Series and its Value Calculate the value of $\sum_1^\infty (\frac{1}{3})^{2n}$.
I can clearly see that after listing a few partial sums this seems to tend toward $\frac{1}{8}$ and each partial sum is greater than the last.
$\sum_1^2 (\frac{1}{3})^{2n} = \frac{10}{81}$
$\sum_1^3 (\frac{1}{3})^{2n} = \frac{91}{729}$
$\sum_1^4 (\frac{1}{3})^{2n} = \frac{820}{6561}$
$ \frac{10}{81} \lt \frac{91}{729} \lt \frac{820}{6561} \lt {...}\lt \frac{1}{8} $
My question is how do I show that the answer is indeed $\frac{1}{8}$? I considered using a theorem in my book that says that $\sum_1^\infty x_n=\lim(s_k) = sup\{s_k :k \in N\}$ where $s_k$ are partial sums. The issue I am coming into is showing that $\frac{1}{8}$ is indeed the supremum. Any ideas would be appreciated. I also don't have to use this theorem, it was just a thought.
| The formula for the sum of a geometric series is $\dfrac{a}{1-r}$ where $a$ is the first term in the series and $r$ is the ratio you're multiplying by.
Here, $a = (\dfrac{1}{3})^{2\times1} = \dfrac{1}{9}$, and $r = \dfrac{1}{9}$.
So, we have $\dfrac{\dfrac{1}{9}}{1-\frac{1}{9}} = \dfrac{1}{9-1} = \dfrac{1}{8}$.
| {
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Evaluating $\int \cos^{-1}\left(\frac{x^2+a^2}{x^2-b^2}\right)x^2 dx$ How to evaluate the integral
$$\int \cos^{-1}\left(\frac{x^2+a^2}{x^2-b^2}\right)x^2 dx$$$a<b.$
I posted a similar question here.
Thanks in advance.
| Hint:
$\int\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)x^2~dx$
$=\int\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)d\left(\dfrac{x^3}{3}\right)$
$=\dfrac{x^3}{3}\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)-\int\dfrac{x^3}{3}d\left(\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)\right)$
$=\dfrac{x^3}{3}\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)-\int\dfrac{2(a^2+b^2)x^4}{3(x^2-b^2)\sqrt{(x^2-b^2)^2-(x^2+a^2)^2}}dx$ (according to http://www.wolframalpha.com/input/?i=d%2Fdx%28arccos%28%28x%5E2%2Ba%5E2%29%2F%28x%5E2-b%5E2%29%29%29)
$=\dfrac{x^3}{3}\cos^{-1}\left(\dfrac{x^2+a^2}{x^2-b^2}\right)-\int\dfrac{2(a^2+b^2)x^4}{3(x^2-b^2)\sqrt{b^4-a^4-2(a^2+b^2)x^2}}dx$
| {
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Solving a combintorical problem using generating function
In how many ways $n$ balls can be divided into groups where each group may contain only one ball or two.
Each ball can be a singleton, or coupled with one of the $n-1$ options. Hence,
$a_0 = 1$
$a_1 = 1$
$a_n = a_{n-1} + (n-1)a_{n-2}$
The correspond generating function:
$$F(x) = 1 + x + \sum\nolimits_{n \ge 2} {\left( {{a_{n - 1}} + (n - 1) \cdot {a_{n - 2}}} \right)} \cdot {x^n}$$
$$F(x) = 1 + x + \sum\nolimits_{n \ge 2} {{a_{n - 1}}} {x^n} + \sum\nolimits_{n \ge 2} {(n - 1)} \cdot {a_{n - 2}} \cdot {x^n}$$
$$F(x) = 1 + x + x \cdot \sum\nolimits_{n \ge 1} {{a_n}} {x^n} + {x^2} \cdot \sum\nolimits_{n \ge 0} {(n - 1)} \cdot {a_n} \cdot {x^n}$$
$$F(x) = 1 + x + x\left( {F(x) - 1} \right) + {x^2}\left( ? \right)$$
How to treat the last expression? $$\sum\nolimits_{n \ge 0} {(n - 1)} \cdot {a_n} \cdot {x^n}$$
| Though you did not state it in the question, from your recurrence I guess you are dealing with distinguishable (labelled) balls. Thus for instance, $\{\{1, 2\}, 3\}$ and $\{\{1, 3\}, 2\}$ are distinct groupings. So your recurrence $a_n = a_{n-1} + (n-1)a_{n-2}$ is correct: given $n$ balls labelled $1$ to $n$, you put the ball labelled $n$ either by itself (and group the rest in $a_{n-1}$ ways) or with one of the other $(n-1)$ balls (and group the rest in $a_{n-2}$ ways).
There is a direct way of solving this, using exponential generating functions. Any such grouping is a set of labelled parts, where each part is either a ball by itself, or two labelled balls. In notation, calling the class of all groupings as $\mathcal{G}$, we have
$$\mathcal{G} = \operatorname{S\scriptstyle ET}(\mathcal{P})$$
where $\mathcal{P}$ denotes "either one ball or two balls" and has exponential generating function (as there is only one of either) $\displaystyle P(z) = z + \frac{z^2}{2!}$, giving the exponential generating function to be
$$
\begin{align}
G(z)
&= \exp\left(z + \frac{z^2}{2}\right) \\
&= 1 + z + 2\frac{z^2}{2!} + 4\frac{z^3}{3!} + 10\frac{z^4}{4!} + 26\frac{z^5}{5!} + 76\frac{z^6}{6!} + 232\frac{z^7}{7!} + 764\frac{z^8}{8!} + \dots
\end{align}$$
| {
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"source": "stackexchange",
"question_score": "2",
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Limit involving logarithms I have to solve the limit,
$ \displaystyle \lim_{x \to 0} \frac{ \log ((k+x)^{b} - (k-x)^{b} )}{\log x}$
where $k \in (0,1)$ and $b \in (0,1)$ are constant . I have tried using Taylor expansion but it does not work. Thank you.
| You could use Newton's Binomial Theorem to say that
$$
\begin{align*}
(k+x)^b-(k-x)^b&=k^b\left[\left(1+\frac{x}{k}\right)^b-\left(1-\frac{x}{k}\right)^b\right]\\
&=k^b\sum_{n=0}^{\infty}\binom{b}{n}\left[\left(\frac{x}{k}\right)^n-\left(-\frac{x}{k}\right)^n\right]\\
&=k^b\sum_{m=0}^{\infty}\binom{b}{2m+1}2\left(\frac{x}{k}\right)^{2m+1}\\
&=2k^b\sum_{m=0}^{\infty}\binom{b}{2m+1}\frac{x^{2m+1}}{k^{2m+1}}\\
&=2k^b\left(\frac{b}{k}x+O(x^3)\right)\\
&=2bk^{b-1}x+O(x^3)\\
&=2bk^{b-1}x(1+O(x^2)),
\end{align*}
$$
where the series statement holds provided $\lvert x\rvert<k$ and the asymptotic bound holds as $x\to0^+$. (And clearly, as $x\to0^+$, we eventually do have $\lvert x\rvert<k$.) But, this says that
$$
\begin{align*}
\log[(k+x)^b-(k-x)^b]&=\log(2k^{b-1}bx)+\log(1+O(x^2))\\
&=\log(2k^{b-1}b)+\log(x)+O(x^2),
\end{align*}
$$
so that
$$
\begin{align*}
\frac{\log[(k+x)^b-(k-x)^b]}{\log x}=\frac{\log(2k^{b-1}b)+\log(x)+O(x^2)}{\log(x)}\to1\text{ as }x\to0^{+}.
\end{align*}
$$
| {
"language": "en",
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finding bases for row space and null space of matrix. My problem is:
For the matrix
$$A = \begin{bmatrix}
1& 4& 5& 6& 9\\
3& −2& 1& 4& −1\\
−1& 0& −1& −2& −1\\
2& 3& 5& 7& 8\end{bmatrix}$$
(a) Find a basis for the row space of A.
(b) Find a basis for the null space of A.
(c) Find the rank and nullity of A.
I tried searching online and I became more confused, take the example here.
http://www2.kenyon.edu/Depts/Math/Paquin/PracticeExam1Solns.pdf
As you can see for the column space he takes the columns of the original matrix instead of the rref of A, which I don't understand.
| The row space is the span of the rows of $A$. (or the column space of $A^{T}$).
$$
A^{T} = \begin{bmatrix} 1 & 3 & -1 & 2\\
4 & -2 & 0 & 3 \\
5 & 1 & -1 & 5 \\
6 & 4 & -2 & 7\\
9 & -1 & -1 & 8 \end{bmatrix} \vec{x} = \begin{bmatrix} 1\\ 4 \\ 5 \\ 6 \\ 9\end{bmatrix}x_1 + \begin{bmatrix} 3\\ -2 \\ 1 \\ 4 \\ -1\end{bmatrix}x_2 + \begin{bmatrix} -1\\ 0 \\ -1 \\ -2 \\ -1\end{bmatrix}x_3 + \begin{bmatrix} 2\\ 3 \\ 5 \\ 7 \\ 8\end{bmatrix}x_4
$$
So find a maximally linear independent subset $R$ of:
$$
\left\{\begin{bmatrix} 1\\ 4 \\ 5 \\ 6 \\ 9\end{bmatrix} , \begin{bmatrix} 3\\ -2 \\ 1 \\ 4 \\ -1\end{bmatrix} , \begin{bmatrix} -1\\ 0 \\ -1 \\ -2 \\ -1\end{bmatrix} , \begin{bmatrix} 2\\ 3 \\ 5 \\ 7 \\ 8\end{bmatrix} \right\}
$$
Now the null space is the set: $$\{\vec{x}: A\vec{x} = \vec{0} \}$$
$$
A = \begin{bmatrix}
1& 4& 5& 6& 9\\
3& −2& 1& 4& −1\\
−1& 0& −1& −2& −1\\
2& 3& 5& 7& 8\end{bmatrix}\vec{x} = \vec{0} \longrightarrow \; ?
$$
So you must find a maximally linear independent subset $S$ of
$$\{\vec{x}: A\vec{x} = \vec{0} \}$$ Now once you find these two things,
$$
\text{rank}(A) = |R|, \; \; \text{nullity}(A)= |S|
$$
| {
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How to solve system of two equation with two unknown and using substitution? $$\begin{align}
a^2 - b^2 = 3\\
a \cdot b = 2
\end{align}$$
In aforementioned equations, we can mentally find out the value of $a = 2, b = 1$. But what is the general way to solve this system algebraically?
I tried to use substitution but I got stuck. Rearranging two equations to $a^2=b^2+3$ and $a^2 \cdot b^2 = 2^2$ we will get:
$$ (b^2 + 3) \cdot b^2 = 4\\
\rightarrow b^4 + 3b^2 = 4\\
\rightarrow b^2(b^2 + 3) = 4\\
$$
Then what?
Alternatively can we solve by elimination?
| Your last equation, $b^2+3b^2-4=0$ is a quadratic in $b^2$. You can define $c=b^2$ and find $c^2+3c-4=0$. Maybe you can factor this, if not you can use the quadratic formula. Solve for $c$, then take the square root to get $b$. This works for quartics that only have the even power terms.
| {
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Find complicated Taylor Series According to some software, the power series of the expression,
$$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}}$$
around $x=0$ is
$$\sqrt{x}-x^{3/2}+\mathcal{O}(x^{5/2}).$$
When I try to do it I find that I can't calculate Taylor because there are divisions by zero. Also I do not understand how Taylor could give non integer powers.
Does anybody know how this expression is calculated?
| Close to $x=0$,$$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}} \simeq \frac{1}{2} \sqrt{-1+ (1+4x)}\simeq \sqrt x$$ So, this must be the start of the development (in order that, locally, your expansion looks like the formula)
If you start using $$\sqrt{1+8 x} \simeq 1+4 x-8 x^2+32 x^3+O\left(x^4\right)$$ then $$\frac{1}{2} \sqrt{-1+\sqrt{1+8 x}} \simeq \frac{1}{2}\sqrt {4 x-8 x^2+32 x^3+O\left(x^4\right)}=\sqrt {x-2 x^2+8 x^3+O\left(x^4\right)}$$ Now, extract $\sqrt x$ and you get $$\sqrt x \sqrt {1-2x+8 x^2}$$ Develop the second square root and you arrive to your wanted result.
Suppose we change the problem to $$\frac{1}{2} \sqrt{-1+\sqrt{1+8 \sqrt{x}}}$$ For the same reasons, the first term should be $x^{1/4}$ and the development would be $$x^{1/4}-x^{3/4}+\frac{7 x^{5/4}}{2}-\frac{33 x^{7/4}}{2}+O\left(x^{9/4}\right)$$
| {
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What's the easiest way to factor $5^{10} - 1$? What's the easiest way to factor $5^{10} - 1$?
I believe $5 - 1$ is a factor based off the binomial theorem.
From there I do not know.
We are using congruence's in this class.
| Note that
$$
5^{10} - 1 = (5^5+1)(5^5-1)
$$
And that
$$
5^5-1 = (5 - 1)(5^4 + 5^3 + 5^2 + 5 + 1)\\
5^5+1 = (5+1)(5^4 - 5^3 + 5^2 - 5 + 1)
$$
That should give you a fairly good start. We cannot factor this further using formal factorization alone.
| {
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Solve the equation $(x-3)(x+9)(x+5)(x-7)=385$ I have one question please...I solved it in this way..so I am not sure..that is it right or not? and If I solved it in the wrong way..so would like to know about the correct way and method to solve it..so please :
Solve the equation (x-3) (x+9) (x+5) (x-7) = 385
and here is I solved it in this way :
Solve the equation:
(x-3) (x+9) (x+5) (x-7) = 385
x ( (1-3) (1+9) (1+5) (1-7)) = 385
x ( (-2) (10) (6) (-6) ) = 385
x ( (-20) (6) (-6) ) = 385
x ( (-120) (-6) ) = 385
x ( 720 ) = 385
720x = 385
x = 385/720
x = 0.5347222
waiting for your replies...
| Let $y = x + 1$ (I got this by taking the average of the four). Then the problem becomes:
$$(y-4)(y+8)(y+4)(y-8) = 385$$
which can be simplified nicely (by multiplying out similar terms) into:
$$\begin{align}(y^2 - 16)(y^2 - 64) &= 385 \\&= 7\cdot55\\&=-7\cdot-55\end{align}$$
The difference between $7$ and $55$ is $48$. But this is exactly the difference between $y^2 - 16$ and $y^2 - 64$. Hence we equate them according to their magnitudes ($y^2 - 16 = 55$) to deduce that:
$$\begin{align}&y^2 - 16 = 55 \\\implies&y^2 = 71\\\implies &(x+1)^2 = 71 \\ \implies &x = -1\pm\sqrt{71}\end{align}$$
Now, we deal with $-7$ and $-55$ in a similar manner:
$$\begin{align}&y^2 - 16 = -7 \\\implies&y^2 = 9\\\implies &(x+1)^2 = 9 \\ \implies &x = -1\pm3\end{align}$$
We conclude with with our four solutions for $x$:
$$x = -1\pm\sqrt{71}, -1\pm3$$
The best part: No quadratic formula used!
| {
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Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$ Last decimal digit of any perfect square must be $0,1,4,5,6$ or $9$
My Proof:
Ten cases exist, yielding the following equalities:
$$(1\mod{10})^2 = 1\mod{10}$$
$$(2\mod{10})^2 = 4\mod{10}$$
$$(3\mod{10})^2 = 9\mod{10}$$
$$(4\mod{10})^2 = 6\mod{10}$$
$$(5\mod{10})^2 = 5\mod{10}$$
$$(6\mod{10})^2 = 6\mod{10}$$
$$(7\mod{10})^2 = 9\mod{10}$$
$$(8\mod{10})^2 = 4\mod{10}$$
$$(9\mod{10})^2 = 1\mod{10}$$
$$(0\mod{10})^2 = 0\mod{10}$$
Since the proposition holds for all possible cases, the proposition holds.
Is this an acceptable proof for the proposition?
What is the simplest proof for this?
Note: This is not a homework question, just a question from a weekly tutorial sheet.
| For a more "clever" solution (but yours looks perfect):
As $10=5\times 2$, let us consider
$$
x^2 = 10a + 5b+ r
\\ 0\le 5b+r< 10
\\ 0\le r< 5
$$
then reducing modulo 5:
$$
r = x^2 \in \{ 0,1,4
\}\mod 5\\
5b+r \in \{ 0,1,4,0+5=5,1+5=6,4+5=9
\}\mod 5\\
$$
| {
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$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $ and $ \ x+y+z \ = \ xyz \ $ Consider the system of equations in real numbers $ \ x,y,z \ $ satisfying
$$ \frac{x}{4 \ \sqrt{x^2+1}} \ = \ \frac{y}{5 \ \sqrt{y^2+1}} \ = \ \frac{z}{6 \ \sqrt{z^2+1}} $$
and $ \ x+y+z \ = \ xyz \ $. Find $ \ x,y,z \ . $
I substituted $ \ x,y,z \ $ as $ \ \tan(\theta_1) \ , \ \tan(\theta_2) \ , \ \tan(\theta_3) \ $ , where
$$ \theta_1 \ + \ \theta_2 \ + \ \theta_3 \ = \ 180º \ \ . $$
| If you have $$15\sin(\theta_1)=12\sin(\theta_2)=10\sin(\theta_3)$$ as suggested by ruler501, you can extract $\theta_2$ and $\theta_3$ as a function of $\theta_1$. This leads to $$\theta_2=\sin ^{-1}\left(\frac{5 \sin (\theta_1)}{4}\right)$$ $$\theta_3=\sin ^{-1}\left(\frac{3 \sin (\theta_1)}{2}\right)$$ So, the equation to solve is $$\theta_1+\sin ^{-1}\left(\frac{5 \sin (\theta_1)}{4}\right)+\sin ^{-1}\left(\frac{3 \sin (\theta_1)}{2}\right)=\pi$$ which only has three solutions. There is an obvious solution which is $\theta_1=\pi$. The other are $\theta_1=0.722734$ and $\theta_1=5.56045$ which I evaluated using numeric methods.
If you use the more elegant solution proposed by LAcarguy while I was typing my answer, the solution is quite simple since you arrive to $$A=\pm \cos ^{-1}\left(\frac{3}{4}\right)=\pm 0.722734$$
| {
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Find closed formula $f(n)$ from generating function I'm asked to find a closed formula for
$f(n)=6f(n-1)-9f(n-2)$ for $n>1$ with $f(0)=-1. f(1)=0$,
using the ordinary generating function $F(X)$.
I found $F(X)=-1/(1-3x)^2$ but from there I don't manage to get a satisfactory formula for $f(n)$. Can anyone give me a hint ?
Thanks
| Write the recurrence as:
$$
f(n + 2) = 6 f(n + 1) - 9 f(n) \qquad f(0) = - 1, f(1) = 0
$$
Multiply by $z^n$ and sum over $n \ge 0$. Recognize:
\begin{align}
\sum_{n \ge 0} f(n + 1) z^n &= \frac{F(z) - f(0)}{z} \\
\sum_{n \ge 0} f(n + 2) z^n &= \frac{F(z) - f(0) - f(1) z}{z^2}
\end{align}
and get:
$$
\frac{F(z) + 1}{z^2} = 6 \frac{F(z) + 1}{z} - 9 F(z)
$$
Solving for $F(z)$, and splitting into partial fractions:
$$
F(z) = \frac{1 - 6 z}{1 - 6 z + 9 z^2} = \frac{1}{(1 - 3 z)^2} - \frac{2}{1 - 3 z}
$$
This tells us:
$$
f(n) = \binom{-2}{n} (-1)^n 3^n - 2 \cdot 3^n
= \left( \binom{n + 1}{1} - 2 \right) \cdot 3^n
= \frac{n - 3}{2} \cdot 3^n
$$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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About asymptotic behaviour of a divergent integral. I have the function $f(x) = x \tanh(\pi x) \log (x^2 +a^2)$ where $a$ is some positive real number. For the logarithm I am assuming a branch-cut along the positive imaginary axis starting at $x = ia$. So the branching effect on the "log" is that its imaginary part is $\pi$ on the right and above the point $ai$ on the positive imaginary axis and $-\pi$ on the left and above.
I wanted to understand how the value of the integral of this function behaves when $x$ is integrated from $0$ to $A >0$. Using the parity symmetry of the function I extended it on a semi-circle in the upper half plane of radius $A$ about the origin.
Then I did a large $A$ expansion ("Series" about $A=\infty$ on Mathematica) and found that $xf(x)$ (that would practically be the integrand when converted into a $\phi$ integral with $x = Ae^{i\phi}$), this function asymptotes as, $e^{2i\phi} \log(e^{2i\phi})A^2 + 2A^2 \log(A) + a^2$ (+ terms which go to zero as $A$ goes to infinity). Now on this asymptotic form the $\phi$ is integrable from $0$ to $\pi$ giving a finite answer.
*
*So relying on this Mathematica result can I say that this integral therefore diverges in the large $A$ limit as $ \text{(number)} A^2 + \text{(number)} A^2 \log(A)$ ?
*Further I chose $a=2$ and it seems to me that on using "NIntegrate" Mathematica says that the integral of this function on a circle centered at $2i$ goes to zero as the radius goes to 0.
Is this correct?
Thinking analytically this seems to be right - if one does a small radius expansion of integrand about the branch point $2i$ then the integrand smoothly goes to $0$ in the small radius limit. One might fear that the small radius limit does not commute with the integral - to check that I also plotted the value of the numerical integration of the function around this circle as a function of the radius - and the graph does seem to go to zero smoothly in the zero radius limit.
In doing the above Mathematica based checking I am trusting that Mathematica knows how to contour integrate a function around its branch-cut. Though thinking analytically I get that while going around the point $2i$ the function needs to be modified as follows (to fit with the branching convention stated earlier at the top) - that for $0 < \phi = arg(z - 2i = \epsilon e^{i \phi}) < \frac{\pi}{2}$ one adds a $-\frac{i3\pi}{2}$ to the $log (z^2 + 4)$ and add a $\frac{i\pi}{2}$ for $\frac{\pi}{2} \leq \phi \leq 2\pi$.
| You're interested in the asymptotic behavior of
$$
\begin{align}
I_a(A) &= \int_0^A x \tanh(\pi x) \log(x^2+a^2)\,dx \\
&= \int_1^A x \tanh(\pi x) \log(x^2+a^2)\,dx + \int_0^1 x \tanh(\pi x) \log(x^2+a^2)\,dx \tag{1}
\end{align}
$$
as $A \to \infty$. Let's rewrite the integrand as
$$
\begin{align}
&x \tanh(\pi x) \log(x^2+a^2) \\
&\qquad = x \log(x^2+a^2) + x \log(x^2+a^2) \Bigl(\tanh(\pi x) - 1\Bigr) \\
&\qquad = x\log(x^2) + x\log(1+a^2/x^2) + x \log(x^2+a^2) \Bigl(\tanh(\pi x) - 1\Bigr). \tag{2}
\end{align}
$$
Now
$$
|\tanh(\pi x) - 1| = \frac{2}{1+e^{2\pi x}} \leq 2e^{-2\pi x} = O(e^{-2\pi x}) \qquad \text{as } x \to \infty, \tag{3}
$$
$$
\log(x^2 + a^2) = O(x) \qquad \text{as } x \to \infty, \tag{4}
$$
and
$$
\log(1+a^2/x^2) = \frac{a^2}{x^2} + O(x^{-4}) \qquad \text{as } x \to \infty, \tag{5}
$$
so using $(3)$, $(4)$, and $(5)$ in $(2)$ we have
$$
\begin{align}
x \tanh(\pi x) \log(x^2+a^2) &= x\log(x^2) + \frac{a^2}{x} + O(x^{-3}) + O(x^2 e^{-2\pi x}) \\
&= 2x\log x + \frac{a^2}{x} + O(x^{-3})
\end{align}
$$
as $x \to \infty$. Therefore, if we write
$$
\begin{align}
&\int_1^A x \tanh(\pi x) \log(x^2+a^2)\,dx \\
&\qquad = \int_1^A \left(2x\log x + \frac{a^2}{x}\right)\,dx + \int_1^A \left(x\tanh(\pi x)\log(x^2+a^2) - 2x\log x - \frac{a^2}{x}\right)\,dx,
\end{align}
$$
$$
\tag{6}
$$
then the last integral converges as $A \to \infty$, and the second to last integral is
$$
\int_1^A \left(2x\log x + \frac{a^2}{x}\right)\,dx = A^2 \log A - \frac{1}{2}A^2 + a^2 \log A + \frac{1}{2}. \tag{7}
$$
Thus, defining
$$
\begin{align}
C(a) = &\frac{1}{2} + \int_1^\infty \left(x\tanh(\pi x)\log(x^2+a^2) - 2x\log x - \frac{a^2}{x}\right)\,dx \\&\qquad + \int_0^1 x \tanh(\pi x) \log(x^2+a^2)\,dx
\end{align}
$$
and combining $(1)$, $(6)$, and $(7)$ we get
$$
I_a(A) = A^2 \log A - \frac{1}{2}A^2 + a^2 \log A + C(a) + \epsilon_a(A),
$$
where $\epsilon_a(A) \to 0$ as $A \to \infty$ for fixed $a \in \mathbb R$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
modular arithmetic problem (when solving elliptic curves) Given E: (elliptical curve)
$y^2 = x_3+2x+2 \bmod 17$
Recall: $y^2 = x^3+ax+b$
point $P=(5,1)$
Compute:
$2P = P+P = (5,1)+(5,1)= (x_3,y_3)$
Now the formula used here is
slope $m = \dfrac{3x_1^2 +a}{2y_1}$
and what they got is $s = (2 · 1)^{−1} (3 · 5^2 + 2) = 2^{−1} · 9 ≡ 9 · 9 ≡ 13 \bmod 17$
this is what confuses me.
should it not be $(2·1)^{-1}(3 · 5^2 + 2)= 38.5 \bmod 17$ which is 4.5?
Sorry if this may seem trivial, I'm doing a research for a high school project.
The next steps are:
$x_3 = m^2 − x_1 − x_2 = 13^2 − 5 − 5 = 159 ≡ 6 \bmod 17$
$y_3 = m(x_1−x_3) − y_1 = 13(5 − 6) − 1= −14 ≡ 3 \bmod 17$
$(x_3,y_3) = (6,3)$
which I don't understand as well
If someone could enlighten me on this I would be very happy!
| $$2^{-1}=9\pmod{17}\;\;,\;\;3\cdot 5^2=3\cdot 8=7\pmod{17}\implies$$
$$(2\cdot 1)^{-1}(3\cdot 5^2+2)=9(7+2)=9^2=13\pmod{17}$$
BTW, this is exactly what you got, since
$$38.5=4.5=\frac92=9\cdot2^{-1}=9\cdot9=13\pmod{13}\ldots$$
| {
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"source": "stackexchange",
"question_score": "1",
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} |
How to solve $x^4-8x^3+24x^2-32x+16=0$ How can we solve this equation?
$x^4-8x^3+24x^2-32x+16=0.$
| As $x\ne0,$ dividing either sides by $x^2$
$$x^2+\left(\frac4x\right)^2-8\left(x+\frac4x\right)+24=0$$
Now as $\displaystyle x^2+\left(\frac4x\right)^2=\left(x+\frac4x\right)^2-2\cdot x\cdot\frac4x$
Setting $x+\dfrac4x=y,$ we get $\displaystyle y^2-8-8y+24=0\implies(y-4)^2=0\iff y=4$
So, we have $\displaystyle x+\frac4x=4\iff(x-2)^2=0$
| {
"language": "en",
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"source": "stackexchange",
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Gaussian Quadrature -Deriving a Formula- The following is an exercise in the problem section of the Gaussian Quadrature chapter.
The theorem:
Derive a formula of the form $$\int_{a}^{b} f(x)dx \approx w_0f(x_0) + w_1f(x_1) + w_2f'(x_2) + w_3f'(x_3)$$
I don't really know how to even start this. I have the Gaussian Quadrature theorem, but...I suppose I don't know how to use it. Can anyone show me how to do this?
We have four weights, $w_0, w_1,w_2,w_3$ and four nodes $x_0, x_1,x_2,x_3$. So our polynomial $q(x)$ will be of degree $4$. (Will the number of weights always determine the degree of the polynomial?). So from $$ \int_{a}^{b} x^kq(x)dx = 0$$ where $0 \leq k \leq n$ and the degree of $q(x)$ is $n+1$, we can write $$\int q(x)dx = \int xq(x)dx = \int x^2q(x) dx = 0$$ right?
From here I don't know what to do.
| Let's look first at $\int_{-1}^{1} f(t) \ dt$.
We want the formula to evaluate $\int_{-1}^{1} dt, \int_{-1}^{1} t \ dt, \int_{-1}^{1} t^{2} \ dt, \int_{-1}^{1} t^{3} \ dt, \int_{-1}^{1} t^{4} \ dt, \int_{-1}^{1} t^{5} \ dt, \int_{-1}^{1} t^{6} \ dt $ and $\int_{-1}^{1} t^{7} \ dt$ exactly.
That leads to the following system of equations:
$$ 2 = \omega_{0} + \omega_{1}$$
$$ 0 = \omega_{0} x_{0} + \omega_{1} x_{1}+ \omega_{2}+\omega_{3}$$
$$ \frac{2}{3} = \omega_{0} x_{0}^{2} + \omega_{1} x_{1}^{2} + 2 \omega_{2} x_{2} + 2 \omega_{3} x_{3}$$
$$ 0 = \omega_{0} x_{0}^{3} + \omega_{1} x_{1}^{3}+ 3 \omega_{2} x_{2}^{2} + 3 \omega_{3} x_{3}^{2} $$
$$ \frac{2}{5} = \omega_{0} x_{0}^{4} + \omega_{1} x_{1}^{4}+ 4 \omega_{2} x_{2}^{3} + 4 \omega_{3} x_{3}^{3} $$
$$0 = \omega_{0} x_{0}^{5} + \omega_{1} x_{1}^{5} + 5 \omega_{2} x_{2}^{4} + 5 \omega_{3} x_{3}^{4} $$
$$\frac{2}{7} = \omega_{0} x_{0}^{6} + \omega_{1} x_{1}^{6} + 6 \omega_{2} x_{2}^{5} + 6\omega_{3} x_{3}^{5} $$
$$ 0 = \omega_{0} x_{0}^{7} + \omega_{1} x_{1}^{7} + 7 \omega_{2} x_{2}^{6} + 7\omega_{3} x_{3}^{6}$$
Solve the system using a numerical solver.
Then use the the fact that $$\int_{a}^{b} f(x) \ dx = \frac{b-a}{2} \int_{-1}^{1} f \Big(a+(1+t)\frac{b-a}{2} \Big) \ dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/732830",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proof by induction that $ 169 \mid 3^{3n+3}-26n-27$ $ 169$ | $3^{3n+3}-26n-27$ ?
Fulfilled for $n=0$. Induction to $n+1$:
An integer $x$ exists so that
$ 169x= 3^{3n+6}-26n-27-26$
$ 169x= 27*3^{3n+3}-26n-27-26$
$ 169x= 26*3^{3n+3}+3^{3n+3}-26n-27-26$
An integer $m$ exists so that
$ 169x= 26*3^{3n+3}+169m-26$
($ 13x= 2*(3^{3n+3}-1)+13m$)
Now I'm stuck although it looks simple.
Thanks for any input in advance.
| The same, but using modules.
Let's prove by induction that $3^{3n+3} - 26n - 27 \equiv 0 \pmod {169}$
Working with module $169$:
*
*Base case. If $n = 1$, then $3^3 - 27 \equiv 0$.
*Induction. Let fix an integer $m \geq 0$ and supose that (induction hypothesis):
$$3^{3m+3}-26m -27 \equiv 0 \pmod{169}$$
or, with other words:
$$3^{3m+3} \equiv 27 + 26m \pmod{169}$$
We want to prove that
$$3^{3m+6}-26(m+1) -27 \equiv 0 \pmod{169}$$
Prove:
$$\begin{align}3^{3m+6}-26(m+1)-27&\equiv 3^3 \cdot 3^{3m+3} - 26m - 26 -27\\
&\equiv 27 \cdot 3^{3m+3} - 26m -26-27\quad(*)\\
&\equiv 27(27+26m) - 26m - 26 -27\\
&\equiv (702-26)m + (729-27-26)\\
&\equiv 4\cdot 169m + 4 \cdot 169\\
&\equiv 0m + 0 \\
&\equiv 0 \end{align}$$
We have applied induction hypothesis at $(*)$.
So, as we have proved that $3^{3n+3} - 26n - 27 \equiv 0 \pmod {169}$, then, $169 | 3^{3n+3} - 26n - 27$.
| {
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"source": "stackexchange",
"question_score": "1",
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Fibonacci Proof Using Induction
$$f(n) = \left\{\begin{matrix}
0 & n=1\\
1 & n=2\\
f_{n-1} + f_{n-2} &
n\geqslant 2\end{matrix}\right.$$
How can I prove by induction that $$f_{n} \leq \left ( \frac{1+\sqrt{5}}{2} \right )^{n-1}$$ for all$$ n\geq l_{a}$$, I have to find the smallest value for $$l_{a}$$
| $(1)$ $f_1 \leq (\frac{1+\sqrt{5}}{2})^{1-1} \Rightarrow 0 \leq 1,f_2 = 1 \leq \frac{1+ \sqrt{5}}{2} \approx 1.618 $. $\textbf{True}$
$(2)$ Next: Suppose for some $\textbf{k}$ $\geq n$ we have $f_k \leq (\frac{1+\sqrt{5}}{2})^{k-1}$
$(3)$ We want to show: $f_{k+1} \leq (\frac{1+\sqrt{5}}{2})^{k}$
$(4)$ Well, $f_{k+1} = f_{k-1}+f_{k-2}$
$(5)$ By hypothesis we have bounds on both of the things on right of the inequality.
$(6)$ Thus: $f_{k+1} \leq \ 2(\frac{1+\sqrt{5}}{2})^{k-1} \leq (\frac{1+\sqrt{5}}{2})^{k-1} (\frac{1+\sqrt{5}}{2})^{k} = (\frac{1+\sqrt{5}}{2})^{k}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/733282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of subsets without consecutive numbers Consider $S=\{1,2,\ldots,15\}$. Let $X$ denote the number of subsets of $S$ of four elements which contain no consecutive numbers.
The claim is that $X$ equals the coefficient of $x^{14}$ in $\dfrac{x^6}{(1-x)^5}$. How would I prove this?
| Call the elements of your subset $a_1$, $a_2$, $a_2$, $a_4$ in order. Your restrictions translate into:
\begin{align}
x_1 &= a_1 \ge 1 \\
x_2 &= a_2 - a_1 \ge 2 \\
x_3 &= a_3 - a_2 \ge 2 \\
x_4 &= a_4 - a_3 \ge 2 \\
x_5 &= 15 - a_4 \ge 0
\end{align}
The definitons amount to $x_1 + x_2 + x_3 + x_4 + x_5 = 15$. Set up generating functions for each $x_i$ variable:
\begin{align}
x_1: &\quad z + z^2 + \cdots = \frac{z}{1 - z} \\
x_2 \text{ to } x_4: &\quad z^2 + z^3 + \cdots = \frac{z^2}{1 - z} \\
x_5: &\quad 1 + z + \cdots = \frac{1}{1 - z}
\end{align}
Multiplying all, and remembering we want the term whose exponent is 15:
\begin{align}
[z^{15}] \frac{z^7}{(1 - z)^5}
&= [z^8] (1 - z)^{-5} \\
&= \binom{-5}{8} (-1)^8 \\
&= 495
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Need help to solve taylor series of $e^{\sin x}$ How to derive the taylor series of $e^{\sin x}$, up to $x^5$?
i just don't know how to get the answer
$$f(x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} -\frac{x^5}{15}$$
really need some help. Thanks
| If
$$
f(x) = a_1x + \frac{a_2}{2}x^2 + \frac{a_3}{6} x^3 + \frac{a_4}{24} x^4 + \cdots + \frac{a_n}{n!} x^n + \cdots
$$
then
$$
e^{f(x)} = 1 + a_1 x + \frac{a_2+a_1^2}{2!} + \frac{a_3 + 3a_2a_1+a_1^3}{6} x^3 + \frac{a_4+4a_3a_1 + 3a_2^2 + 6a_2a_1 + a_1^4}{24} x^4 + \cdots
$$
The pattern is this: Consider the last case shown above, where $n=4$. There are several ways to partition the number $4$:
$$
\begin{array}{r|c}
\text{integer partition} & \text{number of set partitions} \\
\hline
4 & 1 \\
3+1 & 4 \\
2+2 & 3 \\
2+1+1 & 6 \\
1+1+1+1 & 1
\end{array}
$$
The numbers of set partitions are the coefficients. See this page: http://en.wikipedia.org/wiki/Exponential_formula
Now try this with
$$
a_1 = 1,\quad a_2=0,\quad a_3=-1,\quad a_4=0,\quad a_5=1,\quad\ldots\ldots
$$
$$
e^{\sin x} = 1 + x + \frac 1 2 x^2 + \frac{(-1)+0+1^3}{3} x^3 + \cdots
$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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} |
Taylor expansion of $\frac{1}{2-z-z^2}$ The problem is:
Find the Taylor expansion of $f(z):= \dfrac{1}{2-z-z^2}$ on the disc $|z| < 1$
So far I have used partial fractions to obtain $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{2+z}\right)$ which I then rewrite as $f(z) = \dfrac{1}{3}\left(\dfrac{1}{1-z} + \dfrac{1}{1-((-1)(1+z))}\right)$.
My problem is that we are told specifically to find the expansion on the disc $|z| < 1$, and whilst $\dfrac{1}{1-z}$ is valid for $|z| < 1$, $\dfrac{1}{1-((-1)(1+z))}$ is valid for $|1+z| < 1$. I am a guessing that we cannot just say $f(z) = \dfrac{1}{3}\left(\displaystyle\sum_{n=0}^\infty z^n + (-1)^n(1+z)^n\right)$ but am not sure as my notes don't have any examples of this sort of difficulty.
| $\frac{1}{2-z-z^2}=\frac{ 1 }{ 2} \frac{1}{1-\frac{1}{2}z-\frac{1}{2}z^2}=\frac {1}{2}\sum _{ n = 0 } ^ \infty(\frac { 1 } { 2 } (z+z^2))^{n}$.
Done.
Note that Since $|z|\lt 1 $, it follows that $|\frac {1}{2}(z + z ^ 2 )|\lt \frac {1}{2}(|z|+|z ^ 2|)\lt1$ so it is justified to use geometric series above.
| {
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Compute summation with a relative error of O(n^-2) $a(n) = \sum_{i \geq 0} a_i n^{-i}$, how can we compute the value of $a(n)^n$ with a relative error of $O(n^{-2})$?
| Since
$$
a(n) = a_0 + a_1 n^{-1} + a_2 n^{-2} + O(n^{-3})
$$
we have
$$
\begin{align}
\log a(n) &= \log a_0 + \log\left(1 + \frac{a_1}{a_0} n^{-1} + \frac{a_2}{a_0} n^{-2} + O(n^{-3})\right) \\
&= \log a_0 + \frac{a_1}{a_0} n^{-1} + \frac{a_2}{a_0} n^{-2} + O(n^{-3}) \\
&\qquad - \frac{1}{2} \left(\frac{a_1}{a_0} n^{-1} + \frac{a_2}{a_0} n^{-2} + O(n^{-3})\right)^2 + O(n^{-3}) \\
&= \log a_0 + \frac{a_1}{a_0} n^{-1} + \left(\frac{a_2}{a_0} - \frac{a_1^2}{2a_0^2}\right)n^{-2} + O(n^{-3})
\end{align}
$$
so
$$
\begin{align}
a(n)^n &= \exp\left[n \log a(n)\right] \\
&= \exp\left[n \log a_0 + \frac{a_1}{a_0} + \left(\frac{a_2}{a_0} - \frac{a_1^2}{2a_0^2}\right)n^{-1} + O(n^{-2})\right] \\
&= \exp\left[n \log a_0 + \frac{a_1}{a_0} + \left(\frac{a_2}{a_0} - \frac{a_1^2}{2a_0^2}\right)n^{-1}\right] \exp\Bigl[O(n^{-2})\Bigr] \\
&= \exp\left[n \log a_0 + \frac{a_1}{a_0} + \left(\frac{a_2}{a_0} - \frac{a_1^2}{2a_0^2}\right)n^{-1}\right] \Bigl(1+O(n^{-2})\Bigr),
\end{align}
$$
as desired.
| {
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I'm stuck. Can somebody help me solve this problem. The sum of our present ages is $41$. If I live $17$ more years, after doubling my present age, I will be $9$ years less than your present age. What are our present age?
| $y=\text{Your age, and }x=\text{Mine}$.
$$x+y=41\\
\text{And also }17+2x=y-9\\
\implies 2x-y=-26$$
Let me do this a bit differently - probably not meant for precalculus, but here it goes anyway.
$$\begin{pmatrix}1&1\\2&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}41\\-26\end{pmatrix}$$
Then,
$$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\dfrac{1}{3}&\dfrac{1}{3}\\ \dfrac{2}{3}&-\dfrac{1}{3}\end{pmatrix}\begin{pmatrix}41\\-26\end{pmatrix}\\
\implies \begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}5\\26\end{pmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate $\int\frac{x}{\sqrt{x^2+x+1}} \, dx$ using trigonometric substitution? I am pretty sure that my answer is correct but given answer for the exercise from textbook Calculus James Steward was slightly different. Any idea to solve this:
$$\int\frac{x}{\sqrt{x^2+x+1}} \, dx$$
The given answers: $\sqrt{x^2+x+1}-\frac{1}{2}\ln(\sqrt{x^2+x+1}+x+\frac{1}{2})+c$
Thanks in advance.
| First split off the bit that does not need a trigonometric substitution:
$$\int\frac{x\,dx}{\sqrt{x^2+x+1}}=\int\frac{x+\frac{1}{2}}{\sqrt{x^2+x+1}}dx
-\frac{1}{2}\int\frac{dx}{\sqrt{x^2+x+1}}\ .$$
You should see that the first integral is now easy. For the second, complete the square:
$$\int\frac{dx}{\sqrt{x^2+x+1}}=\int\frac{dx}{\sqrt{(x+\frac{1}{2})^2+\frac{3}{4}}}\ .$$
If you really want to use a trig substitution, try
$$x+\frac{1}{2}=\frac{\sqrt3}{2}\tan\theta\ ,$$
which still leaves you with a slightly tricky integral. Better, use the hyperbolic substitution
$$x+\frac{1}{2}=\frac{\sqrt3}{2}\sinh\theta\ .$$
If you want to directly get the answer from the book, do some algebra:
$$\frac{1}{\sqrt{x^2+x+1}}
=\frac{\displaystyle\frac{x+\frac{1}{2}}{\sqrt{x^2+x+1}}+1}{\sqrt{x^2+x+1}+x+\frac{1}{2}}\ .$$
With a bit of thought you will see why this is easy to integrate.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proof polynomial has only one real root. I need to prove that this polynomial equation:
$$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2}).$$
has only one root. That it has one real root is obvious because it is of odd degree. But Descartes rules here fails to bound the number of roots to one.
| I try to factorize your formula :
$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=0\quad\text{ for }\quad a\in(0,\frac{1}{2})$
$x^5-(3-a)x^4+(3-2a)x^3-ax^2+2ax-a=x^5+(3-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a$
$=x^5+(1-a)(x^3-x^4)-ax^{3}-ax^{2}+2ax-a+2x^{3}-2x^{4}$
$=x^5-1+1+(1-a)(x^3-x^4)-a(x^{3}-x^{2})+2ax+2ax^{2}-a+2x^{3}-2x^{4}$
$=x^5-1+(1-a)(x^3-x^4)-a(x^{3}-x^{2})+2a(x-1)+2x^{3}(1-x)+1+a+2ax^{2}$
$=(x-1)(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})+1+a+2ax^{2}$$=x(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})+1+a-(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2}-2ax^{2})$
$=x(x^{4}+x^{2}+x+1+ax^{3}-ax^{2}+2a+2x^{2})-(x^{4}+x^{2}+x+ax^{3}-ax^{2}+a+2x^{2}-2ax^{2})$
$\Longrightarrow{x(x^{4}+(a-1)x^{3}+(3-2a)x^{2}+(3a-2)x+2a)=a}$
$\Longrightarrow{f(x)=x(x^{4}+(a-1)x^{3}+(3-2a)x^{2}+(3a-2)x+2a)}$
$\Longrightarrow{f^{'}(x)=0}$
$\Longrightarrow{-10x^{4}=2-4a>0}$
a contradiction shows that $x=0$ is its unique root!
Is it helpful?
| {
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"timestamp": "2023-03-29T00:00:00",
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Invalid subtraction when solving system of equations? I'm trying to solve these two equations:
$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$
and I tried to do it by subtracting the first equation from the second, yielding $(4x-4y)(x^2+y^2)=0$. Clearly this is satisfied when $x=y$, which gives $(x,y)=(\frac{1}{2},\frac{1}{2})$, or when $x^2+y^2=0$, which gives $(x,y)=(0,0)$.
But when I plug them back in, it's obvious that $(x,y)=(0,0)$ is not a solution. Where did I go wrong?
|
$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$
Let $S$ the set of solutions.
You can summarise the logic steps:
*
*If $(x,y)$ is a solution:
$$4x(x^2+y^2)=4y(x^2+y^2)
\implies x^2+y^2=0 \text{ or }x=y
\\
\implies (x,y)=(0,0) \text{ or }
\\
x=y=\frac{1}{4(x^2+y^2)}
=\frac{1}{8x^2}\implies x=y=\frac 1{8^{1/2}}=\frac 12
$$
You have proved $S\subset\{(1/2,1/2),(0,0)\}$.
*If $x=(1/2,1/2)$ the equation is true. If $x=(0,0)$ the equation is false.
You have proved $S=\{(1/2,1/2)\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$ I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.
$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
Is there some other more effective way for this limit?
| Taylor expansions yield the result without much difficulty. We have
$$\sqrt[3]{1+y} = 1 + \frac{y}{3} + O(y^2),$$
and
$$\cos z = 1 - \frac{z^2}{2} + O(z^4),$$
hence
$$\sqrt[3]{\cos x} = \sqrt[3]{1 - x^2/2 + O(x^4)} = 1- \frac{x^2}{6} + O(x^4),$$
so
$$\frac{1-\sqrt[3]{\cos x}}{1-\cos \sqrt[3]{x}} = \frac{\frac{x^2}{6} + O(x^4)}{\frac{x^{2/3}}{2} + O(x^{4/3})} = \frac{x^{4/3}}{3} + O(x^2) \to 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/739919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.
| Alternatively, $a^2+a^2+a^2+a^2+b^2+c^2 \geq 6 \sqrt[6]{a^8b^2c^2} = 6 \sqrt[6]{a^6} = 6|a| \geq 6a$ by AM-GM. Adding the analogous inequalities $a^2+4b^2+c^2 \geq 6b$ and $a^2+b^2+4c^2 \geq 6c$ gives the result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Extra help on inequality Someone very helpfully provided an answer to an inequality. See Hard Olympiad Inequality However I don't get part of their answer. How did they get the last factorization??? Thanks so much for any help.
Would Wolfram Alpha help? I plugged it in it is still loading
| Below is a rather long proof, but every step is straightforward. For any function $\phi$ in three variables $a,b,c$, let us denote
$$
\begin{array}{lcl}
\Sigma_{cyc}\phi&=&\phi(a,b,c)+\phi(b,c,a)+\phi(c,a,b) \\
\Sigma_{all}\phi&=&\phi(a,b,c)+\phi(a,c,b)+\phi(b,a,c)+\phi(b,c,a)+\phi(c,a,b)+\phi(c,b,a)
\end{array}\tag{1}
$$
Our job is to rewrite
$$
A_1= (a+b+c)^2-4(ab+ac+bc)\bigg(\sum_{cyc}\frac{a^2}{(a+b)(a+c)}\bigg) \tag{2}
$$
First, letting $p(a,b,c)=(a+b)(a+c)(b+c), q(a,b,c)=\sum_{all}a^2b$ and $A_2=p(a,b,c)A_1$, we equivalently
have to rewrite $A_2$, where
$$
A_2=(a+b+c)^2p(a,b,c)-4(ab+ac+bc)q(a,b,c) \tag{3}
$$
We can do it as follows, by breaking the symmetry and separating $a$ from the pair $(b,c)$ :
$$
\begin{array}{lcl}
A_2 &=& (a+b+c)^2p(a,b,c)-4(ab+ac+bc)q(a,b,c) \\
&=& 4(a+b+c)^2p(a,b,c)-3(a+b+c)^2p(a,b,c) \\
& & -8(ab+ac+bc)q(a,b,c)+
4(ab+ac+bc)q(a,b,c) \\
&=& \bigg(\sum_{all} 2a(a+b+c)p(a,b,c)\bigg)
+\bigg(\sum_{all} \frac{1}{2}(a(b^2+c^2)-4a^2(b+c)-2abc)(a+b+c)^2\bigg)
\\
& & -\bigg(\sum_{all} 2a(b+c)q(a,b,c)\bigg)+
\bigg(\sum_{all} 2a^2(b+c)(ab+ac+bc)\bigg)
\\
&=& \sum_{all} aA_3(a,b,c),\\
\end{array}\tag{4}
$$
where
$$
\begin{array}{lcl}
A_3 &=& 2(a+b+c)p(a,b,c)+\frac{1}{2}((b^2+c^2)-4a(b+c)-2bc)(a+b+c)^2 \\
& & -2(b+c)q(a,b,c)+2a(b+c)(ab+ac+bc) \\
&=& 2(a+b+c)(q(a,b,c)+2abc)+\frac{1}{2}((b-c)^2-4a(b+c))(a+b+c)^2 \\
& & -2(b+c)q(a,b,c)+2a(b+c)(ab+ac+bc) \\
&=& 2a(q(a,b,c)+2abc)+2(b+c)(2abc)+\frac{1}{2}((b-c)^2-4a(b+c))(a+b+c)^2 \\
& & +2a(b+c)(ab+ac+bc) \\
&=& 2ap(a,b,c)+2a(b+c)(ab+ac+3bc)+\frac{1}{2}((b-c)^2-4a(b+c))(a+b+c)^2 \\
&=& 2ap(a,b,c)+2a(b+c)(ab+ac+3bc-(a+b+c)^2)+\frac{(b-c)^2}{2}(a+b+c)^2 \\
&=& 2a\bigg(p(a,b,c)+(b+c)(bc-(ab+ac)-a^2-b^2-c^2)\bigg)+\frac{(b-c)^2}{2}(a+b+c)^2 \\
&=& 2a(b+c)\bigg((a+b)(a+c)+bc-(ab+ac)-a^2-b^2-c^2)\bigg)+\frac{(b-c)^2}{2}(a+b+c)^2 \\
&=& 2a(b+c)\bigg(-(b-c)^2\bigg)+\frac{(b-c)^2}{2}(a+b+c)^2 \\
&=& (b-c)^2\bigg(-2a(b+c)+\frac{(a+b+c)^2}{2}\bigg)\\
&=& (b-c)^2\frac{(b+c-a)^2}{2}
\end{array}\tag{5}
$$
and we are done.
| {
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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.