Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
$x^a - 1$ divides $x^b - 1$ if and only if $a$ divides $b$ Let $x > 1$ and $a$, $b$ be positive integers. I know that $a$ divides $b$ implies that $x^a - 1$ divides $x^b - 1$. If $b = qa$, then
$$x^b - 1 = (x^a)^q - 1^q = (x^a - 1)((x^a)^{q-1} + \ldots + x^a + 1).$$
I'm interested in the converse of the statement. If $x^a - 1$ divides $x^b - 1$, does this imply that $a$ divides $b$?
| Let $b=a \cdot q+r$, where $0 < r < a$.
Then
$$x^b-1=x^b-x^r+x^r-1=x^r(x^{aq}-1)+x^r-1 \,.$$
Use that $x^a-1$ divides $x^{aq}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
"answer_count": 2,
"answer_id": 1
} |
Matrix multiplication proof: $(A+B)^k$ for matrices such that $A^2=B^2=0$ and $AB=BA$ I have the following details about the matrices $A,B$:
$$A_{n\times n},B_{n\times n}$$ $$A^2 = B^2 = 0$$ $$AB=BA$$
I need to find $x \in \mathbb{N}$ so $(A+B)^x=0$
What implications can I make from the given details?
*
*$A,B$ doesn't have to equal $0$ so $A^2 = B^2 = 0$
*Can I imply from the 2nd and 3rd given details that $A=B$?
*Any other leads?
| $$\mbox{ let } x > 2\mbox{ and } x\in \mathbb{N}$$
$$\textbf{(A+B)}^2 = \textbf{A}^2+\textbf{B}^2+2\textbf{AB} = \textbf{2AB}$$
$$\textbf{(A+B)}^x = \textbf{(A+B)}^2\textbf{(A+B)}^{x-2}$$
$$\Rightarrow \textbf{(A+B)}^x =2\textbf{AB}\textbf{(A+B)}^{x-2}$$
$$\Rightarrow \textbf{(A+B)}^x =2\textbf{AB}\textbf{(A+B)}\textbf{(A+B)}^{x-3}$$
$$\Rightarrow \textbf{(A+B)}^x =2(\textbf{ABA} + \textbf{ABB})\textbf{(A+B)}^{x-3}$$
$$\Rightarrow \textbf{(A+B)}^x =2(\textbf{AAB} + \textbf{ABB})\textbf{(A+B)}^{x-3}\quad(\because \textbf{BA = AB})$$
$$\Rightarrow \textbf{(A+B)}^x = \textbf{0} (\because\hspace{2pt}\textbf{A}^2 = \textbf{B}^2 = 0)$$
$$\text{If x = 2, I can not show }\textbf{(A+B)}^2 = 0$$
$$\text{ can anyone to prove that or give a counter example}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
| Funny brute force approach. Just show:
$$n^5-n = 5!\binom{n+2}{5} + 5\cdot 3!\binom{n+1}{3}=120\binom{n+2}{5}+30\binom{n+1}{3}$$
In general, if $d$ is odd, then you can write:
$$n^d-n = \sum_{i=0}^{\lfloor d/2\rfloor} a_i \binom{n+i}{2i+1}$$
where the $a_i$ are multiples of $(2i+1)!$.
Since the left side is divisible by $(n+1)n(n-1)$, we have that $a_0=0$. We also have $a_{\lfloor d/2\rfloor}=1$. So we have $\lfloor d/2\rfloor-1$ different $a_i$ to determine.
But you can actually determine them by induction, since $\binom{n+i}{2i+1}=0$ when $2i+1>n$. So choose $n=2$ and we get $2^{d}-2 = a_1\binom{3}{3}$ so $a_1=(2^{d}-2)$.
Then take $n=3$, and we get:
$$3^d-3 = a_2\binom{3+2}{5} + a_1\binom{3+1}{3}=a_2 + 4(2^d-2)$$
So $a_2 = 3^d-3 - 4(2^d-2)$.
In general, for $a_k$, we take $n=k+1$ and:
$$a_k = (k+1)^d - (k+1) - \sum_{i=0}^{k-1} a_i\binom{k+1+i}{2i+1}$$
By induction, you can prove that if $D\mid n^d-n$ for all integers $n$, then $D$ must divide $a_i$ for all $i$. And obviously, any factor of all $a_i$ is a common factor of $n^d-n$ for all $n$. So the general solution is the greatest common divisor of $a_1,\dots,a_{\lfloor d/2\rfloor}$. You can do this sequentially, so you compute $3^d-3$ modulo $2^d-2$, and then compute the GCD, etc.
For example, for $d=7$, $2^7-2=126$. Next compute $3^7-3\bmod {126}=42$. Since $42\mid 126$, we get $\gcd(2^7-2,3^7-3)=42$. Then compute $4^7-4$. This is actually obviously divisible by $2(2^6-1)=a_1$. So we get:
$$\gcd(2^7-2,3^7-3,4^7-4)=42$$
This actually works with any odd integer polynomial $g(n)$ of degree $d$ - the greatest common factor of the values of $g(n)$ is:
$$\gcd(g(1),g(2),\dots,g(\lceil d/2\rceil))$$
Now, if $a^d\equiv a\pmod D$ and $b^d\equiv b\pmod D$ then $(ab)^d \equiv ab\pmod{D}$. So $g(n)=n^d-n$ has the particular property that if $D\mid g(a)$ and $D\mid g(b)$ then $d\mid g(ab)$. So you only have to compute the cases in the set where $n$ is prime. So, for example, when $d=19$, you get:
$$\gcd(2^{19}-2,3^{19}-3,5^{19}-5,7^{19}-7)$$
This has all eschewed using Fermat, only using some divisibility theorems and knowledge about binomials and congruences.Indeed, the above argument
But with Fermat's little theorem, we can get much further.
It's pretty obvious that $D$ has to be square-free, because for any prime $p$, $p^d-p$ has only one factor of $p$.
Also, if $n^d-n$ is divisible by $p$ for all $n$ then for any if $g$ is a generator modulo $p$, then $g^d\equiv g\pmod p$ means that $p-1$ divides $d-1$. And visa versa, if $p-1\mid d-1$, then $p$ is a common factor.
So what this really means is that the greatest common factor of $n^d-n$ with $d$ odd is exactly the product of the primes $p$ such that $p-1\mid d-1$.
The same argument works for $d$ even, so that the common factor of $n^d-n$ when $d$ is even is always just $2$, since for any odd prime $p$, $p-1$ is not a factor of the odd $d-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
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Solving $\lim\limits_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}}$ How to get from
$$\lim_{n\to\infty} \frac{\cot{\frac{2}{n}}+n\csc{\frac{3}{n^3}}}{\csc{\frac{3}{n}} + n\cot{\frac{2}{n^2}}} = \lim_{n\to\infty} \frac{\frac{\frac{2}{n}}{\tan{\frac{2}{n}}}\cdot\frac{1}{2n^2}+\frac{\frac{3}{n^2}}{\sin{\frac{3}{n^2}}}\cdot\frac{1}{3}}{\frac{\frac{3}{n}}{\sin{\frac{3}{n}}}\cdot \frac{1}{3n^2}+\frac{\frac{2}{n^2}}{\tan{\frac{2}{n^2}}}\cdot\frac{1}{2}}=...=\frac{2}{3}$$
? It doesn't appear like l'Hôpital's Rule was applied here?
| I think you have to check your question because i don't think answer is $\frac{2}{3}$. But i am given a solution of that question which you are write here.
We know that $\displaystyle \lim_{x\to 0}\frac{x}{\sin x}=1$ and $\displaystyle \lim_{x\to 0}\frac{x}{\tan x}=1$.
$\displaystyle \lim_{n\to\infty}\frac{\cot \frac{2}{n}+n\csc \frac{3}{n^{3}}}{\csc \frac{3}{n}+n\cot \frac{2}{n^{2}}}$
$=\displaystyle \lim_{n\to\infty}\frac{\frac{1}{\tan \frac{2}{n}}+n\frac{1}{\sin \frac{3}{n^3}}}{\frac{1}{\sin \frac{3}{n}}+n\frac{1}{\tan\frac{2}{n^2}}}$
$=\displaystyle \lim_{n\to\infty}\frac{\frac{\frac{2}{n}}{\tan \frac{2}{n}}.\frac{n}{2}+n\frac{\frac{3}{n^3}}{\sin \frac{3}{n^3}}.\frac{n^3}{3}}{\frac{\frac{3}{n}}{\sin \frac{3}{n}}.\frac{n}{3}+n\frac{\frac{2}{n^2}}{\tan\frac{2}{n^2}}.\frac{n^2}{2}}$
$=\displaystyle \lim_{n\to\infty}\frac{\frac{n}{2}+n.\frac{n^3}{3}}{\frac{n}{3}+n.\frac{n^2}{2}}$
$=\displaystyle \lim_{n\to\infty}\frac{3n+2n^{4}}{2n+3n^3}$
$=\displaystyle\lim_{n\to\infty}\frac{\frac{3}{n^3}+2}{\frac{2}{n^3}+\frac{3}{n}}$
$=2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the linear combinations of two vectors I am studying for my finals and I'm trying to answer the following question:
Consider the following two vectors in $\mathbb{R}^3$: $a=(1,2,3)$ and $b=(2,3,1)$. Decide
whether it is possible to express the vector $c=(2,4,5)$ as a linear
combination of $a$ and $b$.
I have used the following row operations to find what $C_1$,$C_2$ and $C_3$ are equal to:
\begin{align*}
\left[ \begin{array}{ccc|c} 1 & 2 &2 & 2\\
2 & 3 & 4 & 4 \\
3 & 1 & 5 & 5 \end{array}\right]
&\overset{R_2 = 2R_1 -R_1}{\Longrightarrow}
\left[ \begin{array}{ccc|c}
1 & 2 &2 &2 \\
0 & 1 & 0 & 0 \\
3 & 1 & 5 & 5
\end{array}\right]
\overset{R_3=3R_1-R_3}{\Longrightarrow}
\left[\begin{array}{ccc|c}
1 &2 & 2 & 2 \\
0 &1 & 0 & 0 \\
0 & 5 &1 & 1
\end{array} \right]
\\
&\overset{R_3=5R_2+R_3}{\Longrightarrow}
\left[\begin{array}{ccc|c}
1 & 2 & 2 & 2 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1
\end{array}\right]
\end{align*}
Is the following argument correct?
By looking at the last row of the matrix, we see that it says $0 = 1$, which is impossible, and the system of equations therefore has no solutions. This means that we can not find the values for $C_1$ and $C_2$, and so $c$ can not be written as a linear combination of $a$ and $b$.
Thanks in advance!
| You can also do your example in following way,
Let $(2,4,5)=\alpha.(1,2,3)+\beta.(2,3,1)$
$(2,4,5)=(\alpha+2\beta,2\alpha+3\beta,3\alpha+\beta)$
$\therefore 2=\alpha+2\beta,4=2\alpha+3\beta,5=3\alpha+\beta$
Now, solving the first two equation, we get, $\alpha=2,\beta=0$ but it does not satisfied last equation therefore $c$ can not be expressed as a linear combination of $a$ and $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Trapezoid area and wrong answer Knowing that the acute angles of the trapezoid are $60^\circ$ and $45^\circ$ and the difference of the squares of base lenghts is equal to 100, calculate the area of this trapezoid.
Here's my solution: Let a be the shorter and b the longer base. By drawing two lines from the ends of a perpendicular to b, we form two right triangles. One of them has angles $45^\circ-45^\circ-90^\circ$ and the second is $60^\circ-30^\circ-90^\circ$. Using the properties of these triangles, we can see that the side of the first triangle lying on b is equal to h=height (both lie between 45 and 90) and the side of the latter which is lying on b is equal to $\frac{h\sqrt{3}}3$. Then, $b-a=h+\frac{h\sqrt{3}}3$ so $h=\frac{3(b-a)}{3+\sqrt{3}}=(b-a)(1-\frac{\sqrt{3}}{3})$. We just plug it into the initial equation and we get that the total area is $50(1-\frac{\sqrt{3}}{3})$.
And that's the problem. From what I read on the page I got this problem from, the result should be $25(3-\sqrt{3})$. Why is that so? What's wrong in my solution?
| I noted things in the following order:
*
*$b^2-a^2=(b-a)(b+a)=100$ so $A=\frac{h(b+a)}{2}=\frac{50h}{b-a}$.
*Then I reasoned that $b-a=h+\frac{h}{\sqrt{3}}$.
*Combining them yields $A=\frac{50h}{h+\frac{h}{\sqrt{3}}}=\frac{50\sqrt{3}}{\sqrt{3}+1}=\frac{50\sqrt{3}(\sqrt{3}-1)}{2}=25(3-\sqrt{3})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/140479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Derivative of $\ln(x\sqrt{x^2-1})$ I am trying to find the derivative of $\ln(x\sqrt{x^2-1})$ but I can not get what the book gets.
I get $$\frac{1}{x \sqrt{x^2-1}} \cdot \sqrt{x^2-1} + x\cdot\frac{1}{2}(x^2-1)^\frac{-1}{2}\cdot2x$$
which I reduce to
$$\begin{align}
&\frac{1}{x\sqrt{x^2-1}}\sqrt{x^2-1} + x^2(x^2-1)^\frac{-1}{2}=\\
&\frac{\sqrt{x^2-1}}{x\sqrt{x^2-1}} + \frac{x^2(x^2-1)^\frac{-1}{2}}{\sqrt{x^2-1}}=\\
&\frac{1}{x} + \frac{x^2}{x^2-1}= \frac{x^2 - 1 +x^3}{x^3 - x}
\end{align}$$
From here I am not sure what to do. This is not the right answer and I do not know what to do.
| You forgot a bracket:
$$\frac{1}{x \sqrt{x^2-1}} * \left[ \sqrt{x^2-1} + x*\frac{1}{2}(x^2-1)^\frac{-1}{2}2x \right]$$
Also, might be much easier to use properties of Log:
$$\ln(x\sqrt{x^2-1}) = \ln(x) +\frac{1}{2} \ln(x^2-1) \,$$
This is much easier to differentiate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/143550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$ Let $0 <a < 1$ and let $b,c \in \mathbb{N}$, evaluate $$\int_0^a \frac{x^{b}+x^{c}}{(1+x)^{b+c+2}}$$
How to evaluate in terms of $a$,$b$ and $c$?
| Write the integral as
$$I(a,b,c)=\int_0^a {{{\left( {\frac{x}{{1 + x}}} \right)}^b}{{\left( {\frac{1}{{1 + x}}} \right)}^c}\frac{{dx}}{{{{\left( {1 + x} \right)}^2}}}} + \int_0^a {{{\left( {\frac{x}{{1 + x}}} \right)}^c}{{\left( {\frac{1}{{1 + x}}} \right)}^b}\frac{{dx}}{{{{\left( {1 + x} \right)}^2}}}} $$
Since $$\frac{x}{{1 + x}} = 1 - \frac{1}{{x + 1}}$$
Let $$u = \frac{1}{{x + 1}}$$
We get
$$ -I(a,b,c)= \int_1^\alpha {{{\left( {1 - u} \right)}^b}{u^c}du} + \int_1^\alpha {{{\left( {1 - u} \right)}^c}{u^b}du} $$
Where $$\alpha = \frac{1}{{a + 1}}$$
Now in any of the two let $u=1-u'$, to get(I'll keep the $u$ variable)
$$-I(a,b,c)=\int_1^\alpha {{{\left( {1 - u} \right)}^b}{u^c}du} - \int_0^{1 - \alpha } {{u^c}{{\left( {1 - u} \right)}^b}du} $$
Now go back to $\alpha = \frac{1}{a+1}$. You should get something like this
$$I\left( {a,b,c} \right) = \int_0^1 {{u^c}{{\left( {1 - u} \right)}^b}du} + \int_{\frac{1}{{a + 1}}}^{\frac{a}{{a + 1}}} {{u^c}{{\left( {1 - u} \right)}^b}du} $$
$$I\left( {a,b,c} \right) = \frac{{\left( {c + 1} \right)!\left( {b + 1} \right)!}}{{\left( {b + c + 2} \right)!}} + \int_{\frac{1}{{a + 1}}}^{\frac{a}{{a + 1}}} {{u^c}{{\left( {1 - u} \right)}^b}du} $$
The last integral might be harder to compute, but given the conditions on $a$, one has
$$0 < \frac{a}{{a + 1}} < \frac{1}{{a + 1}} < 1$$
so the Binomial expansion can be used.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/144982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $ \int\limits_{0}^{2 \pi} \cos( n \arctan( R_0 \sin(\vartheta))) \cos( n \arctan( R_0 \cos(\vartheta))) \; d \vartheta.$ Let $R_0$ be a real number and $n$ an arbitrary integer
$$ \int\limits_{0}^{2 \pi} \cos( n \arctan( R_0 \sin(\vartheta))) \cos( n \arctan( R_0 \cos(\vartheta))) \; d \vartheta.$$
I have tried using $2 \cos(a) \cos(b) = \cos(a+b) +\cos(a-b)$, and also some substitutions. I would like to know how this integral behave with respect to $R_0$.
E.g. you can get something like
$$\int\limits_{0}^{2 \pi} \cos( n \vartheta) \cos( n \arctan( \sqrt{ R_0^2 -\tan^2(\vartheta)})) \frac{1+\tan^2(\vartheta)}{\sqrt{ R_0^2 -\tan^2(\vartheta))}} d\; \vartheta.$$
I expect Bessel integrals to turn up, but I am not sure about this.
This question is related, but less specific:
An integral involving trigonometric functions and its inverse
| Note that, using invariance of the integrand under $\theta \to \frac{\pi}{2} + \theta$ for $0 < \theta < \frac{\pi}{2}$:
$$
\mathcal{I}_n(\rho)= \int_0^{2 \pi} \cos\left( n \arctan(\rho \sin(\theta) ) \right) \cos\left( n \arctan(\rho \cos(\theta) ) \right) \mathrm{d} \theta = \\4 \int_0^{\pi/2} \cos\left( n \arctan(\rho \sin(\theta) ) \right) \cos\left( n \arctan(\rho \cos(\theta) ) \right) \mathrm{d} \theta
$$
Now change the variables, with $\sin(\theta) = x$:
$$
\mathcal{I}_n(\rho)=4 \int_0^1 \cos\left(n \arctan(\rho x) \right) \cos\left(n \arctan(\rho \sqrt{1-x^2}) \right) \frac{\mathrm{d} x}{\sqrt{1-x^2}}
$$
Now, we use, for $0<\rho<1$, $\arctan(\rho x) = \arccos\left( \frac{1}{\sqrt{1+ x^2 \rho^2}}\right)$, $\arctan(\rho \sqrt{1-x^2}) = \arccos\left( \frac{1}{\sqrt{1+ \rho^2 - x^2 \rho^2}}\right)$, and, for $n\in \mathbb{Z}_{\geqslant 0}$:
$$
\cos\left(n \arccos\left( \frac{1}{\sqrt{1+ x^2 \rho^2}}\right) \right) = T_{n}\left(\frac{1}{\sqrt{1+ x^2 \rho^2}}\right)
$$
Now using explicit expression for Chebyshev polynomial $T_n(z)$:
$$
\cos\left(n \arccos\left( \frac{1}{\sqrt{1+ x^2 \rho^2}}\right) \right) = \frac{1}{2} \left( 1 + x^2 \rho^2 \right)^{n/2} \sum_{k=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2k} \left(-x^2 \rho^2\right)^k
$$
This allows to compute the integral for even values of $n$, relatively easily:
$$
\mathcal{I}_{2n}(\rho)= \sum_{k=0}^{n} \sum_{m=0}^n \int_0^1 \frac{(1+x^2 \rho^2)^{n} (1+(1-x^2) \rho^2)^{n}}{\sqrt{1-x^2}} (-x^2 \rho^2)^{k} ( (x^2-1) \rho^2)^{m} \mathrm{d} x
$$
Here are values, obtained for few low even values of $n$:
$$
\mathcal{I}_2(\rho) = \frac{\pi \left(\rho ^4+\left(3-4 \sqrt{\rho ^2+1}\right) \rho ^2+2\right)}{2
\left(\rho ^4+3 \rho ^2+2\right)}
$$
$$
\mathcal{I}_4(\rho) = -\frac{4 \pi \rho ^2 \left(\rho ^6-2 \rho ^4+4 \rho ^2+8\right)}{\left(\rho^2+1\right)^{3/2} \left(\rho ^2+2\right)^3}
$$
$$
\mathcal{I}_6(\rho) = -\frac{2 \pi \rho ^2 \left(3 \rho ^{12}-26 \rho ^{10}+137 \rho ^8+152 \rho ^6-152
\rho ^4+144\right)}{\left(\rho ^2+1\right)^{5/2} \left(\rho ^2+2\right)^5}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/145603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Finding the integral of $(1+2^{2x})/2^x$
Evaluate the integral:
$$\int\frac{1+2^{2x}}{2^x}\,dx = \int \frac{ 1 + (2^x)^2}{2^x}\,dx$$
Let $u = 2^x$. Then $du = 2^x\ln2\,dx$, which yields $\frac{du}{2^x\ln2} = dx$ so
$$ \int \frac{ 1 + (2^x)^2}{2^x}\,dx = \int \frac{1+u^2}{u}du =
\left( x+ \frac{u^3}{3} \right)\ln u+C$$
$$=\left(x+ \frac{(2^x)^3}{3} \right)\ln 2^x +C$$
I'm not sure if I integrated this correctly. Any help would be appreciated.
| All you need is that $$\int a^x dx = \int \exp(x \log a) dx = \frac{\exp(x \log a)}{\log a} +C= \frac{a^x}{\log(a)}+C$$
$$\begin{align*}
\int \frac{1+2^{2x}}{2^{x}} dx &= \int \left(\left(\frac12 \right)^x + 2^x \right)dx\\
&= \left( \frac{\left(\frac12 \right)^x}{\log(1/2)} + \frac{2^x}{\log(2)} \right) + C\\
&= \left( -\frac{\left(\frac12 \right)^x}{\log(2)} + \frac{2^x}{\log(2)} \right) + C\\
&= \frac{\left(2^x - \frac1{2^x} \right)}{\log(2)} + C\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If an integer number is a square and a cube, then it can be writen as $5n,5n+1$, or $5n+4$ I want to show that if an integer number is a square and a cube, then it can be writen as $5n,5n+1$, or $5n+4$.
I tried the following. There are integers numbers $x,y$ such that $n=x^{2}=y^{3}.$ By using Euclidean division, then $x$ and $y$ can be writen as $5k,5k+1,5k+2,5k+3$ or $5k+4.$ If $x=5k$ and $y=5l$ we have $n=5(5k^{2})=5(5^{2}l^{3})$. If $x=5k$, then we must have $y=5l.$ I don't know what to do. Any hint?
| Already if an integer is a square it must be of the shapes you have described.
It is easiest to prove this using congruence notation. Any integer is congruent to $0$, $1$, $2$, $3$, or $4$ modulo $5$. Note now that $0^2$, $1^2$, $2^2$, $3^2$, and $4^2$ are respectively congruent to $0$, $1$, $4$, $4$, and $1$ modulo $5$, so never to $2$ or to $3$ modulo $5$.
If we do not want to use congruence notation, calculate the remainder when $(5k)^2$, $(5k+1)^2$, $(5k+2)^2$, $(5k+3)^2$, and $(5k+4)^2$ are divided by $5$. As a sample, let's do it for $(5k+3)^2$. This is $25k^2+30k+9$, which is $5(5k^2+6k+1)+4$, so it has remainder $4$ on division by $5$.
Out of curiosity, let us explore the cubes modulo $5$. Note that $0^3$, $1^3$, $2^3$, $3^3$, and $4^3$ are congruent respectively to $0$, $1$, $3$, $2$, and $4$ modulo $5$. So, unlike squares, cubes can take on any value modulo $5$.
By looking at the numbers $0$, $1$, and $64$, which are all perfect squares and perfect cubes, we can see that indeed a number which is a perfect square and a perfect cube can be of any of the shapes described.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/147909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
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} |
Linear transform of polynomials The following question is from Golan's linear algebra book. I have posted a solution in the answers.
Problem: Let $F$ ba field and let $V$ be a vector subspace of $F[x]$ consisting of all polynomials of degree at most 2. Let $\alpha:V\rightarrow F[x]$ be a linear transformation satisfying
$\alpha(1)=x$
$\alpha(x+1)=x^5+x^3$
$\alpha(x^2+x+1)=x^4-x^2+1$.
Determine $\alpha(x^2-x)$.
| The idea here is to determine the action on each of the basis elements $1$, $x$, and $x^2$.. By linearity we see
$\alpha(x)=\alpha((x+1)-1)=\alpha(x+1)-\alpha(1)=x^5+x^3-x$
$\alpha(x^2)=\alpha((x^2+x+1)-(x+1))=\alpha(x^2+x+1)-\alpha(x+1)=-x^5+x^4-x^3-x^2+1$
Using linearity again we see
$\alpha(x^2-x)=\alpha(x^2)-\alpha(x)=-x^5+x^4-x^3-x^2+1-x^5-x^3+x=-2x^5+x^4-2x^3-x^2+x+1$
and if the field is of characteristic 2, the $x^5$ and $x^3$ terms disappear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How can we produce another geek clock with a different pair of numbers? So I found this geek clock and I think that it's pretty cool.
I'm just wondering if it is possible to achieve the same but with another number.
So here is the problem:
We want to find a number $n \in \mathbb{Z}$ that will be used exactly $k \in \mathbb{N}^+$ times in any mathematical expresion to produce results in range $[1, 12]$. No rounding, is allowed, but anything fancy it's ok.
If you're answering with an example then use one pair per answer.
I just want to see that clock with another pair of numbers :)
Notes for the current clock:
1 o'clock: using 9 only twice, but it's easy to use it 3 times with many different ways. See comments.
5 o'clock: should be $\sqrt{9}! - \frac{9}{9} = 5$
| For $n=2$ and $k=12$ here is a solution:
$1=\left(2 \times \left(2 \times \left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$
$2=\left(2+\left(2 \times \left(2+\left(2+\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$
$3=\left(2 \times \left(2+\left(2 \times \frac{2}{\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)\right)$
$4=\frac{2}{\left(2 \times \left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)\right)}$
$5=\frac{2}{\left(2 \times \frac{2}{\left(2-\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$
$6=\left(2-\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$
$7=\frac{2}{\left(2 \times \frac{2}{\left(2-\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$
$8=\left(2-\left(2+\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$
$9=\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)}\right)}$
$10=\left(2+\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)}\right)}\right)$
$11=\left(2+\left(2+\frac{2}{\left(2 \times \frac{2}{\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)}\right)}\right)\right)$
$12=\left(2-\left(2+\left(2+\left(2-\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+\left(2+2\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 10,
"answer_id": 4
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simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $? How can I simplify $ (-2 + 2\sqrt3i)^{\frac{3}{2}} $ to rectangular form $z = a+bi$?
(Note: Wolfram Alpha says the answer is $z=-8$. My professor says the answer is $z=\pm8$.)
I've tried to figure this out for a couple hours now, but I'm getting nowhere.
Any help is much appreciated!
| $\:\sqrt{-2+2\sqrt{-3}}\:$ can be denested by a radical denesting formula that I discovered as a teenager.
Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $
Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $
and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$
Here $ {-}2+2\sqrt{-3}\:$ has norm $= 16.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = -4\ $ yields $\ 2+2\sqrt{-3}\:$
and this has $\rm\ \sqrt{trace}\: =\: 2,\ \ hence\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:1+\sqrt{-3}.$
Checking we have
$\ \smash[t]{\displaystyle \left(1+\sqrt{-3}\right)^2 =\ 1-3 + 2\sqrt{-3}\ =\ -2 + 2 \sqrt{-3}}$
Therefore $\quad\ \begin{eqnarray}\rm\:(-2 + 2\sqrt{-3})^{3/2} &=&\ (-2+2\sqrt{-3})\ (-2+2\sqrt{-3})^{1/2} \\ &=&\ -2\,(1-\sqrt{-3})\ (1+\sqrt{-3}) \\
&=&\ -8\rm\ \ \ (up\ to\ sign)
\end{eqnarray}$
See this answer for general radical denesting algorithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/152989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I compute mean curvature in cylindrical coordinates? If I have a surface defined by $ z=f(r, \theta) $, does anyone know the expression for the mean curvature? There is a previous post dealing with Gaussian instead of mean curvature, the answer I'm looking for is similar to that given by J.M. on that post.
The mentioned post:
How do I compute Gaussian curvature in cylindrical coordinates?
Many thanks in advance for your help,
| It's similar to the Gaussian Curvature. J.M. actually hide his/her calculation.
You start with the parametrization:
\begin{align}
x &= \rho(\vartheta, z)\cos\vartheta \\
y &= \rho(\vartheta, z)\sin\vartheta \\
z &= z \\
\end{align}
you need to find the values of $L$, $M$ and $N$ of the Second fundamental form and then to find the trace of the matrix:
\begin{pmatrix}
L & M \\
M & N
\end{pmatrix}
and divide it by 2. Practically you only need $L$ and $N$.
Let $\mathbf{r}\colon [0, 2\pi)\times \mathbb{R}^2 \to \mathbb{R}^3$ defined by the parametrization, then:
\begin{align}\mathbf{n} &= \frac{\mathbf{r}_{\vartheta} \times \mathbf{r}_z}{\lvert\mathbf{r}_{\vartheta} \times \mathbf{r}_z\rvert} \\
L &= \mathbf{r}_{\vartheta\vartheta}\cdot \mathbf{n} \\
N &= \mathbf{r}_{zz}\cdot \mathbf{n} \\
\end{align}
You have:
\begin{align}\mathbf{r}_{\vartheta} &= \bigl(\rho_{\vartheta} \cos\vartheta - \rho \sin\vartheta, \rho_{\vartheta}\sin\vartheta + \rho\cos\vartheta, 0\bigr) \\
\mathbf{r}_{z} &= \bigl(\rho_{z} \cos\vartheta, \rho_{z}\sin\vartheta, 1\bigr) \\
\mathbf{r}_{\vartheta\vartheta} &= \bigl(\rho_{\vartheta\vartheta} \cos\vartheta - \rho_{\vartheta} \sin\vartheta - \rho_{\vartheta} \sin\vartheta - \rho \cos\vartheta, \rho_{\vartheta}\sin\vartheta + \rho\cos\vartheta, 0\bigr) \\
&= \bigl(\rho_{\vartheta\vartheta} \cos\vartheta - 2\rho_{\vartheta} \sin\vartheta - \rho \cos\vartheta, \rho_{\vartheta\vartheta}\sin\vartheta + \rho_{\vartheta}\cos\vartheta + \rho_{\vartheta}\cos\vartheta - \rho\sin\vartheta, 0\bigr) \\
\mathbf{r}_{zz} &= \bigl(\rho_{zz} \cos\vartheta, \rho_{zz}\sin\vartheta, 0\bigr) \\
&= \rho_{zz}\bigl(\cos\vartheta, \sin\vartheta, 0\bigr)
\end{align}
If I define $\mathbf{k} = (0,0,1)$, $\mathbf{u} = (\cos\vartheta,\sin\vartheta,0)$ and $\mathbf{v} = \mathbf{u}_{\vartheta} = (-\sin\vartheta,\cos\vartheta,0)$, I can rewrite them as:
\begin{align}\mathbf{r}_{\vartheta} &= \rho_{\vartheta}\mathbf{u} + \rho\mathbf{v} \\
\mathbf{r}_{z} &= \rho_{z}\mathbf{u} + \mathbf{k} \\
\mathbf{r}_{\vartheta\vartheta} &= \rho_{\vartheta\vartheta}\mathbf{u} + 2\rho_{\vartheta}\mathbf{v} - \rho\mathbf{u} \\
&= (\rho_{\vartheta\vartheta} - \rho)\mathbf{u} + 2\rho_{\vartheta}\mathbf{v} \\
\mathbf{r}_{zz} &= \rho_{zz}\mathbf{u}
\end{align}
\begin{align}\mathbf{u} \times \mathbf{v} &= (0,0, cos^2\vartheta + \sin^2\vartheta) = \mathbf{k}\\
\mathbf{u} \times \mathbf{k} &= (\sin\vartheta, -\cos\vartheta, 0) = -\mathbf{v}\\
\mathbf{v} \times \mathbf{k} &= (\cos\vartheta, \sin\vartheta, 0) = \mathbf{u}\\
\mathbf{u} \cdot \mathbf{u} &= 1\\
\mathbf{v} \cdot \mathbf{v} &= 1\\
\mathbf{k} \cdot \mathbf{k} &= 1\\
\mathbf{u} \cdot \mathbf{v} &= -\cos\theta\sin\theta + \sin\theta\cos\theta = 0\\
\mathbf{u} \cdot \mathbf{k} &= 0\\
\mathbf{v} \cdot \mathbf{k} &= 0\\
\end{align}
Now I can calculate $\mathbb{n}$, $L$ and $N$.
\begin{align} \mathbf{r}_{\vartheta} \times \mathbf{r}_z &= \bigl(\rho_{\vartheta}\mathbf{u} + \rho\mathbf{v}\bigr)\times \bigl(\rho_{z}\mathbf{u} + \mathbf{k}\bigr) \\
&= \rho_{\vartheta}\mathbf{u}\times\rho_{z}\mathbf{u} + \rho\mathbf{v}\times\rho_{z}\mathbf{u} + \rho_{\vartheta}\mathbf{u}\times\mathbf{k} + \rho\mathbf{v}\times\mathbf{k} \\
&= \rho_{\vartheta}\rho_{z}\mathbf{u}\times\mathbf{u} + \rho\rho_{z}\mathbf{v}\times\mathbf{u} + \rho_{\vartheta}\mathbf{u}\times\mathbf{k} + \rho\mathbf{v}\times\mathbf{k} \\
&= -\rho\rho_{z}\mathbf{k} - \rho_{\vartheta}\mathbf{v} + \rho\mathbf{u} \\
\lvert\mathbf{r}_{\vartheta} \times \mathbf{r}_z \rvert &= \rho^2 + \rho_{\vartheta}^2 + \rho^2\rho_{z}^2 = \rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2\\
\mathbf{n} &= \frac{\rho\mathbf{u} - \rho_{\vartheta}\mathbf{v} -\rho\rho_{z}\mathbf{k} }{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
\end{align}
\begin{align}
L &= \bigl[(\rho_{\vartheta\vartheta} - \rho)\mathbf{u} + 2\rho_{\vartheta}\mathbf{v}\bigr]\cdot \frac{\rho\mathbf{u} - \rho_{\vartheta}\mathbf{v} -\rho\rho_{z}\mathbf{k} }{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
&= \frac{(\rho_{\vartheta\vartheta} - \rho)\rho + 2\rho_{\vartheta}^2}{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
N &= \rho_{zz}\mathbf{u}\cdot \frac{\rho\mathbf{u} - \rho_{\vartheta}\mathbf{v} -\rho\rho_{z}\mathbf{k} }{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2} \\
&= \frac{\rho\rho_{zz}}{\rho^2\bigl(1 + \rho_{z}^2\bigr) + \rho_{\vartheta}^2}
\end{align}
Now, the mean curvature is:
\begin{align}
H &= \frac{L + N}{2} \\
&= \frac{(\rho_{\vartheta\vartheta} - \rho)\rho + 2\rho_{\vartheta}^2 + \rho\rho_{zz}}{2\rho^2\bigl(1 + \rho_{z}^2\bigr) + 2\rho_{\vartheta}^2}
\end{align}
I suggest you to check my calculations. Anyway this is the general methods to find the mean curvature when you have a parametrization of a suface (in $\mathbf{R}^3$).
P.S. I abbreviate $\frac{\partial f}{\partial x}$ with $f_x$ and $\frac{\partial f}{\partial x\partial y}$ with $f_{xy}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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Integral of $\int \frac {\sqrt {x^2 - 4}}{x} dx$ I am trying to find $$\int \frac {\sqrt {x^2 - 4}}{x} dx$$
I make $x = 2 \sec\theta$
$$\int \frac {\sqrt {4(\sec^2 \theta - 1)}}{x} dx$$
$$\int \frac {\sqrt {4\tan^2 \theta}}{x} dx$$
$$\int \frac {2\tan \theta}{x} dx$$
From here I am not too sure what to do but I know I shouldn't have x.
$$\int \frac {2\tan \theta}{2 \sec\theta} dx$$
I also know I shouldn't have dx anymore.
$$dx = 2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int \frac {2\tan \theta}{2 \sec\theta} 2\sec \theta \tan \theta \; \mathrm d\theta$$
$$\int {2\tan^2 \theta} \; \mathrm d\theta$$
$$2\int {\tan^2 \theta} \; \mathrm d\theta$$
I have no idea how to find the integral of $\tan^2 \theta$
So I use Wolfram Alpha:
$$\tan \theta - \theta + c$$
Now I need to replace theta with x.
$$x = 2 \sec\theta$$
With same mathmagics I produce
$$ \frac {x}{2} = \sec \theta$$
$$ \theta = \operatorname {arcsec} \left(\frac{x}{2}\right)$$
$$\tan \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) - \left(\operatorname {arcsec} \left(\frac{x}{2}\right)\right) + c$$
This is wrong but I am not sure why.
| You are correct. First note that you have not carried a factor of $2$, since your integral is $2 \int \tan^2(\theta) d \theta$.
Hence your solution should read $$2 \tan(\text{arsec}(x/2)) - 2 \text{arcsec}(x/2) + c$$
You may want to rewrite your solution to match with the solution in your text. For instance, $$\text{arsec}(x/2)) = \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right)$$
So why is the above identity true? If we let $\text{arsec}(x/2)) = \theta$, then we get that $\sec(\theta) = x/2$ i.e. $\sec^2(\theta) = \dfrac{x^2}{4}$. We have that $\tan^2(\theta) = \sec^2(\theta) - 1 = \dfrac{x^2}{4} - 1 = \dfrac{x^2-4}{4}$ i.e. $\tan(\theta) = \dfrac{\sqrt{x^2 - 4}}{2}$. Hence, $$\theta = \arctan \left( \dfrac{\sqrt{x^2-4}}{2}\right)$$
Hence, we have the identity, $$\text{arsec}(x/2)) = \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right)$$
If you use this, then your solution will read $$\sqrt{x^2 - 4} - 2 \, \text{arsec}(x/2)) + c = \sqrt{x^2 - 4} - 2 \, \arctan \left( \dfrac{\sqrt{x^2-4}}{2} \right) + c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/153553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Simplify this expression with nested radical signs My question is-
Simplify:
$$\frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}$$
| $$
\begin{align} & {}\quad \frac1{\sqrt{12-2\sqrt{35}}}-\frac2{\sqrt{10+2\sqrt{21}}}-\frac1{\sqrt{8+2\sqrt{15}}}\\[10pt]
& =\frac {1}{\sqrt{ 12-2 \sqrt {35}}} \frac {\sqrt{ 12+2 \sqrt {35}}}{\sqrt{ 12+2 \sqrt {35}}}- \frac {2 }{\sqrt{ 10+2 \sqrt {21}}} \frac {\sqrt{ 10-2 \sqrt {21}}}{\sqrt{ 10-2 \sqrt {21}}}- \frac{1}{\sqrt {8+2\sqrt {15}}}\frac{\sqrt {8-2\sqrt {15}}}{\sqrt {8-2\sqrt {15}}}\\[10pt]
& =\frac {\sqrt{ 12+2 \sqrt {35}}}{2}-\frac {\sqrt{ 10-2 \sqrt {21}}}{2}-\frac {\sqrt {8-2\sqrt {15}}}{2}\\[10pt]
& =\frac {{\sqrt{ 12+2 \sqrt {35}}}- {\sqrt{ 10-2 \sqrt {21}}}- {\sqrt {8-2\sqrt {15}}}}{2}\\[10pt]
& = \frac {{{|\sqrt 5 + \sqrt 7|}}- {{|\sqrt 3 - \sqrt 7|}}- {{|\sqrt 3 - \sqrt 5|}}}{2}=\sqrt 3
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Putting ${n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ in a closed form As the title says, I'm trying to transform $\displaystyle{n \choose 0} + {n \choose 5} + {n \choose 10} + \cdots + {n \choose 5k} + \cdots$ into a closed form. My work:
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = \displaystyle\sum_{p=0}^{n}\binom{n}{p}\exp\left(\frac{p\cdot2i\pi}{5} \right)$
$\displaystyle=\binom{n}{0} + \binom{n}{1}\exp\left(\frac{1\cdot2i\pi}{5} \right) + \binom{n}{2}\exp\left(\frac{2\cdot2i\pi}{5} \right) + \binom{n}{3}\exp\left(\frac{3\cdot2i\pi}{5} \right) + \binom{n}{4}\exp\left(\frac{4\cdot2i\pi}{5} \right) + \binom{n}{5} + \cdots = \left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] + \exp\left(\frac{2i\pi}{5} \right)\left[\binom{n}{1} + \binom{n}{6} + \binom{n}{11} \right ] + \exp\left(\frac{4i\pi}{5} \right)\left[\binom{n}{2} + \binom{n}{7} + \binom{n}{12} \right ] + \exp\left(\frac{6i\pi}{5} \right)\left[\binom{n}{3} + \binom{n}{8} + \binom{n}{13} \right ] + \exp\left(\frac{8i\pi}{5} \right)\left[\binom{n}{4} + \binom{n}{9} + \binom{n}{14} \right ] + \cdots$
I'll recall $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = k$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = u$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = v$, $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = w$ and $\left[\binom{n}{0} + \binom{n}{5} + \binom{n}{10} + \cdots\right ] = z$. Thus
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = k + u\cdot\exp\frac{2i\pi}{5} + v\cdot\exp\frac{4i\pi}{5} + w\cdot \exp\frac{6i\pi}{5} + z\cdot\exp\frac{8i\pi}{5} = k + u\cdot\left (\cos\frac{2\pi}{5} + i\cdot\sin\frac{2\pi}{5} \right ) + v\cdot\left (\cos\frac{4\pi}{5} + i.\sin\frac{4\pi}{5} \right ) + w\cdot\left (\cos\frac{6\pi}{5} + i.\sin\frac{6\pi}{5} \right ) + z\cdot\left (\cos\frac{8\pi}{5} + i.\sin\frac{8\pi}{5} \right )$
Noting that $\cos\frac{2\pi}{5} = \cos\frac{8\pi}{5}$, $\cos\frac{4\pi}{5} = \cos\frac{6\pi}{5}$, $\sin\frac{2\pi}{5} = -\sin\frac{8\pi}{5}$ and $\sin\frac{4\pi}{5} = -\sin\frac{6\pi}{5}$:
$\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = k + \left(u + z\right)\cos\frac{2\pi}{5} + i\cdot\left(u - z \right)\sin\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5} + i\cdot\left(v - w \right)\sin\frac{4\pi}{5} = \left(k + \left(u + z\right)\cos\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5}\right) + i\cdot\left(\left(u - z \right)\sin\frac{2\pi}{5} + \cdot\left(v - w \right)\sin\frac{4\pi}{5} \right)$
But $\displaystyle\left(1 + \exp\frac{2i\pi}{5} \right )^n = \left(2\cos\left(\frac{\pi}{5} \right)\cdot\exp\left(\frac{i\pi}{5}\right)\right)^n = \left(2^n\cos^n\left(\frac{\pi}{5} \right)\cdot\exp\left(\frac{ni\pi}{5}\right)\right) = \left(2^n\cos^n\frac{\pi}{5} \right)\left(\exp\left(\frac{ni\pi}{5}\right)\right) = \left(2^n\cos^n\frac{\pi}{5} \right)\left(\cos\frac{n\pi}{5} + i.\sin\frac{n\pi}{5} \right ) = \left(2^n\cos^n\frac{\pi}{5}\cos\frac{n\pi}{5} \right) + i\cdot \left(2^n\cos^n\frac{\pi}{5}\sin\frac{n\pi}{5} \right)$
So,
$\displaystyle k + \left(u + z\right)\cos\frac{2\pi}{5} + \left(v + w\right)\cos\frac{4\pi}{5} = 2^n\cos^n\frac{\pi}{5}\cos\frac{n\pi}{5}$
and
$\displaystyle\left(u - z \right)\sin\frac{2\pi}{5} + \left(v - w \right)\sin\frac{4\pi}{5} = 2^n\cos^n\frac{\pi}{5}\sin\frac{n\pi}{5}$
and I'm stuck here. I noted that $k + u + v + w + z = 2^n$ but I couldn't isolate $k$. So, any help finishing this result will be fully appreciated. Thanks.
| Hint: with $\omega=\exp(2\pi i /5)$ a primitive $5$th root of unity, we have
$$\sum_{r=0}^4 \omega^{rk}=\begin{cases}5 & k\equiv 0 \bmod 5 \\ 0 & k\not\equiv 0\bmod 5.\end{cases}$$
So then what is
$$(1+\omega^0)^n+(1+\omega^1)^n+(1+\omega^2)^n+(1+\omega^3)^n+(1+\omega^4)^n~? $$
(Combine the binomial expansions...)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this string ascending or descending? I need help in checking that string is ascending or descending and I have problems with it.
so it's my "$a_n$"
$$
\sqrt {n+5} - \sqrt{n}
$$
and I need to use $a_{n+1} - a_n$ or $\frac {a_n+1}{a_n}$.
EDIT:
it would be something like that
$$
\sqrt {n+6} - \sqrt{n+1} -\sqrt{n+5}+\sqrt{n}
$$
and what's next?
| \begin{align*}
a_n &= \sqrt{n+5}-\sqrt{n} = \left(\sqrt{n+5}-\sqrt{n}\right)\frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+5}+\sqrt{n}} \\
&= \frac{\left(\sqrt{n+5}-\sqrt{n}\right)\left(\sqrt{n+5}+\sqrt{n}\right)}{\sqrt{n+5}+\sqrt{n}} \\
&= \frac{n+5 - n}{\sqrt{n+5}+\sqrt{n}} \\
&= \frac{5}{\sqrt{n+5}+\sqrt{n}} \\
\end{align*}
This is clearly a decreasing sequence. As $n$ increases, the denominator increases so the sequence as a whole decreases.
If you insist on using $\frac{a_{n+1}}{a_n}$, use the same approach above to show that:
\begin{align*}
\frac{a_{n+1}}{a_n} &= \frac{\sqrt{n+6}-\sqrt{n+1}}{\sqrt{n+5}-\sqrt{n}} \\
&= \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}}
\end{align*}
Now, notice that:
$$
n+1 \gt n \Rightarrow \sqrt{n+1} \gt \sqrt{n}
$$
And:
$$
n+6 \gt n+5 \Rightarrow \sqrt{n+6} \gt \sqrt{n+5}
$$
Add the inequalities side by side to get:
$$
\sqrt{n+6} + \sqrt{n+1} \gt \sqrt{n+5} + \sqrt{n}
$$
Divide both sides by $\sqrt{n+6} + \sqrt{n+1}$ to get:
$$
\frac{a_{n+1}}{a_n} = \frac{\sqrt{n+5}+\sqrt{n}}{\sqrt{n+6}+\sqrt{n+1}} < 1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What other substitutions could I use to evaluate this integral? Consider the integral
$$ \int x^2\sqrt{2 + x} \, dx$$
I need to find the value of this integral, yet all its (seemingly) possible substitutions don't allow me to cancel appropriate terms. Here are three substitutions and their outcomes, all of which cover both terms in the integrand, as well as the composed function inside the root.
$\boxed{\text{Let }u = (2 + x)}$
$$u = 2 + x$$
$$du = 1 \space dx$$
$$dx = du$$
$$\int x^2 \sqrt{u} \space du$$
$\boxed{\text{Let } u = \sqrt{2+x}}$
$$ u = \sqrt{2 + x}$$
\begin{align*} du &= \frac{1}{2}(2 + x)^{-\frac{1}{2}} \space dx \\
&= \frac{1}{2\sqrt{2 + x}} \space dx
\end{align*}
$$ dx = 2\sqrt{2 + x} \space du$$
$$ 2\int ux^2\sqrt{2 + x} \space du $$
$\boxed{\text{Let }u = x^2}$
$$ u = x^2$$
$$ du = 2x \space dx$$
$$ dx = \frac{1}{2x} \space du$$
$$ \frac{1}{2} \int \frac{u\sqrt{2 + x}}{x} \space du$$
As you can see, none of these allow me to move forward into integrating with respect to $u$. What other substitutions can I use that would help me move on with this integral?
| If $u=\sqrt{2+x}$ then $u^2 = 2+x$, so $2u\,du=dx$, and $x=u^2-2$. Then you have
$$
\int x^2\sqrt{2+x}\,dx = \int(u^2-2)^2 u\, 2u\,du
$$
After expanding the polynomial and getting its antiderivative, put $\sqrt{2+x}$ in place of $u$ wherever it appears.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this quadratic congruence equation How to solve $x^2+x+1\equiv 0\pmod {11}$ ?
I know that in some equations like $ax\equiv b\pmod d$ if $(a,d)=1$ then the equation has one and only one solution $x\equiv ba^{\phi(d)-1}\pmod d$. Any help will be appreciated. ;)
| You can use the technique of Completing the Square.
Suppose that we want to solve the congruence $ax^2+bx+c\equiv 0 \pmod p$, where $a\not\equiv 0\pmod{p}$, and $p$ is an odd prime. The congruence is equivalent to
$$4a^2x^2+4abx+4ac\equiv 0\pmod{p}.\tag{$1$}$$
Completing the square, we see that the congruence is equivalent to
$$(2ax+b)^2-b^2+4ac\equiv 0\pmod{p},$$
or equivalently to
$$(2ax+b)^2\equiv b^2-4ac\pmod{p}.\tag{$2$}$$
To solve this last congruence we (i) Try to solve the congruence $y^2 \equiv b^2-4ac\pmod{p}$. There may be no solution, in which case the original congruence has no solution. There may be a single solution, if $b^2-4ac\equiv 0\pmod{p}$. In all other cases there will be two solutions. (ii) After having found such $y$, we solve the linear congruences $2ax+b\equiv y\pmod{p}$.
Now we turn to $x^2+x+1\equiv 0\pmod{p}$, and use the basic technique, although inspection would work faster.
Rewrite as $4x^2+4x+4\equiv 0\pmod{11}$, then as $(2x+1)^2-1+4\equiv 0\pmod{11}$, then as $(2x+1)^2\equiv -3\equiv 8\pmod{11}$.
So we first try to solve the congruence $y^2\equiv -3\equiv 8\pmod{11}$.
It turns out there are no solutions. There are powerful methods to deal with such questions for large primes $p$. For $11$, just calculate $1^2$, $2^2$, $3^3=2$, $4^2$, and $5^2$ modulo $11$. None of them work. The rest cannot work, for they are essentially the negatives of the numbers we have already checked, so their squares are, modulo $11$, the same as the squares of the numbers $1$ to $5$. For example, $6\equiv -5\pmod{11}$, so $6^2\equiv (-5)^2\equiv 5^2\pmod{11}$, and similarly $7^2\equiv 4^2$, and so on.
Remarks: $1.$ The congruence $(2)$ has a solution if and only if the congruence $y^2\equiv b^2-4ac\pmod{p}$ has a solution, that is, iff $b^2-4ac$ is a square modulo $p$. This $b^2-4ac$ is the familiar discriminant of elementary algebra.
$2.$ The general idea can be used even when we are solving a quadratic congruence modulo $m$, where $m$ is not prime. But significant modifications need to be made. The congruence $ax^2+bx+c\equiv 0\pmod{m}$ is equivalent to $4a^2x^2+4abx+4ac\equiv 0 \pmod{4am}$, so we end up looking at
$$(2ax+b)^2\equiv b^2-4ac\pmod{4am}.$$
The study of the congruence $y^2\equiv b^2-4ac\pmod{4am}$ is more complicated than when the modulus is $p$, and there may be many solutions. But after that we are solving linear congruences $2ax+b\equiv y\pmod{4am}$, and we can use the Extended Euclidean Algorithm.
| {
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"timestamp": "2023-03-29T00:00:00",
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the probability of picking certain amount of items from a sack. I need a help on calculating the probability. Here is the task that I can "translate" it into a probability task.
Let's assume we have a sack with 12 items. There are 1 X item and 2 Y items and other staff. Every time we pick an item from the sack and throw it back into the sack. So what is the smallest n such that if we perform the experiment n times, the probability of having picked 2X and 3Y items at least once is no more than m?
| Ok then. Your sack of $12$ items has $1$ item of type 'X', $2$ items of type 'Y', and $9$ other items. Suppose you draw an item from the sack $n$ times.
To not have at least $2$ X and at least $3$ Y, you must have picked either:
*
*$0$ X and any number of $Y$ (ways: $11^n$),
*$1$ X and any number of $Y$ (ways: ${n \choose 1}11^{n-1}$),
*$0$ Y and any number of $X$ (the case of $0$ or $1$ X is already counted, so ways: $10^n - 9^n - {n \choose 1}9^{n-1}$),
*$1$ Y and any number of $X$ (the case of $0$ or $1$ X is already counted, so ways: ${n \choose 1}10^{n-1} - {n \choose 1}9^{n-1} - {n \vphantom{-1} \choose 1}{n-1 \choose 1}9^{n-2}$), or
*$2$ Y and any number of $X$ (the case of $0$ or $1$ X is already counted, so ways: ${n \choose 2}2^210^{n-2} - {n \choose 2}2^29^{n-2} - {n\vphantom{-2}\choose 2}{n-2\choose 1}2^29^{n-3}$).
So putting all these together, the probability of getting at least two X and at least three Y is:
$$1 - (11^n + n11^{n-1} + (10^n - 9^n -n9^{n-1}) + (n10^{n-1} - n9^{n-1} -n(n-1)9^{n-2}) + (\frac{n(n-1)}{2}2^210^{n-2} - \frac{n(n-1)}{2}2^n9^{n-2} - \frac{n(n-1)(n-2)}{2}2^29^{n-3}))/{12^n}$$
To understand the behaviour of this number for large $n$ (of course it gets very close to $1$, since the probability of picking so few X or so few Y is close to $0$), you may want to collect exponential terms with the same base together:
$$\frac{12^n - 11^n(1 + \frac{n}{11}) - 10^n(1 + \frac{n}{10} + \frac{n(n-1)}{50}) + 9^n(1 + \frac{2n}{9} + \frac{3n(n-1)}{81} + \frac{2n(n-1)(n-2)}{729})}{12^n}$$
In particular, for $n = 100$, the probability that you will have at least two X and at least three Y is about $0.998318$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two algebra questions I have two questions that I need help with:
1) How many single digit even natural number solutions are there for the equation $A+B+C+D = 24$ such that $A+B > C+D$
A)20 B)11 C)16 D)24
2) Three positive real numbers $x,y,z$ are such that $x+y+Z = 1$. which of the following inequalities best discribe the relation between $XY,YZ,ZX$.
A) $xy+yz+zx > 1/3$
B) $xy+yz+zx <1/3$
C) $xy+yz+zx \le 1/3$
D) $xy+yz+zx \le 2/3$
| For the inequality note that:
$$(x-y)^2 + (y-z)^2+(z-x)^2 = 2(x^2+y^2+z^2)-2(xy+yz+zx) \geq 0$$
so that $$x^2+y^2+z^2 \geq xy+yz+zx$$
Now
$$1=(x+y+z)^2=(x^2+y^2+z^2) + 2(xy+yz+zx)\geq 3(xy+yz+zx)$$
I think that is pre-calculus algebra, and not nearly as well known a trick as it should be.
| {
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How to find $ \lim\limits_{ x\to 100 } \frac { 10-\sqrt { x } }{ x-100 }$ Find $ \lim\limits_{ x\to 100 } \dfrac { 10-\sqrt { x } }{ x-100 }$
(without using a calculator and other machines...?)
| Recognizing that $\sqrt{x} \approx 10$ makes the numerator vanish, we may be inspired to use a differential approximation:
$$ \sqrt{x} = 10 + \frac{1}{20}(x - 100) + r(x) (x - 100) $$
where $r(x)$ has the property that $\lim_{x \to 100} r(x) = 0$. Therefore,
$$ \lim_{x \to 100} \frac{10 - \sqrt{x}}{x - 100}
= \lim_{x \to 100} \frac{-\frac{1}{20} (x - 100) - r(x) (x - 100)}{x - 100}
= \lim_{x \to 100} -\frac{1}{20} - r(x) = -\frac{1}{20}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$ $X = \log_{12} 18$ and $Y= \log_{24} 54$. Find $XY + 5(X - Y)$
I changed the bases to 10, then performed manual addition/multiplication but it didn't yield me any result except for long terms. Please show me the way.
All I'm getting is $$\frac{\lg 18\lg54 + 5 \lg18\lg24 - 5\lg54\lg12}{\lg12\lg24} $$
| $ X = \log_{12} 18 $ and $ Y= \log_{24} 54 $.
$ X = \frac {\log_{2}18}{\log_{2}12}=\frac{\log_{2}9\cdot2}{\log_{2}4\cdot3}=\frac{\log_{2}3^2\cdot 2}{\log_{2}2^2\cdot3}=\frac{2\log_{2}3 + 1}{\log_{2}3 + 2}=\frac{2A+1}{A+2}$
$ Y = \frac {\log_{2}54}{\log_{2}18}=\frac{\log_{2}27\cdot2}{\log_{2}8\cdot3}=\frac{\log_{2}3^3\cdot 2}{\log_{2}2^3\cdot3}=\frac{2\log_{2}3 + 1}{\log_{2}3 + 2}=\frac{3A+1}{A+3}$, where $A=\log_{2}3$
By X have:
$X(A+2)=2A+1\Rightarrow AX+2X=2A+1 \Rightarrow A(X-2)=1-2X \Rightarrow A=\frac{1-2X}{X-2}$
courses, by Y have:
$Y(A+3)=A+1\Rightarrow Ay+3X=3A+1 \Rightarrow A(Y-3)=1-3XY\Rightarrow A=\frac{1-3Y}{Y-3}$
Here we have to:
$A=\frac{1-2X}{X-2}$,
$A=\frac{1-3Y}{Y-3}$
where
$\frac{1-2X}{X-2}=\frac{1-3Y}{Y-3}$
$(1-2X)(Y-3)=(1-3Y)(X-2)$
$Y-2XY-3+6X=X-3XY-2+6Y$
$Y-2XY-3+6X-X+3XY+2-6Y=0$
$XY+5(X-Y)-1=0$
$XY+5(X-Y)=1$
| {
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All integer solutions for $x^4-y^4=15$ I'm trying to find all the integer solutions for $x^4-y^4=15$.
I know that the options are $x^2-y^2=5, x^2+y^2=3$, or $x^2-y^2=1, x^2+y^2=15$, or $x^2-y^2=15, x^2+y^2=1$, and the last one $x^2-y^2=3, x^2+y^2=5$.
Only the last one is valid. $x^2+y^2=15$ is not solvable since the primes which have residue $3$ modulo $4$ is not of an equal power.
One particular solution for $x^2-y^2=3, x^2+y^2=5$, is $x_0=2, y_o=1$. How do I get to an expression of a general solution for this system?
Thanks!
| Consider $f(x) = (x+1)^4 - x^4 = 4x^3 + 6x^2 + 4x + 1$. Then $f'(x) = 12x^2 + 12x + 4 = 3(2x+1)^2 + 1$, which is always positive. So f(x) is always increasing, meaning that for y more than 1 we don't have any valid solutions. Similarly you can go into the negatives and see that y cannot be less than -1. Now you only have to check a very limited set of values, for y going from -1 to 1 and x going from -2 to 2. It is easy to see now that the only solutions are $x=\pm 2, y = \pm 1.$
| {
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Recurrence relation $C_n = n + 1 + \dfrac{2}{n}\sum\limits_{k=0}^{n-1}C_k$. A Discrete Mathematics book from which I'm self-studying ("Discrete Mathematics and Its Applications", by Kenneth Rosen) asks me to do the following:
Given the following recurrence relation:
$$C_n = n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k$$
The book asks me to show that the sequence $\{C_n\}$, with base case $C_0 = 0$, also satisfies the recurrence relation $nC_n=(n+1)C_{n-1}+2n$ for $n=1,2,\cdots$.
I tried to solve it by induction. For this, I wrote the second recurrence relation for $n+1$:
$$(n+1)C_{n+1} = (n+2)C_{n} + 2n + 2$$
Then, assuming that the first recurrence relation holds for $n$, I tried to substitute $C_n = n + 1 + \frac{2}{n}\sum\limits_{k=0}^{n-1}C_k$ in the above equation, to see if I obtain $C_{n+1} = n + 2 + \frac{2}{n+1}\sum\limits_{k=0}^{n}C_k$.
$$\begin{align*}
(n+1)C_{n+1} &= (n+2)C_{n} + 2n + 2\\
(n+1)C_{n+1} &= (n+2)\left( n + 1 + \frac{2}{n}\sum_{k=0}^{n-1}C_k \right ) + 2n + 2\\
(n+1)C_{n+1} &= n(n+2) + n + 2 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k + 2n + 2\\
(n+1)C_{n+1} &= n^2 + 5n + 4 + \frac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
(n+1)C_{n+1} &= (n+1)(n+4) + \dfrac{2(n+2)}{n}\sum_{k=0}^{n-1}C_k\\
C_{n+1} &= n+4 + \dfrac{2(n+2)}{n(n+1)}\sum_{k=0}^{n-1}C_k
\end{align*}$$
From this point, I'm not sure how to proceed.
| $C_n=n+1+(2/n)\sum^{n-1}$; $nC_n=n^2+n+2\sum^{n-1}$; $$nC_n-(n+1)C_{n-1}=n^2+n-((n-1)C_n-2\sum^{n-2})=n^2+n-((n-1)^2+(n-1))=2n$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fastest way to compare fractions Which is the fastest method to compare the below fractions with minimum calculation possible and finding which is greatest and which the smallest??
$$\frac{26}{686},\quad \frac{48}{874},\quad \frac{80}{892},\quad \frac{27}{865}$$
| $\dfrac{1}{30} < \dfrac{26}{686} < \dfrac{1}{20}$ because $30 \times 26 = 780 > 686$ while $20 \times 26 = 520 < 686$.
$\dfrac{1}{20} < \dfrac{48}{874} < \dfrac{1}{12}$ because $20 \times 48 = 960 > 874$ while
$12 \times 48 = 144 \times 4 = 576 < 874$
$\dfrac{1}{12} < \dfrac{80}{892}$ because $12 \times 80 = 960 > 892$.
$\dfrac{27}{865} < \dfrac{1}{30}$ because $30 \times 27 = 810 < 865$.
So $\dfrac{27}{865} < \dfrac{26}{686} < \dfrac{48}{874}< \dfrac{80}{892}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$? Is there a sequence in $(0,1)$ such that the product of all its terms is $\frac{1}{2}$?
| Recall either of Euler's two famous expressions for $\sin x$:
$$\sin x=x\prod_{n=1}^\infty \cos\left(\frac{x}{2^n}\right),$$
or
$$\sin x=x\prod_{n=1}^\infty\left(1-\frac{x^2}{\pi^2n^2}\right).$$
Now let $x=\dfrac{\pi}{6}$.
Or else use the following formula of Viète
$$\frac{2}{\pi}=\frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2}\cdots,$$
and multiply both sides by $\dfrac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/169128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Showing that the last digit of $a$ and $a^{13}$ are the same
For $a \in \mathbb N$, show that the last digit of $a$ and $a^{13}$
are the same.
For example: $2^{13} = 8,192$
$7^{13} = 96,889,010,407$
| Claim: For all integers $a$ we have $10|a^5-a$.
This follows from Fermat Little Theorem, but can also be proven elementary.
Proof of the claim
$$a^5-a=a(a-1)(a+1)(a^2+1) \,.$$
Now, one of $a$ or $a+1$ is even thus $a^5-a$ is divisible by $2$.
Also
$$a^5-a=a(a-1)(a+1)(a^2-4+5)=a(a-1)(a+1)(a^2-4)+5a(a-1)(a+1)$$
$$a^5-a=(a-2)(a-1)a(a+1)(a+2)+5a(a-1)(a+1)$$
The product of 5 consecutive integers is always divisible by 5, hence so is $a^5-a$.
This proves the claim.
Proof of your exercise
$$a^{13}-a=a^{13}-a^9+a^9-a^5+a^5-a=(a^5-a)(a^8+a^4+1) \,.$$
| {
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Finding a number when its remainder is given. I am trying to solve this problem
W is a positive integer when divided by 5 gives remainder 1 and when divided by 7 gives remainder 5. Find W.
I am referring back to an earlier post I made. Now I am attempting to solve it that way.
We know that
$$w\equiv1(mod~5)$$
$$w\equiv5(mod~7)$$
$w=7r+5$
$w=5r+5+2r$
Since $5r+5$ is divisible by 5
$w=5(r+1)+2r$ this shows the remainder is $2r$
Now $2r$ divided by 5 gives a remainder 1 , thus giving the equation
$2r = 5k + 1$ or $r=\frac{5k+1}{2}$
Putting r back in $w=7r+5$ we get
$2w = 35k + 12$
So I guess $w= \frac{35+12}{2} = 23.5$
This is wrong and the answer is suppose to be 26. Any suggestions what I might be doing wrong ? or anything that I might be missing ?
Edit:
The problem was in calculation
$w= \frac{35+17}{2} = 26$
| w is of the form 5a+1=7b+5 where a,b are integers.
=>5a=7b+4
=>5a+10=7b+14
=>5(a+10)=7(b+2)
=>5 divides (b+2) as (5,7)=1
=>b is of the form 5c-2 where c is any integer.
w=7b+5=7(5c-2)+5=35c-9=35d+26 where d=c-1
Alternatively, according to Euclid's GCD algorithm,
there exists integers c,d such that cx+dy=(x,y).
As (5,7)=1 => 5c+7d=1.
(i)By observation, one set of values of (c,d) = (3,-2).
Or (ii)$\frac{7}{5} = 1 + \frac{2}{5} = 1 + \frac{1}{2+\frac{1}{2}} $,
The 2nd convergent = $ 1 + \frac{1}{2} = \frac{3}{2}$
Then, 3(5)-2(7) must be ±1 is actually 1. So, (c,d) = (3,-2)
Then, 5a+1=7b+5 =>5a=7b+4
=>5a=7b+4(5(3)+7(-2))
=>5(a-12)=7(b-8)
=>5=7$\frac{b-8}{a-12}$ an integer.
=>7|(a-12)
=>$\frac{b-8}{5}$=$\frac{a-12}{7}$=h(some integer)
=>b=5h+8
=>w=7b+5=7(5h+8)+5=35h+61=35k+26 if k=h+1
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluate :$\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx$ How to evaluate
$$
\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}} dx
$$
I know that $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ but after that I have no idea, so please help me. Thanks in advance.
I tried this way,
$$
\int \frac{\sin^{-1} \sqrt{x} -\cos^{-1} \sqrt{x}}{\frac{\pi}{2}}dx
$$
then I put the value $\sin^{-1} \sqrt{x} +\cos^{-1} \sqrt{x}=\frac{\pi}{2}$ so
$$
\frac{2}{\pi}\int\left(\sin^{-1} \sqrt{x} -\left(\frac{\pi}{2}-\sin^{-1} \sqrt{x}\right)\right)dx
$$
Is this right?
after that I integrate by part and get,
$$ \int \frac{\sqrt{x}}{\sqrt{1-x}}$$ now,what can i do?
| *
*Using the relation $\arcsin(\sqrt{x}) + \arccos(\sqrt{x}) = \frac{\pi}{2}$ (which is valid for $0 \leqslant x \leqslant 1$, so this must an implicit assumption in your problem) solve for $\arccos(\sqrt{x})$ and substitute that into the integrand.
*After that make a $u$-substitution $u = \arcsin(\sqrt{x})$. This should lead to $\int \left( \frac{4 u}{\pi} - 1\right) \sin(2u) \mathrm{d} u$. This can be integrated by parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/173866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Looking for multiple solutions to show my students. Let ABC be a right triangle with B the right angle. X,Y and Z are on BC, CA and AB respectively such that BXYZ is a square. If the square is of side length m, AY = r and YC = s, find m in terms of r and s.
I have two solutions for my students (I instruct a math team class): One involves proportions and the other involves the angle bisector theorem (BY is an angle bisector). I am just curious of there are any other creative solutions for this problem.
| The solution for someone who really prefers algebra to geometry:
Let $AZ=p$, $XC=q$. Then the Pythagorean theorem tells us that
\begin{eqnarray}
p^2+m^2&=&r^2\\
q^2+m^2&=&s^2\\
(p+m)^2+(q+m)^2&=&(r+s)^2 \, .
\end{eqnarray}
Subtracting the first two equations from the third and simplifying yields
$p+q=\frac{rs}{m}$, while subtracting the second equation from the first yields $p^2-q^2=r^2-s^2$. So $p-q=\frac{m}{rs}(r^2-s^2)$, and thus
$$p=\frac{1}{2}\left(\frac{mr}{s}+\frac{rs}{m}-\frac{ms}{r}\right) \, .$$
Substituting this value into the first Pythagorean relation, we have
$$\frac{1}{4}\left(\frac{m^2r^2}{s^2}+\frac{r^2s^2}{m^2}+\frac{m^2s^2}{r^2}+2r^2-2s^2-2m^2\right)+m^2-r^2=0 \\
\frac{1}{4}\left(\frac{m^2r^2}{s^2}+\frac{r^2s^2}{m^2}+\frac{m^2s^2}{r^2}-2r^2-2s^2+2m^2\right)=0 \\
\left(\frac{mr}{s}-\frac{rs}{m}+\frac{ms}{r}\right)^2=0 \, ;$$
isolating $m$ gives
$$m^2=\frac{rs}{\frac{r}{s}+\frac{s}{r}} \, .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? What function satisfies $x^2 f(x) + f(1-x) = 2x-x^4$? I'm especially curious if there is both an algebraic and calculus-based derivation of the solution.
| Given, $x^2f(x)+f(1−x)=2x−x^4$
$f(-1) = 0$
$f(0) = 1$
$f(1) = 0$
$f(2) = -3$
$f(3) = - 8$
$f(4) = -15$
By careful observation one can see that the difference between the terms vanishes after the second stage
i.e. $f(x)$ is of the form $a{x^2} + b{x} +c$
i.e. $f(x) = a{x^2} + b{x} + c$ -------(1)
i.e. $f(0) = c = 1$
i.e. $c = 1$ -------(2)
$f(1) = a + b + c = 0$
i.e. $f(1) = a + b + 1= 0$
i.e. $a + b = -1$ -------(3)
$f(2) = 4a + 2b + c = -3$
i.e. $4a + 2b + 1 = -3$
i.e. $4a + 2b = -2$
i.e. $2a + b = -2$ -------(4)
Solve equations $(3)$ and $(4)$ to get $a = -1$ and $b = 0$
i.e. $f(x) = {-1}{x^2} + {0}{x} + 1$
i.e. $f(x) = {-x^2} + 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$ Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$.
Can anyone help me with this? Thank You!
| Here's a linear algebraic route: from this answer, we find that the eigenvalues of the $4\times4$ min-matrix
$$\mathbf M=\begin{pmatrix}
1 & 1 & 1 & 1 \\
1 & 2 & 2 & 2 \\
1 & 2 & 3 & 3 \\
1 & 2 & 3 & 4
\end{pmatrix}$$
are $\lambda_k=\dfrac14\sec^2\left(\dfrac{k\pi}{9}\right)$ for $k=1,\dots,4$. From this, we have that the eigenvalues of $4\mathbf M-\mathbf I$ are $\nu_k=\tan^2\left(\dfrac{k\pi}{9}\right)$, and since the sum of the eigenvalues is equal to the trace of the matrix,
$$\tan^2\frac{\pi}{9}+\tan^2\frac{2\pi}{9}+\tan^2\frac{4\pi}{9}=4(1+2+3+4)-4-\tan^2\frac{\pi}{3}=33$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 1
} |
Compute $ I_{n}=\int_{-\infty}^\infty \frac{1-\cos x \cos 2x \cdots \cos nx}{x^2}\,dx$ I'm very curious about the ways I may compute the following integral. I'd be very glad to know your approaching ways for this integral:
$$
I_{n} \equiv
\int_{-\infty}^\infty
{1-\cos\left(x\right)\cos\left(2x\right)\ldots\cos\left(nx\right) \over x^{2}}
\,{\rm d}x
$$
According to W|A, $I_1=\pi$, $I_2=2\pi$, $I_3=3\pi$, and one may be tempted to think that it's about an arithmetical progression here, but things change (unfortunately) from $I_4$ that is $\frac{9 \pi}{2}$. This problem
came to my mind when I was working on a different problem.
| First note that
$$\int_{-\infty}^{\infty} \frac{1-\cos ax}{x^2} \; dx
= \left[ -\frac{1-\cos ax}{x}\right]_{-\infty}^{\infty} + a \int_{-\infty}^{\infty} \frac{\sin ax}{x} \; dx
= \pi \, |a|,$$
by the Dirichlet integral. Also, by mathematical induction we can easily prove that
$$ \prod_{k=1}^{n} \cos \theta_k = \frac{1}{2^n} \sum_{\mathrm{e}\in S} \cos\left( e_1 \theta_1 + \cdots + e_n \theta_n \right),$$
where the summation runs over the set $S = \{ -1, 1\}^n$. Thus we have
$$ \begin{align*}
I_n = \int_{-\infty}^{\infty} \frac{1-\cos x \cdots \cos nx}{x^2} \; dx
&= \frac{1}{2^n} \sum_{\mathrm{e}\in S} \int_{-\infty}^{\infty} \frac{1-\cos(e_1 x + \cdots + e_n nx)}{x^2} \; dx \\
&= \frac{\pi}{2^n} \sum_{\mathrm{e}\in S} \left|e_1 + \cdots + e_n n\right|.
\end{align*}$$
For example, if $n = 3$, we have $\left|\pm 1 \pm 2 \pm 3\right| = 0, 0, 2, 2, 4, 4, 6, 6$ and hence
$$I_3 = \frac{\pi}{8}(0 + 0 + 2 + 2 + 4 + 4 + 6 + 6) = 3\pi.$$
Let the summation part as
$$ A_n = \sum_{\mathrm{e}\in S} |e_1 + \cdots + e_n n|.$$
The first 10 terms of $(A_n)$ are given by
$$ \left(A_n\right) = (2, 8, 24, 72, 196, 500, 1232, 2968, 7016, 16280, \cdots ), $$
and thus the corresponding $(I_n)$ are given by
$$ \left(I_n\right) = \left( \pi ,2 \pi ,3 \pi ,\frac{9 \pi }{2},\frac{49 \pi }{8},\frac{125 \pi }{16},\frac{77 \pi }{8},\frac{371 \pi }{32},\frac{877 \pi
}{64},\frac{2035 \pi }{128} \right).$$
So far, I was unable to find a simple formula for $(A_n)$, and I guess that it is not easy to find such one.
p.s. The probability distribution of $S_n = e_1 + \cdots + e_n n$ is bell-shaped, and fits quite well with the corresponding normal distribution $X_n \sim N(0, \mathbb{V}(S_n))$. Thus it is not bad to conjecture that
$$ \frac{A_n}{2^n} = \mathbb{E}|S_n| \approx \mathbb{E}|X_n| = \sqrt{\frac{n(n+1)(2n+1)}{3\pi}},$$
and hence
$$ I_n \approx \sqrt{\frac{\pi \, n(n+1)(2n+1)}{3}}.$$
Indeed, numerical experiment shows that
I was able to prove a much weaker statement:
$$ \lim_{n\to\infty} \frac{I_n}{n^{3/2}} = \sqrt{\frac{2\pi}{3}}. $$
First, we observe that for $|x| \leq 1$ we have
$$ \log \cos x = -\frac{x^2}{2} + O\left(x^4\right).$$
Thus in particular,
$$
\sum_{k=1}^{n} \log\cos\left(\frac{kx}{n}\right)
= \sum_{k=1}^{n}\left[-\frac{k^2 x^2}{2n^2} + O\left(\frac{k^4x^4}{n^4}\right)\right]
= -\frac{nx^2}{6} + O\left(x^2 \vee nx^4\right).$$
Now let
$$ \begin{align*}\frac{1}{n^{3/2}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n}\cos (kx)}{x^2} \; dx
&= \frac{1}{\sqrt{n}} \int_{-\infty}^{\infty} \frac{1 - \prod_{k=1}^{n}\cos \left(\frac{kx}{n}\right)}{x^2} \; dx \qquad (nx \mapsto x) \\
&= \frac{1}{\sqrt{n}} \int_{|x|\leq 1} + \frac{1}{\sqrt{n}} \int_{|x| > 1}
=: J_n + K_n.
\end{align*}$$
For $K_n$, we have
$$ \left|K_n\right| \leq \frac{1}{\sqrt{n}} \int_{1}^{\infty} \frac{2}{x^2}\;dx = O\left(\frac{1}{\sqrt{n}}\right).$$
For $J_n$, the substitution $\sqrt{n} x \mapsto y$ gives
$$ \begin{align*}
J_n
&= \frac{1}{\sqrt{n}} \int_{|x|\leq 1} \left( 1 - \exp\left( -\frac{nx^2}{6} + O\left(x^2 \vee nx^4\right) \right) \right) \; \frac{dx}{x^2} \\
&= \int_{|y|\leq\sqrt{n}} \left( 1 - \exp\left( -\frac{y^2}{6} + O\left(\frac{y^2}{n}\right) \right) \right) \; \frac{dy}{y^2} \\
&\xrightarrow[]{n\to\infty} \int_{-\infty}^{\infty} \frac{1 - e^{-y^2/6}}{y^2} \; dy \\
&= \left[-\frac{1-e^{-y^2/6}}{y}\right]_{-\infty}^{\infty} + \frac{1}{3} \int_{-\infty}^{\infty} e^{-y^2/6} \; dy
= \sqrt{\frac{2\pi}{3}}.
\end{align*}$$
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/175843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 0
} |
Prove $\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)$ if $A+B+C=180$ degrees I most humbly beseech help for this question.
If $A+B+C=180$ degrees, then prove
$$
\sin^2(A)+\sin^2(B)-\sin^2(C)=2\sin(A)\sin(B) \cos(C)
$$
I am not sure what trig identity I should use to begin this problem.
| This may not be the shortest way, but is pretty systematic and doesn't involve any real trickery. First, we eliminate the $C$ angles from the equation, using that $C = 180 - A - B$. We write $\sin(C) = \sin(180 - A - B) = \sin(A + B)$, and $\cos(C) = \cos(180 - A - B) = -\cos(A + B)$ and what you need to prove is
$$\sin^2A + \sin^2B - \sin^2(A + B) = -2\sin A\sin B \cos(A+B)$$
Inserting the sine and cosine addition formulas in this, what you need to prove becomes
$$\sin^2A + \sin^2B - (\sin A\cos B + \cos A \sin B)^2 = -2\sin A \sin B(\cos A \cos B - \sin A \sin B)$$
Writing this out, it becomes
$$\sin^2A + \sin^2B - \sin^2 A\cos^2 B - 2\sin A \cos A \sin B \cos B -\cos^2 A \sin^2B$$
$$ = -2\sin A\sin B \cos A \cos B + 2\sin^2 A \sin^2 B$$
Cancelling terms, your goal is to prove
$$\sin^2A + \sin^2B - \sin^2 A\cos^2 B-\cos^2 A \sin^2B = 2\sin^2 A \sin^2 B$$
Equivalently,
$$\sin^2A(1 - \cos^2B) + \sin^2B(1 - \cos^2A) = 2\sin^2 A \sin^2 B$$
This last equation is true since $1 - \cos^2 = \sin^2$. So going backwards in the above steps, one obtains the original equation $\sin^2A + \sin^2B - \sin^2C = 2\sin A \sin B \sin C$ as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/177208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
a question about an infinite series calculation. I want to prove that for $y >0$, $ x \in \mathbb R$,
$$ \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \frac{1}{2} \frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}$$
| Let's use J.M.'s excellent hint :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\ \Im{\sum_{n=-\infty}^\infty \frac 1{x+iy+n}}$$
Setting $z:=x+iy\ $ we will evaluate :
$$\sum_{n=-\infty}^\infty \frac 1{z+n}=\frac 1z+\sum_{n=1}^\infty \frac 1{z+n}+\frac 1{z-n}=\frac 1z+\sum_{n=1}^\infty \frac {2z}{z^2-n^2}$$
The series at the right may be simplified (this was proved many times at S.E. for example here or here or here) getting :
$$\pi\cot(\pi z)=\frac1{z}+\sum_{n=1}^{\infty}\frac{2z}{z^2-n^2}$$
and the simple result :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\;\Im{(\cot\pi (x+iy))}$$
Your remaining problem is that this simple answer doesn't appear related to your more complicated result. Worse your L.H.S and R.H.S. terms don't numerically correspond so that it seems that you have a problem in your question !
Let's add that your R.H.S. term may be rewritten as half :
$$\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}=\frac{\sinh(2\pi y)}{\cosh(2 \pi y) - \cos ( 2 \pi x )}$$
UPDATE: In fact your equation should read (simply multiplying your R.H.S. by $2\pi$) :
$$\boxed{\displaystyle \sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2} = \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}}$$
We want :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}= -\pi\Im{(\cot \pi (x+iy))}$$
Let's observe that $\ \sin a-\sin b=2\sin\frac{a-b}2\cos\frac{a+b}2$ and $\ \cos a-\cos b=-2\sin\frac{a+b}2\sin\frac{a-b}2$
implies (dividing these expressions) :
$$\frac {\sin a-\sin b}{\cos a-\cos b}=-\cot \frac{a+b}2$$
Set $a:=2\pi x,\ b:=2\pi i y\ $ to get :
$$-\cot\pi (x+iy)=\frac {\sin 2\pi x-\sin 2\pi i y}{\cos 2\pi x-\cos 2\pi i y}=\frac {\sin 2\pi x-i\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$
so that :
$$\sum_{n=-\infty}^\infty \frac{y}{(x+n)^2 + y^2}=-\pi\frac {\sinh 2\pi y}{\cos 2\pi x-\cosh 2\pi y}$$
$$= \pi\frac{1 - e^{-4 \pi y }}{1 - 2 e^{-2 \pi y} \cos ( 2 \pi x ) + e^{-4 \pi y}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/177388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Number of bit strings with 3 consecutive zeros or 4 consecutive 1s I am trying to count the number of bit-strings of length 8 with 3 consecutive zeros or 4 consecutive ones. I was able to calculate it, but I am overcounting. The correct answer is $147$, I got $148$.
I calculated it as follows:
Number of strings with 3 consecutive zeros = $2^5+5\times2^4 = 112$, because the 3 zeros can start at bit number 1, 2, 3, .., 6
Number of strings with 4 consecutive ones = $2^4+4\times2^3 = 48$, I used the same reasoning.
Now I am trying to count the number of bit-strings that contain both 3 consecutive zeros and 4 consecutive 1s. I reasoned as follows:
the strings can be of the following forms: 0001111x, 000x1111, x0001111..thus there are $2+2+2 = 6$ possibilities for bit-strings where the 3 consecutive zeros come first. Symmetrically there are $6$ bit-strings where the 4 consecutive ones come first.
Thus the answer should be = $112+48-12 = 148$.
clearly there's something wrong with my reasoning, if someone could point it out, that would be awesome. Thanks
| Let $n$ be a nonnegative integer. Let $a_n$ be the number of bit strings of length $n$ with at least 3 consecutive zeros. Clearly $a_0 = 0$, $a_1 = 0$, $a_2 = 0.$ Let $n \geq 3.$ Denote by $S$ the set of all bit strings of length $n$ with at least 3 consecutive zeros. The set $S$ is a disjoint union of the following four sets:
*
*$S_3$ the set of all bit strings of length $n$ that end with $000$,
*$S_2$ the set of all bit strings in $S$ that end with $100$,
*$S_1$ the set of all bit strings in $S$ that end with $10$,
*$S_0$ the set of all bit strings in $S$ that end with $1$.
These sets have the following cardinalities: $|S_3| = 2^{n-3}$, $|S_2| = a_{n-3}$, $|S_1| = a_{n-2}$, $|S_0| = a_{n-1}$. Since $|S| =a_n$, by the sum rule
$$
a_n = 2^{n-3} + a_{n-3} + a_{n-2} + a_{n-1}.
$$
Since $a_0 = 0$, $a_1 = 0$, $a_2 = 0$, we have
\begin{align*}
a_3 & = 2^0 + 0 + 0 + 0 = 1\\
a_4 & = 2^1 + 0 + 0 + 1 = 3\\
a_5 & = 2^2 + 0 + 1 + 3 = 8\\
a_6 & = 2^3 + 1 + 3 + 8 = 20\\
a_7 & = 2^4 + 3 + 8 + 20 = 47\\
a_8 & = 2^5 + 8 + 20 + 47 = 107.
\end{align*}
For bit strings with at least 4 consecutive ones, the same reasoning leads to the recursion
$$
b_0 = b_1 = b_2 = b_3 = 0, \quad b_n=2^{n-4}+b_{n-4}+ b_{n-3}+ b_{n-2}+ b_{n-1}, \ n\geq 4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/178605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Proving $\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$ I would like to know how to prove the following assertion :
For every $n>0$: $$\sqrt{n}+\frac{1}{\sqrt{n+1}} \geq \sqrt{n+1}$$
| There are many approaches. One of them is to try to prove the equivalent assertion that $\sqrt{n+1}-\sqrt{n} \le \frac{1}{\sqrt{n+1}}$.
Note that
$$\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}.\tag{$1$}$$
It is clear that for $n \gt 0$, the right-hand side of $(1)$ is $\lt \frac{1}{\sqrt{n+1}}$.
The calculation gives a stronger estimate. The right-hand side of $(1)$ is actually close to $\frac{1}{2\sqrt{n+1}}$ when $n$ is large.
Another way: Equivalently, we want to prove that $\sqrt{n+1}-\frac{1}{\sqrt{n+1}} \le \sqrt{n}$.
We have
$$\sqrt{n+1}-\frac{1}{\sqrt{n+1}}=\frac{(n+1)-1}{\sqrt{n+1}}=\frac{n}{\sqrt{n+1}}.$$
Since $\sqrt{n+1} \gt \sqrt{n}$, we conclude that if $n \gt 0$ then $\frac{n}{\sqrt{n+1}}\lt \sqrt{n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/179194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Evaluating $\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$ using $\frac{dy}{dx}=\frac{2}{(x+3)^2}$ where $y=\frac{x+1}{x+3}$ Find derivative of $$y= \frac{ax+b}{cx+d}$$
I found it to be $$\frac{dy}{dx}=\frac{a}{cx+d}-\frac{c(ax+b)}{(cx+d)^2}$$
Use it to evaluate:
$$\int_0^1{\frac{1}{(x+3)^2}}\ln\left(\frac{x+1}{x+3}\right)dx$$
I figured that here $y=\frac{x+1}{x+3}$ and $$\frac{dy}{dx}=\frac{1}{x+3}-\frac{(x+1)}{(x+3)^2}$$
and using the technique I learned from my last question I did this:
$$\frac{dy}{dx}=\frac{(x+3)}{(x+3)^2}-\frac{(x+1)}{(x+3)^2}=\frac{2}{(x+3)^2}$$
which I could then substitute back, having changed the limits by substituting $1$ into $y$ and then $0$ into $y$:
$$y|_{x=1}=\frac{x+1}{x+3}=\frac{1}{2}$$
$$y|_{x=0}=\frac{1}{3}=\frac{1}{3}$$
$$2\int_0^1{\frac{dy}{dx}}\ln(y)dx=2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$
This gives me:
$$2\int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$
$$=2\left[y(\ln(y)-1)\right]_\frac{1}{3}^\frac{1}{2} = 2\left[\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-\frac{1}{2}\right)\right]-\frac{1}{3}\left[\ln\left(\frac{1}{3}\right)-\frac{1}{3}\right]\\$$
$$\ln\left(\frac{1}{2}\right)-1-\frac{2}{3}\ln\left(\frac{1}{3}\right)+\frac{2}{9}$$
The problem is I am supposed to end up with something else. Can anyone spot any issues with this?
EDIT: This is the answer I am supposed to be getting:
$$\frac{1}{6}\ln(3)-\frac{1}{4}\ln(2)-\frac{1}{12}$$
| *
*You have:
$$\dfrac{dy}{dx}=\dfrac{(x+3)}{(x+3)^2}-\dfrac{(x+1)}{(x+3)^2}=\dfrac{2}{(x+3)^2}$$
But the integral is $I=\int_0^1{\dfrac{1}{(x+3)^2}}\ln\left(\dfrac{x+1}{x+3}\right)dx$ where ${\dfrac{1}{(x+3)^2}}$ is actually $\dfrac 12 \times \dfrac{2}{(x+3)^2}$. Therefore:
$$I= \dfrac 12 \int_0^1{\frac{dy}{dx}}\ln(y)dx=\dfrac 12 \int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}$$
*
*The other issue might be:
$$\dfrac 12 \int_\frac{1}{3}^\frac{1}{2}{\ln(y)dy}=\dfrac 12 \bigg[y(\ln(y)-1\bigg]_\frac{1}{3}^\frac{1}{2}
=\\\frac 12 \left(\frac{1}{2}\left(\ln\left(\frac{1}{2}\right)-1\right)-\frac{1}{3}\left(\ln\left(\frac{1}{3}\right)-1\right)\right)
\\=\frac 12 \left(-\frac 12 \ln 2 - \frac 12 +\frac 13 \ln 3+ \frac 13\right)\\=\frac 16 \ln 3 -\frac 14 \ln 2 -\frac 1{12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/179475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality $\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$? Let $a,b,c>0$ and $a+b+c= 1$, how to prove the inequality
$$\frac{\sqrt{a}}{1-a}+\frac{\sqrt{b}}{1-b}+\frac{\sqrt{c}}{1-c}\geq \frac{3\sqrt{3}}{2}$$?
| This is as far as I got...
$\frac{(1-b)\sqrt{a}}{(1-b)(1-a)}$ + $\frac{(1-a)\sqrt{b}}{(1-a)(1-b)}$ + $\frac{\sqrt{c}}{(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$
$\frac{(1-b)\sqrt{a} + (1-a)\sqrt{b}}{(1-b)(1-a)}$ + $\frac{\sqrt{c}}{(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$
$\frac{(1-c)(1-b)\sqrt{a}+(1-c)(1-a)\sqrt{b}+(1-b)(1-a)\sqrt{c}}{(1-a)(1-b)(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$
$\frac{\sqrt{a}(1-b-c+bc)+\sqrt{b}(1-a-c+ac)+\sqrt{c}(1-a-b+ab)}{(c+ab)((1-c)} $ $\geq$ $\frac{3\sqrt{3}}{2} $
$\Leftrightarrow$
$\frac{\sqrt{a}(bc+a)+\sqrt{b}(ac+b)+\sqrt{c}(ab+c)}{(ab+c)(1-c)}$ $\geq$ $\frac{3\sqrt{3}}{2} $
Can anyone explain the rest to me? Can you say that (bc+a),(ac+b),(ab+c) $\leq$ 1 ? Also how would you formally say that, I mean I know that two fractions times each other is less than 1, but I don't know how to formally state that. Thanks!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
Compute $\lim\limits_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{{1}/{2}}\cdots\left(1+\frac{n}{n}\right)^{{1}/{n}}$ Compute the limit:
$$\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots\left(1+\frac{n}{n}\right)^{\frac{1}{n}}$$
| We want to calculate $$\lim_{n \to \infty} \prod_{1 \leqslant k \leqslant n} \left(1 + \frac {k} {n}\right)^{\frac {1} {k}}.$$ Denote it by $\ell$. Taking logarithms we have
$$\begin {eqnarray}
\log \ell & = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \frac {1} {k} \log \left(1 + \frac {k} {n}\right) \nonumber \\
& = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \frac {1} {k} \sum_{m \geqslant 1} (-1)^{m + 1} \frac {k^m} {m n^m} \nonumber \\
& = & \lim_{n \to \infty} \sum_{1 \leqslant k \leqslant n} \sum_{m \geqslant 1} (-1)^{m + 1} \frac {k^{m - 1}} {m n^m} \nonumber \\
& = & \lim_{n \to \infty} \sum_{m \geqslant 1} \sum_{1 \leqslant k \leqslant n} (-1)^{m + 1} \frac {k^{m - 1}} {m n^m} \nonumber \\
& = & \lim_{n \to \infty} \sum_{m \geqslant 1} \left(\frac {(-1)^{m + 1}} {m^2} + O \left(\frac {1} {n}\right) \right) \nonumber \\
& = & \sum_{m \geqslant 1} \frac {(-1)^{m + 1}} {m^2} \nonumber \\
& = & \frac {\pi^2} {12}.
\end {eqnarray}$$
Hence, $\ell = \exp \left(\frac {\pi^2} {12}\right)$.
Note that in step 2 we changed $\log \left(1 + \frac {k} {n}\right)$ with its Taylor expansion, and in step 5 we used the fact that $\sum_{1 \leqslant k \leqslant n} k^p = \frac {n^{p + 1}} {p + 1} + O (n^p)$ for any non-negative integer $p$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 2,
"answer_id": 1
} |
is there a method to find all the solutions to, $y^2-6y+\sqrt{y}+4=0$ is there a method to find all the solutions to the following set of irrational equations,
$\sqrt{x}+y=3$
$x+\sqrt{y}=5$
NOTE: $(4-1)=(2-1)(2+1)=3$ and $(4+1)=(2^2+1^2)=(3^2-2^2)=(3-2)(3+2)=5$
| $(3-y)^2+\sqrt{y}=5$
$y^2-6y+\sqrt{y}+4=0$
$(\sqrt{y}-1)(y\sqrt{y}+y-5\sqrt{y}-4)=0$
i.e. $\sqrt{y}=1$, and $x=4$ is a solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Properties of ellipse x-y form $$5x^2+8xy+5y^2=1$$
$$1\left(\frac{x-y}{\sqrt{2}}\right)^2+9\left(\frac{x+y}{\sqrt{2}}\right)^2=1$$
I know that these two forms are equal, showing that the equation is an ellipse.
I do know what happens when the ellipse is in the form of $ax^2 + by^2 = 1$ but not sure what happens in the combination of x-y cases.
Can anyone help me with this? (e.g., what would be two axis?)
| Using this, if we rotate the axes $\theta $ as the rotation of axes does not change the nature of the curve,
$x=X\cos\theta-Y\sin\theta$, $y=X\sin\theta+Y\cos\theta$,
$5(X\cos\theta-Y\sin\theta)^2+5(X\sin\theta+Y\cos\theta)^2+8(X\cos\theta-Y\sin\theta)(X\sin\theta+Y\cos\theta)=1$
$=>X^2(5+8\sin\theta\cos\theta)+Y^2(5-8\sin\theta\cos\theta)+8XY(\cos^2\theta-\sin^2\theta)=1$
$=>X^2(5+4\sin2\theta)+Y^2(5-4\sin2\theta)+8XY(\cos2\theta)=1$
If we make $\cos2\theta=0=>\sin2\theta=±1$
If $\sin2\theta=1$ i.e, $2\theta=2n\pi+\frac{\pi}{2}=>9X^2+Y^2=1$
If $\sin2\theta=-1$ i.e, $2\theta=2n\pi-\frac{\pi}{2}=>X^2+9Y^2=1$
Alternatively, according to 7.2 of this,
Here a=b=5, h=4, f=g=0
So,
$C=\begin{pmatrix} a & h \\ h & b \end{pmatrix}=5^2-4^2>0$
$Δ= \begin{pmatrix} a & h & g \\ h & b & f \\g & f & c \end{pmatrix}$
$=\begin{pmatrix} 5 & 4 & 0 \\ 4 & 5 & 0 \\0 & 0 & -1 \end{pmatrix}$
$=-(5^2-4^2)=-9<0$
and $aΔ=5(-9)<0$
So, the given curve represents an ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Sum of series $\sin x + \sin 2x + \sin 3x + \cdots $ Please help me compute the sum of the series:
$$\sin(x)+\sin(2x)+\sin(3x)+\cdots$$
| $$2\sin\frac{x}{2}\sin rx=\cos\frac{(2r-1)}{2}x-\cos\frac{(2r+1)}{2}x$$
Putting $r=1,2,\ldots,n-1,n$ we get,
$$2\sin\frac{x}{2}\sin x=\cos\frac{1}{2}x-\cos\frac{3}{2}x$$
$$2\sin\frac{x}{2}\sin 2x=\cos\frac{3}{2}x-\cos\frac{5}{2}x$$
$$\vdots$$
$$2\sin\frac{x}{2}\sin rx=\cos\frac{(2n-3)}{2}x-\cos\frac{(2n-1)}{2}x$$
$$2\sin\frac{x}{2}\sin nx=\cos\frac{(2n-1)}{2}x-\cos\frac{(2n+1)}{2}x$$
Adding we get, $2\sin\frac{x}{2}(\sin x+\sin 2x+...+\sin nx)=cos\frac{1}{2}x-\cos\frac{(2n+1)}{2}x=2\sin\frac{(n+1)x}{2}\sin\frac{nx}{2}$
So, $$\sin x+\sin 2x+\cdots+\sin nx=\frac{\sin\frac{(n+1)x}{2}\sin\frac{nx}{2}}{\sin\frac{x}{2}}$$
As $2\sin B\sin(A+2rB) = \cos(A+(2r-1)B) - \cos(A+(2r+1)B)$, we need to multiply with $2\sin B$ for $\sum_{r}\sin(A+2rB)$.
Here in this problem, $A=0, 2B=x$
Also the way Nerd-Herd has approached the problem,
$\sin rx$ = Imaginary part of $e^{irx}$
So, $\sum_{0 ≤ r ≤n}\sin rx=\sum_{0 ≤ r ≤n}$(Imaginary part of $e^{ix})$=Imaginary part of($\sum_{0 ≤ r ≤n}e^{ix}$)
Now,
$$\sum_{0 ≤ r ≤n}e^{irx}= \frac{e^{(n+1)ix}-1}{e^{ix}-1}= e^{\frac{(n+1)ix}{2}}\frac{(e^{\frac{(n+1)ix}{2}}-e^{-\frac{(n+1)ix}{2}})}{e^{\frac{ix}{2}}(e^{\frac{ix}{2}}-e^{-\frac{ix}{2}})}$$
We know, $\sin y=\frac{e^{iy}-e^{-iy}}{2i}$ So,$e^{iy}-e^{-iy}=2i\sin y$
So,$\sum_{0 ≤ r ≤n}e^{irx}=e^{\frac{inx}{2}}\frac{2i\sin\frac{(n+1)x}{2}}{2i\sin\frac{x}{2}}=(\cos \frac{nx}{2}+i\sin \frac{nx}{2})\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$
So, imaginary part of $\sum_{0 ≤ r ≤n}e^{irx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$
$\sum_{0 ≤ r ≤n}\sin{rx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$
$\sum_{1 ≤ r ≤n}\sin{rx}=\sin \frac{nx}{2}\frac{\sin\frac{(n+1)x}{2}}{\sin\frac{x}{2}}$ as $\sin(0x)=0$
So, the either approach leads to a compact from provided $\sin\frac{x}{2}≠0$
If $\sin\frac{x}{2}=0$ i.e., $\frac{x}{2}=m\pi$ where m is some integer,
$=>x=2m\pi=>\sin sx= 0$ for any integer s.
By observation if $\sin x=0$ i.e., $x=m\pi$ where m is some integer, $\sin sx= 0$
for any integer s.
Then the sum is clearly 0 if $x=m\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Writing 1/3 as a sum of other numbers Is it possible to write $0.3333(3)=\frac{1}{3}$ as sum of $\frac{1}{4} + \cdots + \cdots\frac{1}{512} + \cdots$ so that denominator is a power of $2$ and always different?
As far as I can prove, it should be an infinite series, but I can be wrong.
In case if it can't be written using pluses only, minuses are allowed as well.
For example, $\frac{1}{2}$ is $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$
So, what about $\frac{1}{3}$?
| Archimedes showed that if you have a finite sum in which each term is $1/4$ of the previous term, except that the last term is $1/3$ of the previous term, then the sum does not depend on the number of terms, but is just $4/3$ of the first term. In modern terminology:
$$
1+\frac 1 4 + \frac{1}{16} + \cdots + \frac{1}{4^n} + \frac{1}{4^n}\cdot\frac 1 3\ =\ \frac 4 3.
$$
He deduced from that that the infinite sum
$$
1 + \frac 1 4 + \frac{1}{16} + \cdots = \frac 4 3.
$$
(If the first term is $1/4$ then $4/3$ of that is of course $1/3$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
how can one solve for $x$, $x =\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}$
Possible Duplicate:
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
how can one solve for $x$, $x =\sqrt[]{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2\cdots }}}}}}$
we know, if $x=\sqrt[]{2+\sqrt{2}}$, then, $x^2=2+\sqrt{2}$
now, if $x=\sqrt[]{2+\sqrt{2}}$, then, $(x-\sqrt{2})(x+\sqrt{2})=\sqrt{2}$
| In general, the function $f(x)=\displaystyle\bigg(\small\sqrt{\normalsize x+\small\sqrt{\normalsize x+\sqrt{ x+\sqrt{x}}}}\;\normalsize\bigg)$ is:
$0.5(1+\sqrt[]{1+4x})$
Ref: Math-Integration of nested square roots of x
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
A cube is divided into two cuboids A cube is divided into two cuboids. The surfaces of those cuboids are in the ratio $7: 5$. Calculate the ratio of the volumes.
How can I calculate this?
| Suppose the original cube had side $a$. And let dimensions of the resulting cuboids be $a \times a \times b$ and $a \times a \times c$, where $a = b+c$.
The surface areas of the resulting cuboids is $2\left(a \cdot a + a \cdot b + a \cdot b\right) = 2 a(a+2 b)$ and $2a(a+2 c)$, and volumes $a^2 b$ and $a^2 c$ respectively.
It is known that:
$$
\frac{7}{5} = \frac{a+2b}{a+2c} \quad b+c = a
$$
Solving this gives $b = \frac{2}{3} a$ and $c=\frac{a}{3}$, hence the ratio of volumes is $\frac{a^2 b}{a^2 c} = \frac{b}{c} = 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Min/Max of $f_{n}(x)=\lim_{n\to\infty}\frac{(n^{x+1}+1^{x})(n^{x+1}+2^{x})\cdots(n^{x+1}+n^{x})}{(n^{x+1}-1^{x})(n^{x+1}-2^{x})\cdots(n^{x+1}-n^{x})}$ Let's consider the function $f_{n}(x)$ with $x>0$ defined as:
$$f_{n}(x)=\lim_{n\to\infty}\frac{(n^{x+1}+1^{x})(n^{x+1}+2^{x})\cdots(n^{x+1}+n^{x})}{(n^{x+1}-1^{x})(n^{x+1}-2^{x})\cdots(n^{x+1}-n^{x})}$$
I'd like to know what is the way to follow such that I may find out for what values of $x$ the function reaches its minimum and maximum , and then to compute these values. It's a
problem that came to my mind after studying another limit. However, I have more questions regarding this function, but for the moment it's enough if I find an answer for the part with minimum/maximum. Thanks.
| First write
\begin{align*}
\log f_n(x) &= \sum_{j=1}^n \log(n^{x+1}+j^x) - \sum_{j=1}^n \log(n^{x+1}-j^x) \\
&= \sum_{j=1}^n \log\bigg( 1+\frac{j^x}{n^{x+1}} \bigg) + \sum_{j=1}^n \log\bigg( 1- \frac{j^x}{n^{x+1}} \bigg)^{-1}.
\end{align*}
The following inequalities are valid for all $0<y<\frac12$:
\begin{align*}
y-y^2 &< \log(1+y) < y \\
y &< \log(1-y)^{-1} < y+y^2
\end{align*}
Therefore when $n\ge2$,
\begin{align*}
\log f_n(x) &> \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} - \bigg( \frac{j^x}{n^{x+1}} \bigg)^2 \bigg) + \sum_{j=1}^n \frac{j^x}{n^{x+1}} \\
&> 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \sum_{j=1}^n \frac1{n^2} \\
&= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x - \frac1n
\end{align*}
and
\begin{align*}
\log f_n(x) &< \sum_{j=1}^n \frac{j^x}{n^{x+1}} + \sum_{j=1}^n \bigg( \frac{j^x}{n^{x+1}} + \bigg( \frac{j^x}{n^{x+1}} \bigg)^2 \bigg) \\
&< 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \sum_{j=1}^n \frac1{n^2} \\
&= 2\sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x + \frac1n.
\end{align*}
By the squeeze theorem,
$$
\lim_{n\to\infty} \log f_n(x) = 2 \lim_{n\to\infty} \sum_{j=1}^n \frac1n \bigg( \frac jn \bigg)^x = 2 \int_0^1 u^x\, du = \frac2{x+1},
$$
because the sum is a Riemann sum for the integral. In other words,
$\lim_{n\to\infty} f_n(x) = e^{2/(x+1)}$
for all $x>0$ (since exponentiation is continuous).
| {
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"url": "https://math.stackexchange.com/questions/187367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why $ \lim_{n \rightarrow \infty}\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} =5e$? I have a small exercise and I don’t know who to get the result.
The exercise is: $$ \lim_{n \rightarrow \infty}\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} $$
I did following transformations:
$$
\frac{(5n^3-3n^2+7)(n+1)^{n-2}}{n^{n+1}} \\
(5n^{2-n}-3n^{1-n}+7^{-1-n})(n+1)^{n-2} \\
(\frac{5}{n^{-2+n}} - \frac{3}{n^{-1+n}} + \frac{7}{n^{1+n}})(n+1)^{n-2}
$$
But none of them helped me to see the result.
It would be great if someone could explain it to me.
Edit
@adrian-barquero
Ok. Fist you factories $^n$ and get
$$
\frac{(n+1)^n}{n^n} = (1+\frac{1}{n})^n = e \\
$$
In the other fraction I could extend with $n^3$
$$
\frac{5n^3-3n^2+7}{n(n + 1)^2} = \frac{n^3(5 - \frac{3n^2}{n^3} + \frac{7}{n^3})}{n^3(1 + \frac{2n^2}{n^3} + \frac{n}{n^3})} = 5
$$
| I would recommend rearranging as
$$
\frac{(5n^3-3n^2+7)(n+1)^n}{n^{n+1}(n+1)^2} = \frac{5n^3-3n^2+7}{n(n + 1)^2}\frac{(n+1)^n}{n^n} = \frac{5n^3-3n^2+7}{n(n + 1)^2} \left ( 1 + \frac{1}{n} \right )^n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Permutation across matrices. Matrices may be used to permute the order of elements in a set. For example:
$$
\begin{bmatrix}
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0
\end{bmatrix}
\times
\begin{bmatrix}
x \\
y \\
z \\
w
\end{bmatrix}
=
\begin{bmatrix}
w \\
x \\
y \\
z
\end{bmatrix}
$$
My problem is that I need to permute values in different matrices:
$$
A = \begin{bmatrix}
0 & 0 & x & 0 & 0 \\
x & x & x & x & x \\
x & 0 & 0 & 0 & x \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
$$
B = \begin{bmatrix}
0 & 0 & 0 & 0 & 0 \\
0 & 0 & y & 0 & 0 \\
y & y & y & y & y \\
y & 0 & 0 & 0 & y
\end{bmatrix}
$$
$$
C = \begin{bmatrix}
z & 0 & 0 & 0 & z \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & z & 0 & 0 \\
z & z & z & z & z
\end{bmatrix}
$$
$$
D = \begin{bmatrix}
w & w & w & w & w \\
w & 0 & 0 & 0 & w \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & w & 0 & 0
\end{bmatrix}
$$
Thus given the permutation matrix (from above) and the following pattern matrix how would you change all the $x$ in $A$ to $w$, all the $y$ in $B$ to $x$, all the $z$ in $C$ to $y$, and all the $w$ in $D$ to $z$:
$$
\begin{bmatrix}
0 & 0 & 1 & 0 & 0 \\
1 & 1 & 1 & 1 & 1 \\
1 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
$$
| When you multiply by a permutation matrix on the left, you permute rows (those become single entries in a column vector). When you multiply by a permutation matrix on the right, you permute columns. Here you need to permute rows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find $ \cot^2 a + \frac 1{\cot^2 a}$ if $\cot a + \frac 1{\cot a} = 1$ If $ \cot a + \frac 1 {\cot a} = 1 $, then what is $ \cot^2 a + \frac 1{\cot^2 a}$?
the answer is given as $-1$ in my book, but how do you arrive at this conclusion?
| Taking $x=\cot a$, $x+\frac{1}{x}=1\implies x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2x\frac{1}{x}=1-2=-1$
Alternatively, $x+\frac{1}{x}=1\implies x^2-x+1=0$
$x^2-x+1=0\implies x^3+1=0$
So,
$x^{3m}+(\frac{1}{x})^{3m}=(x^3)^m+\frac{1}{(x^3)^m}=2(-1)^m$
$x^{3m+1}+(\frac{1}{x})^{3m+1}=(x^3)^m\cdot x+\frac{1}{(x^3)^m\cdot x}=(-1)^m(x+\frac{1}{x})=(-1)^m$
$x^{3m+2}+(\frac{1}{x})^{3m+2}=(x^3)^m\cdot x^2+\frac{1}{(x^3)^m\cdot x^2}=(-1)^m(x^2+\frac{1}{x^2})=(-1)^m(-\frac{1}{x}-x)$ as $x^3=-1\implies x^2=-\frac{1}{x}$ and $\frac{1}{x^2}=x$
So,$x^{3m+2}+(\frac{1}{x})^{3m+2}=(-1)^{m+1}\cdot (x+\frac{1}{x})=(-1)^{m+1}$
If we put $m=0$, $x^2+\frac{1}{x^2}=(-1)^1=-1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the probability that $XYZ$ is divisible by $5$? A solution of $X + Y + Z = 20$ in non-negative integers is chosen at random. What is the probability that $XYZ$ is divisible by $5$?
Edit:
This happens to be an exam question. So I can't use calculators or computers and have to get the answer in less than 20 minutes while showing systematic workings. I appreciate the answers below, but can someone instruct me on solving the question given the mentioned constraints?
| There are two questions here:
*
*How many triples of numbers $(X,Y,Z)$ add to 20? We can call this $A$.
*How many of these are divisible by 5? We can call this $B$.
The answer will then be $B/A$.
First we calculate $A$. Let us write $C(x)$ for the total number of ways of choosing $(X,Y,Z)$ so that $X=x$ and $X+Y+Z=20$.
There are 21 possible choices for $X$. For each such $X$, there are $21-X$ possible choices for $Y$ that make $X+Y\le 20$, namely $\{0, 1, \ldots 20-X\}$. The other choices have $X+Y>20$ and therefore $X+Y+Z>20$. Once we have chosen $X$ and $Y$ at random, there is exactly one possible choice for $Z$ that makes $X+Y+Z=20$. So we have $C(x) = 21-x$.
We want the total of $C(x)$ for each $x$ between 0 and 20:
$$\begin{eqnarray}
A&=&\sum_{x=0}^{20} C(x) \\ &=& \sum_{x=0}^{20} (21-x)\\
&=& \sum_{x=0}^{20} 21 - \sum_{x=0}^{20} x \\
&=& 441 - 210 \\ &=& 231.
\end{eqnarray}$$
Now we calculate $B$. 5 is prime, so $XYZ$ is divisible by 5 if and only if one of $X$, $Y$, or $Z$ is divisible by 5. We can use inclusion-exclusion: $B$ is the sum of the cases where (at least) $X$, $Y$, or $Z$ is divisible by 5, minus the cases where (at least) two are divisible by 5, plus the cases where all three are divisible by 5. That is, $$\begin{eqnarray}B&=&D_x + D_{y} + D_z \\ &&- D_{xy} - D_{xz} - D_{yz} \\ &&+ D_{xyz}\end{eqnarray}$$
Where $D_{xy}$ denotes the number of choices of $(X,Y,Z)$ where $5\mid X$ and $5\mid Y$, and similarly for the others.
By symmetry, $D_x = D_y = D_z$, and $D_{xy} = D_{xz} = D_{yz}$. Also, it is impossible to have two of $(X,Y,Z)$ divisible by 5 without the third also being divisible by 5, so $D_{xy} = D_{xyz}$. So the previous equation reduces to:
$$B = 3D_x - 2D_{xyz}$$
We calculate $D_x$: $X$ will be a multiple of 5 whenever $X\in\{0,5,10,15,20\}$, so we want
$$\begin{eqnarray}
&&\sum_{5\mid X} C(X) \\
&=& \sum_{X\in\{0,5,10,15,20\}} (21-X) \\
&=& 21\cdot5 - (0+5+10+15+20) \\
&=& 105 - 50 \\ &=& 55
\end{eqnarray}$$
To calculate $D_{xyz}$ is quick because there are very few such triples, and we can enumerate them by brute force:
$(0,0,20)\ldots (0,20,0), (5,0,15)\ldots (5,15,0),\ldots (20,0,0)$. This is 5+4+3+2+1 = 15.
So we have $B = 3\cdot 55 - 2\cdot 15 = 135$.
Thus the answer is $B/A = 135/231 = 45/77$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int\frac{\cos^{n-1}\frac{x+a}{2}}{\sin^{n+1}\frac{x-a}{2}}\;dx$ Please help me to evaluate the following integral:
$$\int\dfrac{\cos^{n-1}\dfrac{x+a}{2}}{\sin^{n+1}\dfrac{x-a}{2}}\;dx$$
| $$\int\frac{\cos^{n-1}\frac{x+a}{2}}{\sin^{n+1}\frac{x-a}{2}}dx=\int\frac{\cos^{n-1}\frac{x+a}{2}}{\sin^{n-1}\frac{x-a}{2}}\cdot\frac{dx}{\sin^2\frac{x-a}{2}}=\left|\frac{\cos\frac{x+a}{2}}{\sin\frac{x-a}{2}}=t\Rightarrow\frac{dx}{\sin^2\frac{x-a}{2}}=-\frac{2dt}{\cos a}\right|=-\frac{2}{\cos a}\int{t^{n-1}dt}=-\frac{2}{\cos a}\cdot\frac{t^n}{n}=-\frac{2}{n\cos a}\cdot\frac{\cos^n\frac{x+a}{2}}{\sin^n\frac{x-a}{2}}+C.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $ \frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}. $ Prove that $$ \frac{1}{2}\cot^{-1}\frac{2\sqrt[3]{4}+1}{\sqrt{3}}+\frac{1}{3}\tan^{-1}\frac{\sqrt[3]{4}+1}{\sqrt{3}}=\dfrac{\pi}{6}. $$
| Use the following identities:
$$\arctan a \pm\arctan b=\arctan\left(\frac{a\pm b}{1\mp ab}\right)$$
$$\arctan a+\mathrm{arccot} \hspace{2pt}a=\frac{\pi}{2},\hspace{10 pt}\forall a>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factoring the expression $x^6 + 64$ Alright, so apparently I've factored this out wrong...
$x^6 + 64 =$
$x^6 + 2^6$
Then I continued, using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ to get ...
$$(x^2)^3 + 64 =
(x^2)^3 + 4^3 =
(x^2 + 4)(x^4 - 4x^2 + 16)$$
How is this incorrect?
| This is correct but you are probably asked to continue with the identity
$$
x^4-4x^2+16=x^4+8x^2+16-12x^2=(x^2+4)^2-12x^2=a^2-b^2,
$$
for some $a$ and $b$ I will let you discover. The final factorisation of $x^6+64$ over the field $\mathbb R$ is the product of three polynomials $x^2+px+q$ with $p^2\lt4q$.
Recall that, together with the polynomials of degree $1$, these are the only irreducible polynomials over $\mathbb R$, hence every real polynomial is a multiple of the product of some degree $2$ polynomials $x^2+px+q$ with $p^2\lt4q$ and some degree $1$ polynomials $x-c$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Expressing in the form $A \sin(x + c)$
Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$
sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$
Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)
Thanks in advance =]
| Using the appropriate formula for $\sin$ you have $A \sin(x+c) = A \sin x \cos c + A \cos x \sin c$. You need to determine $A,c$ so the formula holds true for a), b).
Equating $A \sin x \cos c + A \cos x \sin c = \sin x + \sqrt{3} \cos x$ gives $A \cos c = 1$, $A \sin c = \sqrt{3}$. This gives $\tan c = \frac{A \sin c}{A \cos c} = \sqrt{3}$. If $\tan c = \sqrt{3}$, then $\sin c = \frac{\sqrt{3}}{2}$ and $\cos c = \frac{1}{2}$. This gives $A \frac{1}{2} = 1$, so $A = 2$. You can check that $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}, \cos \frac{\pi}{3} = \frac{1}{2}$, from which it follows that $\sin x + \sqrt{3} \cos x = 2 \sin ( x + \frac{\pi}{3})$.
Similarly, $A \sin x \cos c + A \cos x \sin c = \sin x - \cos x$ gives $A \cos c = 1$, $A \sin c = -1$. This gives $\tan c = \frac{A \sin c}{A \cos c} = -1$, which in turn gives $\sin c = -\frac{1}{\sqrt{2}}$, $\cos c = \frac{1}{\sqrt{2}}$. Then $A \cos c = A \frac{1}{\sqrt{2}} = 1$ gives $A = \sqrt{2}$. You can check that $\sin (-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}, \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, from which it follows that $\sin x - \cos x = \sqrt{2} \sin ( x - \frac{\pi}{4})$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $ 1 + 2 + 4 + 8 + \dots = -1$
Possible Duplicate:
Infinity = -1 paradox
I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said:
Do I have to?
Okay so, Let $x = 1+2+4+8+\dots$
$2x-x=x$
$2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$
Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$.
Therefore, $1+2+4+8+\dots = -1$.
I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong?
| What is wrong is that ``$1+2+4+8+\cdots$'' is not a number, and so you cannot treat it like one.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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} |
How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots
=
\ln\left(3\right)
$$
| Let $s_N=\sum_{n=1}^N \frac1n$ be the $n$th partial sum of the harmonic series.
Then
$$\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}-\frac2{3n})=\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}+\frac1{3n})-\sum_{n=1}^N\frac1n=s_{3N}-s_N.$$
It is well-known that $s_N=\ln N +\gamma+O\left(\frac1N\right)$, hence
$$\sum_{n=1}^N(\frac1{3n-1}+\frac1{3n-2}-\frac2{3n})=\ln(3N)-\ln N+O\left(\frac1N\right)=\ln3+O\left(\frac1N\right).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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By using properties of determinants show that determinant is equal to $(1+a^2+b^2)^3$ $$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\
2ab&1-a^2+b^2&2a\\
2b&-2a&1-a^2-b^2\end{vmatrix}=(1+a^2+b^2)^3$$
I have been trying to solve the above determinant. But unfortunately my answer is always coming as:
$$1+3a^2+3a^4+a^6+3a^2b^4+3b^2+4a^2b^2+a^4b^2+b^6+3b^4$$
Please help me to solve this problem.
| $$\begin{vmatrix}1+a^2-b^2&2ab&-2b\\
2ab&1-a^2+b^2&2a\\
2b&-2a&1-a^2-b^2\end{vmatrix}$$
$$=\begin{vmatrix}1+a^2-b^2&2ab&-2b\\
2ab-a(2b)&1-a^2+b^2-a(-2a)&2a-a(1-a^2-b^2)\\
2b&-2a&1-a^2-b^2\end{vmatrix}$$ (applying $R'_2=R_2-aR_3$)
$$=\begin{vmatrix}1+a^2-b^2&2ab&-2b\\
0&1+a^2+b^2&a(1+a^2+b^2)\\
2b&-2a&1-a^2-b^2\end{vmatrix}$$
$$=(1+a^2+b^2)\begin{vmatrix}1+a^2-b^2&2ab&-2b\\
0&1&a\\
2b&-2a&1-a^2-b^2\end{vmatrix}$$
$$=(1+a^2+b^2)\begin{vmatrix}1+a^2-b^2+b(2b)&2ab+b(-2a)&-2b+b(1-a^2+b^2)\\
0&1&a\\
2b&-2a&1-a^2-b^2\end{vmatrix}$$ (applying $R'_1=R_1+bR_3$)
$$=(1+a^2+b^2)\begin{vmatrix}1+a^2+b^2&0&-b(1+a^2+b^2)\\
0&1&a\\
2b&-2a&1-a^2-b^2\end{vmatrix}$$
$$=(1+a^2+b^2)^2\begin{vmatrix}1&0&-b\\
0&1&a\\
2b&-2a&1-a^2-b^2\end{vmatrix}$$
$=(1+a^2+b^2)^2$ $\begin{vmatrix}1&0&-b\\
0&1&a\\2b+2a(0)-2b(1)&-2a+2a(1)-2b(0)&1-a^2-b^2+2a(a)-2b(-b)\end{vmatrix}$
(applying $R'_3=R_3+2aR_2-2bR_1$)
$=(1+a^2+b^2)^2$ $\begin{vmatrix}1&0&-b\\0&1&a\\0&0&1+a^2+b^2\end{vmatrix}$
$=(1+a^2+b^2)^3$
| {
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"timestamp": "2023-03-29T00:00:00",
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the least possible value for :$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $ If we know that for every $a,b,c>0$ ,how we can find the least possible value for :
$$ \lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor $$
| I would come up with one line proof starting from the trivial fact that $\left\lfloor x \right\rfloor>x-1$. Then
$$\lfloor \frac{a+b}{c}\rfloor +\lfloor \frac{b+c}{a} \rfloor+\lfloor \frac{c+a}{b} \rfloor>\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}-3\ge3.$$
Since the left side is an integer, then the smallest value is $4$.
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the differential equation using Taylor-series expansion Solve the differential equation using Taylor-series expansion:
$$
\frac{dy}{dx} = x + y + xy \\
y (0) = 1
$$
to get value of $y$ at
$x = 0.1$ and $x = 0.5$. Use terms through $x^5$.
| Suppose that $y$ has the Taylor series expansion about $x=0$ given by
$$y=a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots.$$
Because $y(0)=1$, we have $a_0=1$.
Differentiate. We get
$$\frac{dy}{dx}=a_1+2a_2x+3a_3x^2+4a_4x^3+5a_5x^4+\cdots.\tag{$1$}$$
Also,
$$\begin{align}x+y+xy&=x+(1+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots)\\&+(x+a_1x^2+a_2x^3+a_3x^4+a_4x^5+\cdots).\end{align}$$
In the expression above, gather like powers of $x$ together. We get
$$x+y+xy=1+(2+a_1)x+(a_1+a_2)x^2+(a_2+a_3)x^3+(a_4+a_5)x^4+\cdots.\tag{$2$}$$
The expansions $(1)$ and $(2)$ must be identical. It follows that they have the same constant term, that is, that $a_1=1$.
The coefficients of $x$ in $(1)$ and $(2)$ must match. It follows that $2a_2=2+a_1=3$, and therefore $a_2=\frac{3}{2}$.
The coefficients of $x^2$ must match. It follows that $3a_3=a_1+a_2=\frac{5}{2}$, and therefore $a_3=\frac{5}{6}$.
Continue, finding $a_4$ and $a_5$. You have not been asked to find coefficients beyond $a_5$.
For the numerical calculations, just substitute the given values of $x$ in the expression $1+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5$, using the values of the $a_i$ that we have found.
| {
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Inequality. $2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x)$ How do I prove that :
$$2(x^2+y^2+z^2)^2 \geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x).$$
It seems to be easy , but I have no idea:)
thanks :)
| I assume that $x,y,z>0$, even if not explicit from your post. Tell me if I'm wrong.
Expand the $\operatorname{LHS}$. Then we want to prove that
$$2x^4+2y^4+2z^4+4x^2y^2+4y^2z^2+4z^2x^2\geq 3(x^2y^2+y^2z^2+z^2x^2)+3(x^3y+y^3z+z^3x).$$
It is sufficient to estabilish
$$2x^4+2y^4+2z^4+x^2y^2+y^2z^2+z^2x^2\geq 3(x^3y+y^3z+z^3x).$$
Since $$\begin{split}x^4+x^2y^2&\geq 2x^3y,\\y^4+y^2z^2&\geq 2y^3z,\\z^4+z^2x^2&\geq 2z^3x,\end{split}$$ by the arithmetic-geometric inequality, we are reduced to prove that
$$x^4+y^4+z^4\geq x^3y+y^3z+z^3x.$$
This follows from Muirhead inequality (bunching), however we can give a direct proof again by AM-GM
$$\begin{split}&\frac{x^4}{4}+\frac{x^4}{4}+\frac{x^4}{4}+\frac{y^4}{4}\geq x^3y\\&\frac{y^4}{4}+\frac{y^4}{4}+\frac{y^4}{4}+\frac{z^4}{4}\geq y^3z\\\\&\frac{z^4}{4}+\frac{z^4}{4}+\frac{z^4}{4}+\frac{x^4}{4}\geq z^3x.\end{split}$$
Sum up to conclude.
| {
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$\forall n \in \mathbb{N^*}$, prove that $|{\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}| < \frac{1}{2n^2}$ Let $$x_n = \frac{\sqrt{n^2+2}}{2n},n \in \mathbb{N^*} \ldots$$
Prove that $$|{x_n - \frac{1}{2}}| < \frac{1}{2n^2}$$
Indication: Use the relationship: $\sqrt{1+\gamma} < 1 + \frac{\gamma}{2}$, $ \forall \gamma \in \mathbb{R^{*}_{+}}$
I'd appreciate it very much if any answers could indicate a heuristic or general mindset one should have when proving such a proposition.
| First of all we note that $\left|{\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}\right| = {\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}$ because ${\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}}>0$. In fact $\frac{\sqrt{n^2+2}}{2n}>\frac{1}{2} \Rightarrow \frac{n^2+2}{n^2}>1 \Rightarrow 1+\frac{2}{n^2}>1$.
Now we note that
$\frac{\sqrt{n^2+2}}{2n} - \frac{1}{2}=\frac{\sqrt{\frac{2}{n^2}+1}}{2} - \frac{1}{2}$, and from indication $\sqrt{1+\gamma} < 1 + \frac{\gamma}{2}$, we have:
$\frac{\sqrt{\frac{2}{n^2}+1}}{2} - \frac{1}{2}<\frac{\frac{1}{n^2}+1}{2} - \frac{1}{2}=\frac{1}{2n^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do we solve the equation? How do we solve the following equation in the set of real numbers?
$$(x+1)\cdot \sqrt{x+2} + (x+6)\cdot \sqrt{x+7}=(x+3)\cdot (x+4).$$
I wrote the given equation has the form
\begin{equation*}
(x+1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x+7} - 3) = (x-2)(x+4)
\end{equation*}
This equation is equivalent to
\begin{equation*}
(x-2)\left(\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4\right) = 0.
\end{equation*}
But I can not prove that the equation
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 = 0
\end{equation*}
has no solution. Detail
\begin{equation*}
\dfrac{x+1}{\sqrt{x+2}+2} + \dfrac{x+6}{\sqrt{x+7}+3}-x-4 <0, \forall x \geqslant -2.
\end{equation*}
| try squaring everything and expanding then reducing ...
| {
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How do I integrate $\sqrt{x^2+81}$ using trig substitution? How do I integrate $\sqrt{x^2+81}$ using trig substitution? Please be as specific as possible, thank you!
| Jonathan's answer is good, but he left a lot of the work to you when evaluating
$$81 \int \sec^3 t \ dt$$
So, I'll expand on that since it is frequently proved in elementary calculus courses and is a key part of this problem.
Preword: when you see integrals of the form $$\sqrt{a^2+x^2}$$
the substitution $x = a \tan \theta$ usually cleans up the integral quite nicely and makes it easier to work with. Note that the new integral may require another common integration method (such as integration by parts in this example.)
Similarly, if you see integrals of the form
$$\sqrt{a^2-x^2}$$
try the sub. $x = a \sin \theta$. For integrals of the form $$\sqrt{x^2-a^2}$$ try the sub. $x = a \sec \theta$.
Our original integral is
$$\sqrt{x^2+81}$$
Since it fits our $\sqrt{x^2+a^2}$ format above, I'll sub. $x = 9 \tan \theta$. Then, $dx = 9 \sec^2 \theta$. We need to find what $\sqrt{x^2+81}$ is now.
$$x^2+81 = 81 \tan^2 \theta + 81 = 81(\tan^2 \theta +1) = 81\sec^2 \theta \implies \sqrt{x^2+81} = 9 \sec \theta$$
Now we can rewrite our integral as $$\int 9\sec \theta \cdot 9\sec^2 \theta \ d\theta$$
which is the same as $$81\int\sec^3 \theta \ d\theta$$
For ease later on, I will now define our integral
$$I := \int\sec^3 \theta \ d\theta \tag{1}$$
Note that I DID drop the $81$ here on purpose so I won't have to carry it down for further calculations. I can simply multiply through by $81$ at the end.
Here comes the fun integral! I'll use integration by parts for this guy and it will turn out to not be so bad after all. Recall the integration by parts formula:
$$\int u \ dv = uv - \int v \ du \tag{2}$$
We want to pick a $dv$ so that, upon integrating, we get a simple $v$. Thus, it's wise to choose a $dv$ that is an exact differential (i.e. $dv = \sec^2 \theta \implies v = \tan \theta$ is a good choice).
I will let $u = \sec \theta, dv = \sec^2 \theta \implies du = \sec\theta\tan\theta \ d\theta, v = \tan \theta$. Plugging this into the integration by parts formula, we have
$$I = \sec\theta\tan\theta - \int \tan\theta \cdot \sec\theta\tan\theta \ d\theta \tag{3}$$
$$I = \sec\theta\tan\theta - \int \sec\theta \ \tan^2\theta \ d\theta \tag{4}$$
Since $1 + \tan^2 x = \sec^2 x, \tan^2 x = \sec^2 x - 1$ which allows us to rewrite (4):
$$I = \sec\theta\tan\theta - \int \sec\theta \cdot (\sec^2 \theta - 1) \ d\theta \tag{5}$$
Expand $\sec\theta \cdot (\sec^2 \theta - 1)$ to get $\sec^3 \theta - \sec \theta$.
$$I = \sec\theta\tan\theta - \left( \int \sec^3 \theta \ d\theta - \int \sec \theta \ d\theta \right) \tag{6}$$
Distribute the negative sign through the integrals to get:
$$I = \sec\theta\tan\theta - \int \sec^3 \theta \ d\theta + \int \sec \theta \ d\theta \tag{7}$$
From here, recall that $I = \int \sec^3 \theta \ d\theta$. So, we have:
$$I = \sec\theta\tan\theta - I + \int \sec \theta \ d\theta \tag{8}$$
$$2I = \sec\theta\tan\theta + \int \sec \theta \ d\theta \tag{9}$$
Now, I will go ahead and evaluate $\int \sec \theta \ d\theta$ since in the end, we would like an expression for $I$ that does not involve any other integrals.
$$\int \sec \theta \ d\theta = \ln |\sec\theta + \tan \theta| \tag{10}$$
$$2I = \sec\theta\tan\theta + \ln |\sec\theta + \tan \theta| \tag{11}$$
$$\implies I = \frac{1}{2} \sec\theta\tan\theta + \frac{1}{2} \ln |\sec\theta + \tan \theta| \tag{12}$$
Since I defined $I$ as the $\int \sec^3 \theta \ d \theta$, and we really wanted $81\int\sec^3 \theta \ d\theta$, we need to multiply line (12) through by $81$.
$$81\int\sec^3 \theta \ d\theta = \frac{81}{2} \sec\theta\tan\theta + \frac{81}{2} \ln |\sec\theta + \tan \theta| \tag{13}$$
Are we done yet? Almost! We need to get $\theta$ in terms of $x$ now. Since we let $x = 9 \tan \theta$, then $\frac{x}{9} = \tan \theta$. Draw a right triangle with the sides opposite and adjacent to the angle $\theta$ being $x$ and $9$ respectively. This means that the hypotenuse is equal to $\sqrt{x^2+81}$ by the Pythagorean theorem. Since we want $\sec \theta$, we need to figure out $\cos \theta$ because $\sec \theta = \frac{1}{\cos \theta}$.
$$\cos \theta = \frac{9}{\sqrt{x^2+81}} \implies \sec \theta = \frac{\sqrt{x^2+81}}{9} \tag{14}$$
Putting this all together with line (13), we have that
$$81\int\sec^3 \theta \ d\theta = \frac{81}{2} \frac{\sqrt{x^2+81}}{9} \cdot \frac{x}{9} + \frac{81}{2} \ln |\frac{\sqrt{x^2+81}}{9} + \frac{x}{9}| \tag{15}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Ideas on the ways to integrate $\int \tan^2( x)\sec^3( x) dx$ I would proceed by thus , let $y = [\sec (x)]^2 $
then
$$dy = 2 \cdot \sec(x) \cdot \sec(x) \cdot \tan(x) \cdot dx = 2 \cdot ( \sec (x))^2 \cdot \tan(x) \cdot dx $$
so,
$$
2 \tan^2(x) \sec^2 (x) dx = \sec(x) \cdot \tan(x) \cdot dy = y(y-1)^\frac{1}{2} \cdot dy
$$
since $$\sec(x) = y^{\frac{1}{2}}$$ and by considering
positive square roots only $\tan y = ( \sec^2(x) - 1)^{1/2} = (y - 1)^{1/2}$. Thus the substitution $y = \sec^2 x$ yields
$$
2 \int \tan^2 (x) \sec^3(x) dx = \int (y(y - 1) )^{1/2} dy
$$
and this later form can be reduced to the standard form $\int(z^2 - a^2)^{1/2} dz$ since
$$
y(y-1)=(y-(1/2))^2 - (1/2)^2 .
$$
What are the other ways to integrate this expression, except for the substitution
$\tan^2(x)^2=\sec (x)^2 -1$ which gives
$$
\sec^5(x) - \sec^3 (x)
$$
in the integrand which I quite don't like.
| The answer for $\int \tan^2 (x)\sec^3 (x) dx$:
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a series $n(1 + n + n^2 + n^3 + n^4 +.......n^{n-1})$ I'm trying to sum the following series?
$n(1 + n + n^2 + n^3 + n^4 +.......n^{n-1})$
Do you have any ideas?
| Let $ n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n = Sum $
Then,
$ 1 + n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n = Sum + 1$
$ n \times (1 + n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n ) = n \times (Sum + 1)$
$ n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n + n^{n+1} = n\times Sum + n$
$ (n + n^2 + n^3 + n^4 + \cdot\cdot\cdot + n^n) + n^{n+1} = n\times Sum + n$
$ (Sum)+ n^{n+1} = n\times Sum + n$
$ n^{n+1} = (n-1) \times Sum + n$
$ n^{n+1} -n = (n-1) \times Sum$
$ \frac {n^{n+1} -n}{n-1} = Sum$
Hence,
$ Sum = \sum_{i = 1}^{n} n^i = \frac {n^{n+1} -n}{n-1} $
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Describe all solutions of the linear system system Describe all solutions of the system (the last column is the augmented column)
$-x_1 +2x_2 + x_3 + 4x_4 = 0$
$2x_1 + x_2 -x_3 +x_4 = 1$
$\pmatrix{-1&2&1&4&0 \\
2&1&-1&1&1
}
\sim \pmatrix{1&-2&-1&-4&0 \\
0&5&1&9&1}
\sim \pmatrix{1&0 & -\dfrac{3}{5}& -\dfrac{2}{5}& \dfrac{2}{5} \\
0&5&1&9&1}
$
Solving we get the equations:
$x_1 = \dfrac{3}{5}x_3 + \dfrac{2}{5}x_4 + \dfrac{2}{5} \iff x_1 = 3x_3 + 2x_4 + 2$ (Can I scale like this?)
$5x_2 = -x_3 -9x_4 +1 \iff x_2 = x_3 +9x_4 - 1$ (can I scale like this?)
Now we have:
$\langle x_1, x_2, x_3, x_4\rangle = \langle 3x_3 + 2x_4 +2, -x_3 -9x_4 + 1, x_3, x_4\rangle$
$= x_3\langle 3, -1, 1, 0 \rangle + x_4\langle 2, -9, 0, 1\rangle + \langle 2,1,0,0\rangle
$
Where did I go wrong?
| You need to be careful scaling.
You have the equations:
$$-x_1+\frac{3}{5}x_3 + \frac{2}{5}x_4 = -\frac{2}{5}, \ \ \ 5 x_2 + x_3 + 9x_4 = 1$$
which can be written as
$$x_1 = \frac{3}{5}x_3 + \frac{2}{5}x_4 + \frac{2}{5}, \ \ \ x_2 = -\frac{1}{5} x_3 - \frac{9}{5} x_4 + \frac{1}{5}$$
Thus all the solutions are:
$$x = \frac{1}{5} \pmatrix{3 \\ -1 \\ 5 \\ 0} x_3 + \frac{1}{5} \pmatrix{2 \\ -9 \\ 0 \\ 5} x_4 + \frac{1}{5} \pmatrix{2 \\ 1 \\ 0 \\ 0}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving an inequality concerning arbitary complex numbers $\def\abs#1{\left|#1\right|}$If $y$ and $z$ are any complex numbers then prove that
\[
2 \abs{y+z}\ge \bigl(\abs y + \abs z\bigr)
\abs{\frac y{\abs y} + \frac z{\abs z}} \]
| $\def\abs#1{\left|#1\right|}$If $y=r(\cos A+i\sin A), z=R(\cos B+i\sin B)$
$|y+z|=\sqrt{r^2+R^2+2Rr\cos(A-B)}$
$$ \abs{\frac y{\abs y} + \frac z{\abs z}}=\sqrt{2+2\cos(A-B)}$$
$2|y+z|$ will be $\ge (|y|+|z|)(\abs{\frac y{\abs y} + \frac z{\abs z}}) $
if $2\sqrt{r^2+R^2+2Rr\cos(A-B)} \ge (r+R)\sqrt{2+2\cos(A-B)}$
if $4(r^2+R^2+2Rr\cos(A-B)) \ge (r+R)^2(2+2\cos(A-B))$
if $2(1-\cos(A-B))(R-r)^2\ge 0$ which is true
| {
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"timestamp": "2023-03-29T00:00:00",
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Coefficent of an equation I am suppose to find the coefficent of x^2 in this equation
after doing the calculation I ended up with this
but that is the wrong answer what is that I am missing?
| You have $f(x)=x^2+bx+c$, and you want $f\big(f(x)\big)$. As you say, this is $$\left(x^2+bx+c\right)^2+b\left(x^2+bx+c\right)+c\;.\tag{1}$$
Now
$$\begin{align*}
\left(x^2+bx+c\right)^2=x^4+2bx^3+\left(b^2+2c\right)x^2+2bcx+c^2\;,
\end{align*}$$
so $(1)$ becomes
$$x^4+2bx^3+\left(b^2+2c+b\right)x^2+\left(2bc+b^2\right)x+c^2+bc+c\;,$$
in which the coefficient of $x^2$ is $b^2+2c+b$. If $b=-23$ and $c=15$, this is
$$(-23)^2+2(15)-23=529+30-23=536\;.$$
I did it this way to follow (roughly) your approach. Note, however, that you don’t actually have to multiply everything out: you just need the $x^2$ term from $\left(x^2+bx+c\right)^2$ and the $x^2$ term from $b\left(x^2+bx+c\right)$. The latter is clearly $bx^2$. The former can only come from the terms $\left(x^2\right)(c)$, $(bx)(bx)$, and $(c)\left(x^2\right)$ in $\left(x^2+bx+c\right)^2$, so the $x^2$ term here is $\left(b^2+2c\right)x^2$. Now just add the coefficients to get $b^2+2c+b$, and do the arithmetic as before.
| {
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"timestamp": "2023-03-29T00:00:00",
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In how many ways can I have $1.50 using exactly 50 Coins? In how many ways can I have $1.50 using exactly 50 Coins? The coins may be pennies (1 cent), nickels (5 cents), dimes (10 cents) or quarters (25 cents).
| Suppose that you use $p$ pennies, $n$ nickels, $d$ dimes, and $q$ quarters; then $p,n,d$, and $q$ must satisfy the system
$$\left\{\begin{align*}
&p+n+d+q=50\\
&p+5n+10d+25q=150\;.
\end{align*}\right.\tag{1}$$
Thus, you want to count the solutions to $(1)$ in non-negative integers.
The problem is small enough that you can easily solve by hand mostly using brute force. First, subtracting the first equation in $(1)$ from the second gives us the condition $$4n+9d+24q=100\;,\tag{2}$$
from which it’s clear that $q\le 4$.
Suppose that $q=4$; then $4n+9d=100-24\cdot4=4$, which clearly has only the solution $n=1,d=0$. At this point we’ve used four quarters, no dimes, and one nickel, for a total of $105$ cents from $5$ coins, and setting $p=45$ clearly gives the unique solution with $q=4$.
Now suppose that $q=3$, so that $(2)$ reduces to $4n+9d=28$. Clearly $d\le 3$. But both $4n$ and $28$ are divisible by $4$, while $9d$ is not for $0<d\le 3$, so we must have $d=0$ and $n=7$. We’ve now used $10$ coins that total $110$ cents, and it’s clear that setting $p=40$ gives us the unique solution with $q=3$.
Now suppose that $q=2$, so that $(2)$ reduces to $4n+9d=52$. Then $d\le 5$, and we can argue as in the previous case that $9d$ must be a multiple of $4$, so either $d=0$ and $n=13$, or $d=4$ and $n=4$. In the first case we have $15$ coins that total $115$ cents, and we must set $p=35$; in the second we have $10$ coins that total $110$ cents, and we must set $p=40$. These are clearly the only solutions when $q=2$.
If $q=1$, $(2)$ reduces to $4n+9d=76$, so that $d\le 8$. As in the previous two cases we must have $4\mid d$, so $d$ must be $0,4$, or $8$. These lead to the following solutions: $q=1,d=0,n=19,p=30$; $q=1,d=4,n=10,p=35$; and $q=1,d=8,n=1,p=40$.
Finally, if $q=0$, $(2)$ becomes $4n+9d=100$, the feasible values of $d$ are again $0,4$, and $8$, and the resulting solutions are: $q=0,d=0,n=25,p=25$; $q=0,d=4,n=16,p=30$; and $q=0,d=8,n=7,p=35$.
Thus, there are precisely $10$ solutions, and we’ve not just counted them, but also found them.
| {
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"timestamp": "2023-03-29T00:00:00",
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rotation of conic sections In the discriminant test of conic sections(rotations), why we're checking with $B^2-4AC$.
How $B^2-4AC=B'^2-4A'C'$, where $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is changed to $A'x^2+C'y^2+D'x+E'y+F'=0$ using rotations by angle alpha.
| We can find from here, the condition for the general 2nd degree equation to represent a pair of straight lines.
If that condition is not satisfied we can apply the following logic.
A conic is the locus of a point $(x,y)$ which maintains constant ratio(called eccentricity$(e)$) of the distances from a fixed point(called focus $(h,k)$ (say),) and a fixed line(called directrix) $lx+my+n=0$(say).
So, $$e=\frac{(x-h)^2+(y-k)^2}{\frac{lh+mk+n}{\sqrt{l^2+m^2}}}$$
On squaring and rearrangement we get,
$$\{m^2+(1-e^2)l^2\}x^2-2lme^2xy+\{l^2+(1-e)^2m^2\}y^2+(..)x+(..)y+(..)=0$$
Comparing with the original equation,
$A=m^2+(1-e^2)l^2,B=-2lme^2, C=l^2+(1-e)^2m^2$
So, $B^2-4AC=4(e^2-1)(l^2+m^2)^2$
Now,
for ellipse $0\le e<1, B^2-4AC=4(e^2-1)(l^2+m^2)^2<0$
for parabola $e=1, B^2-4AC=4(e^2-1)(l^2+m^2)^2=0$
for hyperbola $e>1, B^2-4AC=4(e^2-1)(l^2+m^2)^2>0$
for circle, $A=C\implies m^2+(1-e^2)l^2=l^2+(1-e)^2m^2\implies e=0$,the $xy$ must be absent in the general equation and the focus co0incides with the centre.
Using Rotation of axes,
$A'=A\cos^2\alpha+B\sin\alpha \cos\alpha +C\sin^2\alpha$
$\implies 2A'=(A+C)+(A-C)\cos2\alpha+B\sin2\alpha$
$B'=B(\cos^2\alpha-\sin^2\alpha)-2(A-C)\sin\alpha \cos\alpha=B\cos2\alpha-(A-C)\sin2\alpha$
$C'=A\sin^2\alpha-B\sin\alpha \cos\alpha +C\cos^2\alpha$
$\implies 2C'=A+C-\{B\sin2\alpha+(A-C)cos2\alpha\}$
$2A'\cdot 2C'-B'^2$
$=(A+C)^2-\{B\sin2\alpha+(A-C)cos2\alpha\}^2-\{B\cos2\alpha-(A-C)\sin2\alpha\}^2$
$=(A+C)^2-B^2\{\sin^22\alpha+\cos^22\alpha\}-2B(A-C)\{2\sin2\alpha\cos2\alpha-2\sin2\alpha\cos2\alpha\}-(A-C)^2\{\sin^22\alpha+\cos^22\alpha\}$
$=(A+C)^2-(A-C)^2-B^2$
$$\implies 4A'C'-B'^2=4AC-B^2$$ (This can be established by eliminating $\sin2\alpha, \cos2\alpha$ from the three equations.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Invertible Matrix Let $A$ be the matrix
$$ A=\left(\begin{array}{cccc}1&0&1&2\\2&3&\beta&4\\4&0&-\beta&-8\\ \beta&0&\beta&\beta \end{array}\right). $$
For what values of $\beta$ is the matrix invertible?
| Compute the determinat of $A$ with laplace expansion expanding at second column it will ease the computations because there are there many zeros
$|A|=(-1)^{2+2}3 \left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ \beta&\beta&\beta \end{array}\right|$
For $\beta=0 \Rightarrow |A|=0$
Now for $\beta \not =0$
$$|A| \not= 0 \Leftrightarrow \left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ 1&1&1 \end{array}\right| \not =0$$
$$\left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ 1&1&1 \end{array}\right|=\left|\begin{array}{ccc}1&1&2\\4&-\beta&-8\\ 0&0&-1 \end{array}\right|=\left|\begin{array}{ccc}1&1&2\\0&-\beta-4&-16\\ 0&0&-1 \end{array}\right|=\beta +4$$
Therefore for $\beta \not = 4,0$ $A$ is invertible.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Integrating $\frac{\log(1+x)}{1+x^2}$
Possible Duplicate:
Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} dx$
I am a bit stuck here in evaluating the following integral:$$\int_{0}^{1}\frac{\log(1+x)}{1+x^2}\,\mathrm dx$$.Your help is appreciated.
| A thought: Notice $$\int\limits_0^1 \frac{\log(1 + x)dx}{1 + x^2} = \int\limits_0^1 \frac{\log(1 + x)dx}{1 + 2x -2x + x^2} = \int\limits_0^1 \frac{\log(1 + x)dx}{(x+1)^2 - 2x } \int\limits_0^1 \frac{\log(1 + x)dx}{(x+1)^2 - 2(x+1) + 2}$$ Now put $u = x + 1$ and so youll have
$$\int\limits_1^2 \frac{\log(u)du}{u^2 - 2u + 2} = \int\limits_1^2 \frac{\log(u)du}{(u-1)^2 + 2} = \int\limits_1^2 \log(u)d[\arctan(u-1)]$$
Now use integration by parts..
| {
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What is the range of the function $f:[-1,1] \rightarrow \mathbb{R}$ defined by $f(x) = 12x^2 + 5x\sqrt{1-x^2} - 10$? I was navigating through math exercises about functions and I got with this question
If $f:[-1,1] \rightarrow \mathbb{R}$ defined by $f(x) = 12x^2 + 5x\sqrt{1-x^2} - 10$.
Give the range of $f(x)$ in the reals.
Any help will be appreciated
| Finding the derivative and solving $f'(x)=0$ may be unpleasant. So it is natural to let $x=\cos t$ where $0\le t\le \pi$. Our function is then $12\cos^2 t+5\cos t\sin t-10$. If we are in a calculus mood, differentiate.
But if we are in a trigonometric mood, we can use the identities $\cos 2t=2\cos^2 t-1$ and $\sin 2t=2\sin t\cos t$ to rewrite our expression as $6\cos 2t+\frac{5}{2}\sin 2t -4$, or equivalently
$$\frac{13}{2}\left(\frac{12}{13}\cos 2t+\frac{5}{13}\sin 2t\right)-4.$$
Let $\varphi$ be an angle whose sine is $\frac{12}{13}$ and whose cosine is $\frac{5}{13}$. Then our function is
$$\frac{13}{2}\sin(2t+\varphi) -4.$$
For $t$ in our interval, $\sin(2t+\varphi)$ has maximum value $1$ and minimum value $-1$. From this we see that the maximum and minimum values of $f(x)$ are $\dfrac{13}{2}-4$ and $-\dfrac{13}{2}-4$.
| {
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Prove the following relation: I must prove the relation $$\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k}=2\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}.$$
I got this far before I got stuck:
$\begin{eqnarray*}
\sum_{k=0}^{n+1}\binom{n+k+1}{k}\frac1{2^k} & = & \sum_{k=0}^{n+1}\left\{\binom{n+k}{k}+\binom{n+k}{k-1}\right\}\frac1{2^k}\\
& = & \sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}+\sum_{k=0}^n\binom{n+k}{k-1}\frac1{2^k}+\binom{2n+1}{n}\frac1{2^k}.
\end{eqnarray*}$
If I can combine the second and third terms and get something same as first term, I am done but I could not do that.
| By way of enrichment here is another algebraic proof using basic
complex variables.
Suppose we are trying to show that
$$\sum_{k=0}^{n+1} {n+1+k\choose k} \frac{1}{2^k}
= 2 \sum_{k=0}^n {n+k\choose k} \frac{1}{2^k}.$$
Introduce the integral representations
$${n+1+k\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+1+k}}{z^{k+1}} \; dz.$$
This gives for the LHS the integral
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+1}}{z}
\sum_{k=0}^{n+1} \frac{(1+z)^k}{z^k\times 2^k}\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+1}}{z}
\frac{1-((1+z)/z/2)^{n+2}}{1-(1+z)/z/2} \; dz
\\ = 2\times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+1}}{z}
\frac{1-((1+z)/z/2)^{n+2}}{2-(1+z)/z} \; dz
\\ = 2\times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{n+1}
\frac{1-((1+z)/z/2)^{n+2}}{2z-(1+z)} \; dz
\\ = 2\times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
(1+z)^{n+1}
\frac{1-((1+z)/z/2)^{n+2}}{z-1} \; dz.$$
Now the first term in the numerator clearly gives a zero contribution
so we get
$$2\times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{n+1}}{1-z}
\frac{(1+z)^{n+2}}{z^{n+2} 2^{n+2}} \; dz
\\ \frac{1}{2^{n+1}} \times \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{2n+3}}{1-z}
\frac{1}{z^{n+2}} \; dz.$$
Extracting the residue we find
$$\frac{1}{2^{n+1}} \sum_{q=0}^{n+1} {2n+3\choose q}
= \frac{1}{2^{n+2}} \sum_{q=0}^{2n+3} {2n+3\choose q}
= \frac{1}{2^{n+2}} 2^{2n+3} = 2^{n+1}.$$
Now the RHS is just twice the LHS with $n+1$ replaced by $n$ so we get
$$2\times 2^n = 2^{n+1}$$
and the two are indeed the same as claimed.
We have not made use of the properties of complex integrals here so
this computation can also be presented using just algebra of
generating functions.
Apparently this method is due to Egorychev although some of it is
probably folklore.
| {
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"url": "https://math.stackexchange.com/questions/228339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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} |
The quadratic form $x^2 + ny^2$ via prime factors Elementary algebra shows that the product of two numbers in the form $x^2 + ny^2$ again has the same form, since if $p = (a^2 + nb^2)$ and $q = (c^2 + nd^2)$,
$$pq = (a^2 + nb^2)(c^2 + nd^2) = (ac \pm nbd)^2 + n(ad \mp bc)^2$$
My question is: Assuming that a number $z$ can be factored into primes of the form $x^2 + ny^2$, does every representation of $z$ in this form arise from repeated applications of this formula to the prime factors?
| Well, no. It is a fair question, though. Even with class number one, we can begin with $1 + 3 = 4,$ although $x^2 + 3 y^2$ does not represent $2.$ The way Dickson would have talked about this is the imprimitive form of the same discriminant, namely $2 x^2 + 2 x y + 2 y^2.$
Staying with one class per genus, we have $1 + 5 = 6,$ although $x^2 + 5 y^2$ does not represent $2,3.$ This is a different phenomenon called Gauss composition. The trick for this one is that $2 x^2 + 2xy+3 y^2$ does represent $2,3,$ and there is a different identity that allows $(2 a^2 +2ab+ 3 b^2)(2 c^2 +2cd+ 3 d^2) = x^2 + 5y^2. $ You should probably be able to find such an identity by hand.
One of the simplest ones where identifying the primes involved becomes a mess is $x^2 + 11 y^2.$ It is a bit of a problem (although solved) to say which primes can be expressed as $x^2 + 11 y^2$ and which by $3 x^2 + 2 x y + 4 y^2,$ among those primes $p$ for which the Jacobi symbol $(-11 | p) = 1.$ But, seeing as the latter form does represent $3,5$ integrally, we are not surprised to see $4 + 11 = 15.$ In this case, you ought to involve the "opposite" form in $(3 a^2 + 2 a b + 4 b^2)(3 c^2 - 2 c d + 4 d^2) = x^2 + 11 y^2.$ The presence of the opposite form is what makes the set (of three "classes") into a group.
Well, that's a start.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Proving with factorials Let x and y be the postive integers.
Show that :
$\displaystyle\frac{(x + y)!}{ (x + y)^{(x + y)}} < \frac{x! y!}{ (x^x + y^y)}$
Are there any identities we can use to easily prove this?
| Edit: The statement in itself is not correct: If $x=y=1$, both sides are $\frac{1}{2}$.
Nevertheless, the statement is correct if $x > 1$ or $y > 1$. I'll first prove it for $x,y$ both $>1$, and then for the case where one of them is 1.
I will first assume that $x > 1$, $y > 1$. First, note that $\binom{x+y}{x} = \frac{(x+y)!}{x!y!}$ , so we can rewrite equivalently prove $\binom{x+y}{x} < \frac{(x+y)^{x+y}}{(x^x+y^y)}$.
Next, by the binomial theorem, $(x+y)^{x+y} = \sum_{i=0}^{x+y} \binom{x+y}{i} x^i y^{x+y-i}$. Clearly, $x^x + y^y \le x^x \cdot y^y$, since $x,y > 0$ and hence, $x^x, y^y \ge 1$. Now, $S := \sum_{i=0}^{x+y} \binom{x+y}{i} x^i y^{x+y-i} \ge \binom{x+y}{0} x^0 y^{x+y-0} + \binom{x+y}{x} x^x y^{x+y-x}$ by considering the $0$-th and $x$-th term of the sum (all terms are $\ge 0$). Simplifying this, we get $S \ge 1 \cdot 1 \cdot y^{x+y} + \binom{x+y}{x} x^x y^y > \binom{x+y}{x} x^x y^y \ge \binom{x+y}{x} (x^x + y^y)$. Here, we used that $y^{x+y} \ge 1$.
Dividing by $x^x+y^y$ gives $$ \frac{S}{x^x+y^y} = \frac{(x+y)^{x+y}}{x^x+y^y} > \binom{x+y}{x}. $$
As mentioned above, this proves the claim.
Now assume that $x = 1$ or $y = 1$. I'll do it quickly for $x=1$, the $y=1$ case is symmetric.
We have to prove $$ \frac{(1+y)y!}{(1+y)(1+y)^y} \frac{(1+y)!}{(1+y)^{1+y}} < \frac{1!y!}{1^1+y^y} = \frac{y!}{y^y+1}.$$ Dividing out common factors, this boils down to $$\frac{1}{(1+y)^y} < \frac{1}{y^y+1},$$ which is clear by the
binomial theorem and $y > 1$.
| {
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"url": "https://math.stackexchange.com/questions/235598",
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"source": "stackexchange",
"question_score": "2",
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How to integrate $\int 1/(x^7 -x) dx$? How should I proceed about this integral?
$$\int {1/(x^7 -x)} dx$$
I've tried integration by parts or substitution but I can't seem to solve it. Can I have some hints on how should I get started?
These are some of the things I've tried:
IBP: $u = \frac {1}{x^6-1}$, $du = \frac {-5x^6}{x^6-1}$, $dv = \frac 1x dx$, $v = \ln|x|$
Tried substitution method, but not successful.
| As in this example it is useful to extract something of the form
$$u=x^a\pm\frac{1}{x^a}$$
$$\int\frac{dx}{x^{7}-x}=\int\frac{1}{x^{4}}\frac{dx}{x^{3}-\frac{1}{x^{3}}}=\int\frac{\left(x^{2}+\frac{1}{x^{4}}\right)dx}{x^{3}-\frac{1}{x^{3}}}-\int\frac{x^{2}dx}{x^{3}-\frac{1}{x^{3}}}=I_1-I_2$$
Now in $I_1$ we may let
$$u=x^{3}-\frac{1}{x^{3}}$$
$$du=3\left(x^{2}+\frac{1}{x^{4}}\right)dx$$
So that
$$I_1=\frac{1}{3}\int\frac{du}{u}=\frac{1}{3}\ln|u|=\frac{1}{6}\ln u^2=\frac{1}{6}\ln\frac{(x^6-1)^2}{x^6}$$
In $I_2$ apparently $v=x^3$
$$I_2=\frac{1}{3}\int\frac{vdv}{v^{2}-1}=\frac{1}{6}\ln\left(v^{2}-1\right)=\frac{1}{6}\ln|x^6-1|$$
Finally,
$$I=I_1-I_2=\frac{1}{6}\ln\frac{x^6-1}{x^6}=\frac{1}{6}\ln\left(1-\frac{1}{x^6}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/237469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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How can I calculate the limit of exponential divided by factorial? I suspect this limit is 0, but how can I prove it?
$$\lim_{n \to +\infty} \frac{2^{n}}{n!}$$
| To prove $\lim_{n\to \infty} \,\frac{n!}{a^n} = +\infty$ it is suffice to prove that for all $k>0$ it is possible $\frac{n!}{a^n} > K$ for almost $n$, that is $n!>K a^n$. This means
$1\cdot 2\cdot\ldots\cdot\frac{n}{2}\cdot\ldots \cdot n > K \cdot a\cdot \stackrel{\stackrel{n}{\smile}}{\ldots}\cdot a,$
that is
$\frac{n/2+1}{a}\cdot\ldots\cdot \frac{n}{a} > K\cdot \frac{a}{1}\cdot\frac{a}{2}\cdot \stackrel{\stackrel{n/2}{\smile}}{\ldots}\cdot \frac{a}{n/2}.
$
Note that
$\frac{n/2+1}{a}\cdot\ldots\cdot \frac{n}{a} >
\left(\frac{n/2+1}{a}\right)^{n/2} > K \cdot a^{n/2} >
K\cdot \frac{a}{1}\cdot\frac{a}{2}\cdot \stackrel{\stackrel{n/2}{\smile}}{\ldots}\cdot \frac{a}{n/2}$
which is true for $n$ verifying $\frac{n/2+1}{a} > 2 a,$ because $K^{2/n}=\sqrt[n]{K^2}\to 1$ and $2a>K^{2/n}a$ for $n$ large enough.
| {
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"source": "stackexchange",
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We define a sequence of rational numbers {$a_n$} by putting $a_1=3$ and $a_{n+1}=4-\frac{2}{a_n}$ for all natural numbers. Put $\alpha = 2 + \sqrt{2}$ We define a sequence of rational numbers {$a_n$} by putting $a_1=3$ and $a_{n+1}=4-\frac{2}{a_n}$ for all $n \in \mathbb N$. Put $\alpha = 2 + \sqrt{2}$
(a) Prove by induction on n, that $3 \le a_n \lt 4$ for all $n \in \mathbb N$.
(b) Show that $3\lt \alpha \lt 4$ and $\alpha=4-\frac{2}{\alpha}$
(c) Show that $a_{n+1}-\alpha=\frac{2(a_n-\alpha)}{\alpha a_n}$ for all $n \in \mathbb N$.
(d) Prove, by induction on n, that |$a_n-\alpha$|$\le\frac{|a_1-\alpha|}{4^{n-1}}$ for all $n \in \mathbb N$.
(e) Deduce that $a_n \rightarrow \alpha$ as $n \rightarrow \infty$
$a_2=3.33333$
$a_3=3.4$
$a_4=3.411764706$
$a_5=3.413793103$
$a_6=3.1414141414$
$\alpha=3.1414213562$
| Certainly $a_1 \le 3 \lt 4$, since $a_1=3$.
Suppose that for the specific integer $k$, we know that $3\le a_k \lt 4$.
We want to show that similar inequalities hold for the "next" term $a_{k+1}$. That is, we want to show that $3\le a_{k+1}\lt 4$.
There are two inequalities to prove. Let us deal with them separately. First we show that (i) $3\le a_{k+1}$ and then that (ii) $a_{k+1}\lt 4$.
(i) Note that $a_{k+1}=4-\frac{2}{a_k}$. We know that $a_k \ge 3$. So $\frac{2}{a_k}\le \frac{2}{3}$. It follows that $4-\frac{2}{a_k} \ge 4-\frac{2}{3}=\frac{10}{3} \ge 3$.
(ii) Because $a_k$ is positive, $4-\frac{2}{a_k}\lt 4$.
So we have shown that our inequalities hold at $n=1$, and that if they hold for some integer $k$, they hold for the next integer $k+1$. By the principle of mathematical induction, the inequality $3\le a_n \lt 4$ holds for every positive integer $n$.
Added: Since the above answer was written, the question has quintupled in length. I hope that the above will at least help with the mechanics of induction. We make a few comments about the added parts.
The added question (b) is undoubtedly something you can handle. Showing that $3\lt \alpha \lt 4$ can be done by observing that $1\lt \sqrt{2}\lt 2$. To show that $\alpha=4-\frac{2}{\alpha}$, note that $\frac{2}{\alpha}=\frac{2}{2+\sqrt{2}}$. Multiply top and bottom by $2-\sqrt{2}$, and everything will collapse. Alternately, we want to show that $\alpha^2=4\alpha -2$. Solve the quadratic equation $x^2-4x+2=0$ using the Quadratic Formula. You will find that the roots are $2\pm\sqrt{2}$.
For (c), the natural thing is to take the left-hand side $a_{n+1}-\alpha$, and replace $a_{n+1}$ by its value in terms of $a_n$. Also, we use the result of (b) as a hint, and replace $\alpha$ by $4-\frac{2}{\alpha}$. We get
$$a_{n+1}-\alpha=\left(4-\frac{2}{a_n}\right)-\left(4-\frac{2}{\alpha}\right).$$
There is some nice cancellation: The expression on the right simplifies to $-\frac{2}{a_n}+\frac{2}{\alpha}$. Bring to a common denominator and we get
$$a_{n+1}-\alpha=\frac{2(a_n-\alpha)}{a_n \alpha}.$$
Now you be able to show (d). For the above formula gives us information about how close $a_{n+1}$ is to $\alpha$ in terms of how close $a_n$ is to $\alpha$. Use the fact that since $a_n \gt 3$ and $\alpha\gt 3$, we have $\frac{2}{a_n \alpha}\lt \frac{2}{9}\lt \frac{1}{4}$.
| {
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$2n+1$ and $n^2+1$ are always coprime or their gcd is $5$ Using a spreadsheet, it can be inferred that when $n≡2[5]$, then $\gcd(n^2+1,2n+1)=5$, else $\gcd(n^2+1,2n+1)=1$.
Indeed, when $n≡2[5]$, $n^2+1$ and $2n+1$ can easily be shown to be multiples of $5$, so their gcd is at least $5$. But then, I can't see how to complete the proof.
| Clearly when $n=0$ the polynomials are relatively prime (gcd = 1).
notice that $2n+1$ is always odd, while $n^2 + 1$ is always even. Thus the gcd can never contain $2$ as a factor.
Let $d | n^2 + 1$ and $d | 2n + 1$, then [because $a|b$ implies $a|kb$] we have
*
*$d|2n^2 + 2$
*$d|2n^2 + n$
and [because $a|b$ and $a|c$ imply $a|c-b$] this gives us
*
*$d|n-2$
we can use this (since it's degree 1) to cancel out the $n$ in our assumption $d | 2n + 1$ to get
*
*$d|n + 3$
and again to get
*
*$d|5$
Since $5$ is prime, we've found that the GCD of the polynomials for some $n$ must be either $1$ or $5$.
We would now like to find out exactly when the GCD is 5 (i.e. when $5|2n+1$ and $5|n^2+1$ and when it's one. For this we can use the equivalence [$a|b$ iff $b \equiv 0 \pmod a$]. So we only need to check 5 values.
n | 2n+1 | n^2+1
----------------
0 | 1 | 1
1 | 3 | 2
2 | 0 | 0 !
3 | 2 | 0
4 | 4 | 2
therefore we have proved the result:
$$\gcd(2n+1,n^2+1) = \begin{cases}5 & \text{when } n \equiv 2 \mod 5 \\ 1 & \text{otherwise}\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why ${ \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent , but ${ \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent? I don't understand why ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent, but ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent and its limit is equal to ${ \displaystyle\frac{\pi^2}{6} }$. In both cases, ${n^{th}}$ term tends to zero, so what makes these series different?
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + }$ ...
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} =1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + }$ ...
| The following example may be easier to grasp. Consider the series
$$1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\cdots. $$
So we have $2$ copies of $\frac{1}{2}$, $4$ copies of $\frac{1}{4}$, $8$ copies of $\frac{1}{8}$, $16$ copies of $\frac{1}{16}$, and so on like that forever.
The $2^k$ copies of $\frac{1}{2^k}$ add up to $1$, so partial sums get arbitrarily large, and therefore our series diverges.
Now look at the series of the squares of the above numbers. We get
$$1+\frac{1}{4}+\frac{1}{4}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\frac{1}{64}+\cdots $$
The first entry is $1$. The next $2$ entries add up to $\frac{1}{2}$. The next $4$ entries add up to $\frac{1}{4}$. The next $8$ entries add up to $\frac{1}{8}$. And so on forever. So the full sum is
$$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$$ This geometric series has sum $2$.
| {
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Prove that if $ |a_n-a_{n-1}| < \frac{1}{2^{n+1} }$ and $a_0=\frac12$, then $\{a_n\}$ converges to $0I try to solve this question but I don't know how.
given $ a_0 = \frac12 $ and for each $n\geq 1$:
$$ |a_n-a_{n-1}| < \frac{1}{2^{n+1}} $$
show that $\{a_n\}$ converges and the limit is $a$ such that $0<a<1$
Update (Edited):
I showed by cauchy that $ |a_m-a_n| < |a_m-a_{m-1}+a_{m+1}-...+a_{n+1}-a_n| < \frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}}$
by the sum of Geometric series, $q=2, a_1=\frac{1}{2^{m+1}}$ then $s_n=\frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}]$, so $$\frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}} = \frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}] = \frac{1}{2^{n+1}}-\frac{1}{2^{m+1}}\leq\frac{1}{2^{n+1}}$$
now, it converges!
Can someone help me please to show that the limit is $a$ with $0<a<1$?
Thank you!
| Hint: $\vert a_n - a_0 \vert = \vert a_n - a_{n-1} + a_{n-1} - \dots - a_1 + a_1 - a_0 \vert \leq \sum_{k=1}^n \vert a_k - a_{k-1} \vert$
| {
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"source": "stackexchange",
"question_score": "2",
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How to convert this into a single closed form? I have a loop that iterates a variable $i$ from 0 to some boundary $b$. Let $p=4i+4$. Each iteration of the loop I add $(p^3+5p)/3+2$ to a sum variable $s$ which is initially set to 0.
Question: Instead of this loop, how can I convert this all into a closed form expression that gives me the same $s$ based on the value of $b$?
| In the end of the loop, if I understand correctly, you have:
$$\begin{align*}s&=\sum_{i=0}^b \left(\frac{(4i+4)^3+5(4i+4)}{3}+2\right)=\sum_{i=0}^b \left(4\frac{16i^3+48i^2+53i+21}{3}+2\right)\\&=\frac{64}{3}\sum_{i=0}^b i^3+64\sum_{i=0}^b i^2+\frac{212}{3}\sum_{i=0}^b i+30\sum_{i=0}^b 1
\end{align*}$$
Using the formulas (which you can prove by induction, if you want):
$$\sum_{i=0}^b 1=b+1, \hspace{10pt} \sum_{i=0}^b i=\frac{b(b+1)}{2}, \hspace{10pt} \sum_{i=0}^b i^2=\frac{b(b+1)(2b+1)}{6}, \hspace{10pt} \sum_{i=0}^b i^3=\frac{b^2(b+1)^2}{4}$$
You can get:
$$\begin{align*}s&=\frac{64}{3}\frac{b^2(b+1)^2}{4}+64\frac{b(b+1)(2b+1)}{6}+\frac{212}{3}\frac{b(b+1)}{2}+30(b+1)\\&=\frac{16}3 b^4+32 b^3+\frac{218}3 b^2+76 b+30=\frac23(b+1)(8 b^3+40 b^2+69 b+45)\end{align*}$$
| {
"language": "en",
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Numerical linear algebra (pseudoinverse of a matrix) Let $A$ be the matrix:
$$\left(\begin{matrix}
\alpha I_{n} \\
\beta I_{n}
\end{matrix}\right)$$
where $\alpha,\beta\in\Bbb C$ are not both zero. Derive (a) the (reduced) QR factorization of $A$ and (b) the pseudoinverse of $A$.
Any help for the second question about pseudoinverse ?
Thanks in advance
NEW
I know that if 1) rank(A)=n then $A^{+} = (A^{T} A)^{-1} A^{T}$
and if 2) rank(A)=n=m then
$A^{+} = A^{-1}$.
I use the 1) and I found :
$A^{+} = (a^{2} I_{n} + b^{2} I_{n})^{-1} $$\left(\begin{matrix}
\alpha I_{n} \
\beta I_{n}
\end{matrix}\right)$
note the second brackets is a matrix (1x2).
How could I solve this ?
Any help
| Example $\mathbf{Q} \mathbf{R}$
$$
\begin{align}
\mathbf{A} &= \mathbf{Q} \, \mathbf{R} \\
% A
\left[
\begin{array}{ccc}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a \\
b & 0 & 0 \\
0 & b & 0 \\
0 & 0 & b \\
\end{array}
\right]
%
&=
% Q
\left( a^{2}+b^{2} \right)^{-\frac{1}{2}}
\left[
\begin{array}{ccc}
a & 0 & 0 \\
0 & a & 0 \\
0 & 0 & a \\
b & 0 & 0 \\
0 & b & 0 \\
0 & 0 & b \\
\end{array}
\right]
% R
\left( a^{2}+b^{2} \right)^{\frac{1}{2}}
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right]
%
\end{align}
$$
Example SVD
The singular value decomposition will use both $\color{blue}{range}$ and $\color{red}{null}$ spaces.
$$
\begin{align}
\mathbf{A} &=
\left[ \begin{array}{c|c}
\color{blue}{\mathbf{U}_{\mathcal{R}}} &
\color{red} {\mathbf{U}_{\mathcal{N}}}
\end{array} \right]
\, \Sigma \, \color{blue}{\mathbf{V}^{*}} \\
%
&=
% U
\left( a^{2}+b^{2} \right)^{-\frac{1}{2}}
\left[
\begin{array}{ccc|ccc}
\color{blue}{0} & \color{blue}{0} & \color{blue}{a} & \color{red}{0} & \color{red}{0} & \color{red}{-a} \\
\color{blue}{0} & \color{blue}{a} & \color{blue}{0} & \color{red}{0} & \color{red}{-a} & \color{red}{0} \\
\color{blue}{a} & \color{blue}{0} & \color{blue}{0} & \color{red}{-a} & \color{red}{0} & \color{red}{0} \\
\color{blue}{0} & \color{blue}{0} & \color{blue}{b} & \color{red}{0} & \color{red}{0} & \color{red}{b} \\
\color{blue}{0} & \color{blue}{b} & \color{blue}{0} & \color{red}{0} & \color{red}{b} & \color{red}{0} \\
\color{blue}{b} & \color{blue}{0} & \color{blue}{0} & \color{red}{b} & \color{red}{0} & \color{red}{0} \\
\end{array}
\right]
% S
\left( a^{2}+b^{2} \right)^{\frac{1}{2}}
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\\hline
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{array}
\right]
% V
\left[
\begin{array}{ccc}
\color{blue}{0} & \color{blue}{0} & \color{blue}{1} \\
\color{blue}{0} & \color{blue}{1} & \color{blue}{0} \\
\color{blue}{1} & \color{blue}{0} & \color{blue}{0} \\
\end{array}
\right]
%
\end{align}
$$
The pseudoinverse matrix is
$$
\begin{align}
\mathbf{A}^{+} &=
\color{blue}{\mathbf{V}} \,
\Sigma^{+}
\left[ \begin{array}{c}
\color{blue}{\mathbf{U}_{\mathcal{R}}^{*}} \\
\color{red} {\mathbf{U}_{\mathcal{N}}^{*}}
\end{array} \right]
%
=
\left( a^{2}+b^{2} \right)^{\frac{1}{2}}
\left[
\begin{array}{cccccc}
a & 0 & 0 & b & 0 & 0 \\
0 & a & 0 & 0 & b & 0 \\
0 & 0 & a & 0 & 0 & b \\
\end{array}
\right]
%
\end{align}
$$
Derivation $\mathbf{Q}\, \mathbf{R}$
The $\mathbf{Q}\, \mathbf{R}$ is computationally cheaper than the SVD. It provides an orthonormal basis for the column space. The column vectors of the target matrix are already orthogonal; they just need normalization:
$$
\mathbf{Q} = \left( a^{2} + b^{2} \right)^{-\frac{1}{2}} \mathbf{A}.
$$
We can bypass the usual Gram-Schmidt process.
The $\mathbf{R}$ matrix is upper-triangular. The the supradiagonal terms describe the projections of the column vectors of $\mathbf{Q}$ on the column vectors of $\mathbf{A}$. For example
$$
\mathbf{R}_{1,2} = \mathbf{Q}^{*}_{1} \mathbf{A}_{2}
$$
Because the columns or $\mathbf{A}$ were already orthogonal, the supradiagonal terms are $0$. The diagonal terms hold the lengths of the column vectors of $\mathbf{A}$
$$
\mathbf{Q}_{k,k} = \lVert \mathbf{A}_{k} \rVert, \quad k = 1, n.
$$
The target matrix $\mathbf{A}$ has column vectors of uniform length $\sqrt{a^{2} + b^{2}}$.
Derivation SVD
The SVD starts by resolving the eigensystem of the product matrix:
$$
\mathbf{A}^{*} \mathbf{A} =
\left( a^{2} + b^{2} \right)
\left[
\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right]
$$
The eigenvalue spectrum is
$$
\lambda_{k} = a^{2} + b^{2}, \quad k = 1, n
$$
There are no zero eigenvalues. There is no need to order the spectrum. We can harvest the singular values directly
$$
\sigma_{k} = \sqrt{\lambda_{k}\left( \mathbf{A}^{*} \mathbf{A} \right)}
= \sqrt{ a^{2} + b^{2} }, \quad k=1,n
$$
The algebraic multiplicity $(n)$ matches the geometric multiplicity for the eigenvalue. For the eigenvectors we choose the simplest set, the unit vectors in the identity matrix. Using the relationship
$$
\mathbf{A} \, \mathbf{V} = \mathbf{U} \, \Sigma
$$
we find that
$$
\color{blue}{\mathbf{U}_{\mathcal{R}}} = \left( a^{2} + b^{2} \right)^{-\frac{1}{2}} \mathbf{A}
$$
To summarize the SVD
$$
\mathbf{V} = \mathbf{I}_{n}, \quad \Sigma = \left( a^{2} + b^{2} \right)^{\frac{1}{2}}
\left[ \begin{array}{c}
\mathbf{I}_{n} \\
\mathbf{0}
\end{array} \right], \quad
\color{blue}{\mathbf{U}_{\mathcal{R}}} = \left( a^{2} + b^{2} \right)^{-\frac{1}{2}} \mathbf{A}
$$
The psuedoinverse
$$
\mathbf{A} = \mathbf{V} \, \Sigma^{+} \, \color{blue}{\mathbf{U}_{\mathcal{R}}} = \left( a^{2} + b^{2} \right)^{\frac{1}{2}} \mathbf{A}^{*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/245633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do you compute the limit with multiple variables without fail? $$\lim\limits_{(x,y)\rightarrow (0,0)} \dfrac{(x^3-y^3) \sin (2x^2+3y^2)}{x^2+2y^2}$$
I don't know what to do doesn't it give always $0$?
Whether $x=0$ $x=y$ or $y=0$ or $x=y^2$ it always give $0$ since it goes to $(0,0)$
| HINT
Note that $$-2 \leq \dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \leq 2$$
Hence, $$(x^3 - y^3)\dfrac{\sin(2x^2 + 3y^2)}{x^2 + 2y^2} \in \left[-2\vert(x^3-y^3) \vert, 2\vert(x^3-y^3)\vert \right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/251884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to multiply polynominals I can't figure out how to multiply these polynominals $$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$
I tried multiplying like this
$$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)$$
$$3x^4-7x^3+(5x^2)(2x^2)(-6x^2)+(5x)(7x)(-4x)+(8)(9)$$
$$3x^4-7x^3-60x^2-140x+72$$
It says the answer is $$-12x^6-19x^5-14x^4-68x^3+28x^2+69x+72$$ but how did they get it?
| You have to multiply every term by every other term. A good way to make sure you don't miss any is to use a table.
First, combine like terms within each group of parentheses:
$$(5x^2+3x^4-7x^3+5x+8)(2x^2-4x+9-6x^2+7x)\\
=(3x^4-7x^3+5x^2+5x+8)(-4x^2+3x+9)$$
Then form a table and multiply each term by multiplying the coefficients and adding the exponents:
$$
\begin{array}{c|cc}
\text{} & 3x^4 & -7x^3 & 5x^2 & 5x & 8 \\
\hline
-4x^2 & -12x^6 & -28x^5 & -20x^4 & -20x^3 & -32x^2 \\
3x & 9x^5 & -21x^4 & 15x^3 & 15x^2 & 24x \\
9 & 27x^4 & -63x^3 & 45x^2 & 45x & 72 \\
\end{array}
$$
Now take the new polynomial from the table and combine like terms:
$$-12x^6-28x^5-20x^4-20x^3-32x^2+9x^5-21x^4+15x^3\\
+15x^2+24x+27x^4-63x^3+45x^2+45x+72\\$$$$
=-12x^6+(9-28)x^5+(27-21-20)x^4+(-63+15-20)x^3\\+(45+15-32)x^2+(45+24)x+72\\$$$$
=-12x^6-19x^5-14x^4-68x^3+28x^2+69x+72$$
This method will also work with negative and non-integer exponents, as it is not restricted to polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/253522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the sum of an alternating series?
Find whether the series diverges and its sum:
$$\sum_{n = 1}^\infty (-1)^{n+1} \frac{3}{5^n}.$$
I found that the series converges using the Alternating Series test because the absolute value of each $n$ decreases while the value of $n$ increases. Then I took the limit as $x$ approaches infinity of $3/5^x$ which is $0$.
But I am not sure how to go about finding the sum at this point. This is a practice exam, so the solution is more important then the actual answer.
| Notice that$$(-1)^{n+1}\frac{3}{5^n}=-3\frac{(-1)^{n}}{5^{n}}=-3\left(\frac{-1}{5}\right)^{n}$$
Since $\sum_{k=1}^{\infty}ar^{k}=\frac{ar}{1-r}$ (iff $|r|<1$), $$\sum_{n=1}^{\infty}-3\left(\frac{-1}{5}\right)^{n}=\frac{-3\cdot\frac{-1}{5}}{1-\frac{-1}{5}}=\frac{\frac{3}{5}}{\frac{6}{5}}=\frac{1}{2}$$ and the sum converges because $\left|\frac{-1}{5}\right|=\frac{1}{5}<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/253997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Integrating a function over a domain How could you integrate the function $f(x,y) = x^2 + y^2$ over the triangle with vertices $(0,0)$, $(0,1)$ and $(1,0)$?
I define the set $D = \{(x,y)\; |\; 0\leq x\leq 1 \text{ and } 0\leq y\leq x\}$ and then calculate
$$\int_0^1 \int_0^x x^2 + y^2 \; \mathrm{d}y \; \mathrm{d}x = \frac{1}{3},$$
but apparantly the answer is $\frac{1}{6}$.
| I think you have the limits changed and wrong. I think it must be
$$\int_0^1dx\int_0^{1-x}(x^2+y^2)dy=\int_0^1\left.\left(x^2y+\frac{1}{3}y^3\right|_0^{1-x}\right)dx=\int_0^1\left(x^2(1-x)+\frac{1}{3}(1-x)^3\right)dx=$$
$$=\frac{1}{3}-\frac{1}{4}-\frac{1}{12}(0-1)=\frac{2}{12}=\frac{1}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/256514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove that for any triple $a_{i},a_{j},a_{k} $ are three edge lengths of some triangle $ a_{1},a_{2},....,a_{n}(n\geq 3) $ are positive numbers that :
$(a_{1}+a_{2}+....+a_{n})^2 >\frac{3n-1}{3}(a_{1}^2+a_{2}^2+....+a_{n}^2)$
Prove that for any triple $a_{i},a_{j},a_{k} $ are three edge lengths of some triangle, where natural numbers $ i,j,k $ satifying $ 0< i< j< k\leq n $
It seems a Interesting problem , i tried to prove it in case $n=3$ starting an induction on but i failed at that level
please help me. Thank
Maybe can use Induction
| Since the condition is symmetric about $a_1,\dots,a_n$, it suffices to show that $a_1+a_2>a_3$.
When $n=3$, by Cauchy-Schwarz inequality,
$$2\left((a_1+a_2)^2+a_3^2\right)\ge \left((a_1+a_2)+a_3\right)^2 \quad{and}\quad a_1^2+a_2^2\ge \frac{1}{2}(a_1+a_2)^2.$$
Combining the two inequalities above with your condition for $n=3$, we have $(a_1+a_2)^2>a_3^2$, so the conclusion follows.
When $n>3$, note that $\frac{3n-1}{3}=\frac{8}{3}+n-3$ and
$$a_1+\dots+a_n=\sqrt{\frac{8}{3}}\times\sqrt{\frac{3}{8}}(a_1+a_2+a_3)+1\times a_4+\dots+1\times a_n.$$
Then by Cauchy-Schwarz inequality,
$$\frac{3n-1}{3}\left(\frac{3}{8}(a_1+a_2+a_3)^2+a_4^2+\dots+a_n^2\right)\ge(a_1+\dots+a_n)^2.$$
Combining the inequality above with your condition, we can obtain the condition for $n=3$, which completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the upper and lower limits of $xf(x)$, as $x\rightarrow \infty$ Define $$f(x)=\int_{x}^{x+1}\sin(t^2)dt$$
Find the upper and lower limits $xf(x)$, as $x\rightarrow \infty$.
I find the answer as $+1, -1$ since $|\sin(x)| \le 1$. (Of course I calculated that function)
Is that right or did I miss something?
==========================================================
I solved this way.
$2xf(x)=\cos(x^2)-\cos[(x+1)^2]+r(x)$ where $r(x)=\frac{\cos(x+1)^2}{x+1}-2x\int_{x^2}^{(x+1)^2}\frac{cos(u)}{4u^{3/2}}du$
Therefore $xf(x)=\frac{1}{2}{\cos(x^2)-\cos(x+1)^2}+\frac{r(x)}{2}$
Using trigonomeric formula: $2\sin(a)\sin(b)=\cos(a-b)-\cos(a+b)$
Rewrite $xf(x)=±\sin(x^2+x+\frac{1}{2})\sin(x+\frac{1}{2})+\frac{r(x)}{2}$.
As $x\rightarrow \infty, r(x) \rightarrow 0$.
Suppose $x^2=2k\pi$ for integer $k$.
To achieve $±1$, we have to show that $x+\frac{1}{2} \rightarrow 2n\pi+\frac{\pi}{2}$ for some $n$ as $x\rightarrow \infty$.
For each $n$, there exists $k$ such that $\sqrt{2\pi k}+\frac{1}{2} < 2n\pi+\frac{\pi}{2} <\sqrt{2\pi (k+1)} +\frac{1}{2} $
Distance between $\sqrt{2\pi k}+\frac{1}{2}$ and $2n\pi+\frac{\pi}{2}$ is at most $\sqrt{2\pi (k+1)} +\frac{1}{2} -\sqrt{2\pi k}+\frac{1}{2}$.
As $k \rightarrow \infty$ the distance becomes arbitrary small.
Therefore $ x \rightarrow \infty$, $xf(x)=±\sin(2n\pi+\frac{\pi}{2})=±1$.
| Here is a clarified version of your proof. (In fact, it was a part of my original solution, but I modified it by following your idea.)
Step 1. Estimation of $xf(x)$
Let $t = \sqrt{u}$. Then
\begin{align*}
xf(x)
&= x\int_{x}^{x+1} \sin (t^2) \, dt = x \int_{x^2}^{(x+1)^2} \frac{\sin u}{2\sqrt{u}} \, du \\
&= x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du
\end{align*}
Since
\begin{align*}
x \left[ \frac{1-\cos u}{2\sqrt{u}} \right]_{x^2}^{(x+1)^2}
&= \frac{x}{2} \left(\frac{1-\cos\big((x+1)^2\big)}{x+1} - \frac{1 - \cos\big(x^2\big)}{x} \right) \\
&= \frac{1}{2} \left(\cos \left( x^2 \right) - \cos\left((x+1)^2\right) \right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} \\
&= \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)},
\end{align*}
we have
$$xf(x) = \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du. $$
It is easy to observe that
$$ - \frac{1 - \cos\left((x+1)^2\right)}{2(x+1)} + x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du = O\left(\frac{1}{x}\right). $$
Indeed, it follows from the estimation
\begin{align*}
\left| x \int_{x^2}^{(x+1)^2} \frac{1-\cos u}{4u^{3/2}} \, du \right|
&\leq x \int_{x^2}^{(x+1)^2} \frac{\left| 1-\cos u \right| }{4u^{3/2}} \, du \\
&\leq x \int_{x^2}^{(x+1)^2} \frac{1}{2x^3} \, du
\leq \frac{2x+1}{2x^2}
= O\left(\frac{1}{x}\right).
\end{align*}
Step 2. Evaluation of limit superior and limit inferior
The estimation above in particular implies that
$$ \limsup_{x\to\infty} xf(x) = \limsup_{x\to\infty} \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right)$$
and likewise for the liminf. To find the limsup, note from the identity above that
$$\limsup_{x\to\infty} xf(x) \leq 1. $$
We claim that it is indeed 1. To this end, it suffices to find a subsequence $(x_n)$ such that $x_n \to \infty$ and
$$ \sin \left(x_n +\frac{1}{2}\right)\sin\left(x_n^2+x_n+\frac{1}{2}\right) \to 1. $$
Let $x = \sqrt{2\pi k}$. Then we have
$$ \sin \left(x+\frac{1}{2}\right)\sin\left(x^2+x+\frac{1}{2}\right) = \sin^2 \left(\sqrt{2\pi k}+\frac{1}{2}\right) $$
As $\sqrt{2\pi k} \to \infty$ and $\sqrt{2\pi(k+1)} - \sqrt{2\pi k} \to 0$ as $k \to \infty$, we can find a subsequence $(k_n)$ such that
$$ \min \left\{ \left| \sqrt{2\pi k_n} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} \to 0 \quad \text{as} \quad n \to \infty. \tag{1}$$
Thus for $x_n = \sqrt{2\pi k_n}$ we obtain $ x_n f(x_n) \to 1$ as $n \to \infty$ and therefore $\limsup_{x\to\infty} xf(x) = 1$. A slight modification of this argument also proves that $\liminf_{x\to\infty} xf(x) = -1$.
Proof of the claim $(1)$
Let $\epsilon > 0$. Then there exists $N$ such that whenever $k \geq N$ we have $0 < r_k < \epsilon$, where $r_k$ denotes $r_k = \sqrt{2\pi(k+1)} - \sqrt{2\pi k}$ for simplicity. Then for the sequence of open balls
$$ B_k = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi k} + r_k\right) = \left(\sqrt{2\pi k} - r_k, \sqrt{2\pi (k+1)}\right), $$
we have $B_k \cap B_{k+1} \neq \varnothing $ and hence
$$ \bigcup_{k=N}^{\infty} B_k = \left(\sqrt{2\pi N} - r_{N}, \infty \right). $$
Here, we used the fact that $\sqrt{2\pi k} \to \infty$ as $k \to \infty$. Then for any sufficiently large integer $m$ we have
$$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in \left(\sqrt{2\pi N} - r_{N}, \infty \right). $$
Thus we can pick some $k \geq N$ satisfing
$$ \left(m+\frac{1}{2}\right)\pi - \frac{1}{2} \in B_k \subset \left(\sqrt{2\pi N} - \epsilon, \sqrt{2\pi N} + \epsilon \right). $$
This proves the following proposition: For any $\epsilon > 0$ there exists a positive integer $k$ such that
$$ \min \left\{ \left| \sqrt{2\pi k} + \frac{1}{2} - \left( m+\frac{1}{2}\right)\pi \right| : m \in \Bbb{Z} \right\} < \epsilon.$$
Then the claim $(1)$ immediately follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/259313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
A question regarding case notation Consider the parameter $P$ and a property $S$ associated with this parameter represented as $P_S$. For example, consider that the height of a tree (“H”) is my parameter and the property is the tree’s name. Therefore, $H_{T1}$ refers to the height of the tree named $T1$.
Now, consider that $H_{T1}$ depends on 2 inputs, named $A$ and $B$ respectively. So, for the above example we have:
$$
H_{T1}^{A,B}
$$
My first concern is whether the above is a valid notation or should I have written something else like $H_{T1}(A,B)$?
I am then using the cases environment in order to define the different value that $H_{T1}$ takes for a number of differetn $A, B$ inputs:
$$
\begin{eqnarray}
H_{T1}^{A,B} =
\begin{cases}
H_{T1}^{5,8} & \text{if $A=5$, $B=8$}\\
H_{T1}^{9,1} & \text{if $A=5$, $B=1$}\\
H_{T1}^{3,3} & \text{if $A=3$, $B=3$}\\
\end{cases}
\end{eqnarray}
$$
For a particular set of $A,B$ inputs, e.g. for the set $(A=3, B=3)$, $H_{T1}$ depends on $2$ extra inputs, named $C$ and $D$ respectively like that:
$$
\begin{eqnarray}
H_{T1}^{3,3,C,D} =
\begin{cases}
H_{T1}^{3,3,1,1} & \text{if $A=3$, $B=3$, $C=1$, $D=1$}\\
H_{T1}^{3,3,4,1} & \text{if $A=3$, $B=3$, $C=4$, $D=1$}\\
H_{T1}^{3,3,6,2} & \text{if $A=3$, $B=3$, $C=6$, $D=2$}\\
\end{cases}
\end{eqnarray}
$$
Since only for $H_{T1}^{3,3}$ I have the 2 extra inputs, what would be the best way of writing that? I thought of writing something like that:
$$
\begin{eqnarray}
H_{T1}^{A,B} =
\begin{cases}
H_{T1}^{5,8} & \text{if $A=5$, $B=8$}\\
H_{T1}^{9,1} & \text{if $A=5$, $B=1$}\\
H_{T1}^{3,3,C,D} & \text{if $A=3$, $B=3$}\\
\end{cases}
\end{eqnarray}
$$
where
$$
\begin{eqnarray}
H_{T1}^{3,3,C,D} =
\begin{cases}
H_{T1}^{3,3,1,1} & \text{if $A=3$, $B=3$, $C=1$, $D=1$}\\
H_{T1}^{3,3,4,1} & \text{if $A=3$, $B=3$, $C=4$, $D=1$}\\
H_{T1}^{3,3,6,2} & \text{if $A=3$, $B=3$, $C=6$, $D=2$}\\
\end{cases}
\end{eqnarray}
$$
but I am not sure whether it’s correct or not.
Thanks in advance.
| You could use $H_{T1,A,B}$, or $H_{T1;A,B}$. Your idea for your function is almost right, but you've included some superfluous information. Try
$H_{T1,A,B,C,D}=\cases{a, \text{if } A=1,B=1,∀C,D,\\b, \text{if }A=2,B=2, ∀ C,D,\\c, \text{if } A=3,B=3,C=1,D=2,\\d, \text{if } A=3,B=3,C=1,D=3}$
or use your simplifying trick and say
$H_{T1,A,B,C,D}=\cases{a, \text{if } A=1,B=1,\\b, \text{if }A=2,B=2, \\ H_{T1,3,3,C,D}, \text{if } A=3,B=3, \text{where}}$
$H_{T1,3,3,C,D}=\cases{c, \text{if } C=1,D=2,\\d, \text{if } C=1,D=3.}$
However, you can make clearer what is going on by presenting $H_{T1,A,B},A,B\not=3$ as a special case of $H_{T1,A,B,C,D}. Thus,
$H_{T1,A,B,C,D}=\cases{H_{T1,A,B}, \text{if } (A,B)\not=(3,3),\\c, \text{if } A=B=3,C=1,D=2,\\d, \text{if } A=B=3,C=1,D=3.}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/261039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
How to solve system of equations? I want to sove the system of equations $$\begin{cases}
x^3 y-y^4=7,\\
x^2 y+2 xy^2+y^3=9.
\end{cases}
$$
I tried divide these two equations we obtain
$$\dfrac{x^3 - y^3}{(x+y)^2 } = \dfrac{7}{9}$$
From here, I don't know how to solve.
| $$\begin{cases}
x^3 y-y^4=7,\\
x^2 y+2 xy^2+y^3=9.
\end{cases}$$
$x^3y-y^4=y(x^3-y^3)$, while $x^2y+2xy^2+y^3=y(x+y)^3$.
We try the transformation $x+y=z$.
Then $y(x^3-y^3)=y(x-y)(x^2+xy+y^2)=y(z-2y)(z^2-xy)=y(z-2y)(z^2-(z-y)y)=$
$-2y^4+3y^3z-3y^2z^2+yz^3=7$
On the other hand, the second equation gives $yz^2=9$, so $y=\frac{9}{z^2}$. We expand and then notice that it is actually equal to $9(z^3-18)(z^6-9z^3+81)=7z^8$. Then we get $z=3$ as a root.
The problem is that I do not know how do I show that there are no other roots, but probably someone who knows calculus can be able to resolve this? or probably by a weird factoring?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/262449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove among two integers $2^n-1$ and $2^n+1$, at most one is prime. For any $n>2$, among $2^n-1$ and $2^n+1$, at most one is prime.
I have no clue :(
| As $(2^n-1),2^n,(2^n+1)$ are three consecutive integers, exactly one of them is divisible by $3$
Now, $(2,3)=1\implies 3\mid(2^n-1)(2^n+1)$
which can also proved as $(2^n-1)(2^n+1)=4^n-1^n$ which is divisible by $4-1=3$ as $(a-b)\mid(a^m-b^m)$ where $m$ is any natural number.
So, exactly one of them is divisible by $3$, hence is composite as $n>2, 2^n-1>3$ so is $2^n+1$
In fact $3\mid(2^n+1)$ if $n$ is odd as $(a+b)\mid(a^{2m+1}+b^{2m+1})$ (where $m$ is any natural number)
and $3\mid(2^n-1)$ if $n$ is even as we have just proved $3\mid (4^m-1)\implies 3\mid(2^{2m}-1)$
A little generalization: for $(a^n-1)a^n(a^n+1)$ where natural number $n>1$
Now,
(i)
$(a-1)\mid(a^n-1)$ so $(a^n-1)$ will be composite if $a-1>1\implies a>2$
Again, as Marvis has observed, if $n$ is composite $=c\cdot d$(say), $(a^c-1)\mid(a^n-1)$
So, the necessary condition for $a^n-1$ to be prime is $a=2,n$ is prime $=p$(say), so that $a^n-1$ becomes $2^p-1$(Mersenne numbers)
(ii)
$a^n+1$ will be even, hence composite if $a>1$ is odd.
Again, if $n$ is odd $=e(2f+1)$(say) where $e,f$ are natural numbers, $(a^e+1)\mid(a^n+1)$
So, the necessary condition for $a^n+1$ to be prime is $a$ is even, $n$ does not have any odd factor, so that $a^n+1$ becomes $(2m)^{2^r}+1$
If $m=1,$ it becomes $2^{2^r}+1$(Fermat number)
So, $a^n+1,a^n-1$ both can be prime if $a=2, p=2^r\implies r=1,p=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/262812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Compute $\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx-\sum_{k=1}^n \frac{1}{k}\right)$ Compute the limit
$$\lim_{n\to\infty} \left(\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx-\sum_{k=1}^n \frac{1}{k}\right)$$
| Let's suppose that $\ \displaystyle f(n):=\int_0^{\pi} \frac{\sin^2 n x}{\sin x} \ dx\ $ then :
\begin{align}
f(n+1)-f(n)&=\int_0^{\pi} \frac{\sin^2((n+1) x)-\sin^2(n x)}{\sin x} \ dx\\
&=\int_0^{\pi} \frac{\cos(2n x)-\cos(2(n+1) x)}{2\sin x} \ dx\\
&=\int_0^{\pi} \frac{\cos(2n x)(1-\cos(2x))+\sin(2nx)\sin(2x)}{2\sin x}dx\\ &=\int_0^{\pi} \frac{\cos(2n x)2\sin(x)^2)+\sin(2nx)2\sin(x)\cos(x)}{2\sin x} dx\\
&=\int_0^{\pi} \cos(2n x)\sin(x)+\sin(2nx)\cos(x)\;dx\\
&=\int_0^{\pi} \sin((2n+1) x)\;dx\\
&=\frac 2{2n+1} \\
\end{align}
So that your limit (as $\ (n+1)\to +\infty$) is the series :
$$f(0)+\sum_{n=0}^\infty \left(\frac 2{2n+1}-\frac 1{n+1}\right)=2\sum_{n=0}^\infty \left(\frac 1{2n+1}-\frac 1{2n+2}\right)=2\,\log(1+1)=\log(4)$$
(using the expansion of $\;\log(1+x)\,$ at $\,x=1$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/263767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
sum of this series: $\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$ I am trying to calculate the sum of this infinite series after having read the series chapter of my textbook: $$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}$$
my steps:
$$\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\sum_{n=1}^{\infty}\frac{2}{4n^2-1}-\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=..help..=sum$$
I am lacking some important properties, I feel I am coming to the right step and cannot spit that out..
| Note $\frac{1}{4n^2-1}=\frac{1}{(2n+1)(2n-1)}={\frac{1}{2}}\times\frac{(2n+1)-(2n-1)}{(2n+1)(2n-1)}={\frac{1}{2}}\times[\frac{1}{2n-1}-\frac{1}{2n+1}]$ for $n\in\mathbb N.$
Let for $k\in\mathbb N,$ $S_k=\displaystyle\sum_{n=1}^{k}\frac{1}{4n^2-1}$ $\implies S_k={\frac{1}{2}}\displaystyle\sum_{n=1}^{k}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right].$ Thus for $k=1,2,...$
$S_1={\frac{1}{2}}\displaystyle\sum_{n=1}^{1}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}(1-\frac{1}{3})$
$S_2={\frac{1}{2}}\displaystyle\sum_{n=1}^{2}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})]=\frac{1}{2}(1-\frac{1}{5})$
$S_3={\frac{1}{2}}\displaystyle\sum_{n=1}^{3}\left[\frac{1}{2n-1}-\frac{1}{2n+1}\right]=\frac{1}{2}[(1-\frac{1}{3})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{7})]=\frac{1}{2}(1-\frac{1}{7})$
...
$S_k=\frac{1}{2}(1-\frac{1}{2k+1})$
$\implies\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n^2-1}=\displaystyle\lim_{k\to\infty}S_k=\frac{1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/265277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 5,
"answer_id": 4
} |
Compute the series Compute the series
$$1)\space\sum_{n=1}^{\infty}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
$$2)\space\sum_{n=1}^{\infty}\left(1-\frac{1}{2}+\frac{1}{3}-\cdots(-1)^{n}\frac{1}{n+1}\right)\frac{1}{n(n+1)}$$
| In both of these, we will use the telescoping sum
$$
\begin{align}
\sum_{k=n}^\infty\frac1{k(k+1)}
&=\sum_{k=n}^\infty\frac1k-\frac1{k+1}\\
&=\frac1n
\end{align}
$$
$1)$ Here is one way:
$$
\begin{align}
\sum_{n=1}^\infty\sum_{k=0}^n\frac1{k+1}\frac1{n(n+1)}
&=1+\sum_{n=1}^\infty\sum_{k=1}^n\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty\sum_{n=k}^\infty\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty\frac1{k+1}\frac1k\\[6pt]
&=1+1\\[12pt]
&=2
\end{align}
$$
$2)$ Here is a similar way:
$$
\begin{align}
\sum_{n=1}^\infty\sum_{k=0}^n(-1)^k\frac1{k+1}\frac1{n(n+1)}
&=1+\sum_{n=1}^\infty\sum_{k=1}^n(-1)^k\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty\sum_{n=k}^\infty(-1)^k\frac1{k+1}\frac1{n(n+1)}\\
&=1+\sum_{k=1}^\infty(-1)^k\frac1{k+1}\frac1k\\
&=1+\sum_{k=1}^\infty(-1)^k\frac1k-\sum_{k=1}^\infty(-1)^k\frac1{k+1}\\
&=1+\sum_{k=1}^\infty(-1)^k\frac1k+\sum_{k=1}^\infty(-1)^{k+1}\frac1{k+1}\\[6pt]
&=1-\log(2)+(1-\log(2))\\[12pt]
&=2-2\log(2)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/267088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
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