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How to prove $ \sum_{k=0}^n \frac{(-1)^{n+k}{n+k\choose n-k}}{2k+1}=\frac{-2\cos\left(\frac{2(n-1)\pi}{3}\right)}{2n+1}$ How to prove
$$\sum_{k=0}^n \binom{n+k}{n-k}\frac{(-1)^{n+k}}{2k+1}=-\frac{2}{2n+1}\,\cos\left(\frac{2(n-1)\pi}{3}\right)\;\text{?}$$
I have a proof by induction for it, but it isn't simple! I want to seek an interesting proof for it. (Sorry for my English study is very bad)
Can you help me re-open this question?
Thank you very much!
| Let's try a generating function approach.
Multiply by $x^ky^n$ and sum in both $k$ and $n$:
$$
\begin{align}
\sum_{k,n}\binom{n+k}{n-k}x^ky^n
&=\sum_{k,n}\binom{n}{2k}\frac{x^k}{y^k}y^n\tag{1}\\
&=\frac1{1-y}\sum_k\frac{x^ky^k}{(1-y)^{2k}}\tag{2}\\
&=\frac{1-y}{(1-y)^2-xy}\tag{3}
\end{align}
$$
Explanation:
$(1)$: $\binom{n+k}{n-k}=\binom{n+k}{2k}$, then substitute $n\mapsto n-k$
$(2)$: $\sum\limits_n\binom{n}{k}x^n=\frac{x^k}{(1-x)^{k+1}}$
$(3)$: $\sum\limits_kx^k=\frac1{1-x}$
Substitute $x\mapsto-x^2$, $y\mapsto y^2$ and integrate $(3)$ in $x$ from $0$ to $1$:
$$
\begin{align}
\sum_{k,n}\frac{(-1)^k}{2k+1}\binom{n+k}{n-k}y^{2n}
&=\int_0^1\frac{1-y^2}{(1-y^2)^2+x^2y^2}\,\mathrm{d}x\tag{4}\\
&=\frac1y\,\tan^{-1}\!\left(\frac{y}{1-y^2}\right)\tag{5}\\[6pt]
&=\frac1y\left[\tan^{-1}\left(e^{i\pi/3}y\right)+\tan^{-1}\left(e^{-i\pi/3}y\right)\right]\tag{6}\\
&=2\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\cos\left(\frac{(2n+1)\pi}3\right)y^{2n}\tag{7}
\end{align}
$$
Explanation:
$(4)$: substitute $x\mapsto-x^2$, $y\mapsto y^2$ and integrate $(3)$ in $x$ from $0$ to $1$
$(5)$: perform the integration
$(6)$: use the formula for tangent of a sum
$(7)$: use the series for arctangent
Equation $(7)$ implies that
$$
\begin{align}
\sum_{k=0}^n\frac{(-1)^{n+k}}{2k+1}\binom{n+k}{n-k}
&=\frac2{2n+1}\cos\left(\frac{(2n+1)\pi}3\right)\tag{8}\\
&=-\frac2{2n+1}\cos\left(\frac{(2n-2)\pi}3\right)\tag{9}
\end{align}
$$
Explanation:
$(8)$: equate identical powers of $y$ in $(7)$
$(9)$: $\cos(x+\pi)=-\cos(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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Specific approach to integral Problem: Evaluate $$\int_{0}^{\infty} \dfrac{\sin^3{x}}{x \cdot e^x} dx=\dfrac{A\pi}{B}-\dfrac{\tan^{-1} (C)}{D},$$
Attempt through Differentiation under the Integral Sign:
Using $\sin^3x=\frac{3\sin x-\sin 3x}{4}$, rewrite as
$$ \frac{3}{4}\int_0^\infty \frac{e^{-x}\sin x}{x}\,dx-\frac{1}{4}\int_0^\infty \frac{e^{-x}\sin 3x}{x}\,dx $$
$$ I(b)=\int_0^\infty \frac{e^{-x}\sin bx}{x}\,dx\qquad,\qquad\mbox{where}\,\,b>0 $$
$$ \begin{align} I'(b)&=\int_0^\infty e^{-x}\cos bx\,dx\\ \end{align} $$
How to proceed, please?
| We may use:
$$ \int_{0}^{+\infty}\frac{f(x)}{x\,e^{x}}\,dx = \int_{1}^{+\infty}(\mathcal{L}\,f)(t)\,dt \tag{1}$$
so we just need to find the Laplace transform of $\sin^3(x)$.
Pretty easy task through the triplication formulas, since $\mathcal{L}(\sin x)=\frac{1}{1+t^2}$ implies:
$$ \mathcal{L}(\sin^3 x) = \frac{3}{4}\left(\frac{1}{t^2+1}-\frac{1}{t^2+9}\right)=\frac{6}{(1+t^2)(9+t^2)}\tag{2}$$
then:
$$ \int_{0}^{+\infty}\frac{\sin^3 x}{x\,e^{x}}\,dx = \frac{3\pi}{16}-\frac{3}{4}\int_{1}^{+\infty}\frac{dt}{t^2+9}=\color{red}{\frac{3\pi}{16}-\frac{\arctan 3}{4}}.\tag{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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I think I can complete the square of any quadratic, is it true? (Any reason to ever use Quad. Formula?) I was taught that you could only complete the square of a quadratic if the coefficient on the $x^2$ term is 1.
However, playing a little bit with other quadratics, I've found that it's just not true. Based on the CTS algorithm, you just need to divide the coefficient of the $x$ term by twice the square root of the coefficient of the $x^2$ term.
So, if you have $ax^2 + bx + c$, your perfect square would be $(\sqrt{a}x + \frac{b}{2 \sqrt{a}})^2$
If $a$ is not a perfect square it could get nasty, but then you can just square the whole quadratic and go from there.
For example:
In the equation $5x^2 + 6x + 5 = 0$, we could do:
$25x^2 + 30x + 25 = 0$
$(5x+3)^2 = -16$
$5x+3 = \pm4i$
$x = \pm \frac{4i}{5} - \frac{3}{5}$
My questions are:
-Is this correct?
-Is there ever an advantage to using the quadratic formula?
-Are there quadratics that are unsolvable this way?
| If $a$ is not a “perfect square” there's no problem either. If your equation is
$$
ax^2+bx+c=0
$$
then it's equivalent to
$$
4a^2x^2+4abx+4ac=0
$$
and completing the square is more evident:
$$
4a^2x^2+4abx+b^2-b^2+4ac=0
$$
or
$$
(2ax+b)^2-(b^2-4ac)=0
$$
If you just want to factor the polynomial $ax^2+bx+c$ (with $a\ne0$, of course, just do the same:
$$
ax^2+bx+c=\frac{1}{4a}(4a^2x^2+4abx+4ac)=
\frac{1}{4a}\bigl((2ax+b)^2-(b^2-4ac)\bigr)
$$
If $b^2-4ac<0$ there's nothing else to do, because the polynomial is irreducible over the reals; if $b^2-4ac=0$ it is
$$
\frac{1}{4a}(2ax+b)^2
$$
and, if $b^2-4ac>0$ you get
$$
ax^2+bx+c=\frac{1}{4a}(x+2a-\sqrt{b^2-4ac})(x+2a+\sqrt{b^2-4ac})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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Riccati Differential Equation Help Show that $y(x)=\frac{1}{x}$ is a solution of the Ricatti differential equation:
$$y'+\frac{1-3xy}{x^2}+3y^2=0$$
Hence using a suitable transformation of the dependant variable, solve the differential equation when $y(1)=3$.
Any help appreciated.
| here is another way to solve this problem. we will make a change of variable $$v = \frac1{xy}, y =\frac1{xv}, y' = -\frac1{x^2v}-\frac{v'}{xv^2}\tag 1 $$
subbing $(1)$ in $$y'+\frac{1-3xy}{x^2}+3y^2=0, $$ we find that
$$-\frac1{x^2v}-\frac{v'}{xv^2} +\frac{1-3/v}{x^2}+\frac3{x^2v^2} = 0$$ multiplying by $x^2v^2$ gets us $$ -v-xv'+v^2-3v+3=0\to xv' = (v-1)(v-3)$$
leading to $$\int_{1/3}^v \frac{dv}{(v-1)(v-3)}= \frac12\int_{1/3}^v \frac{dv}{v-3} - \frac{dv}{v-1} = \int_1^x\frac{dx}x = \ln (x) $$
that is $$\ln\left(\frac{3-v}{4(1-v)}\right) = 2\ln x \to \frac{3-v}{1-v} =4x^2\to v = \frac{3-4x^2}{1-4x^2}$$ and finally $$y = \frac{1-4x^2}{x(3-4x^2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}$ with and without L'Hopital's Rule. So, we have to find $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}$ with and without L'Hopital's Rule.
My Work: Let $\lim\limits_{x\to 0}(\frac{\sin(x)}{x})^{\frac{\sin(x)}{x-\sin(x)}}=L$
Taking $\ln$ of both sides and bringing the exponent down.
$\lim\limits_{x\to 0}{\frac{\sin(x)}{x-\sin(x)}\ln(\frac{\sin(x)}{x})}=\ln(L)$
But it changes to undefined form? The answer (in my textbook) is $$\boxed{L=\frac1{e}}$$
| Take logarithm as you did then write what follows
\begin{equation*}
\frac{\sin x}{x-\sin x}\ln \left( \frac{\sin x}{x}\right) =\frac{\ln \left(
\frac{\sin x}{x}\right) }{\frac{x-\sin x}{\sin x}}=\frac{\ln \left( 1+\left[
\frac{\sin x}{x}-1\right] \right) }{\frac{x-\sin x}{x}\times \frac{x}{\sin x}%
}=-\frac{\ln \left( 1+\left[ \frac{\sin x}{x}-1\right] \right) }{\left[
\frac{\sin x}{x}-1\right] }\times \frac{\sin x}{x}
\end{equation*}
Now we use classic limits
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\sin x}{x} &=&1 \\
\lim_{u\rightarrow 0}\frac{\ln (1+u)}{u} &=&1
\end{eqnarray*}
it follows that
\begin{equation*}
\lim_{x\rightarrow 0}\frac{\sin x}{x-\sin x}\ln \left( \frac{\sin x}{x}%
\right) =-\lim_{x\rightarrow 0}\frac{\ln \left( 1+\left[ \frac{\sin x}{x}-1%
\right] \right) }{\left[ \frac{\sin x}{x}-1\right] }\times \frac{\sin x}{x}%
=-1\times 1=-1.
\end{equation*}
Therefore, $L=e^{-1}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof of $\sum_{x = 1}^\infty \frac{1}{x}$'s divergence by absurdity? (From this site.)
The following argument purports to show that the series $1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \dots = 0$. It begins with the harmonic series.
$$
\begin{aligned}
\sum \frac{1}{x} &= \sum \left( \frac{1}{2x - 1} + \frac{1}{2x} \right)\\
&= \sum \left( \frac{2(2x - 1)}{2x(2x - 1)} + \frac{1}{2x(2x - 1)} \right)\\
&= \sum \left(\frac{1}{x} + \frac{1}{2x(2x - 1)}\right)\\
&= \sum \left(\frac{1}{x}\right) + \sum\left(\frac{1}{2x(2x - 1)}\right)\\
\end{aligned}
$$
So $\sum \frac{1}{x}$ equals itself plus something, which implies that the second series is $0$. Of course, this argument is false because the harmonic series diverges.
My question: suppose we didn't know the harmonic series diverged, but we did know that $\sum\left(\frac{1}{2x(2x - 1)}\right)$ converged to a nonzero value -- it is in fact $\ln{(2)}$.
Would we then be able to conclude, from the above argument, that the harmonic series diverges? That is, if we arrive at $A = A + c$, where $c$ is constant, can we conclude that $A$ is infinite?
| $$ A = A + c \\
1 = 1 + c/A \\
0 = c/A \text{ only if A is infinite } $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\lim\limits_{x\to\ 0}\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}$ without L'Hospital's rule I have no idea how to solve this limit. I tried transforming the expression
$$\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}$$
in
$$e^{\ln\frac{1-\cos x\times (\cos2x)^{\frac{1}{2}}\times (\cos3x)^{\frac{1}{3}}}{x^2}}$$
but that didn't work.
| You can write $$1-\cos x(\cos 2x)^{1/2}(\cos3x)^{1/3}=1-\cos x+\cos x(1-(\cos 2x)^{1/2}(\cos3x)^{1/3})=$$$$1-\cos x+\cos x\left((1-(\cos 2x)^{1/2}+(\cos 2x)^{1/2}\left(1-(\cos3x)^{1/3}\right)\right),$$ therefore $$\frac{1-\cos x(\cos 2x)^{1/2}(\cos3x)^{1/3}}{x^2}=$$$$\frac{1-\cos x}{x^2}+\cos x\frac{1-(\cos 2x)^{1/2}}{x^2}+\cos x(\cos 2x)^{1/2}\frac{1-(\cos3x)^{1/3}}{x^2}.$$ But, for all $n\in\mathbb N$, $$\frac{1-(\cos nx)^{1/n}}{x^2}=\frac{1}{1+(\cos nx)^{1/n}+(\cos nx)^{2/n}+\dots+(\cos nx)^{1-1/n}}\frac{1-\cos nx}{x^2}=$$$$\frac{n^2}{1+(\cos nx)^{1/n}+(\cos nx)^{2/n}+\dots+(\cos nx)^{1-1/n}}\frac{1-\cos nx}{(nx)^2}\xrightarrow{x\to 0}\frac{n^2}{2n}=\frac{n}{2}.$$ Therefore, the requested limit is equal to $$\frac{1}{2}+\frac{2}{2}+\frac{3}{2}=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313900",
"timestamp": "2023-03-29T00:00:00",
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equations of sides of triangles; find largest angle If the sides of a triangle are $2x+3$, $x^2 + 3x + 3$, and $x^2 + 2x$, find the greatest interior angle of a triangle.
The answer is $120$ degrees.
I was hoping to find a formula to relate all the three sides, then use cosine law to find an angle. If I use the sine law, I don't have an angle.... If I use the cosine law, I also don't have an angle (and you also have to square terms; takes more time?). Pythagorean doesn't assume a right triangle.
Do I assume a value for $x$? How so? I am totally lost. Any hint?
| Use cosine rule. It's really quite trivial when you keep the algebra simple.
The longest side is $x^2 + 3x + 3$, so the greatest angle lies opposite this. To show that that's the longest side, you simply need to sketch the curves for $x>0$, which is the region that matters, since for $x \leq 0$, the side of length $x^2 + 2x$ becomes zero or negative, which is impossible.
Apply the cosine rule:
$(x^2 + 3x + 3)^2 = (2x+3)^2 + (x^2 + 2x)^2 - 2(2x+3)(x^2+2x)\cos\theta$
When simplifying, avoid opening up the brackets immediately. Use identities like $a^2 - b^2 = (a+b)(a-b)$ to keep your life simple.
Working in this careful fashion, we get:
$(x+3)(2x^2 + 5x + 3) - (2x+3)^2= -2(2x+3)(x^2 + 2x)\cos \theta$
$(x+3)(2x +3)(x+1) - (2x+3)^2= -2(2x+3)(x^2 + 2x)\cos \theta$
It is permitted to cancel $2x+3$ from both sides as this cannot be zero (since $x>0$).
So:
$(x+3)(x+1) - 2x - 3 = -2(x^2 + 2x)\cos\theta$
$x^2 + 4x + 3 - 2x - 3 = -2(x^2 + 2x)\cos\theta$
$(x^2 + 2x) = -2(x^2 + 2x)\cos\theta$
giving $\cos \theta = -\frac 12$ and $\theta = 120^{\circ}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying the product $\prod\limits_{k=2}^n \left(1-\frac1{k^2}\right)$ Can we simplify the given product to a general law?
$$\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{n^2}\right)$$
| Let us show by induction you tagged that for $n\ge 2$,
$$\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)=\frac{n+1}{2n}.$$
For $n=2$, it is true since $1-\frac{1}{2^2}=\frac{3}{4}=\frac{2+1}{2\cdot 2}.$
Supposing that $\prod_{k=2}^{n}\left(1-\frac{1}{k^2}\right)=\frac{n+1}{2n}$ gives you
$$\begin{align}\prod_{k=2}^{n+1}\left(1-\frac{1}{k^2}\right)&=\frac{n+1}{2n}\cdot\left(1-\frac{1}{(n+1)^2}\right)\\&=\frac{n+1}{2n}\cdot\frac{(n+1)^2-1}{(n+1)^2}\\&=\frac{n(n+2)}{2n(n+1)}\\&=\frac{(n+1)+1}{2(n+1)}\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the generating function
What is the generating function for ${a_k}$, where $a_k$ is the number of solutions of $x_1 + x_2 + x_3 = k$ when $x_1,x_2,x_3$ are integers with $x_1 \geq 2$, $0 \leq x_2 \leq 3$, and $2 \leq x_3 \leq 5$?
I know that the answer of this problem is $x^4(1+x+x^2+x^3)^2/(1-x)$. I want to know how to find the generating function in detail. Please help me!
| Let me change the names of the variables in hopes of reducing confusion later: instead of $x_1,x_2$, and $x_3$, I’ll use $y_1,y_2$, and $y_3$. Eventually my generating function will use the indeterminate $x$, but temporarily I’m going to split it into three indeterminates $x_1,x_2$, and $x_3$: for $k=1,2,3$ the exponent on $x_k$ will keep track of a possible value of $y_k$. The possible values of $y_1$ are the integers greater than or equal to $2$, represented by
$$x_1^2+x_1^3+x_1^4+\ldots=\sum_{n\ge 2}x_1^n\;.$$
The only possible values of $y_2$ are $0,1,2$, and $3$, represented by
$$x_2^0+x_2^1+x_2^2+x_2^3\;.$$
And the only possible values of $y_3$ are $2,3,4$, and $5$, represented by
$$x_3^2+x_3^3+x_3^4+x_3^5\;.$$
Multiply these three expressions together to get
$$(x_1^2+x_1^3+x_1^4+\ldots)(x_2^0+x_2^1+x_2^2+x_2^3)(x_3^2+x_3^3+x_3^4+x_3^5)\;,$$
and imagine actually carrying out the multiplication. Each term of the product will have the form $x_1^kx_2^\ell x_3^m$, where $k\ge 2$, $0\le\ell\le 3$, and $2\le m\le 5$. Now drop the subscripts on $x_1,x_2$, and $x_3$, making them all simply $x$: each term of the product will have the form $x^{k+\ell+m}$ for some integers $k,\ell$, and $m$ such that $k\ge 2$, $0\le\ell\le 3$, and $2\le m\le 5$. Moreover, every possible combination of $k,\ell$, and $m$ satisfying those constraints will occur. Thus, when you collect like terms, you’ll have one copy of $x^n$ for each combination of $k,\ell$, and $m$ such that $k+\ell+m=n$, $k\ge 2$, $0\le\ell\le 3$, and $2\le m\le 5$. That is, the coefficient of $x^n$ will be the number of solutions to $y_1+y_2+y_3=n$ satisfying the given conditions.
This means that the desired generating function is
$$(x^2+x^3+x^4+\ldots)(x^0+x^1+x^2+x^3)(x^2+x^3+x^4+x^5)\;,$$
and we need only simplify it. First, $x^2+x^3+x^4+x^5=x^2(1+x+x^2+x^3)$, so
$$(x^0+x^1+x^2+x^3)(x^2+x^3+x^4+x^5)=x^2(1+x+x^2+x^3)^2\;.$$
Then
$$x^2+x^3+x^4+\ldots=\sum_{n\ge 2}x^n=x^2\sum_{n\ge 0}x^n=\frac{x^2}{1-x}$$
from the formula for a geometric series. Now just put the pieces togeter to get
$$\frac{x^2}{1-x}\cdot x^2(1+x+x^2+x^3)=\frac{x^4(1+x+x^2+x^3)}{1-x}\;.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a little Diophantine equation:$(n-1)!+1=n^m$ How can I solve this Diophantine equation:
$$(n-1)!+1=n^m$$
with $n,m$ positive integers?
From Wilson's theorem we can note that $n$ is a prime number. I proved to rewriting the equation as:$$(n-2)!=n^{m-1}+n^{m-2}+....+1$$
but in vain. I proved to solve also through the theorem LTE bur I analysed only some cases without obtain a general solution.
| If $1\le n\le 5$, then $(n,m)=(2,1),(3,1),(5,2)$. Let $n\ge 6$.
$$(n-1)!+1=n^m$$
$$\iff (n-2)!=\frac{n^m-1}{n-1}=$$
$$=1+n+n^2+\cdots+n^{m-2}+n^{m-1}$$
$$\iff (n-2)!-m=$$
$$=(1-1)+(n-1)+(n^2-1)+\cdots+(n^{m-1}-1)$$
$2$, $\frac{n-1}{2}$ are different and $\le n-2$, so
$2\cdot \frac{n-1}{2}=n-1\mid (n-2)!$.
For all $a,b\in\mathbb R$, $k\ge 2$, $k\in\mathbb Z$, $$a^k-b^k=(a-b)(a^{k-1}b^0+a^{k-2}b^1+\cdots+a^0b^{k-1})$$
If $k$ is odd, $c=-b$, then
$$a^k+c^k=(a+c)(a^{k-1}c^{0}-a^{k-2}c^1+\cdots+a^0c^{k-1})$$
Let $a=n$, $b=1$. Then $n-1\mid n^k-1$ for all $k\ge 0$, $k\in\mathbb Z$ because $n-1\mid n^0-1=1-1=0$ and $n-1\mid n^1-1=n-1$.
$n-1\mid m$, so $m\ge n-1$ (because $m>0$) and $$ n^m = (n-1)! + 1 = 1 \cdot 2 \cdots (n-1) + 1 < $$
$$<\underbrace{(n-1)(n-1)\cdots (n-1)}_{\text{n-1 times}}=$$
$$ = (n-1)^{n-1} < n^{n-1}, $$
contradiction. Answer: $(n,m)=(2,1),(3,1),(5,2)$.
| {
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Hint for solving a definite integral $\int_{-a}^{a}\frac{xdy}{(x^{2}+y^{2})^{\frac{3}{2}}}$ Can anyone provide a hint for solving this definite integral:
$\int_{-a}^{a}\frac{xdy}{(x^{2}+y^{2})^{\frac{3}{2}}}$
| Another way, in this particular example, is to rewrite your expression as
$$
\begin{aligned}
\frac{x}{(x^2+y^2)^{3/2}}&=\frac{x^2+y^2-y^2}{x(x^2+y^2)^{3/2}}=\frac{1}{x(x^2+y^2)^{1/2}}-\frac{y^2}{x(x^2+y^2)^{3/2}}\\
&=\Bigl(\frac{d}{dy} y\Bigr)\cdot \frac{1}{x(x^2+y^2)^{1/2}}+ y\cdot \frac{d}{dy}\frac{1}{x(x^2+y^2)^{1/2}}\\
&=\frac{d}{dy}\Bigl(\frac{y}{x(x^2+y^2)^{1/2}}\Bigr)
\end{aligned}
$$
Thus, you have a simple primitive function that you can put into the fundamental theorem of calculus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A unusual inequality about function $\ln$ These day,I met a unusual inequality when I solve a difficult problem, and proving the inequality means I have done the work!
Could you show me how to prove it or deny it? By the way, I believe that it's true!
Prove that, for all $t > 0$,
\begin{align*}
&4\ln t\ln (t + 2) - \ln t\ln (t + 1) - 3\ln t\ln (t + 3)\\
+ &4\ln (t + 1)\ln (t + 3) - 3\ln (t + 1)\ln (t + 2) - \ln (t + 2)\ln \left( {t + 3} \right)>0.
\end{align*}
Let
$$f\left( t \right) = 4\ln t\ln \left( {t + 2} \right) - \ln t\ln \left( {t + 1} \right) - 3\ln t\ln \left( {t + 3} \right) + 4\ln \left( {t + 1} \right)\ln \left( {t + 3} \right) - 3\ln \left( {t + 1} \right)\ln \left( {t + 2} \right) - \ln \left( {t + 2} \right)\ln \left( {t + 3} \right),$$
We have
$$f'\left( t \right) = \frac{{2\left[ {{t^2}\ln t - 3{{\left( {t + 1} \right)}^2}\ln \left( {t + 1} \right) + 3{{\left( {t + 2} \right)}^2}\ln \left( {t + 2} \right) - {{\left( {t + 3} \right)}^2}\ln \left( {t + 3} \right)} \right]}}{{t\left( {t + 1} \right)\left( {t + 2} \right)\left( {t + 3} \right)}}.$$
Let
$$g\left( t \right) = {t^2}\ln t - 3{\left( {t + 1} \right)^2}\ln \left( {t + 1} \right) + 3{\left( {t + 2} \right)^2}\ln \left( {t + 2} \right) - {\left( {t + 3} \right)^2}\ln \left( {t + 3} \right),$$
we got
$$g'\left( t \right) = 2\left[ {t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right)} \right].$$
And let
$$h\left( x \right) = t\ln t - 3\left( {t + 1} \right)\ln \left( {t + 1} \right) + 3\left( {t + 2} \right)\ln \left( {t + 2} \right) - \left( {t + 3} \right)\ln \left( {t + 3} \right),$$
we have
\begin{align*}
h'\left( x \right) &= \ln t - 3\ln \left( {t + 1} \right) + 3\ln \left( {t + 2} \right) - \ln \left( {t + 3} \right)\\
&= \ln \frac{{t{{\left( {t + 2} \right)}^3}}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}} = \ln \left[ {1 - \frac{{2t + 3}}{{{{\left( {t + 1} \right)}^3}\left( {t + 3} \right)}}} \right] < 0.
\end{align*}
However, it seems that there are no use!
| It's not true.
Try plugging in a large value for t such as $10^9$ (tried in R and sage)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
How to integrate$ I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy $ I am stuck with the integration
$$
I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$$
I got this from the question from the book
"Field and wave electromagnetics, Cheng, 2nd, Problem 3-18.
I tried to solve this equation using method of integration by parts, but my equation got worse.
I know the answer by Wolfram Alpha, but I can't get how.
| This is incomplete.
But it's close.
We want
$I=\int\ln\left(\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$
If we look at
$\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}
=\frac{L+\sqrt{L^2+4(y^2+z^2)}}{2}
$,
this is one of the roots
of the quadratic
$x^2-Lx-(y^2+z^2)
=0$.
Since the product of the roots is
$-(y^2+z^2)
$,
if we define
$J=\int\ln\left(\frac{L}{2}-\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$,
then
$I+J
=\int\ln(-(y^2+z^2))dy
$.
However,
the inner expression is negative,
so I'll define
$J=\int\ln\left(-\frac{L}{2}+\sqrt{\frac{L^2}{4}+y^2+z^2}\right)\ \mathrm dy
$.
We now get
then
$I+J
=\int\ln(y^2+z^2)dy
$.
This is,
according to
Wolfram Alpha,
$y (log(y^2+z^2)-2)+2 z tan^{-1}(y/z)
$.
What we need now
is $I-J$.
The expression in the $\ln$
would be
$\begin{array}\\
\frac{L+\sqrt{L^2+4(y^2+z^2)}}{-L+\sqrt{L^2+4(y^2+z^2)}}
&=\frac{L+\sqrt{L^2+4(y^2+z^2)}}{-L+\sqrt{L^2+4(y^2+z^2)}}\\
&=\frac{L+\sqrt{L^2+4(y^2+z^2)}}{-L+\sqrt{L^2+4(y^2+z^2)}}
\frac{L+\sqrt{L^2+4(y^2+z^2)}}{L+\sqrt{L^2+4(y^2+z^2)}}\\
&=\frac{L^2+L^2+4(y^2+z^2)+2L\sqrt{L^2+4(y^2+z^2)}}{4(y^2+z^2)}\\
&=\frac{L^2}{2(y^2+z^2)}+1+\frac{L\sqrt{L^2+4(y^2+z^2)}}{2(y^2+z^2)}\\
\end{array}
$
Again,
according to WA,
$\int \frac{\sqrt{x^2+a}}{x^2+b} dx
= \frac{\sqrt{a-b} \tan^{-1}\left(\frac{x \sqrt{a-b}}{\sqrt{b} \sqrt{a+x^2}}\right)}{\sqrt{b}}+\ln(\sqrt{a+x^2}+x)
$
This would be enough to get
$I-J$,
and, from these,
$I$ amd $J$.
However, it's late
and I'm tired,
so I'll stop here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
How should I find Splitting Field of $x^3-2$ over $\mathbb Q$. How should I find Splitting Field of $x^3-2$ over $\mathbb Q$.
**My try **:
$x^3-2=(x-2^\frac{1}{3})(x^2+2^\frac{1}{3}x+2^\frac{2}{3})$
On solving I am getting the roots as $2^\frac{1}{3},\dfrac{2^\frac{1}{3}[-1+\sqrt 3 i]}{2},\dfrac{2^\frac{1}{3}[-1-\sqrt 3 i]}{2},$
But I dont know the answer is given that roots are $2^\frac{1}{3},2^\frac{1}{3}\omega ,2^\frac{1}{3}\omega ^2$ where $\omega $ is cube root of unity
| Splitting field will be $\mathbb{Q}(2^{1/3},\omega)$ where w is cube roots of unity.
where $\omega =\dfrac{-1+\sqrt 3 i}{2}~\&~\omega ^2=\dfrac{-1-\sqrt 3 i}{2},$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x} \, dx$? Find error in integration of $\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$?
The answer is supposed to be ($\arctan \tan^2 x + C$), but I obtained ($-\arctan \cos2x + C$) as follows. Please identify the error.
$$\int \frac {\sin 2x}{\sin^4 x + \cos^4 x}dx$$
$$= \int \frac {\sin 2x}{(\sin^2 x+ \cos^ 2x)^2 - 2\cos^2 x\sin^2 x}dx$$
$$= \int \frac {\sin 2x}{1- \frac{\sin^2 2x}{2}}dx$$
$$= \int \frac {2\sin 2x}{2- \sin^2 2x}dx$$
$$= \int \frac {2\sin 2x}{1 + \cos^2 2x}dx$$
$1 + \cos^2 2x = t ; dt = 2(\cos 2x)(-\sin 2x)(2)dx; 2\sin 2x.dx= \frac{-dt}{2\cos 2x} = \frac{-dt}{2(\sqrt{t-1})};$
$$\int \frac{\frac{-dt}{2\sqrt {t-1}}}{t} = -1/2\int \frac{dt}{t.\sqrt{t-1}}$$
$\sqrt{t-1} = u; du=\frac{dt}{2\sqrt{t-1}}; 2u.du = dt$;
$$= -1\int \frac{du}{u^2+1} = -\arctan{u} = -\arctan \sqrt{t-1} = -\arctan \sqrt{1+\cos^2 2x-1} = -\arctan \cos2x$$
Also, is there an easier method to this problem?
| $$\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}dx=\int\frac{4\sin(2x)}{\left(1-\cos(2x)\right)^2+(1+\cos^2(2x))^2}dx=-\int\frac{\,d\cos(2x)}{1+\cos^2(2x)}\\
=-\arctan(\cos(2x))+C.$$
As
$$\tan^2(x)=\frac{\sin^2(x)}{\cos^2(x)}=\frac{1-\cos(2x)}{1+\cos(2x)}=\tan\left(\frac\pi4-\arctan(\cos(2x))\right),$$
$\arctan(\tan^2(x))$ and $-\arctan(\cos(2x))$ only differ by a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Does the series: $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? does $\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}$ converge? I think yes, it does, because the $a_n$ in the series converges to zero. but I'm trying to prove this by the help of the fact that:
$\frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}$
any suggestions?
| $$\frac{2n+1}{n^2(n+1)^2}=\frac{1}{n^2}-\frac{1}{(n+1)^2}$$
$$\sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2}=\frac{1}{1}-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}+....=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
Identity on Fibonacci numbers: $F_{2n}^2=F_{2n+2}F_{2n-2}+1$? Let $F_n$ be the Fibonacci Sequence ($F_1=F_2=1, F_{n+2}=F_{n+1}+F_{n}$).
Prove that $F_{2n}^2=F_{2n+2}F_{2n-2}+1$.
I've tried everything from induction to telescoping series but I haven't got close. They seem to make the identity even more messy. The main problem is with the squared term and the 1 at the end.
Does Pell's Equation or continued fractions have anything to do with this?
| The difference equation $F_{n+2} = F_{n+1} + F_{n}$ admits the solution
\begin{align}
F_{n} = \frac{\alpha^{n} - \beta^{n}}{\alpha - \beta}
\end{align}
where $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$. Now,
\begin{align}
F_{2n-2} F_{2n+2} &= \frac{1}{5} \left(\alpha^{2n-2} - \beta^{2n-2} \right) \left( \alpha^{2n+2} - \beta^{2n+2} \right) \\
&= \frac{1}{5} \, \left( \alpha^{4n} - \alpha^{2n-2} \beta^{2n+2} - \alpha^{2n+2} \beta^{2n-2} + \beta^{4n} \right) \\
&= \frac{1}{5} \, \left( \alpha^{4n} - \alpha^{4} - \beta^{4} + \beta^{4n} \right) \\
&= \frac{1}{5} \left( L_{4n} - L_{4} \right) = \frac{L_{4n} - 7}{5} \tag{1}
\end{align}
where $L_{n}$ are the Lucas numbers and have the form $L_{n} = \alpha^{n} + \beta^{n}$. Consider the square of $F_{2n}$.
\begin{align}
F_{2n}^{2} &= \frac{1}{5} \, \left( \alpha^{4n} - 2 \alpha^{2n} \beta^{2n} + \beta^{4n} \right) = \frac{L_{4n} - 2}{5}. \tag{2}
\end{align}
By comparison of equations (1) and (2) it is evident that
\begin{align}\tag{3}
F_{2n}^{2} = F_{2n-2} F_{2n+2} + 1
\end{align}
which is the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
The asymptotic behavior of an integral The integral in hand is
$$
I(n) = \frac{1}{\pi}\int_{-1}^{1} \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}\, dx
$$
I dont know whether it has closed-form or not, but currently I only want to know its asymptotic behavior. Setting $x=\cos\theta$, then
$$
I(n) = \frac{1}{\pi}\int_{0}^{\pi/2} \Big[(1+2\cos\theta)^{2n}+(1-2\cos\theta)^{2n}\Big]\, d\theta
$$
The second term can be neglected, therefore
$$
I(n) \sim \frac{1}{\pi}\int_{0}^{\pi/2}(1+2\cos\theta)^{2n}\, d\theta
$$
How can I move on?
| To compute the asymptotic expansion of the integral, we split it into two pieces
$$\frac{1}{\pi}\int_{-1}^{1} \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx
= \frac{1}{\pi}\left(\int_{-1}^0 + \int_0^1\right)\frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx
$$
Over the interval $[-1,0]$, we have $|1+2x|\le 1$, so the contribution there is bounded.
$$\mathcal{I}_1 \stackrel{def}{=} \left|\frac{1}{\pi}\int_{-1}^0 \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx\right| \le
\frac{1}{\pi}\int_{-1}^0 \frac{dx}{\sqrt{1-x^2}} = \frac12
$$
Over the interval $[0,1]$, introduce variable
$$1 + 2x = 3 e^{-t} \quad\iff\quad x = \frac{3e^{-t}-1}{2}$$
We have
$$\mathcal{I}_2 \stackrel{def}{=} \frac{1}{\pi}\int_{0}^1 \frac{(1+2x)^{2n}}{\sqrt{1-x^2}}dx =
\frac{3^{2n+1}}{2\pi}\int_{0}^{\log 3} e^{-(2n+1)t} \left[1 - \left(\frac{3e^{-t}-1}{2} \right)^2 \right]^{-1/2} dt$$
Notice near $t = 0$, the complicated mess of the square bracket has following Taylor series expansion:
$$\left[1 - \left(\frac{3e^{-t}-1}{2} \right)^2 \right]^{-1/2}
= \frac{1}{\sqrt{3t}}\left(
1+ \frac{5}{8}t+ \frac{49}{384}t^2-\frac{29}{3072}t^3 + \cdots
\right)\tag{*1}$$
The whole expression is in the form which we can apply Watson's Lemma and read off the asympotic expansion:
$$\begin{align}
\mathcal{I}_2 \;\approx &\;
\frac{3^{2n+1}}{2\sqrt{3}\pi}
\left(
\frac{\Gamma\left(\frac12\right)}{\sqrt{2n+1}}
+ \frac{5}{8}\frac{\Gamma\left(\frac32\right)}{\sqrt{2n+1}^3}
+ \frac{49}{384}\frac{\Gamma\left(\frac52\right)}{\sqrt{2n+1}^5}
- \frac{29}{3072}\frac{\Gamma\left(\frac72\right)}{\sqrt{2n+1}^7}
+ \cdots
\right)\\
\;\approx &\;
\frac{3^{2n}\sqrt{3}}{2\sqrt{\pi(2n+1)}}
\left( 1 + \frac{5}{16(2n+1)} + \frac{49}{512(2n+1)^2}
- \frac{145}{8192(2n+1)^3} + \cdots\right)
\end{align}
$$
Since $\mathcal{I}_1$ is always of $O(1)$, above asymbolic expansion for $\mathcal{I}_2$ is also the one for $\mathcal{I}_1 + \mathcal{I}_2$. i.e. the one you are looking for.
If one want more terms for the asymptotic expansion, one just need to throw the LHS of $(*1)$ to an CAS, crank out more terms of the Taylor expansion and repeat above process.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
What is the exact value of $\eta(6i)$? Let $\eta(\tau)$ be the Dedekind eta function. In his Lost Notebook, Ramanujan played around with a related function and came up with some of the nice evaluations,
$$\begin{aligned}
\eta(i) &= \frac{1}{2} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(2i) &= \frac{1}{2^{11/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(3i) &= \frac{1}{2\cdot 3^{3/8}} \frac{1}{(2+\sqrt{3})^{1/12}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(4i) &= \frac{1}{2^{29/16}} \frac{1}{(1+\sqrt{2})^{1/4}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(5i) &= \frac{1}{2\sqrt{5}}\left(\tfrac{1+\sqrt{5}}{2}\right)^{-1/2}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(6i) &=\; \color{red}{??}\\
\eta(7i) &= \frac{1}{2\sqrt{7}}\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)^{{1/4}}\, \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(8i) &= \frac{1}{2^{73/32}} \frac{(-1+\sqrt[4]{2})^{1/2}}{(1+\sqrt{2})^{1/8}} \frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\\
\eta(16i) &= \frac{1}{2^{177/64}} \frac{(-1+\sqrt[4]{2})^{1/4}}{(1+\sqrt{2})^{1/16}} \left(-2^{5/8}+\sqrt{1+\sqrt{2}}\right)^{1/2}\,\frac{\Gamma\big(\tfrac{1}{4}\big)}{\pi^{3/4}}\end{aligned}$$
with the higher ones $>4$ added by this OP. (Note the powers of $2$.)
Questions:
*
*Similar to the others, what is the exact value of $\eta(6i)$?
*Is it true that the function,
$$F(\sqrt{-N}) = \frac{\pi^{3/4}}{\Gamma\big(\tfrac{1}{4}\big)}\,\eta(\sqrt{-N}) $$
is an algebraic number only if $N$ is a square?
P.S. It seems strange there is a function that yields an algebraic number for square input $N$ and a transcendental number for non-square $N$. (Are there well-known functions like that?) For an example of non-square $N$, we have,
$$\eta(\sqrt{-3}) = \frac{3^{1/8}}{2^{4/3}} \frac{\Gamma\big(\tfrac{1}{3}\big)^{3/2}}{\pi} = 0.63542\dots$$
and $F(\sqrt{-3})$ seems to be transcendental.
| While giving this and this answer I got all the ingredients necessary to solve the current problem.
Consider the nome $q=e^{-\pi} $ so that elliptic modulus is $k=2^{-1/2}$ and the function $f(q) $ defined by $$f(q)=q^{1/6}\prod_{n=1}^{\infty}(1-q^{4n})\tag{1}$$ which is expressed in terms of modulus $k$ and complete elliptic integral $K=K(k) $ as $$f(q) =2^{-2/3}\sqrt{\frac{2K}{\pi}}k^{1/3}k'^{1/12}\tag{2}$$ (see first answer linked above).
If $l$ is the elliptic modulus corresponding to nome $q^{3}$ then we have $$\eta(6i)=f(q^{3})=2^{-2/3}\sqrt{\frac{2L}{\pi}}(l^2\sqrt{l'})^{1/6}\tag{3}$$ The values $$\sqrt{\frac{2L}{\pi}} =\frac{(27+18\sqrt{3})^{1/4}\Gamma(1/4)} {3\sqrt{2}\pi^{3/4}} ,l=\frac{(\sqrt{3}-1)(\sqrt{2}-\sqrt[4]{3})}{2},\sqrt{l'}=\frac{\sqrt{2}+\sqrt[4]{3}(\sqrt{3}-1)}{2^{5/4}}$$ are available from the second answer linked above which help us to get the value of $\eta(6i)$ in explicit form. The calculations are somewhat lengthy but can be performed using pen and paper.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 5,
"answer_id": 1
} |
Finding $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx$
How can I integrate $$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx\ \ \ ?$$
I have made the integral into the form of $\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}$, but been able to go no further.
| Let $I$ be your integral.
Setting $u=\frac{\pi}{2}-x$ and using $\tan\left(\frac{\pi}{2}-x\right)=\frac{1}{\tan x}$, we have
$$I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{1}{1+\sqrt{\tan x}}dx=\int_{\frac{\pi}{3}}^{\frac{\pi}{6}}\frac{-1}{1+\sqrt{\frac{1}{\tan u}}}du=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\sqrt{\tan u}}{1+\sqrt{\tan u}}du=J.$$
Now note that
$$I+J=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.$$
Hence, $$I=\color{red}{\frac{\pi}{12}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Remainder when divided by 9 I'd like help with this question :
What is the remainder when
$$2^{2} + 22^{2} + 222^{2}+ \ldots + \underbrace{2222...22^{2}}_{49 \text{ times}} $$
is divided by $9$
| In mod $9$, we have
$$2^2+22^2+\cdots+22\ldots 2^2$$
$$\equiv 2^2+4^2+6^2+8^2+1^2+3^2+5^2+7^2+0^2+2^2+\cdots$$
$$\equiv 5(2^2+4^2+6^2+8^2+1^2+3^2+5^2+7^2+0^2)+2^2+4^2+6^2+8^2$$
$$\equiv 5\cdot\frac{8\cdot 9\cdot 17}{6}+4+7+0+1\equiv 5\cdot 12\cdot 17+3\equiv 5\cdot 3\cdot (-1)+3\equiv 6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $ $$ \cos ^2\left(x\right)+\cos ^2\left(2x\right)+\cos ^2\left(3x\right)=\frac{3}{2} $$
How can I solve this one, I mean I get something like this:
$-3+\left(-1+2\cos ^2\left(x\right)\right)^22+2\left(-3\cos \left(x\right)+4\cos ^3\left(x\right)\right)^2+2\cos ^2\left(x\right)=0$
This equation seems rather hard to solve from here, any tips or other ways to come to an solution?
| Doing it the way you originally started gives us $$-3+\left(-1+2\cos ^2\left(x\right)\right)^22+2\left(-3\cos \left(x\right)+4\cos ^3\left(x\right)\right)^2+2\cos ^2\left(x\right)=0$$ Now let $\cos x = u$ and simplify to get $$32u^6 - 40u^4 + 12u^2 - 1 = 0$$ which is simply a cubic in $u^2$ and can be factorised as $$4\left(\frac{1}{2}+u\right)\left(-\frac{1}{2}+u\right)\left(8u^4-8u^2+1\right)=0$$ giving us $u = \pm \frac{1}{2}$ and we need to solve the remaining quartic(which is a quadratic in $u^2$) to get $$u=-\frac{\sqrt{2-\sqrt{2}}}{2},\:u=\frac{\sqrt{2-\sqrt{2}}}{2},\:u=-\frac{\sqrt{2+\sqrt{2}}}{2},\:u=\frac{\sqrt{2+\sqrt{2}}}{2}$$ as our other solutions.
So we have:
$$\cos \left(x\right)=\frac{1}{2}\implies x=\frac{\pi }{3}+2\pi n,\:x=\frac{5\pi }{3}+2\pi n$$
$$\cos \left(x\right)=-\frac{1}{2}\implies x=\frac{2\pi }{3}+2\pi n,\:x=\frac{4\pi }{3}+2\pi n$$
$$\cos \left(x\right)=-\frac{\sqrt{2+\sqrt{2}}}{2} \implies x=2\pi n-\arccos \left(-\frac{\sqrt{2+\sqrt{2}}}{2}\right), \\ x=2\pi n+\arccos \left(-\frac{\sqrt{2+\sqrt{2}}}{2}\right)$$
$$\cos \left(x\right)=-\frac{\sqrt{2-\sqrt{2}}}{2}\implies x=2\pi n-\arccos \left(-\frac{\sqrt{2-\sqrt{2}}}{2}\right),\\ x=2\pi n+\arccos \left(-\frac{\sqrt{2-\sqrt{2}}}{2}\right)$$
$$\cos \left(x\right)=\frac{\sqrt{2-\sqrt{2}}}{2}\implies x=2\pi n-\arccos \left(\frac{\sqrt{2-\sqrt{2}}}{2}\right),\\ x=2\pi n+\arccos \left(\frac{\sqrt{2-\sqrt{2}}}{2}\right)$$
$$\cos \left(x\right)=\frac{\sqrt{2+\sqrt{2}}}{2}\implies x=2\pi n-\arccos \left(\frac{\sqrt{2+\sqrt{2}}}{2}\right),\\ x=2\pi n+\arccos \left(\frac{\sqrt{2+\sqrt{2}}}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Finding an explicit formula for a recursive sequence. How to show that the recurrent formula
$$A_n=A_{n-1} + A_{n-2} +4.$$
gives a sequence of the form $f(n)=cr^n+cr^n$?
The only way we are allowed to solve it, is with the quadratic formula $(r^2-r-1)$...
| Consider $B_n=A_n+4$, to get $B_n=B_{n-1}+B_{n-2}$.
Now you can solve the chactaristic equation $r^2-r-1$.
This gives $r=\frac{1+\sqrt{5}}{2}$, $r=\frac{1-\sqrt{5}}{2}$.
Therefore we have $$B_n= A \left(\frac{1+\sqrt{5}}{2}\right)^n+ B \left(\frac{1-\sqrt{5}}{2}\right)^n$$
Therefore
$$A_n= A \left(\frac{1+\sqrt{5}}{2}\right)^n+ B \left(\frac{1-\sqrt{5}}{2}\right)^n -4$$
Where $A,B$ depend on your initial conditions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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I need help with a Finite Series Problem:
Find the sum to $n$ terms of
\begin{eqnarray*}
\frac{1}{1\cdot 2\cdot 3} + \frac{3}{2\cdot 3\cdot 4} + \frac{5}{3\cdot 4\cdot 5} +
\frac{7}{4\cdot 5\cdot 6}+\cdots \\
\end{eqnarray*}
Answer:
The way I see it, the problem is asking me to find this series:
\begin{eqnarray*}
S_n &=& \sum_{i=1}^{n} {a_i} \\
\text{with } a_i &=& \frac{2i-1}{i(i+1)(i+2)} \\
\end{eqnarray*}
We have:
\begin{eqnarray*}
S_n &=& S_{n-1} + a_n \\
S_n &=& S_{n-1} + \frac{2n-1}{n(n+1)(n+2)} \\
\end{eqnarray*}
I am tempted to apply the technique of partial fractions
but I believe there is no closed formula for a series of the of the form:
\begin{eqnarray*}
\sum_{i=1}^{n} \frac{1}{i+k} \\
\end{eqnarray*}
where $k$ is a fixed constant. Therefore I am stuck. I am hoping that somebody
can help me.
Thanks Bob
| You can write it as $\sum_{k\ge0}\frac{2k+1}{(k+3)_3}=\sum_{k\ge0}(2k+1)(k)_{-3}$ and now it seems easy to solve by summation by parts:
$$\sum (2k+1)(k)_{-3}\delta k=(2k+1)\frac{(k)_{-2}}{-2}+\sum(k+1)_{-2}=\frac{2k+1}{-2(k+2)_2}-\frac{1}{k+2}=\frac{4k+3}{-2(k+2)_2}$$
And taking limits we have that the series converges to $\sum_{k\ge 0}\frac{2k+1}{(k+3)_3}=\frac{3}{4}$
The partial sum will be
$$\sum_{k=0}^{n}\frac{2k+1}{(k+3)_3}=\frac{4n+3}{-2(n+2)_2}+\frac{3}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
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} |
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$?
What is the remainder when $6\times7^{32} + 7\times9^{45}$ is divided by $4$ ?
$7 \equiv 3 \pmod 4$
$7^2 \equiv 9 \pmod 4\equiv 1 \pmod 4$
$(7^2)^{16} \equiv 1^{16} \pmod 4$
i.e $7^{32} \equiv 1 \pmod 4$
Similarly $9 \equiv 1 \pmod 4$ implies $9^{45} \equiv 1 \pmod 4$.
But the problem arise with the coefficients and addition sign.
what to do?
| You're on the right track. Note that $9 = 1 \mod 4$, so $9^{45} = 1 \mod 4$. Then just apply normal rules. You have $6\times 1+7\times 1=13=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Generating functions of bills Using generating functions, find the number of ways to make change for a $\$$100
bill using only dollar coins and $\$$1, $\$$5, and $\$$10 bills.
My answer: I had $1/(1-x)^2*(1-x^5)*(1-x^{10})=1/(1-x)^2*(1-x^5)^2*(1+x^5)$.
I know I need to find the coefficient of $x^{100}$. What should I do next? My guess is partial fractions but the computation looks very long. So is there an easier way to determine the coefficient?
| Rewrite the generating function $G(x)$ as
\begin{align*}
G(x) &= \frac{1}{\left(1-x\right)^2\left(1-x^5\right)\left(1-x^{10}\right)}\\
&= \frac{\left(1+x+x^2+x^3+x^4\right)^2\left(1+x^5\right)^3}{\left(1-x^{10}\right)^4}\\
&= \left(1+x+x^2+x^3+x^4\right)^2\left(1+x^5\right)^3\sum_k \binom{k+3}{3} x^{10\, k}
\end{align*}
Now, extracting $x^{10k}$ gives
$$[x^{10k}]G(x) = \binom{k+3}{3} + 15 \binom{k+2}{3}+4 \binom{k+1}{3}$$
For $k=10$, we get
$$[x^{100}]G(x) = \binom{13}{3} + 15 \binom{12}{3}+4 \binom{11}{3} = 4246$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How do i find this : $\int \frac{1}{(x+a) \sqrt{x+b}}\ dx$, where $a > b > 0$? Is there someone show me how do I find : $$\int \frac{1}{(x+a) \sqrt{x+b}}\ dx$$, where $$a > b > 0$$ ?
I tried to make it as sum of fraction to be easier but sorry i didn't up
Thank you for any help .
| Use the substitution $u^2 = x + b$.
Let
$I = \int\frac{1}{(x+ a)\sqrt{x + b}}\,dx
$
Let
\begin{align*}
&u^2 = x + b\implies x + a = u^2 + a - b\\
&2u\,du = dx
\end{align*}
Then
\begin{align*}
I &= 2\int\frac{1}{u^2 + a - b}\,du\\
\\
&= \frac{2}{\sqrt{a - b}}\tan^{-1}\left(\frac{u}{\sqrt{a - b}}\right) + C\\
\\
&= \frac{2}{\sqrt{a - b}}\tan^{-1}\left(\frac{\sqrt{x + b}}{\sqrt{a - b}}\right) + C
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Find maximum of a function I want to find the maximum of a function.
$$ d = \frac{35}{3} + \frac{7}{3}\sin( \frac{2\pi}{365}t ) $$
I don't know if I applied the chain rule correctly.
$$ w = \frac{2\pi}{365}t $$
$$ w' = \frac{2\pi}{365} $$
$$ g = \sin(w) $$
$$ g' = \cos(w) $$
$$ f = \frac{35}{3} + \frac{7}{3}g $$
$$ f' = \frac{7}{3}g $$
$$ d' = \frac{7}{3}\sin( \frac{2\pi}{365}t ) \times \cos( \frac{2\pi}{365}t ) \times \frac{2\pi}{365} $$
I set it to $0$
$$ a\times b \times c = 0 $$
And it seems that I can't solve this because $x = 0$
$$ \arcsin(a) = \arcsin(0) \\
\frac{2\pi}{365}x = 0 \\
x = 0$$
| Since $\sin(s)\in [-1,1]$ for any $s$, we have
$$d=\frac{35}{3} + \frac{7}{3}\sin\Big( \frac{2\pi}{365}t \Big)\leq \frac{35}{3} + \frac{7}{3}=\frac{42}{3}=14$$
for every $t$. Now, you have to find $t$ such that $\sin\Big( \frac{2\pi}{365}t \Big)=1$. By periodicity of $\sin$ this is true whenever $$\frac{2\pi}{365}t = \frac{\pi}{2}+2k\pi \iff t = \frac{365}{2\pi}\Big(\frac{\pi}{2}+2k\pi \Big)=\frac{365}{4}+k365$$ with $k\in \Bbb Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Separate real and imaginary part of $\arccos(z)$ Beginning with
$$i \cos \left[ \frac{1}{n} \arccos \left( \frac{i}{\epsilon} \right) + \frac{m \pi}{n} \right]$$
where $m,n \in \mathbf{Z}$, $\epsilon >0$, $\epsilon \in \mathbf{R}$ and $i$ is the imaginary unit, I would like to obtain separately the real and imaginary part of the cosine argument:
(1)
$$\frac{1}{n} \arccos \left( \frac{i}{\epsilon} \right) + \frac{m \pi}{n} = x + iy$$
By simply applying the definition:
(2)
$$\frac{1}{n} \arccos \left( \frac{i}{\epsilon} \right) = \frac{\pi}{2n} + i \frac{1}{n} \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
Similarly, if we apply the definition of complex $\arcsin$, we obtain:
$$\arcsin \left( \frac{i}{\epsilon} \right) = -i \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
Remembering (as stated here) that $$\mathrm{arsinh}(z) = -i\arcsin(iz)$$ we obtain
$$\mathrm{arsinh} \left( \frac{1}{\epsilon} \right) = - \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
So we can write
$$x + iy = \frac{m \pi}{n} + \frac{\pi}{2n} + i\frac{1}{n} \ln \left[ \frac{1}{\epsilon} (\sqrt{1 + \epsilon^2} - 1) \right]$$
$$x + iy = \frac{\pi}{2} \left( \frac{2m + 1}{n} \right) - i\frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right) = \frac{\pi}{2} \left( \frac{2m - 1}{n} \right) - i\frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right)$$
But the correct result should be
(3)
$$x + iy = \frac{\pi}{2} \left( \frac{2m - 1}{n} \right) + i\frac{1}{n} \mathrm{arsinh} \left( \frac{1}{\epsilon} \right)$$
So, are there any errors? How to obtain (3) from (1)?
| $$i \cos \left( \frac{1}{n} \cos^{-1} \left( \frac{i}{\epsilon} \right) + \frac{m \pi}{n} \right)=i\cos\left(\frac{\cos^{-1}\left(\frac{i}{\epsilon}\right)+m\pi}{n}\right)=$$
$$i\cos\left(\frac{\sec^{-1}\left(-i\epsilon\right)+m\pi}{n}\right)=i\cos\left(\frac{\frac{\pi}{2}-i\sinh^{-1}\left(\frac{1}{\epsilon}\right)}{n}+ \frac{m\pi}{n}\right)=$$
$$i\cos\left(\frac{\pi}{2n}-\frac{i\sinh^{-1}\left(\frac{1}{\epsilon}\right)}{n}+ \frac{m\pi}{n}\right)=$$
$$i\cos\left(\frac{\pi+2m\pi}{2n}-\frac{i*csch^{-1}(\epsilon)}{n}\right)=$$
$$i\cos\left(\frac{\pi+2m\pi}{2n}-\frac{csch^{-1}(\epsilon)}{n}i\right)$$
$$\frac{\pi+2m\pi}{2n}-\frac{csch^{-1}(\epsilon)}{n}i=\frac{\pi}{2}\left(\frac{2m-1}{n}\right)-\frac{\sinh^{-1}\left(\frac{1}{\epsilon}\right)}{n}i=\frac{\pi}{2}\left(\frac{2m-1}{n}\right)-i\frac{1}{n}\sinh^{-1}\left(\frac{1}{\epsilon}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the general values of $x$ satisfying the trigonometric equation
Find the general values of $x$ satisfying
$$
\frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}=\frac{3}{2}
$$
It seems to me just some equality case of an inequality. But I am unable to find the inequality. Thanks.
| I would get everything
in terms of sin and cos,
which I will write as
s and c cause I'm lazy.
$\begin{array}\\
\frac{\tan^2 x \sin^2 x}{1-\sin^2 x \cos2x}+\frac{\cot^2 x \cos^2 x}{1-\cos^2 x \cos2x}+\frac{2\sin^2 x}{\tan^2 x+\cot^2 x}
&=\frac{(s^2/c^2)s^2 }{1-s^2(c^2-s^2)}+\frac{(c^2/s^2) c^2 }{1-c^2 (c^2-s^2)}+\frac{2s^2}{(s^2/c^2)+(c^2/s^2)}\\
&=\frac{s^4 }{c^2-s^2c^2(c^2-s^2)}+\frac{c^4 }{s^2-c^2s^2 (c^2-s^2)}+\frac{2s^4c^2}{s^4+c^4}\\
&=\frac{s^4 }{c^2-s^2c^4+s^4c^2}+\frac{c^4 }{s^2-c^4s^2+c^2s^4}+\frac{2s^4c^2}{s^4+c^4}\\
&=\frac{s^4 }{c^2-s^2c^4+s^4c^2}+\frac{c^4 }{s^2-c^4s^2+c^2s^4}+\frac{2s^4c^2}{s^4+c^4}\\
\end{array}
$
At this point, I gave up,
and went to Wolfy,
which said
x ~~ -5.40621425424778,
0.631651356268605,
0.876072957166250,
...
and a bunch of other roots.
Interestingly,
I was able to paste the
MathJax expression
unedited
and it was interpreted correctly.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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What is wrong in my $f'(x)$? We have $f:\mathbb{R}\rightarrow\mathbb{R}, f(x)=\frac{x^2-x+1}{x^2+x+1}$ and we need to find $f'(x)$.
Here is all my steps:
$$\begin{align}f'(x)&=\frac{(2x-1)(x^2+x+1)-(x^2-x+1)(2x+1)}{(x^2+x+1)^2}\\&=\frac{(2x-1)(x^2+1)-(2x+1)(x^2+1)}{(\cdots)^2}\\&=\frac{(x^2+1)(2x-1-2x-1)}{(\cdots)^2}\\&=\frac{2(1-x)(1+x)}{(\cdots)^2},\forall x\in\mathbb{R}\end{align}$$
But in my book they say that $f'(x)=\frac{2(x-1)(x+1)}{(x^2+x+1)^2}$. What is wrong in my method ?
| You were quick in cancelling entire $ x ( 2 x -1) $ and $ x ( 2 x +1) $ . For the first term your logic is ok, not the next term. Else all fine.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{ 5^{125}-1}{ 5^{25}-1}$ is a composite number Prove that $\dfrac {\left( 5^{125}-1\right)}{\left( 5^{25}-1\right)}$ is composite number using number theory. Do not use calculator or Wolfram alpha or anything like that.
| This is problem $87$ of Putnam and Beyond. Here is the solution:
$\dfrac{5^{125}-1}{5^{25}-1}=1+a+a^2+a^3+a^4$ where $a=5^{25}$.
We have $1+a+a^2+a^3+a^4+a^5=(a^2+3a+1)^2-(5^{13}(a+1))^2=(a^2+3a+1+5^{13}(a+1))(a^2+3a+1-5^{13}(a+1))$
The reason the second factor is larger than $1$ is $a^2=5^{50}>5^{39}>5^{13}(a+1)$
This problem was also part of the 1992 imo shortlist and was proposed by Korea, here is a link with the same solution but shorter: https://mks.mff.cuni.cz/kalva/short/soln/sh9216.html
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "13",
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solve $\sqrt{x+7}
solve $\sqrt{x+7}<x$
I tried $\sqrt{x+7}<x\\
x+7<x^2\\
x^2-x-7>0\\
x\in \left(-\infty, \dfrac{1-\sqrt{29}}{2}\right) \cup \left( \dfrac{1+\sqrt{29}}{2},+\infty\right) $
I m not sure, if this is correct method and if the solution is correct .
I look for a simple and short way.
I have studied maths up to $12$th grade.Thanks.
| First of all, we need to have $x+7\ge 0$.
Then, note that $\sqrt{x+7}$ is non-negative. So, since we have
$$0\le\sqrt{x+7}\lt x,$$
we have
$$0\lt x.$$
Hence, we have
$$x+7\ge0\ \ \ \text{and}\ \ \ x+7\lt x^2\ \ \ \text{and}\ \ \ 0\lt x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant Question. Show that if $A=\begin{bmatrix}a & b\\c & d\end{bmatrix}$, then
$\det(A)=\frac{1}{2}\det\left(\begin{bmatrix}1 & 1\\tr(A^2) & (tr(A))^2\end{bmatrix}\right)$.
I tried finding the determinant using the formula but it didn't come out to be the same as when I found determinant using cofactor expansion.
| We have that
$$det(A) = ad-bc$$
$$A^{2} = \begin{bmatrix} a^{2} + bc & ab + bd \\ ca + dc & cb + d^{2}\end{bmatrix}$$
Then
\begin{align*}
\frac{1}{2} \cdot det(\begin{bmatrix}1 & 1\\tr(A^2) & (tr(A))^2\end{bmatrix}) &=\frac{1}{2} \cdot det(\begin{bmatrix}1 & 1\\a^{2} + 2bc + d^{2} & (a+d)^{2}\end{bmatrix}) \\
&= \frac{1}{2}(a^{2} + 2ad + d^{2} - a^{2} - 2bc - d^{2}) \\
&= ad - bc
\end{align*}
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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trigonometry expression simplification with inverse cosine While working on a problem, I ended up with this expression for y:
$$
y=x\sin\left(\arccos\left(\frac{\sqrt{x^2-y^2}}x\right)\right)
$$
Is there any way to express $y$ in terms of $x$ only, with no $y$ on RHS?
Thanks.
| $$y=x\sin\left(\cos^{-1}\left(\frac{\sqrt{x^2-y^2}}x\right)\right)=$$
$$y=x\left(\sin\left(\cos^{-1}\left(\frac{\sqrt{x^2-y^2}}x\right)\right)\right)=$$
$$y=x\left(\sqrt{1-\left(\frac{\sqrt{x^2-y^2}}x\right)^2}\right)=$$
$$y=x\sqrt{1-\left(\frac{\sqrt{x^2-y^2}}x\right)^2}=$$
$$y=x\sqrt{1-\frac{x^2-y^2}{x^2}}$$
$$y=x\sqrt{1-\frac{x^2-y^2}{x^2}}\Longleftrightarrow$$
$$y^2=x^2\left(1-\frac{x^2-y^2}{x^2}\right)\Longleftrightarrow$$
$$y^2=x^2-\frac{x^2\left(x^2-y^2\right)}{x^2}\Longleftrightarrow$$
$$y^2=x^2-\frac{x^4-x^2y^2}{x^2}\Longleftrightarrow$$
$$y^2=x^2-\left(x^2-y^2\right)\Longleftrightarrow$$
$$y^2=x^2-x^2+y^2\Longleftrightarrow$$
$$y^2=0+y^2\Longleftrightarrow$$
$$y^2=y^2\Longleftrightarrow$$
$$y=y$$
So it's true foe every value of $x$ and $y$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347989",
"timestamp": "2023-03-29T00:00:00",
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why $\lim_{x \to \frac{\pi}{4}} \frac{\cos 2x}{\sin x-\cos x}=-\sqrt{2}$? I have this very simple limit to find
$$\lim_{x \to \frac{\pi}{4}} \frac{\cos (2x)}{\sin x-\cos x}$$
which is equal to $-\sqrt{2}$. However I can get the outcome as mentioned, or $\sqrt{2}$ in the following way:
$$\lim_{x \to \frac{\pi}{4}} \frac{\cos (2x)}{\sin x-\cos x}=\lim_{x \to \frac{\pi}{4}} \frac{\cos ^2x-\sin^2x}{\sin x-\cos x}=\lim_{x \to \frac{\pi}{4}} \frac{(\cos x-\sin x)(\cos x+\sin x)}{\sin x-\cos x}$$
$$=\lim_{x \to \frac{\pi}{4}} (\sin x+\cos x)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}=\sqrt{2}$$
Why the mentioned solution method is wrong?
Is limit outcome dependent on the expression rearrangement?
| Note that $$\cos x-\sin x\not=\sin x-\cos x.$$
We have
$$\frac{(\cos x-\sin x)(\cos x+\sin x)}{\sin x-\cos x}=\frac{-(\color{red}{\sin x-\cos x})(\cos x+\sin x)}{\color{red}{\sin x-\cos x}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Radius of Convergence Is the radius of convergence of
$$\frac{n(x+3)^n}{4^n}$$
equals 4?
I got $|x+3|\lt 4$ as the final result. How do you know, what is the radius from here?
| The radius of convergence of
$$
\sum_{n=0}^\infty a_n(x-a)^n\tag{1}
$$
is the number $R$ such that (1) converges for $\lvert x-a\rvert<R$ and diverges for $\lvert x-a\rvert>R$. In our case, we can apply the Ratio Test to the power series
$$
\sum_{n=0}^\infty \frac{n}{4^n}(x+3)^n\tag{2}
$$
Doing so gives
$$
\lim_{n\to\infty}\left\lvert\frac{\frac{n+1}{4^{n+1}}(x+3)^{n+1}}{\frac{n}{4^n}(x+3)^n}\right\rvert
= \lim_{n\to\infty} \frac{n+1}{n}\frac{4^n}{4^{n+1}}\left\lvert\frac{(x+3)^{n+1}}{(x+3)^n}\right\rvert
= \frac{1}{4}\lim_{n\to\infty}\frac{n+1}{n}\left\lvert x+3\right\rvert
=\frac{1}{4}\left\lvert x+3\right\rvert
$$
It follows that (2) converges when
$$
\frac{1}{4}\left\lvert x+3\right\rvert<1
$$
and diverges when
$$
\frac{1}{4}\left\lvert x+3\right\rvert>1
$$
Equivalently, (2) converges when
$$
\left\lvert x+3\right\rvert<4
$$
and diverges when
$$
\left\lvert x+3\right\rvert>4
$$
Thus the radius of convergence of (2) is $R=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Three variable, second degree diophantine equation I am trying to solve this diophantine equation:
$x^2 + yx + y^2 = z^2$
In other words, I am trying to find integers $x$ and $y$ such that $x^2 + yx + y^2$ is a perfect square.
So far, the only methods to solve quadratic diophantine equations I am familiar with are Pythagorean triples and Pell equations. The $yx$ term has a coefficient of one, so I can't complete the square to reduce it to a Pell equation somehow, and I am wondering if there are other methods to solve this kind of equation.
Some insight would be highly appreciated.
| Let $z$ be a fixed integer. We have $x^2+y^2+xy=z^2$ is equivalent to
$(x+y)^2-xy=z^2$ and $(x-y)^2+3xy=z^2.$ Therefore, $(x+y)^2=z^2+xy\geq 0$ and $(x-y)^2=z^2-3xy\geq 0.$ Then, we have $z^2+xy$ and $z^2-3xy$ are squares.
So, $|x+y|=\sqrt{z^2+xy}$ and $|x-y|=\sqrt{z^2-3xy}.$ This implies that
$x+y=\sqrt{z^2+xy}$ or $x+y=-\sqrt{z^2+xy}$ and $x-y=\sqrt{z^2-3xy}$ or $x-y=-\sqrt{z^2-3xy}.$ You use after that the basic tools of calculus to find $x$ and $y$ separatly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Formula for calculating a progressive sum If we say that initially the addition is $1$, the sum $0$ and $d$ is constant of $5$.
step 1)
sum = previous sum + addition sum = 0 + 1 = 1
addition = previous addition + d addition = 1 + 5 = 6
step 2)
sum = previous sum + addition sum = 1 + 6 = 7
addition = addition + d addition = 6 + 5 = 11
step 3)
sum = previous sum + addition sum = 7 + 11 = 18
addition = addition + d addition = 11 + 5 = 16
step 4)... the same as above
What I want to achieve here is to find a formula for the calculations shown above from which I can find the sum if I know the rest.
The closest formula found so far is arithmetic progression. But still it's not the one I'm looking for.
What would be the formula for that?
| First let's label the sums as a sequence $S_n$ and your "additions" as $A_n$.
It appears that $A_0$, the initial condition, is $1$, and $S_0=0$.
Then we have the system: $$S_{n+1} = S_n + A_n$$ $$A_{n+1} = A_n + D.$$
Here you have $D=5$. We can solve the recurrence for $A_{n+1}$ first, since it does not involve $S_n$.
Well, $A_0=1$, $A_1 = 1+D$, $A_2 = A_1 + D = 1 + 2D$, etc. Thus $A_{n} = 1 + nD$.
Plugging this back into $S_{n+1}$ gives:
$$S_{n+1} = S_n + 1 + nD.$$
Now $S_1 = 0 + 1 + D$, $$S_2 = S_1 + 1 + 2D = (0 + 1 + D) + 1 + 2D = 2 + (1+2)D.$$
$$S_3 = S_2 + 1 + 3D = (2 + (1+2)D) + 1 + 3D = 3 + (1+2+3)D$$ we can see the general form taking shape here:
$$S_{n} = n + (1+ 2+ 3+ \cdots + n)D= n + \frac{n(n+1)}{2} \cdot D.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$n$ dimensional determinant using recurrence relations Find determinant
$$D_n(a,b,c)=
\begin{vmatrix}
a & b & 0 & 0 & \cdots & 0 & 0 & 0 \\
c & a & b & 0 & \cdots & 0 & 0 & 0 \\
0 & c & a & b & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & c & a & b \\
0 & 0 & 0 & 0 & \cdots & 0 & c & a
\end{vmatrix}
$$
for $a,b,c \in \mathbb{R}$.
I used recurrence relations, $$D_n=aD_{n-1}-bcD_{n-2}$$
Now, we must consider these cases:
*
*If $b=0 \lor c=0$ then $$D_n=a^{n-1}D_1=a^n\frac{1}{a}\times a=a^n$$
*If $b\neq 0 \land c\neq 0$ then we solve quadratic equation $$x^2-px-r=0$$ where $p=a$ and $r=-bc$, so equation is $$x^2-ax+bc=0$$
Roots of this equation are $$x_1=\frac{a+\sqrt{a^2-4bc}}{2},x_2=\frac{a-\sqrt{a^2-4bc}}{2}$$
2.1. If $x_1\neq x_2$ then $$D_n=k_1x_1^n+k_2x_2^n$$ where $$k_1=\frac{D_2-x_2D_1}{x_1(x_1-x_2)},k_2=-\frac{D_2-x_1D_1}{x_2(x_1-x_2)}$$
Solving for $k_1$ and $k_2$, $$k_1=\frac{a^2-2bc+a\sqrt{a^2-4bc}}{a^2-4bc+a\sqrt{a^2-4bc}}$$ $$k_2=\frac{2bc-a^2-a\sqrt{a^2-4bc}}{4bc+a\sqrt{a^2-4bc}-a^2}$$
This gives $D_n=\frac{a^2-2bc+a\sqrt{a^2-4bc}}{a^2-4bc+a\sqrt{a^2-4bc}}\left(\frac{a+\sqrt{a^2-4bc}}{2}\right)^n+\frac{2bc-a^2-a\sqrt{a^2-4bc}}{4bc+a\sqrt{a^2-4bc}-a^2} \left(\frac{a-\sqrt{a^2-4bc}}{2}\right)^n$
2.2. If $x_1=x_2$ then $$D_n=x_1^{n-1}D_1+(n-1)x_1^{n-2}(D_2-x_1D_1)$$
$$D_n=a\left(\frac{a+\sqrt{a^2-4bc}}{2}\right)^{n-1}+(n-1)\left(\frac{a+\sqrt{a^2-4bc}}{2}\right)^{n-2}\times \frac{2bc-a^2-a\sqrt{a^2-4bc}}{2}$$
Could someone check this? Are there any more cases to consider?
Thanks for replies.
| Your solution looks ok to me.
You have all cases covered.
The determinant $D_n$ is a continuous function of $a,b,c$, and it is a good idea to check that your formulas obey this.
If you let $c\to0$ in your last formula of 2.1, you get $D_n=a^n$ as you should.
In the case $x_1=x_2$ you have $a^2-4bc=0$ and $x_1=x_2=a/2$, which simplifies your formula significantly:
$$
D_n=a(a/2)^{n-1}+(n-1)(a/2)^{n-2}\times\frac{-a^2}4
=
a^n2^{-n}(3-n).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Find coefficients of polynomials $f(x)=x^2+ax+b$ and $g(x)=x^2+cx+d$ $(a,b,c,d \in \mathbb{R})$ Roots of polynomial $f(x)=x^2+ax+b$ are cubes of the roots of polynomial $g(x)=x^2+cx+d$. Sum and product of roots of polynomial $g(x)$ are equal. Find coefficients $a,b,c,d$ so that polynomial $f(x)$ has double root, and $g(x)$ doesn't have double root.
From the condition that $f(x)$ has double root, discriminant of quadratic equation $x^2+ax+b=0$ must be zero, $$D=a^2-4b=0,b=\frac{a^2}{4}$$
Sum and product of roots of polynomial $g(x)$ are equal, $$x_1^3+x_2^3=x_1^3x_2^3$$
Vieta's formulas for two polynomials,
$$x_1+x_2=-a$$
$$x_1x_2=b$$
$$x_1^3+x_2^3=-c$$
$$x_1^3x_2^3=d$$
If we cube first equation, $$(x_1^3+x_2^3)+3x_1x_2(x_1+x_2)=-a^3$$
$$-c-3ba=-a^3$$
From $b=\frac{a^2}{4}$ $$a^3-4c=0,c=\frac{a^3}{4}$$
From $x_1^3+x_2^3=x_1^3x_2^3$ implies $d=-c$
If we cube second equation, $$x_1^3x_2^3=b^3=\frac{a^6}{64}=d=-c$$
I don't know how to find these coefficients.
Thanks for replies.
| Let the two roots of $g(x)$ be $p$ and $q$. Since roots of $f(x)$ are cubes of the roots of $g(x)$,
$$\begin{align*}
p+q &= -c\tag1\\
pq &= d\tag2\\
p^3 + q^3 &= -a\tag3\\
p^3q^3 &= b\tag4
\end{align*}$$
From $(3)$,
$$\begin{align*}
-a &= p^3 + q^3\\
&= (p+q)(p^2 - pq+q^2)\\
&= (p+q)(p^2 + 2pq + q^2 - 3pq)\\
&= -c[(-c)^2 -3d]\\
a &= c^3 -3cd\tag5
\end{align*}$$
And from $(2)$ and $(4)$, $$b = p^3 q^3 = d^3\tag6$$
Since the sum and the product of roots of $g(x)$ are equal, $$d = -c \tag7$$
Writing $a$ and $b$ in terms of $c$,
*
*Substitute $(7)$ into $(5)$, $$a = c^3 + 3c^2\tag8$$
*Substitute $(7)$ into $(6)$, $$b = -c^3\tag{9}$$
Since $f(x)$ has double roots, by using $(8)$ and $(9)$,
$$\begin{align*}
\Delta_f = a^2 - 4b &= 0\\
(c^3 + 3c^2)^2 + 4c^3 &= 0\\
c^4 (c+3)^2 + 4c^3 &= 0\\
c^3 (c^3 + 6c^2 + 9c + 4) &= 0\\
c^3 (c+1)^2(c+ 4) &= 0\\
\end{align*}$$
If $c \in \{0, -4\}$ and using $(7)$,
$$\begin{align*}
\Delta_g &= c^2 - 4d\\
&= c^2 + 4c\\
&= c(c+4)\\
&= 0
\end{align*}$$ which means $g(x)$ has a double root, contradicting to the given conditions.
Hence $c = -1$. Then $a = 2$, $b=1$ and $d = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Help with this limit? I am trying to focus on the limits of functions with similar series expansions and I stumbled on this.
$$\lim_{x\to\infty}\left({\left(\frac{x^2+5}{x+5}\right)}^{1/2}\sin{\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)}-(x-5)^{1/2}\sin{\left({\left(x^2-5x+25\right)}^{1/2}\right)}\right)=0$$
I heard the mean value is possible but the entire function is not bounded. I can take the taylor series at infinity however the terms would be undefined. I could use substitution with the taylor series but it would become a complicated mess.
$$\lim_{x\to\infty}\left(\left(\frac{x^2+5}{x+5}\right)^{1/2}{\left(\frac{x^3+5}{x+5}\right)}^{1/2}-\frac{\left(\frac{x^2+5}{x+5}\right)^{1/2}\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)^3}{3!}+\frac{\left(\frac{x^2+5}{x+5}\right)^{1/2}\left({\left(\frac{x^3+5}{x+5}\right)}^{1/2}\right)^5}{5!}...-\left(\left({x-5}\right)^{1/2}{\left(x^2-5x+25\right)}^{1/2}-\frac{\left(x-5\right)^{1/2}\left({\left(x^2-5x+25\right)}^{1/2}\right)^3}{3!}+\frac{\left(x-5\right)^{1/2}\left({\left(x^2-5x+25\right)}^{1/2}\right)^5}{5!}....\right)\right)$$
I only have limited knowledge of series expansion so I am not so sure how to approach this. Is their an easier way?
| We apply the following scheme $$AB-CD=\frac {\frac {1}{CD}-\frac {1}{AB}}{\frac{1}{ABCD}}$$ in which one has a shape which allows the application of the Hôpital´s rule.
The numerator is $$N=\frac {1}{(\sqrt {x-5})sin\sqrt{x^2-5x+25}}-\frac {1}{\sqrt \frac {x^2+5}{x+5}sin\sqrt{\frac{x^3+5}{x+5}}}$$ and the denominator is $$D=\frac{1}{\sqrt\frac {(x-5)(x^2+5)}{(x+5)^2}sin\frac{x^3+5}{x+5}sin\sqrt{x^2-5x+25}}$$
The derivatives give (we use for short $\alpha$ and $\beta$ for the angles):
$$N’=-\frac{1}{2(x-5)^{\frac {3}{2}}sin(\beta)} -\frac{(2x-5)}{2\sqrt{(x-5)(x^2-5x+25)}sin(\beta)}-\frac{1}{2\sqrt{(x+5)(x^2+5)}sin(\alpha)}+\frac{x\sqrt{x+5}}{(x^2+5)^{\frac {3}{2}}sin(\alpha)}+\frac{x^2+10x-5}{2(x+5)\sqrt{(x^2+5)(x^3+5)}}\frac{cos(\alpha)}{(sin(\alpha))^2}$$
$$D’ =-\frac{(3x^2-5x+5)\sqrt{(x+5)}}{2(x-5)(x^2+5)^{\frac{3}{2}}sin(\alpha)sin(\beta)}+\frac{1}{2\sqrt{(x^2-25)(x^3+5)} sin(\alpha)sin(\beta)}- \frac{(2x-5)\sqrt{x+5}}{2\sqrt{(x-5)(x^2+5)(x^2-5x+25)}} \frac{cos(\beta)}{(sin(\beta))^2sin(\alpha)}
–\frac{(x+5)(2x^3+15x^2-5)}{2\sqrt{(x-5)(x^2+5)(x^3+5)}}\frac{cos(\alpha)}{(sin(\alpha))^2sin(\beta}$$
See the “degrees” in $N’$ and $D’$.
In $N’$ the second term has “degree” $-\frac 12$ and all of the other ones has “degree” $-\frac 32$. Hence $N’\to 0$ because the five terms tend to zero.
In $D’$ the first term has“degree” $-\frac 32$; the second one has “degree” $-\frac 52$, the third one has "degree" $-1$ and the fourth one has “degree” $1$.
Hence $D’\to\infty$ because the fourth term tends to $\infty$.
So we have got the $\frac{0}{\infty}$ shape that ends the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there another simple way to solve this integral $I=\int\frac{\sin{x}}{\sin{x}+\cos{x}}dx$? The integral I want to find is$$I=\int\frac{\sin{x}}{\sin{x}+\cos{x}}dx$$
The way I learnt is to introduce$$J=\int\frac{\cos{x}}{\sin{x}+\cos{x}}dx$$
Then $J+I=x+C_1$ and $J-I=\ln|\sin{x}+\cos{x}|+C_2$.
Is there some simple way to solve this integral $I$? For example, do not introduce other integrals?
Any hints will be appreciated. Thank you.
| As the given
$$I=\int \frac{sinx}{sinx+cosx}dx$$
$$= \int \frac{tanx}{tanx+1}dx$$
Put $ tanx=t \implies sec^2x.dx=dt$
$\implies (1+tan^2x)dx=dt \implies dx=\frac{dt}{1+t^2}$
So we have
$$I=\int\frac{t}{t+1}.\frac{dt}{1+t^2} \to (1)$$
Now by using partial fraction
$$\frac{t}{(1+t)(1+t^2)}=\frac{A}{1+t}+\frac{Bt+C}{1+t^2} \to (2) $$
$$t=A(1+t^2)+(Bt+C)(1+t) \to(3)$$
Put $1+t=0 \implies t=-1$ in above eq
$$-1=A(1+1)+0 \implies A=\frac{-1}{2}$$
Now from eq $(3)$
$$t=A+At^2+Bt+Bt^2+C+Ct$$
Comparing coefficients of $t^2$ and $t$.
Coefficient of $t^2$
$$0=A+B \implies B=-A$$
$$B=\frac{1}{2}$$
Coefficient of $t$
$$1=B+C \implies C=1-B $$
$$C=1-\frac{1}{2} \implies C=\frac{1}{2} $$
Putting these values in eq $(2)$
$$\frac{t}{(1+t)(1+t^2)} = \frac{\frac{-1}{2}}{1+t} + \frac{\frac{1}{2}t+\frac{1}{2}}{1+t^2}$$
Putting this value in eq $(1)$
$$I=\int (\frac{\frac{-1}{2}}{1+t} + \frac{\frac{1}{2}t+\frac{1}{2}}{1+t^2})dt $$
$$I= \frac{1}{2} \int \frac{-1}{1+t}dt + \frac{1}{2} \int \frac{t+1}{t^2+1}dt $$
$$I= - \frac{1}{2} \ln (1+t) + \frac{1}{4}\int \frac{2t}{t^2+1}dt + \frac{1}{2} \int \frac{1}{t^2+1}dt$$
$$I= - \frac{1}{2} \ln (1+t) + \frac{1}{4} \ln (t^2+1) + \frac{1}{2} \arctan(t)+C $$
As $t=tanx$. So we have
$$I= - \frac{1}{2} \ln (1+tanx) + \frac{1}{4} \ln (tan^2x+1) + \frac{1}{2} \arctan(tanx)+C $$
$$I= - \frac{1}{2} \ln (1+tanx) + \frac{1}{4} \ln (tan^2x+1) + \frac{1}{2} x +C$$
$$=-\frac{1}{2}\ln(1+tanx)-\frac{1}{2}\ln(cosx)+\frac{1}{2}x+C$$
$$=\frac{1}{2}x-\frac{1}{2}\ln(sinx+cosx)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
} |
Points on Surface, Distance Optimized How do I find the points on the surface:
$$x^3+y^3+z^3=1$$
such that the distance to the origin is minimized?
My Thoughts:
Perhaps we can minimize the distance squared? Not sure.
| You want to minimise the function: $f(x,y,z) = x^2+y^2+z^2$ with the condition that $g(x,y,z) = 1- x^3-y^3-z^3 = 0$
Note that if $x^2+y^2+z^2$ is minimal so is $\sqrt{x^2+y^2+z^2}$
This is best done using Lagrange multipliers. You define $ F(x,y,z,\lambda ): = f(x,y,z) + \lambda\cdot g(x,y,z)$
Then the set of equations
$$ \frac{\partial F}{\partial x} = 0,\ \frac{\partial F}{\partial y} = 0,\ \frac{\partial F}{\partial z} = 0,\ \frac{\partial F}{\partial \lambda}= 0
$$
will give us the answer. We get $x=\lambda 3x^2$ same for $y,z$ plus $g(x,y,z) = 0$
(i) $x,y,z \ne 0$
We get: $2x=\lambda 3x^2 \Rightarrow \lambda = \frac{2}{3x}$
Inserting into the second and third equation: $x = y = z$
With the last equation: $3x^3 = 1 \Rightarrow x =y=z = \frac{1}{\sqrt[3]{3}}$
(ii) $x,y\ne 0$ and $z = 0$
$ 2x^3 = 1 \Rightarrow x =y = \frac{1}{\sqrt[3]{2}}, z = 0$
(iii) $x\ne 0, y,z = 0$
$ x^3 = 1 \Rightarrow x = 1, y = z = 0$ and $\lambda = \frac23$.
Because of symmetry we will get the last two cases three times for the permutations of $x,y,z$
Now we calculate the distances:
(i) $d = 3^{\frac{1}{6}} = 1,2...$
(ii) $d = 2^{\frac{1}{6}} = 1,1...$
(iii) $d = 1$
So the points with minimal distance are case (1) $ x = 1, y=z=0$ (2) $ y = 1, x=z=0$ (3) $ z = 1, x =y=0$.
To check if we have a local minimum we could also calculate the Hessian matrix on the tangent space of $\nabla g(x,y,z) = (-3, 0, 0)^T$ for say case (1):
\begin{align}
\frac{\partial^2F}{\partial y^2} = \frac{\partial^2F}{\partial z^2} &= 2 - 6\lambda y = 2 \\
\frac{\partial^2F}{\partial x \partial y} &= 0
\end{align}
That is
$$ \nabla^2_{(y,z)} f(1,0,0) = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}.$$
This is obviously positive define, thus $x=1,y=z=0$ is a strict local minimum.
Here is an alternative solution for the minima argument, not sure if this is correct though:
If I vary $y,z$ to be slightly different from zero: $y = \epsilon$, $z = \delta$ then x is given by $x = (1-\epsilon^3 -\delta^3)^{\frac{1}{3}}$. Now checking the distance: $ d= (1-\epsilon^3 -\delta^3)^{\frac{2}{3}} + \epsilon^2 +\delta^2$ tayor expanding the first term $d= 1 - \frac{2}{3}\epsilon^3 - \frac{2}{3}\delta^3 + \epsilon^2 +\delta^2 = 1 + \epsilon^2(1- \frac{2\epsilon}{3}) +\delta^2(1- \frac{2\delta}{3}) > 1$
similarly for just varying one component. Thus the points are indeed local minima.
| {
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"url": "https://math.stackexchange.com/questions/1363573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Arc length of $f(x)=x^2-\ln x$ over [1,e] This is how I learned to solve for arc length:
$$ \frac d{dx} f(x)=2x - \frac 1x$$
$$ \left(\frac d{dx}f(x)\right)^2 = \left(2x - \frac 1x\right)^2 = 4x^2-4+\frac1{x^2} $$
$$ 1+\left(\frac d{dx}f(x)\right)^2 = 4x^2-3+\frac1{x^2} $$
$$ \sqrt{1+\left(\frac d{dx}f(x)\right)^2} = \sqrt{4x^2-3+\frac1{x^2}}$$
So to solve for arc length, I have:
$$\int_1^e \sqrt{4x^2-3+\frac1{x^2}} ~dx $$
I already have the answer (from Wolfram alpha) but I cannot figure out how to solve this integral. It seems complicated since Wolfram cannot generate the steps for this integral.
Any suggestion is highly appreciated.
| I think that the statement contains a misprint. Verify if it is not $x^{2}-%
\frac{1}{8}\ln x$ instead. If it is the case then
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&1+(2x-\frac{1}{8x})^{2} \\
&=&1+(2x)^{2}-2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&=&1+(2x)^{2}-\frac{1}{2}+(\frac{1}{x})^{2} \\
&=&(2x)^{2}+(1-\frac{1}{2})+(\frac{1}{x})^{2} \\
&=&(2x)^{2}+\frac{1}{2}+(\frac{1}{x})^{2} \\
trick\ here &\rightarrow &=(2x)^{2}+2(2x)(\frac{1}{8x})+(\frac{1}{8x})^{2} \\
&=&(2x+\frac{1}{8x})^{2}.
\end{eqnarray*}
EDIT:
Assume that $f(x)$ is $ax^{2}-b\ln x,$ and ask whether $1+(f^{\prime
}(x))^{2}$ is a perfect square.
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&1+(2ax-\frac{b}{x})^{2} \\
&=&1+(2ax)^{2}-2(2ax)(\frac{b}{x})+(\frac{b}{x})^{2} \\
&=&(2ax)^{2}+(1-4ab)+(\frac{b}{x})^{2}
\end{eqnarray*}
this is a perfect square if
\begin{equation*}
1-4ab=4ab
\end{equation*}
and in this case
\begin{eqnarray*}
1-4ab &=&4ab \\
&=&2(2ax)(\frac{b}{x})
\end{eqnarray*}
then
\begin{eqnarray*}
1+\left( f^{\prime }(x)\right) ^{2} &=&(2ax)^{2}+2(2ax)(\frac{b}{x})+(\frac{b%
}{x})^{2} \\
&=&(2ax+\frac{b}{x})^{2}.
\end{eqnarray*}
So $ab=\frac{1}{8}.$ That is why, probably for your case $a=1$ then $b$
should be $\frac{1}{8}.$ But it is also possible that $ab=\frac{1}{8}$
for other combinations. $(a=\frac{1}{2},\ b=\frac{1}{4},\ ....)$
If you browse some textbook (for calculus courses), you will find that each
related exercise is constructed in this way. I mean each time $(1+\left(
f^{\prime }(x)\right) ^{2})$ is a perfect square!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Problem using trigonometric substitution: Domain of $\theta$ I'm studying calculus from Rogawski's Calculus. In trigonometric substitution $x=a\sec \theta$ he made a note:
In the substitution $x = a \sec θ$ , we choose $0\le θ \le π$
2 if $x \ge a$ and $π \le θ < \frac{3π}2$ if
$x \le −a$. With these choices, $a \tan \theta$ is the
positive square root $\sqrt{x^2 − a^2}$.
When I work on the integral:
$$\int \frac {\mathrm{d}x}{x\sqrt{x^2-9}}$$
Using the substitution of $x=3\sec\theta$ with the domain of $\theta$ shown above, the integration will be :
$$\int \frac {dx}{x\sqrt{x^2-9}}= \int \frac {3\sec\theta\tan\theta d\theta}{(3\sec\theta)\sqrt{9\sec^2-9}}=\int \frac {\tan\theta d\theta}{3\sqrt{\tan^2 \theta}}$$
$$= \int \frac{d\theta}{3}= \frac\theta 3+ \mathrm{C}$$
$$\int \frac {dx}{x\sqrt{x^2-9}}= \frac 13 \sec^{-1}\left(\frac x3\right)+ \mathrm{C}$$
Which is very wrong in the negative part of the domain of $x$ as shown in this graph:
Notice that the slope of the blue function in the negative domain should be positive not negative!.
The problem of this substitution is $2$ things:
$1)$ The domain of $\theta$ is chosen so that inverse cannot be done, since $\theta =\sec^{-1}x\notin (\pi,\frac{3\pi}2) $ which is the domain chosen for $\theta $.
$2)$ Depending on first problem, we should choose $\theta \in (0,\pi)-\{\frac\pi 2\}$ which makes $\sqrt{\tan^2 \theta}=|\tan \theta|$. The integral then should be re-written as a piecewise function:
$$\int \frac {dx}{x\sqrt{x^2-9}}= \int \frac {3\sec\theta\tan\theta d\theta}{(3\sec\theta)\sqrt{9\sec^2-9}}$$
$$\int \frac {\tan\theta d\theta}{3\sqrt{\tan^2 \theta}} = \begin{cases} = \int \frac {d\theta}{3} = \frac 13 \sec^{-1}(\frac x3)+ C & \text{if $\theta \in (0,\frac \pi 2)$} \equiv x>3 \\= \int \frac {-d\theta}{3} = \frac {-1}3 \sec^{-1}(\frac x3)+ C & \text{if $\theta \in (\frac \pi 2,\pi)$}\equiv x<-3 \end{cases}$$
My questions are:
Is this thinking right ? there is mistake to choose the domain of $\theta$ as mentioned in the book ?
Mathematica gives me the answer of $-\dfrac {1}{3} \tan^{-1} \left(\dfrac{3}{\sqrt{x^2-9}}\right) +\mathrm{C}$ which is right when graphed. But how to construct this integral ?
Thanks for help.
| To get to the answer Mathematica gives, you can let$$ u = \sqrt{x^2 -9}.$$ You should get $$ \frac{1}{3} \arctan \left( \frac{\sqrt{x^2-9}}{3} \right) + \text{constant} $$ and use the fact that $$\arctan(\frac{1}{x}) = \frac{\pi}{2} - \arctan(x)~~~,x\gt 0$$ and
$$\arctan(\frac{1}{x}) = -\frac{\pi}{2} - \arctan(x)~~~,x\lt 0 $$
| {
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Proof by induction that $3^{2n} + 7$ is divisible by $4$ Demonstrate by induction: $3^{2n} + 7 = 4k$ is true, for any $n\in \mathbb N$. I need to demonstrate this using the induction principle.
So far I have:
$n = 1$
$$3^{2\cdot 1} + 7 = 4\cdot k $$
$$9 + 7 = 4k$$
$$16 = 4k$$
$$k = 4$$
So it checks for $n=1$.
$n = h$
$$3^{2\cdot h} + 7 = 4\cdot k$$
$n = h +1$
$$3^{2\cdot (h + 1)} + 7 = 4\cdot k'$$
(I use $k'$ to note that it's not the same $k$ as in $n = h$)
And I don't know how to continue.
| Observe that $3^{2n} = 9^n = (9^n-1^n) + 1 = 8\cdot(...)+1=0+1 =1\pmod 4$, and $7 = 3 \pmod 4$, hence $3^{2n} + 7 = 1+3 = 4 = 0 \pmod 4$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $(\frac{n}{k})^k\leq{{n}\choose{k}}\leq\frac{n^k}{k!}$? How to prove that $(\frac{n}{k})^k\leq{{n}\choose{k}}\leq\frac{n^k}{k!}$?
I can only manage to see the second inequality, could any one give a hint about the first one?
| It is easy to prove $$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}$$ if we write it as $$\binom{n}{k}=\frac{n}{k}\binom{n-1}{k-1}=\\ \frac{n(n-1)}{k(k-1)}\binom{n-2}{k-2}=\\\frac{n(n-1)(n-2)}{k(k-1)(k-2)}\binom{n-3}{k-3}=\\..\\\frac{n(n-1)(n-2)...(n-(k-1))}{k(k-1)(k-2)...(k-(k-1)))}\binom{n-k}{k-k}=\\ \frac{n(n-1)(n-2)...(n-(k-1))}{k(k-1)(k-2)...(k-(k-1)))}\binom{n-k}{0}\\=\frac{n(n-1)(n-2)...(n-(k-1))}{k(k-1)(k-2)...(k-(k-1)))} $$so $$\frac{n}{k}\frac{n}{k}...\frac{n}{k}\leq \frac{n(n-1)(n-2)...(n-(k-1))}{k(k-1)(k-2)...(k-(k-1)))}\leq\frac{n(n)(n)...(n)}{k(k-1)(k-2)...(k-(k-1)))}\\ \frac{n^k}{k^k}\leq \frac{n(n-1)(n-2)...(n-(k-1))}{k(k-1)(k-2)...(k-(k-1)))}\leq \frac{n^k}{k!} $$
| {
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If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then.. If $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ then
(A) $a=c$
(B) either $a=c$ or $a+b+c+d=0$
(C) $a+b+c+d=0$
(D) $a=c$ and $b=d$
I solved $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ and got $a(a+b+d)=c(c+b+d)$ and so I thought that (A) is the correct option. But the correct answer is (B).
I'm how $\frac{a+b}{b+c}=\frac{c+d}{d+a}$ if $a+b+c+d=0$. Please help.
| You made a small error when you solved the equation. You should have gotten
$$a(a+b+d)=c(c+b+d)$$
Note the $c$ in $c+b+d$ on the right. Can you take it from there?
Added after the OP's edit: Now write the above as
$$a((a+b+c+d)-c)=c((a+b+c+d)-a)$$
which becomes
$$(a-c)(a+b+c+d)=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Showing a basis for a polynomial I am having trouble with this basic basis problem. Need to show that $\{z^4,z^4-z^3,z^4-z^3+z^2,z^4-z^3+z^2-z,z^3-1\}$ is basis for $P_4$. I figured out that it is linearly independent but having problem show that it spans $P_4$
| From $z^4$ and $z^4-z^3$, you get $z^3$ by a linear combination.
From $z^4-z^3+z^2$, you get $z^2$ since you have $z^4$ and $z^3$.
From $z^4-z^3+z^2-z$, you get $z$ since you have $z^4$, $z^3$, $z^2$.
From $z^3-1$, you get $1$ since you have $z^3$.
So, $z^4$, $z^3$, $z^2$, $z$, $1$ are all in the subspace generated by $z^4$, $z^4-z^3$, $z^4-z^3+z^2$, $z^4-z^3+z^2-z$, $z^3-1$. Therefore, this subspace must be the whole space $P_4$.
More systematically, write the given polynomials with respect to the basis $z^4$, $z^3$, $z^2$, $z$, $1$; you'll get a triangular matrix that is invertible because it has $\pm1$ in the diagonal.
| {
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"source": "stackexchange",
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Integration of $\int \frac{(1 + x)\sin x}{(x^2 +2 x)\cos^2 x-(1 + x)\sin2x}dx$ The integral is $$\int \dfrac{(1 + x)\sin x}{(x^2 + 2x)\cos^2 x-(1 + x)\sin2x}dx.$$I've tried the problem by first multiplying both the numerator and denominator by $\sec^2 x$ but couldn't do justice. Can anyone help, please?
| Let $$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x)\cos^2 x-(1+x)\sin 2x}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x+1)\cos^2 x-(1+x)\sin 2x-\cos^2 x}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x\right]^2-2(x+2)\sin x\cdot \cos x-(1-\sin^2 x)}dx$$
So $$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x\right]^2-2(x+1)\sin x\cdot \cos x+\sin^2 x-1}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x-\sin x\right]^2-1^2}dx$$
Now Let $(x+1)\cos x-\sin x = t\;,$ Then $(x+1)\sin xdx = -dt$
So Integral $$\displaystyle I = -\int\frac{1}{t^2-1}dt = -\frac{1}{2}\int\left[\frac{1}{t-1}-\frac{1}{t+1}\right]dt$$
So we get $$\displaystyle I = \frac{1}{2}\left[\ln|t+1|-\ln|t-1|\right]+\mathcal{C} = \frac{1}{2}\ln\left|\frac{t+1}{t-1}\right|+\mathcal{C}$$
So we get $$\displaystyle I = \frac{1}{2}\ln \left|\frac{(x+1)\cos x-\sin x+1}{(x+1)\cos x-\sin x-1}\right|+\mathcal{C}$$
| {
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Proof that $\sqrt6 - \sqrt2 - \sqrt3$ is irrational. I want to prove that: $$\sqrt6 - \sqrt2 - \sqrt3$$ is irrational. I have tried using squares, the $p/q$ definition of rationality and the facts that
1)rational$\times$ irrational=irrational (unless rational=0),
2)rational$+$irrational=irrational.
However, I haven't been able to reach some conclusion. Things seem harder than when you have two square roots. Any help would be appreciated!
| Let us assume $\sqrt{6}-\sqrt{3}-\sqrt{2}$ to be a rational number.
Then from the definition of rational no it can be expressed as :
$\sqrt{6}-\sqrt{3}-\sqrt{2}=\frac{p}{q}$, where p and q are co-primes and $q\ne0.$
Squaring both sides we get:
$11-2\sqrt{18}-2\sqrt{12}+2\sqrt{6}=\frac{p^2}{q^2}$
$2\sqrt{6}-6\sqrt{2}-4\sqrt{3}=\frac{p^2-11q^2}{q^2}$
To prove L.H.S is irrational:
According to our assumption,
$ \sqrt{6}-\sqrt{3}-\sqrt{2}$ is a rational no.
$ \implies 2\sqrt{6}-2\sqrt{3}-2\sqrt{2}$ is rational no.
Now, $2\sqrt{6}-6\sqrt{2}-4\sqrt{3}= (2\sqrt{6}-2\sqrt{2}-2\sqrt{3})-(4\sqrt{2}+2\sqrt{3})$
You can easily prove that $(4\sqrt{2}+2\sqrt{3})$ is an irrational no.
So, $2\sqrt{6}-6\sqrt{2}-4\sqrt{3}= (2\sqrt{6}-2\sqrt{2}-2\sqrt{3})-(4\sqrt{2}+2\sqrt{3})$= a rational no.(from assumption)- an irrational no.
Now the L.H.S is an irrational no whereas the R.H.S is a rational no. which clearly is a contradiction.
So this contradicts our assumption that $\sqrt{6}-\sqrt{3}-\sqrt{2}$ is a rational no.
Hence it follows $\sqrt{6}-\sqrt{3}-\sqrt{2}$ is irrational.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Infinitely nested radical problem? I became interested in this nested radical from another question and thought I would have a go at trying to come up with a formula for it. It is $$G(0)=\sqrt{1+\sqrt{2+\sqrt{4+\sqrt{8+\cdots}}}}.$$ This can be written as a recurrence relation when evaluated at $x=0$: $$G(x)^2=2^x+G(x+1).$$ Due to Ramanujan, we know that for some function $F(x)$ such that $$F(x)^2=ax+(a+n)^2+xF(x+n)$$ we have $$F(x)=a+n+x.$$ We can put $G(x)$ and $F(x)$ into equatable forms by making $x=1, n=1, a=\frac{\sqrt{13}-3}{2}.$ Summing these and correcting for $G(0)$ we should have $$G(0)=\sqrt{1+\frac{\sqrt{13}-3}{2}+1+1}=\sqrt{\frac{\sqrt{13}+3}{2}}.$$ this gives $1.817354021023971.$ However the correct value for $G(0)$ is around $1.78316580926410.$ What is the error in my reasoning?
| Your mistake was assuming that $F(x)$ and $G(x)$ are the same for any $x$ simply because they are the same for $x=1$:
$$G(1)=\sqrt{2^1+F(1+1)}=\sqrt{2+F(2)} \\
F(1)=\sqrt{\left(\frac{\sqrt{13}-3}{2}\right)(1)+\left(\left(\frac{\sqrt{13}-3}{2}\right)+1\right)^2+(1)F(1+1)}=\sqrt{2+F(2)}$$
But look at $x=2$:
$$G(2)=\sqrt{2^2+F(2+1)}=\sqrt{4+F(3)} \\
F(2)=\sqrt{\left(\frac{\sqrt{13}-3}{2}\right)(2)+\left(\left(\frac{\sqrt{13}-3}{2}\right)+1\right)^2+(2)F(2+1)} \approx \sqrt{2.3028+2F(3)}$$
So if you replace $G(x+1)$ with $F(x+1)$ in your definition of $G$ (as you did to solve), then $G(0)$ actually looks like this when expanded:
$$G(0)=\sqrt{\frac{\sqrt{13}+3}{2}} \approx\sqrt{1+\sqrt{2+\sqrt{2.3028+2\sqrt{2.6055+\dots}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Boundary Value Problem $y''+uy=0$ Consider the boundary value problem
$$y''+uy=0 \qquad y(0)=y(\pi/2)=0$$
(a) For what values of $u$ does this problem have the trivial solution $y \equiv 0$?
(b) For what values of $u$ does the problem have nontrivial solutions?
I would really appreciate any help on this one.
| Here is your error: you say "What i had done was sub into the equation $r$, such that $y'' + uy$ becomes: $r^2 + u = 0 r= \pm (-u)^{1/2}$ Then since the root is a complex number: $r= (u)^{1/2}i$".
The root will be a complex number only if u is positive and you do not know that.
Rather, consider the separate cases.
1) $u= 0$. In this case, the differential equation is $y''= 0$ which can be solved by integrating twice: $y'= A$, a constant, so $y(x)= Ax+ B$. Now, $y(0)= B= 0$ and $y(\frac{\pi}{2})= A\cdot\frac{\pi}{2}= 0$ so $A= 0$. Since $A$ and $B$ are both $0$, $y(x)$ is identically $0$, the "trivial" solution.
2) $u< 0$. Let $u= -a^2$ where $a$ can be any non-zero number. Then the equation is $y''- a^2y= 0$ which has characteristic equation $r^2- a^2= 0$. That has real roots $a$ and $-a$. The general solution to the equation is $y(x)= Ae^{ax}+ Be^{-ax}$. $y(0)= A+ B= 0$ and $y(\frac{\pi}{2})= Ae^{a\cdot\frac{\pi}{2}}+ Be^{-a\cdot\frac{\pi}{2}}$. From $A+ B= 0, B= -A$ so that so that $A- B= 0$ and $A= B$. Then $Ae^{a\frac{\pi}{2}}+ Be^{-a\frac{\pi}{2}}= Ae^{a\frac{\pi}{2}}- Ae^{-a\frac{\pi}{2}}= A(e^{a\frac{\pi}{2}}- e^{-a\frac{\pi}{2}})= 0$. Since $a$ is not $0$, $e^{a\frac{\pi}{2}}- e^{-a\frac{\pi}{2}}$ is not $0$ and we must have $A= 0$ and $B= -A= 0$. Again $A$ and $B$ are both $0$ so the $y(x)$ is identically $0$, the "trivial" solution.
3) $u> 0$. Let $u= a^2$ where a can be any non-zero number. Then the equation is $y''+ ay^2= 0$ which has characteristic equation $r^2+ a^2= 0$. That has imaginary roots $ai$ and $-ai$. The general solution to the equation is $y(x)= A \cos(ax)+ B \sin(ax)$. $y(0)= A= 0$ and $y(\frac{\pi}{2})= B \sin(a\frac{\pi}{2})= 0$. Now, we must have either $B= 0$ so that we have the "trivial" solution again or $\sin(a\frac{\pi}{2})$ which happens if and only if $a$ is an odd number: $a= 2n+ 1$ for some integer $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.
1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.
|
ie. $(1 - x)^2(1 + 2x + 3x^2 + ...) = 1$
| {
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Solve $\begin{cases} x + y + z = 2 \\ 2xy - z^2 = 4 \\ \end{cases} $ for x, y, z. It came to my mind to rewrite the expression above as
$$\begin{cases}
x + y = 2 - z \\
2xy = (2 - z)^2 + 4z \\
\end{cases}
$$
and see if there any restrictions on the values of the variables occur.
What I can see is that $(2 - z)^2 + 4z \gt 0$ for all $z$ (since $z^2 > -4$) but when I express $y$ in terms of $z$ and $x$
$$\begin{cases}
y = 2 - z - x\\
x(x - (2 - z)) = (2 - z)^2 + 4z \\
\end{cases}
$$
I see that $x(x - (2 - z)) \gt 0$ not for all $z$. This is all I got for now.
The answer is $(2, 2, -2)$.
| $\bf{My\; Solution::}$ Given $$x+y+z = 2$$ and $$2xy-z^2 = 4$$
Now Put $z=2-(x+y)$ into $2xy-z^2=4\Rightarrow 2xy-[2-(x+y)]^2=4$
So we get $$2xy-[4+(x+y)^2-4(x+y)]=4$$
So we get $$2xy-4-x^2-y^2-2xy+4x+4y=4$$
So we get $$x^2+y^2-4x-4y+8=0$$
So we get $$(x-2)^2+(y-2)^2 = 0\;,$$ Which is possible for real no. when
$(x-2)=0$ and $(y-2) = 0$ , So we get $x=2$ and $y=2$ and $z=-2$
| {
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Evaluate the integral $\int \frac{x}{a+bx^3}\ dx$ How do I solve integral at this form
$\displaystyle\int \frac{x}{a+bx^3}\ dx$ ?
| $$
a+bx^3 = (\sqrt[3]a +x\sqrt[3]b)(\sqrt[3]a^2 - x\sqrt[3]a\sqrt[3]b + x^2\sqrt[3]b^2)
$$
So use partial fractions:
$$
\frac x {a+bx^3} = \frac C {\sqrt[3]a +x\sqrt[3]b} + \frac {Dx+E} {\sqrt[3]a^2 - x\sqrt[3]a\sqrt[3]b + x^2\sqrt[3]b^2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$ $\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$
I let, $x = \frac{1}{n}$, then as
$\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$
L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$
Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.
| By Taylor's theorem with Peano's remainder, we have
\begin{align*}
& \left(1 + \frac{1}{n}\right)^n = \exp\left(n\log\left(1 + \frac{1}{n}\right)\right)\\
= & \exp\left(n\left(\frac{1}{n} - \frac{1}{2n^2} + o\left(\frac{1}{n^2}\right)\right) \right)\\
= & \exp\left(1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right)
\end{align*}
Now expand $\exp\left(1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right)$ at $1$:
$$\exp\left(1 - \frac{1}{2n} + o\left(\frac{1}{n}\right)\right) = e + e \times \left(-\frac{1}{2n} + o\left(\frac{1}{n}\right)\right) + o\left(\frac{1}{n}\right)$$
Therefore,
$$n\left[\left(1 + \frac{1}{n}\right)^n - e\right] = -\frac{1}{2}e + o(1)\to -\frac{1}{2}e$$
as $n \to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find equation of curve ${dy \over dx}= (3x^2-a)^2$, where $a$ is a constant. Given that the curve has a stationary point at $(3,2)$, find the equation of the curve.
I managed to get the equation $y=3x^3+3ax^2+xa^2$+c. I'm not sure if I integrated correctly though. If I did how do I proceed?
| We see that $\frac{\partial y}{\partial x} = (3x^2-a)^2= 9 x^4 - 6x^2a+a^2$. Taking the anti-derivative(integration) we see that $y=\frac{36}{5}x^2 - 2x^3a+a^2x + c$ where c is a constant. We know that the stationary point is $(3,2)$ and the derivative equals zero there(definition stationary point). So $\frac{\partial y}{\partial x}(3,2)=(3\cdot 3^2-a)^2= 0$ We can easily see that $a=3\cdot 3^2=27$. Thus $y=\frac{36}{5}x^2 - 54x^3+729x + c$. Filling in the point $(3,2)$ we see $2=\frac{36}{5} 3^2 - 54 \cdot3^3+729 \cdot 3 + c=\frac{324}{5}- 486+ 2187+c=\frac{8181}{5}+c$.
Thus $c=2-\frac{8181}{5}=\frac{-8171}{5}=-1634.2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove an equality If $a+b+c=0$
prove that
$\frac {(a^4 +b^4 +c^4)}{2}=\frac {(a^2+b^2+c^2)}{2^2}^2$
I have expanded the right side and have got this far:
$a^4+b^4+c^4+2(a^2b^2+a^2c^2+b^2c^2)$
I need $a^2=b^2=c^2$ to prove the equality.
Any ideas?
| Following is a more systematic approach uses Newton's identities.
Let $e_1, e_2, e_3$ be the elementary symmetric polynomials associated with $a, b, c,$ i.e.
$$\begin{cases}
e_1 &= a + b + c\\
e_2 &= ab + bc + ca\\
e_3 &= abc
\end{cases}
\quad\iff\quad (x-a)(x-b)(x-c) = x^3 - e_1 x^2 + e_2 x - e_3
$$
and let $p_k = a^k + b^k + c^k, k \in \mathbb{Z}_{+}$ be the corresponding power sums.
We are given $e_1 = a + b + c = 0$. Newton identities tell us
$$\require{cancel}
\newcommand{\xxx}[1]{\color{red}{\cancelto{0}{\color{gray}{#1}}}}
\begin{array}{rlclr}
p_1 -\xxx{e_1} &= 0 &\implies& p_1 = 0 \\
p_2 -\xxx{e_1} p_1 + 2 e_2 &= 0 &\implies& p_2 = -2e_2 & (*1a) \\
p_3 -\xxx{e_1} p_2 + e_2 \xxx{p_1} - 3 e_3 &= 0 &\implies& p_3 = 3 e_3\\
p_4 -\xxx{e_1} p_3 + e_2 p_2 - e_3\xxx{p_1} &= 0 &\implies& p_4 = -e_2 p_2 & (*1b)
\end{array}
$$
Combine $(*1a)$ and $(*1b)$, we have
$$a^4 + b^4 + c^4 = p_4 = -e_2 p_2 = \frac12 p_2^2 = \frac12 (a^2 + b^2 + c^2)^2\tag{*2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
find x in $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
Which one satisfies the equation $\sqrt[3]{6+\sqrt x} + \sqrt[3]{6-\sqrt x} = \sqrt[3] {3}$
(A)$27$ (B)$32$ (C)$45$ (D)$52$ (E)$63$
let $a = 6+\sqrt x , b=6-\sqrt x$
cube each side
\begin{align}
(\sqrt[3]a + \sqrt[3]b)^3 &= (\sqrt[3]3)^3 \\
(\sqrt[3]{a^2} + 2\sqrt[3]{ab} + \sqrt[3]{b^2})(\sqrt[3]a + \sqrt[3]b) &= 3 \\
\sqrt[3]{a^3} + \sqrt[3]{3a^2b} + \sqrt[3]{3ab^2} + \sqrt[3]{b^3} &= 3 \\
a + b + 3\sqrt[3]{a^2b} + 3\sqrt[3]{ab^2} &= 3
\end{align}
There's still had cube root, how do I remove it?
| Start with your idea $a=6+\sqrt{x}$ and $b=6-\sqrt{x}$. You have $$ab=36-x \text{, } \sqrt[3]{a} + \sqrt[3]{b} = \sqrt[3]{3} \text{ and } a+b=12$$ Raise the middle equation to power $3$ you get $$a+b +3\sqrt[3]{ab}(\sqrt[3]{a} + \sqrt[3]{b})=3$$ and using the initial assumption $$12 +3\sqrt[3]{ab}\sqrt[3]{3}=3 \text{ or } \sqrt[3]{ab}\sqrt[3]{3}=-3$$ and finally $$ab=-9$$ which allows to conclude to $x=45$ as $ab=36-x$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Evaluating the indefinite integral $\int\sqrt{16-9x^2}\,dx$ I need to solve the integral below, but I just can't figure how.
$$\int \sqrt{16-9x^2}\,dx$$
I have tried to replace $9x^2$ with $16\sin^2\theta$. I get to a point where I have the function below. Please let me know whether I'm on the right track, and please explain to me how to finish it...
$$
\frac {16}3 \int \cos^2\theta \,d\theta\
$$
| $$\int\sqrt{16-9x^2}dx=$$
(Substitute $x=\frac{4\sin(u)}{3}$ and $dx=\frac{4\cos(u)}{3}du$ then $\sqrt{16-9x^2}=\sqrt{16-16\sin^2(u)}=4\cos(u)$ and $u=\sin^{-1}\left(\frac{3x}{4}\right)$):
$$\frac{4}{3}\int 4\cos^2(u)du=$$
$$\frac{16}{3}\int \cos^2(u)du=$$
$$\frac{16}{3}\int \left(\frac{1}{2}\cos(2u)+\frac{1}{2}\right)du=$$
$$\frac{8}{3}\int \cos(2u)du+\frac{8}{3}\int 1 du=$$
(Substitute $s=2u$ and $ds=2du$):
$$\frac{4}{3}\int \cos(s)ds+\frac{8}{3}\int 1 du=$$
$$\frac{4\sin(s)}{3}+\frac{8}{3}\int 1 du=$$
$$\frac{4\sin(s)}{3}+\frac{8u}{3}+C=$$
$$\frac{4\sin(2u)}{3}+\frac{8u}{3}+C=$$
$$\frac{8u}{3}+\frac{8}{3}\sin(u)\cos(u)+C=$$
$$\frac{8u}{3}+\frac{8}{3}\sin(u)\sqrt{1-\sin^2(u)}+C=$$
$$\frac{8\left(\sin^{-1}\left(\frac{3x}{4}\right)\right)}{3}+\frac{8}{3}\sin\left(\left(\sin^{-1}\left(\frac{3x}{4}\right)\right)\right)\sqrt{1-\sin^2\left(\left(\sin^{-1}\left(\frac{3x}{4}\right)\right)\right)}+C=$$
$$\frac{1}{2}\sqrt{16-9x^2}x+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+C$$
So:
$$\int\sqrt{16-9x^2}dx=\frac{1}{2}\sqrt{16-9x^2}x+\frac{8}{3}\sin^{-1}\left(\frac{3x}{4}\right)+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 2
} |
Solve for $x$ - Logarithm Equation $\ln x+\ln(x+1)=\ln 2$ My attempt:
$\ln x(x+1)=\ln 2$
$e^{\ln x(x+1)}=e^{\ln 2}$
$x(x+1)=2$
$x^2+x-2=0$
$(x-1)(x+2)=0$
therefore $x=1, -2$
| We have, $$\ln x+\ln x(x+1)=\ln 2$$ $$\implies \ln(x(x+1))=\ln 2$$
$$\implies \frac{\ln(x(x+1))}{\ln 2}=1$$ $$\implies \log_2(x(x+1))=1$$ $$\implies x(x+1)=2$$ $$\implies x^2+x-2=0$$ now, solving above quadratic equation for $x$ as follows $$\implies x=\frac{-1\pm\sqrt{(1)^2-4(1)(-2)}}{2(1)}$$ $$\implies x=\frac{-1\pm\sqrt{9}}{2}$$ $$\implies x=\frac{-1\pm 3}{2}$$ $$\implies x=\frac{-1+3}{2}=\color{}{1}$$ &$$\implies x=\frac{-1-3}{2}=\color{}{-2}$$
Edit: Since log is defined for positive number i.e. $x>0$ hence we have $x=\color{blue}{1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
how to solve $\displaystyle \frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geq\frac{3}{4}$
$a,b,c>0,a+b+c=3,$
prove that:
$$\frac{a^3}{a^2+2b+c}+\frac{b^3}{b^2+2c+a}+\frac{c^3}{c^2+2a+b}\geq\frac{3}{4}$$
| With $\sum $ denoting cyclic sum, using Cauchy Schwarz inequality:
$$\sum \frac{a^4}{a(a^2+2b+c)} \ge \frac{(a^2+b^2+c^2)^2}{\sum a^3+3\sum ab}$$
So it is enough to show
$$4(a^2+b^2+c^2)^2 \ge (a+b+c)\sum a^3+(a+b+c)^2\sum ab$$
$$\iff 3\sum a^4 +6\sum a^2b^2 \ge 2\sum ab(a^2+b^2)+5abc\sum a$$
which follows from Schur $\sum a^4 + abc\sum a\ge \sum ab(a^2+b^2)$ and the AM-GMs $\sum a^2b^2 \ge \sum a^2bc$ and $2\sum a^4 \ge \sum ab(a^2+b^2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1379937",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Trigonometry problem Okay..this one simple problem but I am really stuck and have no idea how to start..
$\cos(a-b)+\cos(b-c)+\cos(c-a)=-\frac32$ we need to prove $\cos(a)+\cos(b)+\cos(c)=\sin(a)+\sin(b)+\sin(c)=0 $
| $\cos(a−b)+\cos(b−c)+\cos(c−a)=−3/2$
$\cos a \cos b+\sin a \sin b+\cos b \cos c+\sin b \sin c+\cos c \cos a+\sin c \sin a=-3/2 $
$\frac{1}{2}\cos a(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2a+\frac{1}{2}\sin a(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 a+
\frac{1}{2}\cos b(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2b+\frac{1}{2}\sin b(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 b+
\frac{1}{2}\cos c(\cos a + \cos b +\cos c)-\frac{1}{2}\cos^2c+\frac{1}{2}\sin c(\sin a+\sin b+\sin c) - \frac{1}{2}\sin^2 c=-\frac{3}{2}$
$\cos a(\cos a + \cos b +\cos c)+\sin a(\sin a+\sin b+\sin c) - (sin^2 a+\cos^2a)+
\cos b(\cos a + \cos b +\cos c)+\sin b(\sin a+\sin b+\sin c) - (sin^2 b+\cos^2b)+
\cos c(\cos a + \cos b +\cos c)+\sin a(\sin a+\sin b+\sin c) - (sin^2 c+\cos^2c)=-3$
$(\cos a + \cos b +\cos c)^2+(\sin a+\sin b+\sin c)^2=0$
then $\cos a + \cos b +\cos c=0$ and $\sin a+\sin b+\sin c=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1382047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove by induction: $\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}$
Prove
$$\frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{n}{(n+1)!}=\frac{n!-1}{n!}.$$
My problem with this is that it doesn't hold for the base case: $n=1$. This question is from the book "Abstract Algebra" by Charles Pinter (Page 212, exercise 7). When I go about proving it, I can't reach my goal which is $\frac{(n+1)!-1}{(n+1)!}=1-\frac{1}{(n+1)!}$, for $n=n+1$.
| Here's an alternate proof, with generating functions instead of induction.
Note that
$$
\sum_{i=1}^n \frac{i}{(i+1)!}
$$
is the sum of the first $n$ terms of the series
\begin{align*}
\sum_{i=1}^\infty \frac{i x^i}{(i+1)!}
&= x \sum_{i=1}^\infty \frac{i x^{i-1}}{(i+1)!} \\
&= x \frac{d}{dx} \sum_{i=1}^\infty \frac{x^i}{(i+1)!} \\
&= x \frac{d}{dx} \frac{1}{x} \sum_{i=1}^\infty \frac{x^{i+1}}{(i+1)!} \\
&= x \frac{d}{dx} \frac{1}{x} \left[ e^x - x - 1\right] \\
&= x \frac{d}{dx} \left[ \frac{e^x}{x} - 1 - \frac{1}{x} \right] \\
&= x \left[ \frac{xe^x - e^x}{x^2} + \frac{1}{x^2} \right] \\
&= \frac{xe^x - e^x + 1}{x}.
\end{align*}
To get the generating function for the sum of the first $n$ terms,
we then multiply by $\frac{1}{1 - x}$:
\begin{align*}
\frac{xe^x - e^x + 1}{x(1-x)}
&= \frac{(1 - e^x)(1 - x) + x}{x(1-x)} \\
&= \frac{1}{1-x} - \frac{e^x - 1}{x} \\
\end{align*}
The $n$th term of $\frac{1}{1-x}$ is $1$ and the $n$th term of $\frac{e^x - 1}{x}$ is $\frac{1}{(n+1)!}$, so the result is
$$
\boxed{1 - \frac{1}{(n+1)!}}
$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Is it possible to factor a quadratic equation when $a$, $b$, and $c$ are all equal? I have the equation $4x^2+4x+4$ to factor. I know that need to start with
$$(2x \quad )(2x \quad )$$
to make $4^2$, but I can't seem to factor the rest of the way. What should I do?
| We have $$4x^2+4x+4=4(x^2+x+1)$$ Let's consider $$f(x)=x^2+x+1$$ $$x=\frac{-1\pm\sqrt{1^2-4(1)(1)}}{2}=\frac{-1\pm i\sqrt3}{2}$$ Thus we have $$4x^2+4x+4=4\left(x-\frac{-1+i\sqrt3}{2}\right)\left(x-\frac{-1-i\sqrt3}{2}\right)$$ $$=4\left(x+\frac{1-i\sqrt3}{2}\right)\left(x+\frac{1+i\sqrt3}{2}\right)$$ $$=\left(2x+1-i\sqrt3\right)\left(2x+1+i\sqrt3\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1383691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the general formula for $a_{n+1}=2^n a_n +4$, where $a_1=1$. Problem:
*
*Find the general formula for $a_{n+1}=2^n a_n +4$, where $a_1=1$.
*Find the sum of its first $2n$ terms with odd subscript.
My effort:
*
*It seems to me that $a_{n+1} / 2^{(n+1)^2/2}=\dfrac{1}{\sqrt{2}}a_n/2^{n^2/2} +4/ 2^{(n+1)^2/2}$, which is $b_{n+1}=\dfrac{1}{2^{1/2}} b_{n} + \dfrac{1}{2^{(n+3)(n-1)/2}}$, where $b_n=a_n/2^{n^2/2}$. But it seems hard to deal with the last term.
*The first ten $a_n$ is {1, 6, 28, 228, 3652, 116868, 7479556, 957383172, 245090092036, 125486127122436}, which follows no immediate rule.
*Write the sequence in binary form, I find it {1, 110, 11100, 11100100, 111001000100, ...} which is generally in a 1 2*0 1 3*0 1 4*n ... pattern (apart from the first few). So I highly suspect that there is not closed form expression. But how to prove this?
| Here is a solution by substitution:
\begin{align}
a_1&=1\\
a_2&=2^1+4\\
a_3&=2^{1+2}+4(1+2^2)\\
a_{1+3}&=2^{1+2+3}+4(1+2^3+2^{3+2})\\
a_{1+4}&=2^{1+2+3+4}+4(1+2^4+2^{4+3}+2^{4+3+2})\\
a_{1+5}&=2^{1+2+3+4+5}+4(1+2^5+2^{5+4}+2^{5+4+3}+2^{5+4+3+2})\\
a_{1+6}&=2^{1+2+3+4+5+6}+4(1+2^6+2^{6+5}+2^{6+5+4}+2^{6+5+4+3}+2^{6+5+4+3+2})\\
...\\
a_{1+n}&=2^{\sum_{j=1}^nj}+4(1+\sum_{j=1}^{n-1}2^{\sum_{k=j+1}^nk})
\end{align}
I should be simplifying this last expression ... but I could not ... :-)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Harmonic progression based question
If $a,b,c,d$ are distinct positive numbers in harmonic progression, then
(A) $a+b>c+d$
(B) $a+c>b+d$
(C) $a+d>b+c$
(D) none of these
I tried $\frac{1}{a},\frac{1}{b},\frac{1}{c},\frac{1}{d}$ are in AP so either $\frac{1}{a}>\frac{1}{b}>\frac{1}{c}>\frac{1}{d}$ or $\frac{1}{a}<\frac{1}{b}<\frac{1}{c}<\frac{1}{d}$.That means either $a>b>c>d$ or $a<b<c<d$.But i could not judge correct answer based on this.Is my approach not correct?What is correct answer?Can someone help me in solving this question?
| The correct option is $(C)$.
We have
$$\frac 1a+\frac 1c=\frac 2b\quad\text{and}\quad \frac 1b+\frac 1d=\frac 2c$$$$\Rightarrow c=\frac{ab}{2a-b}\quad\text{and}\quad d=\frac{ab}{3a-2b}$$
Here, since $c,d$ are positive, we have $2a-b\gt 0,3a-2b\gt 0$.
By the way, we have
$$\begin{align}a+d-(b+c)&=a+\frac{ab}{3a-2b}-b-\frac{ab}{2a-b}\\&=\frac{(a-b)(3a-2b)(2a-b)+ab(2a-b)-ab(3a-2b)}{(3a-2b)(2a-b)}\\&=\frac{(a-b)(3a-2b)(2a-b)-ab(a-b)}{(3a-2b)(2a-b)}\\&=\frac{(a-b)((3a-2b)(2a-b)-ab)}{(3a-2b)(2a-b)}\\&=\frac{(a-b)\cdot 2(3a^2-4ab+b^2)}{(3a-2b)(2a-b)}\\&=\frac{2 (a-b)^2 (3 a-b)}{(3a-2b)(2a-b)}\end{align}$$
This is positive because $3a-2b\gt 0\Rightarrow 3a-b\gt 0$.
Hence, $a+d\gt b+c$ holds.
(By the way, options $(A),(B)$ are wrong. Take $a=2,b=5/2$. So, the correct option is $(C)$ only.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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determine type of singularities and compute residue of a function Determine the type of singularities and residue of $$\frac{1}{\sin^2(z)}$$
For this problem, this is the way I approach this:
we have :
$$\sin(z) = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$$
$$= z\left(1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots\right)$$
Let $h(z) = 1 - \frac{z^2}{3!} + \frac{z^4}{5!} - \cdots$. Then, we have:
$$\sin(z) = z\cdot h(z)$$
$$\sin^2(z) = z^2 \cdot h^2(z)$$
Therefore,
$$\frac{1}{\sin^2(z)} = \frac{1}{z^2} \cdot \frac{1}{h^2(z)}$$
Thus, it has a simple pole at $z_0 = 0$.
But, I don't know how to calculate the residue of this function. Can someone please how me how to compute its residue.
| You are on the right track. For the pole at $z=0$, let's write
$$\begin{align}
\frac{1}{\sin^2 z}&=\frac{1}{\left(z-\frac16 z^3++O(z^5)\right)^2}\\\\
&=\frac{1}{z^2\left(1-\frac16 z^2++O(z^4)\right)^2}\\\\
&=\frac{\left(1+\frac16 z^2+O(z^4)\right)^2}{z^2}\\\\
&=\frac{1}{z^2}+\frac13+O(z^2)
\end{align}$$
Thus, we find the singularity is a pole of order $2$.
We can perform a similar expansion for around any of the zeros of the sine function and see that there are singularities of $\csc^2 z$ that are poles of order $2$ at $z=n\pi$ for all integer values of $n$.
Another way to see this is to recall the infinite product representation of the sine function. Then,
$$\sin z=z\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)$$
implies that
$$\csc^2 z=\frac{1}{z^2 \prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2\pi^2}\right)^2}$$
which clearly shows the second order poles at $z=n\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proof of inequality without calculus
So we are given the equation $3x+4y+xy=2012$ where $x$ and $y$ are positive integers. Prove that $x+y\geq83$.
Using calculus optimisation methods, this can be proved. However, it requires a lot of difficult calculations, and students are not expected to know calculus to answer this. How else could you prove this? You could always go through each possible value, but that would take a very long time.
| You can rearrange the equation $3x+4y+xy=2012$:
\begin{align}
3x+4y+xy\quad =\quad 2012\\ 3x+4y+xy+12\quad =\quad 2024\\ x(y+3)+4y+12\quad =\quad 2024\\ x(y+3)+4(y+3)\quad =\quad 2024\\ (y+3)(x+4)\quad =\quad 2024\\
\end{align}
And from the AM-GM inequality:
$\frac{a+b}{2}\ge\sqrt{ab}$
And using $a = x+4$ and $b = y+3$ we get:
\begin{align}
\frac{x+4+y+3}{2}&\ge\sqrt{(x+4)(y+3)}\\
\frac{x+y+7}{2}&\ge\sqrt{2024}\\
x+y+7&\ge2\sqrt{2024}\\
x+y&\ge2\sqrt{2024}-7\\
x+y&\ge83
\end{align}
Note:
\begin{align}
{(2\sqrt{2024})}^{2}-{89}^{2} &= 8096-7921>0 \\
2\sqrt{2024}&>89 \\
2\sqrt{2024}-7&>82 \\
x+y&>82\\
x+y&\ge 83 \text{ (because x, y are integers) }
\end{align}
In general we have for $a,b,c\in \mathbb{{C}^{*}}$:
\begin{align}
ax+by+cxy&=x(a+cy)+by\\&=cx\left( y+\frac { a }{ c } \right) +by\\&=cx\left( y+\frac { a }{ c } \right) +by+\frac { ab }{ c } -\frac { ab }{ c } \\ &=cx\left( y+\frac { a }{ c } \right) +b\left( y+\frac { a }{ c } \right) -\frac { ab }{ c } \\&=\left( y+\frac { a }{ c } \right) \left( b+cx \right) -\frac { ab }{ c }
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Different approaches for evaluating a series I was playing around with the Fourier series of $\cos x , \; x \in (0, \frac{\pi}{2})$ and a series came up. First of all the fourier series of $\cos x$ at the given interval is:
$$\cos x =\frac{8}{\pi} \sum_{n=1}^{\infty} \frac{n \sin 2nx}{4n^2-1}$$
Calculations are pretty straightforward. By applying Parseval we get that:
$$\sum_{n=1}^{\infty} \frac{n^2}{(4n^2-1)^2}= \frac{\pi^2}{64}$$
What other approaches can we use in order to evaluate this series? Perhaps partial decomposition? One notes that:
$$\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\left [ \frac{1}{\left ( 2n+1 \right )^2}- \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
Hence:
$$\sum_{n=1}^{\infty}\frac{n^2}{\left ( 4n^2-1 \right )^2}= \frac{1}{16}\sum_{n=1}^{\infty}\left [ \frac{1}{\left ( 2n+1 \right )^2}- \frac{1}{(2n-1)^2} \right ]+ \frac{1}{16}\sum_{n=1}^{\infty}\left [ \frac{1}{2n-1} - \frac{1}{2n+1} \right ]$$
However, we all know where the first two sums evaluate to. They are quite famous:
$\displaystyle {\color{gray} \bullet} \;\;\; \sum_{n=1}^{\infty}\frac{1}{(2n+1)^2}= \frac{\pi^2}{8}- 1$
$\displaystyle {\color{gray} \bullet} \;\;\; \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2}= \frac{\pi^2}{8}$
Hence the first bracket is evaluated to $-\frac{1}{16}$. The second bracket, is easily noted that it can be expressed in terms of digamma. However, I am having a difficult time converting to the digamma.
Any help on that point would be nice. Also I am interest in other approaches, such as contour integration.
| I would define (valid when $x>1$) $$S(x)=\sum_{n=1}^{\infty} \frac1{x^2n^2-1}=\frac1{x^2}\sum_{n=1}^{\infty}\frac{1}{n^2-\frac1{x^2}}\\
=\frac12-\frac{\pi}{2x}\cot(\frac{\pi}{x})$$
This follows from logarithmically differentiating the Weierstrass product form of th sine function $$\frac{\sin(\pi z)}{\pi z}=\prod_{n=1}^{\infty}\left(1-\frac{z^2}{n^2}\right).$$
Differentiation yields $$S'(x)=\sum_{n=1}^{\infty} \frac{-2xn^2}{(x^2n^2-1)^2}=\frac{\pi}{2x^2}\cot(\frac{\pi}{x})-\frac{\pi^2}{2x^3}\csc^2(\frac{\pi}{x})$$
and letting $x=2$ gives the desired value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\displaystyle\lim_{x\to 0}\tfrac{\sqrt{x+25}-5} {\sqrt{x+16}-4}$ \begin{eqnarray}
\\&\lim_{x\to 0}\frac{\sqrt{x+25}-5} {\sqrt{x+16}-4}
\end{eqnarray}
Undefined limit
\begin{eqnarray}
\frac{0} {0}
\end{eqnarray}
| Given $\displaystyle \lim_{x\to 0}\frac{\sqrt{x+25}-5}{\sqrt{x+16}-4} \times \frac{\sqrt{x+16}+4}{\sqrt{x+16}+4} \times \frac{\sqrt{x+25}+5}{\sqrt{x+25}+5}$
We Get $\displaystyle \lim_{x\rightarrow 0} \frac{x}{x} \times \frac{\sqrt{x+16}+4}{\sqrt{x+25}+5} = \frac{8}{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Diagonalize the matrix A (complex numbers) Diagonalize the matrix A
$A=\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}$
So, i began the problem by finding the characteristic polynomial which was
$λ^3-7λ^2-15λ-27$
using long division i got $(λ-9)(λ^2+2λ+3)$
so i used the quadratic formula and got
$λ_1=-1+i\sqrt{2}$ and $λ_2=-1-i\sqrt{2}$ and $λ_3=9$
I decided to start with $λ_1$
$A-\left(-1+i\sqrt{2}\right)I=\begin{pmatrix}2-i\sqrt{2} & 2 & 4 \\3 & 6-i\sqrt{2} & 2 \\2 & 6 & 2-i\sqrt{2}\end{pmatrix}$
Now i understand how to diagonalize when i have all numbers but once i get these $i$'s in the equation it's like my brain doesn't comprehend the steps i need to take the get the diagonal form..
I would like to think i'm supposed to start by getting the inverse of this new matrix but fail to see how to do that with $i$'s involved.
| A strange way to find the eigenvectors is to use the Cayley-Hamilton Theorem,
$$
(A-9I)(A^{2}+2A+3I) = 0.
$$
Because of this, the columns of $A^{2}+2A+3I$ are eigenvectors of $A$ with eigenvalue $9$. For example,
$$
\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}
\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}+
2\begin{pmatrix}1 & 2 & 4 \\3 & 5 & 2 \\2 & 6 & 1\end{pmatrix}+
3\begin{pmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1\end{pmatrix} \\
=\begin{pmatrix}15 & 36 & 12 \\ 22 & 43 & 24 \\22 & 40 & 21 \end{pmatrix}
+ \begin{pmatrix}2 & 4 & 8 \\ 6 & 10 & 4 \\ 4 & 12 & 2\end{pmatrix}
+ \begin{pmatrix}3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3\end{pmatrix} \\
= \begin{pmatrix}20 & 40 & 20 \\ 28 & 56 & 28 \\ 26 & 52 & 26\end{pmatrix}
$$
Similarly, the columns of the following are eigenvectors with eigenvalue $-1-\sqrt{2}i$
$$
(A-9I)(A-(-1+\sqrt{2}i)I)=A^{2}+(-8-\sqrt{2}i)A+(-9+9\sqrt{2}i)I
$$
You only need the first column:
$$
\begin{pmatrix} 15 \\ 22 \\ 22\end{pmatrix}+(-8-2\sqrt{2}i)\begin{pmatrix} 1 \\ 3 \\ 2\end{pmatrix} +
(-9+9\sqrt{2}i)\begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix} \\
=\begin{pmatrix} -2+7\sqrt{2}i \\ -2-6\sqrt{2}i \\ 6-4\sqrt{2}i\end{pmatrix}
$$
And the conjugate of this vector is going to have the conjugate of its eigenvalue. So a basis of eigenvectors is
$$
\begin{pmatrix}10 \\ 14 \\ 13\end{pmatrix},
\begin{pmatrix} -2+7\sqrt{2}i \\ -2-6\sqrt{2}i \\ +6-4\sqrt{2}i\end{pmatrix},
\begin{pmatrix} -2-7\sqrt{2}i \\ -2+6\sqrt{2}i \\ +6+4\sqrt{2}i\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1392317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Number of Interesting Quadruples
Define an ordered quadruple of integers $(a, b, c, d)$ as interesting if $1 \le a<b<c<d \le 10$, and a+d>b+c. How many interesting ordered quadruples are there?
This is a bit of trouble here actually, I am to use $a + d \gt b + c$ as a constraint.
Without any restrictions (no $a + d \gt b + c$) there are: $\binom{10}{4} = 210$ possible values for $a, b, c, d$.
We could have three cases: $a + d \gt b + c$ or $a + d < b + c$ or $a + d = b + c$.
We need to take out $a + d = b + c$ cases first: It is possible that:
$a + b = \{1 + \sum_{k=4}^{10}k, 2 + \sum_{k=5}^{10}, 3 + \sum_{k=6}^{10}, 4 + \sum_{k=7}^{10}k, ..., 7 + 10 \}$
Total (incl. overcounting): $7 + 6 + 5 + 4 + 3 + 2 + 1 = 28$ possible.
But I think I have messed the whole problem.
Hints Please!
| Here is my solution:
Restate the inequality as $b-a < d-c$ so that the range of numbers for a and b is strictly less than for d and c.
Now, there are $\binom{10}{4}$ total possible choices of the 4 different integers in the given range. $\binom{8}{2}$ of those choices are such that the pairs (a,b) and (d,c) each have a range of 2 (i.e. when (a,b) are adjacent and (c,d) are adjacent), $\binom{6}{2}$ are such that (a,b) and (c,d) each have a range of 3, $\binom{4}{2}$ where the pairs have a range of 4 and $\binom{2}{2}$ where each have a range of 5.
By subtracting these choices from our total we are left with those choices where $b-a\ne d-c$ and by symmetry exactly $1/2$ of these are of the desired type, hence our require count is: $$\frac{1}{2}\left(\dbinom{10}{4}-\dbinom{8}{2}-\dbinom{6}{2}-\dbinom{4}{2}-\dbinom{2}{2}\right)$$
This gives us:
$$\frac{1}{2}\left(210 - 28 - 15 - 6 - 1\right) = \frac{160}{2} = 80$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $|x-2|<0.001$, then $|\frac{1}{x}-\frac{1}{2}|<3\times 10^{-3}$ I still have difficulties with absolute value, and even if I manage to solve questions and problems, I do that awkwardly.
So, please show me if this is the way to answer this question.
Thank you in advance.
$|x-2|<0.001 \iff -0.001<x-2$ or $x-2<0.001$
$$\begin{align}\\-0.001<x-2 & \iff 2-0.001<x\\ &\iff1.999<x\\ &\iff\frac{1}{x}<\frac{1}{1.999}\\ &\iff\frac{1}{x}-\frac{1}{2}<\frac{0.001}{3.998}\\ &\iff \frac{1}{x}-\frac{1}{2}<3\times 10^{-3}\end{align}$$
$$\begin{align}\\ x-2<0.001 &\iff x<2.001\\ &\iff\frac{1}{x}>\frac{1}{2.001}\\ &\iff\frac{1}{x}-\frac{1}{2}>\frac{3.002}{2}\\ &\iff\frac{1}{x}-\frac{1}{2}>-3\times10^{-3}\end{align}$$
We have: $-3\times 10^{-3}<\frac{1}{x}-\frac{1}{2}<3\times10^{-3}$
Then, $|\frac{1}{x}-\frac{1}{2}|<3\times10^{-3}$
| The way I would answer
(writing out every step):
$\begin{array}\\
|x-2|<0.001
&\iff -0.001<x-2 < .001\\
&\iff 2-0.001<x < 2+.001\\
&\iff 1.999<x < 2.001\\
&\iff \frac1{1.999}>\frac1{x} > \frac1{2.001}\\
&\iff \frac1{1.999}-\frac12 >\frac1{x}-\frac12 > \frac1{2.001}-\frac12\\
&\iff \frac{2-1.999}{2\cdot1.999} >\frac1{x}-\frac12 > \frac{2-2.001}{2\cdot 2.001}\\
&\iff \frac{.001}{3.998} >\frac1{x}-\frac12 > \frac{-.001}{4.002}\\
&\implies |\frac1{x}-\frac12| < \frac{.001}{3.998}\\
\end{array}
$
Note how this generalizes,
using copy, paste, and edit:
If $0 < c < a$,
$\begin{array}\\
|x-a|<c
&\iff c<x-a < c\\
&\iff a-c<x < a+c\\
&\iff \frac1{a-c}>\frac1{x} > \frac1{a+c}\\
&\iff \frac1{a-c}-\frac1{a} >\frac1{x}-\frac1{a} > \frac1{a+c}-\frac1{a}\\
&\iff \frac{a-(a-c)}{a(a-c)} >\frac1{x}-\frac1{a} > \frac{a-(a+c)}{a(a+c)}\\
&\iff \frac{c}{a(a-c)} >\frac1{x}-\frac1{a} > \frac{-c}{a(a+c)}\\
&\implies |\frac1{x}-\frac1{a}| < \frac{c}{a(a-c)}
\quad\text{since }|\frac{c}{a(a+c)}| < |\frac{c}{a(a-c)}|\\
\end{array}
$
To make
$|\frac1{x}-\frac1{a}|
< d
$,
we can choose
$d
> \frac{c}{a(a-c)}
$
or
$c
< da(a-c)
=da^2-cda
$
or
$c(1+da)
< da^2
$
or
$c
<\frac{da^2}{1+da}
$.
Note that
$\frac{da^2}{1+da}
< a
$,
so that
$c < a$
is automatically satisfied.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\frac{1}{1999} < \prod_{i=1}^{999}{\frac{2i−1}{2i}} < \frac{1}{44}$
Prove that $$\dfrac{1}{1999} < \prod_{i=1}^{999}{\dfrac{2i−1}{2i}} < \dfrac{1}{44}$$
from the 1997 Canada National Olympiad.
I have been able to prove the left half of the inequality using induction. Need help with the second part.
Prove $\prod_{i=1}^{n}{(1-\frac{1}{2i})} \ge \frac{1}{2n}$
Define $$p(n)=\prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \tag{1}$$
We wish to show $$p(n)\ge\frac{1}{2n},\quad\forall{n}\in\mathbb{Z^+} \tag{2}$$
For the base case, $n=1$, we have $p(1)=1-\frac{1}{2}=\frac{1}{2}\ge\frac{1}{2}$ which holds.
Assume the induction hypothesis (2) for $n$. Then for $n+1$, we can write:
$$\begin{align}
p(n+1) = \prod_{i=1}^{n+1}{\left(1-\frac{1}{2i}\right)} &= \left(1-\frac{1}{2(n+1)}\right) \cdot \prod_{i=1}^{n}{\left(1-\frac{1}{2i}\right)} \\
&\ge \left(1-\frac{1}{2(n+1)}\right) \cdot \frac{1}{2n} &(\text{by IH}) \\
&\ge \left(\frac{2n+1}{2(n+1)}\right) \cdot \frac{1}{2n} \\
&\ge \left(\frac{1}{2(n+1)}\right) \cdot \frac{2n+1}{2n} \\
&> \left(\frac{1}{2(n+1)}\right) &(\text{for }n\ge1)
\end{align}$$
This proves the induction step, establishing (2) for all positive $n$. Hence, we conclude
$$\boxed{\prod_{i=1}^{999}{\dfrac{2i−1}{2i}} = p(999) \ge \frac{1}{2\times999} = \frac{1}{1998} > \frac{1}{1999}}$$
Prove $\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} < \frac{1}{44}$
For this part, I have applied AM-GM to get an upper bound with a harmonic sum.
$$\left(\prod_{i=1}^{999}{\left(1-\frac{1}{2i}\right)}\right)^{1/999} \le \frac{1}{999} \cdot \sum_{i=1}^{999}{\left(1-\frac{1}{2i}\right)} \tag{3}$$
I can't see a way of evaluating the summation on the RHS.
| Since $\frac{2i-1}{2i}\lt\frac{2i}{2i+1}$, we have
$$
\left(\prod_{i=1}^{999}{\frac{2i−1}{2i}}\right)^2\lt\frac14\prod_{i=2}^{999}{\frac{2i−1}{2i}}\prod_{i=2}^{999}{\frac{2i}{2i+1}}=\frac14\frac3{1999}\lt\frac14\frac1{666}\tag1
$$
Since $\frac{2i-1}{2i}\gt\frac{2i-2}{2i-1}$, we have
$$
\left(\prod_{i=1}^{999}{\frac{2i−1}{2i}}\right)^2\gt\frac14\prod_{i=2}^{999}{\frac{2i−1}{2i}}\prod_{i=2}^{999}{\frac{2i-2}{2i-1}}=\frac14\frac2{1998}=\frac14\frac1{999}\tag2
$$
Therefore, since $32^2=1024\gt999$ and $25^2=625\lt666$, we have
$$
\frac1{64}\lt\frac12\frac1{\sqrt{999}}\lt\prod_{i=1}^{999}{\frac{2i−1}{2i}}\lt\frac12\frac1{\sqrt{666}}\lt\frac1{50}\tag3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1394874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
} |
a integration with two constants I am trying to solve below integration
$$\int\frac{dx}{(x^2+a^2)(x^2+a^2+b^2)^{\frac12}}$$
I tried substituting $x=a \,tan\,u$. Then I ended up with $$\int\frac{du}{(a)(b^2+a^2sec^2\,u)^{\frac12}}$$. I am not able to continue after this. Appreciate a small help on this
| Using rogerl's suggestion, letting $x=c\tan\theta$ where $c=\sqrt{a^2+b^2}$ gives
$\displaystyle\int\frac{1}{(x^2+a^2)\sqrt{x^2+a^2+b^2}}dx=\int\frac{1}{(c^2\tan^2\theta+a^2)(c\sec\theta)}\cdot c\sec^2\theta d\theta$
$\displaystyle=\int\frac{\sec\theta}{a^2\sec^2\theta+b^2\tan^2\theta} d\theta$ $\;$[since $c^2\tan^2\theta+a^2=(a^2+b^2)\tan^2\theta+a^2=a^2\sec^2\theta+b^2\tan^2\theta$]
$=\displaystyle\int\frac{\cos\theta}{a^2+b^2\sin^2\theta} d\theta=\frac{1}{b}\int\frac{1}{a^2+u^2} du$ $\;\;\;$ (letting $u=b\sin\theta$)
$\displaystyle=\frac{1}{ab}\tan^{-1}\frac{u}{a}+C=\frac{1}{ab}\tan^{-1}\frac{b\sin\theta}{a}+C=\frac{1}{ab}\tan^{-1}\bigg(\frac{bx}{a\sqrt{x^2+a^2+b^2}}\bigg)+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1395757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integration by u-substitution. Evaluate $$\int_0^1 x^2 \cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right) \mathrm{d}x$$
My attempt: Let $u = \dfrac{x^3}{3}+1$ so $\mathrm{d}u = x^2\mathrm{d}x$
$$\int_0^1\cos(u)\cos(u)\mathrm{d}u$$
Then I am stuck trying to integrate $\cos^2(x)$. Any help is good. Thanks.
| $$\int_0^1x^2\cos\left(\frac{x^3}{3}+1\right)\cos\left(\sin\left(\frac{x^3}{3}+1\right)\right)\mathrm{d}x$$
Let's let $u = \sin\left(\dfrac{x^3}{3}+1\right)$ so $\mathrm{d}u = x^2\cos\left(\dfrac{x^3}{3}+1\right)$. We also need to change the bounds of integration. So from $u = \sin\left(\dfrac{x^3}{3}+1\right) $, plugging in $\left[0,1\right]$ we get $\left[\sin(1),\ \sin\left(\dfrac43\right)\right]$ as the bounds. So now our integral reduces down to
\begin{align*}
\int_{\sin(1)}^{\sin\left(\frac43\right)} \cos(u) \ \mathrm{d}u &= \bigg[\sin(u)\bigg]_{\sin(1)}^{\sin\left(\frac43\right)} \\
&= \sin\left(\sin\left(\frac43\right)\right) - \sin(\sin(1))
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Limit with parameter "a" If the function is :
$$f(x)=\sqrt[3]{x^3-ax^2+1}-\sqrt[3]{x^3-x^2+1}.$$
Determine the parameter "$a$" so that $\lim_{x\rightarrow \infty} f(x)=\dfrac 13$.
I have tried solving this limit having in mind that "$a$" is some constant and I get $0$ over something, so I calculate that the limit of $f(x)$ is zero. I've started off by multiplying with the conjugate so I get rid of the roots in the numerator, but I am stuck and I don't know how to proceed. Please help...
| $$
f(x) = x\sqrt[3]{1-\frac{a}{x} +\frac{1}{x^3}} - x\sqrt[3]{1-\frac{1}{x} +\frac{1}{x^3}}
$$
expanding the cube radicals we find
$$
f(x) = x\left(1-\frac{1}{3}\left(\frac{a}{x} -\frac{1}{x^3}\right) + O\left(\frac{1}{x^2}\right)\right) - x\left(1-\frac{1}{3}\left(\frac{1}{x} -\frac{1}{x^3}\right) + O\left(\frac{1}{x^2}\right)\right) = -\frac{1}{3}(a-1)+O\left(\frac{1}{x}\right)
$$
thus
$$
\lim_{x\to\infty}f(x) = \lim_{x\to\infty}-\frac{1}{3}(a-1)+O\left(\frac{1}{x}\right) = \frac{1}{3}(1-a)
$$
therefore we require
$$
\frac{1}{3}(1-a) = \frac{1}{3} \implies a = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Why does this "miracle method" for matrix inversion work? Recently, I answered this question about matrix invertibility using a solution technique I called a "miracle method." The question and answer are reproduced below:
Problem: Let $A$ be a matrix satisfying $A^3 = 2I$. Show that $B = A^2 - 2A + 2I$ is invertible.
Solution: Suspend your disbelief for a moment and suppose $A$ and $B$ were scalars, not matrices. Then, by power series expansion, we would simply be looking for
$$ \frac{1}{B} = \frac{1}{A^2 - 2A + 2} = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots$$
where the coefficient of $A^n$ is
$$ c_n = \frac{1+i}{2^{n+2}} \left((1-i)^n-i (1+i)^n\right). $$
But we know that $A^3 = 2$, so
$$ \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A^4}{8}-\frac{A^5}{8} + \cdots = \frac{1}{2}+\frac{A}{2}+\frac{A^2}{4}-\frac{A}{4}-\frac{A^2}{4} + \cdots $$
and by summing the resulting coefficients on $1$, $A$, and $A^2$, we find that
$$ \frac{1}{B} = \frac{2}{5} + \frac{3}{10}A + \frac{1}{10}A^2. $$
Now, what we've just done should be total nonsense if $A$ and $B$ are really matrices, not scalars. But try setting $B^{-1} = \frac{2}{5}I + \frac{3}{10}A + \frac{1}{10}A^2$, compute the product $BB^{-1}$, and you'll find that, miraculously, this answer works!
I discovered this solution technique some time ago while exploring a similar problem in Wolfram Mathematica. However, I have no idea why any of these manipulations should produce a meaningful answer when scalar and matrix inversion are such different operations. Why does this method work? Is there something deeper going on here than a serendipitous coincidence in series expansion coefficients?
| It seems that he is doing this the hard way.
If you know that $A^3 = 2I$, then any infinite sum of powers of $A$, if it converges, should reduce to a sum of the form $xI + yA + zA^2.$ So it seems reasonable to at least try to solve
$$ (xI + yA + zA^2)(A^2 - 2A + 2I) = I$$
for $x, y,$ and $z$.
\begin{align}
I
&= (xI + yA + zA^2)(2I - 2A + A^2)\\
&= 2xI +(-2x + 2y)A + (x - 2y + 2z)A^2 + (y - 2z)A^3 + zA^4\\
&= 2xI +(-2x + 2y)A + (x - 2y + 2z)A^2 + (2y - 4z)I + 2zA\\
&= (2x + 2y - 4z)I + (-2x + 2y + 2z)A + (x - 2y + 2z)A^2
\end{align}
Solve
\begin{align}
2x + 2y - 4z &= 1 \\
-2x + 2y + 2z &= 0 \\
x - 2y + 2z &= 0
\end{align}
and you get $x = \frac 25,\, y = \frac 3{10},\, z = \frac 1{10}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "71",
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} |
Determine all positive integers $n$ which have a divisor $d$ with the property that $dn+1$ is a divisor of $d^2 + n^2$ Determine all positive integers $n$ which have a divisor $d$ with the property that $dn+1$ is a divisor of $d^2 + n^2$.
So i formed the equation that $$\frac{n}{d} = \frac{d^2 + n^2}{dn + 1}$$
And ended with $n = d^3$ which I think is wrong
If I'm wrong can someone please show me the right way.
| Write $n = cd$.
Then
\begin{align*}
1 + dn \mid d^2 + n^2
&\iff 1 + cd^2 \mid d^2 + c^2 d^2 \\
&\iff 1 + cd^2 \mid d^2 (1 + c^2) \\
&\iff 1 + cd^2 \mid 1 + c^2 \quad \quad \quad (\text{since } 1 + cd^2 \text{ and } d \text{ are relatively prime}) \\
&\iff 1 + cd^2 \mid 1 + c^2 - (1 + cd^2) = c^2 - cd^2 \\
&\iff 1 + cd^2 \mid c(c - d^2) \\
&\iff 1 + cd^2 \mid c - d^2. \quad \quad \quad (\text{since } 1 + cd^2 \text{ and } c \text{ are relatively prime}) \\
\end{align*}
Therefore, we restate the problem as follows:
find all positive integers $\boldsymbol{c,d}$ such that $\boldsymbol{1 + cd^2 \mid c - d^2}$.
There are three cases: $c - d^2 = 0$, $c - d^2 > 0$, and $c - d^2 < 0$.
*
*Case 1: $c - d^2 = 0$
Here we get that $n = cd = d^3$, which is indeed one possible solution.
*Case 2: $c - d^2 > 0$.
In this case, $c - d^2$ is a positive multiple of $1 + cd^2$, so
$1 + cd^2 \le c - d^2$, so $cd^2 - c + d^2 - 1 \le -2$,
so $(c + 1)(d^2 - 1) \le -2$.
But this is a contradiction, since $c+1$ and $d^2-1$ are nonnegative.
*Case 3: $c - d^2 < 0$.
In this case, $d^2 - c$ is a positive multiple of $1 + cd^2$, so $1 + cd^2 \le d^2 - c$, so $cd^2 - d^2 + c - 1 \le -2$,
so $(c - 1)(d^2 + 1) \le -2$.
But this is a contradiction, since $c+1$ and $d^2-1$ are nonnegative.
Therefore, the only solution is $n = d^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to resolve $n>(1+\frac{1}{n})^n$? I'm trying to prove that $\forall n\geq 3, n^{n+1}>(n+1)^n$. I came that this is true for $n>(1+\frac{1}{n})^n$. WolphramAlpha gives $n>2.293166...$ but I failed to compute it analytically.
| Crostul's solution is probably the most straightforward way to go about it, but I'm going to do some induction because I probably have nothing better to do.
Base Case: When $n = 3$, $n > \left(1 + \frac{1}{n}\right)^n$ since $3 > \frac{64}{27}$.
Inductive Step: Suppose that $n > \left(1 + \frac{1}{n}\right)^n$. We wish to show that $n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$.
Notice that with $n \ge 3$, we have $1 + \frac{1}{n+1} < 1 + \frac{1}{n}$.
Thus, if $n > \left(1 + \frac{1}{n}\right)^n$, then clearly $n > \left(1 + \frac{1}{n+1}\right)^n$.
It follows that $n\left(1 + \frac{1}{n}\right) > \left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right) \Longleftrightarrow n+1 > \left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right)$
We see that since $1 + \frac{1}{n+1} < 1 + \frac{1}{n}$, we have $\left(1 + \frac{1}{n+1}\right)^n\left(1 + \frac{1}{n}\right) > \left(1 + \frac{1}{n+1}\right)^{n+1}$.
Therefore, $n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$.
Since we have shown a base case for $n = 3$ and have also shown that $n > \left(1 + \frac{1}{n}\right)^n \Longrightarrow n + 1 > \left(1 + \frac{1}{n+1}\right)^{n+1}$, we have proven that for $n \ge 3$, $n > \left(1 + \frac{1}{n}\right)^n$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\displaystyle\int_{0}^{\infty}\left[ne^{-x}\right]dx=\log\frac{n^{n-1}}{(n-1)!}$ $\displaystyle\int_{0}^{\infty}\left[ne^{-x}\right]dx=\log\frac{n^{n-1}}{(n-1)!}, \ $ where $[\,\cdot\,]$ is the greatest integer function.
Put $e^{-x}=t$
$$
\begin{aligned}
\int_{0}^{\infty}\left[ne^{-x}\right]dx
& =\int_{1}^{0}\left(\left[n t\right]\cdot\frac{-1}{t}\right)dt
=\int_{0}^{1}\left(\left[n t\right]\times\frac{1}{t}\right)dt
\\
& =\int_{0}^{\frac{1}{n}}0\,dt+\int_{\frac{1}{n}}^{\frac{2}{n}}\frac{1}{t}\,dt+\int_{\frac{2}{n}}^{\frac{3}{n}}\frac{2}{t}\,dt+\int_{\frac{3}{n}}^{\frac{4}{n}}\frac{3}{t}\,dt+ \dots +\int_{\frac{n-1}{n}}^{1}\frac{n-1}{t}\,dt
\\
&=\log 2+2\log\frac{3}{2}+3\log\frac{4}{3}+ \dots +(n-1)\log\frac{n}{n-1}
\end{aligned}
$$
However I did not the result in the form given in answer.
| $$
\log \frac{2^1 \cdot 3^2 \cdot 4^3 \cdots n^{n-1}}{1^1 \cdot 2^2 \cdot 3^3 \cdots (n-1)^{n-1}}=\log \frac{n^{n-1}}{1 \cdot 2 \cdot 3 \cdots (n-1)}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the longest side of the triangle. The sides $a,b,c$ of a $\triangle ABC$ are in $GP$ whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6\sqrt{\frac{7}{209}}$.Find the longest side of the triangle.
I used law of sines $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
I took $a=a,b=\frac{2}{3}a,c=\left(\frac{2}{3}\right)^2a$
which gives $$\frac{\sin A}{\sin B}=\frac{\sin B}{\sin C}$$
I am stuck now,how to find the longest side $a$.
| The side of the triangle in G.P. with common ratio $\frac{2}{3}$ can be taken as $$a=\frac{3}{2}x, b=x, c=\frac{2}{3}x$$
Where, $x$ is a positive real number.
Hence, using cosine formula we get $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ $$=\frac{\frac{9}{4}x^2+x^2-\frac{4}{9}x^2}{2\cdot\frac{3}{2}x \cdot x }=\frac{101}{108}$$ $$\implies \sin C=\sqrt{1-\cos ^2 A}=\sqrt{1-\frac{101^2}{108^2}}=\frac{\sqrt{1463}}{108}$$ Hence, the circumscribed radius of the triangle is given by the following formula $$\frac{c}{2\sin C}$$ $$=\frac{\frac{2}{3}x}{2\frac{\sqrt{1463}}{108}}=\frac{36}{\sqrt{1463}}x$$ $$\implies \frac{36}{\sqrt{1463}}x=6\sqrt{\frac{7}{209}}$$ $$x=\frac{7}{6}$$ Now, setting the value of $x$, all the sides of the triangle are calculated as follows $$a=\frac{3}{2}x=\frac{3}{2}\cdot \frac{7}{6}=\frac{7}{4}$$ $$b=x=\frac{7}{6}$$ $$c=\frac{2}{3}x=\frac{2}{3}\cdot \frac{7}{6}=\frac{7}{9}$$
longest side is $a=\frac{7}{4}$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Longest side of the triangle},a=\color{blue}{\frac{7}{4}}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $2xy$ divides $x^2+y^2-x$, prove that $x$ is a perfect square This problem is from ( BMO Exam1991 ). I tried to solve but it was difficult.
The problem is:
If $ x^{2} + y^{2} - x $ is a multiple of $ 2xy $ where $x$ & $y$ are integers, prove that $x $ is a perfect square.
Any help or hint please.
| So we are given that $$x^2+y^2-x-2kxy=0.$$
Write $x=a^2b$ with $b$ square-free. Then
$$a^4b^2+y^2-a^2b-2ka^2by=0,$$ so $a^2\mid y^2$ and $a\mid y$, say $y=ac$.
Then $$a^2b^2+c^2-b-2kabc=0,$$ so $b\mid c^2$ and as $b$ is square-fee in fact $b\mid c$, say $c=bd$.
Then
$$a^2b+bd^2-1-2kabd=0,$$
so $b\mid 1$.
Assume $b=-1$. Then we have
$$ a^2+d^2=2kad-1.$$
As the right hand side is odd, exactly one of $a,d$ must be odd, the other even. But then the right hand side is $\equiv -1\pmod 4$ and the left is $\equiv +1\pmod 4$.
We conclude $b\ne -1$, hence $b=+1$ and $$x=a^2.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that the sequence $a_n=\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$ is monotonically decreasing sequence Prove that the sequence $$a_n=\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)}$$ is monotonically decreasing sequence. I tried $a_{n+1} - a_{n} < 0$, but i was not able to do it.
Help will be appreciated.
Thanks!
| Notice, we have $$a_n=\frac{1.3.5\dots (2n-1)}{2.4.6\ldots (2n)}$$
$$=\frac{[1.3.5\dots (2n-1)]\cdot [2.4.6\ldots (2n)]}{[2.4.6\ldots (2n)]\cdot[(2.4.6\ldots (2n)]}$$ $$=\frac{1.2.3.4.5\dots (2n-1)(2n)}{[2.4.6\ldots (2n)]^2}$$ $$=\frac{(2n)!}{[2^n(1.2.3\ldots (n))]^2}$$
$$=\frac{(2n)!}{[2^n(n!)]^2}$$
$$a_n=\frac{(2n)!}{2^{2n}(n!)^2}\tag 1$$ $$\implies a_{n+1}=\frac{(2n+2)!}{2^{2n+2}((n+1)!)^2}$$ $$\implies a_{n+1}=\frac{(2n+2)(2n+1)(2n)!}{4\cdot 2^{2n}((n+1)n!)^2}$$
$$=\frac{2(n+1)(2n+1)(2n)!}{4(n+1)^2\cdot 2^{2n}(n!)^2}$$
$$a_{n+1}=\frac{(2n+1)(2n)!}{2(n+1)\cdot 2^{2n}(n!)^2}\tag 2$$
Now, dividing (2) by (1), we get $$\frac{a_{n+1}}{a_n}=\frac{\frac{(2n+1)(2n)!}{2(n+1)\cdot 2^{2n}(n!)^2}}{\frac{(2n)!}{2^{2n}(n!)^2}}$$ $$=\frac{2n+1}{2(n+1)}$$$$\frac{a_{n+1}}{a_n}=\frac{2n+1}{2n+2}<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all $x,y$ such that $\frac{1}{x} - \frac{1}{y} \in \mathbb{Z}$ Let $C=D=\mathbb{Z} \backslash \{ 0 \}$. Define a relation $S$ from $C$ to $D$ as follows:
For all $(x,y) \in C \times D$,
$$(x,y) \in S \Leftrightarrow \frac{1}{x} - \frac{1}{y} \in \mathbb{Z}$$
Question: Write $S$ as a set of ordered pairs.
Attempt: Clearly the ordered pairs of the form $(x,x)$ and $(-x,-x)$ satisfy the relation. Other than this two trivial pairs, what other pair can we obtain?
| Another answer, not as slick as the others perhaps but I don't see the harm in posting it anyway:
Since $\frac{1}{x}, \frac{1}{y} \in [-1,0)\cup(0,1]$, the difference $\frac{1}{x}-\frac{1}{y}$ can only take on one of $5$ possible values: $0,\pm1,\pm2$.
Case-by-case, we have
$$\begin{align} \frac{1}{x}-\frac{1}{y} = 0 &\Longleftrightarrow x = y \\
\frac{1}{x}-\frac{1}{y} = \pm1 &\Longleftrightarrow x = \frac{y}{1 \pm y} \\ &\Longleftrightarrow y = \mp 2 \hspace{5mm} \text{(since $x,y$ must be integers)} \\ \frac{1}{x}-\frac{1}{y} = \pm 2 &\Longleftrightarrow x = \frac{y}{1 \pm 2y} \Longleftrightarrow y = \mp1 \hspace{5mm} \text{(since $x,y$ must be integers)}\end{align} $$
So our full set of solutions for distinct $x,y$ is $(-1,1),(1,-1),(-2,2),(2,-2)$. If we allow $x=y$, then $(x,x)$ is a solution for any $x \in \mathbb{Z}\backslash\{0\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$ For $a\geq2$,if the value of the definite integral $\int_{0}^{\infty}\frac{dx}{a^2+\left(x-\frac{1}{x}\right)^2}$ equals $\frac{\pi}{5050}$.Find the value of $a$.
Substituting $x-\frac{1}{x}=t$ does not seem to work here,what is the good substitution to make it integrable?Thanks in advance.
| Note
\begin{eqnarray}
I&=&\int_{0}^{\infty}\frac{1}{a^2+\left(x-\frac{1}{x}\right)^2}dx\\
&=&\int_{0}^{\infty}\frac{1}{x^2+\frac{1}{x^2}+(a^2-2)}dx\\
&=&\int_{0}^{\infty}\frac{x^2}{x^4+kx^2+1}dx.
\end{eqnarray}
Here $k=a^2-2$.
Suppose $a>2$. Letting $x^4+kx^2+1=(x^2+r_1^2)(x^2+r_2^2)$ where $r_1,r_2>0$, we have
\begin{eqnarray}
I&=&\frac1{r_1^2-r_2^2}\left(r_1^2\int_{0}^{\infty}\frac{1}{x^2+r_1^2}-r_2^2\int_{0}^{\infty}\frac{1}{x^2+r_2^2}\right)\\
&=&\frac1{r_1^2-r_2^2}\frac{\pi(r_1-r_2)}{2}\\
&=&\frac{\pi}{2}\frac1{r_1+r_2}.
\end{eqnarray}
Noting $r_1^2+r_2^2=a^2-2$ and $r_1^2r_2^2=1$, we have $r_1+r_2=\sqrt{r_1^2+r_2^2+2r_1r_2}=a$. So
$$ I=\frac{\pi}{2a}.$$
Letting $\frac{\pi}{2a}=\frac{\pi}{5050}$ gives $a=2525$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The pattern in mathematical induction proofs When given a statement to be proven by mathmatical induction
the statement tends to look like this
$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
so going about the proof.
1) Prove the base case
$\frac{1\times(1+1)}{2} = 1$
2) Prove the inductive case
Assume
$1 + 2 + 3 + \dots + n = \frac{n(n+1)}{2}$
Now add $n+1$ to both sides of the equation.
\begin{align}
1 + 2 + 3 + \dots + n + (n + 1) & = \frac{n(n+1)}{2} + (n + 1) \\
& = \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \\
& = \frac{(n + 1)(n + 2)}{2} \\
& = \frac{(n + 1)((n + 1) + 1)}{2}
\end{align}
Thus the proof is complete. Now my question is, can we just remove the $1 + 2 + 3 + \dots + n$ side of the equation, and put anything there? for example, if we used $f(n)$, Then the proof would look like this.
1) Prove the base case
$f(1) = \frac{1\times(1+1)}{2} = 1$
2) Prove the inductive case
Assume
$f(n) = \frac{n(n+1)}{2}$
Now add $n+1$ to both sides of the equation.
\begin{align}
f(n) + (n + 1) & = \frac{n(n+1)}{2} + (n + 1) \\
& = \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \\
& = \frac{(n + 1)(n + 2)}{2} \\
& = \frac{(n + 1)((n + 1) + 1)}{2}
\end{align}
Thus we now know $f(n) = \frac{n(n+1)}{2}$. But what pattern represents $f(n)$?
it's not immediately obvious that $f(n)$ is the sum of the first $n$ positive integers.
| You are making the assumption that $f(n) = \text{ sum of the first } $n$ \text{ integers}$ during your induction. When you evaluate $f(1)$ you make this assumption in how you evaluate. And in your induction step you mean to assert:
$$f(n+1) = f(n) + (n+1)$$
before going on with your proof. Implicit in this assumption is a recursive definition of $f$ that forces $f$ to be the sum of the first $n$ integers. That is, there is exactly one function on the natural numbers such that $f(1) = 1$ and $f(n+1) = f(n) + (n+1)$, and that function is the one that sums the first $n$ integers. Indeed there are many ways to define this function, i.e. $f(n) = n(n+1)/2$, but you can prove that they are all the same function.
| {
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Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$ Determine a positive integer $n\leq5$,such that $\int_{0}^{1}e^x(x-1)^ndx=16-6e$.
I tried to solve it.But since $n$ is given to be $\leq$ 5,my calculations went lengthy.
Applying integration by parts repeatedly,we get
\begin{align}
\int e^x(x-1)^n \, dx &= \left[(x-1)^ne^x-n(x-1)^{n-1}e^x+n(n-1)(x-1)^{n-2}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(x-1)^{n-3}e^x+n(n-1)(n-2)(n-3)(x-1)^{n-4}e^x \right. \\
& \hspace{5mm} \left.-n(n-1)(n-2)(n-3)(n-4)e^x\right]
\end{align}
\begin{align}
\int_{0}^{1}e^x(x-1)^n \, dx &= -n(n-1)(n-2)(n-3)(n-4)e-(-1)^n+n(-1)^{n-1}-n(n-1)(-1)^{n-2} \\
& \hspace{5mm} +n(n-1)(n-2)(-1)^{n-3}-n(n-1)(n-2)(n-3)(-1)^{n-4} \\
& \hspace{5mm} +n(n-1)(n-2)(n-3)(n-4) \\
&=16-6e
\end{align}
Now solving this is very difficult,is there another simple and elegant method to find $n=3.$
| HINT:
Notice, the following property of definite integral
$$\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx$$
Now, we have
$$\int_{0}^{1}e^x(x-1)^ndx=16-6e$$
$$\int_{0}^{1}e^{1-x}(1-x-1)^ndx=16-6e$$
$$\int_{0}^{1}e^{1-x}(-x)^ndx=16-6e$$
$$\int_{0}^{1}e \cdot e^{-x}(-1)^nx^n dx=16-6e$$
$$(-1)^n e\int_{0}^{1} e^{-x}x^n dx=16-6e$$
I hope you can proceed further.
| {
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Finding the missing digits of $23!$ It is given $23!=2585201xy38884976640000$. Now it is required to find the value of $x$ and $y$. I know I could find it by using divisibility rules and solving simultaneous equations. Is there any other way to solve it (without computing it by a calculator)? This question is just out of curiosity.
| Given $23!=2585201xy38884976640000$, Now Here $23!$ must Contain a no. $3,9$
So $\bf{R.H.S}$ must be divisible by $3$ and $9$
If no. is Divisible by $3\;,$ Then Sum of Digit on $\bf{R.H.S}$ is divisible by $3$
So $2+5+8+5+2+0+1+x+y+3+8+8+8+4+9+7+6+6+4$ must be divisible by $3$
So $88+x+y$ is divisible by $3$
So $1+x+y$ must be divisible by $3$
So $x+y = 2,5,8,11,14,17$
Similarly If no. is Divisible by $9\;,$ Then Sum of Digit on $\bf{R.H.S}$ is divisible by $9$
So $2+5+8+5+2+0+1+x+y+3+8+8+8+4+9+7+6+6+4$ must be divisible by $9$
So $88+x+y$ is divisible by $9$
So $7+x+y$ must be divisible by $9$
So $x+y = 2,11$
Now also Divisibility test for $11$. If no. is divisibility by $11$
Then $\displaystyle \bf{(Sum \; of odd\; position \; no)-(sum\; of \; evev \; position\; no.)}$ must be divisible by $11$
So $(48+y)-(38+x) = 10-(y-x)$ is divisible by $11$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can i solve these? I have done them but I am unsure if I did them right.How could i solve this?
$\frac{2x^2-x-19}{x^2+3x+2}>1$
$-1<\frac{2x+3}{x-1}<1$
| For two see See Solve this inequality equation with 3 sides?
For one,
$$\dfrac{2x^2-x-19}{x^2+3x+2}<1\iff\dfrac{2x^2-x-19}{x^2+3x+2}-1<0$$
$$\iff\dfrac{x^2-4x-21}{x^2+3x+2}<0$$
$$\iff\dfrac{(x-7)(x+3)}{(x+2)(x+1)}<0$$
$$\iff0>(x-7)(x+3)(x+2)(x+1)=(x-7)\{x-(-1)\}\{x-(-2)\}\{x-(-3)\}$$
So, we need odd number of negative multiplicands
$-1<x<7\implies $ only $x-7<0$
and $-3<x<-2\implies$ only $x+3>0$
| {
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Find number of nonnegative integer solutions to x+3y+3z=n, given n, using generating functions
For every $n,x,y,z\in \mathbb N$, where $x\ge{0}$ and $y,z \ge1$
Find the number of nonnegative integer solutions for
$x+3y+3z = n$
I created a generating function for the problem:
$(1+x+x^2+...)(x^3+x^6+x^9+...)^2$
$(1+x+x^2+...)x^6(1+x^3+x^6+...)^2$
$\frac{1}{1-x} \frac{x^6}{(1-x^3)^2}$
$= \frac{x^6}{(1-x)(1-x^3)^2}$
I can try breaking this expression into partial functions and try to find the coefficient of $x^n$ from there, but I get stuck in that stage.
can you please help me complete this problem?
| Call your variables $a_i$, so that you want $a_1 + 3 a_2 + 3 a_3 = n$ with $a_i \ge 0$ integers.
The values of $a_1$ are represented by:
$$
1 + z + z^2 + \dotsb = \frac{1}{1 - z}
$$
The values added by $a_2$ and $a_3$ are:
$$
z^3 + z^6 + \dotsb = \frac{z^3}{1 - z^3}
$$
In all, you want the coeficient of $z^n$ in:
$\begin{align}
[z^n] \frac{z^6}{(1 - z) (1 - z^3)^2}
&= [z^{n - 6}]
\left(
\frac{1}{9 (1 - z)^3}
+ \frac{2}{9 (1 - z)^2}
+ \frac{7}{27 (1 - z)}
+ \frac{1 + 2 z}{(1 + z + z^2)^2}
+ \frac{8 + 7 z}{1 + z + z^2}
\right)
\end{align}$
The last two terms are troublesome. We can factor more using complex numbers, using the cube root $\omega = \exp(2 \pi \mathrm{i} / 3)$, but it gets quite ugly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}....\cos \alpha_{n}.$
Maximum value of $\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}.$ If it is given that
$\cot \alpha_{1}\cdot \cot \alpha_{2}\cdot \cot \alpha_{3}.......\cot \alpha_{n} = 1$ and $\displaystyle 0 \leq \alpha_{1},\alpha_{2},\alpha_{3},.......,\alpha_{n}\leq \frac{\pi}{2}.$
$\bf{My\; Try::}$ Using $\bf{A.M\geq G.M}$
$$\displaystyle \frac{\sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}}{n}\geq \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}$$
So we get
$$\displaystyle \sin^2 \alpha_{1}+\sin^2 \alpha_{2}+\sin^2 \alpha_{3}+........+\sin^2 \alpha_{n}\geq n\cdot \sqrt[n]{\sin^2 \alpha_{1}\cdot \sin^2 \alpha_{2}...\sin^2 \alpha_{n}}.......(1)$$
Similarly
$$\displaystyle \frac{\cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}}{n}\geq \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
So we get
$$\displaystyle \cos^2 \alpha_{1}+\cos^2 \alpha_{2}+\cos^2 \alpha_{3}+........+\cos^2 \alpha_{n}\geq n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}.......(2)$$
Now Adding $(1)$ and $(2)\;,$ We get $$n\geq 2n\cdot \sqrt[n]{\cos^2 \alpha_{1}\cdot \cos^2 \alpha_{2}...\cos^2 \alpha_{n}}$$
Because above it is given that $$\sin \alpha_{1}\cdot \sin\alpha_{2}\cdot\sin \alpha_{3}.........\sin\alpha_{n}=\cos\alpha_{1}
\cdot\cos\alpha_{2}\cdot \sin\alpha_{3}......\cos\alpha_{n}$$
So we get $$\displaystyle \cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdot \cos \alpha_{4}....\cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$
My question is that can we solve it using any other short method? If yes then please explain it here.
Thanks.
| Using $$\sec^2 \alpha_{1}\cdot \sec^2 \alpha_{2}\cdot \sec^2 \alpha_{3}\cdots \sec^2 \alpha_{n} =(1+\tan^2 \alpha_{1})\cdot (1+\tan^2 \alpha_{2})\cdots \cdots (1+\tan^2 \alpha_{n})$$
Now Apply $\bf{A.M\geq G.M}$
$$(1+\tan^2 \alpha_{1})\cdot (1+\tan^2 \alpha_{2})\cdots \cdots (1+\tan^2 \alpha_{n})\geq 2\tan \alpha_{1}\cdot 2\tan \alpha _{2}\cdot \cdots2\tan \alpha_{n}=2^{n}$$
So $$\sec^2 \alpha_{1}\cdot \sec^2 \alpha_{2}\cdot \sec^2 \alpha_{3}\cdots \sec^2 \alpha_{n}\geq 2^n$$
So $$\cos \alpha_{1}\cdot \cos \alpha_{2}\cdot \cos \alpha_{3}\cdots \cos \alpha_{n}\leq \frac{1}{2^{\frac{n}{2}}}$$
| {
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"url": "https://math.stackexchange.com/questions/1410049",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How many 3 digit numbers can be formed using digits 1,2,3,4 and 5 such that the number is divisible by 6
How many $3$ digit numbers can be formed using digits $1,2,3,4$ and $5$ without repetition such that the number is divisible by $6$
First Approach:
A number is divisible by $6$ if it is divisible by $2$ and $3$.
Now the possible combinations I found are
$(1,3,2)$
$(3,1,2)$
$(2,3,4)$
$(3,2,4)$
$(4,3,2)$
$(3,4,2)$
$(3,5,4)$
$(5,3,4)$
total $8$ ways.
Second Approach
Case1:
unit digit can be filled in only two ways $(2,4)$ for nos $(3,2,4)$
Tens digit can be filled in $2$ ways
Hundred digit can be filled in $1$ ways
the required number is 2*1*2=4 ways
Case2:
unit digit can be filled in only one ways $(2)$ for nos $(1,2,3)$
Tens digit can be filled in $1$ ways
Hundred digit can be filled in $2$ ways
the required number is 2*1*1=2 ways
Case3:
unit digit can be filled in only one ways $(4)$ for nos $(3,4,5)$
Tens digit can be filled in $1$ ways
Hundred digit can be filled in $2$ ways
the required number is 2*1*1=2 ways
So, Total ways=8
Is there still a better way to solve this problem?
| Hint:-Form even numbers such that their sum is divisible by 3.
Divisibility rule by 6 says that the number must be divisible by both 2 and 3.Your second try does not take into account that the number must be divisible by 3.The possible ways are ,as you quoted-(1,3,2)
(3,1,2)
(2,3,4)
(3,2,4)
(4,3,2)
(3,4,2)
(3,5,4)
(5,3,4).There is possibly no better way to do it without using divisibility.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $((a+b)/2)^n\leq (a^n+b^n)/2$ Struggling with this proof.
Prove that $$\left(\frac{a+b}{2}\right)^n≤\frac{a^n+b^n}{2},$$ where $a$ and $b$ are real numbers such that $a+b≥0$ and $n$ is a positive integer.
What technique would you use to prove this (e.g. induction, direct, counter example). How would you go about proving it?
Thanks in advance.
| Solution 1.(Partial solution) $x^n$ is a convex function on $\mathbb{R}^+$ for $n\geq 1$, thus by Jensen:
$$\left(\frac{a+b}{2}\right)^n\leq \frac{a^n+b^n}{2}$$
Solution 2.('Hidden' usage of condition) For $n=1$, true. By induction:
$$\left(\frac{a+b}{2}\right)^n=\left(\frac{a+b}{2}\right)^{n-1}\left(\frac{a+b}{2}\right)\leq \frac{a^{n-1}+b^{n-1}}{2}\cdot\frac{a+b}{2} $$
And:
$$\frac{a^n+b^n}{2}-\frac{a^{n-1}+b^{n-1}}{2}\cdot\frac{a+b}{2}=\frac{(a^{n-1}-b^{n-1})(a-b)}{4}$$
And the factors have the same sign.
Solution 3. Let $\alpha=\frac{a+b}{2}$, then $a=\alpha+x$ and $b=\alpha-x$, then the LHS is $\alpha^n$, and the RHS-LHS is:
$$
\frac{1}{2}(a^n+b^n)-\left(\frac{a+b}{2}\right)^n=\sum_{i=1}^{\lfloor n/2\rfloor} \binom{n}{2i}\alpha^{n-2i}x^{2i}
$$
And by the conditions, all the terms are positive.
| {
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"url": "https://math.stackexchange.com/questions/1410604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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trying to grasp disphenoid tetrahedral honeycomb, what are the dihedral angles? What are the dihedral angles in a disphenoid with four identical triangles, each having one edge of length $2$ and two edges of length $\sqrt{3}$? Tried to look it up, but couldn't find it...
| Let $\alpha$ be the vertex angle of a face, and $\beta$ a base angle. Then, by the Law of Cosines:
$$\cos\alpha = \frac{-2^2 + \sqrt{3}^2 + \sqrt{3}^2}{2\cdot\sqrt{3}\sqrt{3}} = \frac{1}{3} \qquad \cos\beta = \frac{2^2 - \sqrt{3}^2 + \sqrt{3}^2}{2\cdot 2\sqrt{3}} = \frac{\sqrt{3}}{3}$$
Also,
$$\sin\alpha = \frac{2\sqrt{2}}{3} \qquad \sin\beta = \frac{\sqrt{6}}{3}$$
Each vertex of your tetrahedron is surrounded by one $\alpha$ angle and two $\beta$ angles. Let $A$ be the dihedral angle opposite face angle $\alpha$ at a vertex; and let $B$ be opposite a $\beta$. By the Spherical Law of Cosines, we can calculate
$$\begin{align}
\cos A &= \frac{\cos\alpha-\cos^2\beta}{\sin^2\beta} = \frac{1/3-1/3}{2/3} = 0 \\[4pt]
\cos B &= \frac{\cos\beta - \cos\alpha\cos\beta}{\sin\alpha\sin\beta} = \frac{\sqrt{3}/3\,(1-1/3)}{2\sqrt{12}/9} = \frac{1}{2}
\end{align}$$
Therefore,
$$A = \frac{\pi}{2} \qquad\qquad B = \frac{\pi}{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Missing values of the ratio $\frac{(x+y+z)^2}{x^2+y^2+z^2}$ Let $x,y,z$ be some positive integers. Is it true that we cannot find any positive integer $n$ for which
$$
\frac{(x+y+z)^2}{x^2+y^2+z^2}=1+\frac{2}{3n}\,\,?
$$
| Assuming the result for smaller $x+y+z$, it is inductively true in the case that any of $x,y,z$ is divisible by 3; this answer is a work in progress.
Expand and simplify to obtain:
$$\frac{xy+yz+xz}{x^2+y^2+z^2} = \frac{1}{3n}$$
or equivalently
$$3n(xy+yz+zx) = x^2+y^2+z^2$$
But the only way a sum of three squares can be a multiple of 3 is if all of them are multiples of 3 or all of them are not multiples of 3.
Former case: writing $x=3i, y=3j, z=3k$ we obtain
$$3n(ij+jk+ki) = i^2+j^2+k^2$$
We are done by inductive hypothesis: given any solution we may make a strictly smaller solution, but we have assumed for induction that no smaller solution exists. Therefore the only solution can be $x=y=z=0$, which is outside the allowable range.
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.