Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Show that $(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$ for some integers $A,B,C$ How do you solve this question?:$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$ is for all $x$. Here $A$, $B$ and $C$ is constants. I need to know $A$, $B$ and $C$ to pass this. They are integers.
I got this far:
(LS = left side)
$$LS = 49\cos^2(x)+\sin^2(x)+14\sin(x)\cos(x)$$
And then i follow some steps online and got right side to
$$RS = (A+C)\cos^2(x)+(C-A)\sin^2(x)+2B\sin(x)\cos(x)$$
Only problem is that i get C to 49/2.
| Anything forall x implies an identity so you can put any * values of x to get C. for e.g.:
$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$
$$
\left\{
\begin{array}{c}
49=A+C\quad x=0 \\
1=-A+C\quad x=\pi/2\\
18=B+C\\
\end{array}
\right.
$$
Solving these we get:
$$
\left\{
\begin{array}{c}
A=24 \\
C=25\\
B=-7\\
\end{array}
\right.
$$
Alternatively:
$$(7\cos(x)-\sin(x))^2=49\cos^2x+\sin^2x-14\cos x\sin x
\\=49\left(\frac{1+\cos{2x}}2\right)+\left(\frac{1-\cos{2x}}2\right)-7\sin 2x
\\=24\cos 2x-7\sin 2x+25$$
So $A=24,B=-7,C=25$
Use:
$$\cos 2x=2\cos^2x-1=1-2\sin^2x,\sin 2x=2\sin x\cos x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How is $ i^{-1} = -i$ and $i^{-3} = i$? Now I know that with positive powers of $i$ the cycle is: $i , -1 , -i , 1\ldots$
The negative power cycle is: $-i , -1 , i , 1 \ldots$
Can someone explain to me how $\frac 1 {\sqrt{-1}}$ is equal to $-i$ and $\frac 1 {-\sqrt{-1}}$ is equal to $i$?
| $\frac{1}i=i^{-1}$ is true by definition and $n+\frac{1}n=\frac{n^2+1}n$ (since $n=\frac{n^2}n$ and $\frac{n^2}n+\frac{1}n=\frac{n^2+1}n$) is also true. If you plug $i$ in $\frac{i^2+1}i=\frac{-1+1}i=\frac{0}i=0$.
Therefore $i+\frac{1}i=0$ and $\frac{1}i=-i$. By our original statement (namely $i^{-1}=\frac{1}i$) we can conclude that $i^{-1}=-i$.
$i^{-3}=\frac{1}{i^{3}}$ by definition also and $n- \frac{1}{n^3}=\frac{n^4-1}{n^3}$ (since $\frac{n^4}{n^3}-\frac{1}{n^3}$ clearly is $\frac{n^4-1}{n^3}$) when we plug $i$ into this $\frac{i^4-1}i = \frac{1-1}{i} = \frac{0}i = 0$.
Thus, $i - \frac{1}{i^3}$ = 0. This becomes $\frac{1}{i^3}=i$. Therefore $i^{-3} = i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that $\sum\limits_{\mathrm{cyc}}{\frac{1}{(x+2y)^2}} \geq\frac{1}{xy+yz+zx}$ for $x, y, z > 0$
Let $x,y,z>0$. Prove that
$$\frac{1}{(x+2y)^2}+\frac{1}{(y+2z)^2}+\frac{1}{(z+2x)^2} \geq\frac{1}{xy+yz+zx}.$$
I tried to apply Cauchy - Schwarz's inequality but I couldn't prove this inequality!
| Alternative proof without full expanding or uvw:
Let
$$u = \frac{1}{x + 2y}, \quad
v = \frac{1}{y + 2z}, \quad w = \frac{1}{z + 2x}.$$
We have
\begin{align*}
&\mathrm{LHS} - \mathrm{RHS} \\
=\,& u^2 + v^2 + w^2 - \frac{1}{xy + yz + zx}\\
=\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} + uv + vw + wu - \frac{1}{xy + yz + zx}\\
=\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2} + \frac{3(x + y + z)}{(x + 2y)(y + 2z)(z + 2x)} - \frac{1}{xy + yz + zx}\\
=\,& \frac{(u-v)^2 + (v - w)^2 + (w-u)^2}{2}
- \frac{(y-x)(y-z)(x-z)}{(x+2y)(y+2z)(z+2x)(xy+yz+zx)}. \tag{1}
\end{align*}
WLOG, assume that $z = \min(x, y, z)$.
From (1), we only need to prove the case $y \ge x \ge z$.
If $2x \ge y + z$, from (1),
we have
\begin{align*}
\mathrm{LHS} - \mathrm{RHS}
&\ge \frac{(v - u + v - w + w - u)^2}{6} - \frac{(y-x)\cdot (y - z + x - z)^2/4}{(x+2y)(y+2z)(y+2z)xy} \tag{2}\\
&= \frac{2(x + y - 2z)^2}{3(x + 2y)^2(y + 2z)^2} - \frac{(y-x) (x + y - 2z)^2}{4(x+2y)(y+2z)^2xy}\\
&= \frac{(x + y - 2z)^2}{(x+2y)(y+2z)^2}
\left(\frac{2}{3(x + 2y)} - \frac{y-x}{4xy}\right)\\
&\ge 0
\end{align*}
where in (2) we use $(u-v)^2 + (v - w)^2 + (w-u)^2
\ge (v-u + v - w + w - u)^2 /3$, and
$(y - z)(x - z) \le (y - z + x - z)^2/4$, and $z + 2x \ge y + 2z$,
and $xy + yz + zx \ge xy$;
and the last inequality follows from
$\frac{2}{3(x + 2y)} - \frac{y-x}{4xy}
\ge \frac{2}{3(x + 4x)} - \frac{y-x}{4xy} = \frac{15x -7y}{60xy} \ge 0$.
If $2x < y + z$, from (1), we have
\begin{align*}
\mathrm{LHS} - \mathrm{RHS}
&\ge \frac{(v - u + w - v + w - u)^2}{6} - \frac{(x-z)(2y - x - z)^2/4}{(x+2y)(z+2x)(z+2x)xy} \tag{3}\\
&= \frac{2(2y - x - z)^2}{(x+2y)^2(z+2x)^2} - \frac{(x-z)(2y - x - z)^2}{4(x+2y)(z+2x)^2xy}\\
&= \frac{(2y - x - z)^2}{(x + 2y)(z + 2x)^2}
\left(\frac{2}{3(x + 2y)} - \frac{x - z}{4xy}\right)\\
&\ge 0
\end{align*}
where in (3) we use
$(u-v)^2 + (v-w)^2 + (w-u)^2 \ge (v - u + w - v + w - u)^2/3$, and $(y - x)(y - z) \le (y - x + y - z)^2/4$,
and $y+2z \ge z + 2x$, and $xy + yz + zx \ge xy$;
and the last inequality follows from
$\frac{2}{3(x + 2y)} - \frac{x - z}{4xy} = \frac{2xy + 3zx + 6yz - 3x^2}{12(x+2y)xy} \ge 0
$ (using $2xy + 2zx = 2x(y + z) \ge 2x \cdot 2x = 4x^2$).
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/906972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
} |
Use the range for $\theta$ to determine the indicated function value Problem:
$$\sin\theta = \frac {1}{2}; \frac {\pi}{2} \leq \theta \leq \pi, \text{find} \cos\theta$$
So I have that $\sin\theta = \frac {1}{2}$ and that the range for $\theta$ is $\frac {\pi}{2} \leq \theta \leq \pi$.
I'm assuming I need to find the $\theta$ for which $\sin\theta = \frac {1}{2}$ so I can enter that value into the cos function but I'm not sure how to go about it.
I know $\sin\frac {\pi}{2} = 1$ and $\sin\pi$ = 0. So I assumed.. maybe $\sin\frac{3\pi}{4}$ would be $\frac{1}{2}$ but I don't think that's correct. How would I go about solving this?
| Solving $\sin\theta = \frac {1}{2}$, we get $\theta = \frac {\pi}{3}$ or $\frac {2\pi}{3}$.
Edit: my mistake and should be $\theta = \frac {\pi}{6}$ or $\frac {5\pi}{6}$
Note that only the latter fits the given range.
Thus, $\cos\theta = \cos \frac {5\pi}{6} = \cos (\pi - \frac{\pi}{6}) = … = - \cos \frac {\pi}{6} = -\frac {\sqrt 3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/907058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
The general solution for this differential equation? Find the general solution of this differential equation:
$$ \frac{dy}{dx} = \frac{3x^5 y^3}{4} $$
Here's what I've done so far:
$ dy=\frac{3x^5 y^3 dx}{4} $
$ 4dy = 3x^5 y^3 dx $
$ \frac{4dy}{y^3} = 3x^5 dx $
Integrating both sides:
$ \int \frac{4dy}{y^3} = \int 3x^5 dx $
$ \int \frac{4dy}{y^3} = \frac{3x^6}{6} + C $
$ 4\int \frac{dy}{y^3} = \frac{x^6}{2} + C $
To solve the left side, let $ u=y^3 $, therefore $ du=3y^2 dy $, so:
$ \frac{4}{3} \int \frac{3y^2 dy}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^5} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{y^3 y^2} = \frac{x^6}{2} + C $
$ \frac{4}{3} \int \frac{du}{u * y^2} = \frac{x^6}{2} + C $
I'm not sure what to do next. How do I get rid of that $ y^2 $? Or did I do something wrong?
| Hint: $\displaystyle\int\frac{dy}{y^3} = \int y^{-3} dy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Limit of a rational function Calculate the limit
$$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}$$
I divided by the highest degree of x, which is $x^{4}$, further it gave
$$ \frac{-\frac{1}{6}}{\frac{64}{3}} = \frac{-1}{128}$$
which is wrong... what is my error?
| At $0$ we have
$$x^4=o(x^2)\quad\text{and}\quad x^3=o(x^2)$$
so
$$ \lim_{x \to 0} \frac{3x^{2} - \frac{x^{4}}{6}}{(4x^{2} - 8x^{3} + \frac{64x^{4}}{3} )}=\lim_{x \to 0}\frac{3x^2}{4x^2}=\frac34$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/908964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Simplifying a product written in Capital Pi Notation I'm having some trouble figuring out how to simplify Capital Pi Notation.
What I tried was to expand the multiplication with various n and tried to find a pattern. Could someone point me in the right direction on how to approach these problems?
$$\prod_{k=2}^{n} \left(1 - \frac{1}{k^2}\right)$$
| To get this question off the unanswered list, I'll convert my comment into an answer.
Notice that $1 - \dfrac{1}{k^2} = \dfrac{k^2-1}{k^2} = \dfrac{(k-1) \cdot (k+1)}{k \cdot k}$. Therefore, we have:
$\displaystyle\prod_{k = 2}^{n}\left(1-\dfrac{1}{k^2}\right) = \prod_{k = 2}^{n}\dfrac{(k-1) \cdot (k+1)}{k \cdot k} = \dfrac{1 \cdot 3}{2 \cdot 2} \cdot \dfrac{2 \cdot 4}{3 \cdot 3} \cdots \dfrac{(n-2) \cdot n}{(n-1) \cdot (n-1)} \cdot \dfrac{(n-1) \cdot (n+1)}{n \cdot n}$.
Now, just cancel common factors and simplify what is left.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/910479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Infinite geometric progression involving square terms The sum of an infinite geometric progression is 15 and the sum the squares of these terms is 45. Find the series.
The formula for sum of infinite GP is $\frac{a }{1-r} $ and I got two equations $15=\frac {a}{1-r} $ snd $45=\frac {a^2}{1-r^2} $. Solving I got $3=\frac{a}{1+r} $
| Clearly, $a\ne0, r^2\ne1$
We have $\displaystyle\frac{\dfrac{a^2}{1-r^2}}{\left(\dfrac a{1-r}\right)^2}=\frac{1-r}{1+r}$
But, $\displaystyle\frac{\dfrac{a^2}{1-r^2}}{\left(\dfrac a{1-r}\right)^2}=\frac{45}{15^2}=\frac15$
Now apply Componendo and dividendo
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/911153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to divide two inequalities I would like to know if somebody knows how to properly divide one inequality by another, as a resolution method similar to when we divide one equality by another.
Take this as an example: $x^2 - y^2 \lt 8$ and $x + y \gt 3$.
I want to know if it is possible to factor the left hand side of the first equation, to multiply by $-1$ the second equation and divide both equation as to obtain: $y - x \gt -8/3$. I don't really know if the sign must be < or >.
Thanks!
| Suppose $x^2-y^2 < 8$ and $x+y > 3$. If $y - x \le -\dfrac{8}{3}$, then $x-y \ge \dfrac{8}{3}$, and hence $x^2-y^2 = (x-y)(x+y) > \dfrac{8}{3} \cdot 3 = 8$, a contradiction since $x^2-y^2 < 8$.
Thus, any $(x,y)$ which satisfy both $x^2-y^2 < 8$ and $x+y > 3$ will also satisfy $y - x < -\dfrac{8}{3}$.
However, the converse is not true, that is if $(x,y)$ satisfy $y-x < -\dfrac{8}{3}$, then it is not necessarily true that $x^2-y^2 < 8$ and $x+y > 3$. See the other comments for counterexamples.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/911627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Inverse Trigonometric functions - Boyce & Diprima 2.2.19 The problem asks for a solution to the initial value problem:
\begin{align}
&\sin(2x)dx+\cos(3y)dy=0\\
&y\left(\frac{\pi}{2}\right)=\frac{\pi}{3}
\end{align}
The problem is separable and I arrive at the following implicit solution:
\begin{align}
\sin(3y)-3\cos^2(x)=0
\end{align}
If I now take an $\arcsin$ of both sides and solve for $y$ I would naively have:
\begin{align}
y(x)=\frac{1}{3}\arcsin\left(3\cos^2(x)\right)
\end{align}
However, when $x=\frac{\pi}{2}$ we have LHS $=\frac{\pi}{3}$ and RHS $=0 $ which seems to be a problem. The solution provided in the book shifts the RHS by $\frac{\pi}{3}$. Could someone explain what is really going on here? thx
The author's given solution is:
\begin{align}
y(x)=\frac{1}{3}\left(\pi-\arcsin\left(3\cos^2(x)\right)\right)
\end{align}
| After integrating you have:
$$ - \frac{1}{2} \cos{2x} + \frac{1}{3} \sin{3y} = K, \quad K \in \mathbb{R}, $$
which is as nice as an explicit solution. Substitute the initial data to find the value of the constant of integration:
$$ - \frac{1}{2} \cos{\pi} + \frac{1}{3} \sin{\pi} = \frac{1}{2} = K, $$
and you are done. I don't think finding an explicit solution for $y(x)$ before is a good idea.
Cheers!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/912019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving determinants using properties of determinants $$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
= (a-b)(b-c)(c-a)(a^2+b^2+c^2)$$
we have to solve this by using the properties of determinants without actually expanding the determinant. I am Unable to think which calculation to apply so was hoping for an hint.
Edit: just tried the problem and here is how I have done it
$$\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & -(a^2+b^2)+c(a-b) & -(a^3-b^3)\\
1 & c^2-a^2-b(c-a) & c^3-a^3
\end{vmatrix}$$
then
$$\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & -((a-b)(a+b))+c(a-b) & -((a-b)(a^2+ab+b^2))\\
0 & (c-a)(c+a)-b(c-a) & (c-a)(c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & a+b+c & -(a^2+ab+b^2)\\
0 & a-b+c & (c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a)\begin{vmatrix}
a+b+c & -(a^2+ab+b^2)\\
a-b+c & (c^2+ca+a^2)
\end{vmatrix}$$
then
$$(a-b)(c-a) * ( (a+b+c)(c^2+ca+a^2)-(a-b+c)(-(a^2+ab+b^2)) )$$
Here I am Confused on how to multiply them and get the answer
note: typed because I own a very bad handwriting
Thank you every one for your help
| Seeing the column of $1$'s, a first thought would be to subtract the first row from the other two. This turns out to give you a factor $b-a$ in the second row and a factor $c-a$ in the third.
$$
\begin{align}
\begin{vmatrix}
1 & a^2+bc & a^3\\
1 & b^2+ca & b^3\\
1 & c^2+ab & c^3
\end{vmatrix}
&=
\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & b^2-a^2+c(a-b) & b^3-a^3\\
0 & c^2-a^2+b(a-c) & c^3-a^3
\end{vmatrix}
\\{}&
=(b-a)(c-a)
\begin{vmatrix}
1 & a^2+bc & a^3\\
0 & (b+a)-c & a^2+ab+b^2\\
0 & (c+a)-b & a^2+ac+c^2
\end{vmatrix}
\end{align}
$$
Now you can subtract the second row from the third, which makes the latter divisible by $b-c$, so you can factor again. I shouldn't be too difficult to complete the computation.
| {
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"url": "https://math.stackexchange.com/questions/913798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the value of $\frac{\cos^4\beta}{\cos^2\alpha} + \frac{\sin^4\beta}{\sin^2\alpha}$. Trigonometry
$\dfrac{\cos^4 \alpha}{\cos^2 \beta}+ \dfrac{\sin^4\alpha}{\sin^2\beta} = 1$
then the value of
$\dfrac{\cos^4\beta}{\cos^2\alpha}+ \dfrac{\sin^4\beta}{\sin^2\alpha}$ is?
NOTE: can somebody help me
$\cos^2\alpha \left(\frac{\cos^2 \alpha}{\cos^2 \beta}\right)+ \sin^2\alpha \left(\frac{\sin^2 \alpha}{\sin^2\beta}\right)$
| Let $t=\sin^2(\alpha)$ and $s=\sin^2(\beta)$. Then, multiplying both sides of the given identity by $s(1-s)$ gives:
$$
(1-t)^2s+t^2(1-s)=s(1-s).
$$
Bringing the RHS over to the LHS simplifies to $(s-t)^2=0$ so $s=t$. Now, the expression you want to evaluate is just
$$
\frac{(1-s)^2}{1-s}+\frac{s^2}{s}=1-s+s=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/914519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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An improper integral involving nested square roots I have:
$$I_{a,b}= \int_b^{ + \infty } \left( \sqrt {\sqrt {x + a} - \sqrt x } - \sqrt {\sqrt x - \sqrt {x - b} } \right)dx$$
with $a>0$ and $b>0$.
I should determine whether this is a convergent or divergent integral. The problem is that I don't know how to start.
| We have:
$$\sqrt{x+a}-\sqrt{x}=\frac{a}{\sqrt{x}+\sqrt{x+a}},\qquad \sqrt{x}-\sqrt{x-b}=\frac{b}{\sqrt{x}+\sqrt{x-b}}$$
and:
$$f_{a,b}(x)=\sqrt{\sqrt{x+a}-\sqrt{x}}-\sqrt{\sqrt{x}-\sqrt{x-b}}=\frac{\frac{a}{\sqrt{x}+\sqrt{x+a}}-\frac{b}{\sqrt{x}+\sqrt{x-b}}}{\frac{\sqrt{a}}{\sqrt{\sqrt{x}+\sqrt{x+a}}}+\frac{\sqrt{b}}{\sqrt{\sqrt{x}+\sqrt{x-b}}}}$$
hence $f_{a,b}(x)$ behaves like
$$ x^{1/4}\left(\frac{a}{\sqrt{x}+\sqrt{x+a}}-\frac{b}{\sqrt{x}+\sqrt{x-b}}\right) $$
as $x\to +\infty$. If $a\neq b$, $I_{a,b}$ cannot converge, since $x^{-1/4}\not\in L^1((1,+\infty))$.
If $a=b$ then $f_{a,b}(x)$ behaves like
$$ x^{1/4}\left(\frac{1}{\sqrt{x}+\sqrt{x+a}}-\frac{1}{\sqrt{x}+\sqrt{x-a}}\right)=x^{1/4}\left(\frac{\sqrt{x-a}-\sqrt{x+a}}{(\sqrt{x}+\sqrt{x+a})(\sqrt{x}+\sqrt{x-a})}\right)$$
that is $\Theta(x^{1/4-3/2})$ as $x\to +\infty$, hence $I_{a,b}$ converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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The sequence $(a_0,a_1,a_2,\cdots,)$ satisfies $ a_{n+1}=a_n+2a_{n−1}$. What is $a_5$? Assume that the sequence $(a_0,a_1,a_2,\cdots,)$ satisfies the recurrence $\displaystyle a_{n+1}=a_n+2a_{n−1}$. We know that $a_0=4$ and $a_2=13$. What is $a_5$?
I got $a_1=5, a_3=23, a_4=49, a_5=95$
| For just the 5th term this is overkill, but anyway.
Define the generating function:
$\begin{align*}
A(z)
&= \sum_{n \ge 0} a_n z^n
\end{align*}$
Write the recurrence shifted (subtraction in indices gets nasty), multiply by $z^n$, sum over $n \ge 0$, recognise the sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 2} z^n
&= \sum_{n \ge 0} a_{n + 1} z^n + 2 \sum_{n \ge 0} a_n z^n \\
\frac{A(z) - a_0 - a_1 z}{z^2}
&= \frac{A(z) - a_0}{z} + 2 A(z)
\end{align*}$
Solve for $A(z)$, using initial values, write as partial fractions:
$\begin{align*}
A(z)
&= \frac{4 + 9 z}{(1 + z) (1 - 2 z)} \\
&= \frac{17}{3 (1 - 2 z)}- \frac{5}{3 (1 + z)}
\end{align*}$
We are interested in the coefficients:
$\begin{align*}
a_n
&= [z^n] A(z) \\
&= \frac{17}{3} \cdot 2^n - \frac{5}{3} (-1)^n \\
&= \frac{17 \cdot 2^n - 5 \cdot (-1)^n}{3}
\end{align*}$
Thus:
$\begin{align*}
a_5
&= \frac{17 \cdot 2^5 - 5 \cdot (-1)^5}{3} \\
&= 183
\end{align*}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve such a nonlinear ODE, the analytical solution of which is known! I have the following ODE with initial/boundary value conditions:
$$\left. \begin{aligned}
\left(x^2-10 x-y^2\right)y\, y'(x)+(x-5) y^2 y'(x)^2-(x-5) y^2=0
;\qquad (\text{ODE})\\
y(0)^2=25;\qquad y'(0)^2=\frac{3-\sqrt{5}}{2} \qquad\qquad\qquad (\text{IBCs})
\end{aligned}
\right\}
$$
How to solve such a nonlinear ODE?
Additionally, how can I verify whether a special function is a potential solution or not, e.g., the one in implicit form as below:
$$\left(\sqrt{5}-1\right)\left(x-5\right)^2+2 y^2=25\left(\sqrt{5}+1\right)$$
UPDATE$^{(1)}$
By substituting $y^2$ and $y\,y'(x)$ obtained from the special solution into the original ODE and its IBCs, it seems this is the solution to the original nonlinear ODE problem.
So the only question remaining is how to solve it to obtain the known solution.
UPDATE$^{(2)}$
Below is another solution to the same ODE with the same IBCs:
$$\left(\sqrt{5}+1\right)\left(x-5\right)^2-2 y^2=25\left(\sqrt{5}-1\right)$$
The problem has double solutions:
| Let $U = x^2 + y^2$ and $V = (x-10)^2 + y^2$.
Multiply the equation by $20$ and look at individual terms, we have:
$$\begin{align}
20(x-5) (yy')^2
&= \left[(x^2+y^2) - ((x-10)^2+y^2)\right] (yy')^2\\
&= (U-V)(yy')^2\\
\\
20(x^2 - 10x - y^2) yy'
&= 2yy'\left[(x-10)\left(x^2+y^2\right) - x\left((x-10)^2+y^2\right)\right]\\
&= 2yy'\left[(x-10)U - xV\right]\\
\\
-20(x-5) y^2
&= ((x-10)^2 - x^2)y^2\\
&= (x-10)^2 U - x^2 V
\end{align}$$
So the ODE is equivalent to
$$U ( (yy')^2 + 2(x-10) + (x-10)^2 ) -
V ( (yy')^2 + 2x + x^2 ) = 0\\
\iff U (x - 10 + yy')^2 - V(x + yy')^2 = 0$$
Since $U' = 2(x + yy')$ and $V' = 2(x - 10 + yy')$, this leads to
$$U (V')^2 - V(U')^2 = 0
\quad\iff\quad \frac{V'}{\sqrt{V}} \pm \frac{U'}{\sqrt{U}} = 0
\quad\iff\quad (\sqrt{V} \pm \sqrt{U})' = 0
$$
and hence the solution of the ODE is given by either
$$
\sqrt{(x-10)^2 + y^2} + \sqrt{x^2+y^2} = \text{const}\\
\text{OR}\\
\sqrt{(x-10)^2 + y^2} - \sqrt{x^2+y^2} = \text{const}\\
$$
i.e. an ellipse or a hyperbola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/916768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Taylor polynomial of $\frac{1}{2-x}$ Can someone show how to find the Taylor polynomial of $\frac{1}{2-x}$?
I tried this: $\frac{1}{2-x}=\frac{1}{1-(x-1)}$ and then use that $ \ T_n(\frac{1}{1-x})=1+x+\dots +x^n.$ But this gives $1+(x-1)+\dots +(x-1)^n$ which is the wrong answer. Why?
| Hint:
Try this:
$$\frac{1}{2-x} = \frac{\frac{1}{2}}{1-\frac{x}{2}} = \frac{1}{2}\sum_{n=0}^\infty \left(\frac{x}{2}\right)^n = \sum_{n=0}^\infty 2^{-(n+1)}x^n$$
Your method gives the Taylor expansion around $x=1$ with radius of convergence of $|x-1| < 1$ so for $x = 0$ you have $|x-1| = |-1| = 1 \not < 1$ you are on the boundary of the radius of convergence and you could end up with a delicate case.
$$\frac{1}{2-x} = \frac{1}{1-(x-1)} = \sum_{n=0}^\infty (x-1)^n = \sum_{n=0}^\infty\sum_{k=0}^n \binom{n}{k}(-1)^{n-k}x^k \\ \stackrel{*}{=} \lim_{m\to\infty} \sum_{n=0}^m \binom{n}{0}(-1)^n + \sum_{n=1}^m \binom{n}{1}(-1)^{n-1}x + \ldots + \sum_{n=k}^m \binom{n}{k}(-1)^{n-k}x^k + \ldots$$
$(*)$ is alowed when we have convergence (wich we don't at $x=0$ as we will see). Note that at $x=0$ the first sum is divergent and the other ones are $0$. So you can see that we don't have convergence for $x=0$ in the Taylor expansion around $x=1$.
Note that the Taylor expansion at $x=0$ is often called Maclaurin series. If that's what you wanted then the method at the top is the one to go, otherwise, your method is a correct Taylor expansion, but centered at $x=1$.
| {
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"answer_count": 4,
"answer_id": 2
} |
Inequality: $x^2+y^2+xy\ge 0$ I want to prove that $x^2+y^2+xy\ge 0$ for all $x,y\in \mathbb{R}$.
My "proof": Suppose wlog that $x\ge y$, so $x^2\cdot x\ge x^2\cdot y\ge y^2\cdot y=y^3$ (because $x^2\ge 0$ so we can multiply both sides by it without changing the inequality sign) giving $x^3\ge y^3$. Substracting we have $x^3-y^3\ge 0$ or $(x-y)(x^2+xy+y^2)\ge 0$. Since $x\ge y$ we can divide by $x-y$ to get $x^2+xy+y^2\ge 0$.
Is it right?
Thanks for your help!
| The question posed was whether the OP's proof is correct. Mark Bennet pointed out one flaw in comments, namely that you can't divide by $x-y$ if $x=y$. It's worth noting another problem with the proof: You can indeed suppose wlog that $x\ge y$, from which $x^2\cdot x\ge x^2\cdot y$ follows (because $x^2\ge0$), but the full inequality, ending in $\ge y^2\cdot y=y^3$, does not. If, for example, $x=2$ and $y=-1$, we do not have $2^2\cdot(-1)\ge(-1)^2\cdot(-1)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
} |
Two variables Diophantine equation $x^3 - 2y^2 =54$ How can we solve $x^3 - 2y^2 =54$ Diophantine equation in natural numbers? Modular arithmetic method is not useful maybe?
| Since $x$ must be even, $x=2z$, we have:
$$ 4z^3- y^2 = 27 \tag{1}$$
or:
$$z^3 = \frac{y+3\sqrt{-3}}{2}\cdot\frac{y-3\sqrt{-3}}{2}=w\cdot\overline{w},\tag{2}$$
where $w$ lies in the ring of integers of $\mathbb{Z}[\sqrt{-3}]$, that is a UFD.
This implies that, if $w$ and $\overline{w}$ are coprime, $w=\frac{y-1}{2}+3\omega$ must be a cube in the Eisenstein integers $\mathbb{Z}[\omega]$ (notice that $y$ must be odd, so $\frac{y-1}{2}\in\mathbb{Z}$), but:
$$ (a+b\omega)^3 = a^3+3a^2 b\omega + 3ab^2\omega^2 +b^3 = (a^3+b^3-3ab^2)+3ab(a-b)\omega\tag{3}$$
and $3ab(a-b)=3$ has no integer solutions. The only possibility left is that $w=\sqrt{-3} u^3$ and $\overline{w}= -3v^3$ with $\gcd(u,v)=1$. This happens for $(x,y)=(6,9)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/920760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Properties of a recursively-defined sequence using induction This is a homework problem. Not expecting the solution, just a nudge in the right direction!
$N$ is a function defined inductively as follows:
$$N(1) = N(2) = N(3) = 1$$
$$N(n) = N(n−1) + N(n−3) \quad \text{for } n > 3.$$
a. Prove by induction on $a$, $$N(n) = N(a+2)N(n−1−a) + N(a)N(n−2−a) + N(a+1)N(n−3− a)$$
for $a > 0$ and $n > a + 3$. (Assume $N(0) = 0$).
Induction on "$a$". :O So lost! :'(
b. Assuming the fact from part (a), prove that
$$N(2k) = N(k)N(k) + 2N(k)N(k−2) + N(k−1)N(k−1)$$ and
$$N(2k−1) = N(k)N(k) + 2N(k−1)N(k−2)$$
c. Find similar formulas for $N(2k+1)$ and $N(2k+2)$ in terms of $N(k)$, $N(k−1)$,
and $N(k−2)$.
| (a.1) Basis ($a=1$): you assume $n>4$ and:
$$
N(3)N(n-2) + N(1)N(n-3) + N(2)N(n-4) = N(n-2) + N(n-3) + N(n-4)
=N((n-1) - 1) + N((n-1)-3) + N(n-3) = N(n-1) + N(n-3) = N(n)
$$
Obs: you can check manually the case $n=4$ where the induction formula is not guaranteed to work using $N(0)=0$.
(a.2) Inductive step: assume the formula works for $a$, and as usual, assume $n>a+3$. We work on $a+1$:
$$
N(a+3)N(n-2-a) + N(a+1)N(n-3-a) + N(a+2)N(n-4-a) = \\
=N((a+2)+1)N((n-1-a)-1) + N((a)+1)N((n-2-a)-1) + N((a+1)+1)N((n-3-a)-1)
$$
which we want to show it's still $N(n)$. We can approach it in parts. From the recurrence of the function, we know that $N(n+1) = N(n)+N(n-2)$ and $N(n-1) = N(n)-N(n-3)$. We have 3 product blocks, so let's open up each one:
1. > $(N(a+2) + N(a))(N(n-a-1)-N(n-a-4))$.
2. > $(N(a) + N(a-2))(N(n-2-a) - N(n-a-5))$.
3. > $(N(a+1) + N(a-1))(N(n-3-a)-N(n-a-6))$.
Note that, when you distribute these products, the first term on each of them adds up to $N(n)$ by the induction hypothesis. So, you're left to prove that the sum of the remaining terms is $0$, and believe me, you can. It's pretty gross, so I'm not doing it here (took one A4 page), but multiply them up and use the two recurrence relations I mentioned above and you can check that they actually cancel each other. It's messy, but not hard.
(b) Pick $a=k-2$, and let's apply the formula from the previous item:
$$
N(2k) = N(k)N(k+1) + N(k-2)N(k) + N(k-1)N(k-1) = \\
= N(k)(N(k) + N(k-2)) + N(k-2)N(k) + N(k-1)N(k-1) = \\
N(k)N(k) + 2N(k)N(k-2) + N(k-1)N(k-1),
$$
as desired. For the second formula, again, pick $a=k-2$, and the expression yields:
$$
N(2k-1) = N(k)N(2k-2-k+2) + N(k-2)N(2k-1-2-k+2) + N(k-1)N(2k-1-3-k+2) = \\
= N(k)N(k) + 2N(k-1)N(k-2).
$$
(c) For $N(2k+1)$, pick $a=k-1$. Applying the formula, we have:
$$
N(2k+1) = N(k+1)N(k+1) + 2N(k-1)N(k) = \\
= (N(k)+N(k-2))^{2} + 2N(k-1)N(k).
$$
Finally, for $N(2k+2)$, note that we have a formula for $N(2k)$. Apply that for $2(k+1)$, and we have:
$$
N(2k+2) = N(2(k+1)) = N(k+1)N(k+1) + 2N(k+1)N(k-1) + N(k)N(k) = \\
=(N(k)+N(k-2))^{2} + 2N(k-1)(N(k) + N(k-2)) + (N(k))^{2}
$$
which are all in terms of $N(k),N(k-1)$ or $N(k-2)$, and you can expand it further if you want.
| {
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Prove $3(\sin x-\cos x)^4 + 6(\sin x+ \cos x)^2 + 4(\sin^6 x + \cos^6 x) -13 = 0$
Q) Prove that $3(\sin \theta-\cos \theta)^4 + 6(\sin \theta+ \cos \theta)^2 + 4(\sin^6 \theta + \cos^6 \theta) -13 = 0$
Source: Trigonometric Functions, Page 5.9, Mathematics XI - R.D. Sharma
My Attempt::
For writing convenience, let $\color{red}{s} = \sin \theta$ and $\color{blue}{c} = \cos \theta$
$
\begin{align}
\text{To Prove }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) - 13 = 0\\
\equiv \text{TP }&:\space 3(\color{red}{s}-\color{blue}{c})^4 + 6(\color{red}{s}+\color{blue}{c})^2 + 4(\color{red}{s^6} + \color{blue}{c^6}) = 13
\end{align}
$
$$
\require{cancel}
\begin{align}
\text{LHS }
&= 3\left[(\color{red}{s}-\color{blue}{c})^2\right]^2
+ 6(\color{red}{s} +\color{blue}{c})^2
+ 4(\color{red}{s^6} + \color{blue}{c^6})
\\
&= 3(\color{limegreen}{s^2 + c^2} - 2\color{red}{s}\color{blue}{c})^2
+ 6(\color{limegreen}{s^2 + c^2} + 2\color{red}{s}\color{blue}{c})
+ 4\left[(\color{red}{s^2})^3 + (\color{blue}{c^2})^3\right]
\\
&= 3(1 - 2\color{red}{s}\color{blue}{c})^2 + 6(1+2\color{red}{s}\color{blue}{c})
+ 4(\color{limegreen}{s^2 + c^2})(\color{red}{s^4}
- \color{red}{s^2}\color{blue}{c^2}
+ \color{blue}{c^4})
\\
&= 3(1 - 4\color{red}{s}\color{blue}{c} + 4\color{red}{s^2}\color{blue}{c^2}) + 6 (1+2\color{red}{s}\color{blue}{c}) + 4(\color{red}{s^4} - \color{red}{s^2} \color{blue}{c^2}+\color{blue}{c^4})
\\
&= 3
\cancel{- 12\color{red}{s}\color{blue}{c}}
+12\color{red}{s^2}\color{blue}{c^2}
+ 6
\cancel{+ 12\color{red}{s}\color{blue}{c}} + 4\color{red}{s^4}
-4\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4}\\
&= 4\color{red}{s^4} + 8\color{red}{s^2}\color{blue}{c^2} + 4\color{blue}{c^4} + 9\\
&= 4(\color{red}{s^4} + 2\color{red}{s^2}\color{blue}{c^2} + \color{blue}{c^4}) + 9\\
&= 4(\color{limegreen}{s^2 + c^2})^2 + 9\\
&= 4 + 9 = 13 = \text{RHS} \tag{Q.E.D.}
\end{align}
$$
Thanks to @mathlove, I found and corrected the mistake in my attempt.
$\Huge\color{lightgrey}{☺}$ Although, a quicker alternate way will always be nice.
| Here is another way to prove such identities:
Get the groebner basis of all the relevant polynomial equalities for what you are dealing with. In this case it's very simple since we only have $\sin\theta=s$ and $\cos\theta=c$ and the relevant polynomial equality is $s^2+c^2-1=0$. This is your groebner basis - no extra work required. However if you additionally had terms like $\sin2\theta=s_2$ and $\cos2\theta=c_2$ you would need $s_2^2+c_2^2-1=0$ and equalites that relate $s,c,s_2$ and $c_2$.
Now once you have this groebner basis reduce your polynomial (the equality you want to prove) by the groebner basis. This yields 0 if and only if the polynomial is in the ideal of the polynomial system. If it is a member of the ideal your equality is proved.
The nice thing about this approach is that you can do it in four lines of sympy code:
from sympy import *
s, c = symbols('s, c')
G = groebner([s**2+c**2-1])
print G.contains(3*(s-c)**4+6*(s+c)**2+4*(s**6+c**6)-13)
| {
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"url": "https://math.stackexchange.com/questions/921345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 4
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Find $\lim_{x\to-\infty}\frac{x}{\sqrt{x^2+2}}$ Find $\displaystyle \lim_{x\to-\infty}\frac{x}{\sqrt{x^2+2}}$ without a calculator.
L'Hopital's rule only gives me the limit for one side. Not two. How would I solve this problem?
| One definition of the absolute value function is $|x| = \sqrt{x^2}$. For negative values of $x$, observe that
$$\frac{x}{|x|} = \frac{x}{-x} = -1$$
Thus,
\begin{align*}
\lim_{x \to -\infty} \frac{x}{\sqrt{x^2 + 2}} & = \lim_{x \to -\infty} \frac{x}{\sqrt{x^2\left(1 + \dfrac{2}{x^2}\right)}}\\
& = \lim_{x \to -\infty} \frac{x}{|x|\sqrt{\left(1 + \dfrac{2}{x^2}\right)}}\\
& = \lim_{x \to -\infty} \frac{x}{|x|} \lim_{x \to -\infty} \frac{1}{\sqrt{\left(1 + \dfrac{2}{x^2}\right)}}\\
& = -1 \cdot \frac{1}{\sqrt{1 + 0}}\\
& = -1 \cdot 1\\
& = -1
\end{align*}
| {
"language": "en",
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"question_score": "3",
"answer_count": 5,
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How to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$. I am trying to Integrate $\int\frac{(x^2)}{\sqrt{7-x^2}}dx$ and I have worked this problem a couple times and keep getting the same answer. So I will show my process and please point my errors out.
$$a=\sqrt{7}\hspace{10pt} x=\sqrt{7}\sin\theta\hspace{10pt}dx=\sqrt{7}\cos\theta d\theta \hspace{7pt}7-x^2=7\cos^2\theta$$
$$\int\frac{7\sin^2\theta \sqrt{7}\cos\theta d\theta}{\sqrt{7\cos^2\theta}}$$
$$7\int\sin^2\theta d\theta\rightarrow \frac{7}{2}\int1-\cos(2\theta)d\theta$$
$$\frac{7}{2}\left(\theta-\frac{\sin(2\theta)}{2} \right) \rightarrow \frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{2\sin\theta\cos\theta}{2} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7-x^2}}\frac{\sqrt{7-x^2}}{\sqrt{7}} \right)$$
$$\frac{7}{2}\left(\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-\frac{x}{\sqrt{7}} \right)$$
But apparently the answer is
$$\frac{1}{2}\left(7\sin^{-1}\left(\frac{x}{\sqrt{7}}\right)-x\sqrt{7-x^2} \right)$$
So how do I go about getting that answer? Thanks for all the help in advance.
| You look like you are on the right path, and have the triangle drawn out correctly. I recommend using the substitution $\frac{x}{\sqrt{1-x^2}}=\tan(\theta)$. This should convert your integral into $$7\int \! \tan(\theta)\sin(\theta)\cos(\theta) \, \mathrm{d}\theta=7\int \!\sin^2(\theta) \, \mathrm{d}\theta$$ Now using the identity $sin^2(\theta)=\frac{1-\cos(2\theta)}{2}$ our integral becomes much easier. It should be obvious that $$7\int \! \frac{1-\cos(2\theta)}{2} \, \mathrm{d}\theta=\frac{7}{2}\left[\theta-\frac{1}{2}\sin(2\theta)+C \right]=\frac{7}{2}\left[\theta-\sin(\theta)\cos(\theta)+C \right]$$ Now we can go back to our substitutions from earlier and solve for $\theta$. Using the substitution $x=\sqrt{7}\sin(\theta)$ and $\sqrt{\frac{7-x^2}{7}}=\cos(\theta)$ it shouldn't be too difficult to get $\theta=\arcsin(\frac{x}{\sqrt{7}})$ and $\theta=\arccos\left(\sqrt{\frac{7-x^2}{7}}\right)$. Now we may plug the appropriate $\theta$ back into our indefinite integral to find $$\frac{7}{2}\left[\arcsin\left(\frac{x}{\sqrt{7}}\right)-\sin\left(\arcsin\left(\frac{x}{\sqrt{7}}\right)\right)\cdot\cos\left(\arccos\left(\sqrt{\frac{7-x^2}{7}}\right)\right)+C \right]$$ $$=\frac{7}{2}\left[\arcsin\left(\frac{x}{\sqrt{7}}\right)-\left(\frac{x}{\sqrt{7}}\right)\cdot\left(\sqrt{\frac{7-x^2}{7}}\right)+C \right]$$ This should be identical to your result above, once you factor in the ${7}$ out front.
| {
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Testing convergence of the series of $n^p((n-1)^{-1/2}-n^{-1/2})$ Exercise 8.15 (l) of Analysis by Apostol states:
Test for convergence: $$\sum\limits_{n = 2}^\infty n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}} \right)$$
The solution I have states as the first step: "Note that
$$ n^p\left(\frac{1}{\sqrt{n-1}}-\frac{1}{\sqrt{n}} \right) = \frac{1}{n^{3/2-p}}\left( \sqrt{\frac{n}{n-1}} \frac{1}{1+\sqrt{\frac{n-1}{n}}} \right)"$$
This step is not very self evident for me, could some one help me understand how we would obtain this equality.
| You start with (ignoring the $n^p$ for the time being)
$$\frac{1}{\sqrt{n-1}} - \frac{1}{\sqrt{n}} = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n-1}\sqrt{n}},$$
then you rationalise the numerator, using $\sqrt{a}-\sqrt{b} = \frac{a-b}{\sqrt{a}+\sqrt{b}}$, getting
$$\frac{n-(n-1)}{\sqrt{n-1}\sqrt{n}(\sqrt{n}+\sqrt{n-1})}.$$
Now the numerator is $1$, and when you extract $(\sqrt{n})^3$ from the denominator, you get
$$\frac{1}{n^{3/2}}\frac{1}{\sqrt{\frac{n-1}{n}}\left(1 + \sqrt{\frac{n-1}{n}}\right)},$$
from which the given form is easily obtainable.
| {
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"question_score": "2",
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Solve the following inequality I have the following inequality $\frac {2x^2}{x+2} < x-2$. I tried to solve it the with the following steps.
step 1 $\frac {2x^2}{x+2} < x-2$
step 2 $\frac {2x^2}{x+2} - (x-2) < 0$
step 3 $\frac {2x^2}{x+2} - \frac {(x-2)(x+2)}{1(x+2)} < 0$
step 4 $\frac {2x^2}{x+2} - \frac {x^2-2^2}{x+2} < 0$
step 5 $\frac {2x^2 - x^2 + 4}{x+2} < 0$
step 5 $\frac {x^2 + 4}{x+2} < 0$
step 6 I used character study to get the result x > 2. But this is incorrect.
Where did I go wrong with this?. I feel that I made a mistake somewhere in step 2 but not sure what I did wrong.
Thanks!
| Your inequality holds good if $x>-2$, not $2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Show that $n^3+2n$ is divisible by 3 for all $n\ge 1$ i want to prove it with mathematical induction :
first i am tried with n=0
then it is divisible by zero then i move to next step change all n with K then i get this product :
$$P(K)=K^3+2K = 3m$$
Note: $3k$ because we multiply any no. with $3$ is divisible by $3$
now , next step i am increase $+1$ in $k$ so i get this step :
$$(K+1)^3 + 2(K+1) = 3m$$
so now next step i am not able to solve please help.
| I know you want to prove by induction, but anothter proof would be to observe that every integer is congruent to either 0, 1, or 2 (mod 3). Then we want to see if for every integer n:
$n^3 + 2n \equiv 0\pmod{3}$
Well, take all integers n such that $n \equiv 0\pmod{3}$. Then $n^3 \equiv 0\pmod{3}$ and $2n \equiv 0\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
Take all integers n such that $n \equiv 1\pmod{3}$. Then $n^3 \equiv 1\pmod{3}$ and $2n \equiv 2\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
Take all integers n such that $n \equiv 2\pmod{3}$. Then $n^3 \equiv 2\pmod{3}$ and $2n \equiv 4\equiv1\pmod{3}$ $\implies n^3 + 2n \equiv 0\pmod{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/923765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 6
} |
Show that $ \sum_{k=0}^{r} \binom{r-k}{m} \binom{s+k}{n} = \binom{r+s+1}{m+n+1} $? I can't resolve this exercise and I need a tip.
$$ \sum_{k=0}^{r} \binom{r-k}{m} \binom{s+k}{n} = \binom{r+s+1}{m+n+1} $$
where $ n \geq s $.
| Suppose we seek to evaluate
$$\sum_{k=0}^r {r-k\choose m} {s+k\choose n}$$
where $n\ge s$ and $m\le r.$
Introduce
$${r-k\choose m}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-k-m+1}} \frac{1}{(1-z)^{m+1}} \; dz.$$
Note that this is zero when $k\gt r-m$ so we may extend the sum in $k$
to $k=\infty.$
Introduce furthermore
$${s+k\choose n}
= \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{s+k}}{w^{n+1}} \; dw.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-m+1}} \frac{1}{(1-z)^{m+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{s}}{w^{n+1}}
\sum_{k\ge 0} z^k (1+w)^k
\; dw\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-m+1}} \frac{1}{(1-z)^{m+1}}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{s}}{w^{n+1}}
\frac{1}{1-(1+w)z}
\; dw\; dz.$$
For the geometric series to converge we must have $|z(1+w)|\lt 1$,
which also ensures that the inner pole is not inside the contour.
Observe that $|z(1+w)| = \epsilon |1+w| \le \epsilon (1+\gamma).$ So
we need to choose $1+\gamma \lt 1/\epsilon$ with $\epsilon$ in a
neighborhood of zero. The choice $\epsilon=1/2$ and $\gamma=1/2$ will
work.
Continuing we find
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-m+1}} \frac{1}{(1-z)^{m+2}}
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{(1+w)^{s}}{w^{n+1}}
\frac{1}{1-wz/(1-z)}
\; dw\; dz.$$
Extracting the inner residue we get
$$\sum_{q=0}^n {s\choose n-q} \frac{z^q}{(1-z)^q}.$$
Now
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{r-m-q+1}} \frac{1}{(1-z)^{m+q+2}} \; dz
= {r+1\choose m+q+1}$$
which yields for the sum
$$\sum_{q=0}^n {s\choose n-q} {r+1\choose m+q+1}.$$
Continue by re-indexing for
$$\sum_{q=0}^s {s\choose q} {r+1\choose m+n-q+1}$$
where we have lowered the upper limit to $s$ since the first binomial
coefficient is zero when $q\gt s.$
Using
$${r+1\choose m+n-q+1} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{r+1}}{z^{m+n-q+2}} dz$$
we thus obtain for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{r+1}}{z^{m+n+2}}
\sum_{q=0}^s {s\choose q} z^q dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{(1+z)^{r+s+1}}{z^{m+n+2}}
= {s+r+1\choose n+m+1}.$$
Remark. This can be done using formal power series only.
We have for the sum
$$\sum_{k=0}^r {r-k\choose m} {s+k\choose n}
= \sum_{k=0}^r
[z^{r-k-m}] \frac{1}{(1-z)^{m+1}}
[w^n] (1+w)^{s+k}
\\ = [z^{r-m}] \frac{1}{(1-z)^{m+1}}
[w^n] (1+w)^{s}
\sum_{k=0}^r z^k (1+w)^k.$$
Now we may certainly extend the sum to infinity as there is no
contribution to the coefficient extractor when $k\gt r-m$ (recall that
$r\ge m$) getting
$$[z^{r-m}] \frac{1}{(1-z)^{m+1}}
[w^n] (1+w)^{s}
\sum_{k\ge 0} z^k (1+w)^k
\\ = [z^{r-m}] \frac{1}{(1-z)^{m+1}}
[w^n] (1+w)^{s} \frac{1}{1-z(1+w)}
\\ = [z^{r-m}] \frac{1}{(1-z)^{m+1}}
[w^n] (1+w)^{s} \frac{1}{1-z-wz}
\\ = [z^{r-m}] \frac{1}{(1-z)^{m+2}}
[w^n] (1+w)^{s} \frac{1}{1-wz/(1-z)}.$$
Now with $n\ge s$ we get for the inner coefficient
$$\sum_{q=0}^s {s\choose q} \frac{z^{n-q}}{(1-z)^{n-q}}.$$
Substitute into the outer coefficient extractor to get
$$[z^{r-m}] \frac{1}{(1-z)^{m+2}}
\sum_{q=0}^s {s\choose q} \frac{z^{n-q}}{(1-z)^{n-q}}
= [z^{r-m}]
\sum_{q=0}^s {s\choose q} \frac{z^{n-q}}{(1-z)^{n+m+2-q}}
\\ = \sum_{q=0}^s {s\choose q}
[z^{r-m-n+q}] \frac{1}{(1-z)^{n+m+2-q}}
= \sum_{q=0}^s {s\choose q}
{r+1\choose n+m+1-q}
\\ = \sum_{q=0}^s {s\choose q}
[z^{n+m+1-q}] (1+z)^{r+1}
= [z^{n+m+1}] (1+z)^{r+1}
\sum_{q=0}^s {s\choose q} z^q
\\ = [z^{n+m+1}] (1+z)^{r+1} (1+z)^s
= [z^{n+m+1}] (1+z)^{r+s+1}
= {r+s+1\choose n+m+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/928271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Closed-forms for several tough integrals These integrals came up in the process of finding solution to Vladimir Reshetnikov's problem. I wonder if there are closed-forms for the following integrals:
\begin{array}{1,1}
&[\text{1}] &\quad\int_0^1\frac{\operatorname{Li}_3(ax)}{1+2x}\ dx\\[12pt]
&[\text{2}] &\quad\int_0^1\frac{\operatorname{Li}_2(ax)\ln x}{1+2x}\ dx\\[12pt]
&[\text{3}] &\quad\int_0^1\frac{\ln(1-ax)\ln^2 x}{1+2x}\ dx
\end{array}
I have tried many substitutions, integration by parts, or differentiation under integral sign method, but without success so far. I do not need a complete or rigorous answer and your answer can be only Mathematica's or Maple's output since I don't have those software packages in my computer or links of related papers. I'd be grateful for any help you are able to provide.
| $\def\Li{{\rm Li}}$I think I know how Mr. Tunk-Fey's approach to find solution to Mr. Vladimir Reshetnikov's problem:
\begin{equation}
I=\int_0^1\frac{\ln(1-x)\,\ln(1+x)\,\ln(1+2x)}{1+2x}dx
\end{equation}
He used integration by parts method. Here might be the way he evaluated $I$:
Let
\begin{equation}
u=\ln(1+2x)\quad\Rightarrow\quad du=\dfrac{2\ dx}{1+2x}
\end{equation}
and
\begin{equation}
v=\int\frac{\ln(1-x)\ln(1+x)}{1+2x}\,dx
\end{equation}
In order to evaluate $v$, he might be using his technique in Mr. Vladimir Reshetnikov's another problem:
\begin{equation}
\int_0^1\frac{\ln^2x}{\sqrt{x^2-x+1}}dx
\end{equation}
See also David H's answer. I will just use that technique to evaluate $I$ without providing the complete steps. After several substitutions, we will arrive at the following equation
\begin{align}
v&=\frac{1}{6}\int\frac{\ln^2x}{1-\frac{2}{3}x}\ dx-\frac{1}{2}\int\frac{\ln^2x}{1-2x}\,dx-\frac{1}{2}\int\frac{\ln^2x}{1+x}\ dx-\frac{1}{6}\int\frac{\ln^2x}{1-\frac{1}{3}x}\,dx\\
&=\frac{1}{6}J_1-\frac{1}{2}J_2-\frac{1}{2}J_3-\frac{1}{6}J_4
\end{align}
where
\begin{align}
J_n=\int\frac{\ln^2x}{1-ax}\ dx=\frac{2}{a}\Li_3(ax)-\frac{2}{a}\Li_2(ax)\ln x-\frac{1}{a}\ln(1-ax)\ln^2x+C
\end{align}
Using $J_n$ we get
\begin{align}
J_1&=3\Li_3\left(\frac{2}{3}x\right)-3\Li_2\left(\frac{2}{3}x\right)\ln x-\frac{3}{2}\ln\left(1-\frac{2}{3}x\right)\ln^2x\\
J_2&=\Li_3\left(2x\right)-\Li_2\left(2x\right)\ln x-\frac12\ln\left(1-2x\right)\ln^2x\\
J_3&=-2\Li_3(-x)+2\Li_2(-x)\ln x+\ln(1+x)\ln^2x\\
J_4&=6\Li_3\left(\frac{1}{3}x\right)-6\Li_2\left(\frac{1}{3}x\right)\ln x-3\ln\left(1-\frac{1}{3}x\right)\ln^2x\\
\end{align}
Hence, we have
\begin{align}
uv&=\left.\left[\frac{1}{6}J_1-\frac{1}{2}J_2-\frac{1}{2}J_3-\frac{1}{6}J_4\right]\ln(1+2x)\right|_{x=0}^1\\
&=\left[\frac{1}{2}\Li_3\left(\frac23\right)-\frac{1}{2}\Li_3\left(2\right)+\Li_3\left(-1\right)-\Li_3\left(\frac{1}{3}\right)\right]\ln3\\
&=\left[\frac{1}{2}\Li_3\left(\frac{2}{3}\right)-\frac{1}{2}\Li_3\left(2\right)-\frac{3}{4}\zeta(3)-\Li_3\left(\frac{1}{3}\right)\right]\ln3
\end{align}
The next step is to evaluate $\displaystyle\int v\,du$ of which consists of three general form of integrals:
\begin{align}
\frac{\Li_3(ax)}{1+2x}\,dx\tag1\\[10pt]
\int_0^1\frac{\Li_2(ax)\ln x}{1+2x}\,dx\tag2\\[10pt]
\int_0^1\frac{\ln(1-ax)\ln^2 x}{1+2x}\,dx\tag3
\end{align}
from which the OP follows. To evaluate $I$, the corresponding values of $a$ are $\displaystyle\frac{2}{3},\,2,\,-1,\,$and $\displaystyle\frac{1}{3}$.
I wish I could evaluate $(1),\,(2),\,$and $(3)$ or, at least, evaluate the integrals with the corresponding values of $a$, but I couldn't (Sorry...). Luckily, Mr. Kirill has provided the results for $a=\displaystyle\frac{2}{3}$ and $a=-1$. I hope he will be generous to provide the results for $a=\displaystyle\frac{1}{3}$ and $a=2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 3,
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} |
How to prove the sequence $x_{n+1} = \frac{x_n}2 + \frac 1{x_n}$ is a Cauchy sequence Here is a question I do not know how to prove. Thanks for your helping！
Prove that the sequence $$x_{n+1} = \dfrac{x_n}2 + \dfrac 1{x_n}, x_0 = 1$$is a Cauchy sequence.
| Another method:
Firstly a bit of pedantry. The sequence is decreasing only after the second element. The first three elements of the sequence are $ 1, \frac{3}{2}, \frac{17}{12} $.
Now consider the function $f$ which takes $ x_n \to x_{n + 1} $.
$$ f(x) = \frac x 2 + \frac 1 x \implies f'(x) = \frac{x^2 - 2}{ 2x^2 }$$
$ f' \gt 0 $ if $ x \gt \sqrt 2 $ and
$ f' \lt 0 $ if $ x \lt \sqrt 2 $. So for $ x \gt \sqrt 2 $ the function $f$ is increasing.
We will prove by induction the statement:
$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; P(n)$: $\;\; \sqrt 2 \lt x_{n + 1} \lt x_n \;\;$ for $ n \ge 1 $.
It must first be verified that $ x _1 = \frac 3 2 $ and $ x_2 = \frac{17}{12} $ are both numbers greater than $\sqrt 2$ which can be done by squaring both sides. Then the statement is true for $n = 1$. Suppose $P(n)$ is true for arbitrary $n \in \Bbb N$. Then,
$$ \sqrt 2 \lt x_{n + 1} \lt x_{n} \implies f(x_{n + 1}) \lt f(x_{n }) \implies \sqrt 2 \lt x_{n + 2} \lt x_{n + 1} $$
which requires another small observation (to claim $ f(x_{n + 1}) \gt \sqrt 2 $) $$f(x) = \dfrac{x^2 + 2}{2x} = \dfrac{(x - \sqrt 2)^2 + 2 \sqrt 2 x}{2x} = \dfrac{(x - \sqrt 2)^2}{2x} + \sqrt 2 \gt \sqrt 2 \; \text {if} \; x \gt \sqrt 2$$
Hence $P(n)$ is true for all $n \in \Bbb N$. Hence $(x_n)$ is decreasing and is bounded below by $\sqrt 2$. Hence
$$ (x_n) \; \text{converges} \iff (x_n) \; \text{is a Cauchy sequence} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/930484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Multiplying both sides of an equation when there's a limit on one side?
Determine the value of $a\in\mathbb{R}$, such that $\displaystyle\lim_{x\to 1}\dfrac{x^2+(3-a)x+3a}{x-1}=7$
My attempt:
\begin{align*}
&\lim_{x\to 1} \dfrac{x^2+(3-a)x+3a}{x-1}=7\\
&\implies\lim_{x\to 1} x^2+(3-a)x+3a=7x-7\\
&\implies\lim_{x\to 1} x^2-4x-a(x-3)=-7\\
&\implies1^2-4(1)-a(1-3)=-7\\
&\implies a=-2
\end{align*}
My question is can I multiply both sides of my first line by $x-1$, even though the equation has a $\lim_{x\to 1}$ on the left side, as opposed to a normal equation?
| There is a safer way to think. If you try to plug $1$ into the limit, you'll get a division by zero. So, your only hope for this limit to make sense, is if $x^2 + (3 - a)x + 3a$ is divisible by $x - 1$, so you can simplify and eliminate the division by zero. This means that $1$ must be a root of $x^2 + (3-a)x + 3a$. So: $$1 + (3-a) + 3a = 0 \implies 4 +2a = 0 \implies a = -2.$$
Fortunately, we indeed have: $$\lim_{x \to 1} \frac{x^2 + 5x - 6}{x - 1} = \lim_{x \to 1}\frac{(x+6)(x-1)}{x-1 } = \lim_{x \to 1} x + 6 = 7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/932973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 11,
"answer_id": 0
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Lucas Numbers Proof $L_n = \alpha^n + \beta^n$ Proof by Induction:
Lucas numbers are recursively defined as:
$L_n = L_{n-1} + L_{n-2}$ where $L_1 = 1$ and $ L_2 = 3 $for $n \ge 3$
Show that: $L_n = \alpha^n + \beta^n$ for $\alpha = {1+\sqrt{5}\over2}$ and $\beta = {1-\sqrt{5}\over2}$
Is this the correct approach?
base case:
n = 3
$L_3 = L_2 + L_1 = 4$
$L_3 = \alpha^3 + \beta^3 = 4$
inductive hypothesis:
Assume true: $L_k = L_{k-1} + L_{k-2} = \alpha^k + \beta^k$ for $k\ge3$
inductive step:
$L_{k+1} = L_k + L_{k-1} = \alpha^{k+1} + \beta^{k+1}$
$(\alpha^k + \beta^k) + L_{k-1} = \alpha^{k+1} + \beta^{k+1}$
$L_{k-1} = (\alpha^{k+1} - \alpha^k) + (\beta^{k+1} - \beta^k) = \alpha^{k-1} + \beta^{k-1}$
| Our induction assumption is that the formula holds for all $i\le n+1$, and we show that the formula holds for $i=n+2$. So we are using what is called strong induction.
Actually, it is enough to have as induction assumption that the formula holds at $n=k$ and $n=k+1$, and show it holds at $n=k+2$.
So we know that for a certain $k$ we have $L(k)=\alpha^k+\beta^k$ and $L(k+1)=\alpha^{k+1}+\beta^{k+1}$. We show that $L(k+2)=\alpha^{k+2}+\beta^{k+2}$.
By the recurrence, and the induction assumption, we have
$$L(k+2)=\alpha^k+\alpha^{k+1}+\beta^k+\beta^{k+1}.$$
But it is easy to verify that $\alpha^k+\alpha^{k+1}=\alpha^{k+2}$, with a similar result for the powers of $\beta$.
Detail: We have $\alpha^2=\alpha+1$, basically by the definition of $\alpha$. Multiply both sides by $\alpha^k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving for $x$ given $y = 2x - 9$ and $y = 5$. When does the 5 come into play? I am doing some problems. It states "Isolate the left side for each of the following equations. Then solve for x, assuming the value of y is 5 in a all cases." I noticed that I can get two different answers, depending on when I decide to plug in the value of y. For example, this is what I get:
$$ y = 2x - 9 $$
$$ 9 + y = 2x -9 +9 $$
$$ 9 + y = 2x \frac{1}{2} $$
$$ \frac{1}{2} * 9 + y = x $$
$$ x = \frac{1}{2} * 9 + 5 $$
$$ x = \frac{9}{2} + 5 $$
$$ x = \frac{19}{2} $$
But, if 5 is plugged into y in the beginning, I've seen it done like this:
$$ 5 = 2x - 9 $$
$$ 5 + 9 = (2x - 9) + 9 $$
$$ 14 = 2x -9 + 9 $$
$$ 14 * \frac{1}{2} = 2x * \frac{1}{2} $$
$$ x = 7 $$
Which one is correct?
| You have a mistake in the first. The second has no mistake.
For the first, note that
$$\begin{align}9+y=2x&\iff \frac{1}{2}\left(9+y\right)=\frac 12\times 2x\\&\iff \frac{9+5}{2}=x\\&\iff x=7.\end{align}$$
Note that you need to multiply $9+y$ by $\frac 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral $\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$ I decided to follow a recent trend and ask a question about logarithmic integrals :)
Is there a closed form for this integral?
$$\int_0^1\frac{\log(x)\log^2(1-x)\log^2(1+x)}{x}\mathrm dx$$
| This integral is equal to
$$ -4\big( \zeta(-3,-1,-1,-1) +\zeta(-3,-1,1,-1) +\zeta(-3,1,-1,1) +\zeta(3,-1,-1,-1) +\zeta(3,-1,1,-1) +\zeta(3,1,-1,1) \big) $$
in terms of the multiple zeta function, which can also be simplified to
$$ 2\zeta(-5,-1)-2\zeta(-5,1)+2\zeta(5,-1)+{\textstyle\frac32}\zeta(5,1)+4\zeta(-3,1,1,1), $$
of which only
$$ \begin{aligned}
\zeta(5,1) &= {\textstyle\frac34}\zeta(6)-{\textstyle\frac12}\zeta(3)^2
\\
\zeta(5,-1) &= {\textstyle\frac{111}{64}} \zeta (6)-{\textstyle\frac{9}{32}} \zeta (3)^2-{\textstyle\frac{31}{16}} \zeta (5) \log (2)
\end{aligned}
$$
have a known closed form (see also this article about Euler sums, and also Euler Sums and Contour Integral Representations by Philippe Flajolet and Bruno Salvy).
Update (by editor): Based on MZV reduction of weight $6$, expression above is furtherly simplified to: $$-2 \zeta(\bar5,1)+8 \text{Li}_6\left(\frac{1}{2}\right)+4 \text{Li}_4\left(\frac{1}{2}\right) \log ^2(2)+8 \text{Li}_5\left(\frac{1}{2}\right) \log (2)-\frac{13 \zeta (3)^2}{16}+\frac{7}{6} \zeta (3) \log ^3(2)-\frac{221 \pi ^6}{30240}+\frac{\log ^6(2)}{9}-\frac{1}{12} \pi ^2 \log ^4(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/933977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$ $x,y,z$ are positive real numbers such that $$xyz=1$$ Prove that $\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$. I have no idea how to solve this problem.
I've tried Engel form of Cauchy inequality, but I get $x^2+y^2+z^2$ in denominator, and from condition $xyz=1$ I only find $x^2+y^2+z^2\geq3..$
| $$\dfrac{1}{y(x+y)}+\dfrac{1}{z(y+z)}+\dfrac{1}{x(z+x)} \geqslant \dfrac{3}{2}$$
Since $xyz=1$ we substitute, $x=\frac{a}{b} , y= \frac{b}{c} , z= \frac{c}{a} $
Now, by Engel form(also called Titu lemma)
$$ \sum_{cyc}\frac {a^2}{c^2 + ab }\ge\frac {(a^2 + b^2 +c^2)^2}{a^2b^2 + b^2c^2 + c^2a^2+a^3b + b^3c + c^3a} $$
So it remains to prove
$$ 2(a^2 +b^2+c^2)^2\ge 3\sum_{cyc}a^2b^2 + 3\sum_{cyc}a^3b $$
$$ 2a^4+2b^4+2c^4 +a^2b^2+b^2c^2+a^2c^2 \ge 3(a^3b+b^3c+c^3a) $$
Which is obvious from,
$$ a^4 + b^4 +c^4 \ge a^3b + b^3c + c^3a $$
$$ \sum_{cyc}(a^4 + a^2b^2)\ge \sum_{cyc}2a^3b
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate $\pi$ By Hand? All over the internet the only hand equation i found was
$$\frac\pi4 = 1 - \frac13 + \frac15 - \frac17+\cdots.$$
But this takes something like a thousand iterations to get to four digits, is there a better way to calculate pi by hand?
| One easy-to-understand improvement to your method, which I I don't see used much, is:
$$\pi/6 = \arctan \left ( \frac{\sqrt{3}}{3} \right ) \\
= \int_0^{\frac{\sqrt{3}}{3}} \frac{1}{1+x^2} dx \\
= \sum_{n=0}^\infty \frac{(-1)^{n} \left ( \frac{\sqrt{3}}{3} \right )^{2n+1}}{2n+1} \\
= 3^{-1/2} \sum_{n=0}^\infty \frac{(-1)^{n} 3^{-n}}{2n+1}.$$
Consequently we have
$$\pi = 2 \sqrt{3} \sum_{n=0}^\infty \frac{(-1)^{n} 3^{-n}}{2n+1} \\
\approx 2 \sqrt{3} \sum_{n=0}^N \frac{(-1)^{n} 3^{-n}}{2n+1} \\
= 2 \sqrt{3} \left ( 1 - \frac{1}{9} + \frac{1}{45} - \frac{1}{189} + \dots \right )$$
This gives each decimal digit in slightly fewer than $\log_3(10) \approx 2.1$ steps, provided you can accurately estimate $\sqrt{3}$ to do the final multiplication.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Do we simplify the Proof by Contradiction? Prove the following by contradiction:
Suppose $a,b\in\mathbb{Z}$. If $4|\left(a^2+b^2\right)$, then $a$ and $b$ are not both odd (in other words, $a$ and $b$ are even)
So, I did this:
Assume $a$ and $b$ are odd
Let $a={2k+1}$
Let $b={2l+1}$
$4|\left((2k+1\right)^2+\left(2l+1\right))^2$
$\left(2k+1\right)*\left(2k+1\right)=4k^2+4k+1$
$\left(2l+1\right)*\left(2l+1\right)=4l^2+4l+1$
$4|\left(\left(4k^2+4k+1\right)+\left(4l^2+4l+1\right)\right)$
For this, do we still, simplify it
Or, what do you do after that??
| $$
(2k + 1)^2 + (2l + 1)^2 = 4\left(k^2 + l^2 + k + l\right) + 2
$$
This means that $\left(a^2 + b^2\right) = 4 \lambda + 2$, meaning that $4$ could not possibly divide $\left(a^2 + b^2\right)$, since there's always a remainder of $2$.
...by the way, this means that for $4$ to divide $a^2 + b^2$, both $a$ and $b$ must be even, since certainly if one is odd and the other even, you will get an odd. Further, if they are both even, then $4$ definitely divides them.
| {
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How to solve for X in Cubic poynomial I've been given a Polynomial (Cubic)
$$k=\frac16x\cdot(x+1)\cdot(2x+1)$$
If $k$ is given, is there any way to solve for $x$?
| We solve the problem in a very special case only, where $x$ happens to be a positive integer.
Since $\frac{x(x+1)(2x+1)}{6}\gt \frac{x^3}{3}$, we have $x^3\gt 3k$, and therefore $x\gt (3k)^{1/3}$.
Note that $2(x+1)^3=2x^3+6x^2+6x+2$ while $x(x+1)(2x+1)=2x^3+3x^2+x$. It follows that if $x$ is a positive integer then $\frac{x(x+1)(2x+1)}{6}\lt \frac{(x+1)^3}{3}$, and therefore $x+1\gt (3k)^{1/3}$.
It follows that $x=\lfloor (3k)^{1/3}\rfloor$.
| {
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Is $5^2x^3-x^5 = x^3(x-5)(x+5)$ or $-x^3(5-x)(5+x)$ Geogebra's Factor function says that
$5^2x^3-x^5$
is
$-x^3(x-5)(x+5)$
but from what I do, it is positive, $x^3(5+x)(5-x)$
Note the x isnt in the same position
Am I wrong?
| $5^2x^3-x^5=x^3(5^2-x^2)=x^3(5-x)(5+x)=-x^3(x-5)(x+5)$
| {
"language": "en",
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Help with implicit differentiation simplifications STEP 1: $$ (x+y)^{3} = x^3 + y^3 $$
STEP 2: $$ 3(x+y)^2 (1+ dy/dx) = 3x^2 + 3y^2(dy/dx) $$
STEP 3: $$ 3(x+y)^2 + \frac{dy}{dx}\cdot3(x+y)^2 = 3x^2 + 3y^2(dy/dx)$$
STEP 4: $$\frac{dy}{dx} \cdot 3(x+y)^2 = 3x^2 + 3y^2(dy/dx) - 3(x+y)^2$$
STEP 5: $$\frac{dy}{dx} = (3x^2 + 3y^2(dy/dx))/3(x+y)^2 - 1$$
I do not quite get how to move the dy/dx to one side, and all the others to the
other side to solve for what dy/dx is.
It is one of my worst weaknesses when I try to do an implicit differentiation.
I've searched thorugh wolfram alpha to see how such simplification works,
but some of the problems seems to only have simple step-by-step process shown,
not the whole step-by-step solution.
| Hint $x^3 + y^3 = (x+y)(x^2-xy+y^2)$
so we have the expression
$$(x+y)^3=(x+y)(x^2-xy+y^2)$$
$$(x+y)[(x+y)^2-(x^2-xy+y^2)]=0$$
$$(x+y)(3xy)=0$$
$$x^2y+y^2x=0$$
which when differentiated becomes
$$x^2\displaystyle\frac{dy}{dx}+y(1)+y^2(1)+x\displaystyle\frac{dy}{dx}=0$$
$$\displaystyle\frac{dy}{dx}(x^2+x)=-(y^2+y)$$
$$\displaystyle\frac{dy}{dx}=-\displaystyle\frac{y^2+y}{x^2+x}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $$9^x=5^x+2\sqrt{20^x}+4^x$$
I'm not really sure where to start. I tried simplifying with logarithms and factoring out the x but it ended up looking just as complicated..
| \begin{align*}
9^x&=5^x+2\sqrt{20^x}+4^x\\
9^x-5^x-4^x&=2\sqrt{20^x}\\
\text{At this point, one solution jumps out}&\\
9^2-5^2-4^2\space
=\space 81-25-16\space
=\space 40&=2\sqrt{20^2}\\
\implies x=2
\end{align*}
| {
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Not a perfect square of the form for any integer x. Now a days, I become good fan of this site, as this site making me to learn more math..hahaha.
Okay!
Can we prove that $x^3 + 7$ cannot be perfect square for any positive/negative or odd/even integer of $x$.
I checked with number up to x = 1,...1000. I realized that, not a perfect. But, how do we prove the statement without further checking in place of 1001, ...
| Rewrite our Diophantine equation as $y^2+1=x^3+8$. Note that $x$ cannot be even, for $y^2+1$ cannot be divisible by $4$.
So $x$ must be odd. We have
$$y^2+1=x^3+8=(x+2)(x^2-2x+4).$$
Suppose $x\equiv 1\pmod{4}$. Then $x+2\equiv 3\pmod{4}$, which is impossible, since a positive number of the form $4k+3$ cannot divide $y^2+1$.
Suppose $x\equiv 3\pmod{4}$. Then $x^2-2x+4\equiv 1-2(3)\equiv 3\pmod{4}$, again impossible.
| {
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"source": "stackexchange",
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Prove or disprove $ p^{r+s}\mid q^{ke} - 1 \iff p^s \mid k$. Let $p$ be an odd prime and $q$ be a power of prime. Suppose $e := \min\{\, e \in \mathbb{N} : p \mid q^e - 1 \,\}$ exists. Put $r := \nu_p(q^e - 1)$ (that is, $p^r \mid q^e - 1$ and $p^{r+1} \nmid q^e - 1$). What I want to prove is the following:
$ \forall k \in \mathbb{N},~ \forall s \in \mathbb{N}_0,~p^{r+s} \mid q^{ke} - 1 \iff p^s \mid k.$
I feel that this is true (but perhaps some upper bounds may needed for $k$ and $s$). How can I prove or disprove this statement?
Example:
If $p = 3,~q = 5$ then $e = 2,~r = 1$.
*
*$5^{1 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0}$
*$5^{2 \cdot 2} - 1 = 2^4 \cdot 3^{1 + 0} \cdot 13$
*$5^{3 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 1} \cdot 7 \cdot 31$
*$5^{4 \cdot 2} - 1 = 2^5 \cdot 3^{1 + 0} \cdot 13 \cdot 313$
*$5^{5 \cdot 2} - 1 = 2^3 \cdot 3^{1 + 0} \cdot 11 \cdot 71 \cdot 521$
My attempt:
In case $(\Leftarrow)$,
$$\begin{align}
q^{ke} - 1 &= ((q^e - 1) + 1)^k - 1 \\
&= (q^e - 1)^k + k(q^e - 1)^{k - 1} + \dotsb + \binom{k}{2}(q^e - 1)^2 + k(q^e - 1).
\end{align}$$
I can see that the last term is divisible by $p^{r + s}$ from the assumption. But is it true for the former terms?
In case $(\Rightarrow)$, we may start by introducing the quotient and the remainder $k = ap^s + b~(0 \le b < p^s)$ and calculate similar expansion as before.
| Let $x=q^e-1$. We have $q^{ke}-1 = \exp(k\cdot\ln (1 + x))-1$. Because $\nu_p (x)\geq 1$, we have $\nu_p (\ln (1 + x)) = \nu_p(x)\geq 1$ (look at power series), so that $k\cdot \ln(1+x)$ is in the radius of convergence of the $p$-adic exponential (here we use the fact that $p$ is odd).
Looking again at the power series, we have $\nu_p(q^{ke}-1) = \nu_p (\exp(k\cdot\ln (1 + x))-1) = \nu_p(k\cdot\ln (1 + x)) = \nu_p(k) + \nu_p(q^e-1)$, which is exactly what we wanted to show.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find closed form of summation of Fibonacci Sequence? I created two formulas to prove a binary theory involving the Fibonacci sequence.
(1) $\sum_{i=0}^n F_{2i+1} $
Equation (1) is the sum of all Fibonacci numbers up to $F_n$ where every $i$ in $F_i$ is an odd number.
I had come up with a closed form where (1) = $F_{n+1} -1$
Is this right? If so, how do I prove it?
| The Fibonacci numbers have the form $(\alpha - \beta) F_{n} = \alpha^{n} - \beta^{n}$, where $2 \alpha = 1 + \sqrt{5}$ and $2 \beta = 1 - \sqrt{5}$. Now
\begin{align}
\sum_{k=0}^{n} (-1)^{k} F_{k} &= \frac{1}{\alpha - \beta} \sum_{k=0}^{n} \left( (- \alpha)^{k} - (- \beta)^{k} \right) \\
&= \frac{1}{\alpha - \beta} \left( \frac{1 - (-\alpha)^{n+1}}{1+\alpha} - \frac{1 - (-\beta)^{n+1}}{1+ \beta} \right) \\
&= \frac{1}{\alpha - \beta} \left( (-1)^{n} (\alpha^{n-2} - \beta^{n-2}) - (\alpha^{2} - \beta^{2}) \right) \\
&= (-1)^{n} F_{n-2} - (\alpha + \beta) \\
&= (-1)^{n} F_{n-2} -1.
\end{align}
Since $F_{0} = 0$ the summation can also be seen as
\begin{align}
\sum_{k=1}^{n} (-1)^{k} F_{k} = (-1)^{n} F_{n-2} - 1.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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A Polynomial that Passes through the following four points? I'm trying to do this for practice but I'm just going nowhere with it, I'd love to see some work and answers on it.
Thanks :)
Find a polynomial that passes through the points (-2,-1), (-1,7), (2,-5), (3,-1). Present the answer in standard form.
What I've tried:
| $$f(x) = a x^3 + b x^2 + c x + d$$
\begin{eqnarray*}
-1 &=& a (-2)^3 + b(-2)^2 + c(-2) + d \\
7 &=& a (-1)^3 + b(-1)^2 + c(-1) + d \\
-5 &=& a(2)^3 + b (2)^2 + c (2) + d \\
-1 &=& a(3)^3 + b(3)^2 + c (3) + d
\end{eqnarray*}
or
\begin{eqnarray*}
-8 a + 4b -2c + d &=& -1\\
-{\ }a{\ } +{\ } b -{\ } c + d &=&{\ } 7 \\
8a + 4b + 2c + d&=&-5 \\
27a + 9b + 3c + d&=&-1
\end{eqnarray*}
with solution
\begin{eqnarray*}
a&=&{\ }{\ }1\\
b&=&-2\\
c&=&-5\\
d&=&{\ }{\ }5
\end{eqnarray*}
and so $f(x) = x^3 -2x^2-5x+5$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the order of the elements in the given groups I have to find the order of the following elements in the given groups:
*
*$(1 \ \ 2 \ \ 3) \ (1 \ \ 2\ \ 4) \text{ in } S_5$
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5\\
3 & 4 & 1 & 2 & 5
\end{pmatrix}=\begin{pmatrix}
1 & 3
\end{pmatrix}\begin{pmatrix}
2 & 4
\end{pmatrix}$$ order $=\operatorname{lcm}(2,2)=2$
Is it correct?
*
*$a^{45}$ in $G=\langle a\mid a^{140}=1\rangle$
$$\operatorname{ord}(a^{45})=\frac{140}{\gcd(45, 140)}=\frac{140}{5}=28$$
Is it right?
*
*$\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} \text{ in } GL_2(\mathbb{R})$
$$A=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} \\ A^2=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} =\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix} \\ A^3=\begin{bmatrix}
-1 & 0\\
0 & -1
\end{bmatrix} \begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix} =\begin{bmatrix}
0 & 0\\
0 & 0
\end{bmatrix} $$
But how can I find the order?
| The first two computations are correct.
The matrix $A$ is invertible (because $\det A=1$), so $A^3$ can't be the zero matrix. Once you found that $A^2=-I_2$ ($I_2$ the identity matrix), you know that $A^4=(A^2)^2=I_2^2=\dots$
Thus you know the order of $A$, because…
| {
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For all real $\theta$ prove that $ \cos(\sin\theta) \gt \sin(\cos\theta)$ How do I prove this? Im not able to even start it. Help please!
| Try graphing both of the functions.
Also, this answer may help:
$$
\begin{align}
&\cos (\sin x) - \sin (\cos x) > 0\\
\implies &\cos (\sin x) - \cos ( π/2 - \cos x ) > 0\\
\implies &2 \sin \left[ \frac{\pi}{4} + \frac{1}{2}(\sin x - \cos x) \right]\cdot \sin \left[ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) \right] > 0 \tag{1}
\end{align}
$$
If we could prove that both the factors on the left hand side of $(1)$ are positive then the result obtained above $(1)$ is proved.
Since: $$\left| \sin x - \cos x \right| = \left| √2 \sin (x- \frac{\pi}{4}) \right| ≤ √2 < \frac{\pi}{2} $$
We have,
$$- \frac{\pi}{2} < ( \sin x - \cos x ) < \frac{\pi}{2}\\
\implies - \frac{\pi}{4} < ( \sin x - \cos x )/2 < \frac{\pi}{4} $$
So that, $$0 < \frac{\pi}{4}+ \frac{1}{2}( \sin x - \cos x ) < \frac{\pi}{2}$$
$\therefore \space \sin [ \frac{\pi}{4} + \frac{1}{2} (\sin x - \cos x) ] > 0 \quad\text{(ie, Positive)}$
Similarly we can prove that
$\sin [ \frac{\pi}{4} - \frac{1}{2}(\sin x - \cos x) ] > 0$
Hence $(1)$ is true. QED
Reference: Yahoo Answers
| {
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Does a maximum value exist for this expression? Let $x$, $y$, $z$ be positive real numbers and $x + y + z =3$. Does a maximum value exist for this expression?
$$\displaystyle E = \frac{x}{2 y+3 z}+\frac{4 y}{5 z + 6 x}+\frac{7 z}{8x+9 y}.$$
I tried
Put
$$a=2 y+3 z,\quad b=6 x+5 z,\quad c=8 x+9y.$$
Then
$$x = -\dfrac{45a}{242}+\dfrac{27b}{242}+\dfrac{5c}{121},\quad y = \dfrac{20a}{121}-\dfrac{12b}{121}+\dfrac{9c}{121}, \quad z = \dfrac{27a}{121}+\dfrac{8b}{121}-\dfrac{6c}{121}.$$
The expression $E$ has the form
$$E = \frac{378 a^2 b+160 a^2 c+112a b^2-225 a b c+72 a c^2+27b^2 c+10 b c^2}{242 a b c}.$$
From here, I can't solve it.
| The expression $E$ is not bounded above. Let $x$ be very large, and $y$ and $z$ small.
Edit: The condition $x+y+z=3$ has been added. The expression $E$ is still not bounded above. Let $x$ be very close to $3$, making $y$ and $z$ close to $0$. The term $\frac{x}{2y+3z}$ can be made arbitrarily large.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation Ax=x can be rewritten as (A-I)x = 0 and use this result to solve Ax=x for x. Given matrix A = \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix}
and x = \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix}
Answer: Given $Ax = x$. Subtract from x from both sides to get $Ax - x = 0$. We know $x = I \cdot x$ , where $I$ is the identity matrix.
Simplify $Ax - x = 0$ to get $x(A - I) = 0$.
$$\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} \cdot \left(\begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \right) = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$
$$\begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} \cdot \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$
Do I need to write the system of linear equations?
| First it is important, that $\textbf A \textbf x - \textbf x = \textbf 0 \Rightarrow (\textbf A - \textbf I)\textbf x=0$. The vector $\textbf x$ has to be on the right side of $(\textbf A-\textbf I)$. Otherwise it wouldn´t work. And you have to subtract $\textbf I$ from $\textbf A$.
$$ \left( \begin{bmatrix}2 & 1 & 2 \\ 2 & 2 & -2 \\ 3 & 1 & 1\end{bmatrix} -\begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix} \right) \cdot \begin{bmatrix}x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$
After calculating the brackets you can write a linear system of equations.
| {
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"timestamp": "2023-03-29T00:00:00",
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What are the ranges of triangle angles? Lets say, that $\alpha \le \beta \le \gamma$. As shown here, $60 \le \gamma \lt 180$. What are the minimum and maximum values of $\alpha$ and $\beta$?
The answer:
$$0\lt \alpha \le 60 \\ 0 \lt \beta \lt 90\\ 60 \le \gamma \lt 180$$
| If $\alpha \le \beta \le \gamma$ are triangle angles, then their ranges are:
$$
0^\circ \lt \alpha \le 60^\circ\\
0^\circ \lt \beta \lt 90^\circ\\
60^\circ \le \gamma \lt 180^\circ$$
Proof
The lower $\alpha$, $\beta$ and the upper $\gamma$ bounds can be determined from the case, when $\gamma \rightarrow 180^\circ$, $\alpha \rightarrow 0^\circ$, $\beta \rightarrow 0^\circ$. Since $\alpha + \beta + \gamma = 180^\circ$, $\gamma$ cannot be greater than $180^\circ$. And $\alpha$, $\beta$ cannot be $\le 0^\circ$, because when $\alpha = 0^\circ$ we get a line, not a triangle. Hence, $0^\circ < \alpha$, $0^\circ < \beta$ and $\gamma < 180^\circ$.
The upper $\beta$ bound can be determined from the case, when $\gamma \rightarrow 90^\circ$, $\beta \rightarrow 90^\circ$ and $\alpha \rightarrow 0^\circ$. Since $\alpha \ne 0^\circ$, $\beta \le \gamma \lt 90^\circ$. Hence, $\beta \lt 90^\circ$.
Since $\alpha \le \beta \le \gamma$, we get that the upper $\alpha$ bound is when $\alpha = \beta = \gamma = 60^\circ$. So, $\alpha \le 60^\circ$.
MathLove proved that the lower $\gamma$ bound is $60^\circ$. The proof is:
Let $\alpha \le \beta \le \gamma$ be the inner angles of a triangle. Suppose that $\gamma \lt 60^\circ$. Then, $180^\circ = \alpha + \beta + \gamma \lt 60^\circ + 60^\circ + 60^\circ = 180^\circ$ which is a
contradiction.
Hence, $60^\circ \le \gamma$. $\Box$
| {
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Find $z^{10}+\frac{1}{z^{10}}$ given $z^2+z+1=0$ $z$ is a complex number and $z^2+z+1=0$.
$$z^{10}+\frac{1}{z^{10}}=?$$
For the solution:
*
*the roots of $z^2+z+1$ are: $z_1=-\frac12+\frac{\sqrt3}{2}i$ and $z_2=-\frac12-\frac{\sqrt3}{2}i$
*converting these to their trigonometrical forms, we get: $z_1=\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ and $z_2=\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6}$
*How do I proceed?
| Hints:
*
*Note $z_1z_2=1$
*You could write your (2) as $z_1=e^{i2\pi/3}$ and $z_2=e^{-i2\pi/3}$.
*Taking the $10$th power of an exponential is easy
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/952968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
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} |
How to solve this differential equation: $x^2dy-y^2dx+xy^2(x-y)dy=0$
$$x^2dy-y^2dx+xy^2(x-y)dy=0$$
What I tried:
$$\frac{x^2}{y^2} \frac{dy}{dx}+x(x-y)\frac{dy}{dx}=1\\$$
Let $h=-1/x, \; k=-1/y,\; dh=1/x^2 \, dx, \; dk=1/y^2 \,dy$
$$\frac{dk}{dh}+\frac{(k-h)}{k^2} \frac{dk}{dh}=1\\
\frac{dk}{dh}+\frac hk=1+\frac1k\\
he^{\int-1/k^2\; dk}=\int\left(1+\frac1k\right)e^{\int-1/k^2\; dk}dk\\
he^{-y}=\int\frac{1-y}{y^2}e^{-y}dy=\int\left(\frac 1{y^2}-\frac1y\right)e^{-y}dy$$
Which probably is unsolvable?I tried using IBP on RHS.Dont use Ricatti Eqn(Not in my course)
Answer is:
$$\large\ln\left|\frac{x-y}{xy}\right|=\frac{y^2}2+\mathcal C$$
| The equation is $\displaystyle \frac{dy(x)}{dx}\cdot{\frac{x^2}{y(x)^2}}+\frac{dy(x)}
{dx}\cdot{x(x-y(x))}=1$.
Multiplying by $y(x)^2$ we get $$\displaystyle \frac{dy(x)}{dx}\left[x^2+x(x-y(x))y(x)^2\right]=y(x)^2 \rightarrow \frac{dy(x)}{dx}=\frac{y(x)^2}{x^2+x^2y(x)^2-xy(x)^3}$$
Now, $\displaystyle \frac{dy(x)}{dx}=\large \frac{1}{\frac{dx(y)}{dy}}$, so $$\displaystyle \frac{1}{\frac{dx(y)}{dy}}=\frac{y^2}{x(y)^2+x(y)^2y^2-x(y)y^3} \Rightarrow \frac{dx(y)}{dy}=x(y)^2-yx(y)+\frac{x(y)^2}{y^2}$$
Adding $yx(y)$ and dividing by $-x(y)^2$ we get $\displaystyle -\frac{\frac{dx(y)}{dy}}{x(y)^2}-\frac{y}{x(y)}=-\left(\frac{1}{y^2}+1\right)$.
Let $\displaystyle v(y)=\frac{1}{x(y)} \rightarrow \frac{dv(y)}{dy}=-\frac{\frac{dx(y)}{dy}}{x(y)^2}$.
Multiplying both sides by $\displaystyle e^{\Large \frac{-y^2}{2}}$ we get$$\displaystyle e^{\large -\frac{y^2}{2}}\frac{dv(y)}{dy}-\left(e^{\large-\frac{y^2}{2}}y\right)v(y)=-e^{\large-\frac{y^2}{2}}\left(\frac{1}{y^2}+1\right)$$
Using the reverse product rule to get $$\displaystyle \frac{d}{dy}\left(e^{\large -\frac{y^2}{2}}v(y)\right)=-e^{\large-\frac{y^2}{2}}\left(\frac{1}{y^2}+1\right)$$
Integrate both sides to get $$\displaystyle e^{\large -\frac{y^2}{2}}v(y)=-\int e^{\large -\frac{y^2}{2}}\left(\frac{1}{y^2}+1\right)dy$$
Using the result shown in the link Integrate $e^{-\frac{y^2}{2}}\left(\frac{1}{y^2}+1\right)$ we get $$\displaystyle e^{\large -\frac{y^2}{2}}v(y)=\frac{e^{\large -\frac{y^2}{2}}}{y}+\mathcal{C_1}$$
We get that $\displaystyle v(y)=\frac{1}{y}+\mathcal{C_1}e^{\large \frac{y^2}{2}}$.
Now we can write that $\displaystyle v(y)-\frac{1}{y}=\mathcal{C_1}e^{\large \frac{y^2}{2}}$.
Taking logarithm from both sides we get $\displaystyle \ln\left|v(y)-\frac{1}{y}\right|=\frac{y^2}{2}+\ln(\mathcal{C_1})$.
Substitute $\displaystyle v(y)=\frac{1}{x(y)}$ we find that the final answer is $\displaystyle \ln\left|\frac{1}{x}-\frac{1}{y}\right|=\frac{y^2}{2}+C_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/953648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}$ May I know how can I calculate the following expression?
$$
\lim\limits_{x\to-\infty}\frac{\sqrt{6x^2 - 2}}{x+2}
$$
From calculator, the answer is $-\sqrt{6}$ , my approach is by using dividing numerator and denominator by using $x$. Which is,
$$
\lim\limits_{x\to-\infty}\frac{\frac{\sqrt{6x^2 - 2}}{x}}{\frac{x+2}{x}}
=
\lim\limits_{x\to-\infty}\frac{{\sqrt{\frac{6x^2 - 2}{x^2}}}}{\frac{x+2}{x}}=\sqrt{6}
$$
My answer is $\sqrt{6}$, is my working wrong or there are actually another approach? Thank you.
| it is $\frac{|x| \sqrt{6-\frac{2}{x^2}}}{x(1+\frac{2}{x})}=-\frac{\sqrt{6-\frac{2}{x^2}}}{1+\frac{2}{x}}$ and the searched limit is $-\sqrt{6}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Different methods to multiply The sum and the multiplication in $\mathbb{Z}_p$ correspond to the the sum and the multiplication of powerseries.
Example for sum:
$$(3 \cdot 7^0+4 \cdot 7^1+2 \cdot 7^2)+(5 \cdot 7^0+3 \cdot 7^1)= \\ (3+5) \cdot 7^0+(4+3) \cdot7^1+2 \cdot 7^2= \\ (7+1) \cdot 7^0+7 \cdot 7^1+2 \cdot 7^2= \\ 1\cdot 7^0+(7+1) \cdot 7^1+2 \cdot 7^2= \\ 1 \cdot 7^0+1 \cdot 7^1+(2+1) \cdot 7^2=\\ 1 \cdot 7^0+1 \cdot 7^1+3 \cdot 7^2$$
An other method for the calculation:
(we start from the left side)
Example for multiplication:
$$(3 \cdot 7^0+4 \cdot 7^1+2 \cdot 7^2) \cdot (5 \cdot 7^0+3 \cdot 7^1)= \\ (3 \cdot5) \cdot 7^0+(4 \cdot 5+3 \cdot 3) \cdot7^1+(4 \cdot 3+2 \cdot 5) \cdot 7^2+(2 \cdot 3) \cdot 7^3= \\ 15 \cdot 7^0+29 \cdot 7^1+22 \cdot 7^2+6 \cdot 7^3= \\ (2 \cdot 7+1)\cdot 7^0+(4 \cdot7+1) \cdot 7^1+(3 \cdot 7+1) \cdot 7^2+6 \cdot 7^3= \\ 1 \cdot 7^0+(2+1) \cdot 7^1+(4+1) \cdot 7^2+(6+3) \cdot 7^3=\\ 1 \cdot 7^0+3 \cdot 7^1+5 \cdot 7^2+2 \cdot 7^3+1 \cdot 7^4$$
An other method for the calculation:
(we start from the left side)
I have not understand the other method for the multiplication.
Any help would be appreciated!
| The other method for multiplication is a reflected version of "long multiplication".
You'll forgive me if I start on the left with the highest power of 7 (in your example) and align on the right, because what we're looking at here is working arithmetic in base 7. But that just means you need to read the other way across the page.
The multiplication separates one of the multipliers into its component powers, does two simpler multiplications then adds the results:
$$(2 \cdot 7^2+4 \cdot 7^1+3 \cdot 7^0)\cdot \overbrace{(3 \cdot 7^1+5 \cdot 7^0)}^\text{separate this}= \hspace{2in} \\
(2 \cdot 7^2+4 \cdot 7^1+3 \cdot 7^0)\cdot (3 \cdot 7^1) \hspace{1in} \\
\hspace{1in} +(2 \cdot 7^2+4 \cdot 7^1+3 \cdot 7^0)\cdot (5 \cdot 7^0) \\
\begin{array}{rrrr}
\hline
=&&(2\cdot 3 ) 7^3 &+(4\cdot 3) 7^2 &+(3\cdot 3) 7^1&\\
+&& &(2\cdot 5) 7^2 &+(4\cdot 5) 7^1&+(3\cdot 5)7^0\\
\hline
=&(1 ) 7^4&~_1+(0 ) 7^3 &~_1+(6) 7^2 &~_1+(2) 7^1&\\
+&& (1 ) 7^3&~_1+(6) 7^2 &~_3+(1) 7^1&~_2+(1)7^0\\
\hline
=&(1)7^4&+(2) 7^3 &+(5) 7^2 &+(3) 7^1&+(1)7^0\\
\end{array}$$
The multiplications of each component are resolved in my version right to left. The small subscripts are the multiple of 7 taken out of the lower-power result to their right and to be added on to the results of the multiplication for the next higher power on their left - the "carries".
Note in particular that the numbers in brackets on the last few lines correspond exactly to (a reflected version of) the numbers in the layout that you had problems with.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/956859",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Residue Calculus (Computing an Improper Integral)
Use residue calculus to compute the integral $\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz$
My solution
If we add to the interval $I_{R}=[-R,R]$ add the semicircle $\gamma_{R}$ in the upper half plane with centre at the origin and radius $R$, then we obtain a closed contour $\Gamma_{R}$ (which we consider to be oriented in the positive direction over which the integral $f(z)=\frac{1}{(z^{2}+25)(z^{2}+16)}$ can be computed using the Cauchy's Residue theorem.
\begin{align}
\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz&=\lim_{R \to \infty}\int_{\Gamma_{R}}f(z)dz \\ &=2\pi i \sum_{k=1}^{2}\text{Res}(z_{k}), \quad \Im(z_{k})>0
\end{align}
In this case, we have two simple poles at the points $z_{1}=4i$ and $z_{2}=5i$. The residues are computed as
\begin{align}
\text{Res}(z_{1})&=\frac{1}{(z^{2}+25)D(z^{2}+16)}\big|_{z=4i} \\
&=\frac{1}{(z^{2}+25)2z}\big|_{z=4i} \\
&=\frac{1}{9\cdot8i} \\
&=\frac{1}{72i} \\
&=-\frac{i}{72}
\end{align}
Similarly, we get for $z_{2}$ that
\begin{align}
\text{Res}(z_{2})&=\frac{1}{D(z^{2}+25)(z^{2}+16)}\big|_{z=5i} \\
&=\frac{1}{2z(z^{2}+16)}\big|_{z=5i} \\
&=\frac{1}{10i \cdot (-9)} \\
&=-\frac{1}{90i} \\
&=\frac{i}{90}
\end{align}
Thus, we get
\begin{align}
\int_{-\infty}^{\infty}\frac{1}{(z^{2}+25)(z^{2}+16)}dz &=2\pi i \left(\text{Res}(z_{1})+\text{Res}(z_{2})\right) \\
&=2 \pi i \left(-\frac{i}{360}\right) \\
&=\frac{\pi}{180}
\end{align}
Are there other methods one could use to arrive at this answer?
| Yes. Simply rewrite $$\int \frac{1}{(z^2+16)(z^2+25)} \mathrm{d}z= -\frac{1}{225} \int \frac{1}{\frac{z^2}{25}+1} \mathrm{d}z +\frac{1}{144} \int \frac{1}{\frac{z^2}{16}} \mathrm{d}z$$ and substitute $u=\frac{z}{5}$ for the first integral and $v=\frac{z}{4}$ for the second.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/957271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Chinese Remainder Theorem RSA I want to solve the following modular quadratic equation:
$x^2 \equiv 188 \pmod {437}$ using the fact that $437$ can be factorized by the primes as: $19⋅23$.
So far I have done:
$$x^2 \equiv 188 \pmod{19} \Rightarrow x^2 \equiv 17 \pmod {19}\\
x^2 \equiv 188 \pmod {23} \Rightarrow x^2 \equiv 4 \pmod {23}$$
$$x^2 \equiv 7 \pmod{ 19} \Rightarrow x \equiv \pm 6 \pmod {19}\\
x^2 \equiv 4 \pmod {23} \Rightarrow x \equiv \pm 2 \pmod {23}$$
And the last step. But I do not understand how to combine:
$$x \equiv 6 \pmod {19} \text{ and } x \equiv 2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv 2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod{ 437}\\
x \equiv 6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}$$
Please help me fill the "insert-her".
| In general, we have distinct primes $p,q$; we have integers $a \in [0,p)$ and $b \in [0,q)$; and we want $x$ such that $x \equiv a$ mod $p$ and $x \equiv b$ mod $q$.
To solve this equation, the first step is to calculate $r = p^{-1}$ mod $q$. Assuming we have done that, the rest is easy: just set $x = (b-a)rp + a$. Then $x \equiv a$ mod $p$, obviously; and $rp \equiv 1$ mod $q$, so $x \equiv (b-a)rp + a \equiv (b-a)+a \equiv b$ mod $q$.
In practice, you will usually want $x$ to satisfy $0 \le x < pq$; this can be done by reduction mod $pq$.
So the only thing left is how to calculate $r = p^{-1}$ mod $q$. This is a well-known application of the Euclidean algorithm, which you can read about in this Wikipedia article.
In your example, we have $p=19, q=23$; and applying the Euclidean algorithm will give you $r=19^{-1}$ mod $23 = 17$. So, for instance, if $x \equiv 6$ mod $19$ and $x \equiv 2$ mod $23$, we have $x = (2-6)\cdot 17 \cdot 19 + 6 = -1286 \equiv 25$ mod $437$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/958072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
An equality about sums of Bernstein polynomials: $\sum_{k=0}^n \left(x - \frac kn\right)^2 p_{nk}(x) = \frac{x(1-x)}{n}$ We defined the Bernstein polynomials as following
$$
p_{nk}
\ = \
\frac{n!}{k!(n-k)!} x^k(1-x)^{n-k}$$
I have to show this:
$$
\sum_{k=0}^n \left(x - \frac kn\right)^2 p_{nk}(x) \ = \ \frac{x(1-x)}{n}
$$
My own work
It is easy to show that
$$
(1-x+tx)^n \ = \ \sum_{k=0}^n p_{nk}(x)t^k
$$
From this I have deduced that
$$
\sum_{k=0}^n p_{nk}(x) = 1
\qquad \text{and} \qquad
\sum_{k=0}^n kp_{nk}(x) = nx
\qquad \text{and} \qquad
\sum_{k=0}^n k(k-1)p_{nk}(x) = n(n-1)x^2
$$
Now by dividing and plugging in I got that:
$$
\frac{1}{n}(x-x^2)
\ = \
\frac{1}{n^2(n-1)} \left((n-1)\sum_{k=0}^n kp_{nk}(x) + \sum_{k=0}^n k(k-1)p_{nk}(x)\right)
$$
Which equals
$$
\frac{1}{n^2(n-1)} \sum_{k=0}^n k p_{nk}(x)\left( (n-1) - (k-1)\right)
\ = \ \frac{1}{n^2(n-1)} \sum_{k=0}^n k p_{nk}(x)(n-k)
$$
The left hand side is equal to this though:
$$
\sum_{k=0}^n \left(x^2 - \frac {2k}n+ \frac{k^2}{n^2}\right) p_{nk}(x)
$$
And these things don't look equal. Please help me!
| It is simpler to start from $\sum_{k=0}^n\left(x-\frac kn\right)^2p_{n,k}(x)$, using the fact that $\sum_{k=0}^nk^2p_{n,k}(x)=n(n-1)x^2+nx$. This gives
\begin{align}
\sum_{k=0}^n\left(x-\frac kn\right)^2p_{n,k}(x)&=x^2-\frac{2x}n\sum_{k=0}^nkp_{n,k}(x)+\frac 1{n^2}(n(n-1)x^2+nx)\\
&=x^2-\frac {2x}n\cdot nx+\frac 1{n^2}(n(n-1)x^2+nx)\\
&=x^2-2x^2+\frac{(n-1)x^2+x}n \\
&=-x^2+x^2+\frac{x-x^2}n\\
&=\frac{x(1-x)}n.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Mathematical Induction Question, Proof Help Prove using Mathematical Induction that for all natural numbers ($n>0$):
$$
\frac 1 {\sqrt{1}} + \frac 1 {\sqrt{2}} + \cdots + \frac 1 {\sqrt{n}} \ge \sqrt{n}.
$$
Proof by Induction:
Let P(n) denote 1/ √1 + 1/ √2 + … + 1/ √n ≥ √n
Base Case: n = 1, P(1) = 1/√1 ≥ √1
The base cases holds true for this case since the inequality for P(1) holds true.
Inductive Hypothesis: For every n = k > 0 for some integer k
P(k) = 1/ √1 + 1/ √2 + … + 1/ √k ≥ √k, p(k) holds true for any integer k
Inductive Step:
P(k + 1)) = 1/ √1 + 1/ √2 + … + 1/ √k + 1/ √(k + 1) ≥ √k + √(k+1)
√k + √(k+1) > √(k+1) (this is where I got stuck)
| Assuming $\sum_{i=1}^n \frac{1}{\sqrt{i}} \geq \sqrt{n}$, we have
\begin{align*}
\sum_{i=1}^{n+1} \frac{1}{\sqrt{i}} &= \sum_{i=1}^n \frac{1}{\sqrt{i}} + \frac{1}{\sqrt{n+1}} \\
&\geq \sqrt{n} + \frac{1}{\sqrt{n+1}} \\
&= \frac{\sqrt{n^2+n} + 1}{\sqrt{n+1}} \\
&\geq \frac{\sqrt{n^2} + 1}{\sqrt{n+1}} = \frac{n+1}{\sqrt{n+1}} = \sqrt{n+1}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/959654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
$ \lim\limits_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$ Find the limit: $$ \lim_{x \rightarrow \ + \infty}(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})$$
I did the following:
\begin{align}
(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x}) = \frac{(\sqrt{x^2 + 2x} - \sqrt{x^2 - 7x})}{1} \cdot \frac{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}{(\sqrt{x^2 + 2x} + \sqrt{x^2 - 7x})}
\end{align}
I know the final answer is $\frac{9}{2}$. After multiplying by the conjugate, I see where the $9$ in the numerator comes from. I just can't remember how I solved the rest of the problem.
| $$\lim_{x\to \infty}\frac{(\sqrt{x^2+2x}-\sqrt{x^2-7x})(\sqrt{x^2+2x}+\sqrt{x^2-7x})}{\sqrt{x^2+2x}+\sqrt{x^2-7x}}$$
$$=\lim_{x\to \infty}\frac{9x}{\sqrt{x^2+2x}+\sqrt{x^2-7x}}=\lim_{x\to \infty}\frac{\frac{9x}{x}}{\frac{\sqrt{x^2+2x}}{x}+\frac{\sqrt{x^2-7x}}{x}}$$$$=\lim_{x\to \infty}\frac{9}{\sqrt{\frac{x^2+2x}{x^2}}+\sqrt{\frac{x^2-7x}{x^2}}}=\lim_{x\to \infty}\frac{9}{\sqrt{1+\frac 2x}+\sqrt{1-\frac 7x}}=\frac{9}{1+1}=\frac 92.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/962458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Proving by induction $5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$ is divisible by $4$ I want to prove the following twice. Once by induction then again by any other method.
$$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2$$ is a multiple of 4 for all nonnegative integers n.
Let n=0 , since it is the first nonnegative integer
$$5^{3(0)}+2*5^{2(0)}-5^{0}-2 = 0 $$
Factoring gives us $(5-1)(5+1)(5+2)$
| $$5^{3n} + 2 \cdot 5^{2n} - 5^{n} - 2=(5^n+2)(5^{2n}-1)=(5^n+2)(5^n-1)(5^n+1)$$
Do you see middle term is divisible by $4$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/967197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Proving a Binomial Identity Problem $\boldsymbol{25}$ [$\boldsymbol{5}$ Points]: Show that
$$
\sum_{k=0}^n\binom{n+k}{k}\frac1{2^k}=2^n
$$
Hint: Denote the left hand side by $f(n)$ and prove that $f(n+1)=2f(n)$.
Original Image
Can you please help me with problem 25. I need to prove that $f(n+1)=2 f(n)$, where $f(n)$ is the LHS of the expression, from there on I can do it my self. I have tried using the binominal theorem and using different summation identities but i just cant get there. Please help me.
| Showing the hint is just a straightforward calculation. If $f(n) = \sum \binom{n+k}{k} \frac{1}{2^k}$, then
\begin{align*}
f(n+1) &= \sum_{k=0}^{n+1} \binom{n+k+1}{k} \frac{1}{2^k} \\
&= \sum_{k=1}^{n+1} \binom{n+k}{k-1} \frac{1}{2^k} + \sum_{k=0}^{n+1} \binom{n+k}{k} \frac{1}{2^k} \\
&= \frac{1}{2} \sum_{k=0}^n \binom{n+k+1}{k} \frac{1}{2^k} + \sum_{k=0}^{n+1} \binom{n+k}{k} \frac{1}{2^k} \\
&= \frac{1}{2}\left[ \sum_{k=0}^{n+1} \binom{n+k+1}{k} \frac{1}{2^k} - \binom{2n+2}{n+1}\frac{1}{2^{n+1}}\right] \\
&\quad + \left[ \sum_{k=0}^{n} \binom{n+k}{k} \frac{1}{2^k} + \binom{2n+1}{n+1} \frac{1}{2^{n+1}} \right] \\
&= \frac{1}{2} f(n+1) + f(n) + \frac{1}{2^{n+2}} \left[ 2\binom{2n+1}{n+1} - \binom{2n+2}{n+1} \right] \\
&= \frac{1}{2} f(n+1) + f(n),
\end{align*}
and so $f(n+1) = 2f(n)$. Now, showing the result follows by induction.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Evaluate a linear system of three equations Solve for $x, y\ \text{and}\ z\ $:
$x-3z=10\\
-x+y+2z=7\\
2x+2y-5z=-8$
My working:
$$\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
-1 & 1 & 2 & 7 \\
2 & 2 & -5 & -8
\end{array}\right) =
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 3 & 0 & -11
\end{array}\right) \begin{array}{l}
\\
R_1 + R_2 \\
R_3 + R_2 - R_1
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & -3 & 10 \\
0 & 1 & -1 & 17 \\
0 & 0 & 3 & -62
\end{array}\right) \begin{array}{l}
\\
\\
R_3 - 3R_2
\end{array}=
\left(\begin{array}{ccc|c}
1 & 0 & 0 & -52 \\
0 & 1 & 0 & -7 \\
0 & 0 & 1 & -24
\end{array}\right) \begin{array}{l}
R_1 + R_3 \\
R_2 + R_3/3 \\
R_3/3
\end{array}$$
Yet plugging these solutions into the original equation does not work:
$-52+0-3 \times (-24)=20 \ne 10$ and
$-(-52)-7+2\times(-24)=-3\ne7$
What did I do wrong?
| You seem to have 62/3 = 24. That gives you errors in both R2+R3/3 and just R3/3 in the last line. The rest seems to be correct. Fix that and you end up with (-52, -11/3, -62/3) which you can check does fit your original system.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/972269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Quadratic equation $4x^2+4x=7$ using quadratic formula Solve using quadratic formula. $4x^2+4x=7$
So $4x^2+4x-7=0$
$A=4$ $b=4$ $c=-7$
$$x=\frac{-4\pm\sqrt{(4)^2-4(4)(-7)}}{2(4)}=\frac{-4\pm\sqrt{16+112}}{8}=\frac{-4\pm\sqrt{128}}{8}$$
What's next?
| Next is simplifying:
$$x=\frac{-4\pm\sqrt{4^2-4\cdot4\cdot-7}}{2\cdot4}=\frac{-4\pm\sqrt{16+112}}{8}=-\frac12\pm\frac18\cdot8\sqrt{2}=-\frac12\pm\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/975016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An inequality, which is supposed to be simple Let $x,y,z\in\mathbb{R}$.Let $xy+yz+xz=1$.
Prove:$\displaystyle \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\leq \frac{3}{2}$
| Hint: $$x^2 + 1 = x^2 + xy + yz + zx = (x+y)(x+z)$$
Also, you will need the inequality
$$
\sqrt{AB}\le \frac12(A+B)
$$
Solution:
$$ \frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}
\\= \frac{x}{\sqrt{(x+y)(x+z)}}+\frac{y}{\sqrt{(x+y)(x+z)}}
+\frac{z}{\sqrt{(x+y)(x+z)}}
\\
\le \frac12\left[
\frac x{x+y} + \frac x{x+z} +
\frac y{y+z} + \frac y{x+y} +
\frac z{z+y} + \frac z{z+x}
\right]=3/2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/977358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the derivatives to $f(x)=4/x^2$ and $g(t)=(t-5)/(1+\sqrt{t}\,)$ I have these two assignments:
Find the derivatives to (a) $f(x)=4/x^2$ and b) $g(t)=(t-5)/(1+\sqrt{t}\,)$ by using the definition $$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$$
a) This is my attempt at (a); am I correct?
$$\lim_{h \to 0} \frac{\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\displaystyle\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)$$
Then I found the common denominator
$$\begin{align*}
\frac{4}{(x+h)^2}-\frac{4}{x^2}
&=\frac{4x^2}{(x+h)^2x^2}-\frac{4(x+h)^2}{(x+h)^2x^2} \\[6pt]
&=\frac{4x^2-4x^2-4h^2-8xh}{(x+h)^2x^2} \\[6pt]
&=\frac{-4h^2-8xh}{(x+h)^2x^2}
\end{align*}$$
Then I expand the denominator to
$$\frac{-4h^2-8xh}{x^4+2x^3h+h^2x^2}
=\frac{\left(-4h^2-8h\right)}{x^3+2x^3h+h^2x^2}\frac{1}{h}
=\frac{\left(-4h^2-8h\right)}{hx^3+2x^3h+h^2x^2}$$
And $h$ goes out with:
$$\frac{-4h^2-8}{x^3+2x^3h+h^2x^2}
= \lim_{h \to 0}\frac{-4\cdot0^2-8}{x^3+2x^3\cdot 0+0^2\cdot x^2}
= -\frac{8}{x^3}$$
I dont know how to do part (b)
| *
*First function
$$f'(x)=\lim_{h \to 0} \frac{\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)}{h}=\frac{1}{h}\left(\frac{4}{(x+h)^2}-\frac{4}{x^2}\right)=\lim_{h \to 0}\frac{1}{h}\frac{-4h^2-8xh}{(x+h)^2x^2}=\lim_{h \to 0}\frac{1}{h}\frac{-4h(h+2x)}{(x+h)^2x^2}= \lim_{h \to 0}\frac{-4(h+2x)}{(x+h)^2x^2}= \frac{-8x}{x^4}= -\frac{8}{x^3}, $$
for all $x\neq 0$.
*Second function (in what follows we consider $t\geq 0$)
$$f'(t)=\lim_{h \to 0} \frac{1}{h}\left(\frac{t+h-5}{1+\sqrt{t+h}}-\frac{t-5}{1+\sqrt{t}}
\right) =
\lim_{h \to 0} \frac{1}{h}\left(\frac{(t+h-5)(1+\sqrt{t})-(t-5)(1+\sqrt{t+h})}{(1+\sqrt{t+h})(1+\sqrt{t})}\right);$$
Now we simplify the numerator arriving at
$$f'(t)=
\lim_{h \to 0}
\frac{1}{h}\left(\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})+h(1+\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}\right)= \lim_{h \to 0} \frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}+\frac{1}{h}\frac{h(1+\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}=
\lim_{h \to 0} \frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}+\frac{1}{(1+\sqrt{t+h})};
$$
All we need is to use the "trick":
$$\frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}=
\frac{1}{h}\frac{-(t-5)(\sqrt{t+h}-\sqrt{t})}{(1+\sqrt{t+h})(1+\sqrt{t})}\frac{(\sqrt{t+h}+\sqrt{t})}{(\sqrt{t+h}+\sqrt{t})}=
\frac{1}{h}\frac{-(t-5)h}{(1+\sqrt{t+h})(1+\sqrt{t})}\frac{1}{(\sqrt{t}+\sqrt{t+h})}=
-\frac{(t-5)}{(1+\sqrt{t+h})(1+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}.
$$
In summary:
$$f'(t)= \lim_{h \to 0} -\frac{(t-5)}{(1+\sqrt{t+h})(1+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}+\frac{1}{(1+\sqrt{t+h})}= \\
-\frac{(t-5)}{2(1+\sqrt{t})^2\sqrt{t}}+\frac{1}{(1+\sqrt{t})}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/979347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that the element $z=i \cos \frac{\pi}{3}+\sin \frac{\pi}{3} = i( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ belongs to $U_{12}$ Show that the element $\displaystyle z=i \cos \frac{\pi}{3}+\sin \frac{\pi}{3} = i( \cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ belongs to U12
What I don't understand:
*
*In what way $i \cos \frac{\pi}{3} + \sin \frac{\pi}{3}$ is different from $i(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})$ corresponding to the polar solutions of U12
2.U12 has 12 polar elements and from Cayley's table we see there are 4 elements, how this is different?
The 4 elements are the generators and others are not (primes), how's this is different for the polar elements?
Thanks for the help and tips
| I might be misunderstanding your question, but here goes nothing!
It seems like we want to prove that $z^{12} = 1$. Now
$$ z = i(\cos\frac{\pi}{3} - i\sin\frac{\pi}{3}) = i(\cos(-\frac{\pi}{3}) + i\sin(-\frac{\pi}{3})) = ie^{-i\frac{\pi}{3}} = ie^{-i\frac{2\pi}{6}} $$
$$z^6 = i^6(e^{-i\frac{2\pi}{6}})^6 = -1$$
$$z^{12} = (z^6)^2 = (-1)^2 = 1$$
For your second question, notice that $z$ is the $12$-th root of $1$. If another element $\alpha$ satisfies $\alpha^{12} = 1$, see if you can figure out how $z$ relates to $\alpha$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does one evaluate $\int \frac{\sin(x)}{\sin(5x)} \ dx$
*
*The below problem is taken from Joseph Edwards book Integral Calculus for beginners.
How does one show:
$$5 \int \frac{\sin(x)}{\sin(5x)} \ dx= \sin\left(\frac{2\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{2\pi}{5}\right)}{\sin\left(x+\frac{2\pi}{5}\right)}\right\} -\sin\left(\frac{\pi}{5}\right) \cdot \log\left\{\frac{\sin\left(x-\frac{\pi}{5}\right)}{\sin\left(x+\frac{\pi}{5}\right)}\right\} $$
*
*Splitting $\sin{(5x)}$ as $\sin{(4x+x)}$ doesn't seem to be of much help since then we have a big term in the denominator after expansion.
| First of all, we use De Moivres’ Theorem to express $\sin 5x $ in terms of $\cos x$ and $\sin x$.
$$
\begin{aligned}
&\cos 5 x+i \sin 5 x\\=&(\cos x+i \sin x)^{5} \\
=& \cos^{5} x+5 i \cos ^{4} x \sin x-10 \cos^{3} x \sin ^{2} x-10 i \cos ^{2} x \sin ^{3} x+5 \cos x \sin ^{4} x+i \sin ^{5} x
\end{aligned}
$$
Comparing the imaginary parts on both sides yields
$$
\sin 5 x=5 \cos ^4 \sin x-10 \cos ^{2} x \sin ^{3} x+\sin ^{5} x
$$
$$
\begin{aligned}
\because \int\frac{\sin x}{\sin 5 x} d x &=\int \frac{1}{5 \cos ^{4} x-10 \cos ^{2} x \sin ^{2} x+\sin ^{4} x} d x \\
&=\int \frac{\sec ^{4} x d x}{5-10 \tan ^{2} x+\tan ^{4} x} \\
&=\int \frac{1+t^{2}}{t^{4}-10 t^{2}+5}dt, \quad \textrm{ where }t=\tan x.
\end{aligned}
$$
Playing a small trick on the integrand yields
$$
\begin{aligned}
\int \frac{\sin x}{\sin 5 x}dx&=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{5}{t^{2}}-10} d t\\
&= \int \frac{\frac{\sqrt{5}+1}{2}\left(1+\frac{\sqrt{5}}{t^{2}}\right)+\frac{\sqrt{5}-1}{2}\left(1-\frac{\sqrt{5}}{t^{2}}\right)}{t^{2}+\frac{5}{t}-10} d t\\
&=\frac{\sqrt{5}+1}{2} \int \frac{d\left(t-\frac{\sqrt{5}}{t}\right)}{\left(t-\frac{\sqrt{5}}{t}\right)^{2}-(10-2 \sqrt{5})}+\frac{\sqrt{5}-1}{2} \int \frac{d\left(t+\frac{\sqrt{5}}{t}\right)}{\left(t+\frac{\sqrt{5}}{t}\right)^{2}-(10+2 \sqrt{5})}\\&=\frac{\sqrt{5}+1}{2 \sqrt{10-2 \sqrt{5}}} \tan ^{-1}\left(\frac{t-\frac{\sqrt{5}}{t}}{\sqrt{10-2 \sqrt{5}}}\right)+\frac{\sqrt{5}-1}{4 \sqrt{10+2 \sqrt{5}}} \ln \left|\frac{t+\frac{\sqrt{5}}{t}-\sqrt{10+2 \sqrt{5}}}{t+\frac{\sqrt{5}}{t}+\sqrt{10+2 \sqrt{5}}}\right|+C\\&=\frac{\sqrt{5}+1}{2 \sqrt{10-2 \sqrt{5}}} \tan ^{-1}\left(\frac{\tan^2 {x}-\sqrt{5} }{\tan {x}\sqrt{10-2 \sqrt{5}}}\right)\\&+\frac{\sqrt{5}-1}{4 \sqrt{10+2 \sqrt{5}}} \ln \left| \frac{\tan ^{2} x-\sqrt{10+2 \sqrt{5}} \tan x+ \sqrt{5}}{\tan ^{2} x+\sqrt{10+2 \sqrt{5}} \tan x+\sqrt{5}}\right|+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/980961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 4,
"answer_id": 3
} |
Assume that 495 divides the integer $\overline{273x49y5}$ where $x,y \in \{0,1,2...9\}$. Find $x$ and $y$. So, I know that $495 = 5\times 9\times 11$. So then, if that's the case, then the number $\overline{273x49y5}$ must be divisible by $495$ if and only if it is divisible by $5$ and $9$ and $11$.
Then, I know that $5$ divides it for any number $x$ and $y$ because it only has to divide the digits number which is $5$.
To see if it divides by $9$, the sum of the digits must be divisible by $9$, and as for $11$, the alternating sum of digits must be divisible by $11$.
So, to test it by $9$, then, $2+7+3+x+4+9+y+5 = 30+x+y$ has be divisble by $9$. So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are $x= \{0,1,2,3,4,5,6\}$ and $y=\{0,1,2,3,4,5,6\}$.
But $11$ is what gets me. To make it divisible by $11$, then the alternating sums must be divisible by $11$. So then, $-2+7-3+x-4+9-y+5= 12+x-y$ must be divided by $11$. Then, the only numbers $x$ and $y$ that make it divisible by $11$ must be $x=\{0,1,2,\ldots,8\}$ and $y=\{1,2,\ldots,9\}$, if I did my math right.
Next, I assume we must then find numbers in $x$ and $y$ that satisfy the division by $9$ and $11$. So, if that is the case, then I know that since the number must be divisible by $9$ and $11$, then the numbers $x$ and $y$ must fit between $0$ to $6$ because that's $6$ is the highest number $x$ or $y$ can be in the sequence to divide the number. So, how do I check to see which numbers both satisfy division of $9$ and $11$?
| I do not follow "So, the only numbers $x$ and $y$ that make $30+x+y$ divisible by $9$ are..." and I think that is the source of trouble.
If $9$ divides $30+x+y$, then either $30+x+y=36$ or $30+x+y=45$.
If $11$ divides $12+x-y$, then for sure $12+x-y=11$.
Add one of the first two equations to the second. Either $42+2x=47$ or $42+2x=56$. The first can't be, since $47$ is odd. So it's the second situation, and $x=7$. Then $y$ must be $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/981544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Finding an angle $\theta$ in a complex number If we know that $z = \frac{1}{\sqrt2}(\cos\theta+i\cdot\sin\theta)$ and also that $z = \frac{(\sqrt3-1)+i(\sqrt3+1)}{4}$
How can I find $\cos\theta$ and $\sin\theta$? Using a calculator it gives me the angle $\frac{5\pi}{12}$. I tried the following:
$$\cos\theta = \frac{\Re z}{|z|} = \frac{\frac{1}{4}(\sqrt3-1)}{\frac{1}{\sqrt2}} = \frac{\sqrt3-1}{2\cdot\sqrt2}$$ How would I even proceed here, seems near-impossible without a calculator to me.
| $$
\cos(\frac{5\pi}{12}) = \cos(\frac{3\pi}{4}-\frac{\pi}{3})=\cos(\frac{3\pi}{4})\cos(\frac{\pi}{3})+\sin(\frac{3\pi}{4})\sin(\frac{\pi}{3})=\frac{-\sqrt{2}}{2}\cdot\frac{1}{2}+\frac{\sqrt{2}}{2}\frac{\sqrt{3}}{2} = $$ $$\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{3}-1}{2\sqrt{2}}
$$
Now work backwards, amigo.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/984547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Evaluation of $\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$ How to evaluate the following integral
$$\int_0^{\pi/4} \sqrt{\tan x} \sqrt{1-\tan x}\,\,dx$$
It looks like beta function but Wolfram Alpha cannot evaluate it. So, I computed the numerical value of integral above to 70 digits using Wolfram Alpha and I used the result to find its closed-form. The possible candidate closed-form from Wolfram Alpha is
$$\pi\sqrt{\frac{1+\sqrt{2}}{2}}-\pi$$
Is this true? If so, how to prove it?
| To continue the work of Anastasiya-Romanova but not using complex analysis
For $I_1$:
Notice:
$p(z)=1+2z^2+2z^4=2\Big(z^2+az+b\Big)\Big(z^2-az+b\Big)$
where $a=\sqrt{\sqrt{2}-1}$ and $b=\dfrac{\sqrt{2}}{2}$
Therefore:
$f(z)=\displaystyle \dfrac{2z^2}{1+2z^2+2z^4}=\dfrac{z}{2a\Big(z^2-az+b\Big)}-\dfrac{z}{2a\Big(z^2+az+b\Big)}$
$f(z)=\dfrac{2z-a}{4a\Big(z^2-az+b\Big)}-\dfrac{2z+a}{4a\Big(z^2+az+b\Big)}+\dfrac{1}{4\Big(z^2-az+b\Big)}+\dfrac{1}{4\Big(z^2+az+b\Big)}$
Let $c=b-\dfrac{a^2}{4}$, $c>0$
Therefore:
$f(z)=\dfrac{2z-a}{4a\Big(z^2-az+b\Big)}-\dfrac{2z+a}{4a\Big(z^2+az+b\Big)}+\dfrac{1}{4\Big(\big(z-\tfrac{a}{2}\big)^2+c\Big)}+\dfrac{1}{4\Big(\big(z+\tfrac{a}{2}\big)^2+c\Big)}$
So a primitive of $\displaystyle \dfrac{2z}{1+2z^2+2z^4}$ is:
$\dfrac{1}{4a}\log\Big(\dfrac{z^2-az+b}{z^2+az+b}\Big)+\dfrac{1}{4\sqrt{c}}\arctan\Big(\dfrac{z-\tfrac{a}{2}}{\sqrt{c}}\Big)+\dfrac{1}{4\sqrt{c}}\arctan\Big(\dfrac{z+\tfrac{a}{2}}{\sqrt{c}}\Big)$
(think about derivative of $\log(u(x))$ )
Therefore:
$\displaystyle \int_{-\infty}^{+\infty}\dfrac{2x^2dx}{1+2x^2+2x^4}=\dfrac{\pi}{2\sqrt{c}}=\pi\sqrt{\sqrt{2}-1}$
To compute $I_2$ start performing change of variable $u=\dfrac{1}{x}$ , the function to integrate becomes $\dfrac{x^2}{x^4p\Big(\dfrac{1}{x}\Big)}$
$q(x)=x^4p\Big(\dfrac{1}{x}\Big)$
$q(x)=2x^4\Big(\dfrac{1}{x^2}+\dfrac{a}{x}+b\Big)\Big(\dfrac{1}{x^2}-\dfrac{a}{x}+b\Big)$
$q(x)=2(1+ax+bx^2)(1-ax+bx^2)$
$q(x)=2b^2\Big(x^2+\dfrac{a}{b}x+\dfrac{1}{b}\Big)\Big(x^2-\dfrac{a}{b}x+\dfrac{1}{b}\Big)$
The new $a,b$ are respectively $\dfrac{a}{b},\dfrac{1}{b}$
and there is a new $c$.
Therefore:
$\displaystyle \int_{-\infty}^{+\infty}\dfrac{dx}{1+2x^2+2x^4}=\dfrac{2}{b^2}\times \dfrac{\pi}{2\sqrt{c}}=\dfrac{\pi}{b^2\sqrt{c}}=\pi\dfrac{\sqrt{2}}{2}\sqrt{\sqrt{2}-1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/989021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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"answer_id": 1
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Prove that $\int_0^\infty \frac{\ln x}{x^n-1}\,dx = \Bigl(\frac{\pi}{n\sin(\frac{\pi}{n})}\Bigr)^2$ This question inspired me to ask the following.
Prove that
$$I_n = \int_0^\infty \frac{\ln x}{x^n-1}\,dx = \left(\frac{\pi}{n\sin\left(\frac{\pi}{n}\right)}\right)^2,$$
for $\Re(n)>1$.
For some cases there is a nice specific form of $I_n$. For example
$$\begin{align}
I_2 & = \frac{\pi^2}{4} \\
I_3 & = \frac{4\pi^2}{27} \\
I_4 & = \frac{\pi^2}{8} \\
I_5 & = \frac{2\left(5+\sqrt 5\right)\pi^2}{125} \\
I_6 & = \frac{\pi^2}{9} \\
I_7 & = \frac{2\pi^2}{49\left(1-\sin\left(\frac{3\pi}{14}\right)\right)} \\
I_8 & = \frac{\left(2+\sqrt 2\right)\pi^2}{32}
\end{align}$$
| We can use a contour integral in the complex plane, as I showed here for the case $n=3$. Now, however, we use
$$\oint_C dz \frac{\log^2{z}}{z^n-1}$$
where $C$ is the modified keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$.
Let's evaluate this integral over the contours. As before, there are $8$ pieces to evaluate, as follows:
$$\int_{\epsilon}^{1-\epsilon} dx \frac{\log^2{x}}{x^n-1} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (1+\epsilon e^{i \phi}\right )}}{(1+\epsilon e^{i \phi})^n-1} \\ + \int_{1+\epsilon}^R dx \frac{\log^2{x}}{x^n-1} + i R \int_0^{2 \pi} d\theta \, e^{i \theta} \frac{\log^2{\left (R e^{i \theta}\right )}}{R^n e^{i n \theta}-1} \\ + \int_R^{1+\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^n-1} + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \frac{(\log{\left (1+\epsilon e^{i \phi}\right )}+i 2 \pi)^2}{(1+\epsilon e^{i \phi})^n-1} \\ + \int_{1-\epsilon}^{\epsilon} dx \frac{(\log{x}+i 2 \pi)^2}{x^n-1} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log^2{\left (\epsilon e^{i \phi}\right )}}{\epsilon^n e^{i n \phi}-1} $$
As $R \to \infty$, the fourth integral vanishes as $\log^2{R}/R^{n-1}$. As $\epsilon \to 0$, the second integral vanishes as it is $O(\epsilon^n)$, while the eighth integral vanishes as $\epsilon \log^2{\epsilon}$. This leaves the first, third, fifth, sixth and seventh integrals, which in the above limits, become
$$PV \int_0^{\infty} dx \frac{\log^2{x} - (\log{x}+i 2 \pi)^2}{x^n-1} + i \frac{4 \pi^3}{n}$$
The residue computation is a little more involved, because we now have $n-1$ poles at which we need to evaluate residues. The contour integral is thus
$$i 2 \pi \sum_{k=1}^{n-1} \frac{-4 \pi^2 k^2/n^2}{n e^{i 2 (n-1) \pi k/n}} = -i \frac{8 \pi^3}{n^3} \sum_{k=1}^{n-1} k^2 \, e^{-i 2 (n-1) \pi k/n} $$
The sum is doable, but the algebra is a bit hideous. The result is
$$\sum_{k=1}^{n-1} k^2 \, e^{-i 2 (n-1) \pi k/n} = \frac12 \left ( \frac{n}{\sin^2{\frac{\pi}{n}}} - n^2\right ) - i \frac12 n^2 \cot{\frac{\pi}{n}} $$
Equating real and imaginary parts of both equations for the contour integral yields
$$\int_0^{\infty} dx \frac{\log{x}}{x^n-1} = \frac{\pi^2}{n^2 \sin^2{\frac{\pi}{n}}} $$
$$PV \int_0^{\infty} dx \frac{1}{x^n-1} = -\frac{\pi}{n} \cot{\frac{\pi}{n}} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
} |
Prove the $n$th Fibonacci number is the integer closest to $\frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n$
Prove that the $n$th Fibonacci number $f_n$ is the integer that is closest to the number $$\frac{1}{\sqrt{5}}\left( \frac{1 + \sqrt{5}}{2} \right)^n.$$
Hi everyone, I don't really understand the problem. I have the following hint, but I don't know how to work it.
Hint: Show that the absolute value of $\frac{1}{\sqrt{5}}\left( \frac{1 - \sqrt{5}}{2} \right)^n$ is less than $1/2$.
| We prove the hint by induction on $n$. For $n = 1$, $LHS < \dfrac{1}{\sqrt{5}} < \dfrac{1}{2}$, so the statement is true for $n = 1$. Assume it is true for $n = k$, that is: $|R_k|=\left|\dfrac{1}{\sqrt{5}}\cdot \left(\dfrac{1-\sqrt{5}}{2}\right)^k\right| < \dfrac{1}{2}$, we have:
$|R_{k+1}| = \dfrac{1}{\sqrt{5}}\cdot \left(\dfrac{\sqrt{5}-1}{2}\right)^{k+1} < |R_k| < \dfrac{1}{2}$, and the hint is proved. From this the answer follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/992811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove this formula for $\pi$ I have to use a certain approximation for $\pi$ for my computer science class, but I don't really understand what's going on, other than that this is related to the Taylor polynomial for arctangent. Could someone please prove that this is equal to $\pi$?
It's on the first formula under the subsection "Middle Ages" on this page: Approximations of π:
$$ \pi = \sqrt{12}\sum^\infty_{k=0} \frac{(-3)^{-k}}{2k+1} = \sqrt{12}\sum^\infty_{k=0} \frac{(-\frac{1}{3})^k}{2k+1} = \sqrt{12}\left(1-{1\over 3\cdot3}+{1\over5\cdot 3^2}-{1\over7\cdot 3^3}+\cdots\right)$$
| To give an explanation:
It is well known that $\tan'(x)=1+\tan^2(x)$. By deriving $\tan(\arctan(x))=x$ we get: $$\arctan'(x)\cdot\tan'(\arctan(x))=\arctan'(x)\cdot\left(1+(\tan(\arctan(x)))^2\right)=\arctan'(x)\cdot\left(1+x^2\right)=1\iff\arctan'(x)=\frac{1}{1+x^2}$$ and by the standard formula of geometric series we have $\arctan'(x)=\sum_{k=0}^{\infty} \left(-x^2\right)^k$ for $|x|<1$. Therefore we have:
$$
\arctan(x)=\int_{0}^{x}\left(\sum_{k=0}^{\infty} \left(-t^2\right)^k\right)dt=\sum_{k=0}^\infty \left(\int_{0}^x \left(-t^2\right)^k dt\right)=\sum_{k=0}^\infty \left((-1)^k\int_{0}^x t^{2k} dt\right)=\sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{2k+1}={x\sum_{k=0}^\infty \frac{(-1)^kx^{2k}}{2k+1}}=x\sum_{k=0}^\infty \frac{\left(-x^2\right)^k}{2k+1}
$$
And therefore:
$$
\arctan\left(\frac{1}{\sqrt3}\right)=\frac{1}{\sqrt3}\sum_{k=0}^\infty \frac{\left(-\frac{1}{3}\right)^{k}}{2k+1}
$$
Taking a look at the basic definitions of the trigonometric functions on the unit circle we see that $\tan\left(\frac{\pi}{6}\right)$ is a leg of the right triangle which is the half of a regular triangle, while the other leg is $1$. Since $\tan\left(\frac{\pi}{6}\right)$ is the half of the sides of the regular triangle, the hypothenuse is $2\tan\left(\frac{\pi}{6}\right)$. So by the Pythagorean theorem we have:
$$
1+\tan^2\left(\frac{\pi}{6}\right)=4\tan^2\left(\frac{\pi}{6}\right)\iff\tan\left(\frac{\pi}{6}\right)=\frac{1}{\sqrt3}\iff 6\arctan\left(\frac{1}{\sqrt3}\right)=\pi
$$
And therefore:
$$
\pi=6\cdot\frac{1}{\sqrt3}\sum_{k=0}^\infty \frac{\left(-\frac{1}{3}\right)^{k}}{2k+1}=\sqrt{12}\sum_{k=0}^\infty \frac{\left(-\frac{1}{3}\right)^{k}}{2k+1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/993841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The minimum value of $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$ Problem :
The minimum value of $$\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$$
Can I use this in numerator and denominator :
The minimum value of $a +\frac{1}{a}$
Using A.M and G.M inequality :
$a +\frac{1}{a} \geq 2\sqrt{a \times \frac{1}{a}}$
$\Rightarrow a +\frac{1}{a} \geq 2$ .....(1)
By putting the minimum value of (1) in $\frac{(x+\frac{1}{x})^6-(x^6+\frac{1}{x^6})-2}{(x+\frac{1}{x})^3+x^3+\frac{1}{x^3}}$
we get ; $\frac{2^6-2-2}{2^3+2}$ but I think this is wrong especially denominator as we need to find the maximum value of denominator to get the minimum value.
Please suggest ,thanks.
| hint: let
$$f(x)=\dfrac{\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2}{\left(x+\dfrac{1}{x}\right)^3+x^3+x^{-3}}=3\left(x+\dfrac{1}{x}\right)\ge 6$$
because
$$\left(x+\dfrac{1}{x}\right)^6-\left(x^6+\dfrac{1}{x^6}\right)-2=\left(x+\dfrac{1}{x}\right)^6-\left(x^3+\dfrac{1}{x^3}\right)^2$$
so
$$f(x)=\left(x+\dfrac{1}{x}\right)^3-\left(x^3+\dfrac{1}{x^3}\right)=3\left(x+\dfrac{1}{x}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/994458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
On the determinant of a matrix The matrix
$$\left[\begin{array}{ccc}
30&20&30\\
40&50&20\\
30&30&20
\end{array}\right]$$
I tried solving it for myself and got $12000$, but math way tells me its $-1000$.
I dont understand how you get a negative, Where did I mess up my calculations?
I did
30 x ((50x20)-(30x20)) = 12000
20 x ((40x20)-(30x20)) = 4000
30 x ((40*30)-(30x50)) = -9000
12000 - 4000 + (-9000) is -1000
oh I see! Sorry guys I messed up ):
Thx <3
| We want to find the determinant of
$$\begin{pmatrix} 30 & 20 & 30 \\
40 &50 &20 \\
30 &30 &20 \end{pmatrix}.$$
There are many ways of writing this down, but let's do it in the expand-by-minors way. First, since everything is divisible by $10$, let's factor out a $10$. Since it's a $3 \times 3$ matrix, this affects the overall determinant by $10^3$. So we look for
$$\begin{align}
10^3 \begin{bmatrix} 3&2&3\\4&5&2\\3&3&2 \end{bmatrix} &= 10^3 \left( 3 \begin{bmatrix} 5&2\\3&2\end{bmatrix} - 2\begin{bmatrix} 4&2\\3&2\end{bmatrix} + 3\begin{bmatrix} 4&5\\3&3 \end{bmatrix}\right) \\
&= 10^3 \left( 3(10 - 6) - 2(8-6) + 3(12-15)\right) \\
&= 10^3 (12 - 4 + -9) \\
&= -1000,
\end{align}$$
which is just as was claimed. $\diamondsuit$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/995727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$ 1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqrt{6})... $ Converges By comparison test? For what value of real constant $a$ does the following series converge?
$$
1+(1/\sqrt{3})-(a/\sqrt{2})+(1/\sqrt{5})+(1/\sqrt{7})-(a/\sqrt{4})+(1/\sqrt{9})+(1/\sqrt{11})-(a/\sqrt{6})...
$$
I do not have a clue on how to proceed. Suggest a possible route.
can we do it by comparison test somehow? Can we comapre it with $c/\sqrt{(n)}$ for some $c$ and some large enough $n$?
| Okay, I think I can guess at the pattern with the most recent addition
$$
\begin{align}
&\sum_{n=0}^\infty\left(\frac1{\sqrt{4n+1}}+\frac1{\sqrt{4n+3}}-\frac{a}{\sqrt{2n+2}}\right)\\
&=\sum_{n=0}^\infty\frac{\sqrt{(4n+3)(2n+2)}+\sqrt{(4n+1)(2n+2)}-a\sqrt{(4n+1)(4n+3)}}{\sqrt{(4n+1)(4n+3)(2n+2)}}\\
&=\sum_{n=0}^\infty\frac{\sqrt{(1+\frac3{4n})(1+\frac1n)}+\sqrt{(1+\frac1{4n})(1+\frac1n)}-a\sqrt2\sqrt{(1+\frac1{4n})(1+\frac3{4n})}}{2\sqrt{n}\sqrt{(1+\frac1{4n})(1+\frac3{4n})(1+\frac1n)}}\\
&=\sum_{n=0}^\infty\frac{\left[1+O(\frac1n)\right]+\left[1+O(\frac1n)\right]-a\sqrt2\left[1+O(\frac1n)\right]}{2\sqrt{n}\left[1+O(\frac1n)\right]}\\
&=\sum_{n=0}^\infty\frac{2-a\sqrt2+O(\frac1n)}{2\sqrt{n}}\tag{1}
\end{align}
$$
where $O(\cdot)$ is Landau big-O notation. Since convergence concerns the tail of the sequence, we can replace the term for $n=0$ by it value, $1+\frac1{\sqrt3}-\frac{a}{\sqrt2}$, and start using $O(\frac1n)$ for $n\ge1$.
The sum in $(1)$ will converge if and only if $a=\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Equivalence of summations Show that $$\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2}\equiv \frac{1}{n}\sum^{n}_{i=1}x_{i}^{2} - \bar{x}^{2}.$$ Note that $\bar{x} = \frac{1}{n}\sum^{n}_{i=1} x_{i}$. So I have started by:
\begin{align}
\frac{1}{n}\sum^{n}_{i=1} (x_{i} - \bar{x})^{2} &= \frac{1}{n}\sum^{n}_{i=1} (x_{i}^{2} - 2x_{i}\bar{x} + \bar{x}^{2}) \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n}\sum_{i=1}^{n}x_{i}\bar{x} + \frac{1}{n}\sum^{n}_{i=1}\bar{x}^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n}\sum_{i=1}^{n}x_{i}\left(\frac{1}{n}\sum^{n}_{i=1}x_{i}\right) + \frac{1}{n}\sum_{i=1}^{n}\left(\frac{1}{n}\sum^{n}_{i=1}x_{i}\right)^{2} \\
&= \frac{1}{n}\sum^{n}_{i=1} x_{i}^{2} - \frac{2}{n^{2}}\left(\sum_{i=1}^{n}x_{i}\right)^{2} + \frac{1}{n^{3}}\left(\sum^{n}_{i=1}x_{i}\right)^{2}
\end{align}
Not really sure what to do next, any hints?
| \begin{align}
\sum_{i=1}^n (x_i-\bar x)^2 & = \sum_{i=1}^n (x_i^2 -2\bar x x_i + \bar x^2) \\[10pt]
& = \left(\sum_{i=1}^n x_i^2\right) -2\bar x \left( \sum_{i=1}^n x_i\right) + n\bar x^2 \\[10pt]
& = \left(\sum_{i=1}^n x_i^2\right) -2\bar x \left( n\bar x\right) + n\bar x^2 = \cdots\cdots\cdots
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/997473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $ \lim_{n\to \infty} \sum_{r=1}^{n}\frac{1}{n+r}$ by expressing as Riemann sum $$ \lim_{n\to \infty} \left(\frac{1}{n+1} + \frac{1}{n+2} + ..+ \frac{1}{2n}\right)$$
How do i find the limit by expressing it as a definite integral of an appropriate function via Riemann Sums?
I do know that for riemann sums
$\Delta x = \frac{b-a}{n}$ and $ x_i^* = a + \frac{b-a}{n}$
So i would need something like this
$$\lim_{n\to \infty}\sum_{i=1}^{n}\frac{b-a}{n}\cdot f\left(a+i\frac{b-a}{n} \right)$$
But how should i go about solving the question?
| Notice That : $$\frac{1}{n+r}=\frac{1}{n}\frac{1}{1+\frac{r}n}$$
So our some can be written as
$$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r} =
\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+\frac{r}n}$$
Now by replacing $\frac{1}{n}$ with $dx$,$\frac{r}{n}$ with $x$ and $\sum$ with $\int$
$$\lim_{n\to \infty}\frac{1}{n}\sum_{r=1}^{n} \frac{1}{1+\frac{r}n}=
\lim_{n\to \infty}\frac{1-0}{n}\sum_{r=1}^{n} \frac{1}{1+\left(0+1\cdot\frac{r}{n}\right)}$$
Now comparing this with the form you want we have $b=1$ , $ a=0$ and $f\left(x\right)=\frac{1}{1+x}$
$$\lim_{n\to \infty}\frac{1-0}{n}\sum_{r=1}^{n} \frac{1}{1+\left(0+1\cdot\frac{r}{n}\right)}=\int_0^1f(x)dx=\int_0^1\frac{1}{1+x}dx$$
And resultant integral evaluates to
$$\lim_{n\to \infty}\sum_{r=1}^{n} \frac{1}{n+r}=\int_0^1\frac{1}{1+x}dx=\ln\left(\frac{2}{1}\right)=\ln(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Construct a triangle given an angle and two medians Construct, with ruler and compass, a triangle $ABC$ knowing the angle $\widehat{A}$ and $m_a$ and $m_b$, where $m_a$ and $m_b$ are the medians relative to the vertices $A$ and $B$, respectively.
| If we set $m_a=3x,m_b=3y$ and consider the following construction:
We have that:
$$GU = 2x\sin\varphi,\quad GV=x\sin\varphi,\quad UV=M_BW=3x\sin\varphi,$$
$$M_B V^2 = y^2-x^2\sin^2\varphi,\quad AU=2x\cos\varphi,\quad AW=AU-M_BV, $$
hence:
$$\cot A = \frac{AW}{M_B W}=\frac{2x\cos\varphi-\sqrt{y^2-x^2\sin^2\varphi}}{3x\sin\varphi}$$
and we have:
$$3\cot A = 2\cot\varphi-\sqrt{-1+\frac{y^2}{x^2}(1+\cot^2\varphi)},\tag{1}$$
so it is possible to get the value of $\cot\varphi$ (then build the triangle $GAB$ and the triangle $ABC$) by solving the quadratic equation:
$$\left(4-\frac{y^2}{x^2}\right)\cot^2\varphi - 12\cot A\cot\varphi +\left(9\cot^2 A+1-\frac{y^2}{x^2}\right)=0.\tag{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1002418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Minimum of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}$ What is the minimum of $$f(a,b,c):=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}$$ where $a,b,c$ are positive real numbers?
When $a=b=c$, we have $f(a,b,c)=\dfrac{3}{\sqrt{2}}\approx 2.12$
When $a=1,b=c\rightarrow\infty$, we have $f(a,b,c)\rightarrow 2$. So the minimum is at most $2$.
| Hint: this is also
$$
\min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}
=\min_{a,b,c\ge 0, a+b+c=1} \sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\sqrt{\dfrac{c}{1-c}}
$$And then you can for instance use the Lagrange multiplicators.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
How would you show that $\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$? I've recently seen a Highschool problem and I was wondering, how would you show that
$$\sqrt{14+4\sqrt{10}} - \sqrt{14-4\sqrt{10}} = 4$$
Thank you for your time,
| Since $14+4\sqrt{10}\gt14-4\sqrt{10}\gt0$, the difference of the square roots is equal to something positive, i.e.,
$$\sqrt{14+4\sqrt{10}}-\sqrt{14-4\sqrt{10}}=u\gt0$$
If we square both sides, we have
$$\begin{align}
u^2&=(14+4\sqrt{10})-2\sqrt{14+4\sqrt{10}}\sqrt{14-4\sqrt{10}}+(14-4\sqrt{10})\\
&=28-2\sqrt{14^2-4^2\cdot10}\\
&=28-2\sqrt{36}\\
&=16
\end{align}$$
Since $u\gt0$ was already established, we can conclude $u=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1003964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Poisson Integral is equal to 1
Show
$$
\int_{-\pi}^{\pi}P(r, \theta)d\theta = 1
$$
Let $\alpha(r) = \frac{r^2 - 1}{2r}$ and $\gamma(r) = -\big(\frac{r^2 + 1}{2r}\big)$.
Then
$$
\frac{1}{2\pi}
\int_{-\pi}^{\pi}\frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}d\theta
= \frac{\alpha}{2\pi}
\int_{-\pi}^{\pi}\frac{1}{\cos(\theta) + \gamma}d\theta
$$
where
$$
\frac{r^2 - 1}{2r}\frac{1}{\cos(\theta) - \frac{1}{2r} - \frac{r^2}{2r}} = \frac{1 - r^2}{1 - 2r\cos(\theta) + r^2}
$$
Next, let $z = e^{i\theta}$. Then $d\theta = \frac{-i}{z}dz$. Since $\cos(\theta) = \frac{e^{i\theta} + e^{-i\theta}}{2}$, $\cos(\theta) = \frac{z + z^{-1}}{2}$.
$$
= \frac{-i\alpha}{\pi}\int_C\frac{1}{z^2 + 2z\gamma + 1}dz
$$
and $C$ is the contour oriented counter clockwise with simple poles at $z = -\gamma\pm\sqrt{\gamma^2 - 1}$. Let $f(z) = \frac{1}{z^2 + 2z\gamma + 1}$. Then
$$
= \Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
\tag{1}
$$
The only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$.
Then
$$
2\pi i\lim_{z\to z_j}\Bigg[\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)
\frac{1}{\big(z_j + \gamma + \sqrt{\gamma^2 - 1}\big)
\big(z + \gamma - \sqrt{\gamma^2 - 1}\big)}\Bigg] =
\frac{\pi i}{\sqrt{\gamma^2 - 1}}
$$
Now, we can substitute $\frac{\pi i}{\sqrt{\gamma^2 - 1}}$ for $2\pi i\sum\text{Res}$ in equation (1).
\begin{align*}
\Big(\frac{-i\alpha}{\pi}\Big)2\pi i\sum\text{Res}_{z = z_j}f(z)
&= \frac{\alpha}{\sqrt{\gamma^2 - 1}}\\
&= \frac{r^2 - 1}{2r\sqrt{\frac{(r^2 + 1)^2}{4r^2} - 1}}\\
&= \frac{r^2 - 1}{\sqrt{(r^2 + 1)^2 - 4r^2}}\\
&= \frac{r^2 - 1}{\sqrt{r^4 - 2r^2 + 1}}\\
&= \frac{(r - 1)(1 + r)}{(r - 1)(r + 1)}\\
&= 1
\end{align*}
I have been unable to convince myself that the only pole in $\lvert z\rvert < 1$ is $z_j = -\gamma + \sqrt{\gamma^2 - 1}$. I know it is the case because if I use the other pole, the integral becomes $-1$
| For $\boldsymbol{r\in(-1,1)}$
$$
\begin{align}
1
&=\frac1{2\pi i}\oint\frac1{z-r}\mathrm{d}z\tag{1}\\
&=\frac1{2\pi i}\int_{-\pi}^\pi\frac1{e^{i\theta}-r}\mathrm{d}e^{i\theta}\tag{2}\\
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-re^{i\theta}}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{3}\\
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r\cos(\theta)}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{4}\\
&=\frac12+\frac1{2\pi}\int_{-\pi}^\pi\frac{\frac12-\frac12r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{5}\\
1
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta\tag{6}
\end{align}
$$
Explanation:
$(1)$: $\frac1{z-r}$ has residue $1$ at $z=r$ (inside the unit circle)
$(2)$: parametrize the unit circle
$(3)$: multiply integrand by $\frac{e^{-i\theta}-r}{e^{-i\theta}-r}$
$(4)$: take the real part of both sides
$(5)$: add $\frac12$ to and subtract $\frac12$ from the integral
$(6)$: multiply both sides by $2$ and subtract $1$
For $\boldsymbol{r\not\in(-1,1)}$
If $r^2=1$, the integrand is $0$, so assume $r\not\in[-1,1]$.
If $r\not\in[-1,1]$, then the left side of $(1)$ starts at $0$ and step $(6)$ changes the left side to $-1$.
Alternatively, if $r\not\in[-1,1]$, then $\frac1r\in(-1,1)$, and therefore, $(6)$ says
$$
\begin{align}
\frac1{2\pi}\int_{-\pi}^\pi\frac{1-r^2}{1-2r\cos(\theta)+r^2}\mathrm{d}\theta
&=\frac1{2\pi}\int_{-\pi}^\pi\frac{\frac1{r^2}-1}{\frac1{r^2}-\frac2r\cos(\theta)+1}\mathrm{d}\theta\\
&=-\frac1{2\pi}\int_{-\pi}^\pi\frac{1-\frac1{r^2}}{1-\frac2r\cos(\theta)+\frac1{r^2}}\mathrm{d}\theta\\
&=-1\tag{7}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1005102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
A few inequality problems I can't seem to get; Cauchy and the Mean Inequality Chain
*
*Prove that $ \sqrt{\frac{2x^2-2x+1}{2}}\geq\frac{1}{x+\frac{1}{x}} $ for $ 0 < x < 1. $
This one seems reminiscent of the quadratic mean on the left, maybe $\sqrt{\frac{(x-1)^2+x^2}{2}}$, but I can't find a way to compare it to the LHS, which loos a bit like the harmonic mean.
*Prove $ \frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}$ for positive reals $x,y,z$
This one seems a LOT like Cauchy-Schwarz, but I can't quite figure out which lists to use.
I need to solve these problems, but I am pretty stuck. Hints or even solutions would be greatly appreciated. Thanks for the help in advance!
| For the first question, you can try as follows.
Since
\begin{align*}
\sqrt{\frac{2x^2-2x+1}{2}}-\frac{x}{x^2+1}=\frac{1}{2}\cdot \frac{\sqrt{4x^2-4x+2}\cdot (x^2+1)-2x}{x^2+1}=:f(x),
\end{align*}
if we set
\begin{gather*}
a(x):=\sqrt{4x^2-4x+2}=\sqrt{4\left(x-\frac{1}{2}\right)^2+1},
\end{gather*}
the numerator of $f(x)$ is
\begin{align*}
a(x) \cdot x^2-2x+a(x)&=a(x)\cdot \left(x-\frac{1}{a(x)}\right)^2+a(x)-\frac{1}{a(x)}\\
&=a(x)\cdot \left(x-\frac{1}{a(x)}\right)^2+\frac{4\left(x-\frac{1}{2}\right)^2}{a(x)}\\
&> 0,
\end{align*}
from which we can conclude that $f(x)\geq 0, $ that is,
\begin{gather*}\tag{$\star$}
\sqrt{\frac{2x^2-2x+1}{2}}> \frac{x}{x^2+1}, \qquad \forall x\in\mathbb{R}.
\end{gather*}
From $(\star)$ we have
\begin{gather*}
\sqrt{\frac{2x^2-2x+1}{2}}>\frac{1}{x+\frac{1}{x}}, \qquad \forall x\in\mathbb{R}\backslash\{0\}.
\end{gather*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify $(\sqrt{x}) + x + 2 = (\sqrt{y}) + y + 2$ $\sqrt{x} + x + 2 = \sqrt{y} + y + 2$
I've simplified as follows:
$\sqrt{x} + x = \sqrt{y} + y$, square both sides
$x + x^2 = y + y^2$
It seems obvious that $x = y$ but I can’t get to that solution by algebraic means.
| This is the correct formula if you want to square: $ (\sqrt{x} +x)^2 = x^2 + 2x\sqrt{x} +x$.
| {
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Integration question does this substitution work? $$\int_{0}^{3} \frac{1}{\sqrt{3-x}}dx$$
I'm a little lost here do i let $u = 3-x$, so that $du/dx = 1$, $du=dx$
$$\int_0^3 \frac{1}{\sqrt{u}}du = u^{-\frac{1}{2}} du$$
$$ \left[\frac{1}{\frac{1}{2}} u^{\frac{1}{2}}\right]_0^3 = \left[2(3-x)^{\frac{1}{2}}\right]_0^3$$
$$ = 2\sqrt{0}-2\sqrt{3} = -2\sqrt{3}$$
hi sorry i meant 3-x
| The integral is undefined at 3. We must use an improper integral.
$$
\int_0^3 \frac{1}{\sqrt{3-x}} dx = \lim_{x \to 3^-}\int_0^b \frac{1}{\sqrt{3-x}} dx
$$
using a u substitution where u = 3 - x, then du = -dx which gives the following:
$$
\int \frac{-1}{\sqrt{u}}du = -2\sqrt{u}
$$
Putting this back in terms of x gives us
$$ -2\sqrt{3-x}
$$
Thus we get
$$
-2\lim_{x \to 3^-}[\sqrt{3-b} - \sqrt{3-0}] = (-2)(-\sqrt{3}) = 2\sqrt{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1009444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Show that $\sqrt [3]{2}-\sqrt [3]{4}$ is algebraic How do I show, step by step, that $\sqrt [3]{2}-\sqrt [3]{4}$ is a root of $x^3+6x+2$?
Start with $x=\sqrt [3]{2}-\sqrt [3]{4}$ do not use the cubic, the cubic is given for convenience.
( This is example 4.1.3 from Introductory ANT by Alaca/Williams )
| Here is a "by hand" method.
Note that $\sqrt [3] 4 =(\sqrt [3] 2)^2$ and $(\sqrt [3] 4)^2=2\sqrt[3] 2$.
This means that the powers of $y=\sqrt [3] 2-\sqrt [3] 4$ can all be written as polynomials in $x=\sqrt [3] 2$.
But since $x^3=2$ we can always reduce the polynomial to at most quadratic.
$$y=x-x^2$$
$$y^2=x^2-2x^3+x^4=x^2-4+2x=x^2+2x-4$$
$$y^3=x^3-3x^4+3x^5-x^6=2-6x+6x^2-4=6x^2-6x-2$$
Now eliminate $x^2$ from the final two equations using the first, which tells us that $x^2=x-y$
This gives $$y^2 =x-y+2x-4: \text { or } y^2+y=3x-4$$
and $$y^3=6(x-y)-6x-2=-6y-2: \text { or } y^3+6y+2=0$$
It was possible that we would have had to eliminate $x$ between these two equations to get a cubic in $y$, but we were "lucky" that $x$ disappeared of its own accord. See @Stephanos's answer for why this happens.
We were always going to get expressions for $y, y^2, y^3$ as quadratics in $x$ - i.e. as expressions with a constant term, a term in $x$ and a term in $x^2$. If we treat $\{1, x, x^2\}$ like a basis in linear algebra, with three equations we can systematically eliminate $x$ and $x^2$ to leave a cubic in $y$ (with just the constant term).
Applying the techniques of linear algebra in cases like this is one way in which more general problems of this kind can be tackled - and indeed is a tool in proving powerful general theorems.
| {
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"url": "https://math.stackexchange.com/questions/1013589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving Trigonometric Limits compute the following limit:
$$\lim \frac{\cos(a+2x) - 2cos(a+x) + \cos(a)}{x^2}\ \mbox{as}\ x\to0.$$
How would I go about solving this problem? I have attempted to use trig identities (addition of angles) to try to simplify the problem, but it only seemed to make it worse...
but without used ´Hospital
| Note that the numerator of the expression under limit operation can be expressed as $$\begin{aligned}f(a, x) &= \cos (a + 2x) - 2\cos(a + x) + \cos a\\
&= \cos (a + 2x) + \cos a - 2\cos(a + x)\\
&= 2\cos(a + x)\cos x - 2\cos(a + x)\\
&= 2\cos(a + x)(\cos x - 1)\end{aligned}$$ and hence the desired limit $L$ can be evaluated as $$\begin{aligned}L &= \lim_{x \to 0}\frac{f(a, x)}{x^{2}}\\
&= \lim_{x \to 0}\frac{2\cos(a + x)(\cos x - 1)}{x^{2}}\\
&= -2\lim_{x \to 0}\frac{\cos(a + x)(1 - \cos x)}{x^{2}}\\
&= -2\lim_{x \to 0}\frac{\cos(a + x)(1 - \cos^{2} x)}{x^{2}(1 + \cos x)}\\
&= -2\lim_{x \to 0}\cos(a + x)\cdot\frac{\sin^{2}x}{x^{2}}\cdot\frac{1}{(1 + \cos x)}\\
&= -2\cos a\cdot 1\cdot\frac{1}{2} = -\cos a\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1014005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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telescopic series & sum of sequences - how to? The entire web shows the same example basicly(or something extremely similar): $$\sum_{n=2}^\infty \frac{-2}{(n+1)(n+2)}= \frac{-2}{3}$$
Is there a way to find this final answer without doing the pretty long procedure?
In addition to it I want to apply the telescopic series principal on the following sequence: $$(a_{n+1} - a_n)$$
How would I do it? [$a_n$ is sequence and $a_{n+1} - a_n$ is bounded]
| By partial fractions,
$$\frac{-2}{(n+1)(n+2)} \equiv \frac{A}{n+1} + \frac{B}{n+2}$$
For some $A$ and $B$.
Thus,
$$-2 \equiv A(n+2)+B(n+1)$$
By letting $n=-1$ and $n=-2$ in turn, we can see that $A=-2$ and $B=2$. So the sum is now:
$$\sum_{n=2}^{n=k}{\left(\frac{2}{n+2} - \frac{2}{n+1}\right)}=2\sum_{n=2}^{n=k}{\left(-\frac{1}{n+1} + \frac{1}{n+2}\right)}$$
Which is
$$2\left(\left(-\frac{1}{3}+\frac{1}{4}\right)+\left(-\frac{1}{4}+\frac{1}{5}\right)+\left(-\frac{1}{5}+\frac{1}{6}\right)+...+\left(-\frac{1}{k}+\frac{1}{k+1})\right)+\left(-\frac{1}{k+1}+\frac{1}{k+2}\right)\right)$$
Which, by the method of differences, is equivalent to
$$2\left(-\frac{1}{3}+\frac{1}{k+2}\right)$$
If we limit $k$ to $\infty$, $\frac{1}{k+2}$ becomes $0$, thus our sum is $$-\frac{2}{3}$$
More generally, by the same method, $$\sum_{n=x}^{n=y}{(a_{n+1}-a_{n}})=a_{y+1}-a_{x}$$
Hope this helped!
| {
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Find all integers n such that $\;\frac{n^2-9}{n^2-5n+4}$ is an integer.
Find all integers such that $\;\dfrac{n^2-9}{n^2-5n+4}\;$ is an integer.
I am really struggling to figure this out. I can tell that -3,3, and 5 are solutions but I don't know how to show that these are the only solutions or if they even are the only solutions.
| The denominator is $(n-1)(n-4)$. If $n^2\ne 9$, we must have that $n-4$ divides $n^2-9$.
Note that $\frac{n^2-9}{n-4}=n+4+\frac{7}{n-4}$. So the only candidates are those $n$ such that $n-4$ divides $7$. That gives the short list $n=5$, $n=3$, $n=11$, and $n=-3$. Test them all.
| {
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"url": "https://math.stackexchange.com/questions/1015818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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convergence series geometric test Prove if this converges:
$$\sum_{n=1}^\infty \frac{2^n+3}{3^n-1}$$
pf: using geometric
$$0 < \frac{2^n+3}{3^n-1} \leq \frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}} = \cdots $$ and so on
I know how to do the rest but my question is that where in the world did my teacher get
$$\frac{2^n + 2 \times 2^n}{3^n-\frac{3^n}{2}}$$
| A slightly better bound would be
$$\frac{2^n+3}{3^n-1}=\frac{2^n(1+3\cdot 2^{-n})}{3^n(1-3^{-n})}\leq \frac{5/2}{2/3}\frac{2^n}{3^n}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove a limit for $g(x)$ with the definition of limit only We have the function $g(x) = x^3+1$
Prove, with the definition of limit only, that the limit $L=9$ is indeed the limit of the function when $x=2$.
I started with: $\lvert g(x) - L \rvert$ : $\lvert x^3+1-9 \rvert = \lvert x^3-8 \rvert = \lvert (x-2)(x^2+2x+4) \rvert < \epsilon $
From the definition of the limit we know: $ \rvert x-2 \lvert < \delta$, so therefore: $\lvert (x-2)(x^2+2x+4) \rvert < \delta (x^2+2x+4) $
How do I proceed from here? do I need to find a condition for $ x^2+2x+4 $ ?
Thanks guys
| $$|x^2+2x+4|=|x^2-4x+4+6x-12+12|=|(x-2)^2+6(x-2)+12| \leq |\delta^2+6\delta+12|$$
So for small $\delta$ you have (for example) $|x^2+2x+4|<13$
| {
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"source": "stackexchange",
"question_score": "3",
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The inequality $x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+3/4 >0$ holds for all $x\in\mathbb R$
Show
$\forall \ x \in \mathbb{R}:\quad x^{6}-x^{5}+x^{4}-x^{3}+x^{2}-x+\dfrac{3}{4}>0$
My attemps:
Case $x=-1$
That is true for this case
Then for $x \neq - 1$:
$$\dfrac 3 4 - x + x^2 - x^3 + x^4 - x^5 + x^6 = \dfrac{1 + x^7}{1 + x} - \dfrac 1 4$$
let $g(x)=\dfrac{1 + x^{7}}{1 + x}-\dfrac{1}{4} \quad \forall x\in \mathbb{R} \backslash \{-1\}$
then
$
\begin{align*}
g'(x)&=\dfrac {(1+x^7)'(1+x)-(1+x)'(1+x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac {(7 x^6)(1+x)-(1+x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac {7 x^6+7 x^7-1-x^7)}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}\\
g'(x)&=\dfrac{7 x^6+6 x^7-1}{(1 + x)^2} \quad \forall x\in \mathbb{R} \backslash \{-1\}
\end{align*}
$
To determine the sign of the numerator $(7 x^6+6 x^7-1)$
once time let :$ h(x)=7 x^6+6 x^7-1$
then
$$h'(x)=42x^5+42x^6=42(1+x)x^5$$
$$h'(x)=0 \Longleftrightarrow x=-1 \text{or} x=0$$
thus $h$ admits a minimum on the point $x=0$
and a maximum on the point $x=-1$
or $h(-1)=0$ and $h(0)=-1$
$\lim_{x\to -\infty}h(x)=-\infty$ and $\lim_{x\to +\infty}h(x)=+\infty$
as $h(-1)=0$ and $h(0)=-1$ by Intermediate value theorem there is $u \in(-1, 0)$ such that h(u) = 0.
i'm stuck here
*
*am i on my way ?
*is there any other ways to solve it
| Another way: Sum the obvious AM-GMs:
$$\tfrac12 x^6 + \tfrac12 x^4 \ge x^5, \quad \tfrac12 x^4 + \tfrac12 x^2 \ge x^3$$
$$\tfrac12 x^2 + \tfrac34 \ge \sqrt{\tfrac32}|x| \ge x \implies \tfrac12 x^2 + \tfrac34 > x$$
with $\frac12x^6 \ge 0$.
| {
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"url": "https://math.stackexchange.com/questions/1023059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Remainder of a combination Problem from a contest:
What is the remainder when $\binom{169}{13}$ is divided by $13^5$?
I thought that Wolstenholme's/Babbage's would help, but not entirely sure how.
| $$\binom{169}{13}=13\cdot\frac{168\cdot \ldots \cdot 157}{12!}=13\cdot\frac{(13^2-1)\cdot\ldots\cdot(13^2-12)}{12!}$$
$$\frac{1}{12!}\prod_{k=1}^{12}(13^2-k)=\prod_{k=1}^{12}\left(1-\frac{13^2}{k}\right) = 1-13^2 H_{12}+C\cdot13^4$$
and since $H_{12}\equiv 0\pmod{13^2}$ by Wolstenholme's theorem, it follows that:
$$\frac{(13^2-1)\cdot\ldots\cdot(13^2-12)}{12!}\equiv 1\pmod{13^4}$$
and so:
$$\binom{169}{13}\equiv 13\pmod{13^5}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$8x +9y = 5$ where $x,y \in \mathbb{Z}$
Solve the following Diophantine equation algebaically: $$8x+9y=5$$
Give 3 possible solutions for the equation
I have the following:
The Diophantine equation has solutions $x,y \iff 8x=5\mod{9}$ has a solution $x \equiv\mod{9}$
Since $\gcd(8,9)=1$, by Bezout's Lemma, for $r,t \in \mathbb{Z}, \gcd(8,9)=1=r(8)+t(9)$ and $x\equiv r(5)\mod{9}$ is a solution for the linear congruence above.
By Euclid's algorithm for determining $\gcd(8,9)$ we have
\begin{align}9 &= 1(8) +1 \\ 8 &=9(1)+0\end{align}
so $1=(-1)8 + 1(9)$ and $r=-1 \implies x \equiv(-1)5\mod{9}$.
Now \begin{align}[-5]_9 &= \{-5 + 9k \ | k\in\mathbb{Z} \} \\ &= \{ ..., -5,4,13,... \} \\ &=[4]_9\end{align}
$\therefore x \equiv 4 \mod{9}$, that is $x=4+9k$ for all $k \in \mathbb{Z}$ upon which it can be seen that $y= -3 -8k$.
Is this correct?
| Choose $x$ or $y$ ? Let's solve for $x$: $x = \dfrac{5-9y}{8} = \dfrac{5-y}{8} - y \Rightarrow 5-y = 8k \Rightarrow y = 5 - 8k \Rightarrow x = k - (5-8k) = 9k-5$. Thus:
$(x,y) = (9k-5,5-8k), k \in \mathbb{Z}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite integral evaluation of fraction Here's the indefinite integral to evaluate:
$F(x) = \int \left(\frac{3x + 4}{x + 3}\right)dx$
Here are my steps (verbose):
$$\begin{align}
u & = x + 3 \\
(x + 3)dx & = du \\
\left(\frac{d}{dx}(x + 3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + \frac{d}{dx}(3)\right)dx & = du \\
\left(\frac{d}{dx}(x) + 0\right)dx & = du \\
\left( \frac{dx}{dx} + 0 \right)dx & = du \\
(1 + 0)dx & = du \\
(1)dx & = du \\
dx & = \left( \frac{1}{1}\right)du \\
dx & = du \\
F(\ ) & = \int \left( \frac{(3x + 4)}{u} \right)du \\
u & = x + 3 \\
x + 3 & = u \\
x & = u - 3 \\
F(u)& = \int \left( \frac{3(u - 3) + 4}{u} \right)du \\
& = \int \left( \frac{3u - 9 + 4}{u} \right)du \\
& = \int \left( \frac{3u - 5}{u} \right)du \\
& = \int \left( \frac{3u}{u} - \frac{5}{u} \right)du \\
& = \int \left( 3 - \frac{5}{u} \right)du \\
& = \int (3)du - \int \left( \frac{5}{u} \right)du \\
& = \int (3)du - 5*\int \left( \frac{1}{u} \right)du \\
& = \int (3)du - 5*ln \lvert u \rvert \\
& = \int (3*u^0)du - 5*ln \lvert u \rvert \\
& = 3*\int (u^0)du - 5*ln \lvert u \rvert \\
& = 3* \frac{1}{0 + 1}*u^{0 + 1} - 5*ln \lvert u∣ \\
& = 3* \frac{1}{1}*u^1 - 5*ln \lvert u \rvert \\
& = 3u - 5*ln \lvert u \rvert \\
u & = x + 3 \\
F(x) & = 3(x + 3) - 5*ln\lvert x + 3\rvert \\
& = 3x + 9 - 5*ln\lvert x + 3\rvert \\
& = 3x - 5*ln\lvert x + 3\rvert + 9 \\
& = \bf [3x - 5*ln\lvert x + 3\rvert + 9 + C] \\
\end{align}$$
This result agrees with both symbolab and wolfram.
Here's the solution approach provided:
Since the degree of the numerator is not less than the degree of the denominator, we must divide. A clever way to do this is shown below.
$$\begin{align}
\int \left( \frac{3x+4}{x+3} \right)dx &= \int \left( \frac{3x+9−5}{x+3} \right)dx \\
&= \int \left( \frac{3(x+3)}{x+3} − \frac{5}{x+3} \right)dx \\
&= \int \left( 3 − \frac{5}{x+3} \right)dx \\
\end{align}$$
Now the integration is straightforward:
$\int \left(3 −\frac{5}{x+3} \right)dx = 3x−5*ln\lvert x + 3\rvert +C$
#
Although the partial fraction solution appears to be mathematically sound, nonetheless it loses real information: "+9"
Furthermore, I realize that if I were to take the derivative of either solution, I would end up with the same results as both constants "+9" and "C" would resolve to 0.
That stated, my concern is that using a method that is inherently prone to losing information is dangerous as it could result in actual computational error especially if instead of an indefinite integral, one was expected to evaluate a definite integral as in determining the actual area under the functions curve, in which case the missing "+9" would result in an erroneous outcome.
Would my argument above justify contesting the method provided?
| You don't lose any information when you split up the equation; they are equal quantities. The fact that one method nets you a "$9+C$" while another method only nets you another constant $"D"$ is perfectly valid. If you had initial conditions to work with, you would simply find that $9+C = D$, so your integral results are identical as well. When this type of thing happens, you can say that the $9$ was absorbed by the constant $C$. Other than that, very thorough work!
| {
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Evaluation of $\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$ How does one evaluate the following integral?
$$\int_0^1 \frac{x^3}{2(2-x^2)(1+x^2) + 3\sqrt{(2-x^2)(1+x^2)}}\,\mathrm dx$$
This is a homework problem and I have been evaluating this integral for hours yet no success so far. I have tried to rationalize the integrand by multiplying it with
$$\frac{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}{2(2-x^2)(1+x^2) - 3\sqrt{(2-x^2)(1+x^2)}}$$
but the integrand is getting worse. I have tried to use trigonometric substitutions like $x=\tan\theta$ and $x=\sqrt{2}\sin\theta$, but I cannot rid off the square root form. I have also tried to use hyperbolic trigonometric substitutions but the thing does not get any easier neither also substitution $y=x^2$ nor $y=\sqrt{(2-x^2)(1+x^2)}$. Using integration by parts is almost impossible for this one. I have also tried to use the tricks from this thread, but still did not get anything. No clue is given. My professor said, we must use clever substitutions but I cannot find them. Any idea or hint? Any help would be appreciated. Thanks in advance.
Edit :
The answer I got from my Prof is $\dfrac{3-2\sqrt{2}}{6}$.
| If you consider these imaginary parts of elliptic integrals as closed-forms, then
$$\frac{1}{2}-\frac{\sqrt 2}{3}-\frac{\sqrt{2} \log
\left(\sqrt{2}-1\right)}{8}+\frac{\sqrt{2} \log
\left(\sqrt{2}+1\right)}{8}-\frac{\sqrt 2}{8}\Im\left(F\left(i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right)+\frac{\sqrt{2}}{6}\Im\left(E\left(i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right)-\frac{3\sqrt{2}}{8
}\Im\left(\Pi
\left(-2;i \log\left(\sqrt 2+1\right)|-\tfrac{1}{2}\right)\right),$$
where $F\left(x\,|\,m\right)$ is the elliptic integral of the first kind, $E\left(x\,|\,m\right)$ is the elliptic integral of the second kind, and $\Pi\left(n; x \, | \, m\right)$ is the elliptic integral of the third kind with parameters $m=k^2$.
| {
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"answer_id": 5
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Prove that $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$ How can I prove $(a+b-c)^2+(b+c-a)^2+(c+a-b)^2>ab+bc+ca$?
We have
$(a+b-c)^2+=a^2+b^2+b^2+2ab-2bc-2ca$,
$(b+c-a)^2+=b^2+c^2+a^2+2bc-2ca-2ab$
$(c+a-b)^2+=c^2+a^2+b^2+2ca-2ab-2bc$
But I don't know how to show the required result?
Please give hint. Thank you
| We have
$$(a+b-c)^2+(b+c-a)^2+(c+a-b)^2 = 3a^2+3b^2+3c^2-2ab-2bc-2ca$$
From AM-GM, we have $a^2+b^2 \geq 2ab$, $b^2+c^2 \geq 2bc$ and $c^2+a^2 \geq 2ca$. Hence,
$$2(a^2+b^2+c^2) \geq 2(ab+bc+ca) \implies 3(a^2+b^2+c^2) \geq 3ab+3bc+3ca$$
Hence,
\begin{align}
(a+b-c)^2+(b+c-a)^2+(c+a-b)^2 & = 3a^2+3b^2+3c^2-2ab-2bc-2ca\\
& \geq 3ab+3bc+3ca-2ab-2bc-2ca\\
& = ab+bc+ca
\end{align}
which is what you want.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find this limit $\lim_{p\to 0^{+}}\left(\int_{a}^{m-p}f(x)dx+\int_{m+p}^{b}f(x)dx\right)$
Give real numbers $a,b$ such that $0<a<b$ and $m=\dfrac{a+b}{2}<\dfrac{\pi}{4}$,Evaluate
$$\lim_{p\to 0^{+}}\left(\int_{a}^{m-p}f(x)dx+\int_{m+p}^{b}f(x)dx\right)$$
where
$$f(x)=\dfrac{(1+\cos{(2m-2x)})\cos{(a-x)}\cos{(b-x)}}{(1-\sin{(a-x)})(1-\sin{(b-x)})\sin{(2m-2x)}}$$
Now I think follow idea is usefull
Idea:
since
$$\dfrac{1+\cos{(2m-2x)}}{\sin{(2m-2x)}}=\tan{(m-x)}$$
and
$$\dfrac{\cos{(a-x)}}{1-\sin{(a-x)}}=\tan{\left(\dfrac{\pi}{4}+\dfrac{a-x}{2}\right)}$$
$$\dfrac{\cos{(b-x)}}{1-\sin{(b-x)}}=\tan{\left(\dfrac{\pi}{4}+\dfrac{b-x}{2}\right)}$$.
so
$$f(x)=\tan{(m-x)}\cdot \tan{\left(\dfrac{\pi}{4}+\dfrac{a-x}{2}\right)}\cdot \tan{\left(\dfrac{\pi}{4}+\dfrac{b-x}{2}\right)}$$
since
$$m=\dfrac{a+b}{2}$$
so also have
$$\left(\dfrac{\pi}{4}+\dfrac{a-x}{2}\right)+\left(\dfrac{\pi}{4}+\dfrac{b-x}{2}\right)
=\dfrac{\pi}{2}+\dfrac{a+b}{2}-x=\dfrac{\pi}{2}+(m-x)$$
then I fell I will silve it,But can't it deal this integral.
and this problem is interesting
This problem is from The College Mathematics Joutnoal Vol.44.No.3 May 2013 problem,and I can't have this journal.
can see joutnoal
| I think your first identity is wrong:
$$\frac{1+\cos(2m-2x)}{\sin(2m-2x)}=\tan\left(x-m+\frac{\pi}{2}\right)$$
So your function simplifies to:
$$ f(x)=\tan\left(x-m+\frac{\pi}{2}\right)\tan\left(\frac{\pi}{4}+\frac{a-x}{2}\right)\tan\left(\frac{\pi}{4}+\frac{b-x}{2}\right)$$
You can use the triple tangent identity: if $x+y+z=\pi$, then $\tan(x)+\tan(y)+\tan(z)=\tan(x)\tan(y)\tan(z)$.
So:
$$ f(x)=\tan\left(x-m+\frac{\pi}{2}\right)+\tan\left(\frac{\pi}{4}+\frac{a-x}{2}\right)+\tan\left(\frac{\pi}{4}+\frac{b-x}{2}\right)$$
Now the three terms can be integrated separately.
Edit: the further calculation is not complicated conceptually, but still an annoying task. If I did not make mistakes, the result simplifies to:
$$\log\left|\frac{\sin\left(\frac{a-b}{2}\right)}{\sin\left(\frac{b-a}{2}\right)}\right|
+
2\log\left|\frac{\sin\left(\frac{a-b}{2}+\frac{\pi}{4}\right)}{\sin\left(\frac{b-a}{2}+\frac{\pi}{4}\right)}\right|$$
which further simplifies to
$$
2\tan\left(\frac{a-b}{2}+\frac{\pi}{4}\right)$$
I did not thoroughly check this, so it is probable that somewhere I forgot a minus sign or was not careful with taking absolute values.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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$ \int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right)$ How to show the following identity from Wolfram?
$$
\int_0^\infty dx\, x^{-3/4} \ln(1+x) \operatorname{Li}_2 \left( -\frac 1 x \right) =-2\pi \sqrt 2\left[\frac{5 \pi^2}{3} + 16 \left(3 \ln 2 + G - 4 \right) \right].
$$
Here $G$ is Catalan's constant.
I thought of using $\operatorname{Li}_2(-x) + \operatorname{Li}_2(-1/x) = -\zeta(2) - \frac 1 2 \ln(x)^ 2$, but the result is not much easier.
| We start from the integral
$$
\int_{0}^{\infty} \frac{x^u }{(1+s x)(1+t x) }dx= \frac{\pi}{\sin \pi u} \cdot \frac{s^u - t^u}{s^u t^u (s-t)}
$$
for $s$, $t>0$ and $0 < u < 1$.
Integrating with respect to $t$ we get
\begin{eqnarray}
\int_{0}^{\infty} \frac{x^u \log ( 1 + \frac{1}{t x})}{(1+s x)}= - \frac{\pi}{\sin \pi u}\cdot \frac{1}{s^{u+1}} \cdot \left(\, B(s/t;1+u, 0) + \log( 1 - s/t\,)\,\right)
\end{eqnarray}
for $u>0$ and $0 < s < t$, where $B(z;a,b) \colon = \int_0^z v^{a-1}(1-v)^{b-1} d v$ is the incomplete Beta function. Note that we have $B(z;a,0) = z^a \cdot \Phi(z, 1, a)$, where $\Phi(z,s,a) \colon = \sum_{n \ge 0} \frac{z^n}{(n+ a)^s}$ is the Lerch transcendent.
Integrating again with respect to $t$ we get
\begin{eqnarray}
\int_0^{\infty}\frac{x^u}{1 + s x}\, Li_2(-\frac{1}{t x}) = \frac{\pi}{\sin \pi u \cdot (s t)^{1+u}} \cdot \left( s^{1+u} \Phi(s/t, 2, 1+u) - t^{1+u} Li_2(s/t)\right)
\end{eqnarray}
again for $u>0$ and $0 < s < t$. In particular, for $u=1/4$ we get
$$\int_0^{\infty}\frac{x^{1/4}}{1 + s x}\, Li_2(-\frac{1}{t x}) = \frac{\pi \sqrt{2}}{(s t)^{5/4}} \cdot \left( s^{5/4} \Phi(s/t, 2, 5/4) - t^{5/4} Li_2(s/t)\right)$$
Now integrate with respect to $s$ and get, with $t=1$ and $s = v^4$:
\begin{eqnarray}
\int_0^{\infty}x^{-3/4}\log(1 + v^4 x)\, Li_2(-\frac{1}{ x}) =-\frac{4 \pi \sqrt{2}}{v}\left( v^5 \phi(v^4,2,5/4) -Li_2(v^4) - 16 v + 8(1-v) \arctan v + 4 (1+v) \log [(1+v)(1+v^2)] +\\+ 4 (1+v)\log(1-v) + 8(1+v) \textrm{arctanh} (v)\right)
\end{eqnarray}
for $0 < v<1$ and taking $v \to 1$ we get
$$\int_0^{\infty}x^{-3/4}\log(1 + x)\, Li_2(-\frac{1}{ x})=
\frac{2}{3} \sqrt{2} \pi \,( \, 96 + \pi^2 -144 \log 2 - 6\, \zeta(2, 5/4)\,)= -3.339758...$$
and now use that $\zeta(2, 5/4) = \pi^2 + 8 G - 16$, where $\zeta(z, a)$ is the Hurwitz zeta function, and $G$ is the Catalan's constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1029390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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For what $a,b$ such that $ax^2+by^2 = z^2$? This post made me think about this question. What is the criterion on positive integer $a,b$ such that,
$$ax^2+by^2 = z^2$$
can be solved in positive integers $x,y,z$?
(Three broad classes are: 1) $a =\square$; 2) $a+b = \square$; and 3) if the Pell-like eqn $z^2-by^2 = a$ is solvable, but I assume there must be a general approach. This is such a basic question that I'm sure this must have been completely answered already.)
| I'm probably the only proponent of this approach for solving Diophantine equations.
Prefer to understand why the solution is, instead of using modular arithmetic. Especially for complex equations modular arithmetic to use makes no sense. And this approach allows us immediately to find out whether there is a solution and write a direct formula for the solutions of the equation. The meaning is simple. Recorded equation:
$$aX^2+bXY+cY^2=jZ^2$$
Solutions can be written if even a single root.$\sqrt{j(a+b+c)}$ , $\sqrt{b^2 + 4a(j-c)}$ , $\sqrt{b^2+4c(j-a)}$ the solution is, if at least one root of a whole.
Then the solution can be written.
In the case when the root $\sqrt{j(a+b+c)}$ whole.
$$X=(2j(b+2c)^2-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+$$
$$+2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+(j\mp \sqrt{j(a+b+c)})p^2$$
$$Y=(2j(2j-b-2a)(b+2c)-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+$$
$$+2((2j-2a-b)\sqrt{j(a+b+c)}\mp{j(b+2c)})sp+(j\mp\sqrt{j(a+b+c)})p^2$$
$$Z=(2j(b+2c)^2-(b^2+4c(j-a))(a+b+c\pm\sqrt{j(a+b+c)}))s^2+$$
$$+2(b+2c) ( \sqrt{j(a+b+c)} \mp{j})sp + ( a + b + c \mp \sqrt{j(a+b+c)})p^2 $$
In the case when the root $\sqrt{b^2+4c(j-a)}$ whole.
Solutions have the form.
$$X=((2j-b-2c)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+$$
$$+2(4ac+b(2j-b)\pm{(2j-b-2c)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Y=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(2j-b-2a\mp\sqrt{b^2+4c(j-a)}))s^2+$$
$$+2(4ac+b(2j-b)\pm{(b+2a)}\sqrt{b^2+4c(j-a)})sp+(2j-b-2a\pm\sqrt{b^2+4c(j-a)})p^2$$
$$Z=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2+$$
$$+2(4ac+b(2j-b)\pm {(b+2a)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$$
In the case when the root $\sqrt{b^2+4a(j-c)}$ whole.
Solutions have the form.
$$X=(2j^2(b+2a)-j(a+b+c)(2j-2c-b\pm\sqrt{b^2+4a(j-c)}))p^2+$$
$$+2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(2j-2c-b\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Y=(2j^2(b+2a)-j(a+b+c)(b+2a\pm\sqrt{b^2+4a(j-c)}))p^2+$$
$$+2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
$$Z=j(a+b+c)(b+2a\mp\sqrt{b^2+4a(j-c)})p^2+$$
$$+2((a+b+c)\sqrt{b^2+4a(j-c)}\mp{j(b+2a)})ps+ (b+2a\mp\sqrt{b^2+4a(j-c)})s^2$$
The specifics of these equations is that some quadratic form is equivalent to the other. This means that if for example no root is not an integer. You need to find the equivalent quadratic form in which the root is intact. Usually it is enough to do the replacement $X\longrightarrow(X+kY)$ or $Y\longrightarrow(Y+kX)$ .
$p,s - $ any integers which specify.
To find the equivalent form must be substituted in the root and find out what $k$ the root is intact. Usually boils down to the Pell equation or representation of a number as a sum of two squares. If the answer is no it can be seen immediately.
It would be interesting to find these formulas counterexample, but I have not found. I think because it really is a formula in General.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1030037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Range of values of $t$ for which $ 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ Calculation of Range of values of $t$ for which $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\;,$ where $\displaystyle t \in \left[-\frac{-\pi}{2}\;,\frac{\pi}{2}\right]$
$\bf{My\; Try::}$ Given $\displaystyle 2\sin t = \frac{1-2x+5x^2}{3x^2-2x-1}\Rightarrow 6\sin t\cdot x^2-4\sin t \cdot x-2\sin t = 1-2x+5x^2$
$\displaystyle \Rightarrow (6\sin t-5)x^2+2(1-2\sin t)x-(2\sin t+1) = 0$ , Now for calculation of value of $2\sin t\;,$
equation must have real roots. So $\bf{D\geq 0}$.
So $4(1-2\sin t)^2+4(2\sin t+1)\cdot (6\sin t-5)\geq 0$
$\displaystyle (1-2\sin t)^2+(2\sin t+1)\cdot(6\sin t-5)\geq 0$
So we get $16\sin^2 t-8\sin t-4\geq 0\Rightarrow 4\sin^2 t-2\sin t-1\geq 0$
Now HGow can I solve after that, Help me
Thanks
| The final quadratic is an upward drawn parabola. You can solve for $4sin^2 t -2 sin t -1 =0 $ and take the values for t such that sint >0 in [-$\frac{\pi}{2}$,$\frac{\pi}{2}$].
The roots are $\frac{1-\sqrt{5}}{4}$ and $\frac{1+\sqrt{5}}{4}$. So $\frac{1+\sqrt{5}}{4}$ =< sint and sin t <= $\frac{1-\sqrt{5}}{4}$ in [-$\frac{\pi}{2}$,$\frac{\pi}{2}$].Does this help?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1031874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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$ \mathop {\lim }\limits_{n \to + \infty } \frac{{v_{n + 1} }}{{v_n }} = 2$ help me please
true or fulse
(1)$$\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1?
$$
\begin{array}{l}
u_n = \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}}\quad ;u_n > 0 \\
\Leftrightarrow \ln \left( {u_n } \right) = \ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right)^{\frac{1}{n}} \\
\Leftrightarrow \ln \left( {u_n } \right) = \frac{1}{n}\ln \left( {n\left( {n + 1} \right) \cdots \left( {n + n} \right)} \right) \\
\Leftrightarrow \ln \left( {u_n } \right) = \frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n} \\
\Leftrightarrow \mathop {\lim }\limits_{n \to + \infty } \left( {\ln \left( {u_n } \right)} \right) = \mathop {\lim }\limits_{n \to + \infty } \left[ {\frac{{\ln \left( n \right)}}{n} + \frac{{\ln \left( {n + 1} \right)}}{n} + \cdots + \frac{{\ln \left( {n + n} \right)}}{n}} \right] = 0 \\
\mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{n\left( {n + 1} \right) \cdots \left( {n + n} \right)}} = 1 \\
\end{array}
(2)
$$
\begin{array}{l}
t_n = \frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }} \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{1 \times 3 \times \cdots (2n - 1)(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{{1 \times 3 \times \cdots (2n - 1)}} \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \frac{{t_{n + 1} }}{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \frac{{(2n + 1)}}{{\left( {n + 1} \right)^{n + 1} }} \times \frac{{n^n }}{1} \\
= \mathop {\lim }\limits_{n \to + \infty } \left( {2\left( {\frac{n}{{n + 1}}} \right)^{n + 1} + \frac{{n^n }}{{\left( {n + 1} \right)^{n + 1} }}} \right) = 2e^{ - 1} ;\quad \left( {\mathop {\lim }\limits_{n \to + \infty } \left( {1 + \frac{x}{n}} \right)^n = e^x } \right) \\
\Rightarrow \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{t_n }} = \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{{\frac{{1 \times 3 \times \cdots (2n - 1)}}{{n^n }}}} = 2e^{ - 1} \\
\end{array}
$$
(3) $$\mathop {\lim }\limits_{n \to + \infty } \frac{1}{{n^2 }}\sqrt[n]{{\frac{{3n!}}{{n!}}}}$$
$$\left( 4 \right)\mathop {\lim }\limits_{n \to + \infty } \left( {n\sqrt[n]{{\frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^3 }}}}} \right) = !?
$$
| $$\underset{n\rightarrow\infty}{\lim}\frac{v_{n+1}}{v_{n}}=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1\right)\cdots\left(n+1+n-1\right)\left(n+1+n\right)\left(n+1+n+1\right)}{n\cdots\left(n+n\right)}=$$
$$=\underset{n\rightarrow\infty}{\lim}\frac{\left(n+1+n\right)\left(n+1+n+1\right)}{n}=\infty$$ because you have essentialy $n^{2}$ at numerator.$$\underset{n\rightarrow\infty}{\lim}\frac{t_{n+1}}{t_{n}}=\underset{n\rightarrow\infty}{\lim}\frac{1\cdot3\cdots\left(2n-1\right)\left(2n+1\right)}{\left(n+1\right)^{n+1}}\cdot\frac{n^{n}}{1\cdot3\cdots\left(2n-1\right)}=$$
$$=\underset{n\rightarrow\infty}{\lim}\left(2n+1\right)\frac{n^{n}}{\left(n+1\right)^{n+1}}=\underset{n\rightarrow\infty}{\lim}\left(2\left(\frac{n}{n+1}\right)^{n+1}+\frac{n^{n}}{\left(n+1\right)^{n+1}}\right)=2e^{-1}$$
because $\underset{n\rightarrow\infty}{\lim}\left(1+\frac{x}{n}\right)^{n}=e^{x}.$
| {
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"url": "https://math.stackexchange.com/questions/1033546",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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