Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding the limit of a function with ArcTan I've found difficulties finding this limit ( without using Taylor series approximation, as it's intended for the secondary-school ):
$$
\lim_{x\ \to\ \infty}\left[\,
{x^{3} \over \left(\,x^{2} + 1\,\right)\arctan\left(\,x\,\right)} - {2x \over \pi}
\,\right]
$$
Thanks.
| Hint. You may write, as $x$ tends to $+\infty$,
$$
\arctan x=\frac{\pi}{2}-\arctan \frac{1}{x}
$$
and use $$ \arctan u=u-\frac{u^3}{3}+\mathcal{O}(u^4), \quad u \rightarrow 0,$$
to obtain $$
(x^2+1)\arctan x=(x^2+1)\left(\frac{\pi}{2}-\arctan \frac{1}{x}\right)=\frac{\pi x^2}{2}-x+\frac{\pi }{2}+\mathcal{O}\left(\frac{1}{x}\right)
$$
then
$$
\begin{align}
\frac{x^3}{(x^2+1)\arctan x}&=\frac{x^3}{\frac{\pi x^2}{2}-x+\frac{\pi }{2}+\mathcal{O}\left(\frac{1}{x}\right)}\\\\&=\frac{2}{\pi}\frac{x}{1-\frac{2}{\pi x}+\mathcal{O}\left(\frac{1}{x^2}\right)}\\\\&=\frac{2 x}{\pi }+\frac{4}{\pi ^2}+\mathcal{O}\left(\frac{1}{x}\right)
\end{align}
$$ Hence $$ \lim\limits_{x\to +\infty} \left({x^3\over(x^2+1)\arctan(x)}-{2x\over \pi}\right)=\frac{4}{\pi ^2}.$$ Hoping this can help you.
| {
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"url": "https://math.stackexchange.com/questions/1035815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove determinant is zero If
$M = \begin{vmatrix}
1 & a & b+c \\
1 & b & a+c \\
1 & c & a+b \\
\end{vmatrix}$
Show that M = 0 WITHOUT expanding the determinant.
I have tried row operations and haven't had much success. Any tips?
| $$\begin{vmatrix}1&a&b+c\\
1&b&a+c\\
1&c&a+b\end{vmatrix}\stackrel{R_2-R_1\,,\,\,R_3-R_1}\longrightarrow\begin{vmatrix}1&a&b+c\\
0&b-a&a-b\\
0&c-a&a-c\end{vmatrix}= $$
$${}$$
$$=(a-b)(a-c)\begin{vmatrix}1&a&b+c\\
0&-1&1\\
0&-1&1\end{vmatrix}$$
and clearly we reached two equal rows
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Theorem for Equal Sums of Like Powers $x_1^8+x_2^8+x_3^8+\dots$ Kindly see the question at the end of post. Solutions to the system of three equations,
$$\begin{aligned}
a^2+b^2+c^2+d^2\, &= e^2+f^2+g^2+h^2\\
a^4+b^4+c^4+d^4\, &= e^4+f^4+g^4+h^4\\
abcd\, &= efgh
\end{aligned}\tag{1}$$
satisfy two nice consequences.
Consequence 1:
Define $n$ as $n(a^2+b^2+c^2+d^2)^2 = a^4+b^4+c^4+d^4$. Then,
$$32(a^6+b^6+c^6+d^6-e^6-f^6-g^6-h^6)\\(a^{10}+b^{10}+c^{10}+d^{10}-e^{10}-f^{10}-g^{10}-h^{10}) = \\15(n+1)(a^8+b^8+c^8+d^8-e^8-f^8-g^8-h^8)^2$$
This generalizes the Ramanujan 6-10-8 Identity which had the case $d=h=0$.
Consequence 2:
Let $(a+b+c+d)(e+f+g+h) ≠ 0$. Then for $k =2,4,6,8$,
$$(-a+b+c+d)^k + (a-b+c+d)^k + (a+b-c+d)^k + (a+b+c-d)^k + \\(2e)^k + (2f)^k + (2g)^k + (2h)^k\\ =\\ (2a)^k + (2b)^k + (2c)^k + (2d)^k + \\(-e+f+g+h)^k + (e-f+g+h)^k + (e+f-g+h)^k + (e+f+g-h)^k$$
This gives a solution to $x_1^8+x_2^8+\dots+x_8^8 = y_1^8+y_2^8+\dots+y_8^8$.
Question: What are other parameterizations of $(1)$ in terms of binary quadratic forms?
(I know of plenty but I used certain assumptions. It would be good to have a fresh approach.)
| What I write here is not the final answer to this question, but I think it will be helpful.
I notice that the Consequence 1 can be simplified and generalized as below,
Denote
$$R_n=(a^n+b^n+c^n+d^n-e^n-f^n-g^n-h^n)/n$$ for any $n<>0$, and
$$R_0=2(abcd-efgh)/(abcd+efgh)$$
$$m=(a^2+b^2+c^2+d^2)/(a+b+c+d)^2$$
We have
If $R_0=R_1=R_2=0$, then
$$R_3R_5/R_4^2=(m+1)/2$$
$$R_4/R_3=a+b+c+d$$
$$R_3/R_{-1}=-abcd$$
$$\frac{R_{-2}}{R_{-1}}-\frac{R_{-1}}2=\frac1a+\frac1b+\frac1c+\frac1d$$
If $R_1=R_2=R_3=0$, then
$$R_4R_6/R_5^2=(m+1)/2$$
$$R_5/R_4=a+b+c+d$$
$$R_4(1+R_0/2)/R_0=-abcd$$
$$\frac{R_{-1}}{R_0}-\frac{R_{-1}}2=\frac1a+\frac1b+\frac1c+\frac1d$$
More similar identities can be found in my site Algebraic Identities.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is the determinant of this matrix positive or negative? $\left( \begin{array}{ccc}
1 & 1000 & 2 & 3 &4\\
5 & 6 &7&1000 &8\\
1000&9&8&7&6\\
5 & 4&3&2&1000\\
1&2&1000&3&4\\ \end{array} \right)$
When I compute the determinant online, I find that it is positive, but I'm supposed to "see" something about the matrix that allows me to know the determinant is positive. What properties does this specific matrix have that allow you to deduce the determinant will be positive?
| By definition, the determinant is the sum of $60$ positive and $60$ negative terms, one of which is $\mbox{sgn}{\left(
\begin{array}{ccccc}
1 & 2 & 3 & 4 & 5 \\
3 & 1 & 5 & 2 & 4 \\
\end{array}
\right)}(1000)^5=\mbox{sgn}\left((13542)\right)(1000)^5=+(1000)^5$.
Each of the other terms has absolute value at most $9\cdot1000^4$, so the sum of the 60 negative terms is greater than or equal to $-60\cdot9\cdot1000^4$, and the determinant is greater than or equal to $(1000)^5-60\cdot9\cdot1000^4$, which is positive.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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I have proven that $e^x > x^3$ for $x>5$, can I prove that $\lim \frac{x^3}{e^x} = 0$? In order to calculate the limit
$$\lim_{x\to\infty} \frac{x^3}{e^x} = 0$$
I've verified that:
$$f(x) = e^x-x^3\\f'(x) = e^x-3x^2\\f''(x) = e^x-6x\\f'''(x) = e^x-6$$
Note that $x>3 \implies f'''(x)>0$, therefore $f''(x)$ is crescent. Then, I find that $x = 3 \implies e^3-6\cdot3 >0$. Then, $x>3 \implies f'(x)$ is crescent. I just have to find an $x$ for $f'(x)$ such that $f'(x)>0$, then $f$ will be crescent. It happens that $x=4 \implies f'(x)>0$. Then, $x>4 \implies f(x)>0$. But for $x = 5$, $f(x)>0$, and $f$ is crescent, then:
$$x>5 \implies e^x-x^3>0\implies e^x>x^3$$
Therefore, some limits can be calculated easely, like:
$$x>5 \implies e^x>x^3\implies\lim_{x\to\infty}e^x>\lim_{x\to\infty}x^3 = \infty$$
But for the limit I want, we have:
$$x>5 \implies e^x>x^3\implies 1>\frac{x^3}{e^x}\implies \frac{x^3}{e^x}<1$$
I cannot, therefore, simply prove that this limit is equal to $0$, just by this inequality.
Any ideas?
| $$\lim_{x \to 0} \frac{x^3}{e^x} = \lim_{x \to 0} \frac{x^3}{1 + x + x^2/2! + x^3/3! + \dots} = \frac 01 = 0$$
And
$$\begin{align}\lim_{x \to \infty} \frac{x^3}{e^x}
&= \lim_{x \to \infty} \frac{x^3}{1 + x + x^2/2! + x^3/3! + x^4/4!+\dots}
\\&= \lim_{x \to \infty} \frac{1}{x^{-3} + x^{-2} + x^{-1}/2! + 1/3! + x/4!+\dots} \\&= \frac{1}{\infty}
\\&= 0
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Application of Mergesort We have $8$ players and we want to sort them in $24$ hours.
There is one stadium. Each game lasts one hour.
In how many hours can we sort them??
I thought that we could it as followed:
$$\boxed{P1} \ \boxed{P2} \ \boxed{P3} \ \boxed{P4} \ \boxed{P5} \ \boxed{P6} \ \boxed{P7} \ \boxed{P8} \\ \ \ \ \ \boxed{P1} \ \ \ \ \ \ \ \ \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P5} \ \ \ \ \ \ \ \ \ \ \ \boxed{P7} \\ \ \ \boxed{P1} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P5} \\ \ \ \ \boxed{P1}$$
So, the best player is $P1$.
These games took place in $7$ hours.
We know that $P5$ is the best player among the players $P5$, $P6$, $P7$ and $P8$.
$$\boxed{P2} \ \boxed{P3} \ \boxed{P4} \ \boxed{P5}\\ \boxed{P2} \ \ \ \ \ \ \boxed{P5} \\ \ \ \boxed{P2}$$
So, the second best player is $P2$.
These games took place in $3$ hours.
$$\boxed{P3} \ \boxed{P4} \ \boxed{P5}\\ \ \ \ \ \ \boxed{P3}\ \ \ \ \boxed{P5} \\ \ \ \ \ \ \ \boxed{P3}$$
So, the third best player is $P3$.
These games took place in $2$ hours.
$$\boxed{P4} \ \boxed{P5}\\ \ \ \boxed{P4}$$
So, the $4^{th}$ best player is $P4$ and the $5^{th}$ best player is $P5$.
This game took place in $1$ hour.
$$\boxed{P6} \ \boxed{P7} \ \boxed{P8}\\ \ \ \ \ \ \boxed{P6}\ \ \ \ \boxed{P8} \\ \ \ \ \ \ \ \boxed{P6}$$
So, the $6^{th}$ best player is $P6$.
These games took place in $2$ hours.
$$\boxed{P7} \ \boxed{P8}\\ \ \ \ \ \boxed{P7}$$
So, the $7^{th}$ best player is $P7$ and $8^{th}$ best player is $P8$.
This game took place in $1$ hour.
Therefore, we sorted the players $P1 \geq P2 \geq P3 \geq P4 \geq P5 \geq P6 \geq P7 \geq P8$ in $7+3+2+1+2+1=16$ hours.
Is it correct??
Is it maybe an application of Mergesort??
EDIT:
In the case when we have more than one stadium is it as followed??
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P1} \ : \ \boxed{P2} \ \ \ \ \ \ \ \ \boxed{P3} \ : \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ : \ \boxed{P6} \ \ \ \ \ \ \ \ \boxed{P7} \ : \ \boxed{P8} \Rightarrow 4 \text{ Stadiums } \\ \boxed{P1} \ > \ \boxed{P2} \ \ \ \ \ \ \ \ \boxed{P3} \ > \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ > \ \boxed{P6} \ \ \ \ \ \ \ \ \boxed{P7} \ > \ \boxed{P8}$$
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P1} \ : \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P2} \ : \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ : \ \boxed{P7} \ \ \ \ \ \ \ \ \boxed{P6} \ : \ \boxed{P8} \Rightarrow 4 \text{ Stadiums } \\ \boxed{P1} \ > \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P2} \ > \ \boxed{P4} \ \ \ \ \ \ \ \ \boxed{P5} \ > \ \boxed{P7} \ \ \ \ \ \ \ \ \boxed{P6} \ > \ \boxed{P8}$$
Now we know tht $P1 >P2>P4$ and $P1>P3>P4$, that means that we have to know whether $P2>P3$ or $P3>P2$.
Also, we know that $P5>P6>P8$ and $P5>P7>P8$, that means that we have to know whether $P6>P7$ or $P7>P6$.
So, we have the following:
$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \boxed{P2} \ : \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P4} \ : \ \boxed{P5} \ \ \ \ \ \ \ \ \boxed{P6} \ : \ \boxed{P7} \Rightarrow 3 \text{ Stadiums } \\ \boxed{P2} \ > \ \boxed{P3} \ \ \ \ \ \ \ \ \boxed{P4} \ > \ \boxed{P5} \ \ \ \ \ \ \ \ \boxed{P6} \ > \ \boxed{P7}$$
So, we have sorted them $P1>P2>P3>P4>P5>P6>P7>P8$ in $3$ hours.
Is it correct??
| I think that the question is: if we wish to sort $n$ items by comparing them two at a time, and the algorithm is completely determined by the results of those comparisons, what is the minimum number of comparisons that we need to make?
For example, if there are $n = 3$ items, there are 6 possible arrangements, and we need to make some comparisons to see which is the "correct" arrangement. In this case, we will need at least 3 comparisons. The first comparison (say, of P1 and P2) will cut us down to three possible "true" arrangements. But there is no second comparison that, regardless of its outcome, will reduce this to one possible comparison. Therefore, we will need a third comparison in the worst case.
More information about this is written in Wikipedia under Comparison sort.
The exact number of comparison needed to sort $n$ items in the worst case, assuming an optimal algorithm, is apparently unknown. It is sequence A036604 in the OEIS. That web page does show that 16 is the least number possible for $n = 8$ items, as in the question. But the lack of any formula on that page suggests that, in general, the problem is quite difficult. In particular, the merge sort example is not likely to generalize to arbitrary $n$.
| {
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Maximum and minimum of $f(x,y)=xy$ when $x^2 + y^2 + xy =1$ It is asked to find the maximum and minimum points of the function
$$f(x,y)=xy$$
when $x^2 + y^2 + xy=1$
I've tried Lagrange and obtained
$$\lambda = \frac{y}{2x+y}=\frac{x}{2y+x}$$
but what should I do with this? Any other suggestion?
Thanks!
| you can do this without calculus. the $x$-coordinates of the points common to both $xy = k$ and $x^2 + y^2 + xy = 1$ satsfies $ x^2 + k^2/x^2 + k - 1 = 0$ which can be turned into a quadratic equation for $u = x^2$ as $$u^2 + (k-1)u^2 + k^2 = 0$$ whose discriminant $ 1 - 2k - 3k^2 $ is positive for $-1 \le k \le 1/3.$
so the maximum value of $f$ is $2/3$ and the minimum of $f$ is $-1.$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Check: Find all the Sylow $3$-subgroups of $A_4$. Find all the Sylow $3$-subgroups of $A_4$.
The order of $A_4$ is $\frac{4!}{2}=\frac{24}{2}=12=3\cdot 2^2$. So the Sylow $3$-subgroups are
*
*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 2 & 3 & 4 \end{pmatrix}, \begin{pmatrix} 2 & 4 & 3 \end{pmatrix} \}$
*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 3
\end{pmatrix}, \begin{pmatrix} 1 & 3 & 2 \end{pmatrix} \}$
*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 1 & 2 & 4
\end{pmatrix}, \begin{pmatrix} 1 & 4 & 2 \end{pmatrix} \}$
*$\{ \begin{pmatrix} 1 \end{pmatrix}, \begin{pmatrix} 1 & 3 & 4 \end{pmatrix}, \begin{pmatrix} 1 & 4 & 3 \end{pmatrix} \}$
Should I add more details or is this sufficient?
| The highest power of 3 that divides $|A_4|$ is $3^1=3$. So we need to list all subgroups of $A_4$ of order 3. One of them is clearly $\langle (234) \rangle$, as you mention. Now we need to determine all the remaining subgroups of order 3. One way is as follows. Since a group of order 3 is cyclic, a subgroup of order 3 is of the form $\langle (ijk) \rangle$. The number of ways to choose 3 elements from 4 elements is ${4 \choose 3}=4$. Once the three elements are chosen, the three-cycle and its inverse that (along with the identity) form the subgroup of order 3 are uniquely determined.
We could also use the part of Sylow's theorem that says that all Sylow 3-subgroups are conjugate. Hence they all can be obtained by relabeling the points in the first subgroup $\langle (234) \rangle$. This is because conjugation effects a relabeling of the points.
Alternatively, once you list (using any method) the four subgroups you mention, we know these are all the Sylow 3-subgroups. For if a group $G$ has order $p^k m$, where $p$ does not divide $m$, then the number $n_p$ of Sylow $p$-subgroups is congruent to 1 mod $p$ and $n_p$ divides $m$. In this special case, the number $n_3$ of Sylow 3-subgroups is congruent to 1 mod 3 and must divide 4, so $n_3$ must equal 1 or 4. But since you found more than one Sylow 3-subgroup, $n_3$ equals 4.
| {
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How to simplify this expression? $\frac{x+1}{x^2+x-2}-\frac x{x^2-1}$. $$\frac{x+1}{x^2+x-2}-\dfrac x{x^2-1}=\text{ ?}$$
This expression needs to be simplified. I've tried to do it but i couldn't.
| Hint;
$x^2+x-2=(x-1)(x+2)\tag{1}$
$x^2-1=x^2-1^2=(x-1)(x+1)\tag{2}$
| {
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Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$ Prove by induction that
$$
\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}
$$
Let $n=r$, so that
$$
S_r=2-\frac{r+2}{2^r}
$$
Therefore
$$\begin{align}
S_{r+1}=S_r+\frac{r+1}{2^{r+1}}&=2-\frac{r+2}{2^r}+\frac{r+1}{2^{r+1}}\\
&=2-\frac{2^r(r+2)+r+1}{2^{r+1}}\\
&=2-\frac{2^{n-1}(n+1)+n}{2^n}
\end{align}$$
How does $2^{n-1}(n+1)\equiv2$? Or is my method wrong? I've probably made a stupid mistake!
| Note that
$$-\frac{r+2}{2^r}+\frac{r+1}{2^{r+1}}=\frac{-2(r+2)}{2^{r+1}}+\frac{r+1}{2^{r+1}}=-\frac{r+3}{2^{r+1}}=-\frac{(r+1)+2}{2^{r+1}}.$$
| {
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$x^n + y^n = c$ has finitely many integral solutions? Assume $n > 1$ and $n$ is odd because it's easy if $n$ is even.
Please help prove this.
$x^n + y^n = c$ has finitely many integral solutions if $c \neq 0$?
Thank you all for replying. I think I've just found my answer to this one. And I think it's quite simple. Here it goes.
(Non-negative solutions mean solutions $(x,y)$ where $x, y \geq 0;$ and negative solutions mean solutions $(x, y)$ where $x, y < 0$.)
The equation has finitely many non-negative solutions as well as negative solutions. This is quite trivial to show. From here, consider only the solutions $(x, y)$ where $x$ and $y$ have opposite signs. Without loss of generality, assume $x >0$ and $y <0$.
Let $z = -y >0$. So the equation becomes $x^n - z^n = c$.
$\implies |x^n - z^n| = |x-z||x^{n-1}+x^{n-2}z + \dots+xz^{n-2}+z^{n-1}|=|c|$.
Since $|x-z| \geq 1$ (if $|x-z| = 0, c = 0$), $|c| \geq |x^{n-1}+x^{n-2}z + \dots+xz^{n-2}+z^{n-1}| \geq |x^{n-1} + z^{n-1}| = x^{n-1} +z^{n-1}$
Now, it's obvious that there are only finitely many such $(x, z)$. And it's pretty much done here. Please let me know of my any mistake.
| Let's look at the factorization of $x^n+y^n$.
For all odd $n$, $$x^n+y^n=(x+y)\left(x^{n-1}-x^{n-2}y+x^{n-3}y^2-x^{n-4}y^3+\cdots-xy^{n-2}+y^{n-1}\right)$$
Thus, since $x^n+y^n=c$, $$\left(x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1} \right)\Bigg| c$$
This implies that $$\left|x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1}\right|\leq |c|$$
Now here's the key to the argument: we claim that $$x^{n-1}-x^{n-2}y+\cdots-xy^{n-2}+y^{n-1}\geq \min\left(x^{n-1}, y^{n-1} \right)$$
If we can show this, then we will have shown that in order for $x^n+y^n=c$, we must have that $$\min\left(x^{n-1},y^{n-1}\right)\leq |c|$$
This wins the proof for us because this, combined with our restriction that $x^n+y^n=c$, is only true for finitely many $x$ and $y$. So all we have to do now is show the inequality holds. I'll do the $n=5$ case so you can see the general principal at work.
So we're considering the equation $$x^4-x^3y+x^2y^2-xy^3+y^4$$
If $|x|=|y|$ (this implies that $x=y$ since $x\neq -y$ because $x^n+(-x)^n=0$ and $c\neq 0$), then it's very easy to show that the inequality holds. $\checkmark$
Without loss of generality, let's let $|x|>|y|$. We're gonna get a few cases.
Case 1 $x\geq 0$ and $y\geq 0$. Then we have that $x^4-x^3y\geq 0$ and $x^2y^2-xy^3\geq 0$, both since $x>y$. Thus $$x^4-x^3y+x^2y^2-xy^3+y^4\geq y^4= \min\left(x^4,y^4\right) \checkmark$$
Case 2 $x\leq 0$ and $y\leq 0$. This case is extremely similar to Case 1. $\checkmark$
Case 3 $x\geq 0$ and $y\leq 0$. Then we have that $-x^3y\geq 0$ and $-xy^3\geq 0$, both since $y\leq 0$. Thus $$x^4-x^3y+x^2y^2-xy^3+y^4\geq y^4=\min\left(x^4,y^4\right) \checkmark$$
Case 4 $x\leq 0$ and $y\geq 0$. Super similar to Case 3. $\checkmark$
These are all of the possible cases, so that completes the proof for $n=5$. (Woohoo!)
Showing that that inequality holds for all odd $n$ is done using the same general idea as in the $n=5$ case. Then, once you've shown that, there are only finitely many $x$ and $y$ that can satisfy $$\min\left( x^{n-1},y^{n-1}\right)\leq |c|$$ and $$x^n+y^n=c$$ so we're finished!
| {
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Asymptotic development of a recurrent sequence Let $u_0 = 1$ and $u_{n+1} = \frac{u_n}{1+u_n^2}$ for all $n \in \mathbb{N}$.
I can show that $u_n \sim \frac{1}{\sqrt{2n}}$, but I would like one more term in the asymptotic development, something like $u_n = \frac{1}{\sqrt{2n}}+\frac{\alpha}{n\sqrt{n}} + o\bigl(\frac{1}{n^{3/2}}\bigr)$.
Here is the outline of my proof of $u_n \sim \frac{1}{\sqrt{2n}}$:
*
*$(u_n)$ is decreasing, and bounded from below by $0$, hence converges.
*The limit $\ell$ satisifies $\ell = \frac{\ell}{1+\ell^2}$, hence $\ell = 0$.
*A computation gives $v_n = u_{n+1}^{-2} - u_n^{-2} \to 2$.
*Using Cesàro lemma, $\frac{1}{n} \sum_{k=0}^{n-1} v_k \to 2$.
*Hence $\frac{1}{n} u_n^{-2} \to 2$.
| Let $a_n=\dfrac1{u_n^2}$. Then
$$
a_{n+1}-a_n = \left(\frac{1+u_n^2}{u_n}\right)^2 - \frac1{u_n^2} = 2+u_n^2 = 2+\frac1{a_n}.
$$
From this and $a_2=4$ we can find a successive sequence of estimates for $n\ge2$:
\begin{align*}
a_n &\ge 2n; \\
a_n &\le 2n + \sum_{k=2}^{n-1}\frac1{2k}
< 2n+\frac12\log n; \\
a_n &\ge 2n + \sum_{k=2}^{n-1}\frac1{2k+\frac12\log k}
> 2n+\int_2^n \frac{dx}{2x+\frac12\log x}
\\ &> 2n+\int_2^n \frac{dx}{2x}
- \int_2^n \frac{\frac12\log x}{2x(2x+\frac12\log x)}dx
= 2n+\frac12\log n - \mathcal{O}(1).
\end{align*}
Then
$$
u_n = a_n^{-1/2} =
\frac1{\sqrt{2n}} \left(1+\frac{\log n}{4n}+\mathcal{O}\bigg(\frac1n\bigg)\right)^{-1/2} \\
= \frac1{\sqrt{2n}} \left(1-\frac12 \cdot \frac{\log n}{4n} + \mathcal{O}\bigg(\frac1n\bigg) \right) =
\frac1{\sqrt{2n}} - \frac1{8\sqrt2}\cdot \frac{\log n}{n^{3/2}} + \mathcal{O}\left(\frac1{n^{3/2}}\right).
$$
| {
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Differentiation with the quotient rule I have the question:
Given that $$y=\frac{e^x}{\sqrt{1+2x}}$$
Show that $$\frac{dy}{dx} = \frac{2xe^x}{\sqrt{(1+2x)^3}}$$
I've done the question but I got $2xe^x(\sqrt{(1+2x)^3})$. I feel like this is too similar to the sheet's answer, so am I wrong or is the sheet printed wrong?
Working:
Let $u = e^x$, $v = 1+2x$, $w = \sqrt{v}$.
Using the chain rule, $$\frac{dw}{dx} = \frac{dw}{dv} * \frac{dv}{dx}$$
So $$\frac{dw}{dx} = 2(\frac{1}{2}v^{-\frac{1}{2}}) = \frac{1}{\sqrt{v}} = \frac{1}{\sqrt{1+2x}}$$
And $\frac{du}{dx}$ is clearly $e^x$.
Therefore, using the quotient rule, $$\frac{dy}{dx} = \frac{e^x(\sqrt{1+2x})- e^x(\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\sqrt{1+2x}-\frac{1}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{e^x(\frac{2x}{\sqrt{1+2x}})}{(1+2x)^2}$$
$$=\frac{\frac{2xe^x}{\sqrt{1+2x}}}{(1+2x)^2}$$
Using the rule
$$\frac{x^m}{x^n} = x^{m-n}$$
We get
$$\frac{2xe^x}{(1+2x)^{-\frac{3}{2}}}$$
$$=2xe^x(\sqrt{1+2x})^3 $$
Have I done this correctly?
| I suspect
$$\frac{1}{\left(\sqrt{1+2x}\right)^2} \not = \frac{1}{\left({1+2x}\right)^2}$$
and $$\frac{\frac{1}{\sqrt{1+2x}}}{(1+2x)^2} \not =\frac{1}{(1+2x)^{-\frac{3}{2}}}$$
but $$\frac{\frac{1}{\sqrt{1+2x}}}{\left(\sqrt{1+2x}\right)^2} = \frac{1}{\left(\sqrt{1+2x}\right)^3}$$
| {
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Help with two limit problems I have to prove these limits by definition. Can somebody help?
$\lim_{n \to \infty}\left ( \frac{n^2+1}{4n^2+5} \right )=\frac{1}{4}$
$\lim_{n \to \infty}\left ( \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n} \right )=1$
This is what i have done so far:
For the first I got this and i don't know if its right and how to proceed: $\frac{\frac{-1}{\epsilon }-20}{16}<n^2$
For the second: $\frac{-2}{\epsilon+1} < 2^n$.
Thanks.
| For the first one:
$$\left | \frac{n^2+1}{4n^2+5} -\frac{1}{4}\right |=\left | \frac{4n^2+4-4n^2-5}{4(4n^2+5)}\right |= \frac{1}{4(4n^2+5)}<\epsilon \\\iff 4n^2+5>\frac{1}{4\epsilon}\iff 4n^2>\frac{1}{4\epsilon}-5\iff n^2>\frac{1}{16\epsilon}-\frac{5}{4}=\frac{1-5\epsilon}{16\epsilon} \\\iff n>\sqrt{\frac{1-5\epsilon}{16\epsilon} }. $$
Thus, for any $\epsilon >0$ ($\epsilon<1/5$) there exists $N>\sqrt{\frac{1-5\epsilon}{16\epsilon} }$ such that $$n\ge N \implies \left | \frac{n^2+1}{4n^2+5} -\frac{1}{4}\right |<\epsilon.$$
For the first one:
$$\left | \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}-1 \right|=\left|\frac{\frac12-\frac{1}{2^{n+1}}}{1-\frac12}-1\right|=\left|1-\frac{1}{2^n}-1\right|=\frac{1}{2^n}<\epsilon \\ \iff2^n>\frac{1}{\epsilon}\iff n>\log_2 \frac{1}{\epsilon}.$$
Thus for any $\epsilon >0$ there exists $N>\log_2 \frac{1}{\epsilon}$ such that $$n\ge N \implies \left | \frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}-1 \right|<\epsilon.$$
Note that this is more or less what you have got. But you have forgotten the absolute value.
| {
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Triple Integral to Find Volume Question: Use a triple integral to find the volume of the solid enclosed by the parabaloids $y=x^2+z^2$ and $y=8-x^2-z^2$.
My attempt: The best I can figure, this object looks kind of like a football oriented along the $y$-axis from $y=0$ to $y=8$ and is symmetric about the $y$-axis and the plane $y=4$.
It seems best to integrate first with respect to $y$, and $x^2+z^2 \le y \le 8-(x^2+z^2)$.
The widest part of the football is at $y=4$; substitute that into both of the equations above to find that the projection onto the $xz$-plane is $x^2+z^2=4$, or a circle of radius 2, so my bounds for $z$ are $-\sqrt{4-x^2} \le z \le \sqrt{4-x^2}$ and my bounds for $x$ are $-2 \le x \le 2$.
But I believe I can make this easier by integrating from $0\le z \le \sqrt{4-x^2}$ and multiplying by 2 and integrating from $0 \le x \le 2$ and multiplying by another 2. (I actually think I can integrate $y$ from $4 \le y \le 8-(x^2+z^2)$ and multiply by another 2, but that doesn't seem to simplify anything.)
Since I'm finding the volume, the function I integrate is one. I come up with this
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-(x^2+z^2)} 1\ dy\ dz\ dx.$$
(Is this right so far?)
If nothing's wrong yet, I still can't finish this integral
$$4\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{x^2+z^2}^{8-\left(x^2+z^2\right)} 1\ dy\ dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} y \Big|_{x^2+z^2}^{8-\left(x^2+z^2\right)} dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-\left(x^2+z^2\right)\right)-\left(x^2+z^2\right) dz\ dx \\
4\int_0^2 \int_0^{\sqrt{4-x^2}} \left(8-2x^2-2z^2\right)\ dz\ dx \\
4\int_0^2 \left[\left(8-2x^2\right)z-\frac 23 z^3\right]_0^{\sqrt{4-x^2}} \ dx \\
4\int_0^2 \left[ \left(8-2x^2\right)\sqrt{4-x^2} - \frac 23 \sqrt{4-x^2}^3 \right]\ dx \\
-\frac{16}3\int_0^2 \left[ \left(x^2 -4\right)\sqrt{4-x^2} \right]\ dx \\
\vdots \\ ??$$
| HINT:For a start, it is convenient to put $ x^2+z^2 = r^2$, so that you have solids of revolution:
$$ y_1=r^2 ,\ y_2=8-r^2 $$.
| {
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Gemetric representation of complex numbers Find the geometric representation of;
|z-2| - |z+2| < 2
things i know;
| z - 2 | is the distance between a point z and the point (2,0) in the complex plane. It suppose to represent a hyperbola, BUT i have no idea on how to get there. i know the standard form of a hyperbola, but just, no idea on how to manipulate this expression into the standard form.
| If you want to manipulate it algebraically (without considering the geometric definition of the hyperbola), then let $$z=x+iy$$ and note that $$|z|^2=z\overline{z}=x^2+y^2.$$
Edit
First, as someone mentioned in the comments, I am not sure if you want the equation of the hyperbola or the set of points that satisfy your relation. This is how you can algebraically obtain the equation of the hyperbola:
Consider $|z-2|-|z+2|= 2$
Then, letting $z=x+iy$ and using the definition of $|z|$ above, we have
\begin{align*}
|(x-2)+iy|-|(x+2)+iy|&=2 \\
\sqrt{(x-2)^2+y^2}-\sqrt{(x+2)^2+y^2}&=2 \\
(x-2)^2+y^2&=\left(2+\sqrt{(x+2)^2+y^2}\right)^2 \\
&=4+4\sqrt{(x+2)^2+y^2}+(x+2)^2+y^2.
\end{align*}
So
\begin{align*}
\left((x-2)^2-(x+2)^2-4\right)^2&=\left(4\sqrt{(x+2)^2+y^2}\right)^2 \\
\left(-8x-4 \right)^2&=16\left((x+2)^2+y^2\right)\\
16(4x^2+4x+1)&=16\left((x+2)^2+y^2\right) \\
4x^2+4x+1-(x+2)^2-y^2&=0 \\
3x^2-3-y^2&=0 \\
\end{align*}
So the equation of the hyperbola is
$$x^2-\left(\frac{y}{\sqrt{3}}\right)^2=1$$
The geometric way is perhaps simpler:
The equation of the hyperbola is given as
$$|z-z_1|-|z-z_2|=2a,$$ where $z_1,z_2$ are the foci (remember $z=x+iy$ so we can assign the coordinates $(x,y)$ to $z$), and $a$ is half of the major axis. For our example we have $z_1=(2,0)$ and $z_2=(-2,0)$ and the focal length is $2$. Since the focal length is given by $\sqrt{a^2+b^2},$ where b is the minor axis, we have
$$2^2=1+b^2,$$ so $b= \sqrt{3}$. Putting all of this together gives the same equation as above.
Note
Technically we want $$||z-z_1|-|z-z_2||=2a,$$ in order to ensure that the distance does not become negative (if we go to the other branch).
Edit
To solve your inequality, we do
\begin{align*}
x^2-\left(\frac{y}{\sqrt{3}}\right)^2&<1 \\
y^2&>3(x^2-1),
\end{align*}
So
$$y>\sqrt{3(x^2-1)}$$ or
$$y<-\sqrt{3(x^2-1)}$$
| {
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Combinatorics: premutations with repetition? I have following problem from combinatorics:
Let's have set of 8 distinct items: {a,b,c,d,e,f,g,h}
How many ways we can combine 10 of them if we know:
*
*We start with A and end with H
*Every item of the set must be in the result at least once.
I started as such ... we know that A and H are already at positions: (A)()()()()()()()()(H)
So we need to find combinations for the rest of the "fields" and there we have:
*
*6 unique items
*2 repeating items
Hence I arrived to this: 8! + 6*P*(3,1,1,1,1,1) + 2*7*P*(2,1,1,1,1,1,1) + 15*P*(2,2,1,1,1,1)
Is it correct or am I doing it wrong ?
| Since $a$ must appear in the first of the ten positions, $h$ must appear in the last position, and each of the eight letters in the set $\{a, b, c, d, e, f, g, h\}$ must appear at least once, either one letter appears three times or two letters appear twice each.
Consider cases:
*
*The letter $a$ appears three times: Since $a$ appears in the first position and $h$ appears in the last position, we must fill two of the remaining $8$ positions with an $a$, which can be done in $\binom{8}{2}$ ways, then fill the remaining six positions with the remaining letters, which can be done in $6!$ ways. Hence, the number of arrangements in which $a$ appears three times is
$$\binom{8}{2} \cdot 6! = \frac{8!}{6!2!} \cdot 6! = \frac{8!}{2!}$$
*The letter $h$ appears three times: A similar argument shows that this can also occur in $$\frac{8!}{2!}$$ ways.
*A letter other than $a$ or $h$ appears three times: First, we must select one of the six remaining letters to be repeated. Next, we select three of the eight remaining positions for the repeated letter, which can be done in $\binom{8}{3}$ ways, then fill the five remaining positions with the remaining letters, which can be done in $5!$ ways. Thus, the number of arrangements in which a letter other than $a$ or $h$ appears three times is
$$6 \cdot \binom{8}{3} \cdot 5! = 6 \cdot \frac{8!}{5!3!} \cdot 5! = 8!$$
*The letters $a$ and $h$ each appear twice: Each of the letters must appear once in the eight open positions. Thus, the number of arrangements in which $a$ and $h$ each appear twice is $$8!$$
*The letter $a$ appears twice and a letter other than $a$ or $h$ appears twice: We fill two of the remaining positions with a letter other than $a$ or $h$, which can be done in $\binom{8}{2}$ ways, leaving us with six places where we can place the $a$. That leaves us with five positions for the remaining five letters, which can be filled in $5!$ ways. Hence, the number of arrangements in which $a$ appears twice and a letter other than $a$ or $h$ appears twice is
$$\binom{8}{2} \cdot 6 \cdot 5! = \frac{8!}{6!2!} \cdot 6! = \frac{8!}{2!}$$
*The letter $h$ appears twice and a letter other than $a$ or $h$ appears twice: A similar argument shows that this can also occur in $$\frac{8!}{2!}$$ ways.
*Two letters other than $a$ or $h$ each appear twice: There are six ways to choose the first repeated letter and $\binom{8}{2}$ ways to place these letters in the eight open positions. There are five ways to choose the second repeated letter and $\binom{6}{2}$ to place these letters in the six remaining open positions. However, we have counted each possible arrangement of the two pairs of repeated letters twice since filling the second and third positions with a $b$ and then filling the fourth and fifth positions with a $c$ is equivalent to first filling the fourth and fifth positions with a $c$ and then filling the second and third ones with a $b$. Thus, the number of arrangements of the two pairs of repeated letters is $$6 \cdot 5 \cdot \frac{1}{2} \cdot \binom{8}{2} \cdot \binom{6}{2} = 3 \cdot 5 \cdot \binom{8}{2} \cdot \binom{6}{2}$$ Finally, there are $4!$ ways to place the remaining four letters. Hence, there are $$3 \cdot 5 \cdot \binom{8}{2} \cdot \binom{6}{2} \cdot 4! = 3 \cdot 5 \cdot \frac{8!}{6!2!} \cdot \frac{6!}{4!2!} \cdot 4! = \frac{8! \cdot 3 \cdot 5}{2!2!}$$ such arrangements.
Since the seven cases are mutually exclusive, there are
$$4 \cdot \frac{8!}{2!} + 2 \cdot 8! + \frac{8! \cdot 3 \cdot 5}{2!2!}$$
possible arrangements.
| {
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Possible values of $\lim_{x \rightarrow \infty} \left(\frac{p(x)}{q(x)}\right)^x$ I was asked by some freshmen the follwing two questions regarding the limit:
Using the fact that
$$\lim_{x \rightarrow \infty} \left(1 + \frac{1}{x}\right)^x = e$$
evaluate
$$\lim_{x \rightarrow \infty} \left(1 + \frac{3}{x^2}\right)^x$$
$$\lim_{x \rightarrow \infty} \left(\frac{x^2-2x-3}{x^2-3x-28}\right)^x$$
I got the answers $1$ and $e$ respectively. Hope nothing wrong.
In the second question, my attempt is like this:
$$\frac{x^2-2x-3}{x^2-3x-28} = \frac{(x-3)(x+1)}{(x-7)(x+4)}$$
$$\lim_{x \rightarrow \infty} \left(\frac{x-3}{x-7}\right)^x = e^{4}$$
$$\lim_{x \rightarrow \infty} \left(\frac{x+1}{x+4}\right)^x = e^{-3}$$
$$\lim_{x \rightarrow \infty} \left(\frac{x^2-2x-3}{x^2-3x-28}\right)^x = e^4e^{-3} = e$$
We can also do the other way round:
$$\lim_{x \rightarrow \infty} \left(\frac{x-3}{x+4}\right)^x = e^{-7}$$
$$\lim_{x \rightarrow \infty} \left(\frac{x+1}{x-7}\right)^x = e^{8}$$
Indeed, the power of $e$ in the final answer is the difference of the sums of roots of $x^2-2x-3$ in the numerator and $x^2-3x-28$ in the denominator.
From this question, I suspect that
$$\lim_{x \rightarrow \infty} \left(\frac{x^n + \alpha_{n-1} x^{n-1} + \ldots + \alpha_0}{x^n + \beta_{n-1} x^{n-1} + \ldots + \beta_0}\right)^x = e^{\alpha_{n-1} - \beta_{n-1}}$$
is true, but I have no idea how to prove it, if the polynomials are not able to be factorized.
The freshmen had not learnt the rule yet. It would be good if there is a proof without using the rule to let them know an easy way to check their answers and why this way holds.
| Let $c=a_{n-1}-b_{n-1}$. Assume $c\ne 0$.
Let $p(x)=(a_{n-1}-b_{n-1})^{-1}\left((a_{n-1}-b_{n-1}) x^{n-1} + \dots + a_0-b_0\right)$.
Let $q(x)=x^n + b_{n-1} x^{n-1} + \dots + b_0$.
Let $f(x)=\left(1+c\frac{p(x)}{q(x)}\right)^\frac{q(x)}{p(x)}$.
Let $g(x)=\frac{xp(x)-q(x)}{q(x)}$.
$$\left(\frac{x^n + a_{n-1} x^{n-1} + \dots + a_0}{x^n + b_{n-1} x^{n-1} + \dots + b_0}\right)^x = \left(1+\frac{(a_{n-1}-b_{n-1}) x^{n-1} + \dots + a_0-b_0}{x^n + b_{n-1} x^{n-1} + \dots + b_0}\right)^x = \left(1+c\frac{p(x)}{q(x)}\right)^x=\left(\left(1+c\frac{p(x)}{q(x)}\right)^\frac{q(x)}{p(x)}\right)^{1+\frac{xp(x)-q(x)}{q(x)}}=f(x)f(x)^{g(x)}$$
$f(x)\to e^c$ as $x\to\infty$, since $\frac{q(x)}{p(x)}\to\infty$.
For large $x$, $f(x)\in e^c\times [\frac{1}{2},2]$.
For large $x$, $g(x)\in M\times[-\frac{1}{x},\frac{1}{x}]$, for some $M$.
So, for large $x$, $f(x)^{g(x)}\in [2^{-M\frac{1}{x}},2^{M\frac{1}{x}}]$, both of which $\to 1$.
So $f(x)^{g(x)}\to 1$ as $x\to\infty$ by squeeze theorem.
Hence $\left(\frac{x^n + a_{n-1} x^{n-1} + \dots + a_0}{x^n + b_{n-1} x^{n-1} + \dots + b_0}\right)^x\to e^c$ as $x\to\infty$.
| {
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Given a matrix, find a matrix that satisfies Let A be a matrix (3x4)
Prove that there does not exists a matrix X that satisfies
$$
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}X = \begin{pmatrix}
1 & 1 & 1 \\
0 & 2 & 0 \\
2 & 1 & 1 \\
\end{pmatrix}
$$
When I try to peform Gaussian elimination to get the reduced form of A, I always get a row of zeroes, e.g:
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
1 & 1 & 2 & -1 \\
\end{pmatrix}
$$ R_3 - R_1 \to R_3 $$
I get
\begin{pmatrix}
1 & 1 & 2 & -1 \\
0 & 2 & 1 & 3 \\
0 & 0 & 0 & 0 \\
\end{pmatrix}
What can I conclude from the fact that I got a zeroes row?
Does this help solving the problem?
| The rows of the LHS will be given by the rows of $A$, multiplied by $X$. Since the first and third rows of $A$ are the same, the first and third rows of the product will be the same. Therefore the product cannot equal the RHS.
| {
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Find $\iint\limits_\Sigma(x\sqrt{x^2+y^2+z^2}\hat{\imath}+y\sqrt{x^2+y^2+z^2}\hat{\jmath}+z\sqrt{x^2+y^2+z^2}\hat{k})\cdot\hat{n}dS$
Find $\iint\limits_\Sigma(x\sqrt{x^2+y^2+z^2}\hat{\imath}+y\sqrt{x^2+y^2+z^2}\hat{\jmath}+z\sqrt{x^2+y^2+z^2}\hat{k})\cdot\hat{n}dS$ Where $\Sigma$ is the torus $\rho=\sin\phi$ Oriented by the normal directed outward.
I got this as a study question for an upcoming exam of mine, and I've been having some trouble with it.
What I've been able to do so far:
Using the Divergence Theorem, which states that $\iint\limits_\Sigma\vec{F}\cdot\hat{n}dS = \iiint\limits_E div(\vec{F})dV$
$$
div(\vec{F}(x,y,z)=P\hat{\imath}+Q\hat{\jmath}+R\hat{k})=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}
$$
$$
\implies div\left(x\sqrt{x^2+y^2+z^2}\hat{\imath}+y\sqrt{x^2+y^2+z^2}\hat{\jmath}+z\sqrt{x^2+y^2+z^2}\hat{k}\right)
$$
$$ = \frac{2x^2+y^2+z^2}{\sqrt{x^2+y^2+z^2}}+ \frac{x^2+2y^2+z^2}{\sqrt{x^2+y^2+z^2}}+\frac{x^2+y^2+2z^2}{\sqrt{x^2+y^2+z^2}} = \frac{4x^2+4y^2+4z^2}{\sqrt{x^2+y^2+z^2}} = 4(x^2+y^2+z^2)^{3/2}
$$
$$
\implies \iint\limits_\Sigma\vec{F}\cdot\hat{n}dS = 4\iiint\limits_E(x^2+y^2+z^2)^{3/2}dV
$$
What I still can't figure out:
How to get the bounds. I now have this triple integral, which needs bounds. Most of the problems I've done in the past have come out to be the triple integral of some constant, which is just the constant multiplied by the volume of the object, which is usually something simple like a cylindar or sphere. This one seems to actually require evaluating. How do I find the bounds to my new triple integral?
| The equation for the torus is clearly directing you to switch to spherical coordinates $(\rho, \phi, \theta)$. The bounds are then $0 \le \rho \le \sin \phi$, $0 \le \phi \le \pi$, $0 \le \theta \le 2\pi$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1058752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the first two non vanishing maclaurin terms Find the first two nonvanishing terms in the Maclaurin series of $\sin(x + x^3)$.
Suggestion: use the Maclaurin series of $\sin(y)$ and write $y = x + x^3$
Using
this result, find
$\lim\limits_{x\to 0}\frac{\sin(x + x^3)−x}{x^3}$
$\sin(y)= y-\frac{y^3}{3!}+\frac{y^5}{5!}+\frac{y^7}{7!}$
$y= x+x^3$
$x+x^3-\frac{(x+x^3)^3}{3!}+\frac{(x+x^3)^5}{5!}-\frac{(x+x^3)^7}{7!}$
Now what do I do?
| Note that the formula for Maclaurin Series is
$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n.$$
So calculating the first few terms with $y=x+x^3$ gives us
$$\sin(y)=\frac{0}{0!}y^0+\frac{1}{1!}y^1-\frac{0}{2!}y^2-\frac{1}{3!}y^3+\frac{0}{4!}y^4+...\\=y-\frac{y^3}{3!}+\frac{y^5}{5!}-...$$
Now, if we plug $x+x^3$ back into $\sin(y),$ we need to make sure that we have the correct derivatives. So, we have for $f(x)=\sin(x+x^3)$
$$f^{(1)}(0)=(1+3x^2)\cos(x+x^3)=1\\f^{(3)}(0)=6\cos(x+x^3)-18x(1+3x^2)(\sin(x+x^3))-(3x^2+1)(\cos(x+x^3))=6-1=5$$since you only need the first two non vanishing terms. Now we have
$$\sin(x+x^3)=x-\frac{5x^3}{6}.$$
Now we need to find $$\lim\limits_{x\to 0}\frac{\sin(x+x^3)-x}{x^3}.$$Plugging the series in for $\sin(x+x^3)$ we have
$$\lim\limits_{x\to 0}\frac{x-\frac{5x^3}{6}-x}{x^3}=\lim\limits_{x\to 0}\frac{\frac{5x^3}{6}}{x^3}=\lim\limits_{x\to 0}\frac{5x^3}{6x^3}=\frac56.$$
Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $1+2^n+3^n+4^n$ is divisible by $10$ How can I prove
$$1+2^n+3^n+4^n$$ is divisible by $10$ if $$n\neq 0,4,8,12,16.....$$
| We consider it $(\mathrm{mod}\;2)$ and $(\mathrm{mod}\;5)$ separately.
Clearly $1$ and $3^n$ are odd, while $2^n$ and $4^n$ are even, so their sum is even.
Now by Fermat's Little Theorem, when $a$ is not divisible by $5$, $a^5 \equiv a \mod 5$, so $a^4 \equiv 1 \mod 5$. Thus we only need to check $n=1$, $2$, and $3$.
(note that $1$, $2$, $3$, and $4$ are not divisible by $5$.
If $n=1$, we get $1+2+3+4=10$, which is divisible by $5$
If $n=2$, we get $1+4+9+16=30$, which is divisible by $5$.
If $n=3$, we get $1+8+27+64=100$, which is divisible by $5$.
Thus we have proved it $(\mathrm{mod}\;5)$ and we are done.
| {
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"url": "https://math.stackexchange.com/questions/1060746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Compare inequalities in a proof by induction I am solving a proof by induction example. But I ended up with my hypothesis
$$
a_{n-1} \geq \frac{2^n}{2}+n^2-2n+1
$$
and my inductive step
$$
a_{n-1} \geq \frac{2^n}{2}+\frac{n^2}{2}-\frac{n}{2}.
$$
How can I show that if I am assuming that my hypothesis is true then my inductive step is also true?
It seems to me that I can compare both equation's right hand sides as
$$
\frac{2^n}{2}+n^2-2n+1 \geq \frac{2^n}{2}+\frac{n^2}{2}-\frac{n}{2} \Leftrightarrow -\frac{n^2}{2}-\frac{3}{2} n \geq -1 \Leftrightarrow -n^2-3n \geq -2.
$$
But now I just ended up with a neq inequality that I have to proof?
Edit
I am proving that $a_n \geq 2^n + n^2$ for all natural numbers. $a_n$ is defined by the recursive definition
$$
a_n
=
\begin{cases}
1 & n=0 \\
n+2 a_{n-1} & n \geq 1
\end{cases}
$$
| Alternatively, if you prefer a more direct approach, the finite difference method is really useful.
We have $$
\begin{array}{c|l|c|r}
n & a_n & b_n=a_{n+1}-a_{n} & c_n=(a_{n+2}-a_{n+1})-(a_{n+1}-a_{n}) \\
\hline
0 & 1 & 2 & 3\ \\
1 & 3 & 5 & 6\ \\
2 & 8 & 11& 12\ \\
3 & 19 & 23 & 24\ \\
4 & 42 & 47 & \vdots \\
5 & 89 & \vdots & \vdots \\
\vdots&\vdots&\vdots&\vdots
\end{array}
$$
Clearly, $$c_n=3\cdot2^n,$$ whence $$b_n=2+3\sum_{i=0}^{n-1}2^i=2+3(2^n-1)=3\cdot2^n-1.$$
Thus, the general formula for $a_n$ is $$a_n=1+\sum_{i=0}^{n-1}\left(3\cdot 2^i-1\right)=1+3\sum_{i=0}^{n-1}2^i-n=3\cdot2^n-n-2.$$ So we prove by induction $$a_n\ge 2^n+n^2$$$$3\cdot2^n-n-2\ge2^n+n^2$$$$2^{n+1}\ge n^2+n+2.\tag{P(n)} $$
Basis: P($0$): $2\ge2$ is true.
Inductive step: Assume P$(k)$ holds. P$(k+1)$ is equivalent to$$2^{k+2}\ge(k+1)^2+(k+1)+2$$ $$ 2\cdot2^{k+1}\ge k^2+2k+1+k+3$$$$2^{k+1}\ge \frac{k^2+3k+4}{2}.\tag{1}$$ Since $$2^{k+1}\ge k^2+k+2$$ by hypothesis, and $$k^2+k+2\ge \frac{k^2+3k+4}{2}\\ 2k^2+2k+4\ge k^2+3k+4 \\ k^2\ge k,$$ $(1)$ holds, and thus P$(n)$ is true for all $n$.
| {
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"url": "https://math.stackexchange.com/questions/1063506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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To prove $(\sin\theta + \csc\theta)^2 + (\cos\theta +\sec\theta)^2 \ge 9$ I used the following way but got wrong answer
$$A.M. \ge G.M.$$
$$ \frac{\sin \theta + \csc \theta}{2} \ge \sqrt{\sin \theta \cdot \csc \theta}$$
Squaring both sides,
\begin{equation*}
(\sin\theta + \csc\theta )^2 \ge 4 \tag{1}
\end{equation*}
Similarly
\begin{equation*}
(\cos\theta + \sec\theta )^2 \ge 4 \tag{2}
\end{equation*}
Adding equation (1) and (2)
\begin{equation*}
(\sin\theta + \csc\theta )^2+(\cos\theta + \sec\theta )^2 \ge 8
\end{equation*}
What is wrong?
| $$\sin^2\theta+\cos^2\theta+2+2+\csc^2\theta+\sec^2\theta$$
$$=5+\frac1{\sin^2\theta\cos^2\theta}=5+\frac4{(\sin2\theta)^2}=5+4\csc^22\theta$$
Now, $\csc^22\theta=1+\cot^22\theta\ge1$ for real $\theta$
The equality occurs if $\csc^22\theta=1\iff\sin^22\theta=1$
$\iff\cos2\theta=0\implies2\theta=(2n+1)\dfrac\pi2$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1065943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Solve $x^7-5x^4-x^3+4x+1=0$ for $x$
Solve for $x$
$$x^7-5x^4-x^3+4x+1=0$$
This equation has been bugging me since the past few days. I have found, using the Rational Root Theorem that $x=1$ is a root of this equation. However, after dividing, I cannot solve the six degree equation thus generated. I have also tried factorizing the equation, but it's not working.
| Consider the identity
$(x^4-4x-1)^2=x^8-8x^5-2x^4+16x^2+8x+1$
Differentiating both sides w.r.t. $x$, we get,
$x^7-5x^4-x^3+4x+1=(x^4-4x-1)(x^3-1)$
Now, the equation becomes,
$(x^4-4x-1)(x^3-1)=0$
$\implies x=1,\omega, \omega^2$ (where $\omega$ is a non real cube root of unity)
or
$x^4-4x-1=0$
$\implies x^{4}+2x^{2}+1=2x^{2}+4x+2$
$\implies (x^{2}+1)^{2}=2(x+1)^{2}$
$\implies \{x^{2}+1+\sqrt{2} (x+1)\}\cdot \{x^{2}+1-\sqrt{2} (x+1)\}=0$
$\implies x=\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2}$
or
$x=\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$
$\therefore x=1,\omega,\omega^2,\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2},\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1066382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 1
} |
Trigonometric substitution and Integration of $\frac{1}{x^2\sqrt{x^2+1}} $ Regarding the integral
$$
\int \frac{dx}{x^2\sqrt{x^2 + 1}}
$$
I'm not sure what to do about the extra $x^2$ in the denominator. What can I do about it?
| You might have to use the law of tangent which states that
$$ \sec^2{\theta} =\tan^2{\theta} + 1 $$
or
$$ \sec{\theta} = \sqrt{\tan^2{\theta} + 1} $$
so
$$ \int \frac{dx}{x^2\sqrt{x^2 + 1}}d\theta = \int \frac{d \tan \theta}{\tan^2\sqrt{\tan^2 + 1}}d\theta$$
therefore plugging in sec
$$ \int \frac{d \tan \theta}{\tan^2\sqrt{\tan^2 + 1}}d\theta = \int \frac{\sec^2 \theta}{\tan^2\sec \theta}d\theta = \int \frac{\sec \theta}{\tan^2}d\theta $$
Therefore we now get answer
$$ \int \frac{\sec \theta}{\tan^2}d\theta = \int \csc \theta \cot \theta d\theta$$
Final answer
$$ \int \csc \theta \cot \theta d\theta = -\csc \theta + C$$
because
$$ \frac{d}{dx} \csc \theta = -\csc \theta \cot \theta $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1067079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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An exercise from Knuth's book - Proving a formula by induction I would like to find a formula for this sum:
$$
\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4} + \frac{5^3}{5^4+4} - ... + \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}
$$
The answer given (Knuth's book, The Art of Computer Programming Volume 1, Third Edition, Section 1.2.1, Exercise 11) is $$ \frac{(-1^n)(n+1)}{4(n+1)^2+1}$$
Please give me an approach or a solution to get this sum.
I tried to solve by pairing the consecutive terms. The last and the second last terms will correspond to $(2n+1)$ and $(2n-1)$ respectively. But this approach couldn't yield much.
The second part is to prove this sum by induction, i.e we've to prove this equation:
$$
\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4} + \frac{5^3}{5^4+4} - ... + \frac{(-1)^n(2n+1)^3}{(2n+1)^4+4} = \frac{(-1^n)(n+1)}{4(n+1)^2+1}
$$
I approached it like this:
The sum is correct for $k=1$( and $k=0$). (k is a non-negative integer)
Assuming that the sum is correct for $k = 0,1,..,n$
we'll prove that its correct for $k=n+1$.
The sum for $k=n+1$ will be equal to: the sum for k=n + the last term, i.e. $$ \frac{(-1)^n(n+1)}{4(n+1)^2+1} + \frac{(-1)^{n+1}(2(n+1)+1)^3}{(2(n+1)+1)^4+4}$$
This should then be proved equal to $$\frac{(-1)^{n+1}((n+1)+1)}{4((n+1)+1)^2+1}$$
which again I'm not able to do. So, please help in proving this also.
P.S.: Someone please create a 'knuth' or 'taocp' tag for mathematical questions of this book.
| First you have$$ \begin{align} (2n+1)^4+4&= A^4+B^2\\\\ &=(A^4+2A^2B+B^2)-2A^2B\\\\
&=(A^2+B)^2-2A^2B\\\\
&=\left((2n+1)^2+2\right)^2-4\left(2n+1\right)^2\\\\
&=\left((2n+1)^2+2-2(2n+1)\right)\left((2n+1)^2+2+2(2n+1)\right)\\\\
&=\left(4n^2+1\right)\left(4(n+1)^2+1\right) \end{align}$$
Second you have$$ \begin{align} \frac{n}{4n^2+1}+\frac{n+1}{4(n+1)^2+1}&= \frac{n\left(4(n+1)^2+1\right)+(n+1)\left(4n^2+1\right)}{\left(4n^2+1\right)\left(4(n+1)^2+1\right)}\\\\ &=\frac{8n^3+12n^2+6n+1}{(2n+1)^4+4}\\\\
&=\frac{(2n+1)^3}{(2n+1)^4+4}\\\\
\end{align} $$
Now, setting $\displaystyle u_n:=\frac{(-1)^n n}{4n^2+1}$, you see from the preceding lines that $$\begin{align} u_n- u_{n+1}&=\frac{(-1)^n n}{4n^2+1}-\frac{(-1)^{n+1} (n+1)}{4(n+1)^2+1}\\\\&=(-1)^n \left(\frac{n}{4n^2+1}+\frac{n+1}{4(n+1)^2+1}\right)\\\\
&=\frac{(-1)^n(2n+1)^3}{(2n+1)^4+4}\end{align} $$
and by telescoping, you obtain what you want.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $f(x,y,z)=0$ if and only if $(\sqrt {x^2+y^2}-1)^2+z^2=r^2$.
Define $f(x,y,z)=(x^2+y^2+r^2-z^2-1)^2-4(x^2+y^2)(r^2-z^2)$, where $0<r<1$ Show that $f(x,y,z)=0$ if and only if $(\sqrt {x^2+y^2}-1)^2+z^2=r^2$.
Here is what I have tried:
Let $f(x,y,z)=0$:
$(x^2+y^2+r^2-z^2-1)^2-4(x^2+y^2)(r^2-z^2)=0$
$x^4+x^2y^2+x^2r^2-x^2z^2-x^2+x^2y^2+y^4+r^2y^2-y^2z^2-y^2+r^2x^2+r^2y^2+r^4-r^2z^2-r^2-z^2x^2-z^2y^2-z^2r^2+z^4+z^2-x^2-y^2-r^2+z^2+1-4(x^2r^2-x^2z^2+y^2r^2-y^2z^2)=0$
$x^4+x^2y^2+x^2r^2-x^2z^2-x^2+x^2y^2+y^4+r^2y^2-y^2z^2-y^2+r^2x^2+r^2y^2+r^4-r^2z^2-r^2-z^2x^2-z^2y^2-z^2r^2+z^4+z^2-x^2-y^2-r^2+z^2+1-4(x^2r^2-x^2z^2+y^2r^2-y^2z^2)=0$
$1+ r^4+x^4-2y^2 + y^4 + 2 z^2 + 2 y^2 z^2 + z^4 +
2 x^2 (-1 + y^2 + z^2) - 2 r^2 (1 + x^2 + y^2 + z^2)=0$
What should I do next? By the way I think it's absolutely horrible to do it this way, is there any easy way to do?
| Choosing good notation is sometimes one of the keys to solving a problem and avoiding complicated mess. This problem is an excellent illustration of that. So let's take $a$ to be $x^2+y^2$ and $b$ to be $r^2-z^2$. The problem reduces down to showing that $$(a+b-1)^2-4ab=0\iff(\sqrt{a}-1)^2-b=0.$$
We have $$\eqalign{(\sqrt{a}-1)^2-b=0&\iff a-2\sqrt{a}+1-b=0\\
&\iff a-b+1=2\sqrt a\\
&\iff (a-b+1)^2=4a\\
&\iff a^2-2ab+2a+b^2-2b+1=4a\\
&\iff a^2-2ab-2a+b^2-2b+1=0\\
&\iff a^2+2ab-2a+b^2-2b+1=4ab\\
&\iff (a+b-1)^2=4ab\\
&\iff (a+b-1)^2-4ab=0.}$$
| {
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"url": "https://math.stackexchange.com/questions/1071146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$x^2-y^2=2s$, s cannot be an odd integer How can we prove that if $x^2-y^2=2s$ holds, s cannot be an odd integer. What theorem in number theory should we use?
| $x^2-y^2=(x+y)(x-y)$
Let $y=x+a$ then
$(x+y)(x-y)=(2x+a)(2x-a)=4x^2-a^2$
If $4x^2-a^2=2s$ then $a^2$ must be divisible by 2. Since $a$ is an integer $a$ must be divisible by 2. Thus $a=2b$. Thus $2s=4x^2-4b^2=4c$ where $c=x^2-b^2$ is an integer. Thus $s$ is divisible by $2$.
| {
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"url": "https://math.stackexchange.com/questions/1075128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluating $\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$ Evaluating $$\int{\frac{1}{\sqrt{x^2-1}(x^2+1)}dx}$$ using $ux=\sqrt{x^2-1}$
UPDATE 'official' solution
$$u^2x^2=x^2-1$$
$$x^2=\frac{-1}{u^2-1}$$
$$x^2+1=\frac{u^2-2}{u^2-1}$$
$$2xdx=\frac{-2u}{(u^2-1)^2}$$
$$\int{\frac{x}{x\sqrt{x^2-1}(x^2+1)}dx}$$
$$\int{\frac{1}{\frac{-1}{u^2-1}u\frac{u^2-2}{u^2-1}}\left(\frac{-u}{(u^2-1)^2}\right)du} $$
$$\int{\frac{1}{(u^2-2)}du} $$
$$\frac{\log \left(\sqrt{2}-x\right)-\log \left(x+\sqrt{2}\right)}{2 \sqrt{2}}$$
$$\frac{\log \left(\sqrt{2}-\sqrt{x^2-1}/x\right)-\log \left(\sqrt{x^2-1}/x+\sqrt{2}\right)}{2 \sqrt{2}}$$
However at mathematica I get $$\frac{\log \left(-3 x^2-2 \sqrt{2} \sqrt{x^2-1} x+1\right)-\log \left(-3 x^2+2 \sqrt{2} \sqrt{x^2-1} x+1\right)}{4 \sqrt{2}}$$
Where is the error?
| Let us evaluate the general form of the integral
\begin{align}
\int\frac{\mathrm dx}{(x^2+a^2)\sqrt{x^2-b^2}}&=\int\frac{a\sec^2t}{(a^2\tan^2t+a^2)\sqrt{a^2\tan^2t-b^2}}\mathrm dt\tag1\\[7pt]
&=\frac{1}{a}\int\frac{\cos t}{\sqrt{a^2\sin^2t-b^2\cos^2t}}\mathrm dt\tag2\\[7pt]
&=\frac{1}{a}\int\frac{\cos t}{\sqrt{\left(a^2+b^2\right)\sin^2t-b^2}}\mathrm dt\tag3\\[7pt]
&=\frac{1}{a\sqrt{a^2+b^2}}\int\frac{\mathrm dy}{\sqrt{y^2-1}}\tag4\\[7pt]
&=\frac{\operatorname{arccosh} y}{a\sqrt{a^2+b^2}}+C\tag5\\[7pt]
&=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\frac{1}{a\sqrt{a^2+b^2}}\operatorname{arccosh}\left(\frac{x\,\sqrt{a^2+b^2}}{b\left(x^2+a^2\right)}\right)+C}}
\end{align}
Setting $a=1$ and $b=1$, we get
$$\int\frac{\mathrm dx}{(x^2+1)\sqrt{x^2-1}}
=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\frac{1}{\sqrt{2}}\operatorname{arccosh}\left(\frac{x\,\sqrt{2}}{x^2+1}\right)+C}}$$
Explanation :
$(1)\;$ Use substitution $\;\displaystyle x=a\tan t\quad\implies\quad\tan t=\frac{x}{a}$
$(2)\;$ Use identities $\;\displaystyle \sec^2t=1+\tan^2t\;$ and $\;\displaystyle \tan t=\frac{\sin t}{\cos t}$
$(3)\;$ Use identity $\;\displaystyle \cos^2t=1-\sin^2t$
$(4)\;$ Use substitution $\;\displaystyle \sin t=\frac{by}{\sqrt{a^2+b^2}}\,$
$(5)\;$ Use substitution $\;\displaystyle y=\cosh u\,$. We have $\;\displaystyle y=\frac{\sqrt{a^2+b^2}\sin t}{b}\,$ and $\;\displaystyle \sin t=\frac{x}{\sqrt{x^2+a^2}}\,$
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 1
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Liouville sequences We have $(1+2+\cdots+n)^2=1^3+2^3+\cdots+n^3$. We call a finite sequence $a_1,\ldots,a_n$ Liouville if it satisfies \begin{equation}(a_1+\cdots+a_n)^2=a_1^3+\cdots+a_n^3.\end{equation}
Liouville discovered that for all $m\in\mathbb{N}$ the sequence $\tau(d_1),\ldots,\tau(d_n)$ where $d_i$, $i=1,\ldots,n$, are the positive divisors of $m$ and $\tau(k)$ is the number of positive divisors of natural number $k$. $m=27$ has the divisors 1, 3, 9, 27, and we obtain the sequence 1, 2, 3, 4. For $m=10$ we obtain the sequence 1, 2, 2, 4.
$a_1,\ldots,a_n$ with $a_i=n$ for $i=1,\ldots,n$ and 2, 2, 2, 2, 2, 2, 2, 2, 8 are Liouville but are not obtained by Liouville's method.
Are there other such examples?
| I've searched through sequences of length 4, 5, 6 and 7 with entries up to 10, and there appear to be many other examples. I haven't yet spent time analysing the results to find any patterns, so I'll just list some examples for the moment to answer your question.
Firstly, one obvious observation: sequences which don't contain a $1$ cannot come from $ \tau(d_i) $, since $1$ is a divisor of any $ m \in \mathbb{N} $ and $ \tau(1) = 1 $.
Length 4 sequences:
\begin{array}{c}
\{2, 2, 4, 4\}
\end{array}
Length 5 sequences:
\begin{array}{c}
\{3, 3, 3, 3, 6\}, \{3, 3, 3, 4, 6\}, \{1,2,2,3,5\}
\end{array}
The third sequence also does not come from $ \tau(d_i) $. If this was from $ \tau(d_i) $, it would have to come from $ m = p^4 $ since there are $ 5 $ divisors. But this generates $ \{1, 2, 3, 4, 5\} $, so doesn't work.
Length 6 sequences:
\begin{array}{c}
\{1, 1, 1, 2, 2, 5\}, \{1, 1, 1, 4, 4, 5\}, \{1, 1, 2, 4, 5, 5\}, \{1, 1, 4, 5, 5, 5\}, \\
\{1, 2, 2, 4, 4, 6\}, \{1, 4, 4, 4, 6, 6\}, \{2, 2, 2, 2, 2, 6\}, \{2, 2, 4, 4, 6, 6\}, \\
\{2, 4, 4, 5, 5, 7\}, \{2, 4, 4, 6, 6, 6\}, \{3, 3, 3, 3, 5, 7\}, \{3, 3, 3, 6, 6, 6\}, \\
\{3, 4, 5, 5, 6, 7\}, \{3, 5, 5, 5, 6, 7\}, \{4, 5, 5, 6, 6, 7\}
\end{array}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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} |
Evaluation of $\int_{0}^{\infty} \cos(x)/(x^2+1)$ using complex analysis. Evaluate:
$$\int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$
Using only complex analysis.
$$I = \int_{0}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = (\frac{1}{2})\int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$
Consider a contour $C$ with a upper-axis semi-circle $B$ and the axis running from $-R \to R$
We will compute:
$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz$$
First:
$$z^2 + 1 = 0 \implies Z \in \{-i, i\}$$
Only, $z = i$ is in the semi circle region.
$$\text{Res}_{z=i} = \lim_{z \to i} (z-i)(f(i)) = \lim_{z \to i} \frac{\cos(z)}{z + i} = \frac{\cosh(1)}{2i}$$
Applying the residue theorem:
$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = (2\pi i)\cdot \frac{\cosh(1)}{2i} = \pi\cdot\cosh(1)$$
$$\oint_{C} \frac{\cos(z)}{z^2 + 1} dz = \int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx$$
But that is wrong, the answer for the full improper is:
$$\int_{-\infty}^{\infty} \frac{\cos(x)}{x^2 + 1} dx = \frac{\pi}{e}$$
What am I doing wrong?
| we use $$f\left( z \right) = \frac{{e^{iz} }}{{1 + z^2 }}$$
then take real parts of the resulting integral .using the same contour $C$
\begin{array}{l}
i \in C; - i \notin C \\
{\mathop{\rm Re}\nolimits} s\left( {f;i} \right) = \mathop {\lim }\limits_{z \to i} \left\{ {\left( {z - i} \right)\frac{{e^{iz} }}{{1 + z^2 }}} \right\} = \frac{{e^{ - 1} }}{{2i}} \\
\end{array}
\begin{array}{l}
\oint\limits_C {\frac{{e^{iz} }}{{1 + z^2 }}dz} = \int\limits_{ - R}^{ + R} {\frac{{e^{ix} dx}}{{1 + x^2 }}} + \int\limits_{S_R} {\frac{{e^{iz} dz}}{{1 + z^2 }}} =
2\pi i\frac{{e^{ - 1} }}{{2i}} = \frac{\pi }{e} \\
\Rightarrow \int\limits_{ - R}^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}} + i\int\limits_{ - R}^{ + R} {\frac{{\sin \left( x \right)dx}}{{1 + x^2 }} + } \int\limits_{S_R} {\frac{{e^{iz} dz}}{{1 + z^2 }}} = \pi e^{ - 1} \\
2\int\limits_0^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}} = \pi e^{ - 1} ;\quad \left( {\int\limits_{S_R} {\frac{{e^{ix} dx}}{{1 + x^2 }} = 0} ;R \to + \infty \int\limits_{ - R}^{ + R} {\frac{{\sin \left( x \right)dx}}{{1 + x^2 }} = 0} } \right) \\
\mathop {\lim }\limits_{R \to + \infty } \int\limits_0^{ + R} {\frac{{\cos \left( x \right)dx}}{{1 + x^2 }}} = \frac{\pi }{2}e^{ - 1} \\
\end{array}
note:
but if $y = {\mathop{\rm Im}\nolimits} \left( z \right)$ then $\cos \left( z \right) \approx \frac{{e^{\left| y \right|} }}{{\left| z \right|^2 }}$ for large $\left| z \right|$
we have the estimate
$
\left| {\int\limits_{C_R } {\frac{{e^{iz} }}{{z^2 + 1}}dz} } \right| \le \int\limits_{C_R } {\frac{{e^{ - y} }}{{R^2 - 1}}\left| {dz} \right|} \le \frac{{\pi R}}{{R^2 - 1}} \to 0
$ as $R \to \infty $
where $
y = {\mathop{\rm Im}\nolimits} \left( z \right) > 0
$by the residue theorem
$
\int\limits_{ - \infty }^{ + \infty } {\frac{{e^{ix} }}{{x^2 + 1}}dx = } 2\pi i\sum\limits_{{\mathop{\rm Im}\nolimits} a > 0} {{\mathop{\rm Re}\nolimits} s_a } \frac{{e^{iz} }}{{z^2 + 1}} = \frac{\pi }{e}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1077581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Why doesn't squaring the radicand of a square root introduce a plus-minus sign here? The question I have concerns the following problem:
*
*$\sqrt{4x-1} = \sqrt{x+2}-3$
*$(\sqrt{4x-1})^2 = (\sqrt{x+2}-3)^2$
*$\sqrt{4x-1}\times\sqrt{4x-1} = (\sqrt{x+2}-3)\times(\sqrt{x+2}-3)$
*$\sqrt{(4x-1)\times(4x-1)} = (\sqrt{x+2})^2-3\sqrt{x+2}-3\sqrt{x+2}+9$
*$\sqrt{(4x-1)^2} = \sqrt{(x+2)\times(x+2)}-6\sqrt{x+2}+9$
*$\underline{4x-1} = \sqrt{(x+2)^2}-6\sqrt{x+2}+9$
*$4x-1 = \underline{x+2}-6\sqrt{x+2}+9$
How did the underlined expressions in steps 6 and 7 not include $\pm$? I thought that $\sqrt{x^2} = \pm x$.
| I wonder if the problem could be between line $2$ and line $5$. In line $2$, you square while in line $5$ you take the square root of the square. This could introduce some confusion.
Starting from $$\sqrt{4x-1} = \sqrt{x+2}-3$$ just square to get $$4x-1=(\sqrt{x+2}-3)^2=x+2-6\sqrt{x+2}+9=x+11-6\sqrt{x+2}$$ So, now $$3x-12=-6\sqrt{x+2}$$ Now, square again.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1079486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to evaluate residue of $\cot z/z^4$ at $z=0$? How to evaluate residue of $\cot z/z^4$ at $z=0$?
As we know :
$$f(x)=f(0)+f'(0)x+f''(0)x^2/2+...$$
but $\cot(0)\to\infty$ or is undefined? I know that:
$$\tan x=x+x^3/3+2x^5/15+...$$
| Well,
\begin{align}\cot z = \frac{\cos z}{\sin z} &= \frac{1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)}{z - \frac{z^3}{3!} + \frac{z^5}{5!} + O(z^7)}\\
& = \frac{1}{z}\cdot \left(1 - \frac{z^2}{2!} + \frac{z^4}{4!} + O(z^6)\right)\cdot \left(1 + \frac{z^2}{3!} - \frac{z^4}{5!} +\frac{z^4}{3!3!}+ O(z^6)\right)\\
&= \frac{1}{z}\cdot \left(1 + \left(-\frac{1}{2!} + \frac{1}{3!}\right)z^2 + \left(-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!}\right)z^4 + O(z^6)\right).\end{align}
Therefore, the coefficient of $\frac{1}{z}$ in the Laurent expansion of $\frac{\cot(z)}{z^4}$ is
$$-\frac{1}{2!3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{3!3!} = -\frac{1}{45}.$$
Hence, $$\text{Res}_{z = 0}\frac{\cot(z)}{z^4} = -\frac{1}{45}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Why is $x(\sqrt{x^2+1} - x )$ approaching $1/2$ when $x$ becomes infinite? Why is $$\lim_{x\to \infty} x\left( (x^2+1)^{\frac{1}{2}} - x\right) = \frac{1}{2}?$$
What is the right way to simplify this?
My only idea is: $$x((x²+1)^{\frac{1}{2}} - x) > x(x^{2^{0.5}}) - x^2 = 0$$
But 0 is too imprecise.
| Let $x=\tan\theta$.
Then $L=\displaystyle\lim_{x\to \infty} x( (x^2+1)^{\frac{1}{2}} - x ) \\= \displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \tan\theta( \sec\theta - \tan\theta )\\=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \frac{\sin\theta-\sin^2\theta}{\cos^2\theta}\\
=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \frac{\sin\theta-\sin^2\theta}{1-\sin^2\theta}\\
=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}} \frac{(1-\sin\theta)\sin\theta}{(1-\sin\theta)(1+\sin\theta)}\\
=\displaystyle\lim_{{\theta\to \frac{\pi}{2}}^{-}}\frac{\sin\theta}{1+\sin\theta}\\
=\dfrac{1}{2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 5
} |
Why do we have $u_n=\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}}=O(\frac{1}{n^3})$? Why do we have
*
*$u_n=\dfrac{1}{\sqrt{n^2-1}}-\dfrac{1}{\sqrt{n^2+1}}=O\left(\dfrac{1}{n^3}\right)$
*$u_n=e-\left(1+\frac{1}{n}\right)^n\sim \dfrac{e}{2n}$
any help would be appreciated
| For the first one:
$$\begin{align}
\frac{1}{\sqrt{n^2-1}}-\frac{1}{\sqrt{n^2+1}} &= \frac{1}{n}\left(\frac{1}{\sqrt{1-\frac{1}{n^2}}}-\frac{1}{\sqrt{1+\frac{1}{n^2}}}\right) \\
&= \frac{1}{n}\left(\frac{1}{1-\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)}-\frac{1}{1+\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)}\right) \\
&=\frac{1}{n}\left(\left(1+\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)\right)-\left(1-\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)\right)\right) \\
&=\frac{1}{n}\left(\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)+\frac{1}{2n^2}+o\!\left(\frac{1}{n^2}\right)\right) \\
&=\frac{1}{n}\left(\frac{1}{n^2}+o\!\left(\frac{1}{n^2}\right)\right) \\
&=\frac{1}{n^3}+o\!\left(\frac{1}{n^3}\right)
\end{align}$$
using the Taylor expansions:
*
*$\sqrt{1+x} \operatorname*{=}_{x\to0} 1+\frac{x}{2} +o(x)$
*$\frac{1}{1+x} \operatorname*{=}_{x\to0} 1-x+x^2-\cdots+x^k +o(x^k)$
*$\frac{1}{1-x} \operatorname*{=}_{x\to0} 1+x+x^2+\cdots+x^k +o(x^k)$
For the second:
$$\begin{align}
e - \left(1+\frac{1}{n}\right)^n &= e - e^{n\ln\left(1+\frac{1}{n}\right)}
= e - e^{n\left(\frac{1}{n}-\frac{1}{2n^2} + o\!\left(\frac{1}{n^2}\right)\right)} \\
&= e^1 - e^{1-\frac{1}{2n} + o\!\left(\frac{1}{n}\right)} \\
&= e\cdot\left( 1 - e^{-\frac{1}{2n} + o\!\left(\frac{1}{n}\right)}\right) \\
&= e\cdot\left( 1 - \left( 1-\frac{1}{2n} + o\!\left(\frac{1}{n}\right)\right)\right) \\
&= e\cdot\left( \frac{1}{2n} + o\!\left(\frac{1}{n}\right)\right) \\
\end{align}$$
using this time the Taylor expansions:
*
*$\ln(1+x) \operatorname*{=}_{x\to0} x - \frac{x^2}{2} +o(x^2)$
*$e^x \operatorname*{=}_{x\to0} 1+x+o(x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1080858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$? How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?
I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
| Since $2=\sqrt{2}\cdot\sqrt{2}$ you have that $$\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{\sqrt{2}\sqrt{2}}=\frac{1}{\sqrt{2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1082664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 4
} |
3-variable non-symmetric inequality Prove that for $a,b,c > 0$
$$\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b} >0$$
What I did is this:- Let $f(a,b,c)=\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b}$. Therefore
$$\begin{align}f(a,b,c)&=\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b}\\&=(a+b+c)\left(\frac{1}{b+c}+\frac {1}{c+a}+\frac {1}{a+b}\right)-3\left(\frac{c}{b+c}+\frac {a}{c+a}+\frac {b}{a+b}\right)\end{align}$$
The first term is greater than $9/2$ by the Cauchy-Schwarz inequality. How do I prove that the term inside the second bracket is less than $3/2$? Can I use the rearrangement inequality here??
| your inequality is equivalent to
$${\frac {{a}^{3}+{a}^{2}b-2\,{a}^{2}c-2\,a{b}^{2}+a{c}^{2}+{b}^{3}+{b}^
{2}c-2\,b{c}^{2}+{c}^{3}}{ \left( b+c \right) \left( c+a \right)
\left( a+b \right) }}
>0$$ and the numerator can written as
$$a(a-c)^2+b(b-a)^2+c(b-c)^2\geq 0$$ and the equal sign holds if $$a=b=c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1084559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Calculate limit using L'Hopital Rule -> $ \lim_{x \to 0} \left ( 1 + \frac{1}{x^2} \right )^{x^2} $ I have been trying to solve the following problem, but I seem t have some difficulties.
$$ \lim_{x \to 0} \left ( 1 + \frac{1}{x^2} \right )^{x^2} $$
I tryied to solve it, and the answer I got didn't seem to be right, so I tryed to plot it to have an ideia, and I reached the conclusion that it was wrong.
This is what I did on my second attempt:
$$ \lim_{x \to 0} \left ( 1 + \frac{1}{x^2} \right )^{x^2} = \lim_{x \to 0} \left ( \frac{x^2 + 1}{x^2} \right )^{x^2} = \left [ \left ( \frac{(0)^2 + 1}{(0)^2} \right )^{(0)^2} \right ] = \left [ \infty ^0 \right ] \Rightarrow $$
$$ \Rightarrow \lim_{x \to 0} e^{x^2 \ln \left ( \frac{x^2 + 1}{x^2} \right )} = e^{ \lim_{x \to 0} x^2 \ln \left ( \frac{x^2 + 1}{x^2} \right )} $$
Then, to make the calculations easier:
$$ \lim_{x \to 0} x^2 \ln \left ( \frac{x^2 + 1}{x^2} \right ) = \left [ (0)^2 \ln \left ( \frac{(0)^2 + 1}{(0)^2} \right ) \right ] = \left [ 0 \cdot \ln \left ( \frac{ 1}{0} \right ) \right ] $$
How do I proced now? I mean, How can I solve $ \left [ 0 \cdot \ln \left ( \frac{ 1}{0} \right ) \right ] $? There is no signal on the 0. If it was $ 0^{+} $, I would know it would aproximate to $ -\infty $.
Can you help me please?
I do not intend to have the solution to the limit, but some aid to solve this problem.
Thanks in advance, Saclyr.
| $$\left(1+\frac{1}{x^2}\right)^{x^2}=\exp\left(x^2\ln\left(1+\frac{1}{x^2}\right)\right)$$
$$x^2\ln\left(1+\frac{1}{x^2}\right)=\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}$$
therefore
$$\lim_{x\to 0}x^2\ln\left(1+\frac{1}{x^2}\right)=\lim_{x\to 0}\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}$$
$$\lim_{x\to 0^+}\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}=\lim_{u\to\infty }\frac{\ln(1+u^2)}{u^2}\underset{Hop.}{=}\lim_{u\to\infty }\frac{2u}{(1+u^2)2u}=\lim_{u\to\infty }\frac{1}{1+u^2}=0$$
and
$$\lim_{x\to 0^-}\frac{\ln\left(1+\frac{1}{x^2}\right)}{\frac{1}{x^2}}=\lim_{u\to-\infty }\frac{\ln(1+u^2)}{u^2}\underset{Hop.}{=}\lim_{u\to-\infty }\frac{2u}{(1+u^2)2u}=\lim_{u\to-\infty }\frac{1}{1+u^2}=0$$
and thus,
$$\lim_{x\to 0}x^2\ln\left(1+\frac{1}{x^2}\right)=0.$$
Moreover, $x\mapsto e^x$ is continuous in $x=0$, therefore
$$\lim_{x\to 0}\exp\left(x^2\ln\left(1+\frac{1}{x^2}\right)\right)=\exp\left(\lim_{x\to 0}x^2\ln\left(1+\frac{1}{x^2}\right)\right)=e^0=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Inequality $\frac 1{\sqrt{1+xy}}+\frac 1{\sqrt{1+yz}}+\frac 1{\sqrt{1+zx}}\ge \frac 9{\sqrt {10}}$ Let $x,y,z$ be non-negative real numbers such that $x+y+z=1$ , then is the following true? $$\dfrac 1{\sqrt{1+xy}}+\dfrac 1{\sqrt{1+yz}}+\dfrac 1{\sqrt{1+zx}}\ge \dfrac 9{\sqrt {10}}$$
| Here is another approach.
Cauchy-Schwarz says
$$
\begin{align}
(x+y+z)^2
&\le(x^2+y^2+z^2)(1^2+1^2+1^2)\\
&=3(x^2+y^2+z^2)\tag{1}
\end{align}
$$
Therefore,
$$
\begin{align}
xy+yz+zx
&=\tfrac12\left[(x+y+z)^2-\left(x^2+y^2+z^2\right)\right]\\
&\le\tfrac12\left[(x+y+z)^2-\tfrac13(x+y+z)^2\right]\\
&=\tfrac13(x+y+z)^2\tag{2}
\end{align}
$$
Then
$$
\begin{align}
\frac1{\sqrt{1+xy}}+\frac1{\sqrt{1+yz}}+\frac1{\sqrt{1+zx}}
&\ge\frac3{\sqrt[\large6]{(1+xy)(1+yz)(1+zx)}}\tag{3}\\[3pt]
&\ge\frac3{\sqrt{\frac13\left[(1+xy)+(1+yz)+(1+zx)\right]}}\tag{4}\\
&=\frac9{\sqrt{9+3(xy+yz+zx)}}\tag{5}\\[3pt]
&\ge\frac9{\sqrt{10}}\tag{6}
\end{align}
$$
Explanation:
$(3)$: AM-GM
$(4)$: AM-GM in the denominator inside the square root
$(5)$: regrouping
$(6)$: apply $(2)$
| {
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"source": "stackexchange",
"question_score": "5",
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Let $X$ and $Y$ be i.i.d. $\operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$
Let $X$ and $Y$ be i.i.d. $\operatorname{Geom}(p)$, and $N = X + Y$. Find the joint PMF of $X, Y, N$.
I have generally difficulties with such problems, as I get easily confused. Below I detailed my (most probability incorrect) approach. Besides the correct approach, I also would like to have some more general advice on how to think about and tackle such problems.
The random variables $X$ and $Y$ have geometric distributions, so $P(X=x)=(1-p)^xp$ and $P(Y=y)=(1-p)^yp$. I also know that the sum of two i.i.d. geometric random variables has a negative binomial distribution, thus $P(N=n) = \binom{n+r-1}{n} (1-p)^n p^r$, where $r$ is the number of success. In our case $r=2$, as we are dealing with the sum of two random variables. Hence, $P(N=n) = (n+1)(1-p)^n p^2$.
I'm asked to find $P(X=x, Y=y, N=n)$. I don't know how to write that down, so I condition on $N=n$ and write
\begin{align}
P(X=x, Y=y, N=n) &= P(X=x, Y=y \mid N=n)P(N=n) \\
&= P(X=x, Y=y \mid N=x+y)P(N=n),
\end{align}
because I know that $n=x+y$. But conditional on $N=x+y$, $X=x$ and $Y=y$ are the same event. Thus,
\begin{align}
P(X=x, Y=y, N=n) &= P(X=x \mid N=x+y)P(N=n)
\end{align}
I can assume that $X$ is the number of failures till the first of both successes, otherwise I could just consider $P(Y=y \mid N=x+y)$. The probability that there are $X=x$ failures till the frist success, given that there are in total $N=x+y$ failures is the same is having the first success between the $x^{th}$ and $(x+1)^{th}$ failure. As both $X$ and $Y$ have the same probability of success, the probability of this to happen is $\frac{1}{x+y+1}$. Hence,
\begin{align}
P(X=x, Y=y, N=n) &= \frac{1}{x+y+1} (n+1)(1-p)^n p^r \\
&= (1-p)^n p^2.
\end{align}
But this looks like nonsense.
| We have,
$$
P_{X,Y,N}(x,y,n) =
\begin{cases}
P(X=x).P(Y=y) & if x+y=n \\
0 & otherwise
\end{cases}
$$
So,
\begin{align*}
P_{X,Y,N}(x,y,n) = P(X=x, Y=y, N=n)
= P(X=x).P(Y=y)
= pq^x.pq^y
= p^2q^{x+y}
= p^2q^{n}
\end{align*}
Hence, the joint PDF of $X$, $Y$ and $N$ is
$$
P_{X,Y,N}(x,y,n) =
\begin{cases}
p^2q^{n} & if x+y=n \\
0 & otherwise
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How prove $\bigl(\frac{\sin x}{ x}\bigr)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$ How prove $\left(\frac{\sin x}{ x}\right)^{2} + \frac{\tan x }{ x} >2$ for $0 < x < \frac{\pi}{2}$.
Can this be proved with simple way?
| since $\sin(x)>0$ and $x>0$ and $\tan(x)>0$ we have
$$\left(\frac{\sin(x)}{x}\right)^2+\frac{\tan(x)}{x}\geq 2\sqrt{\left(\frac{\sin(x)}{x}\right)^2\cdot \frac{\sin(x)}{x}\cdot\frac{1}{\cos(x)}}$$ Now we have to show that $$\frac{\sin(x)^3}{x^3}>\cos(x)$$ in the given interval, defining $$f(x)=\frac{\sin(x)^3}{x^3}-\cos(x)$$ and we get $$\lim_{x \to 0+}f(x)=0$$ and $$f'(x)=-{\frac {\sin \left( x \right) \left( -{x}^{4}-3\,\sin \left( x
\right) \cos \left( x \right) x+3\, \left( \sin \left( x \right)
\right) ^{2} \right) }{{x}^{4}}}
$$ this is positive since $-\sin(x)<0$ and $$-x^4+3\sin(x)(\sin(x)-x\cos(x))<0$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1085780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Cubic inequality If $x,y,z$ are reals from $[0,1]$, then prove that $$2(x^3+y^3+z^3)-x^2y-y^2z-z^2x \le 3$$ We can assume $x=sin\theta, y=sin\phi,z=sin\gamma$. Therefore inequality can be written as $$\begin{align}&2(sin^3\theta+sin^3\phi+sin^3\gamma)-sin^2\theta.sin\phi-sin^2\phi.sin\gamma-sin^2\gamma.sin\theta\le 3\\&=>4sin^3\theta+4sin^3\phi+4sin^3\gamma\le 6+2sin^2\theta.sin\phi+2sin^2\phi.sin\gamma+2sin^2\gamma.sin\theta\\&=>-sin3\theta-sin3\phi-sin3\gamma\le6+\sum_{cyc-\theta,\phi,\gamma}sin\theta(2sin^2\phi-3)\\&=>\sum_{cyc}sin\theta(2-cos2\phi)\le 6+\sum_{cyc}sin3\theta\end{align}$$ What do I do after this?? Is this approach correct?
| Assume $x\geq y\geq z$. Then $x^2y\geq y^3, y^2 z\geq z^3,$ and $z^2x\geq z^3$. Consequently
$$
\begin{align*}
2(x^3+y^3+z^3)-x^2y-y^2z-z^2x&\leq 2(x^3+y^3+z^3)-y^3-z^3-z^3\\
&=2x^3+y^3\\
&\leq 3,
\end{align*}
$$
since $x,y\leq 1$.
EDIT: There was some consternation about why one is allowed to assume $x\geq y\geq z$. If it helps, here is the same argument phrased without making such an assumption.
Let $X$ denote the largest of $x,y,z$, let $Y$ denote the second largest, and let $Z$ denote the smallest.
$$
\begin{align*}
2(x^3+y^3+z^3)-x^2y-y^2z-z^2x&\leq 2(x^3+y^3+z^3)-Y^3-Z^3-Z^3\\
&=2(X^3+Y^3+Z^3)-Y^3-2Z^3\\
&=2X^3+Y^3\\
&\leq 3.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1086873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^4+3x^3+6x+4=0$... easier way? So I was playing around with solving polynomials last night and realized that I had no idea how to solve a polynomial with no rational roots, such as $$x^4+3x^3+6x+4=0$$
Using the rational roots test, the possible roots are $\pm1, \pm2, \pm4$, but none of these work.
Because there were no rational linear factors, I had to assume that the quartic separated into two quadratic equations yielding either imaginary or irrational "pairs" of roots. My initial attempt was to "solve for the coefficients of these factors".
I assumed that $x^4+3x^3+6x+4=0$ factored into something that looked like this
$$\left(x^2+ax+b\right)\left(x^2+cx+d\right)=0$$
because the coefficient of the first term is one. Expanding this out I got
$$x^4+ax^3+cx^3+bx^2+acx^2+dx^2+adx+bcx+bd=0$$
$$x^4+\left(a+c\right)x^3+\left(b+ac+d\right)x^2+\left(ad+bc\right)x+bd=0$$
Equating the coefficients of both equations
$$a+c = 3$$
$$b+ac+d = 0$$
$$ad+bc = 6$$
$$bd = 4$$
I found these relationships between the various coefficients. Solving this system using the two middle equations:
$$\begin{cases} b+a\left(3-a\right)+\frac4b=0 \\ a\frac4b+b\left(3-a\right)=6 \end{cases}$$
From the first equation: $$a = \frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}$$
Substituting this into the second equation:
$$\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\cdot\frac4b+b\cdot\left(3-\frac{3\pm\sqrt{9+4b+\frac{16}{b}}}{2}\right)=6$$
$$3\left(b-2\right)^2 = \left(b^2-4\right)\cdot\pm\sqrt{9+4b+\frac{16}b}$$
$$0 = \left(b-2\right)^2\cdot\left(\left(b+2\right)^2\left(9+4b+\frac{16}b\right)-9\left(b-2\right)^2\right)$$
So $b = 2$ because everything after $\left(b-2\right)^2$ did not really matter in this case. From there it was easy to get that $d = 2$, $a = -1$ and $c = 4$. This meant that
$$x^4+3x^3+6x+4=0 \to \left(x^2-x+2\right)\left(x^2+4x+2\right)=0$$
$$x = \frac12\pm\frac{\sqrt7}{2}i,\space x = -2\pm\sqrt2$$
These answers worked! I was pretty happy at the end that I had solved the equation which had taken a lot of work, but my question was if there was a better way to solve this?
| Hint:
First check that $0$ is not a solution, hence $x\neq0\,$, so it is legal to divide by $x^2$. We get
$$
x^2+3x+\frac6x+\frac4{x^2}=0. \tag1
$$
Now note that
$$
\left(x+\frac2x\right)^2=x^2+4+\dfrac{4}{x^2}\iff
x^2+\dfrac{4}{x^2}=\left(x+\dfrac2x\right)^2-4.
$$
So $(1)$ can be written as :
$$
\left(x^2+\dfrac4{x^2}\right)+\left(3x+\dfrac6x\right)=0\iff
\left(x+\dfrac2x\right)^2-4+3\left(x+\dfrac2x\right)=0.
$$
Now use the substitution $u=x+\frac2x$ and you get the quadratic :
$$
u^2+3u-4=0.
$$
Answer to a comment:
Awesome, but how did you see that? And is this just then a specific case...?
I remarked that the coefficients of the equation were symmetric in the following sense :
$$
x^4+3x^3+6x+4=0\iff (x^4+\color{#C00}2^2)+3(x^3+\color{#C00}2x)=0.
$$
So I tried to divide by $x^2$, then to find a relation between $x^2+\tfrac4{x^2}$ and $\left(x+\tfrac2x\right)^2$, so that I can convert it into a quadratic.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 4,
"answer_id": 2
} |
Simple limit problem without L'Hospital's rule $$\lim_{x \to 1}\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1}$$
We are not supposed to use any derivatives yet, but I can't find any formula that helps here. It's a $\frac{0}{0}$ indeterminate form, and all I think of doing is
$$\frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} = \frac{\sqrt{x^4 + 1} - \sqrt{2}}{\sqrt[3]{x} - 1} \cdot \frac{\sqrt{x^4 + 1} + \sqrt{2}}{\sqrt{x^4 + 1} + \sqrt{2}} = \frac{x^4-1}{(\sqrt[3]{x}-1)\cdot(\sqrt{x^4+1}+\sqrt{2})}$$
but I don't see if this leads anywhere.
| Note that if you put $y=\sqrt[3] x$ then $$\frac {x^4-1}{\sqrt[3]x-1}=\frac {y^{12}-1}{y-1}= y^{11}+y^{10}+\dots +1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1087458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
How to calculate number of triangles and points after dividing a triangle n times? When having a triangle and dividing it n times, how to get the number of triangles and points?
a
/ \
/ \
f-----d
/ \ / \
/ \ / \
c-----e-----b
edit: This triangle is part of an icosahedron as base geometry for a sphere, left it out due to jitter information. Then the assumptions took place.
| Each division multiplies the number of small triangles by $4$, and you start with one triangle at $0$ divisions, so the number of triangles after $n$ divisions is $4^n$.
The number of distinct horizontal sides of small triangles after $n$ divisions is
$$\sum_{k=1}^{2^n}k=\frac{2^n(2^n+1)}2=2^{n-1}(2^n+1)\;.$$
There are the same number of distinct sides of small triangles parallel to each of the non-horizontal sides of the original triangle, so there are altogether $3\cdot2^{n-1}(2^n+1)$ distinct sides of small triangles after $n$ divisions.
Now let $p_n$ be the number of vertices after $n$ divisions; clearly $p_0=3$. If $n\ge 1$, there are $3\cdot 2^{n-2}(2^{n-1}+1)$ small sides after $n-1$ divisions, and the $n$-th division adds one vertex for each of those sides, so
$$p_n=p_{n-1}+3\cdot2^{n-2}(2^{n-1}+1)\;.$$
Thus,
$$\begin{align*}
p_n&=p_0+\sum_{k=1}^n3\cdot 2^{k-2}(2^{k-1}+1)\\\\
&=3+3\sum_{k=1}^n\left(2^{2k-3}+2^{k-2}\right)\\\\
&=3+\frac32\left(\sum_{k=0}^{n-1}4^k+\sum_{k=0}^{n-1}2^k\right)\\\\
&=3+\frac32\left(\frac{4^n-1}3+2^n-1\right)\\\\
&=2^{2n-1}+3\cdot2^{n-1}+1\;.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1088456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Six of a kind . $$\begin{align}
2^{\color{pink}0}+7^{\color{pink}0}+8^{\color{pink}0}+18^{\color{pink}0}+19^{\color{pink}0}+24^{\color{pink}0}&=3^{\color{pink}0}+4^{\color{pink}0}+12^{\color{pink}0}+14^{\color{pink}0}+22^{\color{pink}0}+23^{\color{pink}0}\\
2^{\color{red}1}+7^{\color{red}1}+8^{\color{red}1}+18^{\color{red}1}+19^{\color{red}1}+24^{\color{red}1}&=3^{\color{red}1}+4^{\color{red}1}+12^{\color{red}1}+14^{\color{red}1}+22^{\color{red}1}+23^{\color{red}1}\\
2^{\color{orange}2}+7^{\color{orange}2}+8^{\color{orange}2}+18^{\color{orange}2}+19^{\color{orange}2}+24^{\color{orange}2}&=3^{\color{orange}2}+4^{\color{orange}2}+12^{\color{orange}2}+14^{\color{orange}2}+22^{\color{orange}2}+23^{\color{orange}2}\\
2^{\color{green}3}+7^{\color{green}3}+8^{\color{green}3}+18^{\color{green}3}+19^{\color{green}3}+24^{\color{green}3}&=3^{\color{green}3}+4^{\color{green}3}+12^{\color{green}3}+14^{\color{green}3}+22^{\color{green}3}+23^{\color{green}3}\\
2^{\color{blue}4}+7^{\color{blue}4}+8^{\color{blue}4}+18^{\color{blue}4}+19^{\color{blue}4}+24^{\color{blue}4}&=3^{\color{blue}4}+4^{\color{blue}4}+12^{\color{blue}4}+14^{\color{blue}4}+22^{\color{blue}4}+23^{\color{blue}4}\\
2^{\color{brown}5}+7^{\color{brown}5}+8^{\color{brown}5}+18^{\color{brown}5}+19^{\color{brown}5}+24^{\color{brown}5}&=3^{\color{brown}5}+4^{\color{brown}5}+12^{\color{brown}5}+14^{\color{brown}5}+22^{\color{brown}5}+23^{\color{brown}5}\\
\end{align}$$
Are there similar examples? Any generalization?
| For those who want the quick version, a particular example of Theorem 5 in the link cited by Zander states that if,
$$a^k+b^k+c^k = d^k+e^k+f^k$$
for $k=2,4$, then,
$$\small(x+a)^k+(x+b)^k+(x+c)^k+(x-a)^k+(x-b)^k+(x-c)^k = \\
\small (x+d)^k+(x+e)^k+(x+f)^k+(x-d)^k+(x-e)^k+(x-f)^k$$
for $k=1,2,3,4,5$. The example by the OP used,
$$5^k + 6^k + 11^k = 1^k + 9^k + 10^k$$
and $x=13$. However, to add a nice twist to this post, note the $6-10-8$ identity,
$$64\big(5^6 + 6^6 + 11^6 -(1^6 + 9^6 + 10^6)\big)\big(5^{10} + 6^{10} + 11^{10} -(1^{10} + 9^{10} + 10^{10})\big) =\\ 45\big(5^8 + 6^8 + 11^8 -(1^8 + 9^8 + 10^8)\big)^2$$
To know why, see this MO post.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1090795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Solve logarithmic equation for $x$ to find the inverse of $f(x)= \ln(x+\sqrt{x^2+1})$
Let $f(x)= \ln(x+\sqrt{x^2+1})$. Find $f^{-1}(x)$.
Here is what I got so far: $y= \ln(x+\sqrt{x^2+1})$, rewrite as $x= \ln(y+\sqrt{y^2+1})$,
then $$e^x= y+\sqrt{y^2+1}$$ $$e^x-y= \sqrt{y^2+1}$$ $$ y^2+ e^{2x}-2(e^x)y= 1$$
So if $e^x= a$, then $a^2-2ay-1= 0$
| here is a trick in this particular situation. that is to recognize
$$(\sqrt{x^2 + 1} +x)(\sqrt{x^2 + 1} - x) = 1 $$ so that
$$\ln(\sqrt{x^2 + 1} + x) = -\ln(\sqrt{x^2 + 1} - x)$$
now we can find the inverse function. suppose $$\ln(\sqrt{x^2 + 1} + x) = y,$$
then $$\ln(\sqrt{x^2 + 1} - x) = -y$$ exponentiating these two equations and subtracting gives you $$x = {e^y - e^{-y} \over 2} = \sinh y $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving a differential equation using Laplace transform? $$y''+2y'+ 10 = b\,δ(t-T),\,\begin{cases}y(0)=3\\ y'(0) = 0\end{cases}$$
I managed to solve this equation. My answer is $$y(t) = 3e^{-t} \cos(3t) - e^{-t}\sin(3t)+\dfrac{1}{3}be^{-(t-T)}\sin(3t-3T)u(t-T)$$
I am asked to find values for $b$ and $T$ such that $y(t) = 0$ for all $t>T$. Answer at the back of the book is $$\begin{cases}b_n=3\sqrt{10} e^{-T_n}\\\\T_n=\dfrac{1}{3} \arcsin\dfrac{3}{\sqrt10}+\dfrac{2}{3}n\pi&n=0,1,...\end{cases}$$ I have no idea how they got that. I would greatly appreciated help.
| Cancelling the $e^{-t}$ in $y(t)=0$ for $t>T$ implies that
$$\begin{align*}
3\cos(3t)-\sin(3t)+ \frac{b}{3}e^T\sin 3(t-T)\equiv 0\qquad (1)
\end{align*}$$
Recall the formula (found in most Diff. Eq. books):
$$
A\sin\theta - B\cos\theta = R\sin(\theta-\alpha),
$$
where $R=\sqrt{A^2+B^2}$ and $\alpha=\arcsin\frac{B}{R}$. We choose $A=1,$ $B=3,$ and $\theta=3t$. Therefore
$$
\sin3t-3\cos 3t=\sqrt{10}\sin\left(3t-\arcsin \frac{3}{\sqrt{10}}\right).
$$
Plugging into equation $(1)$ yields:
$$\begin{align*}
\frac{b}{3}e^T\sin 3(t-T)&\equiv \sqrt{10}\sin\left(3t-\arcsin \frac{3}{\sqrt{10}}\right)\\
\implies b&=3\sqrt{10}e^{-T},\ T=\frac{\arcsin \frac{3}{\sqrt{10}}}{3}.
\end{align*}$$
To be completely correct, when we determined the angle $\alpha$, any integer multiple of $2\pi$ could have been added without affecting the trig functions. Therefore $\alpha$ could take on any of the values
$$
\alpha_n=2\pi n+\arcsin \frac{3}{\sqrt{10}},\qquad n=\cdots,-1,0,1,\cdots
$$
Modifying $T$ and $b$ accordingly yields the answer:
$$
\boxed{T_n=\displaystyle\frac{2\pi n+\arcsin \frac{3}{\sqrt{10}}}{3},\qquad b_n=3\sqrt{10}e^{-T_n}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Integrating factor Can anyone give me some hints as to how to solve the following question?
I have to show that the equation below has an integrating factor of the form $t^2\theta^c$ where $c$ is an integer.
$\theta(\theta^2 + 2t) + 2t(\theta^2+t)\frac{d\theta}{dt} = 0$
I started off by multiplying the integrating factor $t^2\theta^c$ through to obtain
$t^2\theta^{c+3} + 2t^3\theta^{c+1} + (2t^3\theta^{c+2} + 2t^4\theta^c)\frac{d\theta}{dt} = 0 $
and here's where I got a bit stuck-tried a few things and got c = -2 which is not the right answer. I've also tried transforming the equation into a linear one but its not panning out right. Any help would be much appreciated.
| Write your equation as $$t^2\theta^{c+3} + 2t^3\theta^{c+1} + 2t^3\theta^{c+2} \theta'+ 2t^4\theta^c \theta' = \frac 13 (t^3)'\theta^{c+3} + \frac 24 (t^4)'\theta^{c+1} + \frac{2}{c+3} t^3(\theta^{c+3})' + \frac{2}{c+1}t^4(\theta^{c+1})'.$$ Now we want to group the terms with same powers of $t$ and $\theta$. We want to put, of course, $\frac 13 = \frac {2}{c+3}$ and $\frac {2}{4}=\frac{2}{c+1}$. These two conditions are consistent and have a solution $c=3$. we can continue with
$$\frac 13 (t^3 \theta^{6})' + \frac 12 (t^4 \theta^{4})' =0.$$In other words, $t^2\theta^c$ is indeed an integrating factor for $c=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1092971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving $\tan x-\tan(2x)=2\sqrt{3}$ $$\tan x-\tan(2x)=2\sqrt{3}$$
TRY #1
$$\begin{align*}
\tan x-\tan(2x)=2\sqrt{3}&\implies\tan x=2\sqrt{3}+\tan{2x}\\
&\implies \tan^2x=\tan^2(2 x)+4 \sqrt{3} \tan(2 x)+12\\
&\implies\tan^2x=(\frac{2\tan x}{1-\tan^2 x})^2+4\sqrt{3}\frac{2\tan x}{1-\tan^2x}+12
\end{align*}$$
but this will give me an equation with $\tan^4$ which needs quartic formula, too difficult!!
TRY #2
$$\begin{align*}
\tan x-\tan(2x)=2\sqrt{3} &\implies \frac{\sin x}{\cos x}-\frac{\sin 2x}{\cos 2x}=2\sqrt3 \\
&\implies\frac{\sin x\cos 2x-\sin 2x\cos x}{\cos x\cos 2x}=2\sqrt{3}\\
&\implies\frac{-\sin x}{\cos x\cos 2x}=2\sqrt{3}\\
&\implies\frac{-\sin x-2\sqrt{3}\cos x\cos 2x}{1}=0
\end{align*}
$$ then i can't!!
can anyone help me?
| let $u = \tan x, \tan 2x = \dfrac{2u}{1-u^2}$ your equation becomes $$u - \dfrac{2u}{1-u^2} = 2\sqrt 3 $$ which can be simplified $$f(u) = u^3 - 2\sqrt 3 u^2 + u + 2\sqrt 3 = 0 $$ i don't see any simple roots for this. we know that $f(0) = 2\sqrt 3$ and $f(-1) = -2$ so there is at least one $-1 < u < 0, f(u) = 0$
edit: thanks to user winther, we can factor it.
$$u^3 - 2\sqrt 3 u^2 + u + 2\sqrt 3 = (u-\sqrt 3)(u^2 -\sqrt 3 u -2) $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1093507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Limit of $ \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ and probability I begin with this problem:
Calculate the limit of $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx$ when $n\rightarrow \infty$.
It's natural to think of recurrence relation. Let $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx=I_n$. By integration by part: $$\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx=-\sin^{n-1}x\cos x|_{0}^{\pi/2}+(n-1)\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n-2}\cos^2 x xdx$$
or $I_n=(n-1)(I_{n-2}-I_n)$. Thus $$I_n=\frac{n}{n-1}\cdot I_{n-2}$$
So we can find a formula for $I_n$ with odd $n$ : $$I_n=\frac{2\cdot 4\cdot 6\cdots (n-1)}{3\cdot 5\cdot 7\cdots n} $$
From the recurrence relation, we have $(I_n)$ decreases and is bounded below by $0$, then it has a limit $L\ge 0$. We will prove that $L=0$.
Suppose $L>0$. First, consider the series $$1+\frac{1}{2}+\frac{1}{3}+\cdots$$
This series diverges. Suppose the series $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots$ converges. Since $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots>\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots$, the series $\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\cdots$ also converges. This yields that $1+\frac{1}{2}+\frac{1}{3}+\cdots$ converges, which is a contradiction. So $\frac{1}{3}+\frac{1}{5}+\frac{1}{7}\cdots$ diverges.
By Lagrange theorem, there exists $c_n\in [n-1,n]$ ($n\ge 3$) such that
$$\ln (n-1)-\ln n=-\frac{1}{c_n}$$
For all $n\ge 3$: $$\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots>\frac{1}{3}+\frac{1}
{5}+\frac{1}{7}+\cdots$$
so $\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots$ diverges by comparison test. Thus $\left[-\left(\frac{1}{c_3}+\frac{1}{c_5}+\frac{1}{c_7}+\cdots\right)\right]$ also diverges.
Back to our problem. Since $L>0$, then $\lim \ln I_n=\ln L$. This means $\ln(2/3)+\ln(5/4)+\cdots=\ln L$, or $-\frac{1}{c_3}-\frac{1}{c_5}-\cdots=\ln L$, contradiction. Thus $L=0$.
I have some questions for the above problem:
*
*Is there another way to compute the limit of $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^n xdx$? My solution is quite complicated
*Rewrite $I_n$ as $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots \left(1-\frac{1}{n}\right)$. Consider a probability problem: in a test of multiple choice questions, each question has exactly one correct answer. The first question has $3$ choices, the second has $5$ choices, etc. By the above result, if the tests has more and more questions, then the probability for the student to be false at all is smaller and smaller. This sounds interesting, because the questions have more and more choices. Is there any generalized result for this (fun) fact?
| There is a simple way of proving that the limit is zero. Since:
$$\forall x\in[-\pi/2,\pi/2],\qquad \cos x \leq 1-\frac{4x^2}{\pi^2}, $$
we have:
$$ I_n = \int_{0}^{\pi/2}\sin^n(x)\,dx = \int_{0}^{\pi/2}\cos^n x\,dx \leq \frac{\pi}{2}\int_{0}^{1}(1-x^2)^n\,dx\leq\frac{\pi}{2}\int_{0}^{1}e^{-nx^2}\,dx $$
so:
$$ I_n \leq \frac{\pi}{2}\cdot\sqrt{\frac{\pi}{4n}}=O\left(\frac{1}{\sqrt{n}}\right). $$
As an alternative technique, notice that:
$$ I_n^2 = \frac{\pi}{2n}\cdot\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)\Gamma\left(1+\frac{n}{2}\right)}\leq\frac{\pi}{2n}$$
since $\Gamma$ is a log-convex function due to the Bohr-Mollerup theorem.
| {
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Arithmetic pattern $1 + 2 = 3$, $4 + 5 + 6 = 7 + 8$, and so on I am noticing this pattern:
\begin{align}
1+2&=3\\
4+5+6&=7+8\\
9+10+11+12&=13+14+15 \\
16+17+18+19+20&=21+22+23+24 \\
&\vdots
\end{align}
Is there a general formula that expresses a pattern here, such as for the $n$th term, or the closed form of the entire summation? I think the $n$th term starts with $$n^2+(n^2+1)+\cdots=\cdots+[(n+1)^2-1].$$
I am also presuming once this formula is discovered, we can prove it by induction for any $n$.
| Another way to build up the pattern you see.
Start with the $n$th triangle number, $T_n$ (I'll use $n=5$):
$$1 + 2 + 3 + 4 + 5 = 1 + 2 + 3 + 4 + 5.$$
Add $n^2$ to both sides, except distribute the $n^2$ on one side by adding $n$ to each of the $n$ terms:
$$5^2 + 1 + 2 + 3 + 4 + 5 = 6 + 7 + 8 + 9 + 10.$$
Finally, add $n^3$ to both sides by distributing $n^2$ to all terms except the first one on the left:
$$25 + 26 + 27 + 28 + 29 + 30 = 31 + 32 + 33 + 34 + 35.$$
So, both sides are
$$T_n + n^2 + n^3 = \frac{n(n+1)}{2} + n^2 + n^3.$$
| {
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Permutation in discrete math Is the permutation
$$\begin{pmatrix} 1& 2 &3 &4 &5 &6&7 \\ 7 & 4 & 2 & 1 & 3 & 6 & 5 \end{pmatrix}$$
even or odd?
The product of disjoint cycles is $$\begin{pmatrix}1& 7&5&3&2&4\end{pmatrix}\begin{pmatrix}6\end{pmatrix}$$
and the transposition are $$\begin{pmatrix}1 & 7 \end{pmatrix}\begin{pmatrix}1 & 5 \end{pmatrix}\begin{pmatrix}1 & 3\end{pmatrix}\begin{pmatrix}1 & 2 \end{pmatrix}\begin{pmatrix}1 & 4 \end{pmatrix}\begin{pmatrix}6 \end{pmatrix}$$ is it correct?
and the answer is even??
I feel confused about this..
| You only need the decomposition in disjoint cycles. Since $\;(1\;7\;5\;3\;2\;4)\;$ as even length, it is an odd cycle and thus is your original permutation
| {
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How to show that $B^{-1}\cdot A^{-1}=B\cdot A$? I can't find the solution of this problem:
Given two $n\times n$ square matrices $A,B$ such that $A^2\cdot B^2=I_n$, show that $B^{-1}\cdot A^{-1}=B\cdot A$.
Thanks in advance.
| $A^2 \cdot B^2 = I \Rightarrow A, B$ are invertible. The set of all invertible matrices forms a group. So we also have $B^2 \cdot A^2 = I.$ Hence $B \cdot A = B^{-1}\cdot A^{-1}.$
I'm assuming that the entries of the matrices are from a field.
EDIT: $A^2 \cdot B^2 = I \Rightarrow \text{det}(A^2) \cdot \text{det}(B^2) = \text{det}(A^2 \cdot B^2) = 1 \Rightarrow \text{det}(A^2) \neq 0.$ Now $(\text{det}(A))^2 = \text{det}(A^2) \Rightarrow \text{det}(A) \neq 0.$ Similarly for $B.$
| {
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Finding eigenvalues of the linear map of differentiation on polynomial sets I have the following question:
Let $\mathbf{P_2}$ denote the vector space of real polynomials of degree at most two, and let $\mathbf{L}:\mathbf{P_2}\rightarrow\mathbf{P_2}$ be the linear map $\mathbf{L}f=f', $(the derivative of $f$). What are the eigenvalues of $\mathbf{L}$?
I got since the linear map takes $$\begin{pmatrix} \alpha \\ \beta\\ \gamma \end{pmatrix}\rightarrow \begin{pmatrix} 0 \\ 2\alpha \\\beta \end{pmatrix}$$We have $$\begin{pmatrix} 0 \\ 2\alpha \\\beta \end{pmatrix}=\lambda\begin{pmatrix} \alpha \\ \beta\\ \gamma \end{pmatrix}$$
This only will have a solution without $\alpha=\beta=\gamma=0$, when $\lambda=0$
Is this correct? Just making sure my reasoning is sound.
| Hint: A base for $\mathbf{P_2}$ is $\{1,x,x^2\}$ and the derivatives are
$$1\longmapsto 0,$$
$$x\longmapsto 1,$$
$$x^2\longmapsto 2x.$$
Then the matrix of the transformation is
$\left(\begin{array}{ccc}
0&1&0\\
0&0&2\\
0&0&0
\end{array}\right).$
The characteristic matrix is
$\left(\begin{array}{ccc}
x&-1&0\\
0&x&-2\\
0&0&x
\end{array}\right)$, whose determinant is the characteristic polynomial
$$\chi(x)=x^3,$$
with a triple zero $\lambda=0$.
These are the matrix's eigenvalues.
The only non trivial eigenvector is
$\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)$, since
$$\left(\begin{array}{ccc}
0&1&0\\
0&0&2\\
0&0&0
\end{array}\right)
\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)=0
\left(\begin{array}{c}
1\\
0\\
0
\end{array}\right)
.$$
Remark:
Here a polynomial $a+bx+cx^2$ is manipulated by is components conventionally as $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)$.
| {
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How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$? How do you find that: $x^3+4x^2+x-6=(x+2)(x^2+2x-3)$?
I get that you can take out the $x$ like so: $x^3+4x^2+x-6=x(x^2+4x+1)-6$ but how do you get the $2$ from here?
| Observe that $x^3 + 4x^2 + x - 6 = x^3 + 4x^2 + 4x - 3x - 6$. Therefore,
\begin{align*}
x^3 + 4x^2 + x - 6 & = x^3 + 4x^2 + 4x - 3x - 6\\
& = x(x^2 + 4x + 4) - 3(x + 2)\\
& = x(x + 2)^2 - 3(x + 2)\\
& = (x + 2)[x(x + 2) - 3)]\\
& = (x + 2)(x^2 + 2x - 3)
\end{align*}
That said, I recommend using the Rational Roots Theorem. In this case, a rational root must be a factor of $-6/1 = -6$, so the possible rational roots are $\pm 1, \pm 2, \pm 3, \pm 6$. If you test the roots, you will find that $1, -2, -3$ are roots. Once you find that $k$ is a root, you can divide by $x - k$ to find the other factor. Thus, if you discover that $-2$ is a root, you can divide $x^3 + 4x^2 + x - 6$ by $x - (-2) = x + 2$ to discover that $x^2 + 2x - 3$ is the other factor. As you can check, the roots of $x^2 + 2x - 3$ are $1$ and $-3$.
| {
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Limit of $\left(\frac{x^2+x+1}{2x+1}\right)^{1/(x^2-1)}$ when $x\to1$ $$\displaystyle \lim_{x\to 1}\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$
I know that the two-sided limit of $\frac{1}{x^2-1}$ does not exist. I don't know what to do with $\frac{x^2+x+1}{2x+1}$ to get something else than $1^\infty$
| Hint
Consider $$A=\left(\frac{x^2+x+1}{2x+1}\right)^{\frac{1}{x^2-1}}$$ Taking logarithms $$\log(A)=\frac{1}{x^2-1}\log\left(\frac{x^2+x+1}{2x+1}\right)$$ Now, consider the Taylor series built at $x=1$ $$\frac{x^2+x+1}{2x+1}=1+\frac{x-1}{3}+\frac{1}{9} (x-1)^2+O\left((x-1)^3\right)$$ and use the fact that, for small $y$, $$\log(1+y)=y-\frac{y^2}{2}+O\left(y^3\right)$$ Now, make $$y=\frac{x-1}{3}+\frac{1}{9} (x-1)^2$$ in order to get the expansion of $\log\left(\frac{x^2+x+1}{2x+1}\right)$ and simplify.
I am sure that you can take from here and conclude not only what is the limit but also how it is approached.
| {
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Integrate $\int_0^\infty \frac{dx}{(x^2+2x+12)^2}$ using residues I want to find the integral
$$I=\int_0^\infty \frac{dx}{(x^2+2x+12)^2}$$
using contour integration; I am familiar with the trigonometric substitution in real analysis.
There are no branch cuts, there are two second order poles at $z=-1\pm\sqrt{11}i$. I cannot use $\int_{-\infty}^\infty=2\int_0^\infty$ since the function isn't even.
Clearly, the choice of contour is important here (compare with this question). I tried half-circle and quarter-circle contours, but they just give some weird rationals. Is there some better contour here? Am I missing something?
| Yes. There is a standard trick for integrating functions over $[0,\infty)$, and that is to exploit the multi-valuedness of the logarithm about a branch cut. Thus, consider
$$\oint_C dz \frac{\log{z}}{(z^2+2 z+12)^2} $$
where $C$ is a keyhole contour as pictured below:
The radius of the small circle is $\epsilon$ and that of the bigger circle is $R$. The contour integral over $C$ is then
$$\int_{\epsilon}^R dx \frac{\log{x}}{(x^2+2 x+12)^2} + i R \int_0^{2 \pi} d\theta \, e^{i \theta}\frac{\log{\left ( R e^{i \theta} \right )}}{\left (R^2 e^{i 2 \theta} + 2 R e^{i \theta}+12 \right )^2} \\ + \int_R^{\epsilon} dx \frac{\log{x} + i 2 \pi}{(x^2+2 x+12)^2} + i \epsilon \int_{2 \pi}^0 d\phi \, e^{i \phi} \frac{\log{\left ( \epsilon e^{i \phi} \right )}}{\left (\epsilon^2 e^{i 2 \phi}+ 2 \epsilon e^{i \phi}+12 \right )^2}$$
It should be evident that, as $R \to \infty$, the second integral vanishes as $\log{R}/R^3$, and that as $\epsilon \to 0$, the fourth integral vanishes as $\epsilon \log{\epsilon}$. Thus, in this limit, the contour integral is
$$-i 2 \pi \int_0^{\infty} \frac{dx}{(x^2+2 x+12)^2} $$
By the residue theorem, this contour integral is also equal to $i 2 \pi$ times the sum of the residues at the poles $z_{\pm} = -1 \pm i \sqrt{11}$. Therefore we have
$$\begin{align}\int_0^{\infty} \frac{dx}{(x^2+2 x+12)^2} &= - \left [\frac{d}{dz} \frac{\log{z}}{(z-z_-)^2} \right ]_{z=z_+} - \left [\frac{d}{dz} \frac{\log{z}}{(z-z_+)^2} \right ]_{z=z_-}\\ &= -\frac1{z_+ (z_+-z_-)^2} + \frac{2 \log{z_+}}{ (z_+-z_-)^3}\\ &-\frac1{z_- (z_--z_+)^2} + \frac{2 \log{z_-}}{(z_--z_+)^3} \\ &= -\frac{z_++z_-}{z_- z_+ (z_+-z_-)^2} + 2 \frac{\log{z_+}-\log{z_-}}{(z_+-z_-)^3} \end{align}$$
To get this right, it is crucial that we understand how to compute the logs. Understand that, in defining the contour, we demanded that $\arg{z} \in [0,2 \pi)$. Thus,
$$\log{z_+} = \frac12 \log{12} + i \left (\pi- \arctan{\sqrt{11}}\right )$$
$$\log{z_-} = \frac12 \log{12} +i \left (\pi+ \arctan{\sqrt{11}}\right )$$
Putting this altogether, we find, after some arithmetic, that
$$\int_0^{\infty} \frac{dx}{(x^2+2 x+12)^2} = \frac{1}{22 \sqrt{11}} \arctan{\sqrt{11}} - \frac1{264} $$
ADDENDUM
Another way of evaluating this integral is to observe that
$$\int_0^{\infty} \frac{dx}{(x^2+2 x+b)^2} = -\frac{\partial}{\partial b} \int_0^{\infty} \frac{dx}{x^2+2 x +b} $$
then consider
$$\oint_C dz \frac{\log{z}}{z^2+2 z+b} $$
The analysis is the same as above; the result is that
$$-\int_0^{\infty} \frac{dx}{x^2+2 x +b} = \frac{\log{z_+}-\log{z_-}}{z_+-z_-} = -\frac{\arctan{\sqrt{b-1}}}{\sqrt{b-1}}$$
Differentiating, we get
$$\int_0^{\infty} \frac{dx}{(x^2+2 x+b)^2} = \frac{\arctan{\sqrt{b-1}}}{2 (b-1)^{3/2}}-\frac{1}{2 b (b-1)}$$
Plugging in $b=12$ produces the above result.
| {
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check my work on this problem: given tan(2x), find sin x + cos x? $\tan 2x = - 24/7$
$90^\circ < x < 180^\circ$.
Find the value of $\sin x+\cos x$.
What I have so far:
$\tan(2x) = -\frac{24}{7} \Rightarrow \frac{2\tan(x)}{1-\tan^{2}x} = -\frac{24}{7}$.
Cross-multiplying gives us the quadratic $12\tan^{2}x-7\tan(x)-12 = 0$, factorized, it is $(3\tan x-4)(4\tan x+3) \Rightarrow \tan x = \frac{4}{3}, -\frac{3}{4}$. Since $90^\circ < x < 180^\circ$, I would have to pay attention to $\tan x = -\frac{3}{4}$.
$\sin(\arctan(-\frac{3}{4}))+\cos(\arctan(-\frac{3}{4}))$. When calculating this, I got $\frac{1}{5}$, but that was incorrect. Where did I make my mistake?
| You are very close - basically correct in fact. The only issue I see is that in your calculation of $\sin\arctan\left(-\frac{3}{4}\right)$ and $\cos\arctan\left(-\frac{3}{4}\right)$, you forgot the angles needed to be in the range $90^\circ$ to $180^\circ$, so you need a positive $\sin$ and negative $\cos$. If you just plug it into a calculator, you will get the wrong answer because the standard range on $\arctan$ is $-90^\circ$ to $90^\circ$.
| {
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Where does $r = 1 + 2\cos(\theta)$ have tangents? Where does:
$$r = 1 + 2\cos(\theta)$$
Have horizontal and vertical tangent lines?
$x = r\cos(\theta) = \cos(\theta) + 2\cos^2(\theta)$
$y = r\sin(\theta) = \sin(\theta) + 2\sin(\theta)\cos(\theta)$
$$dx/d\theta = -\sin(\theta) - 4\sin(\theta)\cos(\theta)$$
$$dy/d\theta = \cos(\theta) + 2\cos(2\theta)$$
By hand it is very hard to solve $dy/d\theta = 0, dx/d\theta = 0$
How can it be done?
| You have calculated
$$
\frac{dx}{d\theta}=-\sin\theta(1+4\cos\theta)
$$
and
$$
\frac{dy}{d\theta}=\cos\theta+2\cos2\theta.
$$
Since $\cos2\theta=2\cos^2\theta-1$, the $\frac{dy}{d\theta}$ simplifies to
$$
\frac{dy}{d\theta}=\cos\theta+4\cos^2\theta-2.
$$
Vertical and horizontal tangents
In fact, we do not need to find the values of $\theta$ if we want to find the points where we have horizontal/vertical tangents. It is enough to know $\cos\theta$ and $\sin\theta$ and then use the formulas
$$
x=r\cos\theta=(1+2\cos\theta)\cos\theta,\quad\text{and}\quad y=r\sin\theta=(1+2\cos\theta)\sin\theta.
$$
Anyways, I indicate how to solve the equations. First, we can assume that $0\leq \theta<2\pi$, since $r=1+2\cos\theta$ is $2\pi$ periodic.
Now, one should solve $\frac{dx}{d\theta}=0$ to find the vertical tangent lines. For the vertical ones, $dx/d\theta=0$. Since the product $\sin\theta(1+4\cos\theta)=0$ if and only if (at least) one of its factors is zero, we find that either $\sin\theta=0$ (this gives $\theta=0$ and $\theta=\pi$), or $\cos\theta=-1/4$ (this gives $\theta=\arccos(-1/4)$ or $\theta=2\pi-\arccos(-1/4)$. In total, we get four values of $\theta$.
To find the horizontal tangent lines we solve $dy/d\theta=0$. We note that $dy/d\theta$ is a second degree polynomial in $\cos\theta$. Thus, with $t=\cos\theta$, we should solve $4t^2+t-2=0$. This has solutions $t=\frac{1}{8}(-1\pm\sqrt{33})$. So, we find $\theta$ by solving
$$
\cos\theta=\frac{1}{8}(-1+\sqrt{33})\approx 0.59,\quad\text{and},\quad \cos\theta=\frac{1}{8}(-1-\sqrt{33})\approx -0.84.
$$
To calculate $\sin\theta$ you can use the fact that $\cos^2\theta+\sin^2\theta=1$. I leave it to you to fill in the rest of the details. If you succeed, you will probably find something like the picture below, where I have drawn the curve $r=1+2\cos\theta$ together with the points I got by doing these calculations (red for horizontal tangents, green for vertical ones).
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Do I need to use partial fractions to find $\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$? I need to simplify $\frac{2n+1}{n^2(n+1^2)}$ as part of an exam question. The solution states $$\frac{2n+1}{n^2(n+1)^2} = \frac{1}{n^2}-\frac{1}{(n+1)^2}$$
In the solution it does not state how this simplification was made, I figured this could be done in quite a long winded fashion, using partial fractions. But from how it's written in the solutions it seems like this should be an easy simplification.
Is there a simple trick to simplifying fractions like this?
| HINT
Use Partial Fraction Decomposition,
$$\frac{2n+1}{n^2(n+1)^2}=\frac An+\frac B{n^2}+\frac C{n+1}+\frac D{(n+1)^2}$$
$$\iff2n+1=An(n+1)^2+B(n+1)^2+Cn^2(n+1)+Dn^2$$
Now compare the constants & the coefficients of $n,n^2,n^3$ to find $A,B,C,D$
Alternatively by observation,
$$\frac{2n+1}{n^2(n+1)^2}=\frac{(n+1)^2-n^2}{(n+1)^2n^2}=\cdots$$
| {
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Prove that $\sum_{k=1}^{m}\frac{1}{k(k+1)}=1-\frac{1}{m+1}$. I know $\sum_{k=1}^{m}\frac{1}{k(k+1)}=1-\frac{1}{m+1}= H_mH_{m+1}$, for $H_m$ the harmonic sum.
I tried many ways to prove it like this
$$\sum_{k=1}^{m}\left(\frac{1}{k(k+1)} + \frac{1}{m+1}\right)=1 $$
$$ \Rightarrow \sum_{k=1}^{m-1} \left(\frac{1}{k(k+1)}+\frac{1}{m+1} + \frac{1}{m(m+1)}\right) = \sum_{k=1}^{m-1} \left(\frac{1}{k(k+1)} + \frac{m^2 + 1}{m(m+1)}\right) = 1.$$
My questions are:
*
*Could anyone give me any hints to calculate it, but not the solution, please?
Thanks! $\Rightarrow$ it's OK I solved the exercise :) !!
*Is the calculation, I wrote above here, correct? Or not?
| Induction would probably work out here too. For $m=1$ we have $$\frac1{1\cdot 2} = 1-\frac12.$$ Then for $m=2$ we have $$\frac{1}{1\cdot2} + \frac{1}{2\cdot 3} = \left(1 - \frac12\right) + \frac1{2\cdot3}=1-\frac{3-1}{2\cdot 3}=1-\frac{1}{3}.$$
Try to do this in general.
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Optimize over unit circle to prove $|ax + by| \le \sqrt{a^2 + b^2}$ I have the following problem which, straight off the shelf, seems totally approachable. It's been giving me difficulty however:
Let $a,b,x,y \in \mathbb{R}$, and suppose that $x^2 + y^2 =1$. Prove that $|ax + by| \le \sqrt{a^2 + b^2}$.
I have started by considering the function $f(x,y) = ax + by$, and by defining a constraint function $g(x,y) = x^2 + y^2$. Then if I can maximize and minimize $f(x,y)$ subject to the constraint $g(x,y) = 1$, I should be done. By both methods (the method of Lagrange multipliers, and the method of evaluating $f(x,y)$ on the unit circle, finding critical points, etc.) I am having difficulties.
For the Lagrange multiplier method, we know that we will maximize/minimize $f(x,y)$ subject to the constraint $g(x,y) = 1$ if the system $\nabla g = \lambda \nabla f$, $g(x,y) = 1$ is satisfied. In terms of our particular example, we obtain the system of three equations
\begin{align*}
2 \lambda x &= a\\
2 \lambda y &= b\\
x^2 + y^2 &=1.
\end{align*}
I'm not sure how to solve this system. I considered using the function $h(x,y) = (f(x,y))^2 = (ax + by)^2$ instead of $f$, but then things don't work out nicely for the minimum values of $f(x,y)$ and $h(x,y)$.
For the other method, I'm even less sure of what I'm doing. Any help would be greatly appreciated.
| First of all, there is a fairly easy way to show this inequality:
$$|ax + by| = |(a, b)\cdot (x, y)| \leq \|(a, b)\|\|(x, y)\|\cos\theta \leq \|(a, b)\|\|(x, y)\| = \|(a, b)\| = \sqrt{a^2 + b^2}.$$
As for the Lagrange multipliers method, you're almost there. As $2\lambda x = a$ and $2\lambda y = b$, $x = \frac{a}{2\lambda}$ and $y = \frac{b}{2\lambda}$. To determine the values of $\lambda$, we use the fact that $g(x, y) = 1$. Using this, we have
$$1 = g(x, y) = x^2 + y^2 = \left(\frac{a}{2\lambda}\right)^2 + \left(\frac{b}{2\lambda}\right)^2 = \frac{a^2+b^2}{4\lambda^2}$$
so $\lambda = \pm\frac{1}{2}\sqrt{a^2+b^2}$.
If $\lambda = \frac{1}{2}\sqrt{a^2+b^2}$ then $(x, y) = \left(\frac{a}{\sqrt{a^2+b^2}}, \frac{b}{\sqrt{a^2+b^2}}\right)$, and $f(x, y) = \sqrt{a^2+b^2}$.
If $\lambda = -\frac{1}{2}\sqrt{a^2+b^2}$ then $(x, y) = \left(-\frac{a}{\sqrt{a^2+b^2}}, -\frac{b}{\sqrt{a^2+b^2}}\right)$, and $f(x, y) = -\sqrt{a^2+b^2}$.
So $f$ has attains it maximum value of $\sqrt{a^2+b^2}$ at the first point and a minimum value of $-\sqrt{a^2+b^2}$ at the second point. Therefore, for all $(x, y)$ on the unit circle,
$$-\sqrt{a^2+b^2} \leq f(x, y) \leq \sqrt{a^2+b^2}.$$
So $|ax+by| = |f(x, y)| \leq \sqrt{a^2+b^2}$.
| {
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Sum of squares of Fibonacci Numbers: $\sum_{i=0}^{n} (F_{2i+1})^2$ $$ \sum_{i=0}^{n} (F_{2i+1})^2 = \;?$$
I know that sum of squares of first $n$ Fibonacci numbers is $F_{n} \times F_{n+1}$.
| We know that:
$$F_n = \frac{1}{\sqrt{5}}\left(\sigma^n - \bar{\sigma}^n\right)\tag{1}$$
where $\sigma,\bar{\sigma}$ are the roots of $x^2-x-1$. On the other hand,
$$ L_n = \sigma^n+\bar{\sigma}^n, \tag{2}$$
and $\sigma\bar{\sigma}=-1$, so:
$$ F_{2n+1}^2 = \frac{1}{5}\left(\sigma^{4n+2}+\bar{\sigma}^{4n+2}+2\right)=\frac{L_{4n+2}+2}{5}\tag{3}$$
and:
$$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{5}\left(\sigma^2\frac{\sigma^{4(N+1)}-1}{\sigma^4-1}+\bar{\sigma}^2\frac{\bar{\sigma}^{4(N+1)}-1}{\bar{\sigma}^4-1}\right),$$
but since $\sigma^4-1 = (\sigma+1)^2-1 = \sigma(\sigma+2) $ the latter simplifies to:
$$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{5}\left(\sigma\frac{\sigma^{4(N+1)}-1}{\sigma+2}+\bar{\sigma}\frac{\bar{\sigma}^{4(N+1)}-1}{\bar{\sigma}+2}\right),$$
or, since $(\sigma+2)(\bar{\sigma}+2)=5$,
$$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{25}\left(\sigma(\bar{\sigma}+2)(\sigma^{4(N+1)}-1)+\bar{\sigma}(\sigma+2)(\bar{\sigma}^{4(N+1)}-1)\right),$$
$$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{25}\left(\sigma(\bar{\sigma}+2)(\sigma^{4(N+1)})+\bar{\sigma}(\sigma+2)(\bar{\sigma}^{4(N+1)})\right),$$
$$\sum_{n=0}^{N}F_{2n+1}^2 = \frac{2}{5}(N+1)+\frac{1}{25}\left(2L_{4N+5}-L_{4N+4}\right),$$
$$\sum_{n=0}^{N}F_{2n+1}^2 = \color{red}{\frac{2}{5}(N+1)+\frac{1}{25}\left(L_{4N+5}+L_{4N+3}\right)}.\tag{4}$$
| {
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Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$
I have a question about the following solution:
We may write it in the form:
$$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$
Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 < \frac{1}{n} < \ldots < \frac{n}{n}$.
Can you explain to me to reach this conclusion?
| Well you already gave the answer can't figure out what's the problem? You just say: Let $f(x)=\dfrac{1}{1+x^2}$.
$$\lim\limits_{n\to +\infty}\sum\limits_{k=1}^n\frac{n^2}{n^2+k^2}=\lim\limits_{n\to +\infty}\frac{1-0}{n}\sum\limits_{k=1}^nf\left( 0+\frac{1-0}{n}k\right)=\int\limits_0^1f(t)\mathrm dt$$
| {
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Cubes differences and primality In an exercise (Project Euler 131, not to mention it), we are looking for perfect cubes of the form $n^3 + n^2 p$, where p is prime.
I finally got the answer by trial and error but I don't understand why this implies that n and n+p must be perfect cubes. Any proof ?
UPDATE : following up on @miracle173 answer, can it be proved that p doesn't divide n ?
| We have
$$n^2(n+p)=m^3 \tag{1}$$
1) First let us assume that $p$ does not divide $n$.
1.1) Let $q$ be a prime that divides $n$. We have $\gcd(q,p)=1$ and therefore $\gcd(n+p,q)=1$. If $q^e$ is the highest power of $q$ that divides $n$ then $q^{2e}$ is the highest power of $q$ that divides $m^3$. So $3 \mid e$. This holds for all primes dividing $n$ therefore $n$ is a cube.
1.2) If $q$ is a prime that divides $n+p$ (but not $n$) then if $q^e$ is the highest power of $q$ ($\ne p$) that divides $n+p$ then $q^e$ is the highest power of $q$ that divides $m^3$ and again we have $3 \mid e$. So $n+p$ is a prime , too.
2.) Now we assume that $p$ divides $n$ and therefore divides $n+p$, too.
So
$$n=p^ka \tag{2}$$
for some $a$ and $\gcd(a,p)=1$. We substitute $(2)$ in $(1)$ and get
$$p^{2k+1}a^2(ap^{k-1}+1)=m^3 \tag{3}$$
Because $a$ is relatively prime to the both other factors on the lhs of $(3)$ we can conclude that a is relatively prime to the integer $\frac{m^3}{a^2}$ and so $a^2$ is a cube. Therefore $a$ and $\frac{m^3}{a^2}$ are cubes, too, and (3) can be reduced to
$$p^{2k+1}(ap^{k-1}+1)=s^3 \tag{4}$$
if we set $s=\frac{m^3}{a^2}$.
If $k>1$ then then $\gcd(p^{2k+1},ap^{k-1}+1)=1$ and so $p^{2k+1} \mid s^3$.
If $({2k+1}) \not\mid 3$ this is a contradiction, if $({2k+1}) \mid 3$ then $(4)$ reduces to
$$b^3+1=u^3 \tag{5}$$
with $u=\frac{s^3}{p^{2k+1}}$ and $b=a^{\frac{1}{3}} p^{\frac{k-1}{3}}$. Remember that $a$ is a power of 3 and
$$0 \equiv 2k+1 \equiv k-1\pmod{3}$$
If $k=1$ we get the same equation from $(4)$ But $(5)$ is only valid for $b=0, u=1$.
| {
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All matrices which commute with all $2\times 2$ matrices I would like to find all matrices which commute with all $2\times2$ matrices.
I started solving problem in this way:
1) I have this matrix $A$ with real numbers:
$$A=\left[\begin{array}{cc}a &b\\c &d\end{array}\right]$$
2) Matrix which commute with matrix $A$ is matrix $B$:
$$B = \left[\begin{array}{cc}e& f\\ g &h\end{array}\right]$$
3) When i solve the equation $AB=BA$, I get this:
$$\left[\begin{array}{cc}ae+bg&af+bh\\ce+dg&cf+dh\end{array}\right]=\left[\begin{array}{cc}ea+cf&eb+df\\ga+ch&gb+dh\end{array}\right]$$
How I can get the general look of wanted matrix?
| we can try four different scenarios with matrices that have 3 zero entires. We can assumed $\mathbf{A}=\begin{pmatrix} a & b \\ c & d \end{pmatrix}$. We can then try $\mathbf{B}=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$.
Solving for $a,b,c$ & $d$ we get $\mathbf{AB}=\mathbf{BA}$ when $c=0$ and $b=0$.
We can then solve for $\mathbf{B}=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ which gives us $\mathbf{AB}=\mathbf{BA}$ when $b=0$ and $c=0$.
For $\mathbf{B}=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ we get $\mathbf{AB}=\mathbf{BA}$ when $c=0$ and when $a=d$.
The last equation, $\mathbf{B}=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$, give us $\mathbf{AB}=\mathbf{BA}$ when $b=0$, $c=0$, $a=d$.
These equations imply that $a=d$ and $b=c=0$. this means that $\mathbf{A}=\begin{pmatrix} a & 0 \\ 0 & a \end{pmatrix}$ is the form for all possible solutions for $A$ in which every matrix of the form found commute with all matrices $\bold{B}$.
| {
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Implicit Differentiation: $(x/y)+(y/x) =1$ Hi can anyone please tell me where I goes wrong with this question:
Find $ \frac{dy}{dx} $ for the curves defines by this equation:
\begin{align}
\frac{x}{y} + \frac{y}{x} = 1
\end{align}
Here is what I did:
\begin{align}
&\frac{y-xy'}{y^2} + \frac{y'x - y}{x^2} =0 \\
&\frac{yx^2 - x^3 y'+y'y^2 x - y^3}{x^2 y^2} = 0 \\
&yx^2 + (-x^3 +y^2 x)y' -y^3 =0 \\
&\therefore y'= \frac{y^3 -yx^2}{-x^3 +y^2x}
\end{align}
The answer say it should be: $ y' = \frac{y}{x} $ but I had no clue how to proceed from there.
Please help, Thanks.
| Note that $y=y(x)$
So we have that
$$\frac{x}{y(x)} + \frac{y(x)}{x} = 1$$
So,
$$\begin{align} 0 &= \frac{d}{dx}\left( \frac{x}{y(x)} + \frac{y(x)}{x}\right) \\
&=\frac{d}{dx}(x) \frac{1}{y(x)}+x \frac{d}{dx}\left(\frac{1}{y(x)}\right) \\
&= \frac{1}{y(x)}+ x \ \frac{-1}{y^2(x)} \frac{dy}{dx} \\
&= \frac{1}{y(x)}\left(1-\frac{x}{y(x)} \right)\frac{dy}{dx}
\end{align}$$
Assuming $\frac{1}{y(x)} \neq 0$ then
$$0= \left(1-\frac{x}{y(x)} \right)\frac{dy}{dx} $$
$$-1= -\frac{x}{y(x)} \frac{dy}{dx} $$
$$\frac{dy}{dx}=\frac{y(x)}{x}=\frac{y}{x} $$
The only thing I used is the product rule http://en.wikipedia.org/wiki/Product_rule
| {
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$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$ Understand the representation $$\sum _{n=1}^{\infty } \frac{1}{n (n+1) (n+2)}$$=$$\frac{1}{2} \left(-\frac{2}{n+1}+\frac{1}{n+2}+\frac{1}{n}\right)$$
$$s_n=\frac{1}{2} \left(\left(\frac{1}{n+2}+\frac{1}{n}-\frac{2}{n+1}\right)+\left(\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}\right)+\ldots +\left(1 \frac{1}{1}-\frac{2}{2}+\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{2}{3}+\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{2}+\frac{1}{5}\right)+\left(\frac{1}{4}-\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{1}{5}-\frac{1}{3}+\frac{1}{7}\right)+\left(\frac{1}{6}-\frac{2}{7}+\frac{1}{8}\right)\right)$$
Im sorry for not being able to get the term in right order, but as you can see the first two terms in the parenthesis should be the last for a accurate representation.
$$\left(
\begin{array}{cc}
1 & 1-1+\frac{1}{3} \\
2 & \frac{1}{2}-\frac{2}{3}+\frac{1}{4} \\
3 & \frac{1}{3}-\frac{1}{2}+\frac{1}{5} \\
4 & \frac{1}{4}-\frac{2}{5}+\frac{1}{6} \\
5 & \frac{1}{5}-\frac{1}{3}+\frac{1}{7} \\
6 & \frac{1}{6}-\frac{2}{7}+\frac{1}{8} \\
\end{array}
\right)$$
I think that my pure understanding of the representation is what makes me confused. From the table I see that a pattern in the denominators emerge, when n=1, then (1-2+3), when n=2 then, 2-3+4. However given the part of sn where $$\frac{1}{n-1}-\frac{2}{n}+\frac{1}{n+1}$$ I fail to see how the first part of expression right above is not undefined when n=1.
I am convinced that it is my lack of knowledge of what the representation actually means. And I am also struggling to see what cancels to give$$\frac{1}{2} \left(\frac{1}{n+2}-\frac{1}{n+1}+\frac{1}{2}\right)$$
However given that this is true I am able to understand that the answer is (1/4). But as I said it is the representation I do not get. Could someone help me as this would be very useful for my next chapter in my textbook!
| You should write: (define $s_n$ as the sequence of partial sums of the series)
$s_n=\sum _{k=1}^{n} \frac{1}{k (k+1) (k+2)}$
$=\sum _{k=1}^{n} {\frac{1}{2} \left(-\frac{2}{k+1}+\frac{1}{k+2}+\frac{1}{k}\right)}$
$=\frac{1}{2}\left(-2\sum_{k=1}^{n}\frac{1}{k+1}+\sum_{k=1}^{n}\frac{1}{k+2}+\sum_{k=1}^{n}\frac{1}{k}\right)$ because the sums are finite. Observe then that the three sums overlap on most terms.
$=\frac{1}{2}\left(-2\sum_{k=2}^{n+1}\frac{1}{k}+\sum_{k=3}^{n+2}\frac{1}{k}+\sum_{k=1}^{n}\frac{1}{k}\right)$ by just rewriting the sum.
$=\frac{1}{2}\left(\left(-2\sum_{k=3}^{n}\frac{1}{k}-2\cdot\frac{1}{2}-2\frac{1}{n+1}\right)+\left(\sum_{k=3}^{n}\frac{1}{k}+\frac{1}{n+1}+\frac{1}{n+2}\right)+\left(\sum_{k=3}^{n}\frac{1}{k}+\frac{1}{1}+\frac{1}{2}\right)\right)$ by just rewriting the sum again.
$=\frac{1}{2}\left(-2\cdot\frac{1}{2}-2\frac{1}{n+1}+\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{1}+\frac{1}{2}\right)$ by cancelling terms.
$=\frac{1}{2}\left(-\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{2}\right)$ by cancelling terms.
Then only now, can you consider taking the limit $n\to\infty$.
Maybe you are not understanding the $\sum$ operator.
$\sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$ is exactly $\frac{1}{\underline{1}(\underline{1}+1)(\underline{1}+2)}+\frac{1}{\underline{2}(\underline{2}+1)(\underline{2}+2)}+\dots+\frac{1}{\underline{n}(\underline{n}+1)(\underline{n}+2)}$ by definition.
And $\sum_{k=1}^{\infty} \frac{1}{k(k+1)(k+2)}$ is exactly $\lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{k(k+1)(k+2)}$ by definition.
More Examples
$$\sum_{k=1}^n a_{k+1}=a_2+a_3+\dots+a_{n+1}=\sum_{k=2}^{n+1} a_k \quad (Eq1)$$
In general, we have $\sum_{x\in S} f(x)$, which means "take all the distinct values in a finite set $S$, apply $f$ to each of them, and sum the results up". Note the results may not be distinct.
Now we can write $\sum_{k=1}^n f(k)=\sum_{k\in\mathbb{Z}\cap[1,n]} f(k)$.
If $g$ is an injective function on $S$ (distinct inputs give distinct outputs), then
$$\sum_{x\in S} f(x)=\sum_{y\in g(S)} f(g^{-1}(y))\quad\text{, also}\quad\sum_{x\in S} f(g(x))=\sum_{y\in g(S)} f(y)$$
where $g(S)$ are all the outputs of $g$ given inputs in $S$, and $g^{-1}$ is the function which takes as input an output of $g$ and returns the input which gave that output (this is the inverse function).
So in (Eq1), we are taking $g$ to be $k\mapsto k+1$, and $S=\mathbb{Z}\cap[1,n]$. Hence $g(S)=\mathbb{Z}\cap[2,n+1]$.
Of course, you have the distributive rule: $\sum_{x\in S}af(x)=a\sum_{x\in S}f(x)$ (use induction).
| {
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Prove or disprove: $\sum a_n$ convergent, where $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$.
Let $a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}$. Show that $a_n>0\ \forall\ n\ge1$.
Prove or disprove: $\sum\limits_{n=1}^\infty a_n$ is convergent.
I can't show that $a_n > 0\ \forall n\ge1$. I tried using induction but it wouldn't work.
Attempt:
$$
\begin{align}
2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}&=(2\sqrt{n}-(\sqrt{n-1}+\sqrt{n+1}))\cdot {2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\\&={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\end{align}$$
(The calculations are true for sure. No check is desired.)
Denote
$$a_n=2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1}={2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})}\text{ and }b_n={1\over n^{2}}.$$
Then
$$\begin{align*}
\lim_{n\to \infty}{a_n\over b_n} & =\lim_{n\to \infty}n^{2}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=\lim_{n\to \infty}n^{2}\lim_{n\to \infty}{2n-2\sqrt{n-1}\cdot\sqrt{n+1}\over2\sqrt{n}+(\sqrt{n-1}+\sqrt{n+1})} \\
&=0
\end{align*}$$
By the comparison test for series convergence, since $\lim_{n\to \infty}{a_n\over b_n}=0$, then if $b_n$ converges, which it does, so does $a_n$.
| You don't have to see the formula. The $n$th term of the sequence is of the form $f(n+1) - 2f(n) + f(n-1)$, where $f(n) = -\sqrt{n}$.
This is a "difference of differences"; $f(n+1) - 2f(n) + f(n-1) = g(n+1) - g(n)$, where $g(n) = f(n) - f(n-1)$. So the infinite sum is telescoping and will equal $\lim_{n \rightarrow \infty} g(n) - g(1)$ once you show the limit exists.
I thought of a less tricky way to show this. Write
$$a_n = 2\sqrt{n}-\sqrt{n-1}-\sqrt{n+1} = \sqrt{n}\bigg(2 - \sqrt{1 - {1 \over n}} - \sqrt{1 + {1 \over n}}\bigg)$$
Use Taylor series on the square roots to get
$$a_n = \sqrt{n}\bigg[2 - \bigg(1 - {1 \over 2n} + O\bigg({1 \over n^2}\bigg)\bigg) - \bigg(1 + {1 \over 2n} + O\bigg({1 \over n^2}\bigg)\bigg)\bigg]$$
$$= \sqrt{n} O\bigg({1 \over n^2}\bigg)$$
$$= O\bigg({1 \over n^{3/2}}\bigg)$$
Thus the series converges in comparison with $\displaystyle{\sum_n {1 \over n^{3/2}}}$.
| {
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Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ where $f(n+1) = f(f(n))+1$ Consider checking function $\mathbb{N}\to \mathbb{N}$ relationship $f(n+1) = f(f(n))+1$, for any positive integer $n$.
Prove that $1\cdot f(1)+ 2\cdot f(2)+ ...+ n\cdot f(n) \leq n(n+1)(2n+1)/6$ for any positive integer $n$.
| Since $1\cdot 1+ 2\cdot 2+ \cdots + n\cdot n = n(n+1)(2n+1)/6$, it is enough to prove that $f(n)\le n$, for all $n$.
We shall prove that $f(n)\le n$ for all $n$, by full induction.
Assume $f(m)\le m$ for all $m\le n$. Let's prove that $f(n+1) \le n+1$.
Let $m=f(n) \le n$. Then $f(f(n))=f(m) \le m=f(n)$.
So, $f(n+1) = f(f(n))+1 \le f(n)+1 \le n+1$.
All that remains is the base of induction: $f(1)\le 1$.
| {
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How to find all integer solutions of $p^2+q^2=((2q+1)^2+q+1)^2+1$ $$p^2+q^2=((2q+1)^2+q+1)^2+1$$
How do I find integer solutions to this equation? I've already found $(p,q)=(11,1)$. How do I go about finding new ones?
| $$\begin{align}p^2+q^2&=((2q+1)^2+q+1)^2+1\\\implies p^2+q^2&=16q^4+40q^3+41q^2+20q+5 \\\implies p^2&=16q^4+40q^3+40q^2+20q+5\end{align}$$
So for an integer $q$ if $16q^4+40q^3+40q^2+20q+5$ is a perfect square, then you will get integer solutions for your equation.
*
*$q=1\implies16q^4+40q^3+40q^2+20q+5=121=11^2$
*$q=-1\implies16q^4+40q^3+40q^2+20q+5=1=1^2$
So $(11,1)$ and $(1,-1)$ are solutions, others may exist.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Measure of a modified Cantor set Suppose a modified Cantor set:
Starting with $E_0 = [0,1]$, we delete the middle interval of length $1/3$, then we delete the middle intervals of length $1/15$, and so on; in each step we delete from all the intervals of set $E_m$ the middle of length
$$\frac{1}{3\cdot5\cdot\ \space ... \space\cdot(2m+1) \cdot (2m+3)},$$ and we have to prove that the measure of the interesection is between $0$ and $1$.
I found the total length that we toss away but I am not able to bound the sum of the series. Any ideas? The series is actually:
$$\sum^\infty_{m=0}\frac{2^m}{3\cdot5\cdot\ \space ... \space\cdot(2m+1) \cdot (2m+3)}$$
Thank you in advance!
| How about this:
$$\sum_{n=0}^\infty \frac{2^n}{3 \cdot 5 \cdot \ldots \cdot (2n+3)} \ge 1/ 3 $$
For the opposite direction:
$$\sum_{n=0}^\infty \frac{2^n}{3 \cdot 5 \cdot \ldots \cdot (2n+3)} \le \sum_{n=0}^\infty \frac{2^n}{2 \cdot 4 \cdot \ldots \cdot (2n+2)} = \sum_{n=0}^\infty \frac{2^n}{2^{n+1} (n+1)!} = \frac 1 2 \sum_{n=0}^\infty \frac{1}{(n+1)!} \le \frac{e-1}{2} \approx 0.859141$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can we calculate the exponential form of a rotation matrix Considering the rotation matrix:
$$
A(\theta) = \left( \begin{array}{cc}
\cos\space\theta & -\sin \space\theta \\
\sin \space\theta & \cos\space\theta \\
\end{array} \right)
$$
How can I calculate $(A(\theta))^n$ where n ≥ 1 ? I'm not sure what to do nor how to write it.
| $A(\theta)$ is called the rotation matrix simply because it rotates a point in the plane by an angle $\theta$. What happens when I apply this matrix $n$ times? I simply rotate the point $n$ times. How much has it rotated in the end? By an angle $n\cdot\theta$. If you can get this sort of intuitive grasp, you can say the following:
Let $P_n$ be the property
$$[A(\theta)]^n=A(n\theta),\ \ n\ge 1$$
Let's now prove by induction on $n$ that $P_n$ is true.
Base case
$P_1$ is trivial, so let's look at $P_2$:
\begin{align}[A(\theta)]^2&=A(\theta)\cdot A(\theta)=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\\
&=\begin{pmatrix}\cos^2\theta-\sin^2\theta&-2\sin\cos\theta\\2\sin\cos\theta&\cos^2\theta-\sin^2\theta\end{pmatrix}\\
&=\begin{pmatrix}\cos(2\theta)&-\sin(2\theta)\\\sin(2\theta)&\cos(2\theta)\end{pmatrix}\\
&=A(2\theta)
\end{align}
Therefore $P_1$ and $P_2$ are true.
Inductive Step
Assume $P_n$ is true, and consider the case of $P_{n+1}$:
\begin{align}
[A(\theta)]^{n+1}&=[A(\theta)]^n\cdot A(\theta)=A(n\theta)\cdot A(n\theta)\\
&=\begin{pmatrix}\cos(n\theta)&-\sin(n\theta)\\\sin(n\theta)&\cos(n\theta)\end{pmatrix}\cdot\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\\
&=\begin{pmatrix}\cos(n\theta)\cos\theta-\sin(n\theta)\sin(\theta)&-\cos(n\theta)\sin\theta-\sin(n\theta)\cos\theta\\\sin(n\theta)\cos\theta+\cos(n\theta)\sin\theta&-\sin(n\theta)\sin\theta+\cos(n\theta)\cos\theta\end{pmatrix}
\end{align}
Now we use the following four trigonometric identities:
\begin{cases}\cos\theta\cos(n\theta)&=&\frac{1}{2}(\cos(\theta-n\theta)+\cos(\theta+n\theta))\\\sin\theta\sin(n\theta)&=&\frac{1}{2}(\cos(\theta-n\theta)-\cos(\theta+n\theta))\\
\cos\theta\sin(n\theta)&=&\frac{1}{2}(\sin(\theta+n\theta)-\sin(\theta-n\theta))\\
\sin\theta\cos(n\theta)&=&\frac{1}{2}(\sin(\theta+n\theta)+\sin(\theta-n\theta))\end{cases}
Which allow us to write:
$$[A(\theta)]^{n+1}=\begin{pmatrix}\cos(\theta+n\theta)&-\sin(\theta+n\theta)\\\sin(\theta+n\theta)&\cos(\theta+n\theta)\end{pmatrix}=A\left(\left(n+1\right)\theta\right)$$
Therefore $P_{n+1}$ is true.
Conclusion
We have shown that the property $P$ is true for $n=1$ (trivially; but also for $n=2$), and for $n+1$, if $P_n$ holds. Therefore, by the axiom of induction, $P_n$ is true for all $n\ge 1$.
$$[A(\theta)]^n=A(n\theta),\ \ n\ge 1\quad\Box$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Arc length contest! Minimize the arc length of $f(x)$ when given three conditions. Contest: Give an example of a continuous function $f$ that satisfies three conditions:
*
*$f(x) \geq 0$ on the interval $0\leq x\leq 1$;
*$f(0)=0$ and $f(1)=0$;
*the area bounded by the graph of $f$ and the $x$-axis between $x=0$ and $x=1$ is equal to $1$.
Compute the arc length, $L$, for the function $f$. The goal is to minimize $L$ given the three conditions above.
$\mathbf{\color{red}{\text{Contest results:}}}$
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{User} & {} & {} & \text{Arc length} \\ \hline
\text{1} & \text{robjohn $\blacklozenge$} & {} & {} & 2.78540 \\
\text{2} & \text{Glen O} & {} & {} & 2.78567 \\
\text{3} & \text{mickep} & {} & {} & 2.81108 \\
\text{4} & \text{mstrkrft} & {} & {} & 2.91946 \\
\text{5} & \text{MathNoob} & {} & {} & 3.00000 \\\hline
\text{-} & \text{xanthousphoenix} & {} & {} & 2.78540 \\
\text{-} & \text{Narasimham} & {} & {} & 2.78 \\
\end{array}$$
Original question after contest statement: The contest question was inspired by this paper. Can anyone come up with a different entry than those listed in the table below?
$$
\begin{array}{c|ll}
\hline
\text{Rank} & \text{Function} & {} & {} & \text{Arc length} \\ \hline
\text{1} & 1.10278[\sin(\pi x)]^{0.153764} & {} & {} & 2.78946 \\
\text{2} & (8/\pi)\sqrt{x-x^2} & {} & {} & 2.91902 \\
\text{3} & 1.716209468\sqrt{x}\,\mathrm{arccos}(x) & {} & {} & 2.91913 \\
\text{4} & (8/\pi)x\,\mathrm{arccos}(x) & {} & {} & 3.15180 \\
\text{5} & (15/4)x\sqrt{1-x} & {} & {} & 3.17617 \\
\text{6} & -4x\ln x & {} & {} & 3.21360 \\
\text{7} & 10x(1-\sqrt{x}) & {} & {} & 3.22108 \\
\text{8} & -6x^2+6x & {} & {} & 3.24903 \\
\text{9} & 9.1440276(2^x-x^2-1) & {} & {} & 3.25382 \\
\text{10} & (-12/5)(x^3+x^2-2x) & {} & {} & 3.27402 \\
\end{array}$$
| Consider the following:
$$f(x)=\begin{cases}
cx & \text{ if } 0 \le x \le \frac{1}{c} \\
-cx+c & \text{ if } 1-c < x \le 1 \\
1 & \text{ otherwise } \\
\end{cases}$$
If we take $c\to\infty$ we get that it is an arc length of 3.
| {
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"url": "https://math.stackexchange.com/questions/1122929",
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"source": "stackexchange",
"question_score": "65",
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Find all primes $p$ such that $p^2-p+1$ is a perfect cube Find all primes $p$ such that $p^2-p+1$ is a perfect cube.
I found out that p is of the form $18n+1$ and $p=19$ is a solution but I am not getting anything further.
$p^2-p-(m^3-1)=0$
$1+4(m^3-1)=k^2$
$4m^3-3=k^2$
every square is $0,1,4 or 7(\mod9)$ and every cube is $0,1 or 8\mod9$
Using this fact we can conclude that $m^3\equiv{1\mod 9}$
Hence $k$ is either $1$ or $-1$ $\mod9$
But $k\equiv{-1\mod9}$ is not possible since $p$ is a prime. So putting $k\equiv{1\mod9}$ in the quadratic formula we get $p\equiv{1\mod9}$ and since $p$ is odd(as $p=2$ is not a solution), $p\equiv{1\mod18}$
| this is an easier answer:
$$$$we have$$p^2-p=p(p-1)=m^3-1=(m-1)(m^2+m+1)$$we can see $p>m-1$ so we have $$p-1|m-1\to p=1\, MODm-1$$and we have$$m^2+m+1=3\, MODm-1$$we know $$p|m^2+m+1$$ so we have$$m^2+m+1=3p\, or\, m^2+m+1\ge (m+2)p$$if $$m^2+m+1=3p$$ we can see p=19,we have $3(m-1)=p-1$ so we have $m^2+m+1=9m-6$ so m=1 or
,m=7,p=19,$$$$if $$m^2+m+1\ge (m+2)p$$ we have$$(m+2)(m-1)\ge p-1$$ so $$m^2+m-1\ge p=(m^2+m+1)/m+2$$because $m>1$ it is clearly impossible.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$. Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$.
I came up with the formula $\displaystyle\sum_{r=1}^{2n} (-1)^{r+1}(2r-1)^2=-8n^2$ but I got stuck proving it by induction. Should I use another formula or is it correct and I should continue trying to prove it by induction?
| For $n=1$ we have:
$$1^2 - 3^2 = -8 = -8\cdot1^2$$
Now assume that the statement holds for some n. Then we have a sum:
$$1^2-3^2+...+ (4n-3)^2 - (4n - 1)^2= -8n^2$$
Adding the next two terms to both sides we get:
\begin{align}
&1^2-3^2+...+ (4n-3)^2 - (4n - 1)^2 + (4n + 1)^2 - (4n + 3)^2 \\
&= -8n^2 + (4n + 1)^2 - (4n + 3)^2\\
&=-8n^2+16n^2+8n+1-16n^2-24n-9 \\
&=-8n^2-16n-8 \\
&=-8(n+1)^2
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/1123189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Parametrization of solutions of diophantine equation $x^2 + y^2 = z^2 + w^2$ I need integer solutions of $x^2 + y^2 = z^2 + w^2$ parametrized. Can it be done? Thanks.
| \begin{align*}
\left(\dfrac{\alpha^2-\beta^2+\gamma^2-\delta^2}{\alpha^2+\beta^2-\gamma^2-\delta^2}\right)^2+\left(\dfrac{2\left(\alpha\beta-\gamma\delta\right)}{\alpha^2+\beta^2-\gamma^2-\delta^2}\right)^2-\left(\dfrac{2\left(\alpha\gamma-\beta\delta\right)}{\alpha^2+\beta^2-\gamma^2-\delta^2}\right)^2=\large{1}
\end{align*}
...
@Thomas Andrews
| {
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Evaluating $\int\frac{\mathrm dx}{\sqrt{\lfloor 1+ \sqrt{1+x}\rfloor}}$ How can I solve this integral?
$$\int\frac{\mathrm dx}{\sqrt{\lfloor 1+ \sqrt{1+x}\rfloor}}$$
| Notice for any integer $n \ge 2$ and $x \ge 0$,
$$\begin{align}
\left\lfloor 1 + \sqrt{1+x}\right\rfloor = n
&\iff n \le 1 + \sqrt{1+x} < n+1\\
&\iff x \in [ (n-1)^2 - 1, n^2 - 1 )
\end{align}$$
To perform the integral over some interval $[0,X]$ where $X > 0$, you just need to split the interval $[0,X]$ to intervals of the form $[(n-1)^2-1, n^2 - 1)$. Let $N = \left\lfloor 1 + \sqrt{1+X}\right\rfloor$, we have
$$\begin{align}
\int_0^X \frac{dx}{\sqrt{\left\lfloor 1 + \sqrt{1+x}\right\rfloor}}
&= \left( \sum_{n=2}^{N-1}\int_{(n-1)^2-1}^{n^2-1} + \int_{(N-1)^2-1}^X\right)
\frac{dx}{\sqrt{\left\lfloor 1 + \sqrt{1+x}\right\rfloor}}\\
&= \sum_{n=2}^{N-1}\frac{2n-1}{\sqrt{n}} + \frac{X - (N-1)^2 + 1}{\sqrt{N}}
\end{align}
$$
| {
"language": "en",
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"source": "stackexchange",
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Show that $f(x) = \frac{x^3}{1+x^2}$ is bijective This seems like a simple question, but I'm stuck: how do I show that $f: \mathbb{R} \rightarrow \mathbb{R}$ defined as $f(x) = \frac{x^3}{1+x^2}$ is bijective?
I want to demonstrate that it is both injective and surjective. To show that it's injective, I need to show that $f(x) = f(y)$ implies $x = y$. However, I can't see a way to reduce $\frac{x^3}{1+x^2} = \frac{y^3}{1+y^2}$ to $x = y$ (since there are no like terms to combine). I'm also unsure of how to prove surjectiveness.
| you don't need calculus to show that $f(x) = \dfrac{x^3}{1+x^2}$ is 1-1.
suppose $$\dfrac{a^3}{1+a^2} = \dfrac{b^3}{1+b^2} \tag 1$$ we will show that this implies $a = b$ proving $f$ is 1-1.
cleaning up $(1)$ gives $$(a-b)(a^2 + ab + b^2 + a^2b^2) = 0$$ now use the fact that $a^2 + ab + b^2 > 0$ for $a \neq 0, b \neq 0$ to conclude $a = b.$
$\bf Edit:$ To show that $f$ is onto note that $f(x) = \dfrac{x^3}{1+x^2}$ is odd and $\lim_{x \to \infty} \dfrac{x^3}{1+x^2} = \infty$. That is, the range of $f$ is $(-\infty, \infty).$
| {
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Integration problem: $\int x^{2} -x 4^{-x^{2}} dx$ I need to integrate $$\int x^{2} -x 4^{-x^{2}} dx$$
and I know the answer I got is wrong. However, I can't figure out where I went wrong. These steps I took:
Appreciate any help!
| $$\begin{align}
\int x^{2} - x4^{-x^{2}} dx &= \int x^{2} dx - \int x4^{-x^{2}} dx \\
&= \frac{x^{3}}{3} + \int -x4^{-x^{2}} dx \\
\end{align}$$
notice that
$$\frac{d}{dx} 4^{-x^{2}} = -2x\cdot \ln4 \cdot 4^{-x^{2}}$$
So if
$$\begin{align}
u &= 4^{-x^{2}} \\
\implies du &= -2x\cdot \ln4 \cdot 4^{-x^{2}} dx \\
\implies \frac{du}{2 \ln4} &= -x 4^{-x^{2}} dx
\end{align}$$
So our integral becomes
$$\begin{align}
\frac{x^{3}}{3} + \int -x4^{-x^{2}} dx &= \frac{x^{3}}{3} + \int \frac{du}{2 \ln4} \\
&= \frac{x^{3}}{3} + \frac{u}{2 \ln4} + C \\
&= \frac{x^{3}}{3} + \frac{4^{-x^{2}}}{2 \ln4} + C \\
\end{align}$$
| {
"language": "en",
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"source": "stackexchange",
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What does it mean to solve or find solutions in mathematics? Something that has been really confusing me lately is that this equation has four solutions $$3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)$$
But what does that mean? Until now solutions to me has meant, what are the coordinates of $x$ when $y$ equals a given value, normally $y=0$.
But this equation has kind of thrown me off because when plotting this equation on a graph I get only two points at which the line crosses the $x$-axis.
I hope someone understands my point.
Regards
| Definition: Solve. We say an equation is solved if and only if we have listed the set of all objects which make the equation true. We say we have found a solution if we have found an element of the solution set.
Example: If $x= 0$, then
\begin{align}
3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)
\end{align}
becomes
\begin{align}
0 = 0
\end{align}
which is true, and hence $x =0$ solves the equation.
Example: If $x =1$, then
\begin{align}
3x(x+1)(x^2+x+2)=16x(x+1)(2x+1)
\end{align}
becomes
\begin{align}
3\cdot 2 \cdot 4 = 16\cdot 2\cdot 3
\end{align}
which is false, and hence $x=1$ is not a solution.
| {
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"source": "stackexchange",
"question_score": "5",
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Series approximation to $\int_0^1\sqrt{\frac{2x+3}{2u^3-(2x+3)u^2+2x+1}}du$ I have figured out by graphing that, for small $x$:
$$
\int_0^1\sqrt{\frac{2x+3}{2u^3-(2x+3)u^2+2x+1}}du\approx\log(1/x)+\pi/2+O(x)
$$
However, I am unable to prove that this is the case.
As $x\to 0$ the integral diverges at $u=1$. We can check this by factoring the limit of the integrand:
$$
\sqrt{\frac{3}{1+2u}}\frac{1}{1-u}
$$
This looks promising, since the integral of $1/u$ is $\log u$. However, normally the integral converges: taking a series around $u=1$ we get:
$$
\sqrt{\frac{2x+3}{4x}\frac{1}{1-u}}+O(\sqrt{1-x})
$$
This means that when I take the series approximation of $x$ inside the integral, the resulting integral goes like $x^{-1/2}$, which is incorrect. What step am I missing?
| I think the resulting behavior is indeed a log as $x \to 0$. To see why, note that the denominator may be written as
$$f(u,x) = 2 u^3 - 3 u^2 + 1 + 2 x (1 - u^2)$$
This implies that $1-u$ is a factor of $f(u,x)$ independent of $x$. In fact,
$$f(u,x) = (1-u)(1+2 x + (1+2 x) u - 2 u^2) $$
The zeroes of the quadratic are at
$$u_{\pm} = \frac{1}{4} \left(1+2 x \pm \sqrt{9+20 x+4 x^2}\right)$$
The behavior of these zeroes as $x \to 0$ is
$$u_+ = 1+\frac{4 x}{3}+O\left(x^2\right) $$
$$u_- = -\frac{1}{2}-\frac{x}{3}+O\left(x^2\right) $$
Now let's turn our attention to the integral. Let's ignore the numerator, as it is merely a constant. What we have is, as $x \to 0$,
$$\frac1{\sqrt{2}} \int_0^1 \frac{du}{\sqrt{(1-u) \left (\frac{1}{2}+\frac{x}{3}+u \right ) \left (1+\frac{4 x}{3}-u \right )}} $$
Now sub $u=1-v^2$ in the integral. We get, as $x \to 0$ (we can neglect the $x/3$):
$$\sqrt{2} \int_0^1 \frac{dv}{\sqrt{ \left (\frac{3}{2}-v^2 \right )\left (\frac{4 x}{3}+v^2 \right )}} $$
The integral above is in fact an incomplete elliptic integral of the first kind (as defined by Wolfram Mathematica), equal to
$$\sqrt{\frac{3}{x}} \operatorname{F}{\left (\sin ^{-1}\left(\sqrt{\frac{2}{3}}\right),-\frac{9}{8 x} \right )} $$
The leading asymptotic behavior of this result as $x \to 0$, which I did not work out by hand, is
$$\sqrt{\frac{1}{3}} \left (\log{\frac1{x}}-\log {\left[\frac{1}{18} \left(2+\sqrt{3}\right)\right ]}\right ) $$
This seems to have the correct behavior when compared with the numerical integral for small $x$.
ADDENDUM
Interestingly enough, compare the result you proposed:
$$\log{\frac1{x}} + \frac{\pi}{2} \approx \log{\frac1{x}} +1.5708$$
with the analogous result resulting from my illustration:
$$\log{\frac1{x}}-\log {\left[\frac{1}{18} \left(2+\sqrt{3}\right)\right ]} \approx \log{\frac1{x}} +1.5734$$
so your hypothesis was not wildly off.
| {
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Find all continuous functions satisfying $\int_0^xf=(f(x))^2+C$ for some constant $C \neq 0$. Find all continuous functions $f$ satisfying
$$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.
I have a question about the solution to this problem. It says that
Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and
$$f(x)=2f(x)f'(x),$$
so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$
so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.
I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?
Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?
Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?
I would appreciate it if anyone answers my questions.
| putting $x = 0,$ gives $C = -f^2(0)$ differentiating gives you $f = 2ff'$ which has solution $f = 0$ which implies $C = 0$ or $f = \frac{x}{2} + f(0).$
$\int_0^x(f(0) + t/2) \, dt = xf(0) + \frac{x^2}{4}$ and $f^2(x) + C = \frac{x^2}{4} + xf(0) + f^2(0) - f^2(0) = xf(0) + \frac{x^2}{4}$ it verifies that
$$f(x) = \frac{x}{2} + f(0), f(0) \neq 0, C = -f^2(0).$$
| {
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Remainder when $x^{1000000}$ is divided by $x^3 + x +1$ in $\mathbb{Z}_2[x]$ I tried the traditional algorithm of long division hoping to find a pattern, but I was not able to.
I then tried using the root of $x^3 + x + 1$ $\left(x \sim -0.7\right)$ in the equation:
$$x^{1000000} = \left(x^3 + x + 1\right) q\left(x\right) + ax + b.$$
I have a horribly messy numerical result but I would like to solve this cleanly. How can I approach this?
| Just look at the low powers of $x$. Eventually they will cycle.
$x^0 = 1$
$x, x^2$ cannot be simplified, but
$x^3 = 0 - (x+1) = -x-1 = x+1$
$x^4 = x(x+1) = x^2+x$
$x^5 = x(x^2+x) = x^3+x^2 = x+1+x^2 = x^2+x+1$
$x^6 = x(x^2+x+1) = x^3+x^2+x = x+1+x^2+x = x^2+1$
$x^7 = x(x^2+1) = x^3+x = x+1+x = 1$.
So the powers of $x$ cycle with period $7$. You only need to consider the exponent of $x$ modulo $7$, then, and $x^{1000000} = x^{142857*7+1} = x^1 = x$.
| {
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Unknown Taylor expansion I have come across a few apparently related Taylor expansions, as detailed below:
\begin{align}
&\dots\frac{a^7}{140}-\frac{a^6}{80}-\frac{3 a^5}{40}-\frac{a^4}{8}+\frac{a^2}{2}+a+1&&=\exp \left(a-\frac{a^3}{6}\right)&\tag{1}\\
\\
&\dots\frac{11 a^7 b}{1120}+\frac{a^6 b}{32}+\frac{a^5 b}{40}-\frac{a^4 b}{16}-\frac{a^3 b}{4}-\frac{a^2 b}{4}+\frac{b}{2}&&=?&\tag{2}\\
\end{align}
The first was fairly easy to guess the closed form on the RHS, but the bottom one I am stuck on. The extra $b$ is obviously doing something to the coefficients, but I don't know what. Are there any good strategies for finding a closed form for $(2$)?
Unfortunately, I don't have any background information on them, so much is pure guesswork.
Update
All known coefficients for $(2)$:
$$\frac{1}{2},0,-\frac{1}{4},-\frac{1}{4},-\frac{1}{16},\frac{1}{40},\frac{1}{32},\frac{11}{1120},-\frac{1}{1280},-\frac{43}{20160}$$
| $$
\frac{1}{2} (1-a)\exp\left(a-\frac{a^3}{6}\right)=
\frac{1}{2}-\frac{a^2}{4}-\frac{a^3}{4}-\frac{a^4}{16}+\frac{a^5}{40}+\frac{a^6}{32}+\frac{11 a^7}{1120}-\frac{a^8}{1280}-\frac{43
a^9}{20160}-\frac{69 a^{10}}{89600}+O\left(a^{11}\right)
$$
| {
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Integrating $ \int \frac{1}{\sqrt{x^2+4}}\,dx $ using Trigonometric Substitution I'm reviewing integration by trigonometric substitution in anticipation of covering it in class next week. I seem to be a bit rusty and keep catching myself making various mistakes. On this particular problem I keep getting the same answer which is very close to being correct. However, I somehow end up dividing by two where I should not. I'm hoping another set of eyes can quickly set me right so I can stop frustrating myself reworking the problem to the same apparently wrong answer repeatedly! Thanks in advance!
The problem asks to solve:
$$
\int \frac{1}{\sqrt{x^2+4}}\,dx
$$
The answer is given as:
$$
ln\lvert x + \sqrt{x^2+4} \rvert + C
$$
Somehow I keep getting:
$$
ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C
$$
Here's my work:
$$
\int \frac{1}{\sqrt{x^2+4}}\,dx = \int \frac{1}{\sqrt{4(\frac{1}{4}x^2+1)}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{\frac{1}{4}x^2+1}}\,dx = \frac{1}{2}\int\frac{1}{\sqrt{(\frac{1}{2}x)^2+1}}\,dx
$$
At this point I substitute as follows:
$$
\frac{1}{2}x = \tan\theta
$$
$$
x = 2\tan\theta
$$
$$
dx = 2\sec^2\theta
$$
So I continue on with:
$$
\frac{1}{2}\int\frac{2\sec^2\theta}{\sqrt{\tan^2\theta+1}}\,d\theta = \int\frac{\sec^2\theta}{\sqrt{\sec^2\theta}}\,d\theta = \int\frac{\sec^2\theta}{\sec\theta}\,d\theta = \int\sec\theta\,d\theta = ln\lvert\sec\theta + \tan\theta\rvert + C
$$
Finally, to get the answer in terms of x I essentially draw a right triangle and use the fact that $\tan\theta = \frac{x}{2}$. The side opposite $\theta$ I take to be x, the side adjacent $\theta$ is 2, and the hypotenuse is $\sqrt{x^2+4}$. So $\sec\theta = \frac{\sqrt{x^2+4}}{2}$ and $\tan\theta = \frac{x}{2}$.
So, substituting these values back in, as mentioned, I end up with:
$$
ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C
$$
Can anyone help me see where I'm going wrong or failing to understand something?
| You are correct still. Notice that
$$
\begin{align}
\ln\Bigg|\frac{\sqrt{x^2+4}}{2} + \frac{x}{2}\Bigg| + C &= \ln\Bigg|\frac{\sqrt{x^2+4}+x}{2}\Bigg| + C\\
&= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|-\ln(2) + C\\
&= \ln\Bigg|\sqrt{x^2+4}+x\Bigg|+C'
\end{align}
$$
where $C'$ is still an arbitrary constant.
| {
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Calculate the trace of all elements in $F_8$ I got the following exercise where you have to calc the trace of all elements in ${F_8}$ which is constructed as ${F_2}[x]$/(${x^3+x+1}$)${F_2}[x]$.
Up to now I did those steps:
1) Find all elements in ${F_8}$ which are in my opinion: $0,1,x,x+1,x^2,x^2+1,x^2+x,x^2+x+1$
2) Then I found those traces for the elements:
$Tr(0)=0\\
Tr(1)=1\\
Tr(x)=x+x^2+x^4\\
Tr(x+1)=Tr(x)+Tr(1)=x^4+x^2+x+1\\
Tr(x^2)=x^8+x^4+x^2 \text{(must this be reduced!?)}\\
Tr(x^2+1)=Tr(x^2)+Tr(1)=x^8+x^4+x^2+1\\
Tr(x^2+x)=Tr(x^2)+Tr(x)=x^8+x\\
Tr(x^2+x+1)=Tr(x^2)+Tr(x)+Tr(1)=x^8+x+1$
Is this procedure correct? Thank you!
| We have here an extension of dimension three of the prime field $\;\Bbb F_2\;$ , and in fact
$$\Bbb F_{2^3}=\Bbb F_8\cong\Bbb F_2[x]/\langle x^3+x+1\rangle = \text{Span}_{\Bbb F_2}\{\overline 1,\overline x,\overline{x^2}\}\;$$
where for $\;v\in\Bbb F_2[x]\;,\;\;\overline v:=v+\langle x^3+x+1\rangle\in\Bbb F_2[x]/\langle x^3+x+1\rangle$. From now on we remove the lines over the elements and the meaning shall be clear from the context.
We thus have that
$$G:=\text{Aut}\,\left(\Bbb F_8/\Bbb F_2\right)=\{1=Id.\,,\,\sigma\,,\,\sigma^2\}\;,\;\;\sigma x:=x^2$$
So, for example:
$$\begin{align}&\text{Tr.} (x):=Id.x+\sigma x+\sigma^2 x=x+x^2+x^2+x=0\\
&\text{Tr.}(x+1)=x+1+x^2+1+x^2+x+1=1\\\end{align}$$
and etc. You can see in this case the trace is just the sum of the conjugates of each element.
| {
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When $\sqrt{(x+a)^2 -b}$ is an integer? While working on integer factorization problem, I came to this:
How to find for which values of $x$ the next equation is an integer?
$$\sqrt{(x+a)^2 -b}$$
*
*$a,b$ are positive known integers
In order to find a solution to this question, should I test all the values of $x$? Or can I simplify it some how?
| Let that integer be $y^2$ then $b=(x+a)^2-y^2=(x+a+y)(x+a-y)$. So, you only need to consider factorizations of $b=b_1b_2$ and the corresponding solutions of the system $$\begin{align}x+a+y&=b_1\\x+a-y&=b_2\end{align}$$
Example:
Take $a=0$, $b=3$. The factorizations of $b=3$ are $$\begin{align}1&\cdot3\\3&\cdot1\\(-1)&\cdot(-3)\\(-3)&\cdot(-1)\end{align}$$
The first system of equations to consider (the one for the first factorization) would be $$\begin{align}x+y&=1\\x-y&=3\end{align}$$ which has solutions $x=2$, $y=-1$.
The next system would be $$\begin{align}x+y&=3\\x-y&=1\end{align}$$ which has solutions $x=2$, $y=1$.
You can continue the rest of the cases.
Try $a=0$, $b=12$ for an example in which some factorizations give integer solutions and some others don't.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
Find the limit: $\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}}$
My attempt:
$\begin {align}\displaystyle\lim _{n\to \infty}\frac{\sqrt{n+\sqrt{n^2+1}}}{\sqrt{3n+1}} &=
\lim _{n\to \infty}\frac{\sqrt{n+n^4\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\ &=
\lim _{n\to \infty}\frac{n^2\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{n^2\sqrt{3+\frac{1}{n^2}}}\\&=
\lim_{n\to \infty}\frac{\sqrt{1+n^3\sqrt{1+\frac{1}{n^2}}}}{\sqrt{3+\frac{1}{n^2}}}\end{align}$
Which looks like is equal to $\infty$ because of the $n^3\cdot 1$ but it's wrong.
What am I doing wrong here?
| $$\lim_{n\to\infty}\frac{n+\sqrt{n^2+1}}{3n+1}=\lim_{n\to\infty}\frac{1+\sqrt{1+1/n^2}}{3+1/n}$$
$$=\frac{1+\sqrt1}3$$
| {
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summation of a trigonometric series How to evaluate $\tan^2(1) + \tan^2(3) + \tan^2(5) + \tan^2(7) + \ldots + \tan^2(89)$?
Angles are given in degrees.
I tried converting $\tan(89)$ as $\cot(1)$ and then tried combining $\tan(1)$ and $\cot(1)$, but later got stuck.
| For integer $n,\tan(2n+1)90^\circ=\tan90^\circ=\infty$
If $\tan90x=\infty,90x=180^\circ m+90^\circ=90^\circ(2m+1)$ where $m$ is any integer
$\implies x=(2m+1)^\circ$ where $0\le m\le89$
Like Trig sum: $\tan ^21^\circ+\tan ^22^\circ+...+\tan^2 89^\circ = ?$,
$$\tan90x=\frac{\binom{90}1t-\binom{90}3t^3+\cdots+\binom{90}{89}t^{89}}{\binom{90}0-\binom{90}2t^2+\cdots-\binom{90}{90}t^{90}}$$ where $t=\tan x$
For $\tan90x=\infty,$ $$\binom{90}0-\binom{90}2t^2+\cdots+\binom{90}{88}t^{88}-\binom{90}{90}t^{90}=0 \iff t^{90}-\binom{90}{88}t^{88}+\cdots-1=0$$
Now as $\tan(180^\circ-y)=-\tan y\implies\tan^2(180^\circ-y)=\tan^2y,$
So, the roots of $$u^{45}-\binom{90}{88}u^{44}+\cdots-1=0$$ are $\tan(2r+1)^\circ,0\le r\le 44$ or $45\le r\le89$
$$\implies\sum_{r=0}^{44}\tan^2(2r+1)^\circ=\frac{\binom{90}{88}}1=\binom{90}2=\cdots$$
| {
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How can I show that $\prod_{{n\geq1,\, n\neq k}} \left(1-\frac{k^{2}}{n^{2}}\right) = \frac{\left(-1\right)^{k-1}}{2}$? Assume $k$ positive integer. How can I show that
$$ \tag 1 \prod_{{n\geq1,\, n\neq k}} \left(1-\frac{k^{2}}{n^{2}}\right)
= \frac{\left(-1\right)^{k-1}}{2}? $$
I know that
$$ \tag 2 \underset{n\geq1}{\prod}\left(1-\frac{k^{2}}{n^{2}}\right)=\frac{\sin\left(\pi k\right)}{\pi k}$$
and so if $k$ is an integer the product is $0$ but how can I use these information?
| If $N \ge k+1$, then
\begin{align}
\prod_{\underset{n\neq k}{n = 1}}^N \left(1 - \frac{k^2}{n^2}\right)&= \prod_{n = 1}^{k-1}\left(1 - \frac{k^2}{n^2}\right)\prod_{n =k+ 1}^N \left(1 - \frac{k^2}{n^2}\right)\\
&= \prod_{n = 1}^{k-1} \frac{(n-k)(n+k)}{n^2} \prod_{n = k+1}^N \frac{(n-k)(n+k)}{n^2}\\
&= \frac{(-1)^{k-1}(k-1)!(k+1)\cdots (2k-1)}{[(k-1)!]^2}\cdot \frac{(N-k)!(2k+1)\cdots (N+k)}{[(k+1)(k+2)\cdots N]^2}\\
&= \frac{(-1)^{k-1}}{2}\cdot \frac{(N-k)!(k-1)!k(k+1)\cdots (2k-1)(2k)(2k+1)\cdots (N+k)}{[(k-1)!k(k+1)\cdots N]^2}\\
&= \frac{(-1)^{k-1}}{2}\cdot \frac{(N-k)!(N+k)!}{(N!)^2}\\
\end{align}
Since $\lim_{N\to \infty} (N-k)!(N+k)!/(N!)^2 = 1$, we deduce that $$\prod_{\underset{n\neq k}{n \geq 1}} \left(1 - \frac{k^2}{n^2}\right) = \frac{(-1)^{k-1}}{2}.$$
| {
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How to prove that $\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$ How to prove this trigonometric identities ?
$$\cot{\frac{11\pi}{18}} +\cot{\frac{2\pi}{9}}=4\sin{\frac{\pi}{18}}\cot{\frac{2\pi}{9}}$$
Thank you.
| We have to check:
$$ -\tan\frac{\pi}{9}+\cot\frac{2\pi}{9}=4\cos\frac{4\pi}{9}\cot\frac{2\pi}{9} $$
hence, by multiplying both sides by $\sin\frac{2\pi}{9}=2\sin\frac{\pi}{9}\cos\frac{\pi}{9}$:
$$ -2\sin^2\frac{\pi}{9}+\cos\frac{2\pi}{9} = 4\cos\frac{4\pi}{9}\cos\frac{2\pi}{9} $$
so we just have to check that $x=\cos\frac{2\pi}{9}$ is a solution of:
$$ 2x-1 = 4x(2x^2-1) $$
or a root of:
$$ p(x) = 8x^3-6x+1 = 2\cdot T_3(x)-1, \tag{1}$$
where $T_3(x)=4x^3-3x$ is a Chebyshev polynomial. To check that $\cos\frac{2\pi}{9}$ is a root of the RHS of $(1)$ is trivial, since $T_3(\cos\alpha)=\cos(3\alpha)$:
$$ 2\cdot T_3\left(\cos\frac{2\pi}{9}\right)-1 = 2\cdot\cos\frac{2\pi}{3}-1 = 0, $$
hence we're done.
| {
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How to calculate: $\lim \limits_{x \to 0}$ $\frac{-1}{\sin^2x}$ + $\frac{1}{x^2}$ How to calculate: $$\lim \limits_{x \to 0}\frac{1}{x^2} - \frac{1}{\sin^2x} $$
any suggestions what I can do here?
| $$-\frac{1}{\sin(x)^2} + \frac{1}{x^2} = \frac{-x^2+\sin(x)^2}{x^2 \sin(x)^2}.$$
You can use L'Hopital's rule now, but it will take multiple applications since both the numerator and denominator are very small (of order $x^4$, it turns out) near $0$.
A nicer approach is to use the Maclaurin series for $\sin$ to see that
$$\sin(x)^2=(x-x^3/6+O(x^5))(x-x^3/6+O(x^5))=x^2-x^4/3+O(x^6).$$
Consequently
$$-x^2+\sin(x)^2=-x^4/3+O(x^6)$$
and
$$x^2\sin(x)^2=x^4+O(x^6).$$
Now take the ratio and cancel a factor $x^4$ from top and bottom to compute the limit.
| {
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"timestamp": "2023-03-29T00:00:00",
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Induction exercise check-up Prove by induction on $n$ that $13$ divides $2^{4n+2} + 3^{n+2}$ for all natural $n$.
For base case it is divisble by 13, and
$2^{4n+6} + 3^{n+3}$ must be divisble too.
$16 * 2^{4n+2}+ 3* 3^{n+2}$
If we denote $2^{4n+2} + 3^{n+2}$ as some $13y$,
we have $3*13y + 13*2^{4n+2}$
Didn't I make a mistake in assuming that $2^{4n+2} + 3^{n+2}$ is divisble by 13? In induction we have to prove that if something is divisible/whatever by something then n+1 is divisible too which I did
| your assumption is necessary in the induction
the steps of induction is :
1. calculate when n = 1 it is true or not
2. assume when n = k is true then check whether n = k + 1 is also true.
*
*for n = 1,
$2^{4n+2}+3^{n+2} = 2^6+3^3 = 7*13$
*assume $2^{4k+2}+3^{k+2}$ is divisible then
$2^{4(k+1)+2}+3^{(k+1)+2}$
$=2^{4k+6}+3^{k+3}$
$=16*2^{4k+2}+3*3^{k+2}$
$=3*(2^{4k+2}+3^{k+2})+13*2^{4k+2}$
$=13x$ for some real x
So for all natural n, $2^{4k+2}+3^{k+2}$ is divisible by 13. Q.E.D.
| {
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How to prove $\sum_{k=0}^{\infty}k^2x^{k} = \frac{x(1+x)}{(1-x)^3}\text{, }|x| < 1$? How do I prove that the summation $$\sum_{k=0}^{\infty}k^2x^{k} = \dfrac{x(1+x)}{(1-x)^3}\text{, }|x| < 1\text{?}$$
| Write
$$
\frac{x(1+x)}{(1-x)^3}=\frac{x^2-2x+1+3x-1}{(1-x)^3}=\frac{1}{1-x}+3x\frac{1}{(1-x)^3}-\frac{1}{(1-x)^3}.
$$
Let's begin with
$$
\frac{1}{1-x}=\sum_{k=0}^\infty x^k.
$$
Differentiate $2$ times, which gives
$$
\frac{2}{(1-x)^3}=\sum_{k=2}^\infty k(k-1)x^{k-2}=\sum_{k=0}^\infty(k+2)(k+1)x^k.
$$
Thus, your RHS equals to
$$
S\equiv\sum_{k=0}^\infty x^k+\frac{3x}{2}\sum_{k=0}^\infty(k+2)(k+1)x^k-\frac{1}{2}\sum_{k=0}^\infty(k+2)(k+1)x^k.
$$
The coefficient corresponding to $x^0$ is $1-\frac{1}{2}2\times 1=0=0^2$ (i.e. the middle expression above does not contribute.). The coefficient corresponding to $x^k$ for $k\geq 1$ is
$$
1+\frac{3}{2}(k+1)k-\frac{1}{2}(k+2)(k+1)=k^2.
$$
The claim follows.
| {
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Rationalize the denominator: $(\frac{3}{2x^2}) ^{1/4}$ My answer is $\frac{(6x^2)^{1/4}}{ 2x^2}$
However the book's answer is $\frac{(24x^2)^{1/4}}{2x}$
Where did the book get that from?
Here's my work:
$$\sqrt[4]{\frac{3}{2x^2}}= \frac{\sqrt[4]{3}}{\sqrt[4]{2x^2}}\cdot \frac{\sqrt[4]{2x^2}}{\sqrt[4]{2x^2}}$$
$$= \frac{\sqrt[4]{6x^2}}{2x^2}$$
Since the product of two roots with the same index is the root of the product:
$$\sqrt[N]{a} \cdot \sqrt[N]{b} = \sqrt[N]{ab}$$
| Your error is when you go from the first line to the second. We have
$$\sqrt[4]{\frac{3}{2x^2}}= \frac{\sqrt[4]{3}}{\sqrt[4]{2x^2}}\cdot \frac{\sqrt[4]{2x^2}}{\sqrt[4]{2x^2}} = \frac{\sqrt[4]{6x^2}}{\sqrt[4]{4x^4}}$$
You can re-express the denominator as $\sqrt[4]{4}\sqrt[4]{x^4} = |x|\sqrt[4]{4}$, which is almost free of radicals (see my note below for where the absolute value came from). Can you take it from here?
Looking at your work, you seem to be using the (erroneous) rule $\sqrt[N]{a} \cdot \sqrt[N]{a} = a$. This is true for $N=2$, but not otherwise (this problem has $N=4$).
Edit: Technically, $\sqrt[4]{x^4} = x$ only holds if $x \ge 0$; the more general solution is $\sqrt[4]{x^4} = |x|$, since $\sqrt[4]{\cdot} \ge 0$. You may have been given that $x>0$ for the problem, at which point $|x| = x$ and there is no difference, but it's a good idea to keep this fact in mind.
| {
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Factoring Polynomials (x^4) - using completing squares
Exercise 6. By viewing the polynomials as a difference of two squares, factorise the following polymomials.
*
*$x^4+x^2+1$,
*$x^4+3x^2+4$,
*$x^4+4$.
To solve difference of two squares isnt it necessary to have negatives? im confused about how to start solving these questions. I cant seem to factorise by grouping either /:
Information on how to factorise these types of questions would be appreciated. thank You!
| *
*$x^4+x^2+1= (x^2)^2+(x^2)\cdot 1+1^2=(x^2)^2+2\cdot (x^2)\cdot 1+1^2- (x^2)\cdot 1=(x^2+1)^2-x^2=(x^2+1-x)(x^2+1+x)=(x^2+x+1)(x^2-x+1).$
*$x^4+3x^2+4= (x^2)^2+3\cdot (x^2)+2^2=(x^2)^2+4\cdot (x^2)+2^2-x^2=(x^2)^2+2\cdot 2\cdot x^2+2^2-x^2=(x^2+2)^2-x^2=(x^2+2-x)(x^2+2+x)=(x^2+x+2)(x^2-x+2).$
*$x^4+4=(x^2)^2+2^2 =(x^2)^2+2\cdot 2\cdot x^2+2^2-4x^2=(x^2+2)^2-(2x)^2=(x^2-2x+2)(x^2+2x+2).$
The idea is that if you have something like: $a^2+b^2+kab$ you can add and subtract something which in your case gives you a square (the exercise is of such a nature). $a^2+b^2+2ab+(k-2)ab=a^2+b^2+2ab-(2-k)ab=(a+b)^2-(2-k)ab$.
In your examples:
*
*$a=x^2, b=1, k=1$
*$a=x^2, b=2, k=1$
*$a=x^2, b=2, k=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$
Prove that the following inequality
$$\sqrt{a^2+(1-b)^2}+\sqrt{b^2+(1-c)^2}+\sqrt{c^2+(1-a)^2}\geq\frac{3\sqrt{2}}{2}$$
holds for arbitrary real numbers $a$, $b$ and $c.$
Someone says, "It's very easy problem. It can also be proved by AM–GM inequality." But I can't reason out the answer to this question.
| Hint:
$$\sqrt{\frac{a^2+(1-b)^2}{2}}\ge \frac{|a| + |1-b|}{2}$$
Add up, RHS $\ge \frac{|a|+|1-a|+|b|+|1-b|+|c|+|1-c|}{2}\ge \frac{3}{2}$, on LHS it's our sum divided by $\sqrt{2}$.
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