Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How find this sum $\sum_{k=1}^{n}\sqrt[3]{a_{k}}$ let sequence $\{a_{n}\}$ such
$$(6k^2+2-a_{k})^3=27(4k^6+3k^4+3k^2-1)a_{k}$$
for all $k\ge 1$, then compute
$$\sum_{k=1}^{n}\sqrt[3]{a_{k}}$$
My try: I have note
$$(4k^6+3k^4+3k^2-1)=(2k-1)(2k+1)(k^2-k+1)(k^2+k+1)$$
then
$$6k^2+2-a_{k}=3\sqrt[3]{(2k-1)(2k+1)(k^2-k+1)(k^2+k+1)a_{k}}$$
Then I can't.Thank you very much
| This looks rather intractable. It may help to know that:
$$a_k = 2 + 6 k^2 \\ +
3 \left(1 - 3 k + 7 k^3 - 12 k^4 + 15 k^5 - 13 k^6 + 18 k^7 - 12 k^8 + 8 k^9\right)^{1/3} \\+
3 \left(1 + 3 k - 7 k^3 - 12 k^4 - 15 k^5 - 13 k^6 - 18 k^7 - 12 k^8 - 8 k^9\right)^{1/3}$$
And so, developing $\sqrt[3]{a_k}$ into a power series, you can show that for large enough $k$:
$$\sqrt[3]{a_k} \approx 2^{1/3} - 2^{5/3} k^{-2}$$
So for large enough $n$, your sum should be pretty close to:
$$\sum_{k=1}^{n}\sqrt[3]{a_k}\approx2^{-5/3}(4 n-H_{n,2})$$
Of course, you could compute higher orders of the series if you really need the accuracy.
| {
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Sums of central binomial coefficients Are there closed forms for $$\sum^n_{i=0} \binom{2i}{i}$$ and $$\sum^n_{i=0} \binom{2i}{i}^2$$?
Also, how can these sums be approximated?
| What follows is unfortunately not very rigorous, but the approximations obtained do agree with the sums numerically.
We first divide out the last term of the sum, which contains most of its bulk, to get
$$
\sum^n_{i=0} \binom{2i}{i} = \binom{2n}{n} \sum^n_{k=0} \frac{\binom{2(n-k)}{n-k}}{\binom{2n}{n}}.
$$
For fixed $k$ one can verify (using series expansion) that
$$
\frac{\binom{2(n-k)}{n-k}}{\binom{2n}{n}} = 4^{-k} + \frac{k4^{-k}}{2n} + O\left(\frac{1}{n^2}\right).
$$
We might then expect that
$$
\begin{align}
\binom{2n}{n} \sum^n_{k=0} \frac{\binom{2(n-k)}{n-k}}{\binom{2n}{n}} &= \binom{2n}{n}\left[\sum^n_{k=0} 4^{-k} + \frac{1}{2n} \sum^n_{k=0} k4^{-k} + O\left(\frac{1}{n^2}\right)\right] \\
&= \binom{2n}{n}\left[\sum_{k=0}^{\infty} 4^{-k} + \frac{1}{2n} \sum_{k=0}^{\infty} k4^{-k} + O\left(\frac{1}{n^2}\right)\right] \\
&= \binom{2n}{n}\left[\frac{4}{3} + \frac{2}{9n} + O\left(\frac{1}{n^2}\right)\right] \\
&= \frac{4}{3} \binom{2n}{n}\left[1 + \frac{1}{6n} + O\left(\frac{1}{n^2}\right)\right],
\end{align}
$$
where the tails of the sum have been absorbed into the $O(1/n^2)$.
Similarly,
$$
\sum_{i=0}^{n} \binom{2i}{i}^2 = \binom{2n}{n}^2 \sum_{k=0}^{n} \frac{\binom{2(n-k)}{n-k}^2}{\binom{2n}{n}^2},
$$
and
$$
\frac{\binom{2(n-k)}{n-k}^2}{\binom{2n}{n}^2} = 16^{-k} + \frac{k16^{-k}}{n} + O\left(\frac{1}{n^2}\right),
$$
so
$$
\begin{align}
\binom{2n}{n}^2 \sum_{k=0}^{n} \frac{\binom{2(n-k)}{n-k}^2}{\binom{2n}{n}^2} &= \binom{2n}{n}^2\left[\sum_{k=0}^{\infty} 16^{-k} + \frac{1}{n} \sum_{k=0}^{\infty} k16^{-k} + O\left(\frac{1}{n^2}\right)\right] \\
&= \binom{2n}{n}^2\left[\frac{16}{15} + \frac{16}{225n} + O\left(\frac{1}{n^2}\right)\right] \\
&= \frac{16}{15} \binom{2n}{n}^2\left[1 + \frac{1}{15n} + O\left(\frac{1}{n^2}\right)\right].
\end{align}
$$
So the approximations we obtain are
$$
\sum_{i=0}^{n} \binom{2i}{i} = \frac{4}{3} \binom{2n}{n}\left[1 + \frac{1}{6n} + O\left(\frac{1}{n^2}\right)\right]
$$
and
$$
\sum_{i=0}^{n} \binom{2i}{i}^2 = \frac{16}{15} \binom{2n}{n}^2\left[1 + \frac{1}{15n} + O\left(\frac{1}{n^2}\right)\right]
$$
as $n \to \infty$.
| {
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Show that for any $n \in \mathbb{Z}, n^3$ is congruent to 0,1,-1 modulo 9. Having a little difficulties with this one. Tried thinking of going down the line of even/odd proofs, but couldn't get anywhere.
| Modulo 9 table:
$x\equiv 0 \bmod 9 \rightarrow x^2\equiv 0 \bmod 9 \rightarrow x^3\equiv 0 \bmod9$
$x\equiv 1 \bmod 9\rightarrow x^2\equiv 1 \bmod 9 \rightarrow x^3\equiv 1 \bmod9$
$x\equiv 2 \bmod 9\rightarrow x^2\equiv 4 \bmod 9 \rightarrow x^3\equiv 8 \bmod9$
$x\equiv 3 \bmod 9\rightarrow x^2\equiv 0 \bmod 9 \rightarrow x^3\equiv 0 \bmod9$
$x\equiv 4 \bmod 9\rightarrow x^2\equiv 7 \bmod 9 \rightarrow x^3\equiv 1 \bmod9$
$x\equiv 5 \bmod 9\rightarrow x^2\equiv 7 \bmod 9 \rightarrow x^3\equiv 8 \bmod9$
$x\equiv 6 \bmod 9\rightarrow x^2\equiv 0 \bmod 9 \rightarrow x^3\equiv 0 \bmod9$
$x\equiv 7 \bmod 9\rightarrow x^2\equiv 4 \bmod 9 \rightarrow x^3\equiv 1 \bmod9$
$x\equiv 8 \bmod 9\rightarrow x^2\equiv 1 \bmod 9 \rightarrow x^3\equiv 8 \bmod9$
alternatively:
$(3k)^3=27k^3$
$(3k+1)^3=27k^3+3*(9k^2)+3(3k)+1=9(3k^3+3k^2+k)+1,$
$(3k+2)^3=3k^3+3(2*9k^2)+3(4*3k)+8=9(3k^3+6k^2+4k)+8$
The modulo table allways works for proving these things, but the other one is nicer. since $8\equiv -1 \bmod9$ you do get what you want.
| {
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Prove that if $2^{4\times5^k}=x\times5^{k+3}+a,0Let $$2^{4\times5^k}\equiv a \pmod {5^{k+3}},\\2^{4\times5^k}\equiv b \pmod {5^{k+4}},$$
and $0<a<5^{k+3},0<b<5^{k+4},$ prove that $a=b.$$(k>1)$
This is equivalent to this: if $2^{4\times5^k}=x\times5^{k+3}+a,0<a<5^{k+3},$ then $5\mid x.$
ADD: A similar problem:
Prove that if $2^{2\times5^k}=x\times5^{k+4}+a,0<a<5^{k+4},$ then $5\mid x.(k>2)$
| Using this, if ord$\displaystyle _{(p^k)}a = d$ where k is a natural number and $p$ odd prime,
we can show that ord$_{(p^{k+1})}a = d$ or $pd$
Now, $\displaystyle 2^2\equiv-1\pmod5\implies 2^4\equiv1\pmod5\implies$ord$_52=4$
$\displaystyle\implies$ord$_{(5^2)}2=4$ or $4\cdot5=20$ which $=\phi(25)$
Now, $\displaystyle2^4=16\not\equiv1\pmod{25}\implies$ ord$_{(5^2)}2=20$
So, $2$ is a primitive root of $25$
using this, $2$ is a primitive root of $5^k$ for integer $k\ge1$
$\displaystyle\implies2^{4\cdot5^k}\equiv1\pmod{5^{k+1}}\equiv1+c\cdot5^{k+1}\pmod{5^r}$ for integer $r\ge k+1$
where $c$ is some integer not divisible by $5$ as $\displaystyle2^{4\cdot5^k}\not\equiv1\pmod{5^r}$ where $r\ge k+1$
| {
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Indefinite intergral of $\int { \sqrt{ x^2-a^2} \over x } \mathrm dx $ I need to integrate $$\int { \sqrt{x^2-a^2} \over x } \mathrm dx $$ using substition, and show it equals $$ \sqrt{x^2 - a^2} - a(\operatorname{arcsec} ({x\over a })) +c $$
I've tried $x=a\sin t$ but I couldn't finish it out.
Thank in advance for any help.
| Calculating this integral can be done in this way:
$$\int { \sqrt{x^2-a^2} \over x } dx =\int { x^2-a^2 \over x\sqrt{x^2-a^2} } dx= \int { x \over \sqrt{x^2-a^2} } dx-\int { a^2 \over x\sqrt{x^2-a^2} } dx =\sqrt{x^2-a^2}(+-)a\cdot\int {(\frac{x}{a})'\over \sqrt{1-(\frac{x}{a})^2)} } dx=\sqrt{x^2-a^2}(+-)a\cdot\arcsin(\frac{x}{a})+C.$$
| {
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How prove this inequality $\sum\limits_{cyc}\frac{x^2}{(2x+y)(2x+z)}\le\frac{1}{3}$ let $x,y,z>0$,show that
$$\dfrac{x^2}{(2x+y)(2x+z)}+\dfrac{y^2}{(2y+z)(2y+x)}+\dfrac{z^2}{(2z+x)(2z+y)}\le\dfrac{1}{3}$$
My try: $$\Longleftrightarrow\sum_{cyc}\dfrac{4x^2}{4x^2+2x(y+z)+yz}\le\dfrac{4}{3}$$
$$\Longleftrightarrow\sum_{cyc}\left(1-\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\right)\le\dfrac{4}{3} $$
$$\Longleftrightarrow \sum_{cyc}\dfrac{2x(y+z)+yz}{4x^2+2x(y+z)+yz}\ge\dfrac{5}{3}$$
then I can't.Thank you
| Let $x\geq y\ge z$.
Hence,
$$\frac{1}{3}-\sum_{cyc}\frac{x^2}{(2x+y)(2x+z)}=\sum_{cyc}\left(\frac{x}{3(x+y+z)}-\frac{x^2}{(2x+y)(2x+z)}\right)=$$
$$=\sum_{cyc}\frac{x(x-y)(x-z)}{3(x+y+z)(2x+y)(2x+z)}\geq$$
$$\geq\frac{x(x-y)(x-z)}{3(x+y+z)(2x+y)(2x+z)}+\frac{y(y-x)(y-z)}{3(x+y+z)(2y+x)(2y+z)}=$$
$$=\frac{(x-y)^2(2xy(x+y)+(x^2+5xy+y^2)z-(x+y)z^2)}{3(x+y+z)(2x+y)(2x+z)(2y+x)(2y+z)}\geq0.$$
Done!
| {
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Probability - rolling two dices and flipping one coin two times Two six sided fair dice are rolling. If sum of fallen numbers is less than $5$, than coin is flipping two times. Random variable $X$ represents sum of fallen numbers, $Y$ represents number of fallen tails. Find distribution of random variable $(X, Y)$. What is for example $P(X=2, Y=0) = ?$ I little confused, whenever I think of something, sum of all probabilities is less than $1$.
I made table, where $X = \{2,3,4,5,6,7,8,9,10,11,12\}$ $Y=\{0,1,2\}$ Than I filled it like I already told you: $P(X=2, Y=0) = \dfrac{1}{36}\cdot \dfrac{1}{4}$, $P(X=2, Y=1) = \dfrac{1}{36} \cdot \dfrac{2}{4}$, and so on. For $X \gt 5$, there are all zeros.
When I put probabilities for $X$, when $X\gt5$, sum of table is $2.44$ :( But isn't it logic that probabilities when $X \gt 5 = 0$, because you multiply with probability of getting Tails which is $0$ ?
Also, when I assume that if $X\gt5$ coin is flipped $1$ time, I get sum $0.625$, which again is not right.
| The sum of all probabilities should be $1$.
If $X \geq 5$, then $Y = 0$ and so $X + Y = X$.
For example, if $X = 5$, then $Y = 0$, and we get $P(5,0) = (2/36)(1) = 1/18$.
Note also that $P(5, 1) = P(5, 2) = 0$.
For $X < 5$, you have:
$P(2, 0) = (1/36)(1/4), P(2, 1) = (1/36)(1/2), P(2, 2) = (1/36)(1/4)$.
Note the above line sums to $P(2,Y) = 1/36$.
$P(3, 0) = (2/36)(1/4), P(3, 1) = (2/36)(1/2), P(3, 2) = (2/36)(1/4)$.
Note the above line sums to $P(3, Y) = 2/36 = 1/18$.
$P(4, 0) = (3/36)(1/4), P(4, 1) = (3/36)(1/2), P(4, 2) = (3/36)(1/4)$.
Note the above line sums to $P(4, Y) = 3/36 = 1/12$.
If you let $Q(X)$ be the probability that you sum to $X$ after two rolls, then:
$P(2,0) + P(2,1) + P(2,2) = Q(2)$.
Similarly, $P(3,0) + P(3,1) + P(3,2) = Q(3)$ and $P(4,0) + P(4,1) + P(4,2) = Q(4)$.
For $X \geq 5$, we have $Q(X) = P(X,0) + P(X,1) + P(X,2) = P(X,0)$.
Thus, when you sum up all $P(X,Y)$ you get the same total as summing $Q(X)$, which is $1$.
I'm not quite sure how your $2.44$ appeared.
| {
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Factorizing $(x-1)(x-3)(x-5)(x-7)-64$ We need to factorize:
$$(x-1)(x-3)(x-5)(x-7)-64$$
We can, by the rational root theorem, see that there are no roots of this polynomial.Next observation is that $64=(8)^2$.
So this means that if the first part of the polynomial is a square,we can rewrite the whole polynomial as the difference of two squares.But it turns out that the first part of the polynomial is not a square. However,we can note that,
$$(x-1)(x-7)=(x^2)-8x+7$$
$$(x-3)(x-5)=(x^2)-8x+15$$
Therefore,letting $(x^2)-8x+7=p$,we can rewrite the given polynomial as
$$p(p+8)-64=p^2+8p-64$$
which I thought would be factorizable, but I seem to be wrong. So how can I factorize this polynomial? I would appreciate a small hint.
EDIT: The original polynomial seems to have been
$$(x-1)(x-3)(x-5)(x-7)-65$$
But there is no point in changing the whole question.To factorize this polynomial,we would do the exact same things as before and find that
$$(p^2)+8p-65=(p^2)+13p-5p-65=(p+13)(p-5)=(x^2-8x+20)(x^2-8x+2)$$
| $\left(x-1\right)\left(x-7\right)=y-9$ and $\left(x-3\right)\left(x-5\right)=y-1$
for $y=\left(x-4\right)^{2}$ leading to a factorization of $\left(y-1\right)\left(y-9\right)-64=y^{2}-10y-55$
Here:
$y^{2}-10y-55=\left(y-5-4\sqrt{5}\right)\left(y-5+4\sqrt{5}\right)=\left(\left(x-4\right)^{2}-5-4\sqrt{5}\right)\left(\left(x-4\right)^{2}-5+4\sqrt{5}\right)=\left(x-4+\sqrt{5+4\sqrt{5}}\right)\left(x-4-\sqrt{5+4\sqrt{5}}\right)\left(x-4+i\sqrt{4\sqrt{5}-5}\right)\left(x-4-i\sqrt{4\sqrt{5}-5}\right)$
| {
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How to integrate $\int\frac{\sqrt{1-x}}{\sqrt{x}}\ \mathrm dx$ I'm having a bit of trouble solving this integral: $$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx$$
Here is my attempt at a solution:
I multiplied the numerator and the denominator of $\frac{\sqrt{1-x}}{\sqrt{x}}$ by $\sqrt{x}$, yielding $$\int\frac{\sqrt{x-x^2}}{x}dx.$$
Further simplification resulted in $$\int\frac{\sqrt{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}}{x}dx.$$
Using trigonometric substitution, I set $$x-\frac{1}{2}=\frac{1}{2}\sin\theta$$ and solving for the differential $dx$ got $$dx=\frac{1}{2}\cos\theta.$$ Substituting this all back into $\int\frac{\sqrt{\frac{1}{4}-(x-\frac{1}{2})^2}}{x}dx$ (and some simplification later) yielded $$\frac{1}{2}\int{\frac{\cos^2\theta}{\sin\theta+1}}d\theta.$$ By substituting $1-\sin^2\theta$ for $\cos^2\theta$ I obtained $$\frac{1}{2}\int{\frac{1}{\sin\theta+1}-\frac{\sin^2\theta}{\sin\theta+1}d\theta}.$$ The issue I'm having is trying to solve this resultant integral. If there is an easier method to solve the problem, that would be graciously accepted.
| Integrate by parts
$$\int\frac{\sqrt{1-x}}{\sqrt{x}}dx
=\sqrt x \sqrt{1-x}-\int \frac{d(\sqrt x)}{\sqrt{1-x}}
= \sqrt{x(1-x)}-\sin^{-1}\sqrt x$$
| {
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Proving $\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$ for all $n \ge 1$.
$$\frac{(n+1)^4}{4}+(n+1)^3\le\frac{(n+2)^4}{4}$$
For all $n\ge 1$. I thought that I could get rid of the denominators like this:
$$(n+1)^4+4(n+1)^3\le(n+2)^4$$
Then, maybe, take $(n+1)^3$ as common factor:
$$(n+1)^3\cdot((n+1)+4)\le(n+2)^4$$
$$(n+1)^3\cdot(n+5)\le(n+2)^4$$
But I get the feeling that I'm just getting stuck. How can I further prove this?
This came up because I was doing an induction exercise. I needed to prove for all $n \ge 1$:
$$1^3+2^3+3^3+...+n^3\le \frac{(n+1)^4}{4}$$
By the hypothesis I know that
$$1^3+2^3+3^3+...+n^3+(n+1)^3\le \frac{(n+1)^4}{4}+(n+1)^3$$
So what I finally need to prove is that
$$\frac{(n+1)^4}{4}+(n+1)^3 \le \frac{(n+2)^4}{4}$$
| $(a+b)^4=(a^2+2ab+b^2)^2=a^2+4a^2b^2+b^4+4a^3b+4ab^3+2a^2b^2$
$$\frac{(n+2)^4}{4}=\frac{(n+1+1)^4}{4}=\frac{(n+1)^4}{4}+(n+1)^2+\frac{1}{4}+(n+1)^3+(n+1)+\frac{1}{2}(n+1)^2$$
$$\frac{(n+2)^4}{4}=\frac{(n+1)^4}{4}+(n+1)^3+\text { some thing which is positive}$$
So, $$\frac{(n+2)^4}{4}\geq\frac{(n+1)^4}{4}+(n+1)^3 \text{ for all $n\in \mathbb{N}$}$$
| {
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Can Master Theorem be applied on any of these? 1) $T(n) = 6T(n/2) + 2^{3 \log(n)}$
2) $T(n) = 8T(n/2) + \frac{n^3}{(\log(n))^4}$
3) $T(n) = 9T(n/3) + n(\log(n))^3$
Can the complexity for these be calculated with the Master Theorem?
I am not sure how to decide upon which case they fit in.
| For your second problem, lets solve the recurrence
$$T(n) = 8 T(\lfloor n/2 \rfloor) +
\frac{n^3}{(1+\lfloor\log_2 n\rfloor)^4}$$
where $T(0) = 0.$
Let $$n = \sum_{k=0}^{\lfloor\log_2 n\rfloor} d_k 2^k$$ be the binary representation of $n.$
Unrolling the recursion we find the exact formula
$$T(n) = \sum_{j=0}^{\lfloor\log_2 n\rfloor}
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} d_k 2^{k-j}\right)^3.$$
Now to get an upper bound on this consider the case of $n$ being a string of one digits, which gives
$$T(n)\le \sum_{j=0}^{\lfloor\log_2 n\rfloor}
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^{k-j}\right)^3
\\ = \sum_{j=0}^{\lfloor\log_2 n\rfloor}
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(\sum_{k=j}^{\lfloor\log_2 n\rfloor} 2^k\right)^3
= \sum_{j=0}^{\lfloor\log_2 n\rfloor}
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^j)^3
\\ = \sum_{j=1}^{\lfloor\log_2 n\rfloor+1}
\frac{1}{j^4} (2^{\lfloor\log_2 n\rfloor+1} - 2^{\lfloor\log_2 n\rfloor+1-j})^3
= 2^{3(\lfloor\log_2 n\rfloor+1)}
\sum_{j=1}^{\lfloor\log_2 n\rfloor+1}
\frac{1}{j^4} (1 - 2^{-j})^3.$$
The sum term converges rapidly to a constant and hence the asymptotics of the upper bound are
$$\frac{1}{90}
\left(\pi^4 -270 \mathrm{Li}_4(1/2) + 270 \mathrm{Li}_4(1/4) - 90 \mathrm{Li}_4(1/8)\right)
\times 2^{3(\lfloor\log_2 n\rfloor+1)}.$$
For the a lower bound consider the case of a one digit followed by zeros, which gives
$$T(n)\ge \sum_{j=0}^{\lfloor\log_2 n\rfloor}
\frac{8^j}{(1+\lfloor\log_2 n\rfloor - j)^4}
\left(2^{\lfloor\log_2 n\rfloor-j}\right)^3
= 2^{3\lfloor\log_2 n\rfloor}\sum_{j=0}^{\lfloor\log_2 n\rfloor}
\frac{1}{(1+\lfloor\log_2 n\rfloor - j)^4}
\\ = 2^{3\lfloor\log_2 n\rfloor}\sum_{j=1}^{\lfloor\log_2 n\rfloor+1} \frac{1}{j^4} .$$
The sum term once again converges to a constant and the asymptotics of the lower bound are
$$\frac{\pi^4}{90} 2^{3\lfloor\log_2 n\rfloor}.$$
Joining the two bounds we see that
$$T(n) \in \Theta\left(2^{3\lfloor\log_2 n\rfloor}\right) = \Theta(n^3).$$
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Could someone explain the algebra in this sequences problem? $$ \frac{(n^2+n)-n^2}{\sqrt{n^2+n}+n} = \frac{n}{n\left(\sqrt{1+\frac{1}{n}}+1\right)} $$
The advantage here is that the n's cancel, but I don't see why the argument of the sqrt() function ends up becoming $$ 1 + \frac{1}{n} $$
| You have $\sqrt {n^2 + n}$ what you are doing is factoring out $n^2$ from this expression which will give you
$$\sqrt {n^2 \left( 1 + \frac{1}{n} \right)} = \sqrt {n^2 + \frac{n^2}{n}} = \sqrt{n^2 + n}$$
Multiplying out you'll see clearly that you'll end up with the factored expression.
Since it is under the square root, the $n^2$ term simply becomes $n (\sqrt {n^2} = n)$
So that is how you end up with
$$\frac{n}{n \sqrt {1+\frac{1}{n}} +n} = \frac{n}{n \left( \sqrt {1+\frac{1}{n}} +1\right)}$$
In this step you factor out $n$ from $n\sqrt {1+\frac{1}{n}}$ and $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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solving system of 2 equations Are there any nice/elegant ways to solve this system of equations?
$\left\{\begin{matrix} 2x^2+xy+y^2=28 & \\ x^2-xy+2y^2=32 & \end{matrix}\right.$
I can solve it by isolating one of the variables but it's too messy.
Thanks!
| There is a general way to solve such homogeneous equations.
First note that $x=0$ doesn't provide a solution. Now choose $m=\dfrac{y}x$, so that $y=mx$. Your two equations then become,
$x^2(2+m+m^2)=28
\\ x^2(1-m+2m^2)=32$
Dividing the two, we obtain:
\begin{align}
&\dfrac{2+m+m^2}{1 -m+2m^2}=\dfrac78\\
&\iff 2m^2-5m-3=0\\
&\iff (2m+1)(m-3)=0.
\end{align}
For $m=-\dfrac12$, $x^2=16$ while for $m=3$, we get $x^2=2$.
Thus, $(x,y) \in\{\,(4,-2),\,(-4,2),\,(\sqrt2,3\sqrt2),\,(-\sqrt2,-3\sqrt2)\,\}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minium of function Find the minimum of $f(x)=(x-1)(x-2)(x-3)(x-4)$ without using the calculus, I know it's easy to find it using the derivative, but I need to fiugre out how to solve it without it.
I know that the minimum are between $(1,2)\vee(3,4)$ becouse of chart and $f(x)=x^4-10x^3+35x^2-50x+24$, have anyone idea how to solve it?
| $\displaystyle(x-1)(x-2)(x-3)(x-4)=(x-1)(x-4)(x-2)(x-3)$
$\displaystyle=(x^2-5x+4)(x^2-5x+6)=y(y+2)=(y+1)^2-1\ge-1$
(setting $x^2-5x+4=y$)
Now the equality occurs if $\displaystyle x^2-5x+4=y=-1\iff x^2-5x+5=0$
Observe that the discriminant of the last equation is $>0$ which is extremely crucial here(why?)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove this inequality: $\sum{\frac{1}{a^2\sqrt{a^2+2ab}}}\ge\frac{\sqrt{3}}{abc}$ Let $a,b,c>0$. Prove this inquality:
$\frac{1}{a^2\sqrt{a^2+2ab}}+\frac{1}{b^2\sqrt{b^2+2bc}}+\frac{1}{c^2\sqrt{c^2+2ac}}\ge\frac{\sqrt{3}}{abc}$
| Another Hint/way:
analytical soln.: The equality will be satisfied when all terms are equal.
Proof: let $0 < K=a < b < c$ then $\sqrt(a^2+bc)$ is least term and by making b=c=K, we get higher sum.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Riemann Steiljes Integral when alpha changes For example if the question was following
$$ \int_{0}^2 x\,d \alpha $$
where $ \alpha (x) = x $ if $ 0\le x\le 1 $ and $ \alpha(x) = 3x $ when $ 1< x \le 2 $
Is it correct to solve it using
$$ \int_{0}^2 x d \alpha = \int_{0}^1 x d x + \int_{1}^2 x d (3x)? $$
If that is possible, does it mean that the value of $\alpha$ at $1$ is does not change the answer say if it was
$ \alpha (x) = x $ if $ 0\le x< 1 $ and $ \alpha(x)=3x $ when $ 1\le x\le 2 $ ?
| No, you must take account of the jump discontinuity of $\alpha$ at $x = 1$. The correct evaluation should be
\begin{align*}
\int_{0}^{2} x \, d\alpha(x)
&= \int_{0}^{1} x \, dx + 1 \cdot \{ \alpha(1+) - \alpha(1-) \} + \int_{1}^{2} x \, d(3x) \\
&= \left[ \frac{x^{2}}{2} \right]_{0}^{1} + 1 \cdot (3 - 1) + \left[ \frac{3x^{2}}{2} \right]_{1}^{2} \\
&= \frac{1}{2} + 2 + \frac{9}{2}
= 7.
\end{align*}
Or as Mhenni Benghorbal pointed out, you can apply the Riemann-Stieltjes version of the integration by parts as follows:
\begin{align*}
\int_{0}^{2} x \, d\alpha(x)
&= \left[ x \alpha(x) \right]_{0}^{2} - \int_{0}^{2} \alpha(x) \, dx \\
&= 2\alpha(2) - \int_{0}^{1} x \, dx - \int_{1}^{2} 3x \, dx \\
&= 12 - \left[ \frac{x^{2}}{2} \right]_{0}^{1} -\left[ \frac{3x^{2}}{2} \right]_{1}^{2} \\
&= 12 - \frac{1}{2} - \frac{9}{2} = 7.
\end{align*}
| {
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Evaluate: $\int \frac{1}{(x+a)(x+b)}$ Evaluate:
$$\int \frac{1}{(x+a)(x+b)}$$
My attempt:
$$\int \frac{1}{(x+a)(x+b)} = \frac{A}{x+a} + \frac{B}{x+b}$$
$$1 = A(x+b) + B(x+a)$$
$$x = -b$$
$$1 = A(-b + b) + B(-b + a)$$
$$1 = B(-b + a)$$
$$x = -a$$
$$1 = A(-a + b) + B(-a + a)$$
$$1 = A(-a + b)$$
At this point I have no idea how to proceed. Can someone help me with this? Please.
| starting from $1=A(x+b)+B(x+a)=(A+B)x+(Ab+Ba)$, you should get
$$
A+B=0, \quad Ab+Ba=1
$$
and then the solution is
$$
A=1/(b-a),\quad B=1/(a-b)
$$
which means
$$
\frac{1}{(x+a)(x+b)}=\frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right)
$$
| {
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"source": "stackexchange",
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Find maxi,minimum $f(x,y)=x^3+y^3 (\text{where} ~~~ x,y\in \mathbb{R}, x^2+y^2=1)$ I would appreciate if somebody could help me with the following problem
Q: Find maximum and minimum of $f(x,y)$ that
$$f(x,y)=x^3+y^3 (\text{where} ~~~ x,y\in \mathbb{R}, x^2+y^2=1)$$
| Find max:
$x^2 + y^2 = 1 \implies x, y\in[-1,1] \implies x^2 \ge x^3, y^2\ge y^3 \implies 1 = x^2+y^2\ge x^3 + y^3$
Follow that, find min: It is easy to see that $x^3 + y^3$ min $\iff$ $x, y < 0$ and $|x^3 + y^3|$ max
Cleaner:
$$x^2 \ge |x|^3, y^2\ge |y|^3 \implies 1= x^2 + y^2 \ge|x^3|+|y^3| \ge|x^3 +y^3|$$
$$\iff x^3+y^3 \in[-1,1]$$
| {
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Integral $\int_{0}^{3} \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx$ I recently got stuck on evaluating the following integral,
$$ \int_{0}^{3} \frac{e^{-x^2}}{\sqrt{1-x^2}} \,dx. $$
Is it possible to evaluate this integral in a closed form? I am not sure if there is one, but the integrand seems simple enough, so I hope it might exist.
| You can express in terms of Incomplete Bessel Functions:
$\int_0^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx+\int_1^3\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx$
$=\int_0^1\dfrac{e^{-x^2}}{\sqrt{1-x^2}}~dx-i\int_1^3\dfrac{e^{-x^2}}{\sqrt{x^2-1}}~dx$
$=\int_0^\frac{\pi}{2}\dfrac{e^{-\sin^2x}}{\sqrt{1-\sin^2x}}~d(\sin x)-i\int_0^{\cosh^{-1}3}\dfrac{e^{-\cosh^2x}}{\sqrt{\cosh^2x-1}}~d(\cosh x)$
$=\int_0^\frac{\pi}{2}e^\frac{\cos2x-1}{2}~dx-i\int_0^{\ln(3+2\sqrt 2)}e^{-\frac{\cosh2x+1}{2}}~dx$
$=e^{-\frac{1}{2}}\int_0^\pi e^\frac{\cos x}{2}~d\left(\dfrac{x}{2}\right)-ie^{-\frac{1}{2}}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~d\left(\dfrac{x}{2}\right)$
$=\dfrac{e^{-\frac{1}{2}}}{2}\int_0^\pi e^\frac{\cos x}{2}~dx-\dfrac{ie^{-\frac{1}{2}}}{2}\int_0^{2\ln(3+2\sqrt 2)}e^{-\frac{\cosh x}{2}}~dx$
$=\dfrac{\pi e^{-\frac{1}{2}}}{2}I_0\left(\dfrac{1}{2}\right)-\dfrac{ie^{-\frac{1}{2}}}{2}J\left(\dfrac{1}{2},0,2\ln(3+2\sqrt 2)\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{x_1}{1+x_2+x_3+\ldots+x_n}+\frac{x_2}{1+x_1+x_3+\ldots+x_n}+\ldots+\frac{x_n}{1+x_1+x_2+\ldots+x_{n-1}}\ge\frac{n}{2n-1}$. If $x_1,x_2,\ldots,x_n>0$ and $x_1+x_2+\ldots+x_n=1$, prove that $$\frac{x_1}{1+x_2+x_3+\ldots+x_n} + \frac{x_2}{1+x_1+x_3+\ldots+x_n} +\ldots + \frac{x_n}{1+x_1+x_2+\ldots +x_{n-1}} \ge \frac {n}{2n-1}$$
This can easily be simplified: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} \ge \frac {n}{2n-1}$$
I could try using the Cauchy-Schwarz inequality: $$\frac{x_1}{2-x_1} + \frac{x_2}{2-x_2} +\ldots + \frac{x_n}{2-x_n} = \frac{x_1^2}{2x_1-x_1^2} + \frac{x_2^2}{2x_2-x_2^2} +\ldots + \frac{x_n^2}{2x_n-x_n^2} \ge \frac{(x_1+x_2+\ldots+x_n)^2}{2(x_1+x_2+\ldots+x_n)-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{1}{2-(x_1^2+x_2^2+\ldots+x_n^2)}=\frac{n}{2n-(x_1^2+x_2^2+\ldots+x_n^2)n}$$
It's left to prove that $$(x_1^2+x_2^2+\ldots+x_n^2)n \ge 1$$
or $$(x_1^2+x_2^2+\ldots+x_n^2)n=1$$I can continue using the Cauchy-Schwarz inequality again: $$(x_1^2+x_2^2+\ldots+x_n^2)n=(x_1^2+x_2^2+\ldots+x_n^2)(1+1+\ldots+1)\ge(x_1+x_2+\ldots+x_n)^2=1$$
So I've proved it myself. (I've shown this proof after editing, I didn't post the question with the solution in the details. I found the proof a while after putting the question here).
| For any $x,y \in (0,2)$, we have
$$\frac{x}{2-x} - \frac{y}{2-y} = \frac{2(x-y)}{(2-x)(2-y)}
= 2\left(\frac{x-y}{2-y}\right)\left[\left(\frac{1}{2-x} - \frac{1}{2-y}\right) + \frac{1}{2-y}\right]
= 2\left(\frac{x-y}{2-y}\right)\left[\frac{x-y}{(2-x)(2-y)} + \frac{1}{2-y}\right]
= \frac{2(x-y)^2}{(2-x)(2-y)^2} + \frac{2(x-y)}{(2-y)^2}
\ge \frac{2(x-y)}{(2-y)^2}
$$
Substitute $x$ by $x_i$ and $y$ by $\frac{1}{n}$ and then sum over $i$, we get
$$
\sum_{i=1}^n \frac{x_i}{2-x_i} - \frac{n}{2n-1}
= \sum_{i=1}^n \left(\frac{x_i}{2-x_i} -\frac{\frac{1}{n}}{2-\frac{1}{n}}\right)
\ge \frac{2}{(2-\frac{1}{n})^2}\sum_{i=1}^n \left( x_i - \frac{1}{n}\right)
= \frac{2}{(2-\frac{1}{n})^2} \left(\sum_{i=1}^n x_i - 1 \right)
= 0$$
The steps look almost magical. How do I come up with that? The answer is we are using Jensen's inequality from behind.
Jensen's inequality is actually more generic and applicable beyond differentiable functions. If you have any function $f(x)$ whose graph curved up on both side at every point, then Jensen's inequality works. Such functions are called
convex functions. For any convex
function $f(x)$, Jensen's inequality tells us:
$$\frac{1}{n} \sum_{i=1}^n f(x_i)\;\;\ge\;\; f(\frac{1}{n} \sum_{i=1}^n x_i)$$
This means if you want to verify $\sum_{i=1}^n f(x_i) \ge$ some number $M$. You just
need to check what happens when all $x_i$ are equal to each other.
In this case, we are told we cannot use Jensen's inequality. So we expand our
target function $f(x_i)$ around $\frac{1}{n}$, the mean of the $x_i$'s, and
we are sure after we cancel the linear parts, the rest of the expansion will be non-negative.
| {
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If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$. If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$$
Without using calculus.
There are a few ways I've tried solving this:
$1)$ We could try using the Cauchy-Schwarz inequality: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge \frac{(x+y+z)^2}{(x-1)^2+(y-1)^2+(z-1)^2}$$
But it's apparent that nothing's useful here.
$2)$ We could use AM-GM as well: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 3\sqrt[3]{\frac{x^2y^2z^2}{(x-1)^2(y-1)^2(z-1)^2}}=\frac{3}{\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}}$$
So we only have to prove that: $$\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}\le 3$$
We could raise both sides to the power of 3: $$(x-1)^2(y-1)^2(z-1)^2\le 27$$
But this inequality doesn't hold.
$3)$ We could try cleaning the denominators by multiplying both sides by $(x-1)^2(y-1)^2(z-1)^2$. After a bunch of expanding and simplifying we get that: $$x^2y^2+y^2z^2+x^2z^2-6(xy+yz+xz)+2(x+y+z)+9\ge 0$$
I can't tell so easily whether the inequality is true or not. You could help me out on this one. Just don't forget that $x,y,z\in\mathbb R\setminus\{1\}$ and so we can't just simply use AM-GM, unless we're using it for squares that have to be non-negative.
| for your last inequality, note:
$(xy+yz+xz)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)=x^2y^2+y^2z^2+z^2x^2+2(x+y+z)$
let $xy+yx+xz=t \ge 3 \sqrt[3]{(xyz)^2}=3$
LHS$=t^2-6t+9=(t-3)^2\ge 0 $ when $t=3$ hold "="
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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How to solve for $x$ in $\sqrt[4]{x+27}+\sqrt[4]{55-x}=4$? I'm trying to guess a method for getting the values that work on this irrational equation:
$$\sqrt[4]{x+27}+\sqrt[4]{55-x}=4, x\in\mathbb C$$
After using the formula $a^4+b^4=(a+b)(a^3-a^2b+ab^2+b^3)$ and doing some amplifications, I ended in this phase:
$$p=x+27, r=55-x
\\\sqrt[4]{p^3}+\sqrt[4]{r^3}-\sqrt{p}\sqrt[4]{r}+\sqrt[4]{p}\sqrt{r}=\frac{82}{4}$$
, which clearly is complicated than the initial equation.
Also, rising to the power of $4$, it isn't efficient either: you will end up with mixed radicals. Maybe I'm doing something wrong?
| Let $\displaystyle x+27=a^4, 55-x=b^4\implies a^4+b^4=82$ and $a+b=4$
$\displaystyle a^4+b^4=(a^2+b^2)^2-2a^2b^2=\{(a+b)^2-2ab\}^2-2a^2b^2$
$\displaystyle=(16-2ab)^2-2a^2b^2=256+2a^2b^2-64ab$
$\displaystyle\implies 2(ab)^2-64ab+256=82\iff2(ab)^2-64ab+174=0$
Solve the quadratic equation for $ab$ and we already have $a+b=4$
Case $1: ab=3,$ this leads to too simple calculations
Case $2: ab=29,$ then $a,b$ are the roots of $\displaystyle t^2-4t+29=0$
$\displaystyle\implies (i)a,b $ are $2\pm5i$
and $\displaystyle(ii)t^2=4t-29$
$\displaystyle\implies t^4=(4t-29)^2=16t^2-29\cdot8t+29^2$
$\displaystyle=16(4t-29)-29\cdot8t+29^2=29(29-16)-8t(29-8)=377-168t$
$\displaystyle\implies$ the values of $\displaystyle a^4,b^4$ are $\displaystyle377-168(2\pm5i)=41\mp840i$
Find $x$ in either case and check if the values conform
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sin (A-B) = \pm\frac{1}{3}.$ if....... please help me, how can I approach. I can not understand what to do. How can I start. Help me. If $\sqrt{2}\cos A =\cos B +\cos^3 B$ and $\sqrt{2} \sin A = \sin B - \sin^3 B$ show that $\sin (A-B) =\pm\frac{1}{3}$.
| As Kaster has identified,
$(6)\displaystyle\sqrt2\cos(A-B)=2\cos^2B=\cos^2B-\sin^2B+1$
$\displaystyle\implies\cos^2B-\sin^2B=\sqrt2\cos(A-B)-1$
$(3)\displaystyle\cos B\sin B=-\sqrt2\sin(A-B)$
We have to eliminate $B$s keeping $A-B$
Using $\displaystyle(\cos^2B-\sin^2B)^2+4\cos^2B\sin^2B=(\cos^2B+\sin^2B)^2=1,$
$\displaystyle\{\sqrt2\cos(A-B)-1\}^2+4\{-\sqrt2\sin(A-B)\}^2=1$
$\displaystyle\implies 2\cos^2(A-B)-2\sqrt2\cos(A-B)+1+8\sin^2(A-B)=1$
Using $\displaystyle\sin^2(A-B)=1-\cos^2(A-B),$
$\displaystyle\implies -6\cos^2(A-B)-2\sqrt2\cos(A-B)+8=0$
Solve the Quadratic Eqaution for $\displaystyle\cos(A-B)$ and find that only one value lies in $\in[-1,1]$
Use $\displaystyle\sin(A-B)=\pm\sqrt{1-\cos^2(A-B)}$
| {
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integer $a$ for which $x^3-3x+a = 0$ has $3$ distinct real roots in $x\in (0,1)$ If the equation $x^3-3x+a=0$ has $3$ distinct roots between $0$ and $1$ then the integer value of $a$ is
Can we solve it without Calculus.
$\bf{My\; Try}::$ Let $x=2\cos \phi$, where $-1\leq \cos \phi\leq 1\Rightarrow 0\leq \phi \leq \pi$(Principle value, Inverse exists)
$\displaystyle 8\cos^3 \phi-6\cos \phi+a = 0\Rightarrow 4\cos^3 \phi-3\cos \phi = -\frac{a}{2}\Rightarrow \cos 3\phi = -\frac{a}{2}$
Now $-1\leq \cos 3\phi\leq 1$, So $\displaystyle -1 \leq -\frac{a}{2}\leq 1\Rightarrow -2\leq a\leq 2$
So we get $a = \left\{-2,-1,0,1,2\right\}$
But answer is no integer value of $a$
Help me
Thanks
| Also, the product of the roots is $-a$. Since the roots are all between 0 and 1, $a$ can't be an integer.
| {
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Ineqaulity about a convex function. $(e^x-1)\cdot \ln(x+1) >x^2$ $(x \gt 0)$? How can I solve this inequality $(e^x-1)\cdot \ln(x+1) >x^2$ ($x > 0$)?
| Since $e^x-1 \gt0,$ we will prove the following :
$$\ln(x+1)-\frac{x^2}{e^x-1}\gt0.$$
Letting $f(x)$ be the left hand side, we have
$$f^\prime(x)=\frac{1}{x+1}-\frac{2x(e^x-1)-x^2\cdot e^x}{(e^x-1)^2}=\frac{g(x)}{(x+1)(e^x-1)^2}$$
where $g(x)=e^{2x}+(x^3-x^2-2x-2)e^x+2x^2+2x+1.$
Now
$$g^\prime(x)=2e^{2x}+(3x^2-2x-2)e^x+(x^3-x^2-2x-2)e^x+4x+2$$
$$=2e^{2x}+(x^3+2x^2-4x-4)e^x+4x+2$$
$$\gt 2(x+1)^2+(x^3+2x^2-4x-4)(x+1)+4x+2\gt0.$$
Here, I used $e^x\gt x+1$. You will know the last inequality is true with some calculation.
Hence, we now know $g(x)$ is a strictly monotone increasing function. Also, we have $g(0)=0$. Hence, we know $g(x)\gt0.$
Hence, we now know that $f^\prime(x)\gt0.$ So, we know $f(x)$ is a strictly monotone increasing function. Also, we have
$$\lim_{x\to 0+}f(x)=\lim_{x\to 0+}\frac{x^2}{e^x-1}=\lim_{x\to 0+}\frac{2x}{e^x}=0.$$
So, we now know $f(x)\gt0.$ Q.E.D.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/620316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
How to calculate $ \int \frac{x^3+x^2}{x^3-1} \mathrm{d}x$ I was trying to solve this integral:
$$ \int \frac{x^3+x^2}{x^3-1} \mathrm{d}x$$
I made the following steps:
*
*Polynomial division: $$ \frac{x^3+x^2}{x^3-1} = \left(1+\frac{x^2+1}{x^3-1}\right)$$
*Hermite polynomials: $$ \frac{x^2+1}{x^3-1} = \frac{2}{3}\frac{1}{x-1}+\frac{\frac{x}{3}-\frac{1}{3}}{x^2+x+1}$$
So, now the integral becomes: $$x+\frac{2}{3}\log{|x-1|}+\int \dfrac{\frac{x}{3}-\frac{1}{3}}{x^2+x+1} \mathrm{d}x$$
But i think there's something wrong, i can not continue. Any ideas?
| Your work is well-done up to the point you ended.
Now we just need to work with the final integral:
$$\int\frac{x^3+x^2}{x^3-1}dx=
x+\frac{2}{3}\log\left|x-1\right|+\frac{1}{3}\int\frac{x-1}{x^2+x+1}dx$$
$$\dfrac 13\int \frac{x - 1}{x^2 + x + 1}dx = \frac 16\int\frac{2x +1}{x^2 + x + 1}dx - \frac 12\int \frac {dx}{x^2 + x + 1} $$ $$= \frac 16\int\frac{2x +1}{x^2 + x + 1}dx - \frac 12\int \frac {dx}{(x+\frac 12)^2 + \frac 34}$$
In the left-most integral in the last line, we have an integral of the form $$ \int \dfrac{f'(x)}{f(x)} \,dx = \ln (f(x)) + C$$
In the right-most integral in the last line, we have "completed the square" and can now use the substitution $\tan\theta = \frac{\sqrt 3}2(x + \frac 12)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/622296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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$f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$, Then Max $\left\{f(x)-g(x)\right\}$ Let $f(x)=\sqrt{(x-1)^2+(x^2-5)^2}\;\;,\;\; g(x)=\sqrt{(x+2)^2+(x^2+1)^2},\forall x\in \mathbb{R}$.
Find the Minimum of function $\left\{f(x)+g(x)\right\}$ and the maximum of function $\left\{f(x)-g(x)\right\}$.
$\bf{My\; Try}$:: For Minimum of $\left\{f(x)+g(x)\right\}$
Using Minkowski inequality or $\triangle$ Inequality $\sqrt{a^2+c^2}+\sqrt{b^2+d^2}\geq \sqrt{(a+b)^2+(c+d)^2}$
and equality holds when $\displaystyle \frac{a}{b} = \frac{c}{d}$
$\sqrt{(1-x)^2+(5-x^2)^2} + \sqrt{(x+2)^2+(x^2+1)^2}$
$\geq \sqrt{\left(1-x+x+2\right)^2+\left(5-x^2+x^2+1\right)^2} = \sqrt{3^2+6^2} = 3\sqrt{5}$
and equality hold, when $\displaystyle \frac{1-x}{5-x^2} = \frac{x+2}{x^2+1}$
But I did not understand How can I calculate Maximum of $\left\{f(x)-g(x)\right\}$
Help Required
Thanks
| Use the same inequality again, but backwards:
$$\sqrt{(x+2)^2 + (x^2+1)^2} + \sqrt{((x-1)-(x+2))^2 + ((x^2-5)-(x^2+1))^2} \geq \sqrt{(x-1)^2 + (x^2-5)^2}$$
This gives us $f(x)-g(x) \leq 3\sqrt{5}$. It also allows for us to solve for $x$.
One warning: this method is really finding the maximum value of $|f(x)-g(x)|$. So when we solve for $x$, we will get two solutions, one maximizing $f(x)-g(x)$, the other maximizing $g(x)-f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What is the sum of the prime numbers up to a prime number $n$? How to find the sum of prime numbers up to a prime number $n$, that is for example: the sum of prime numbers up to 7 is: 2+3+5+7=17.
So what is the formula for finding: $$\sum_{k=0}^n p_k=????,$$
with $p_k$ being the $k$th prime.
Also if we have the sum of an even number of primes then would it be a new prime? Example: 2+3+5+7=17 and 17 is a prime. 2+3+5+7+11+13=41 and 41 is prime.
Thank you.
| $$
\sum_{p\le x}p = \frac{x^2}{2\log x} + \frac{x^2}{2\log^2x} + \frac{x^2}{4\log^3x} + \frac{3x^2}{8\log^4x} + O\left(\frac{x}{\log^5x}\right).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/623872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Find max: $M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}$ For $a,b,c>0$ and $abc=1$, find the maximum of
$$M=\frac{a}{b^2+c^2+a}+\frac{b}{c^2+a^2+b}+\frac{c}{a^2+b^2+c}.$$
| Take the first term:$$\frac{a}{b^2+c^2+a}=\frac{1}{\frac{1}{a^2}(\frac{b}{c}+\frac{c}{b})+1}$$
Minimum value of $\frac{1}{a^2}(\frac{b}{c}+\frac{c}{b})$ is $\frac{2}{a^2}$ which it takes when $b=c$. By the symmetry of the expression it is maximum when $a=b=c$ and the maximum value is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/624664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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minimal polynomial of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ I have to find the minimal polynomial of of $\alpha+\alpha^2$ where $\alpha^5=1$ and $\alpha\neq 1$ in $\mathbb{Q}$. First I found the minimal polynomial of $\alpha$: $X^5-1=(X-1)(X^4+X^3+X^2+X+1)$ so $X^4+X^3+X^2+X+1$ must be the minimal polynomial of $\alpha$ since its irreducible and has $\alpha$ as a root. I didn't come much further than this. I really need help. Thanks.
| Each row of
$$
M=\left[\begin{array}{r}
1&0&0&0\\
0&1&1&0\\
-1&-1&0&1\\
0&-2&-3&-2\\
3&5&3&0\\
\end{array}\right]
$$
is $(\alpha+\alpha^2)^k$ using the basis $(1,\alpha,\alpha^2,\alpha^3)$ and relation $\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$.
Using the higher dimensional extension of the cross product, we find the vector perpendicular to all the columns of $M$. That is, take the determinant of the matrix deleting successive rows, then alternating signs, we get
$$
\left[\begin{array}{r}
1&3&4&2&1
\end{array}\right]
\left[\begin{array}{r}
1&0&0&0\\
0&1&1&0\\
-1&-1&0&1\\
0&-2&-3&-2\\
3&5&3&0\\
\end{array}\right]=0
$$
Since the determinant of the first four rows is not $0$, the minimal polynomial of $\alpha+\alpha^2$ is $x^4+2x^3+4x^2+3x+1=0$.
Verification
$$
\begin{align}
&(x^2+x)^4+2(x^2+x)^3+4(x^2+x)^2+3(x^2+x)+1\\
&=x^8+4x^7+8x^6+10x^5+11x^4+10x^3+7x^2+3x+1\\
&=(x^4+3x^3+4x^2+2x+1)(x^4+x^3+x^2+x+1)\\
&\equiv0\pmod{x^4+x^3+x^2+x+1}
\end{align}
$$
Computation of $M$
The action of multiplying by $\alpha$ is given by the matrix
$$
A=\left[\begin{array}{r}
0&1&0&0\\
0&0&1&0\\
0&0&0&1\\
-1&-1&-1&-1
\end{array}\right]
$$
Multiplication by $\alpha^2+\alpha$ is given by the matrix
$$
\begin{align}
B
&=A^2+A\\[12pt]
&=\left[\begin{array}{r}
0&1&1&0\\
0&0&1&1\\
-1&-1&-1&0\\
0&-1&-1&-1\\
\end{array}\right]
\end{align}
$$
The $k+1^\text{st}$ row of $M$ is
$$
\left[\begin{array}{r}
1&0&0&0
\end{array}\right]\,B^k
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/627981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solutions for linear equation system Ax = b Two solutions are known for the linear equation system $Ax = b$. These are (for example):
$$
\begin{bmatrix}
1 \\
2
\end{bmatrix},
\begin{bmatrix}
2 \\
1
\end{bmatrix}
$$
How to determine (at least) one extra solution? How to solve problems like this in general?
| You have the following:
$A\begin{pmatrix}
1 \\
2
\end{pmatrix} = b$ and $A\begin{pmatrix}
2 \\
1
\end{pmatrix} = b$ $\Rightarrow $ $A\begin{pmatrix}
1 \\
2
\end{pmatrix} + A\begin{pmatrix}
2 \\
1
\end{pmatrix} = 2b$ $\Rightarrow $ $A\begin{pmatrix}
3 \\
3
\end{pmatrix}=2b$ $\Rightarrow $ $A\begin{pmatrix}
3/2 \\
3/2
\end{pmatrix}=b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/628210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Indefinite integral question. Evaluate the following indefinite integral.
$$ \int { \frac { 8 }{ 81+{ x }^{ 2 } } } dx $$
The answer is
$$ \frac { 8 }{ 9 } \arctan \left(\frac { x }{ 9 } \right)$$
I know that it has something to do with this integral
$$ \int { \frac { 1 }{ { x }^{ 2 }+1 } } dx =\arctan x +C $$
but i can't get it.
I think i can't make the algebra to find the antiderivative.
| For
$$\int\frac{8}{81+x^2} \, dx,$$
If we let
$$
\begin{align*}
x &=9\tan \theta \\
dx&=9\sec^2 \theta \, d \theta \\
81+x^2 &=81+81\tan^2 \theta \\
&=81 \sec^2 \theta.
\end{align*}
$$
We substitute,
$$
\begin{align*}
\int\frac{8}{81+x^2} \, dx &=8 \int \frac{9\sec^2 \theta \, d \theta}{81\sec^2 \theta} \\
&=\frac{8}{9}\int 1 \, d \theta \\
&=\frac{8}{9} \theta +c.
\end{align*}
$$
For the back substitution,
$$x=9\tan \theta \Rightarrow \tan \theta=\frac{x}{9} \Rightarrow \theta=\tan^{-1}\left( \frac{x}{9} \right)$$
Hence our integral,
$$\frac{8}{9}\tan^{-1}\left( \frac{x}{9} \right)+c.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/630703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find $\int\frac{x^2}{\sqrt{1-x^2}}$ I need to compute the following integral:
$$\int\frac{x^2}{\sqrt{1-x^2}}~dx$$
I tried everything I know, including defining $u=x^2$ and $u=\sqrt{1-x^2}$, but it failed.
| Let $x = \sin(u)$, then we have $dx = \cos(u) du$
Substituting in above, we get:
$$\int\frac{x^2}{\sqrt{1 - x^2}} dx = \int \frac{\sin^2(u)}{\sqrt{1 - \sin^2(u)}} \cdot \cos(u) du = \int\frac{\sin^2(u)}{\cos(u)} \cdot \cos(u) du = \int \sin^2(u) du$$
which solves to:
$$\frac{1}{2} \cdot (u - \sin(u) \cos(u))$$
Now you can make the reverse substitution to get the desired result, which is:
$$\frac{1}{2} \cdot \left(\sin^{-1}(x) - \sin(\sin^{-1}(x)) \cos(\sin^{-1}(x))\right) = \frac{1}{2}\cdot \left(\sin^{-1}(x) - \frac{x}{\sqrt{1 - x^2}}\right)$$
which is the required answer. That is:
$$\color{red}{\int\frac{x^2}{\sqrt{1 - x^2}} dx = \frac{1}{2}\cdot \left(\sin^{-1}(x) - \frac{x}{\sqrt{1 - x^2}}\right)}$$
NOTE: The last step is due to the fact that:
$$\sin(\sin^{-1}(x)) = x\,, \textrm{and}\, \cos(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/631434",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Differentiation d^2y/dx^2 of trigonometric function We have a function $x \sin y = y^2$. Let $P$ be $\displaystyle (\frac{\pi^2}{4}, \frac{\pi}{2})$. Evaluate $\displaystyle \frac{dx}{dy}$ and $\displaystyle \frac{d^2y}{dx^2}$ at $P$.
For $\displaystyle \frac{dx}{dy}$:
$\displaystyle \frac{\pi^2}{4} \sin \left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2$
$\displaystyle D_x \left[\frac{\pi^2}{4}\right] \sin \left(\frac{\pi}{2}\right) + \frac{\pi^2}{4} D_x \left[\sin \frac{\pi}{2}\right] = D_x \left[\left(\frac{\pi}{2}\right)^2\right]$
$\displaystyle \frac{\pi}{2} \sin \left(\frac{\pi}{2}\right) + \frac{\pi^2}{4} \cos \left(\frac{\pi}{2}\right) = \pi$
I'm wondering if I could directly substitute the points to the equation or solving first for $\frac{dx}{dy}$.
$\displaystyle x \sin y = y^2 \rightarrow x' \sin y + x D_x [\sin y] = D_x [y^2] \rightarrow \sin y + x \cos \frac{dx}{dy} = 2y$
I tried to use implicit differentiation. The left side by product rule and right by power rule.
| Here is how you find $\frac{dx}{dy}$
$$ x \sin y = y^2 \implies x\cos y +\sin y\frac{dx}{dy}=2y \implies \frac{dx}{dy}=\frac{2y-x\cos y}{\sin y}. $$
Now, evaluate the derivative at $P$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using substitution to evaluate indefinite integral $ \int{x\sqrt{4x+1}}dx$
Evaluate this indefinite integral.
$$I= \int{x\sqrt{4x+1}}dx$$
Let $u=4x+1$
$$\frac{du}{dx}=4\rightarrow{dx=\frac{du}{4}}$$
$$I=\int{x}\sqrt{u}\frac{1}{4}du=\frac{1}{4}\int{x}\sqrt{u}du$$
Then I got stuck at this point.
| $$I = \int x \sqrt{4x+1} \ dx$$
let $u$ = $4x+1$
$x = \frac{u-1}{4}$
$ \text{d}x = \frac{\text{d}u}{4}$
$$
\begin{align}
I &= \int x \sqrt{4x+1} \ \text{d}x
\\ &= \int \left(\frac{u-1}{4} \right) u^{1/2} \frac{\text{d}u}{4}
\\ &= \frac{1}{16}\int u^{3/2}\text{d}u - \frac{1}{16}\int u^{1/2}\text{d}u
\\ &= \frac{1}{16}\frac{u^{5/2}}{5/2} - \frac{1}{16}\frac{u^{3/2}}{3/2} + \text{C}
\\ &= \frac{1}{40}u^{5/2} - \frac{1}{24}u^{3/2} + \text{C}
\\ &= \frac{1}{40}(4x+1)^{5/2} - \frac{1}{24}(4x+1)^{3/2} + \text{C}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/635259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Integrating $\int\sin^{-2}xdx$ I am trying to prove that
$$
\int\frac{1}{\sin^2(x)}dx = -\cot(x) + C
$$
but I have difficulties, I don't know where to start, I can't substitute anything with $sin(x)$ because I don't have a $cos(x)$ to make it up for, I also know that:
$$
\sin^2x = \frac{1 - \cos(2x)}{2}
$$
so I basically have:
$$
\int\frac{2}{1 - \cos(2x)}dx
$$
and there I am stuck, how do I proceed?
| You can use the tangent half angle substitution $t = \tan \frac{x}{2}$ which yields $\sin(x) = \frac{2 t}{1+t^2}$ and ${\rm d}x = \frac{2}{1+t^2}{\rm d} t$.
$$ \int \frac{1}{\sin^2 x}\,{\rm d}x = \int \frac{1}{\left(\frac{2 t}{1+t^2}\right)^2} \left(
\frac{2}{1+t^2} \right)\,{\rm d}t =\int \frac{1+t^2}{2 t^2} {\rm d}t = -\frac{1-t^2}{2 t}+C$$
Since $\sin x = \frac{2 t}{1+t^2}$ and $\cos x = \frac{1-t^2}{1+t^2}$ then $\tan x = \frac{2 t}{1-t^2}$.
So $$\frac{1-t^2}{2 t} = \frac{\cos x}{\sin x} = \cot x$$
and
$$ \int \frac{1}{\sin^2 x}\,{\rm d}x = -\cot x + C $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/637160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
trigonometry identity I have some problem with proving this identity:
$$2\left(1+\cos\alpha \right)-\sin^2\alpha=4\cos^4\frac{\alpha}{2}$$
I tried to start from the right side rewritting it to $(2\cos^2(2\frac{\alpha}{4}))^2$ but it's not working.
| Let there is a function $f(x) = \cos x + 1$, then $$f^2(x) = \cos^2 x + 2\cos x + 1 = 2(1 + \cos x) - \sin^2 x.$$ While $$f(x) = 2\cos^2\frac{x}{2} = \sqrt{4\cos^4\frac{x}{2}}.$$ Therefore $$ 4\cos^4\frac{x}{2} = 2(1 +\cos x) - \sin^2x.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/637214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
What's the simplest way to solve $\int \frac{1}{4-v^2}dv$? I did a substitution on a DE and ended up with this:
$\int \frac{1}{4-v^2}dv$
I tried a trig substitute but things got a bit hairy. WolframAlpha recommended a far less intuitive substitution. I can't remember what it was.
What is the simplest way to solve this integral?
| I think partial fractions may indeed be the "simplest" way, but you mentioned trigonometric substitution getting a bit hairy, and I figured I would give my best derivation. It is a useful skill. This would be advantageous speed wise only if you knew the integral of the secant function from memory.
Consider the more general integral,
$$\int\frac{1}{a^2-v^2} \, dv.$$
If we let
$$v=a\sin\theta,$$
then
$$dv=a\cos \theta \, d \theta,$$
and
$$a^2-v^2=a^2-a^2\sin^2 \theta=a^2\left( 1-\sin^2 \theta \right)=a^2 \cos^2 \theta.$$
We substitute to get
$$
\begin{align*}
\int \frac{1}{a^2-v^2} \, dv &= \int \frac{a \cos \theta \, d \theta}{a^2 \cos^2 \theta} \\
&=\frac{1}{a}\int\sec \theta \, d \theta \\
&=\frac{1}{a}\ln |\sec \theta + \tan \theta|+c.
\end{align*}
$$
For the back substitution,
$$v=a \sin \theta \Rightarrow \sin \theta = \frac{v}{a}.$$
We form right a triangle with angle $\theta$, side $v$ opposite $\theta$, and hypotenuse $a$. Thus the side adjacent $\theta$ has length $\sqrt{a^2-v^2}$. Reading directly from the triangle,
$$\frac{1}{a}\ln |\sec \theta + \tan \theta|+c=\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c.$$
We are done, but this certainly cleans up well (like in an integral table).
$$
\begin{align*}
&\frac{1}{a}\ln \left| \frac{a}{\sqrt{a^2-v^2}}+ \frac{v}{\sqrt{a^2-v^2}} \right|+c \\
=& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a^2-v^2}} \right|+c \\
=& \frac{1}{a}\ln \left| \frac{a+v}{\sqrt{a+v}\sqrt{a-v}} \right|+c \\
=&\frac{1}{a}\ln \left| \frac{\sqrt{a+v}}{\sqrt{a-v}} \right|+c \\
=&\frac{1}{a}\ln \left| \left( \frac{a+v}{a-v} \right)^{\frac{1}{2}} \right|+c \\
=&\frac{1}{2a}\ln \left| \frac{a+v}{a-v} \right|+c. \\
\end{align*}
$$
Thus your particular integral is
$$\int \frac{1}{4-v^2} \, dv=\frac{1}{4}\ln \left| \frac{a+v}{a-v} \right|+c.$$
It might not seem like the "easiest" way, but it is certainly doable, and in my opinion, fun.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
how to evaluate $\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}\text{d}x$ I was solving a physics problem and eventually the problem boiled down to solving the following integral:
$$\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{\sin x}}\text{d}x$$
I have already tried substitutions like $\sin x=t^2$ , $\sin x=t$ and have tried using the properties of definite integrals given on http://www.sosmath.com/calculus/integ/integ02/integ02.html but I could not solve this integral. Please help!
| $$
\begin{aligned}
\int_0^{\frac{\pi}{2}} \frac{d x}{\sqrt{\sin x}} & =\int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{4}\right)-1 }\theta \cos^{ 2\left(\frac{1}{2}\right)-1 }\theta d\theta\\
& =\frac{1}{2} B\left(\frac{1}{4}, \frac{1}{2}\right) \\
& =\frac{1}{2} \cdot \frac{\Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{3}{4}\right)} \\
& =\frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma\left(\frac{1}{4}\right) \cdot \Gamma\left(\frac{1}{4}\right)}{\Gamma\left(\frac{3}{4}\right) \cdot \Gamma\left(\frac{1}{4}\right)} \\
& =\frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma^2\left(\frac{1}{4}\right)}{\pi \csc \left(\frac{\pi}{4}\right)} \\
& =\frac{1}{2 \sqrt{2 \pi}} \Gamma^2\left(\frac{1}{4}\right)
\end{aligned}
$$
where the last step comes from the reflection property of Gamma function.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of odd numbers always gives a perfect square. $1 + 3 = 4$ (or $2$ squared)
$1+3+5 = 9$ (or $3$ squared)
$1+3+5+7 = 16$ (or $4$ squared)
$1+3+5+7+9 = 25$ (or $5$ squared)
$1+3+5+7+9+11 = 36$ (or $6$ squared)
you can go on like this as far as you want, and as long as you continue to add odd numbers in order like that, your answer is always going to be a perfect square.
But how to prove it?
| You can prove it by induction .
Statement : $ 1+3+5+...+(2n-1)=n^2 $
Base case :
For n =1 , the LHS of statement is 1 and the RHS of the statement is 1 . So the statement is true for n=1 .
Induction step :
Let the statement is true .
$1+3+5+...... + (2n-1)=n^2$
So $ 1+3+5+...+(2n-1)+(2n+1) = n^2+2n+1 = (n+1)^2 $
So the statement is true for all natural values of n .
$1+3+5+...+(2n-1)=n^2 $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
The pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfies $|w1| =1$, $|w2| = 1$ and $\Re\left(w_1\overline{w_2}\right) =0$ If $z_1=a + ib$ and $z_2 = c + id$ are complex numbers such that $\left|z_1\right| = \left|z_2\right| = 1$ and $\Re\left(z_1\overline{z_2})\right)=0$, then Prove that the pair of complex numbers $w_1 = a + ic$ and $w_2 = b + id$ satisfies $\left|w_1\right| =1$, $\left|w_2\right| = 1$ and $\Re\left(w_1\overline{w_2}\right) =0$
| The condition on $z_1, z_2$ says that $a^2 + b^2 = c^2 + d^2 = 1, $ and $z_1 = \pm i z_2.$ So, either $a = d, b = c$ or $a = -d, b = -c.$ This shows that $a^2 + c^2 = a^2 + b^2 = c^2 + d^2 = b^2 + d^2 = 1,$ and that $w_1 = \pm i w_2.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Evaluating indefinite integrals Evaluate the following indefinite integral.
$ \int { { \sin }^{ 6 } } x\quad dx $
My try :
$ \int { ({ \sin^2x } } )^{ 3 }dx\\ \int { (\frac { 1 }{ 2 } } (1-\cos2x))^{ 3 }dx\\ \int { \frac { 1 }{ 8 } } (1-\cos2x)^{ 3 }dx\\ \frac { 1 }{ 8 } \int { (1-\cos2x)^{ 3 } } dx\\ \frac { 1 }{ 8 } \int { 1-3\cos2x+3\cos^{ 2 } } 2x-\cos^{ 3 }2x\quad dx $
Then i got stuck.
| HINT:
$\displaystyle32\sin^6x=2(4\sin^3x)^2$
Using $\sin3A$ formula, $\displaystyle32\sin^6x=2(3\sin x-\sin3x)^2=3(2\sin^2x)+2\sin^23x-3(2\sin x\sin3x)$
Now apply $\displaystyle\cos2B=1-2\sin^2B\implies2\sin^2B=1-\cos2B$
and $\displaystyle2\sin A\sin B=\cos(A-B)-\cos(A+B)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need help with integration Can anyone help me with this:
$$ \int \sqrt\frac{e^x-1}{e^x+1}\, \mathrm{d}x$$
I can't think of a good substitution, any hint is welcome :)
Thanks.
| You can write
$$\frac {e^x-1}{e^x+1}=\frac {e^{x/2}-e^{-x/2}}{e^{x/2}+e^{-x/2}}=\tanh \frac x2$$
Thus, if
$$u=\sqrt{\frac {e^x-1}{e^x+1}}=\sqrt{\tanh \frac x2}$$
$$x = 2\arg \tanh u^2$$
$$\mathrm{d}x = \frac{4u\mathrm{d}u}{1-u^4}$$
Now,
$$\int \sqrt{\frac {e^x-1}{e^x+1}} \mathrm{d}x=\int \frac{4u^2\mathrm{d}u}{1-u^4}$$
And you can expand in partial fractions:
$$\frac{2u^2}{1-u^4}=\frac{2u^2-2+2}{(1+u^2)(1-u^2)}=\frac{2}{(1+u^2)(1-u^2)}-\frac{2}{1+u^2}$$
$$=\frac{1}{1-u^2}+\frac{1}{1+u^2}-\frac{2}{1+u^2}=\frac{1}{1-u^2}-\frac{1}{1+u^2}$$
And
$$\int \frac{1}{1-u^2}-\frac{1}{1+u^2} \mathrm{d}u=\arg \tanh u - \arctan u$$
Thus
$$\int \sqrt{\frac {e^x-1}{e^x+1}} \mathrm{d}x=2\arg \tanh \left( \sqrt{\tanh \frac x2}\right) - 2\arctan \left( \sqrt{\tanh \frac x2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the maximum value Find the maximum value $$F(y)=\int_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$$ with $y\in [0;\: 1]$
This problem here
| The maximum value is $1/3$. First, the integral is equal to $1/3$ when $y=1$, so the maximum is at least that large. To see that the integral is never bigger than $1/3$, note that $$\sqrt{x^4+(y-y^2)^2}\leq x^2 + y-y^2\quad \text{for $y\in [0,1]$},$$
and therefore that
\begin{align*}
\int_0^y \sqrt{x^4+(y-y^2)^2}\,dx &\leq \int_0^y(x^2+y-y^2)\,dx = y^2-{2\over3}y^3.\tag{1}
\end{align*}
But $g(y) = y^2-{2\over3}y^3$ is increasing on $[0,1]$ since $g'(y) = 2(y-y^2)\geq0$ there. So $g(y)\leq g(1) = 1/3$ for all $y\in[0,1]$, and the proof is finished by $(1)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Finding powerseries $\frac{1}{z^2-z}$ Can anyone help me out with finding the powerseries of:
$f(z)=\frac{1}{z^2-z}$ about $a=-1$?
| Let $x=z+1$. Then $z^2-z=z(z-1)=(x-1)(x-2)$, so
$$
f(z)=\frac{1}{(x-1)(x-2)}=\frac{1}{1-x}-\frac{1}{2-x}
=\sum_{k=0}^{\infty} x^k-\frac{1}{2}\sum_{k=0}^{\infty}\big(\frac{1}{2}\big)x^k
=\sum_{k=0}^{\infty} \bigg(1-\frac{1}{2^{k+1}}\bigg)(z+1)^k
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/644047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of the function $\lim_{x\to0}\frac{\sin 2x + a\sin x}{x^3}$? Question
If for some real number $a$, $\lim_{x\to 0}\frac{\sin 2x + a\sin x}{x^3}$ exists, then the limit is equal to:
Here what i have done
since it is of $0/0$ form applying L' Hospital's rule$$\implies\lim_{x\to0}\frac{\sin 2x + a \sin x}{x^3} = \frac{2\cos 2x + a\cos x}{3x^2}$$ now what to do i am stuck here please help
Thanks
Akash
| Using no L'Hospital's Rule or Maclaurin series, but instead trig identities and that $\lim_{x\to0}\frac{\sin(x)}{x}=1$:
$$\begin{align}
\lim_{x\to0}\frac{\sin(2x)+a\sin(x)}{x^3}&=\lim_{x\to0}\frac{2\sin(x)\cos(x)+a\sin(x)}{x^3}\\
&=\lim_{x\to0}\frac{\sin{x}}{x}\frac{2\cos(x)+a}{x^2}\\
&=\lim_{x\to0}\frac{2\cos(x)+a}{x^2}\\
\end{align}$$
The denominator of this approaches $0$, so the only way the limit can exist is if the numerator also approaches $0$. So $a$ would have to be $-2$.
$$\begin{align}
\lim_{x\to0}\frac{2\cos(x)-2}{x^2}&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\\
&=2\lim_{x\to0}\frac{\cos(x)-1}{x^2}\frac{\cos(x)+1}{\cos(x)+1}\\
&=2\lim_{x\to0}\frac{\cos^2(x)-1}{x^2(\cos(x)+1)}\\
&=2\lim_{x\to0}\frac{-\sin^2(x)}{x^2(\cos(x)+1)}\\
&=-2\lim_{x\to0}\frac{\sin^2(x)}{x^2}\frac{1}{\cos(x)+1}\\
%&=-2\lim_{x\to0}\frac{1}{\cos(x)+1}\\
&=\ldots
\end{align}$$
| {
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How to factorize $2x^2+5x+3$? I'm doing pre-calculus course at coursera.org and I'm in trouble with this solution $$2x^2 +5x +3 = (2x+3)(x+1)$$
By trial, using ac-method I got stuck:
$$
ac = (2)(3) = 6\\
6 + ? = 5 \Rightarrow~ ? = 5 - 6 = -1
$$
Then,
$$2x^2+6x-x+3 = 2x(x+3)-x+3$$
At this point I could not get the answer, any help?
| $2x^2+5x+3=2x^2+3x+2x+3=x(2x+3)+(2x+3)=(2x+3)(x+1)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Help with solving this limits question How do I solve this limits question:
$$\lim_{x\to-2}\frac{4-x^2}{\sqrt{x^2-x-2}-\sqrt{2-x}}$$
I've already factorised the top and the bottom as far as I can but can't seem to reach the answer.
$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{(x-2)(x+1)}-\sqrt{2-x}}$$
$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{-(2-x)(x+1)}-\sqrt{2-x}}$$
$$ = \lim_{x\to-2}\frac{(2+x)(2-x)}{\sqrt{2-x}(\sqrt{-(x+1)}-1)}$$
$$ = \lim_{x\to-2}\frac{(2+x)\sqrt{2-x}}{(\sqrt{-(x+1)}-1)}$$
Am I moving in the right direction?
The answer to the question is 4.
| Multiply the numerator and denominator by the conjugate of the denominator: multiply your function by $$\frac{\sqrt{x^2-x-2}+\sqrt{2-x}}{\sqrt{x^2-x-2}+\sqrt{2-x}}$$
Recall how a difference of squares factors: $$(a - b)(a+b) = a^2 - b^2$$
In this case, we see that when we multiply numerator and denominator by the conjugate of the denominator, w have a difference of squares in the denominator, which then simplifies greatly:
$$\begin{align} (\sqrt{x^2-x-2}-\sqrt{2-x})(\sqrt{x^2-x-2}+\sqrt{2-x}) & = (x^2 - x - 2) - (2 - x) \\ \\ &= x^2-4 \\ \\ &= -(4-x^2)\end{align}$$
After canceling the common factor of $4 - x^2$ we are left with $$\lim_{x\to -2} -(\sqrt{x^2-x-2}+\sqrt{2-x}) = -4$$
| {
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How prove this $f(n)\le f(n+1)$ where $f(n)=\sum_{k=1}^{n}\frac{n}{n^2+k^2}$ let $$f(n)=\sum_{k=1}^{n}\dfrac{n}{n^2+k^2}$$
prove or disprove
$$f(n)\le f(n+1)$$
this inequality is found when I deal this follow limit:
$$\lim_{n\to\infty}f(n)=\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=1}^{n}\dfrac{1}{1+(k/n)^2}=\int_{0}^{1}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{4}$$
But I can't prove
$$f(n)\le f(n+1)$$
since
$$f(n+1)=\dfrac{n+1}{(n+1)^2+1}+\dfrac{n+1}{(n+1)^2+2^2}+\cdots+\dfrac{n+1}{(n+1)^2+n^2}+\dfrac{n+1}{(n+1)^2+(n+1)^2}$$
$$f(n)=\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2^2}+\cdots+\dfrac{n}{n^2+n^2}$$
so
$$f(n+1)-f(n)=\left(\dfrac{n+1}{(n+1)^2+1}-\dfrac{n}{n^2+1}\right)+\left(\dfrac{n+1}{(n+1)^2+2^2}-\dfrac{n}{n^2+2^2}\right)+\cdots+\left(\dfrac{n+1}{(n+1)^2+n^2}-\dfrac{n}{n^2+n^2}\right)+\dfrac{1}{2(n+1)}$$
so
$$f(n+1)-f(n)=\sum_{k=1}^{n}\dfrac{k^2-n^2-n}{(k^2+n^2)((n+1)^2+k^2)}+\dfrac{1}{2(n+1)}$$
This problem is my found it,can you help to solve this problem?
| This in fact is an extension of the question initially raised.
The finite series can be expressed in terms of digamma function :
Using the integral definition of the digamma :
The finite series can be expressed as a function defined by an integral:
Then the function with integer n is extended to the positive reals x :
The function $f(x)$ is continuous and can be graphically represented :
This shows that $f(x)$ is continuously increassing. As a consequence, we can extend the initial question $f(n+1)>f(n)$ to an extended one $f(x+r)>f(x)$ where $x$ and $r$ are any positive reals.
Of course, the graphical presentation is not an analytical proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "22",
"answer_count": 7,
"answer_id": 2
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Parity of Binomial Coefficients Do $\binom nr$ and $\binom {2n}{2r}$ always have the same parity? I can see that it's true for $r=1$ since $\binom {n}{1}=n$ and $\binom{2n}{2}=n(2n-1)$, but what about for bigget $r$?
| As per robjohn's suggestion:
$$\begin{align}
\binom{2n}{2r} &= \frac{2n}{2r}\times \left(\frac{2n-1}{2r-1}\right)\times \frac{2n-2}{2r-2}\times \left(\frac{2n-3}{2r-3}\right)\times \cdots \times \frac{2n-2r+2}{2}
\times \left(\frac{2n-2r+1}{1}\right)\\
&= \frac{n}{r}\times \frac{n-1}{r-1}\times\cdots\times \frac{n-r+1}{1} \times
\left(\frac{2n-1}{2r-1}\right)\left(\frac{2n-3}{2r-3}\right)\cdots
\left(\frac{2n-2r+1}{1}\right)\\
&= \binom{n}{r}\times \frac{2p+1}{2q+1}
\end{align}$$
where $p$ and $q$ are integers. Thus we have that
$$(2q+1)\binom{2n}{2r} = (2p+1)\binom{n}{r} \tag{1}$$
where the left side of $(1)$ is even or odd according as $\binom{2n}{2r}$ is
even or odd while the right side of $(1)$ is even or odd according as
$\binom{n}{r}$ is even or odd. Hence, $\binom{2n}{2r}$
and $\binom{n}{r}$ must have the same parity modulo 2.
More strongly, we can also deduce from $(1)$ that
the largest power of 2 that divides $\binom{2n}{2r}$
is the same as the largest power of 2 that divides $\binom{n}{r}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series)
Evaluate $\sum^\infty_{n=1} \frac{1}{n^4} $using Parseval's theorem (Fourier series).
I have , somehow, to find the sum of $\sum_{n=1}^\infty \frac{1}{n^4}$ using Parseval's theorem.
I tried some things that didn't work so I won't post them.
Can you please explain me how do I find the sum of this series using Parseval's identity?
Thanks
| Let $f(x)=x^2$ for $x\in(-\pi,\pi)$. Computing the Fourier coefficients gives
$$a_n=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 e^{i n x} dx=\frac{2 \cos(\pi n)}{n^2}=2\frac{(-1)^n}{n^2}$$
for $n\in\mathbb{Z}$, $n\not=0$, and $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi x^2 dx=\frac{\pi^2}{3}$.
Therefore $|a_n|^2=\frac{4}{n^4}$ for $n\in\mathbb{Z}$, $n\not=0$ and $|a_0|^2=\frac{\pi^4}{9}$.
By Plancherel/Parseval's theorem,
$$\frac{\pi^4}{9}+8\sum_{n=1}^\infty \frac{1}{n^4}=\sum_{n=-\infty}^\infty |a_n|^2=\frac{1}{2\pi}\int_{-\pi}^\pi x^4 dx=\frac{\pi^4}{5}$$
Simplifying, this gives
$$\sum_{n=1}^\infty \frac{1}{n^4}=\frac{\pi^4}{8}\left(\frac{1}{5}-\frac{1}{9}\right)=\frac{\pi^4}{90}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Given $x+y$ and $x\cdot y$, what is $x^3+ y^3$ ? I have been looking at an assortment of high school number sense tests and I noticed a reoccurring problem that states what x+y is and what $x\cdot y$ is then asks for $x^3+ y^3$. I want to know how to work these problems. I have a couple of examples.
$x+y=5$ and $x\cdot y=1$, then $x^3+y^3=?$ [key says 110]
$x+y=-1$ and $x\cdot y=2$, then $x^3+y^3=?$ [key says 5]
$x-y=-1$ and $x\cdot y=2$, then $x^3 -y^3=?$ [key says -7]
$x+y=\frac{1}{3}$ and $x\cdot y=\frac{1}{9}$, then $x^3+y^3=?$ $\left[-\dfrac{2}{27}\right]$
| Use the identity
$$(x+y)^3=x^3+3x^2y+3xy^2+y^3= x^3+y^3+3xy(x+y).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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To prove that the centre of 2 circles and the two points at which the 2 circles cut and the origin lie on a circle. Let the circles
$$x^2+y^2-2cy-a^2=0~~~~and~~~~x^2+y^2-2bx+a^2=0$$
with centres at $A$ and $B$ intersect at $P$ and $Q$. Show that the points $A,B,P,Q$ and $O=(0,0)$ lie on a circle.
My work:
I found out the centre of the 2 circles to be $A(0,c),B(b,0)$ and also found out that the points at the which circle cut, $P,Q$ are the common chord of both the circles. So,I found out the equation of the chord of the circle to be $bx-cy=a^2$. Now, I am stuck. I tried to figure out that these points are concylic by considering two quadrilaterals formed by these points but nothing to progress. Please help!
| Let $C_1$ be the circle defined by the equation $x^2+y^2-2cy-a^2=0$ and $C_2$ be the circle defined by the equation $x^2+y^2-2bx+a^2=0.$ Now $$x^2+y^2-2cy-a^2=0 \\ \implies (x-0)^2+(y-c)^2=\left(\sqrt{a^2+c^2}\right)^2 \text{and}\\ x^2+y^2-2bx+a^2=0\\ \implies (x-b)^2+(y-0)^2=\left(\sqrt{b^2-a^2}\right)^2.$$ This implies that $C_1$ is a circle with it's centre as $A=(0,c)$ and radius $R_1=\sqrt{a^2+c^2}$ and $C_2$ is a circle with it's centre as $A=(b,0)$ and radius $R_2=\sqrt{b^2-a^2}.$
Now in $\Delta AQB$, we have $$AB^2=(b-0)^2+(0-c)^2=b^2+c^2, \\ AQ^2=R_1^2=a^2+c^2 \text{ and } BQ^2=R_2^2=b^2-a^2, \\ \implies AB^2=AQ^2+BQ^2.$$
Therefore $\Delta AQB$ is a right-angled triangle with $\angle AQB=90^0.$
Now we know that $\Delta AQB \cong \Delta APB,$ which implies that $\angle APB=90^0$.
Therefore, in quadrilateral $AQBP$, we have $\angle APB+\angle AQB=180^0,$ which implies that $AQBP$ is a cyclic quadrilateral. This implies that points $A,Q,B,P$ lie on a circle having $AB$ as the diameter.
Now, we also have $AOB=90^0,$ which implies that $O$ lies on the same circle defined by the points $A,Q,B,P$.
Hence, the points $A,B,P,Q$ and $O$ lie on a circle.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this $\frac{1}{b_{k+1}b_{k+2}}+\frac{1}{b_{k+2}b_{k+3}}+\cdots+\frac{1}{b_{2k}b_{2k+1}}>\frac{1}{12345},$ let sequence $\{b_{n}\}$,and $b_{n}>0$,let
$$S_{n}=b_{1}+b_{2}+\cdots+b_{n}\le n^{\frac{3}{2}},\forall n\ge 1$$
show that
$$\dfrac{1}{b_{k+1}b_{k+2}}+\dfrac{1}{b_{k+2}b_{k+3}}+\cdots+\dfrac{1}{b_{2k}b_{2k+1}}>\dfrac{1}{12345},k\ge 1$$
I think this problem maybe
$$b_{n}=S_{n}-S_{n-1}?$$
But this problem I can't it.Thank you
| Since the function $\frac{1}{x^2}$ is a convex function for $x > 0$, for any $k$ positive numbers $x_i, i = 1,\ldots,k$, we can apply Jensen's inequality to conclude
$$\frac{1}{k} \sum_{i=1}^k\frac{1}{x_i^2}
\ge \frac{1}{\left(\frac{1}{k} \sum\limits_{i=1}^k x_i\right)^2}
\quad\implies\quad
\sum_{i=1}^k\frac{1}{x_i^2} \ge \frac{k^3}{\left(\sum\limits_{i=1}^k x_i\right)^2}
$$
Now take $\;\displaystyle x_i = \frac{b_{k+i}+b_{k+i+1}}{2}$ for $i = 1,\ldots, k$
and notice
$$\sum_{i=1}^k x_i = \frac12\left(S_{2k+1}+S_{2k} - S_{k+1} - S_{k}\right) \le S_{2k+1} \le (2k+1)^{3/2}$$
We obtain
$$\sum_{i=1}^k\frac{1}{b_{k+i}b_{k+i+1}}
\ge \sum_{i=1}^k\frac{1}{x_i^2}
\ge \frac{k^3}{\left(\sum\limits_{i=1}^n x_i\right)^2}
\ge \frac{k^3}{S_{2k+1}^2}
\ge \left(\frac{k}{2k+1}\right)^3
\ge \frac{1}{27} > \frac{1}{12345}
$$
because $k \ge 1$.
| {
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How to show $2\sum^{n/2}_{k=0}$ $(\frac{1}{2}-\frac{k}{n})\binom{n}{k} $ = $\frac{1}{2}$ $\binom{n}{n/2}$ How to show:
$$2\sum^{n/2}_{k=0}\left(\frac{1}{2}-\frac{k}{n}\right)\binom{n}{k}=\frac{1}{2}\binom{n}{n/2}$$
n:even
please could you help with this equality.
on page 17: Rivlin, an intro to approximation of functions
| Assuming you mean for $n$ to be even, write $n=2r$; then you want to show that
$$2\sum_{k=0}^r \left(\frac{1}{2}-\frac{k}{2r}\right)\dbinom{2r}{k} = \frac{1}{2}\dbinom{n}{n/2}.$$
Expanding the left-hand side gives
$$\sum_{k=0}^r \dbinom{2r}{k} - 2\sum_{k=0}^r \frac{k}{2r}\dbinom{2r}{k}.$$
The first term is the sum of the first $r+1$ terms in that row of Pascal's Triangle; it evaluates to
$$\frac{2^n-\binom{n}{n/2}}{2} + \dbinom{n}{n/2} = 2^{n-1} + \frac{1}{2}\dbinom{n}{n/2}.$$
(To see this, draw a picture of the row and figure out how to get the sum of those terms given that the sum of all terms in the row is $2^n$.)
The second term simplifies as
$$2\sum_{k=0}^r \frac{k}{2r}\cdot\frac{(2r)!}{k!(2r-k)!} = \sum_{k=1}^r \dbinom{2r-1}{k-1} = 2\sum_{k=0}^{r-1} \dbinom{2r-1}{k} = 2\cdot 2^{2r-2} = 2^{2r-1} = 2^{n-1}. $$
Putting the two together gives
$$2^{n-1} + \frac{1}{2}\dbinom{n}{n/2} - 2^{n-1} = \frac{1}{2}\dbinom{n}{n/2}.$$
| {
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Prove that, $\int_0^{\pi/2}\frac{\sin (2n+1)x}{\sin x}~dx=\frac{\pi}{2}$ Prove that, $$\int_0^{\pi/2}\dfrac{\sin (2n+1)x}{\sin x}~dx=\dfrac{\pi}{2}$$ For every integer $n\ge 0$
I tried using induction without any help. Possibly my method of application was not right. Please help me!
| Firstly, note that $$\sin (z+2)x - \sin zx = 2\sin x \cos (z+1) x \iff \frac{\sin (z+2)x - \sin zx}{\sin x} = 2\cos (z+1)x$$
Set $z=2n-1$ and integrate to get:
$$\int_0^{\frac{\pi}{2}} \frac{\sin (2n+1)x}{\sin x} dx - \int_0^{\frac{\pi}{2}} \frac{\sin (2n-1)x}{\sin x}dx = \int_0^{\frac{\pi}{2}} 2\cos (2n)x\ dx = \frac{2}{2n} \sin \left[(2n)\cdot \frac{\pi}{2}\right] = 0$$
Therefore we get the recurrence relation:
$$\int_0^{\frac{\pi}{2}} \frac{\sin (2n+1)x}{\sin x} dx = \int_0^{\frac{\pi}{2}} \frac{\sin (2n-1)x}{\sin x}dx =\cdots = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x}dx = \frac{\pi}{2}$$
| {
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Equation of a line... This is a really embarrassing question, but could someone remind me why for any (continuous, to make this simple) function defined on $[a,b]$, the lines
$$L_1 = m(x - a) + f(a)$$
and $$L_2 = m(x - b) + f(b)$$
are equivalent? Where $m = \frac{f(b) - f(a)}{b - a}$
| Or for the brute force approach, just put them both in $mx + b$ form:
$$L_1 = m \cdot (x - a) + f(a)$$
$$L_1 = \frac {f(b) - f(a)}{b - a} \cdot (x - a) + f(a)$$
$$L_1 = \frac {f(b) - f(a)}{b - a}x - \frac {f(b) - f(a)}{b - a} a + f(a)$$
$$L_1 = \frac {f(b) - f(a)}{b - a}x + \frac {-f(b)a + f(a)a + f(a)b - f(a)a}{b - a}$$
$$L_1 = \frac {f(b) - f(a)}{b - a}x + \frac {f(a)b-f(b)a}{b - a}$$
$$L_2 = m \cdot (x - b) + f(b)$$
$$L_2 = \frac {f(b) - f(a)}{b - a} \cdot (x - b) + f(b)$$
$$L_2 = \frac {f(b) - f(a)}{b - a}x - \frac {f(b) - f(a)}{b - a} b + f(b)$$
$$L_2 = \frac {f(b) - f(a)}{b - a}x + \frac {-f(b)b + f(a)b + f(b)b - f(b)a}{b - a}$$
$$L_2 = \frac {f(b) - f(a)}{b - a}x + \frac {f(a)b - f(b)a}{b - a}$$
$$L_1 = L_2$$
For the less brute force approach, a function $y = f(x - k) + j$ is the function $f$ shift right by $k$ and up by $j$ (this is very important for understanding functions).
Your function is $y = mx$.
So in case $L_1$ we shifted right by $a$ and up by $f(a)$. In case $L_2$ we shifted right by $b$ and up by $f(b)$.
$L_2$ is $(b - a)$ more to the right than $L_1$. $L_2$ is $f(b) - f(a)$ more up than $L_1$. But that ratio is your slope, so you have the same line.
| {
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Find the least nonnegative residue Find the least nonnegative residue of $5^{18} \mod 11$.
To do this I took $5^2 \equiv 3 \mod 11$. Then I did $(5^2)^5 \equiv 3^5 \mod 11$. And $3^5 \equiv 1 \mod 11$.
So now I have $5^{10} \equiv 1 \mod 11$. Then I multiplied both sides by $5^8$ to get $5^{18} \equiv 5^8 \mod 11$. So I believe $5^8$ is the least nonnegative residue but I am not entirely sure. Can someone please confirm that this is correct?
| Best way to do this is to repeatedly divide the exponent by 2.
$$
18 = 2 \times 9, ~ 9 = 2 \times 4 + 1,~ 4=2 \times 2
$$
So
$$
5^2 = 25 \equiv 3 \mod 11
\\
5^4 \equiv 3^2 = 9 \mod 11
\\
5^8 \equiv 9^2 = 81 \equiv 4 \mod 11
\\
5^9 =5 \cdot 5^8 \equiv 20 \equiv 9 \mod 11
\\
5^{18} \equiv 9^2 = 81\equiv 4 \mod 11
$$
Hence the answer is $4$.
Note you can save a lot of work by noting that by Fermat's theorem $5^{10} \equiv 1\mod11$
Hence
$$5^{18} = 5^{10} 5^8 \equiv 5^8 \equiv 4 \mod 11$$
| {
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Proving an integral using a series If $f:(0,1]\rightarrow \textbf {R}$ is defined by $f(x)=2nx$ for $\frac{1}{n+1}\leq x \leq \frac 1n$ and $n$ is a natural number, assuming that $\sum_{k=1}^{k=\infty}1/k^2=\pi^2/6$, show that $\int_0^1 f(x)dx=\pi^2/6$.
So I wrote the summation of the integral from $x=\frac{1}{n+1}$ to $x=\frac{1}{n}$ of $f(x)dx$. After I integrate I'm left with the infinite series of $\frac{1}{k}-\frac{k}{(k+1)^2}$ from $k=1$ to $k=\infty$? Not sure where to go from here.
| I feel a bit silly, as I missed the clearly correct way to do this.
As you say, you manipulate the integral into the sum
$$ \sum_{k \geq 1} \frac{1}{k} - \frac{k}{(k+1)^2}.$$
We now use the deep fact that $k = k + 1 - 1$ to finish.
$$ \begin{align}
\sum_{k \geq 1} \frac{1}{k} - \frac{k}{(k+1)^2} &= \sum_{k \geq 1} \frac{1}{k} - \frac{k+1}{(k+1)^2} + \frac{1}{(k+1)^2} \\
&= \sum_{k \geq 1} \frac{1}{k} - \frac{1}{k+1} + \frac{1}{(k+1)^2} \\
&= \sum_{k \geq 1} \frac{1}{k^2} = \frac{\pi^2}{6}.
\end{align}$$
| {
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Integration: U-Substitution Question:(part a)
By using the substitution, $ u=2x+7$
So here is what I did, however, halfway through I had to look at wolframalpha, and don't quite understand what they do to get the answer.
$\int$ $x$$\sqrt{2x+7}$ $dx$
$du$=$2 $ $dx$
$=\frac{1}{4}$$\int$$(u-7)$$\sqrt{u}$ $du$
$=\frac{1}{4}$$\int$$(u^3/_2-7\sqrt{u})$ $du$
$=\frac{1}{4}$$\int$$(u^3/_2 $$du$$ -\frac{7}{4})$ $\int$$\sqrt{u}$ $du$
Here where they separate the integrals, I don't understand why they do it and if it is a calculus principle I do not know, and if that's the case, can someone please explain to me? thanks. Also this next step as they integrate, they don't integrate the fraction $\frac{7}{4}$,which reminds me, how do they even get the fraction? It must be some piece of calculus about Integration I am missing out on.
$=$$\frac{u^5/_2}{10}$$-$$\frac{7}{4}$ $\int$$\sqrt{u}$ $du$
I do, however, understand after this point, but of-course I am unsure of the method used above.
$=$$\frac{u^5/_2}{10}$$-$$\frac{7u^3/_2}{6}$ $+$ $C$
$=$$\frac{(2x+7)^5/_2}{10}$$-$$\frac{7(2x+7)^3/_2}{6}$$+$ $C$
$\therefore$$\int$ $x$$\sqrt{2x+7}$ $dx$ = $\frac{\sqrt(2x+7)^5}{10}$$-$$\frac{7\sqrt(2x+7)^3}{6}$$+$ $C$
Which indeed is the correct answer.
Can anyone help me with how wolframalpha split the integral at the middle part and also why they didn't integrate $\frac{7}{4}$.Help is greatly appreciated thank you.
(part B) - I don't have a clue on how to do this one.
It's the same integral however they make, $u^2=2x+7$
if anyone can help me with this as well please?
| If $u^2 = 2x + 7$ then $ x = \frac{u^2 - 7}{2}$ and $dx = udu$
$ \int x \sqrt{2x + 7} dx \ = \ \int \frac{u^2 - 7}{2} \sqrt{u^2} u du \ = \ \frac{1}{2} \int (u^4 - 7u^2) du$
$ \frac{1}{2} [\frac{u^5}{5} - \frac{7u^3}{3} ] \ + \ C \ = \ \frac{ (2x + 7)^2 \sqrt{2x + 7} }{10} \ - \ \frac{ 7(2x + 7) \sqrt{2x + 7} }{6} \ + \ C$
| {
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How to solve for getting average of them I am trying to solve this problem
If x+y= 6, y+z= 7 , and x+z= 9 , what is the average of x , y and z ?
Well how i am trying to do is getting the value of 'X' , 'Y' and 'Z' which are X = 4 , Y = 2 , Z = 5 then i take their average as 4+2+5 then 11/3 , but i this is not giving me the right answer. Any help appriciated
| $$\begin{align} \underbrace{(x+y)}_{=\,6} + \underbrace{(y+z)}_{=\,7} + \underbrace{(x + z)}_{=\,9} & = 2x + 2y + 2z \\ &= 2(x + y + z) = 6+7+9 \\ &\iff x + y + z =\frac 12(6+7+9)=11\end{align}$$
To compute the average, now, calculate $\dfrac{x + y+z}{3} = \dfrac {11}3$.
| {
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Distance between point and plane Find the distance from the Point $A = (1,0,2)$ to the plane passing through the point $(1,-2,1)$ and perpendicular to the line given by the parametric equations:
$$
\begin{align}
x & = 7, \\
y & = 1 + 2t, \\
z & = -3 + t.
\end{align}
$$
The answer is $\sqrt{5}$, but I can't seem to get that. I get that the plane equation ends up being $0x + 2y + z + 3 = 0$, but then when I try to compute the distance it turns out to be $\sqrt{3}/\sqrt{5}$ or something along those lines.
| We will utilize the point-plane distance formula [1].
$$D=\frac{\left\lvert ax_0+by_0+cz_0+d \right\rvert}{\sqrt{a^2+b^2+c^2}}$$
We already know that the point in question is $A(1,0,2)$ which means $x_0=1$, $y_0=0$, and $z_0=2$.
$$D=\frac{\left\lvert a(1)+b(0)+c(2)+d \right\rvert}{\sqrt{a^2+b^2+c^2}}=\frac{\left\lvert a+2c+d \right\rvert}{\sqrt{a^2+b^2+c^2}}$$
Since the plane in question is orthogonal to the given line, we can find a normal vector to the plane that is contained in the line. The line contains the points $P(7,1,-3)$ and $Q(7,3,-2)$ when $t=0$ and $t=1$, respectively. The vector from $P$ to $Q$ can be found easily. It is orthogonal to the plane.
$$\vec{\mathbf{n}}=\langle(7-7),(3-1),(-2+3)\rangle=\langle0,2,1\rangle$$
We can place the components of this vector into the general equation of a plane in $\mathbb{R}^3$ [2].
$$0x+2y+1z+d=0\implies2y+z+d=0$$
This plane contains the point $(1,-2,1)$. We can solve for the constant $d$.
$$2(-2)+(1)+d=0\implies-3+d=0\implies d=3$$
We can now return to the distance formula and finish solving the problem with $a=0$, $b=2$, $c=1$, and $d=3$.
$$D=\frac{\left\lvert a+2c+d \right\rvert}{\sqrt{a^2+b^2+c^2}}=\frac{\left\lvert (0)+2(1)+(3)\right\rvert}{\sqrt{(0)^2+(2)^2+(1)^2}}=\frac{5}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}}=\frac{5\sqrt{5}}{5}=\sqrt{5}$$
| {
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Scratch work for delta-epsilon proof for $\lim_{x \to 13} \sqrt{x-4} = 3$
Prove $\lim_{x \to 13} \sqrt{x-4} = 3$.
We need to show for all $E> 0$ there exists $D > 0$ such that
if $0 < |x - 13| < D$, then $|\sqrt{x-4} - 3| < E$. Let me write D for delta and E for epsilon please.
Scratch-work here: $\color{darkred}{\text{ I understand the formal proof. Ergo just asking about this. } }$
Note that $|\sqrt{x-4} - 3| = |\sqrt{x-4} - 3|\dfrac{|\sqrt{x-4} + 3|}{|\sqrt{x-4} + 3|} = \dfrac{|x - 13|}{|\sqrt{x-4} + 3|}$.
We can bound $|x - 13|$ for any choice of D, but we need a certain D to also bound $|\sqrt{x-4} + 3|$.
1. Ishfaaq's answer says 'the denominator will definitely cause trouble.' How?
Why won't bounding $|x - a|, |x - 13|$ bound the whole statements?
Beneath pp 83 of Spivak claims this too. How does '$|x + a|$ cause trouble'?
Assume that $D < 1.$
2. What sanctioned assuming $D < 1$? How to know if $1$ is too small, too big? $\\$ In pp 83 of Spivak beneath, $D$ looks randomly chosen?
4. Ishfaaq's last paragraph. $\color{darkred}{\left|{x^2 - a^2}\right| \lt \delta^2 + \left|{}2a\right|\delta. \text{ But we can hardly equate this to ϵ... }}$
Why not? $d^2 + \left|{}2a\right|d = e \iff d(d + |2a|) = e$?
Ishfaaq says $a = 1/2, e = d/2$ is a counterexample. But then $d^2 + 2|a|d < e \iff d^2 + d < d/2 \iff d(d - \frac{1}{2}) < 0 \iff 0 < d < 1/2.$ What founders?
If this works, then any smaller D will also work.
Then $|x - 13| < 1 \iff -1 < x - 13 < 1 \iff 12 < x < 14 \iff 8 < x - 4 < 10$
$\implies \sqrt(8) < \sqrt{x - 4} < \sqrt(10) \iff \color{blue}{\sqrt(8) + 3 < \sqrt{x - 4} + 3 < \sqrt(10) + 3} \\ \iff 0 < \color{blue}{\dfrac{1}{\sqrt(10) + 3} < \dfrac{1}{\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}} $
$\implies \color{green}{\frac{1}{|\sqrt{x - 4} + 3|} < \frac{1}{\sqrt(8) + 3}} $
So then $|x - 13|\color{green}{\dfrac{1}{|\sqrt{x - 4} + 3} < \dfrac{1}{\sqrt(8) + 3}}|x - 13|$ and need this $ < E.$
Thus, we need $|x - 13| < E * \color{green}{(\sqrt{8} + 3)}$. So this is our choice of D. But note that this only works when $D < 1$. Thus, we can take care of both conditions by choosing $D = \min\{1, E(\sqrt{8}+3)\}.$ █
$\color{darkred}{ \text{ 3. All this algebra fazed me. How to graph this to view all this algebra and $D$? Thanks. } }$
Also tried http://www.ocf.berkeley.edu/~yosenl/math/epsilon-delta.pdf.
| Note that $\sqrt{x-4}+3\geqslant3$ hence $|\sqrt{x-4}-3|\leqslant\frac13|x-13|$. Thus, if $|x-13|\leqslant3\varepsilon$ then $|\sqrt{x-4}-3|\leqslant\varepsilon$.
Note that the function $x\mapsto\sqrt{x-4}$ is only defined on $x\geqslant4$ hence the condition $|x-13|\leqslant3\varepsilon$ guarantees that $\sqrt{x-4}$ exists, provided $\varepsilon\leqslant3$ (which is more than enough to conclude). A formula valid for every $\varepsilon$ would be $\delta=\min\{9,3\varepsilon\}$.
| {
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How to compute the value of $1 \cdot \frac{2}3 \cdot \frac 45 \cdots$, as $n$ goes to infinity? It's supposed to be $\dfrac 12\sqrt{\dfrac \pi n}$.
| Hint.
$$
1\cdot\frac{2}{3}\cdot\frac{4}{5}\cdots\frac{2n}{2n+1}=\frac{2^{2n+1} n!(n+1)!}{(2n+2)!}
$$
Now we use Stirling's asymptotic approximation
$$
\lim_{n\to\infty}\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n}=1,
$$
and obtain
$$
\frac{2^{2n+1} n!(n+1)!}{(2n+2)!}=\frac{\frac{n!}{\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n}
\cdot\frac{(n+1)!}{\sqrt{2\pi (n+1)}\left(\frac{n+1}{\mathrm{e}}\right)^{n+1}}}{\frac{(2n+2)!}{\sqrt{2\pi (2n+2)}\left(\frac{2n+2}{\mathrm{e}}\right)^{2n+2}}}\cdot
\frac{2^{2n+1}\sqrt{2\pi (2n+2)}\left(\frac{2n+2}{\mathrm{e}}\right)^{2n+2}}{\sqrt{2\pi n}\left(\frac{n}{\mathrm{e}}\right)^n\cdot\sqrt{2\pi (n+1)}\left(\frac{n+1}{\mathrm{e}}\right)^{n+1}}.
$$
The first factor of the right-hand side tends to $1$.
| {
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prove by induction: $3 + 5 + 7 + ... + (2n+1) = n(n+2)$
Use the principle of mathematical induction to prove that $$3 + 5 + 7 + ... + (2n+1) = n(n+2)$$ for all n in $\mathbb N$.
I have a problem with induction. If anyone can give me a little insight it would be helpful.
| If you already know that
$$1+2+3+...+n=\frac{n(n+1)}2$$
we can try the following alternative approach:
$$3+5+7+\ldots+(2n+1)=$$
$$=1+2+3+4+5+\ldots+(2n+1)+(2n+2)-1-2(1+2+3+\ldots+ n+n+1)=$$
$$=\frac{(2n+2)(2n+3)}2-1-2\frac{(n+1)(n+2)}2=(n+1)(2n+3)-1-(n+1)(n+2)=$$
$$=(n+1)\left[2n+3-n-2\right]-1=(n+1)^2-1=n^2+2n=n(n+2)$$
| {
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Factoring the quintic polynomial $x^5+4x^3+x^2+4=0$ I am trying to factor
$$x^5+4x^3+x^2+4=0$$
I've used Ruffini's rule to get
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
But I don't know what to do next.
The solution is $(x+1) (x^2+4) (x^2-x+1) = 0$. I've tried using the completing square method but with no result. Could you give me hints?
| Here I think this will work . Main idea being splitting up of $5x^2$ as $4x^2 + x^2$
$$(x+1)(x^4-x^3+5x^2-4x+4)=0$$
$$(x+1) (x^4-x^3+4x^2+ x^2-4x+4)=0 $$
$$(x+1) (x^4+4x^2+ x^2-x^3-4x+4)=0 $$
$$(x+1)[ x^2(x^2+4)+ (1-x)( x^2 +4)]=0 $$
$$(x+1)[ (x^2+4)(x^2+1-x))=0 $$
$$(x+1)[ (x^2+4)(x^2-x+1))=0 $$
Hence proved
| {
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Find the $z\in \Bbb C$ such that $\bar z=z^3$ Here is what I've done. Let $z\in \Bbb C$ be of the form $a+ib$, lets assume in this case that $a,b\not=0$. We have that $\bar z=a-ib$, and $z^3=a^3-3ab^2+i(3a^2b-b^3)$, so to find the correct $z$, it has to satisfy:
$$\begin{cases}
a= a^3-3ab^2\\
-b= 3a^2b-b^3\\
\end{cases}$$
wich is the same as:
$$\begin{cases}
a=a( a^2-3b^2)\\
-b=b (3a^2-b^2)\\
\end{cases} \Rightarrow \begin{cases}
1=a^2-3b^2\\
1=b^2 -3a^2\\
\end{cases}$$
then, $a^2-3b^2=b^2 -3a^2 \Rightarrow 4a^2=4b^2 \Rightarrow a^2=b^2 \Rightarrow \begin{cases}
a=b \;\text{ or,}\\
a=-b\\
\end{cases}$
I thought this was correct, and to try it out I went to Wolfram, and it turns out that the examples that came to my mind didn't satisfied this. What did I do wrong?
| Ana, take a look at this:
Say we have $a+b=1$ and $a-b=1$ It follows very easily that $a=1$ AND $b=0$. However, with your approach, we could say $a+b=a-b$ and here we have infinite options for $a$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/682620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove a generator of $\mathrm{SL}_2(\mathbb{R})$ Definitions
Let $\mathrm{SL}_2(\mathbb{R}) := \left \{M := \begin{pmatrix} a & b\\ c & d\end{pmatrix} \in \mathbb{R}^{2 \times 2}: \det{M} = 1 \right \}$ with matrix multiplication be a group of $2 \times 2$ matrices.
Let
\begin{align}
A_\lambda :&= \begin{pmatrix} \lambda & 0\\ 0 & \lambda^{-1}\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R}) \text{ with }\lambda \in \mathbb{R} \setminus \{0\}\\
B_t :&= \begin{pmatrix} 1 & t\\ 0 & 1\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R}) \text{ with } t \in \mathbb{R}\\
C :&= \begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R})
\end{align}
be elements of $\mathrm{SL}_2(\mathbb{R})$.
Question
Do $A_\lambda, B_t$ and $C$ generate $\mathrm{SL}_2(\mathbb{R})$?
My try
I think they do. When I could show that
$$\begin{pmatrix} a & b\\ c & d\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R})$$
can be multiplied with a combination of the three matrices $A, B, C$ in such a way that the result is the neutral element of this group
$$\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$$
then I could generate the matrix by multiplying the inverse matrices.
The inverse matrices can be generated because:
\begin{align}
A_\lambda^{-1} &= A_{\frac{1}{\lambda}}\\
B_t^{-1} &= B_{-t}\\
C^{-1} &= C^3
\end{align}
Now the only thing left is to get from
$$M = \begin{pmatrix} a & b\\ c & d\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R})$$
to
$$\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$$
Ok. So let's try it.
Case 1: $a = 0$
As $\det(M)=1 = ad - bc = 0d-bc$ we know that $bc=-1$. Especially is $c \neq 0$.
$$\begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix} \cdot \begin{pmatrix} a & b\\ c & d\end{pmatrix} = \begin{pmatrix} c & d\\ -a & -b\end{pmatrix}$$
Continue with Case 2.
Case 2: $a \neq 0$
Normalize ($M \cdot A_{\frac{1}{a}}$):
$$\begin{pmatrix} a & b\\ c & d\end{pmatrix} \cdot \begin{pmatrix} \frac{1}{a} & 0\\ 0 & a\end{pmatrix} = \begin{pmatrix} 1 & ab\\ \frac{c}{a} & ad\end{pmatrix}$$
Continue with Case 3.
Case 3: $a=1$
Get a $0$ with $M \cdot B_{-b}$:
$$\begin{pmatrix} 1 & b\\ c & d\end{pmatrix} \cdot \begin{pmatrix} 1 & -b\\ 0 & 1\end{pmatrix} = \begin{pmatrix} 1 & 0\\ c & d-bc\end{pmatrix}$$
We know that $\det{M} = 1 = ad - bc = d - bc$.
Continue with Case 4.
Case 4: $a=1$, $b=0, d=1$
At this stage we have matrices that look like
$$\begin{pmatrix} 1 & 0\\ c & 1\end{pmatrix}$$
This is where I'm stuck. Can you help me?
| In Case 4 you have $d=1$ since the determinant is still $1$, then
$$
A_{-1} C B_c C \pmatrix{1 & 0 \\ c & 1} = \pmatrix{1 & 0 \\ 0 & 1}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/683390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Let $p$ be prime and $\left(\frac{-3}p\right)=1$. Prove that $p$ is of the form $p=a^2+3b^2$
Let $p$ be prime and $\left(\frac{-3}p\right)=1$, where $\left(\frac{-3}p\right)$ is Legendre symbol. Prove that $p$ is of the form $p=a^2+3b^2$.
My progress:
$\left(\frac{-3}p\right)=1 \Rightarrow$ $\left(\frac{-3}p\right)=\left(\frac{-1}p\right)\left(\frac{3}p\right)=(-1)^{\frac{p-1}2}(-1)^{\left\lfloor\frac{p+1}6\right\rfloor}=1 \Rightarrow$ $\frac{p-1}2+\left\lfloor\frac{p+1}6\right\rfloor=2k$ I'm stuck here. This is probably not the way to prove that.
Also tried this way:
$\left(\frac{-3}p\right)=1$, thus $-3\equiv x^2\pmod{p} \Rightarrow$ $p|x^2+3 \Rightarrow$ $x^2+3=p\cdot k$
stuck here too.
Any help would be appreciated.
| First part:
$$\left(\frac{-3}{p}\right)=1 \text{ if and only if }\; p\equiv{1}\!\!\!\!\pmod{3}.\tag{1}$$
This can be achieved through the Gauss quadratic reciprocity theorem in the most general form, or through the following lines. If $p=3k+1$, by the Cauchy theorem for groups there is an order-3 element in $\mathbb{F}_p^*$, say $\omega$; from $\omega^3=1$ follows $\omega^2+\omega+1\equiv 0\pmod{p}$, hence:
$$(2\omega+1)^2 = 4\omega^2+4\omega+1 = 4(\omega^2+\omega+1)-3 = -3,$$
and $-3$ is a quadratic residue $\pmod{p}$. On the other hand, if $-3$ is the square of something $\pmod{p}$, say $-3\equiv a^2\pmod{p}$, then:
$$\left(\frac{a-1}{2}\right)^3\equiv\frac{1}{8}(a^3-3a^2+3a-1)\equiv\frac{1}{8}\cdot 8\equiv{1},$$
and $\frac{a-1}{2}$ is an order-3 element in $\mathbb{F}_{p}^*$. From the Lagrange theorem for groups it follows that $3|(p-1)$.
Second part:
$$\text{If }p\equiv 1\pmod{3},\qquad p=a^2+3b^2.\tag{2}$$
Since by the first part we know that $-3$ is a quadratic residue $\pmod{p}$, there exists an integer number $c\in[0,p/2]$ such that:
$$ c^2+3\cdot 1^2 = k\cdot p.\tag{3}$$
The trick is now to set a "finite descent" in order to have $k=1$. Let $d$ the least positive integer such that $c\equiv d\pmod{k}$. Regarding $(3)$ mod $k$, we have:
$$ d^2+3\cdot 1^2 = k\cdot k_1.\tag{4}$$
Since the generalized Lagrange identity states:
$$(A^2+3B^2)(C^2+3D^2)=(AC+3BD)^2 + 3(BC-AD)^2,\tag{5}$$
by multiplying $(3)$ and $(4)$ we get:
$$ (cd+3)^2 + 3(c-d)^2 = k^2 pk_1.$$
Since $cd+3\equiv c^2+3\equiv 0\pmod{k}$ and $c\equiv d\pmod{k}$, we can rewrite the last line in the following form:
$$ \left(\frac{cd+3}{k}\right)^2+3\left(\frac{c-d}{k}\right)^2 = k_1\cdot p.\tag{6}$$
Now a careful analysis of the steps involved in the algorithm reveals that $k_1<k$, so the descent is able to reach $k_i=1$, or:
$$ p = a^2 + 3b^2$$
as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/685958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
"answer_count": 3,
"answer_id": 1
} |
How to determine if $f(z)=\frac{i}{z^8}$ is analytic? How can I prove that a function like $$f(z)=\dfrac{i}{z^8}$$ is analytic or not?
I have to use Cauchy–Riemann equations but I can't find the $u$ and $v$ functions.
| First, remember how complex division works:
$$
\frac{1}{w}=\frac{\overline{w}}{\lvert w\rvert ^2}.
$$
Also, recall that complex conjugates break up over multiplication: $\overline{uw}=\bar{u}\bar{w}$.
So, in all, if $z=x+iy$, then
$$
\frac{i}{z^8}=i\left(\frac{\overline{z^8}}{\lvert z^8\rvert ^2}\right)=i\left(\frac{(\bar{z})^8}{\lvert z\rvert^{16}}\right)=i\frac{(x-iy)^8}{(x^2+y^2)^8}.
$$
Now, by the Binomial Theorem,
$$
(x-iy)^8=\sum_{n=0}^{8}\binom{8}{n}x^{8-n}(-iy)^{n},
$$
from which we get
$$
\begin{align*}
\Re[(x-iy)^8]&=x^8-28x^6y^2+70x^4y^4-28x^2y^6+y^8,\\
\Im[(x-iy)^8]&=-8x^7y+56x^5y^3-56x^3y^5+8xy^7.
\end{align*}
$$
So, in all, you want to take the following for $u$ and $v$:
$$
\begin{align*}
u(x,y)&:=\Re\left[i\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=-\Im\left[\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=\frac{8x^7y-56x^5y^3+56x^3y^5-8xy^7}{(x^2+y^2)^8}
\end{align*}
$$
and
$$
\begin{align*}
v(x,y)&:=\Im\left[i\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=\Re\left[\frac{(x-iy)^8}{(x^2+y^2)^8}\right]\\
&=\frac{x^8-28x^6y^2+70x^4y^4-28x^2y^6+y^8}{(x^2+y^2)^8}.
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/688402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Maclaurin series of $\frac{1}{1+\sin x}$ Find the terms through degree four of the Maclaurin series of $f(x)$.
$$f(x) = \frac{1}{1+\sin x}$$
My work:
The Maclaurin series for $\sin x$ up to degree $4$ is $x - \frac{x^3}{6} + \frac{x^5}{120}$
The Maclaurin series for $\frac{1}{1+x}$ up to degree $4$ is $1 - x + x^2 - x^3 + x^4$
I substituted $x - \frac{x^3}{6} + \frac{x^5}{120}$ for $x$ in $1 - x + x^2 - x^3 + x^4$
Did I do this right?
Plugging this into WolframAlpha, I get this: http://goo.gl/SKddyh
Which doesn't seem like the answer in the text: $1-x+x^2-\frac{5x^3}{6}+\frac{2x^4}{3}$
| Have a look at this:
Your equation, expanded
As you can see, what you plugged into WolframAlpha was the same equation as the answer in the text.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
How can I prove the trigonometric Problem? How can I show the following trigonometric problem :
$$\frac{1}{3}\leq \frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
\theta}\leq 3$$
I have tried in the following way :
$$
\begin{align}
& \phantom{\Rightarrow}
\frac{\sec^2\theta-\tan^2\theta}{\sec^2\theta+\tan^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{1}{\sec^2\theta+\tan^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{\cos^2\theta}{1+\sin^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{1-\sin^2\theta}{1+\sin^2
\theta}-\frac{1}{3} \\[8pt]
& \Rightarrow \frac{2-4\sin^2\theta}{3(1+\sin^2
\theta)} \\[8pt]
& \Rightarrow \frac{2\cos 2\theta}{3(1+\sin^2
\theta)}
\end{align}
$$
How can I show $ \dfrac{2\cos 2\theta}{3(1+\sin^2
\theta)}\geq 0$ ?
| I think you need to find $$\frac13\leq \frac{\sec^2\theta-\tan\theta}{\sec^2\theta+\tan\theta}\leq 3$$
Setting $\tan\theta=x,$ we get $$\frac{\sec^2\theta-\tan\theta}{\sec^2\theta+\tan\theta}=\frac{x^2+1-x}{x^2+1+x}$$
Let $\displaystyle\frac{x^2+1-x}{x^2+1+x}=y$
$\displaystyle\iff (y-1)x^2+(y+1)x+y-1=0$ which is a Quadratic Equation if $x$
As $x$ is real, the discriminant must be $\ge0$
$\displaystyle\implies (y+1)^2\ge4\cdot(y-1)^2\iff 3y^2-10y+3\le0\iff (3y-1)(y-3)\le0 $
As the product of two multiplicand is $\le0$
So, we need $\displaystyle(i)3y-1\ge0\iff y\ge\frac13$ and $\displaystyle y-3\le0\iff y\le3\implies\frac13\le y\le3$
or $(ii)3y-1\le0\iff y\le\frac13$ and $y-3\ge0\iff y\ge3$ which is impossible
Alternatively,
I leave for you the Second Derivative Test of $$f(x)=\frac{x^2+1-x}{x^2+1+x}$$ to find the extreme values of $f(x)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Find the value of a+b+c If $x+1$ is a factor of $ax^4 + bx^2 + c$, find the value of $a + b + c$?
I know that it is equal to zero, but I have to know How to do it.
| Let $p(x)=ax^4+bx^2+c$. We are given $p(x)=(x+1)q(x)$ for $q(x)$ some cubic polynomial. Now evaluate it at $x=-1$ and we get $p(-1)=a(-1)^4+b(-1)^2+c=a+b+c=0\cdot q(x)=0$ so $a+b+c=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/696646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
} |
Solving for $\sum_{n = 1}^{\infty} \frac{n^3}{8^n}$? I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if there is a better way to solve it. Here's how I did it:
First I decomposed the series into the following sums:
$S_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$S_2 = \frac{7}{64} + \frac{7}{512} + \dots = \frac{\frac{7}{64}}{\frac{7}{8}}$
$S_3 = \frac{19}{512} + \frac{19}{4096} + \dots = \frac{\frac{19}{512}}{\frac{7}{8}}$
And deduced that the sum can be written as $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{3n^2 - 3n + 1}{8^n}$
$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ is easy to evaluate -- it's $\frac{1}{7} $by geometric series
$\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ can be evaluated in a whole host of ways to get an answer of $\frac{8}{49}$.
It remains to evaluate $\displaystyle \sum_{n = 1}^{\infty} \frac{n^2}{8^n}$, for which I took a similar approach as the cubics by decomposing it into many sums:
$T_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$
$T_2 = \frac{3}{64} + \frac{3}{512} + \dots = \frac{\frac{3}{64}}{\frac{7}{8}}$
And so forth, coming to the conclusion that it is equal to $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{2n-1}{8^n}$
Now, I used this information and the above values for $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ and $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ to get the sum as $\frac{776}{2401}$, which is confirmed by WA.
So, I would like to reiterate here: Is there a simpler way to compute this sum, and are there any known generalizations for this problem given an arbitrary $a$ in the denominator and arbitrary $k$ as the exponent in the numerator?
| $\newcommand{\+}{^{\dagger}}
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With $p\ \ni\ \verts{p} < 1$:
\begin{align}
\sum_{n = 1}^{\infty}p^{n} &= {p \over 1 - p} = - 1 + {1 \over 1 - p}
\end{align}
Derive respect of $p$ and after that multiply by $p$:
\begin{align}
\sum_{n = 1}^{\infty}np^{n} &= {p \over \pars{1 - p}^{2}}
\\
\sum_{n = 1}^{\infty}n^{2}p^{n} &= -\,{p + p^{2} \over \pars{1 - p}^{3}}
\\
\sum_{n = 1}^{\infty}n^{3}p^{n} &= {p - 4p^{2} + p^{3} \over \pars{1 - p}^{4}}
\end{align}
Set $p = 1/8$:
$$\color{#00f}{\large%
\sum_{n = 1}^{\infty}n^{3}\pars{1 \over 8}^{n} =
\left.{p - 4p^{2} + p^{3} \over \pars{1 - p}^{4}}\right\vert_{p\ =\ 1/8}
= {776 \over 2401}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/698598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 0
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Different solutions of $x+y+z=10$ where $x$, $y$, $z$ are all positive integers and $x, y, z \leq 10$ The number of solutions to the equation $x+y+z=10$ where $x,y,z$ are positive integers, is given by ${k−1 \choose n−1}$, where in this case $k=10,n=3$, giving us ${9 \choose 2} = 36$
Now we have
$x + y + z = 10$ with $x, y, z \leq 10$ (where $x,y,z$ are positive integers and can be the same)
What are the different methods by which we can solve this?
| As Trismegistos says, this is, using $[z^k]$ to denote the coefficient of $z^k$ in what follows:
$$
[z^{10}] (z + z^2 + \cdots + z^{10})^3
$$
Now use the formula for the sum of a geometric series,
then expand the cube in the numerator and the denominator as a power series. Note that the terms in $z^{10}$ and higher in the first factor have no effect on the result:
\begin{align}
[z^{10}] z^3 \left( \frac{1 - z^{10}}{1 - z} \right)^3
&= [z^7] (1 - 3 z^{10} + 3 z^{20} - z^{30})
\cdot \sum_{k \ge 0} \binom{-3}{k} (-1)^k z^k \\
&= [z^7] \sum_{k \ge 0} \binom{-3}{k} (-1)^k z^k \\
&= (-1)^7 \binom{-3}{7} \\
&= \binom{7 + 3 - 1}{3 - 1} \\
&= \binom{9}{2} \\
&= 36
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/700216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find $\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15}}}{1+x^{2+\sqrt{3}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx$ I was challenged to prove this identity
$$\int_0^\infty\frac{\log\left(\frac{1+x^{4+\sqrt{15\vphantom{\large A}}}}{1+x^{2+\sqrt{3\vphantom{\large A}}}}\right)}{\left(1+x^2\right)\log x}\mathrm dx=\frac{\pi}{4}\left(2+\sqrt{6}\sqrt{3-\sqrt{5}}\right).$$
I was not successful, so I want to ask for your help. Can it be somehow related to integrals listed in that question?
| This integral can be evaluated in a closed form for arbitrary real exponents, and does not seem to be related to Herglotz-like integrals.
Assume $a,b\in\mathbb{R}$. Note that
$$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^\infty\frac{\ln\left(\frac{1+x^b}2\right)}{\ln x}\frac{dx}{1+x^2}.\tag1$$
Both integrals on the right-hand side have the same shape, so we only need to evaluate one of them:
$$\begin{align}&\phantom=\underbrace{\int_0^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{split the region}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^\infty\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{change variable}\ y=1/x}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}+\underbrace{\int_1^0\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln\left(y^{-1}\right)}\frac1{1+y^{-2}}\left(-\frac1{y^2}\right)dy}_\text{flip the bounds and simplify}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\underbrace{\int_0^1\frac{\ln\left(\frac{1+y^{-a}}2\right)}{\ln y}\frac{dy}{1+y^2}}_{\text{rename}\ y\ \text{to}\ x}\\&=\underbrace{\int_0^1\frac{\ln\left(\frac{1+x^a}2\right)}{\ln x}\frac{dx}{1+x^2}-\int_0^1\frac{\ln\left(\frac{1+x^{-a}}2\right)}{\ln x}\frac{dx}{1+x^2}}_\text{combine logarithms}\\&=\int_0^1\frac{\ln\left(\frac{1+x^a}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}=\underbrace{\int_0^1\frac{\ln\left(\frac{x^a\left(x^{-a}+1\right)}{1+x^{-a}}\right)}{\ln x}\frac{dx}{1+x^2}}_{\text{cancel}\ \ 1+x^{-a}}\\&=\int_0^1\frac{\ln\left(x^a\right)}{\ln x}\frac{dx}{1+x^2}=a\int_0^1\frac{dx}{1+x^2}=a\,\Big(\arctan1-\arctan0\Big)\\&=\vphantom{\Bigg|^0}\frac{\pi\,a}4.\end{align}\tag2$$
So, finally,
$$\int_0^\infty\frac{\ln\left(\frac{1+x^a}{1+x^b}\right)}{\ln x}\frac{dx}{1+x^2}=\frac\pi4(a-b).\tag3$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Determining all right-angled triangles with sides $a$, $a+k$ and $a+2k$
Determine all triangles that are right-angled which has the sides:
$$a, \quad b = a+k, \quad c = a+2k.$$
(a, k are natural numbers!)
So does this mean I have to take a look at:
*
*$a^2+b^2=c^2 \iff a^2-2ak-3k^2=0$
*$b^2+c^2=a^2 \iff a^2+6ak+5k^2=0$
*$a^2+c^2=b^2 \iff a^2+2ak+3k^2=0$
If so, I get (1.1) $a=3k$ (I did $f(a)=a^2-2ak-3k^2=0$)
in all other cases I get negative $k$ or negative $a$.
What's the best way to deal with this (I'll assume) simple task?
| Yes! Sure, in fact
$$b^2+c^2=a^2 \iff a^2+6ak+5k^2=0$$ means $$ a = -k, a = -5k$$
while
$$a^2+c^2=b^2 \iff a^2+2ak+3k^2=0$$ have a unique integer solution $$ a = k = 0.$$
So $$a^2+b^2=c^2 \iff a^2-2ak-3k^2=0$$ is the unique possible answer: $$ a = -k, a = 3k.$$
Finally the sides of all the possible triangles are of the form $(3k,4k,5k)$, the simplest Pitagorean triple!
| {
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"url": "https://math.stackexchange.com/questions/701729",
"timestamp": "2023-03-29T00:00:00",
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$a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$ $a,b,c$ are positive real numbers such that, $a+b+c\ge abc$. Prove that $a^2+b^2+c^2\ge \sqrt{3}abc$
My work:
I tried using Cauchy-Schwarz inequality to find that,
$(a^2+b^2+c^2)(1^2+1^2+1^2)\ge (a+b+c)^2$
$(a^2+b^2+c^2)\ge \dfrac13(a+b+c)^2$
$\sqrt{(a^2+b^2+c^2)}\ge \dfrac{1}{\sqrt{3}}(a+b+c)\ge \dfrac{1}{\sqrt{3}}abc$
which is not what I need and neither I can use it to prove the required inequality. Please help.
| here is another way
$$(a^2+b^2+c^2)^2=\sum\limits_{cyc} a^4 + \sum\limits_{cyc} 2a^2b^2 \ge abc(a+b+c) + 2abc(a+b+c) =3abc(a+b+c)\ge3a^2b^2c^2$$
inequality $\sum\limits_{cyc} a^4 \ge \sum\limits_{cyc} a^2b^2 \ge abc(a+b+c)$
can be proved using AM-GM or Cauchy-Schwarz.
| {
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Find whether there are integer solutions to $7x^2+9y^2=3932$. Is my solution correct? Find whether there are integer solutions to $7x^2+9y^2=3932$
Here's my attempt, I would like to know if it's correct please:
Lets assume $7x^2+9y^2=3932$ has integer solutions. That mean GCD of $x^2,y^2$ is 3932, or $(x^2,y^2)=3932$.
That means that there are $a,b \in \mathbb Z$ so that $x^2=3932a$, $y^2=3932b$.
And we know that $7x^2+9y^2=3932 \rightarrow 7\cdot 3932a+9\cdot 3932b=3932$
That means $7a+9b=1$, and since $(7,9)=1$, we can find such $a,b \in \mathbb Z$, using Euclid algorithem:
$9=1 \cdot 7 +2$
$7=3 \cdot 2 +1$
$2=2 \cdot 1+0$
Therefore, after calculations, I found:
$1=4 \cdot 7-3\cdot 9$
So $a=4$, $b=-3$. But we defined $y^2=3932b$, so we get $y^2$ to be negative. Therefore contraditon.
Is this a valid contradiction?
Thanks in advance for any assistance!
| Notice that $7x^2 + 9y^2 = 67$ has a solution $(x,y) = (2,1)$. But this does not give that $\gcd(x^2,y^2) = 67$ here. So unfortunately, your reasoning does not hold.
On the other hand, if you reduce mod 3, you get $x^2 \equiv 2 \pmod 3$. The only squares mod 3 are $0$ and $1$, so there no solution.
| {
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Trigonometry Identity homework help Could you please help me prove this:
$${2\cos(\theta/2)-1-\cos\theta\over2\cos(\theta/2)+1+\cos\theta}={1-\cos(\theta/2)\over1+\cos(\theta/2)}$$
| Using the double angle formula we have
$$\cos \theta = 2\cos^2\frac{\theta}{2} - 1$$
Substituting this in, we get
$$\begin{align}\frac{2\cos\frac{\theta}{2} - 1 - \cos\theta}{2\cos\frac{\theta}{2} + 1 + \cos\theta} &= \frac{2\cos\frac{\theta}{2} - 2\cos^2\frac{\theta}{2}}{2\cos\frac{\theta}{2} + 2\cos^2\frac{\theta}{2}}\\
&= \frac{1 - \cos\frac{\theta}{2}}{1 + \cos\frac{\theta}{2}}\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integral $ \int_0^1 \frac{\ln \ln (1/x)}{1+x^{2p}} dx$...Definite Integral Calculate
$$
I_1:=\int_0^1 \frac{\ln \ln (1/x)}{1+x^{2p}} dx, \ p \geq 1.
$$
I am trying to solve this integral $I_1$. I know how to solve a related integral $I_2$
$$
I_2:=\int_0^1 \frac{\ln \ln (1/x)}{1+x^2} dx=\frac{\pi}{4}\bigg(2\ln 2 +3\ln \pi-4\ln\Gamma\big(\frac{1}{4}\big) \bigg)
$$
but I am not sure how to use that result here. In this case I just use the substitution $x=e^{-\xi}$ and than use a series expansion. The result is
$$
I_2=\int_0^\infty \frac{\xi^s e^{-\xi}}{1+e^{-2\xi}} d\xi=\sum_{n=0}^\infty (-1)^n \frac{\Gamma(s+1)}{(2n+1)^{s+1}}=\Gamma(s+1)L(s+1,\chi_4)
$$
where L is the Dirichlet L-Function where $\chi_4$ is the unique non-principal character. This result is further simplified but takes some work. I am interested in the general case above, $I_1$ Thanks
| Just for your information, I used a CAS without any success for the general case. However, I obtained some formulas.
For $p=2$,
$$\frac{1}{8} \left(-\gamma _1\left(\frac{1}{8}\right)+\gamma
_1\left(\frac{5}{8}\right)-\sqrt{2} (\gamma +\log (8)) \left(\pi +2 \log
\left(\cot \left(\frac{\pi }{8}\right)\right)\right)\right)$$
For $p=3$,
$$\frac{1}{36} \left(-2 \gamma _1\left(\frac{1}{12}\right)+\gamma
_1\left(\frac{5}{12}\right)+2 \gamma _1\left(\frac{7}{12}\right)-\gamma
_1\left(\frac{11}{12}\right)+12 \sqrt{3} \log (2) \log \left(\sqrt{3}-1\right)+6
\sqrt{3} \log (3) \log \left(\sqrt{3}-1\right)-12 \sqrt{3} \log (2) \log
\left(1+\sqrt{3}\right)-6 \sqrt{3} \log (3) \log \left(1+\sqrt{3}\right)-2 \gamma
\left(\pi +3 \sqrt{3} \left(\log \left(1+\sqrt{3}\right)-\log
\left(\sqrt{3}-1\right)\right)\right)+\pi \left(-3 \log (3)+\log (16)+12 \log
(\pi )-16 \log \left(\Gamma \left(\frac{1}{4}\right)\right)\right)\right)$$ For $p=4$,$$\frac{1}{16} \left(-\gamma _1\left(\frac{1}{16}\right)+\gamma
_1\left(\frac{9}{16}\right)+16 \log (2) \sin \left(\frac{\pi }{8}\right) \log
\left(\sin \left(\frac{3 \pi }{16}\right)\right)-4 \pi \log (2) \csc
\left(\frac{\pi }{8}\right)-16 \log (2) \sin \left(\frac{\pi }{8}\right) \log
\left(\cos \left(\frac{3 \pi }{16}\right)\right)+16 \log (2) \cos \left(\frac{\pi
}{8}\right) \log \left(\tan \left(\frac{\pi }{16}\right)\right)-\gamma \left(\pi
\csc \left(\frac{\pi }{8}\right)+4 \left(\sin \left(\frac{\pi }{8}\right)
\left(\log \left(\cos \left(\frac{3 \pi }{16}\right)\right)-\log \left(\sin
\left(\frac{3 \pi }{16}\right)\right)\right)+\cos \left(\frac{\pi }{8}\right) \log
\left(\cot \left(\frac{\pi }{16}\right)\right)\right)\right)\right)$$ In these formulas, $\gamma$ is the Euler constant and $\gamma_1$ is the Stieltjes constant.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why does the following nonlinear system have 21 solutions? I am curious as to why the following nonlinear system has 21 solutions (according to Wolfram Alpha).
$$y+xy^2-x^3+2xz^4=0 \\ -x-y^3-3x^2y+3yz^4=0 \\ -\frac{5}{2}y^2z^3-2x^2z^3-\frac{z^7}{2}=0$$
I see that there is one real and twenty complex solutions:
$(0,0,0)$
and
Is there any way to figure out the number of solutions by looking at the degrees of the equations? Like the highest degree of any term is $7$, and there are $3$ variables, so there are $7 \times 3 = 21$ solutions?
| Mark the solution $(0,0,0)$ and remove $z^3$ from the last equation. The reduced systems for $z=0$ has two cubic equations, so at most $3⋅3=9$ solutions.
Then $z$ only occurs as $z^4$. Multiply the first equation by $x$, the second by $y$ and express all equations in $a=x^2$, $b=y^2$, $c=xy$ and $d=z^4$, adding the equation $ab-c^2=0$. Then the system reads as
\begin{align}
c+c^2-a^2+2ad&=0
\\
-a-bc-3ac+3cd&=0
\\
-5b-4a-d&=0
\\
ab-c^2&=0
\end{align}
This system has the Bezout bound $8$ for the number of solutions for $(a,b,c,d)$, each of these will have at most $8$ solutions for $(x,y,z)$. This reduces the estimate to $64+9+1=74$ solutions from the $5⋅5⋅7=175$ or $5⋅5⋅4+3⋅3=109$ for the Bezout bound of the (reduced) original system.
For more precise bounds, use the Bernshtejn-Kushnirenko-Khovanskii or BKK bound for sparse polynomial systems. See also the work on the multi-homgeneous resultant.
| {
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Help with logarithmic equation? Find the value of x if:
$(2^x)^2 + 3(2^x) - 18 = 0$
So far, I have done
$(2^x)^2 + 2^x(3)=18$
$(2^x)^2+2^x=6$
What should i do with $(2^x)^2+2^x$ so
i can have only one $^X$ on the left side of the equation?
| Let $A = 2^x$. Then the equation becomes
$$A^2 + 3A - 18 = 0$$
which is just one of your everyday quadratic equations. Now, solve the quadratic equation for $A$ to find that $A = -6, 3$. Equate both with $2^t$ to find that
$$2^t = -6, 3$$
For the case when $2^t = 3$, we have $2^t = 3 \implies t = \log_2{3}$.
We reject the case for which $2^t = -6$, because $2^t$ is always positive for real $t$.
| {
"language": "en",
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How are the Taylor Series derived? I know the Taylor Series are infinite sums that represent some functions like $\sin(x)$. But it has always made me wonder how they were derived? How is something like $$\sin(x)=\sum\limits_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)!}\cdot(-1)^n = x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}\pm\dots$$ derived, and how are they used? Thanks in advance for your answer.
| This is the general formula for the Taylor series:
$$\begin{align} &f(x) \\ &= f(a) + f'(a) (x-a) + \frac{f''(a)}{2!} (x - a)^2 + \frac{f^{(3)}(a)}{3!} (x - a)^3 + \dots + \frac{f^{(n)}(a)}{n!} (x - a)^n + \cdots \end{align}$$
You can find a proof here.
The series you mentioned for $\sin(x)$ is a special form of the Taylor series, called the Maclaurin series, centered $a=0$.
The Taylor series is an extremely powerful representation because it shows that every function can be represented as an infinite polynomial (with a few disclaimers, such as interval of convergence)! This means that we can differentiate a function as easily as we can differentiate a polynomial, and we can compare functions by comparing their series expansions.
For instance, we know that the Maclaurin series expansion of $\cos(x)$ is $1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$ and we know that the expansion of $\sin(x)$ is $x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}\dots$. If we do term-by-term differentiation, we can clearly confirm that the derivative of $\sin(x)$ is $\cos(x)$ by differentiating its series.
We can also use the Maclaurin series to prove that $e^{i\theta}=\cos{\theta}+i\sin{\theta}$ and thus $e^{\pi i}+1=0$ by comparing their series:
$$\begin{align}
e^{ix} &{}= 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \frac{(ix)^6}{6!} + \frac{(ix)^7}{7!} + \frac{(ix)^8}{8!} + \cdots \\[8pt]
&{}= 1 + ix - \frac{x^2}{2!} - \frac{ix^3}{3!} + \frac{x^4}{4!} + \frac{ix^5}{5!} - \frac{x^6}{6!} - \frac{ix^7}{7!} + \frac{x^8}{8!} + \cdots \\[8pt]
&{}= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \frac{x^8}{8!} - \cdots \right) + i\left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \right) \\[8pt]
&{}= \cos x + i\sin x \ .
\end{align}$$
Also, you can use the first few terms of the Taylor series expansion to approximate a function if the function is close to the value on which you centered your series. For instance, we use the approximation $\sin(\theta)\approx \theta$ often in differential equations for very small values of $\theta$ by taking the first term of the Maclaurin series for $\sin(x).$
| {
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How does $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$? Good afternoon my wonderful friends!
Whenever I do this equation I set it up using the difference of two cubes, which is as follows:
$(a+b)^3 = (a+b)(a^2-ab+b^2) = a^3 + b^3$
Whenever I try to use this formula I always get:
$(x+h) x^2 - xh + h^2$
Simplify:
$x^3 - x^2h + h^2x + x^2h - xh^2 + h^3$
Simplify:
x^3 + h^3$
I don't understand were the three's come from in the final answer: $(x+h)^3 = x^3 + 3x^2h + 3xh^2 + h^3$
| Something awry should have jumped out at you when you wrote:
$$\color{blue}{\bf(a+b)^3} = (a+b)(a^2-ab+b^2) = \color{blue}{\bf a^3 + b^3}\;\;?$$
Just as $(a +b)^2 \neq a^2 + b^2$, it's also true that $(a+b)^3 \neq a^3 + b^3$.
So what went wrong? The difference of cubes does not apply here!
Rather, $(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3.$ You can confirm that this is so by expanding the factors: $$(a + b)^3 = (a+b)(a+b)^2 = (a+b)(a^2 + 2ab + b^2) \cdots$$
| {
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Value of this definite integral $\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $ So I came across this question on brilliant.org and didn't know how to go about it:
$$\int_{0}^{\infty} \frac{ \ln(x)}{x^2+2x+4} dx $$
I tried to complete the squares in the denominator and then use a trigonometric substitution (with tan) but I didn't get anything beyond that.
How do I do it? Do we need to differentiate under the integral sign here?
| \begin{align}I=\int_{0}^{\infty} \frac{\ln(x)}{x^2+2x+4} dx\end{align}
Perform the change of variable $x=2t$,
\begin{align}I&=\frac{1}{2}\int_{0}^{\infty} \frac{\ln\left(2t\right)}{t^2+t+1}\,dt\\
&=\frac{1}{2}\int_{0}^{\infty} \frac{\ln t}{t^2+t+1}\,dt+\frac{\ln 2}{2}\int_{0}^{\infty} \frac{1}{t^2+t+1}\,dt
\end{align}
The first integral is $0$ (perform the change of variable $y=\dfrac{1}{t}$)
\begin{align}I&=\frac{\ln 2}{2}\int_{0}^{\infty} \frac{1}{t^2+t+1}\,dt\\
&=\frac{\ln 2}{2}\int_{0}^{\infty} \frac{1}{\left(t+\frac{1}{2}\right)^2+\frac{3}{4}}\,dt\\
&=\frac{2\ln 2}{3}\int_{0}^{\infty} \frac{1}{\left(\frac{2t+1}{\sqrt{3}}\right)^2+1}\,dt\\
\end{align}
Perform the change of variable $u=\dfrac{2t+1}{\sqrt{3}}$
\begin{align}I&=\frac{2\ln 2}{3}\times \frac{\sqrt{3}}{2}\int_{\frac{1}{\sqrt{3}}}^{\infty} \frac{1}{1+u^2}\,du\\
&=\frac{\ln 2}{\sqrt{3}}\left(\arctan(\infty)-\arctan\left(\frac{1}{\sqrt{3}}\right)\right)\\
&=\frac{\ln 2}{\sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)\\
&=\frac{\pi\ln 2}{3\sqrt{3}}
\end{align}
| {
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Asymptotic behavior of $\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k$ I am looking to show that
$$\lim_{n \rightarrow \infty}\frac{1}{e^n}\sum_{k=1}^n \binom{n}{k} \left(\frac{ck}{n}\right)^k = 0. $$
In my application, $c = (e+1)/2 \approx 1.85914\ldots$. I have been looking all over the place, but I can't seem to find a closed form expression or good upper bound for the sum.
The obvious estimation
$$\sum_{k=1}^n \binom{n}{k} \left(c\frac{k}{n}\right)^k \leq \sum_{k=0}^n \binom{n}{k} \left(c\right)^k \leq \left(1+c\right)^n $$
won't do the trick, since $1+c=1+(1+e)/2 > e$.
Any ideas?
| I was also working with the inequality
$$ \left(1 - \frac{k}{n}\right)^{n-k} \leq \exp \left( -k + \tfrac{k^{2}}{n} \right) = \exp\left\{ -n \cdot \tfrac{k}{n} \left( 1 - \tfrac{k}{n} \right) \right\}.$$
Now note that $q(x) = x(1-x)$ satisfies $q(1-x) = q(x)$ and $q(x) \geq \frac{1}{2}x $ on $[0, \frac{1}{2}]$. Then we get
$$ 0 \leq k \leq \tfrac{1}{2}n \quad \Longrightarrow \quad \exp\left\{ -n \cdot \tfrac{k}{n} \left( 1 - \tfrac{k}{n} \right) \right\} = e^{-nq(k/n)} \leq e^{-k/2}. \tag{*} $$
To utilize this bound, we divide the sum into two part:
\begin{align*}
S_{n} :=
e^{-n} \sum_{k=0}^{n} \binom{n}{k} \left( \frac{k}{n} \right)^{k} c^{k}
&= e^{-n} \sum_{k=0}^{n} \binom{n}{k} \left( 1 - \frac{k}{n} \right)^{n-k} c^{n-k} \\
&\leq e^{-n} \sum_{k=0}^{n} \binom{n}{k} c^{n-k} e^{-nq(k/n)} \\
&\leq e^{-n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} c^{n-k} e^{-nq(k/n)} + e^{-n} \sum_{\frac{n}{2} \leq k \leq n} \binom{n}{k} c^{n-k} e^{-nq(k/n)} \\
&= (c/e)^{n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} c^{-k} e^{-nq(k/n)} + e^{-n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} c^{k} e^{-nq(k/n)}.
\end{align*}
Here, in the last line, we applied the change of index $k \mapsto n-k$. Then we can use $\text{(*)}$ and we get the following crude bound:
\begin{align*}
S_{n}
&\leq (c/e)^{n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} \left( \frac{1}{c\sqrt{e}} \right)^{k} + e^{-n} \sum_{k \leq \frac{n}{2}} \binom{n}{k} \left( \frac{c}{\sqrt{e}} \right)^{k} \\
&\leq \left\{ \frac{c}{e} \left( 1 + \frac{1}{c\sqrt{e}} \right) \right\}^{n} + \left\{ \frac{1}{e} \left( 1 + \frac{c}{\sqrt{e}} \right) \right\}^{n}.
\end{align*}
Now note that $c > 0$ satisfies the following condition
$$ \frac{c}{e} \left( 1 + \frac{1}{c\sqrt{e}} \right) < 1 \quad \text{and} \quad \frac{1}{e} \left( 1 + \frac{c}{\sqrt{e}} \right) < 1
\quad \Longleftrightarrow \quad c < e - \frac{1}{\sqrt{e}} \simeq 2.111751169. $$
Since your $c$ is less than $2$, the claim follows.
A slightly generally, you can introduce a parameter $\delta \in (0, 1)$. Then
\begin{align*}
\begin{array}{cl}
0 \leq k \leq \delta n
& \quad \Longrightarrow \quad \left( 1 - \tfrac{k}{n} \right)^{n-k} \leq e^{-(1-\delta)k}, \\
0 \leq k \leq (1-\delta) n
& \quad \Longrightarrow \quad \left( 1 - \tfrac{k}{n} \right)^{n-k} \leq e^{-\delta k}
\end{array}
\end{align*}
and our estimation is refined as
$$ S_{n} \leq \left( \frac{c}{e} + \frac{1}{e^{2-\delta}} \right)^{n} + \left( \frac{1}{e} + \frac{c}{e^{1+\delta}} \right)^{n}. $$
For both ratios to be less than 1, we must have $c < \min\{ e - e^{\delta-1}. e^{\delta}(e - 1) \}$. Maximizing this bound gives
$$ e^{\delta} = \frac{1 + e^{-1}}{1 + e^{-3}}, $$
Therefore
$$ c < \frac{e^{2}(e^{2} - 1)}{e^{3} + 1}
\quad \Longrightarrow \quad S_{n} \leq \left( \frac{c}{e} + \frac{1}{e^{2}-e+1} \right)^{n} + \left( \frac{1}{e} + \frac{e^{2}-e+1}{e^{3}} c \right)^{n}
\xrightarrow[n\to\infty]{} 0. $$
| {
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How prove this inequality $abc(a^2+b^2+c^2)\le 3$ let $a,b,c>0$,and such $$a+b+c=3$$,show that
$$abc(a^2+b^2+c^2)\le 3$$
My idea: since
$$abc\le\left(\dfrac{a+b+c}{3}\right)^3=1$$
but $$a^2+b^2+c^2\ge \dfrac{1}{3}(a+b+c)^2=3$$
so I can't prove this inequality.Thank you
It is said that can use AM-GM inequality to solve it
| Notice that
$$ 3(a+b+c)abc(a^2+b^2+c^2) \leq (ab+bc+ca)^2(a^2+b^2+c^2) \leq \left(\frac{(a+b+c)^2}{3}\right)^3. $$
The first inequality follows from $a^2b^2+b^2c^2+c^2a^2 \geq abc(a+b+c)$, which is true as $a^2b^2+b^2c^2 \geq 2ab^2c$, $b^2c^2+c^2a^2 \geq 2abc^2$ and $c^2a^2+a^2b^2 \geq 2a^2bc$ by AM-GM.
The second inequality follows from AM-GM applied on $ab+bc+ca$, $ab+bc+ca$ and $a^2+b^2+c^2$.
| {
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Taylor Series of $ \frac{1}{1-x^2} $ about x=2 I am trying to form a taylor series of the following:
$ \frac{1}{1-x^2} $ about $x=2$
I tried factoring the equation such that it becomes the following:
$ \frac{1}{{(1+x)}{(1-x)}} $
I tried to substitute $ x = h + 2 $ into the equation and obtained the following after using partial fractions to decompose the result:
$ \frac{1}{2(h+3)} - \frac{1}{2(h+1)} $
I do not know how to proceed from here.
I know I can just compute all the derivatives of the expression and evaluate them. But this would be non-trivial. Could someone please advise me on how I could solve this question?
| Notice that
$$\frac{1}{1 - x^2} = 1 + x^2 + x^4 + x^6 + \cdots = \sum_{k = 0}^\infty x^{2k}$$
provided $|x| < 1$. In general, however,
$$\frac{1}{1 - x} = \frac{1}{1 - a} + \frac{x - a}{(1 - a)^2} + \frac{(x - a)^2}{(1 - a)^3} + \cdots \quad \text{if} \quad |x - a| < 1.$$
Now, simply replace $x$ with $x^2$ and set $a = 2$.
The equality above follows from the formula
$$f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \frac{f'''(a)}{3!}(x - a)^3 + \cdots$$
| {
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Conjectured closed form of $G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right)$ In my answer to this question, I come across the following case of the Meijer G-function:
$$F(b)=G^{2~2}_{3~3}\left(1\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right), b>0$$
and based on my experiments, $F(b)$ have the following closed form:
$$F(b)
\stackrel?=\frac{\Gamma(b)}{b}\left(-\gamma-\psi(b)+\frac{2^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-1\right)+b{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-1\right)\right)$$
Is there any chance of proving this?
Edit: Using the definition of the Meijer G-Function, $F(b)$ have the integral representation:
$$F(b)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}dx.$$
Edit 2: I've found a further generalization:
$$F(b,z)=G^{2~2}_{3~3}\left(z\middle|\begin{array}c1,1;b+1\\b,b;0\end{array}\right)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}e^{(b/2+ix)\log z}dx\\
\stackrel?=\frac{\Gamma(b)z^b}{b}\left(-\log z-\gamma-\psi(b)+\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\right), b\not\in\{0,-1,-2,\dots\},z\neq0,-1.$$
Edit 3: Further simplified the ${_2F_1}$ part.
| Inspired by GEdgar's answer for the $F(1,1)$ case, I give an answer (to my own question!) for the general case.
We start from the integral representation $$
F(b,z)=\frac{1}{2\pi}\int^{+\infty}_{-\infty}\frac{\Gamma(\tfrac{b}{2}+ix)\Gamma(\tfrac{b}{2}-ix)}{\tfrac{b^2}{4}+x^2}z^{\frac{b}{2}+ix}dx.
$$
We consider the function $G(w)=\frac{\Gamma(\tfrac{b}{2}+iw)\Gamma(\tfrac{b}{2}-iw)}{\tfrac{b^2}{4}+w^2}z^{\frac{b}{2}+iw}dw$. $G(w)$ have double poles at $w=\pm\frac{b}{2}i$, and single poles at $w=\pm(\frac{b}{2}+n)i$ for all positive integer $n$.
Now let $N$ be a large integer, and consider the integral path around the rectancle with vertices $-N,N,N-(N+\frac{b+1}{2})i,-N-(N+\frac{b+1}{2})i$. It is possible to prove that the integrals on the three sides go to zero, and therefore we have
$$F(b,z)=-i(\text{sum of residues of $G(w)$ in the lower half-plane}).$$
It remains to calculate the residues at each pole.
In fact, we have $$
\begin{align*}
-i\operatorname{Res}_{w=-\frac{b}{2}i}G(w)&=\frac{z^b\Gamma(b)}{b}\left(\frac1b-\log z-\gamma-\psi(b)\right)\\
-i\operatorname{Res}_{w=-(\frac{b}{2}+n)i}G(w)&=\frac{(-1)^{n+1}z^{b+n}\Gamma(b+n)}{n~n!(b+n)}.
\end{align*}
$$
Comparing with the conjectured closed form, we only need to prove that $$
\frac1b-\sum_{n=1}^{\infty}\frac{(-z)^nb\Gamma(b+n)}{n~n!\Gamma(b)(b+n)}\stackrel?=\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right).$$
Now note that $$
\frac{(z+1)^{1-b}}{b}{_2F_1}\left(\begin{array}c1,1\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\
=b^{-1}{_2F_1}\left(\begin{array}cb,b\\b+1\end{array}\middle|-z\right)+bz~{_3F_2}\left(\begin{array}c1,1,b+1\\2,2\end{array}\middle|-z\right)\\
=\underbrace{\sum_{m=0}^{\infty}\frac{\Gamma(b+m)^2\Gamma(b+1)}{b\Gamma(b)^2\Gamma(b+m+1)m!}(-z)^m}_{n=m}-\underbrace{\sum_{m=0}^{\infty}\frac{b\Gamma(m+1)^2\Gamma(b+m+1)\Gamma(2)^2}{\Gamma(1)^2\Gamma(b+1)\Gamma(m+2)^2m!}(-z)^{m+1}}_{n=m+1}\\
=\sum_{n=0}^{\infty}\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\sum_{n=1}^{\infty}\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\\
=\frac1b+\sum_{n=1}^{\infty}\left(\frac{\Gamma(b+n)}{\Gamma(b)(b+n)n!}(-z)^n-\frac{\Gamma(b+n)}{n~n!\Gamma(b)}(-z)^{n}\right)\\
=\frac1b+\sum_{n=1}^{\infty}\frac{\Gamma(b+n)(-z)^n}{n!\Gamma(b)}\left(\frac{1}{b+n}-\frac{1}{b}\right)\\
=\frac1b-\sum_{n=1}^{\infty}\frac{b\Gamma(b+n)(-z)^n}{n(b+n)n!\Gamma(b)}.
$$
Edit: Ninja'd by 2 minute.
| {
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Prove that $\frac{a^2}{a + b} + \frac{b^2}{b + c} + \frac{c^2}{c + a} \ge \frac{3}{2}$ Let $a,b,c$ are positive real numbers such that, $a^2+b^2+c^2=3$. Prove that
$$\frac{a^2}{a + b} + \frac{b^2}{b + c} + \frac{c^2}{c + a} \ge \frac{3}{2}$$
| As indicated in tipshoni’s comment, the inequality to be shown is equivalent to
$$
\sum_{cyc}\frac{a^2+b^2}{a+b} \geq 3 \tag{1}
$$
This problem is question number 44 in chapter 1 of Vasile Cirtoaje’s book
"Algebraic Inequalities" (available here).
| {
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I am not sure if the answer for the dividing rational expressions problem should be simplified $$ \frac{x}{x+2} \div \frac{1}{x^2 - 4} $$
-I am not sure if the answer for the problem (which is attached) would be $\frac{x^3-4x}{x+2}$ or if it could be simplified further.
| You can simplify this in two ways.
*
*Dividing by a fraction is the same as multiplying by the flipped fraction, and if we do that we get
$$
\frac{x^2-4}{1}\cdot\frac{x}{x+2} = \frac{x(x^2-4)}{x+2}.
$$
*Next, we need to factor the bottom of the fraction. If you have enough practice with factoring you should recognize immediately that $x^2-4$ is a difference of squares. That is, $x^2-4=x^2-2^2$. So the expression is
$$
\frac{x(x-2)(x+2)}{x+2}=x(x-2)
$$
after we cancel the $x+2$.
| {
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Evaluating $\int \frac{\operatorname d \! x}{\sin^4{x}+\cos^4{x}+\sin^2{x}\cos^2{x}}$ How do you integrate
$$\frac{1}{\sin^4{x}+\cos^4{x}+\sin^2{x}\cos^2{x}}$$
or simply
$$\frac{1}{1-\left(\frac{\sin{2x}}{2}\right)^2}.$$
| Note that $$\frac{1}{\sin ^{4}x+\cos ^{4}x+\sin ^{2}x\cos ^{2}x}\ne \frac{1}{1+\left( \frac{
\sin 2x}{2}\right) ^{2}}.$$
Since
\begin{equation*}
\frac{1}{\sin ^{4}x+\cos ^{4}x+\sin ^{2}x\cos ^{2}x}=\frac{1}{1-\left( \frac{
\sin 2x}{2}\right) ^{2}}=\frac{4}{3+\cos ^{2}2x},
\end{equation*}
we will evaluate
$$I=\int\frac{4}{3+\cos ^{2}2x}\, dx.$$
We will use two substitutions: the trigonometric substitution $u=\cos 2x$ and the Euler substitution$\sqrt{1-u^{2}}=(u-1)t$ (third substitution of Euler) or $t=\frac{\sqrt{1-u^{2}}}{u-1}$. We obtain the following rational function of $t$ which is integrable by partial fractions decomposition. Each fraction is easily integrable by completing the square in the denominator.
$$\frac{t^{2}+1}{t^{4}+t^{2}+1} = \frac{1}{2\left( t^{2}+t+1\right) }+ \frac{1}{2\left(
t^{2}-t+1\right) }.$$
We have that
\begin{eqnarray*}
I &=&-2\int \frac{1}{\left( 3+u^{2}\right) \sqrt{1-u^{2}}}\,du \\[2ex]
&=&\int \frac{t^{2}+1}{t^{4}+t^{2}+1}\,dt =\int \frac{1}{2\left( t^{2}+t+1\right) }dt+\int \frac{1}{2\left(
t^{2}-t+1\right) }dt \\[2ex]
&=&\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}\left( 4t+2\right) }{6}+\frac{
\sqrt{3}}{3}\arctan \frac{\sqrt{3}\left( 4t-2\right) }{6}+C \\[2ex]
&=&\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}\left( \frac{4\sqrt{1-u^{2}}}{u-1}
+2\right) }{6}+\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}\left( \frac{4\sqrt{
1-u^{2}}}{u-1}-2\right) }{6}+C \\[2ex]
&=&\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}\left( \frac{4\sqrt{1-\cos ^{2}2x}
}{\cos 2x-1}+2\right) }{6}+\frac{\sqrt{3}}{3}\arctan \frac{\sqrt{3}\left(
\frac{4\sqrt{1-\cos ^{2}2x}}{\cos 2x-1}-2\right) }{6}+C \\[2ex]
&=&\frac{1}{\sqrt{3}}\arctan \frac{ -2+\tan x }{\sqrt{3}\tan x}-\frac{1}{\sqrt{3}}\arctan \frac{ 2+\tan x }{\sqrt{3}\tan x}+C.
\end{eqnarray*}
| {
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Verify the identity: $\tan^{-1} x +\tan^{-1} (1/x) = \pi /2$ Verify the identity: $\tan^{-1} x + \tan^{-1} (1/x) = \frac\pi 2, x > 0$
$$\alpha= \tan^{-1} x$$
$$\beta = \tan^{-1} (1/x)$$
$$\tan \alpha = x$$
$$\tan \beta = 1/x$$
$$\tan^{-1}[\tan(\alpha + \beta)]$$
$$\tan^{-1}\left
[{\tan\alpha + \tan\beta\over 1 - \tan\alpha \tan\beta}
\right]$$
$$\tan^{-1}\left[
{x + 1/x\over 1- x/x }\right]$$
$$\tan^{-1}\left[{x + (1/x)\over 0} \right]$$
I can't find out what I'm doing wrong..
| Assume $x>0$, then
\begin{align}&\tan^{-1} x +\tan^{-1} \dfrac1x
\\\\=&\tan^{-1} x +\tan^{-1}\dfrac1{\tan\tan^{-1} x}
\\\\=&\tan^{-1} x +\tan^{-1}\cot \tan^{-1} x
\\\\=&\tan^{-1} x +\tan^{-1}\tan(\dfrac{\pi}2- \tan^{-1}x)
\\\\=&\tan^{-1} x +\dfrac{\pi}2- \tan^{-1}x\qquad\qquad\qquad
\left(\because\text{for $x>0$,}\;\dfrac{\pi}2- \tan^{-1}x\in\left(0,\dfrac{\pi}2\right)\right)
\\\\=&\dfrac{\pi}2.
\end{align}
| {
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Prove that $n^7+7$ can never be a perfect square. Prove that for a positive integer $n$, $n^7+7$ cannot be a perfect square.
I managed to show that $n \equiv 5 \pmod{8}$ or $n \equiv 9 \pmod{16}$. But nothing came from that so I presume another approach is needed???
Thanks for any help.
| To show that $n^7 + 7 = x^2$ has no solution we shall use the fact that
$n^7 - 7 = x^2$ has one solution namely $n = 2$ and $x = 11$.
Suppose $n^7 + 7 = x^2$. Adding to $2^7 - 7 = 11^2$ we get
$
n^7 + 2^7 = x^2 + 11^2
$
Call $m = n + 2$. Then
$
n^7 + 2^7 = (m - 2)^7 + 2^7 =
$
$$\sum_{0 \le k \le 7} \binom{7}{k} m^k (-2)^{7-k} + 2^7$$
$$ = \sum_{1 \le k \le 7} \binom{7}{k}(-1)^{k-1} m^k (2)^{7-k} $$
$m$ can be factored out and you get
$$n^7 + 2^7 = m \sum_{1 \le k \le 7} \binom{7}{k} (-1)^{k-1} m^{k-1} (2)^{7-k}$$
Let's write $n^7 + 2^7 = m\cdot M.$
Note that $M = m^6 -7\cdot 2\cdot m^5 + - + -21\cdot 2^5m + 7\cdot 2^6$
Hence $\gcd(m,M)$ is a divisor of $7\cdot 2^6$
Note that in $\Bbb{Z}_4$, the squares are $0$ and $1$. Hence $n^7 + 7 = 0$ or $1$ in $\Bbb{Z}_4$, which implies that $n$ is odd.
This implies that $m$ is odd. Hence $\gcd(m,M)$ is either $1$ or $7$.
Observe that in $\Bbb{Z}_7$, the squares are $0,1,2,4$. Hence the sum of $2$ squares can be a multiple of $7$ only if both squares are themselves multiples of $7$. Since $11$ is not, $x^2 + 11^2$ can't be a multiple of $7$.
Therefore $\gcd(m,M)=1$
Now if $m \cdot M$ is a sum of $2$ squares, its square free part has prime factors only of the form $p = 4k + 1$.
Hence the same is true for m, since $\gcd(m,M) = 1$.
This implies that $m = 1 \pmod 4$ and $n = m - 2 = - 1 . \pmod 4$
Hence $n^7 + 7 = -1 -1 = 2 \pmod 4$ and is not a square.
This yields a contradiction: $n^7 + 7$ can't be a square.
| {
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Prove $2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2)$ for $x, y, z \ge 0$
Let $x,y,z\ge 0$. Show that
$$2(x^4+y^4+z^4)+2xyz+7\ge 5(x^2+y^2+z^2).$$
my idea: let $$x+y+z=p,xy+yz+xz=q,xyz=r$$
since
$$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=p^2-2q$$
and
$$(xy+yz+xz)^2=x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)$$
$$\Longrightarrow x^2y^2+y^2z^2+x^2z^2=q^2-2pr$$and
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+x^2z^2)=(p^2-2q)^2-2(q^2-2pr)$$
so
$$\Longleftrightarrow 2[(p^2-2q)^2-2(q^2-2pr)]+2r+7\ge 5(p^2-2q)$$
$$\Longleftrightarrow 2p^4-8p^2q+4q^2+8pr-5p^2+10q+2r+7\ge 0$$
then I can't
This link has a similar problem:
see this
Maybe this problem can use AM-GM inequality,But I can't.Thank you
| 2(x^4+y^4+z^4) + 2xyz + 7 ≥ 5(x^2+y^2+z^2)
We can see easily that the equality holds for for x = y = z = 1 . And as the equation is symmetric hence we can say that x > y > z
Case 1: when x,y,z<1
Then we can easily see that as on the left side as there is 7 so easily the left hand side is greater than the right hand side
Case 2: when x>1 & y,z<1
So the 2(y^4+z^4) + 2xyz + 5 > 5(y^2+z^2) as the int 5 is there. Now for all x
2(x^4) + 2 > 5(x^2) as 2(x^4) > 5(x^2) for all x > (5/2)^(1/2) but if x < (5/2)^(1/2) then +2 is greater than 5(x^2).
Case 3: when x,y>1 & z<1
2(z^4) + 2xyz + 2 > 5(z^2) now for the rest of the part
2(x^4+y^4) + 5 ≥ 5(x^2+y^2) = 2(x^4+y^4) ≥5(x^2+y^2 - 1)
Now using Tchebychef's Inequality on the (x^2 > y^2 > 1) and (5/2 , 3/2 , 1) we get the following equation
(5/2)x^2+(3/2)y^2 -(1)1 ≥ (x^2+y^2 - 1)(5/2+3/2+1/2)
and 2(x^4+y^4) + 1 ≥(5/2)x^2+(3/2)y^2 ( this can be said as 2(x^4+y^4)≥(5/2)x^2+(3/2)y^2 but when this doesn't satisfy the +1 makes the LHS>RHS) and hence proved.
Case 3: when x,y,z>1
2xyz + 2 ≥ 0 &
2(x^4+y^4+z^4) + 5 ≥ 5(x^2+y^2+z^2) =2(x^4+y^4+z^4) ≥ 5(x^2+y^2+z^2-1) as
by using Tchebychef's Inequality on the (x^2 > y^2 > z^2 > 1) and (4,3,2,1) we get the following equation
(4x^2+3y^2+2z^2-1)/4≥ 10(x^2+y^2+z^2-1)/16 = (4x^2+3y^2+2z^2-1)≥5/2(x^2+y^2+z^2-1)
2(x^4+y^4+z^4)≥(4x^2+3y^2+2z^2-1) = 2(x^4+y^4+z^4)+1≥(4x^2+3y^2+2z^2) and hence proved.
| {
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Two definite integrals These integrals are closely related since $\frac{\pi^2}{8}=\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}$ and $G=\sum_{n\mathop=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^2}$. I'm not able to prove them though.
Show that
$$\int_0^1 \frac{1}{\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =\frac{\pi^2}{8}$$
$$\int_1^{\infty} \frac{1}{x\sqrt{x^2-1}}\log\left(\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}}\right)\,\mathrm{d}x =2G$$
where $G$ is Catalan's constant.
| The first integral is easier than it looks. Write
$$\begin{align}I &= \int_0^1 \frac{dx}{\sqrt{x^2-1}} \log{\left (\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \right )} \\ &= -i \int_0^1 \frac{dx}{\sqrt{1-x^2}} \log{\left (x+i \sqrt{1-x^2}\right )}\\ &= \int_0^1 \frac{dx}{\sqrt{1-x^2}} \arccos{x}\\ &= -\frac12 \left [\arccos^2{x} \right ]_0^1\\ &= \frac{\pi^2}{8}\end{align}$$
The second integral is also straightforward, using the sub $x=\cosh{t}$:
$$\begin{align}J &= \int_1^{\infty} \frac{dx}{x \sqrt{x^2-1}} \log{\left (\frac{\sqrt{x+1}+\sqrt{x-1}}{\sqrt{x+1}-\sqrt{x-1}} \right )} \\ &= \int_0^{\infty} dt \frac{t}{\cosh{t}}\\ &=2 \sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} dt \, t \, e^{-(2 k+1) t}\\ &= 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^2} \\ &= 2 G\end{align}$$
| {
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Prove $\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$ How to prove:
$\frac{\sin\theta}{1+\cos\theta} + \frac{1+\cos\theta}{\sin\theta} = \frac{2}{\sin\theta}$
Please help.
| Let $u=\theta/2$. Then, by using double-angle formulas, $\sin \theta = 2 \cos u \sin u$, $1+\cos \theta = 2\cos^2 u$.
So the left-hand side is equal to
$$
\frac{2\cos u \sin u}{2\cos^2 u} + \frac{2 \cos^2 u}{2\cos u \sin u} = \frac{\cos u}{\sin u} + \frac{\sin u}{\cos u} = \frac{\cos^2 u + \sin^2 u}{\cos u \sin u}=\frac{1}{\cos u \sin u} \, ,
$$
while the right-hand side is equal to
$$
\frac{2}{2 \cos u \sin u}=\frac{1}{\cos u \sin u} \, .
$$
| {
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"answer_id": 2
} |
Real Analysis limits of functions delta epsilon proof Prove from first principles that
$f(x) = \displaystyle\frac{x^2-4}{x-4}$ approaches $-5$ as $x$ approaches $3$.
I am terrible at these proofs. I know we start like this
Fix $\epsilon > 0$. We need to find $\delta$ such that
$0 < |x-3| < \delta \implies |f(x)-(-5)| < \epsilon$
I simplified the abs so that i have
$ \left| \displaystyle\frac{(x+8)(x-3)}{x-4} \right|< \epsilon$ but I dont know where to go from here.
Cheers!
| We start from the inequality you arrived at. So we want to show that by making $x$ close enough to $3$, we can make $ \left| \displaystyle\frac{(x+8)(x-3)}{x-4} \right|< \epsilon$.
We can exercise control over $|x-3|$. But there is a potential problem with $ \left| \displaystyle\frac{x+8}{x-4} \right|$. We must make sure this does not get too big,
First of all, we will insist that $\delta\lt \frac{1}{2}$. Why? That will confine $x+8$ to the interval $(10.5,11.5)$. So in particular we will have $|x+8|\lt 11.5$.
Also, $x-4$ will be confined to the interval $(-1.5,0.5)$. So $|x-4|$ will be greater than $0.5$.
Thus, if $\delta$ is chosen as above, we will have $ \left| \displaystyle\frac{x+8}{x-4} \right|\lt 23$, and therefore we will have
$$\left|\frac{(x+8)(x-3)}{x-4} \right|< 23\delta.$$
Thus we can force $ \left| \displaystyle\frac{(x+8)(x-3)}{x-4} \right|$ to have absolute value $\lt \epsilon$ by picking $\delta\lt \frac{1}{2}$ and $\delta\lt \frac{\epsilon}{23}$.
So let $\delta=\min\left(\frac{1}{2},\frac{\epsilon}{23}\right)$. If $|x-3|\lt \delta$, the desired inequality holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/728607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Finding max/min of multivariable function The following function
$f(x,y) = 3xy + \frac{6}{1 + x^2 + y^2 }$ within $\frac{1}{3} \leq x^2 +y^2 \leq 4$
I do partial differention
$\frac{\partial z}{\partial x} = 3y - \frac{12x}{1 + x^2 + y^2}$
$\frac{\partial z}{\partial y} = 3x - \frac{12y}{1 + x^2+ y^2 }$
I try to simplify and get an expression of just one variable setting
$\frac{\partial z}{\partial x} - \frac{\partial z}{\partial x} = 0$
getting that $x = y$
$3x - \frac{12x}{1 + x^2+x^2 }$
solving for $x = 0$, $x_1 = 0, x_2 = \frac{1}{\sqrt{2}}, x_3 = \frac{-1}{\sqrt{2}}$
which gives me
$f(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) = 3\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}} + \frac{6}{1 + \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 } = \frac{9}{2}$
second part of the problem is to try the outer and inner boundaries. I set for inner boundary:
$x = \frac{1}{\sqrt{3}}\cos{t}$
, $y = \frac{1}{\sqrt{3}}\sin{t}$
and outer
$x = 2\cos{t}$
, $y = 2sin{t}$
I then create a new function:
$h(t) = f(x, y) = f(2\cos{t}, 2\sin{t})$ differentiate and get $\sin{t} = \cos{t}$ for $ x = 0$
$f(2\frac{1}{\sqrt{2}},2\frac{1}{\sqrt{2}}) = 3\frac{2}{\sqrt{2}}\frac{2}{\sqrt{2}} + \frac{6}{1 + \left(\frac{2}{\sqrt{2}}\right)^2 + \left(\frac{2}{\sqrt{2}}\right)^2 } = \frac{36}{5}$
which actually yields the correct maximum.
for the inner boundary I get
$f(\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}},\frac{1}{\sqrt{3}}\frac{1}{\sqrt{2}}) = 5$
but neither 9/2 or 5 gets the correct minimum! I can't see what i'm doing wrong.
| The partial derivatives
$\frac{\partial f}{\partial x} = 3y - \frac{-12x}{1 + x^2 + y^2},$
and
$\frac{\partial f}{\partial y} = 3x - \frac{-12y}{1 + x^2+ y^2 }$
are wrong. One has
$\frac{\partial f}{\partial x} = 3y - \frac{-12x}{(1 + x^2 + y^2)^2}$
and
$\frac{\partial f}{\partial y} = 3x - \frac{-12y}{(1 + x^2 + y^2)^2}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/733127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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