Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Which is the easiest and the fastest way to find the remainder when $17^{17}$ is divided by $64$? Which is the easiest and the fastest way to find the remainder when $17^{17}$ is divided by $64$?
| $17^4=(16+1)^4=16^4+4\cdot 16^3+6\cdot 16^2+4\cdot 16+1 \equiv 1 \pmod {64}$
(Just notice that all numbers $16^4$, $16^3$, $16^2$ and $4\cdot 16$ are multiples of $64$.)
$17^{17} = (17^4)^4\cdot 17 \equiv 1\cdot 17 \pmod {64}$
| {
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Evaluate the limit of $\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$ $$\lim_{x\to\infty}\frac{x - \sqrt{x^2+5x+2}}{x-\sqrt{x^2+0.5x + 1}}.$$
Can't find a means to resolve. The answer is $10$ by graphing.
| Completing the squares:
$x^2+5x+2 = (x + 2.5)^2 + O(1)$
$x^2+0.5x+1 = (x + 0.25)^2 + O(1)$
So:
$\sqrt{x^2+5x+2} = x + 2.5 + o(1)$
$\sqrt{x^2+0.5x+1} = x + 0.25 + o(1)$
Now the answer is immediate.
Updated to address didier's concerns:
We show using elementary methods that $\sqrt{u^2+O(1)}=u+o(1)$, which is the same as saying that for any constant $C$, $\sqrt{u^2+C} \to u$ as $u \to \infty$.
Lemma: If $|\epsilon| < 1$, then $1 - |\epsilon| \le \sqrt{1 + \epsilon} \le 1 + |\epsilon|$.
Proof: If $\epsilon \ge 0$, then
$$1 - \epsilon \le 1 \le \sqrt{1 + \epsilon} \le 1 + \epsilon$$
because $\sqrt a \le a$ if $a \ge 1$.
If $\epsilon < 0$, put $\theta = -\epsilon$. Then
$$1 - |\epsilon| = 1 - \theta < \sqrt{1 - \theta} = \sqrt{1 + \epsilon}$$
because $a < \sqrt a$ if $0 < a < 1$. And
$$\sqrt{1+\epsilon} = \sqrt{1-\theta} < 1 < 1 + |\epsilon|$$
Corollary: If $|\epsilon| < 1$, then $|\sqrt{1+\epsilon} - 1| \le |\epsilon|$.
Now we just put $\epsilon = C/u^2$ in the corollary to get (for $u > \sqrt{|C|}$):
$$|\sqrt{u^2 + C} - u| = u\Big|\sqrt{1 + \frac{C}{u^2}} - 1\Big| \le u\frac{|C|}{u^2} = \frac{|C|}{u}$$
which tends to $0$ as $u \to \infty$.
| {
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Help with a geometry problem The problem says:
A triangle has its lengths in an arithmetic progression,
with difference d. The area of the triangle is t. Find the dimensions.
the solution says:
the notation can be even better if we make it more symmetrical, by making the
side lengths $b − d, b,$ and $b + d$ .
by Heron’s formula we know that $t^2 = s(s − b + d)(s − b)(s − b − d)$ , where $s =
((b − d) + b + (b + d))/2$ is the semi-perimeter;
and after simplification
$$ t^2 = \frac{3b}{2}(\frac{3b}{2} - b + d )(\frac{3b}{2} - b ) (\frac{3b}{2} - b - d ) $$
$$ \implies t^2 = \frac{3b^2(b-2d)(b+2d)}{16} = \frac{3b^2(b^2-4d^2)}{16} $$
then we get $$ 3b^4 − 12d^2b^2 − 16t^2 = 0 $$
and using the quadratic formula :
$$ b^2 = \frac{12d^2 \pm \sqrt{144d^4 + 169t^2} }{6} = 2d^2 \pm \sqrt{4d^4 + \frac{16}{3} t^2} $$
and because b has to be positive , we get
$$ b = \sqrt{2d^2 + \sqrt{4d^4 + \frac{16}{3}t^2}} $$
Which is the part that i have a problem with , my question is :
why should we select only the positive sign solution of the quadratic formula ? is that because $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$ which means that the negative sign solution leads to the square root of a negative number which is not valid? why is the positive sign solution is the right solution ?
In other words :
If $\sqrt{ 4d^4 + \frac{16}{3}t^2} > 2d^2$ , how is that ? how can we prove it ?
thank you
| We have $4d^4=(2d^2)^2$, and therefore $4d^4+\frac{3}{16}t^2>(2d^2)^2$ and therefore
$$\sqrt{4d^4+\frac{3}{16}t^2}>2d^2.$$
It follows that $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}<0$, so $2d^2-\sqrt{4d^4+\frac{3}{16}t^2}$ does not have a (real) square root.
| {
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Proof of an inequality: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}}$
Possible Duplicate:
Proving $\sum\limits_{k=1}^{n}{\frac{1}{\sqrt{k}}\ge\sqrt{n}}$ with induction
How do I prove the following?
$$\sqrt{n} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{n}},$$ for all $n \in\mathbb{Z}$, $n\ge 2$.
| For any $r \in [1, n]$, we have
$$
\frac{1}{\sqrt{r}} \geqslant \frac{1}{\sqrt{n}}.
$$
with strict inequality for $1 \leqslant r \lt n$. Adding all these $n$ inequalities,
$$
\sum_{r=1}^{n} \frac{1}{\sqrt{r}} \gt n \cdot \frac{1}{\sqrt{n}} = \sqrt{n}.
$$
Proof using induction. The OP requested a proof using induction. I am assuming you can handle the base case $n=2$.
Now, assume that the inequality holds for some $n \geqslant 2$; we will verify the inequality for $n+1$:
$$
\begin{align*}
\sum_{r=1}^{n+1} \frac{1}{\sqrt{r}}
&= \sum_{r=1}^{n} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{n+1}}
\\ &\gt \sqrt{n} + \frac{1}{\sqrt{n+1}}
\\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - (\sqrt{n+1} - \sqrt{n})
\\ &= \sqrt{n+1} + \frac{1}{\sqrt{n+1}} - \frac{1}{\sqrt{n+1} + \sqrt{n}}
\\ &\gt \sqrt{n+1} ,
\end{align*}
$$
which is what we want to show.
Notice that out second inequality is a bit too crude. robjohn's answer shows how to get a better bound by being more careful.
| {
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$\sum \limits_{n=1}^{\infty}n(\frac{2}{3})^n$ Evalute Sum
Possible Duplicate:
How can I evaluate $\sum_{n=1}^\infty \frac{2n}{3^{n+1}}$
How can you compute the limit of
$\sum \limits_{n=1}^{\infty} n(2/3)^n$
Evidently it is equal to 6 by wolfram alpha but how could you compute such a sum analytically?
| Just a curioursity.
Using Fourier series one comes to:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx = 6\ \frac{13 \cos x-12}{(13-12\cos x)^2}$$
hence:
$$\sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n = \sum_{n=1}^\infty n\left( \frac{2}{3}\right)^n\ \cos nx \Bigg|_{x=0} =6\ \frac{13 \cos x-12}{(13-12\cos x)^2}\Bigg|_{x=0} =6\; .$$
| {
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Simplifying rational fractions I can't get this one either for whatever reason, spent about 20 minutes on it and I have made no progress at all.
$$\frac{x^2}{(x^2-4)} - \frac{x+1}{x+2}.$$
I know that I can simplify this into one fraction so I make it $$\frac{x^2}{x+2}-\frac{(x+1)(x^2-4)}{(x^2-4)(x+2)}$$
I then can simplify it further making the $(x^2+4)$ into $(x-2)(x+2)$ and the $(x+2)$ into $(x-1)(x+1)$ but this does not help me get the answer. I know I have to manipulate it in some counter intuitive way but I can not make it work.
| As first step we may use the common denominator $(x^{2}-4)=(x-2)(x+2)$
because the $\text{lcm}\left( (x-2)(x+2),(x+2)\right) =(x-2)(x+2)$
$$
\begin{eqnarray*}
\frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2} &=&\frac{x^{2}}{(x-2)(x+2)}-\frac{
\left( x+1\right) (x-2)}{\left( x+2\right) (x-2)} \\
&=&\frac{x^{2}-\left( x+1\right) (x-2)}{(x-2)(x+2)}.\tag{1}
\end{eqnarray*}
$$
Otherwise we would get the equivalent but more more complex fraction
$$
\frac{x^{2}}{(x^{2}-4)}-\frac{x+1}{x+2}=\frac{x^{2}\left( x+2\right) -\left(
x+1\right) (x^{2}-4)}{(x^{2}-4)\left( x+2\right) }.
$$
Expanding the second term of the numerator of $(1)$
$$
\begin{eqnarray*}
\left( x+1\right) (x-2) &=&x(x-2)+(x-2)=x^{2}-2x+x-2 \\
&=&x^{2}-x-2
\end{eqnarray*}
$$
and substituting into the fraction we get
$$
\frac{x^{2}-\left( x^{2}-x-2\right) }{(x-2)(x+2)}=\frac{x^{2}-x^{2}+x+2}{
(x-2)(x+2)}=\frac{x+2}{(x-2)(x+2)},\tag{2}
$$
which for $x+2\neq 0$ simplifies to
$$
\frac{1}{x-2}\tag{3}
$$
Added: In general we transform the sum (or difference) of two given rational fractions (the numerator and denominator consists of polynomials) into a
single equivalent fraction, by using properties such as
*
*$$\frac{A(x)}{B(x)}=\frac{A(x)P(x)}{B(x)P(x)}\qquad\text{for }P(x)\neq 0.$$
*$$\frac{A(x)}{B(x)}\pm \frac{C(x)}{D(x)}=\frac{A(x)D(x)\pm B(x)C(x)}{B(x)D(x)}.$$
*$$\frac{A(x)}{B_1(x)B_2(x)}\pm \frac{C(x)}{B_2(x)}=\frac{A(x)\pm B_1(x)C(x)}{B_1(x)B_2(x)}.$$
| {
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solve complex equation $x^8 = \frac{1+i}{\sqrt{3} - i} = \frac{\sqrt[8]{\frac{2}{\sqrt{2}}}(\cos \frac{\pi}{4} + i \sin{\frac{\pi}{4}})}{2 \cos \frac{\pi}{6} + i \sin \frac{3\pi}{2}}$
What's the way to solve this kind of equation? I think there must be 8 solutions.
I tried to solve the following two equations
$a^6 = 1+i$
$b^6 = \sqrt{3} - i$
I tried the following for the first equation
$-1 = i^2 = (a^6 - 1)^2 = a^{12} - 2a^6 +1$
Then I would need to solve this:
$a^{12} - 2a^6 +2 = 0$
Is this the right approach?
| $(1+i)/\sqrt(2) = e^{i \pi/4}$
$(\sqrt(3)-i)/2= e^{-i \pi/6}$
You should obtain something like $z^8= a e^{i b}$ that should be solvable!
| {
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How to get sine / cosine value out of tangens I know that: $\tan(\alpha) = 1/2$.
How can I get clean values for sine / cosine without the calculator?
Is there a relationship?
I know that $\sin(\arctan(1/2))$ is a way ... But I hope you get the point.
Thank you!
| Here is an algebraic solution.
Recall that $\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$. Now your equation states that
$$\sin \alpha = \frac{1}{2}\cos \alpha.$$
Squaring both sides gives
$$\sin^2 \alpha = \frac{1}{4} \cos^2 \alpha = \frac{1}{4}(1 - \sin^2 \alpha).$$
Or
$$\sin^2 \alpha = \frac{1}{5}.$$
Similarly one sees that
$$\cos^2 \alpha = \frac{4}{5}.$$
Thus the possibilities are $$\sin \alpha = \pm \frac{1}{\sqrt{5}}, \quad \cos \alpha = \pm \frac{2}{\sqrt{5}}.$$
We still have to show that these possibilities are possible. We know that there is $\alpha$ such that $\tan \alpha = \frac{1}{2}$. I.e. the original equation has a solution. Now if $\alpha$ is a solution, so is $\alpha + \pi$. From this we can deduce that both $\sin \alpha = \frac{1}{\sqrt{5}}$ and $\sin \alpha = -\frac{1}{\sqrt{5}}$ are attainable and so are $\cos \alpha = \pm \frac{1}{\sqrt{5}}$. Moreover the signs must be the same for both. Thus the solutions are
$$(\cos \alpha, \sin \alpha) \in \{(\frac{1}{\sqrt{5}}, \frac{1}{\sqrt{5}}), (-\frac{1}{\sqrt{5}}, -\frac{1}{\sqrt{5}})\}.$$
| {
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Integration by partial fraction mistake I am trying to understand my mistake using the partial fraction method to solve $ \displaystyle\int{\frac{2x-5}{x^2-4x+4}} $.
Here's what I have so far:
$$ \frac{2x+5}{x^3-4x^2+4x} = \frac{2x+5}{x(x-2)^2} = \frac{A}{x} + \frac{B}{(x-2)^2}$$
And its easy to see that $A=\displaystyle\frac{5}{4}$ and that $B=\displaystyle\frac{9}{2}$. Somehow I get a completely different result then that of my professor's. She did the following:
$$ \frac{2x+5}{x^3-4x^2+4x} = \frac{2x+5}{x(x-2)^2} = \frac{A}{x} + \frac{B}{(x-2)} + \frac{C}{(x-2)^2}$$
Can anyone please tell me where my mistake is? Thanks
| If the denominator has no repeated factor, then it's easy to express it into partial fraction. For example
$$\frac{-2x+3}{x^3-x} =\frac{-2x+3}{x(x-1)(x+1)}= \frac{A}{x} + \frac{B}{(x-1)} + \frac{C}{(x+1)},$$
$$\frac{7}{x^4-3x^2+2} =\frac{-2x+3}{(x-2)(x+2)(x-1)(x+1)}= \frac{A}{x-2} + \frac{B}{x+2} + \frac{C}{x-1}+\frac{C}{x+1}
.$$
However, if the denominator has repeated factor, it's more complicated. Basically we include all the lower power of the repeated factor. Therefore, in your example, we have
$$\frac{2x+5}{x^3-4x^2+4x} = \frac{2x+5}{x(x-2)^2} = \frac{A}{x} + \frac{B}{(x-2)} + \frac{C}{(x-2)^2}.$$
Let me do two more examples
$$\frac{4x+1}{x^4+2x^3+x^2}=\frac{4x+1}{x^2(x+1)^2}=\frac{A}{x} + \frac{B}{x^2} + \frac{C}{x+1} + \frac{D}{(x+1)^2},$$
$$\frac{-5x+1}{(x+2)(x-1)^3}=\frac{A}{x+2} + \frac{B}{x-1} + \frac{C}{(x-1)^2} + \frac{D}{(x-1)^3}.$$
| {
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How many rationals of the form $\large \frac{2^n+1}{n^2}$ are integers? This was Problem 3 (first day) of the 1990 IMO. A full solution can be found here.
How many rationals of the form $\large \frac{2^n+1}{n^2},$ $(n \in \mathbb{N} )$ are integers?
The possible values of $n$ that i am able to find is $n=1$ and $n=3$, so there are two solutions and this seems to be the answer to this problem.
But now we have to prove that no more of such $n$ exists, and thus the proof reduces to: Proving that $n^2$ does not divides $2^n+1$ for any $n \gt 3$.
Does anybody know how to prove this?
| Andre's modification of a wrong answer :)
If $n=3^k$, then
$$2^n+1=2^{3^k}+1=2^{3 \cdot 3^{k-1}}+1= (2^{3^{k-1}}+1)( 2^{2 \cdot 3^{k-1}}-2^{ \cdot 3^{k-1}}+1) $$
The second bracket is never divisible by $9$, thus by induction one can prove that $3^{2k-1}$ doesn't divide $2^n+1$.
Note: Since Geoff's answer was wrong, and this post doesn't make too much sense anymore, a simple observation:
If $n \neq 1$, then $3|n$.
Indeed let $p$ be the smallest prime divisor of $n$.
Then $2^{p-1} \equiv 1 \mod p$ and $2^{2n} \equiv (-1)^2 \equiv 1 \mod p$.
Thus $2^d \equiv 1 \mod p$ where $d=gcd(p-1,2n)$. But no prime factor of $p-1$ can divide $n$, since $p$ is the smallest one, thus $gcd(p-1,n)=1$. Hence $d |2$.
$2^d \equiv 1 \mod p$ implies now that $p=3$.
This proves that $n=3^km$ for some $k \geq 1$ and $m $ relatively prime to $3$. I wonder if the first argument can be modified for this case:
$$2^n+1=2^{3^km}+1=2^{3 \cdot 3^{k-1}m}+1= (2^{3^{k-1}m}+1)( 2^{2 \cdot 3^{k-1}m}-2^{ \cdot 3^{k-1}m}+1) $$
Since $9$ doesn't divide the second bracket we get that $3^{2k-1}$ must divide $(2^{3^{k-1}m}+1)$ and repeating I think we get $3^{k}$ divides $2^m+1$...
It is easy to prove that $2^m \equiv -1 \mod 9$ implies $3 |m$ (this follows from $2^3 \equiv -1 mod 9$ and $ord(2)=6$).
Hence $k=1$, and we must have $n=3 m$ with $gcd(3,m)=1$...
Now, lets try the same again.
Suppose by contradiction $m \neq 1$. Let $q$ be the smallest prime factor of $m$.
Then
$2^d \equiv 1 \mod q$ where $d=gcd(q-1,2n)$. But no prime factor of $p-1$ can divide $n$, excepting $3$, Hence $d |6$.
This implies that
$$2^6 \cong 1 \mod q \,.$$
Thus, the only possible values of $q$ is $q=7$.
But this is not possible since $2^{3m}+1 \equiv 1+1 \mod 7$, thus $7$ cannot divide $2^n+1$.
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How could we solve $x$, in $|x+1|-|1-x|=2$? How could we solve $x$, in $|x+1|-|1-x|=2$?
Please suggest a analytical way that I could use in other problems too like this $ |x+1|+|1-x|=2$ and of this genre.
Thank you,
| (This isn’t really an answer, but it’s too big for a comment. It’s just continuing from where Nana stopped in their answer, revealing that you can’t find the solution using that approach.)
We have just squared both sides to try to get rid of the square roots.
$$ \left( \sqrt{(x+1)^2} \right)^2 = \left( 2 + \sqrt{(1-x)^2} \right)^2 $$
Continuing from here:
Cancel on the left, expand the square on the right.
$$ (x+1)^2 = 4 + 4 \sqrt{(1-x)^2} + (1-x)^2 $$
Expand squares.
$$ x^2 + 2x + 1 = 4 + 4 \sqrt{(1-x)^2} + 1 - 2x + x^2 $$
Combine terms.
$$ 4x - 4 = 4 \sqrt{(1-x)^2} $$
Divide by coefficient.
$$ x - 1 = \sqrt{(1-x)^2} $$
Square both sides again to get rid of that last square root.
$$ (x - 1)^2 = \left(\sqrt{(1-x)^2}\right)^2 $$
Cancel on the right.
$$ (x - 1)^2 = (1-x)^2 $$
Expand squares.
$$ x^2 - 2x + 1 = 1 - 2x + x^2 $$
Rearrange terms.
$$ x^2 - 2x + 1 = x^2 - 2x + 1 $$
Combine terms.
$$ 0 = 0 $$
The equation $0=0$ doesn’t help us find the correct values of $x$. To solve the problem, we must use another approach.
| {
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How can I solve a complicated induction exercise? (formula for sum of fourth powers) I have an induction exercise:
It is for $n \in \mathbb{N_0}$. Show:
$$ \sum\limits_{k=1}^{n} k^4 = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) $$
As far as I understand it, you have to put in $(n+1)^4$ at the end and you have to resolve it to the old function replacing $n$ with $n+1$. So I have on my paper:
$$ \frac 1 {30} (n+1)((n+1)+1)(2(n+1)+1)(3(n+1)^2+3(n+1)-1) = \frac 1 {30} n(n+1)(2n+1)(3n^2+3n-1) + (n+1)^4 $$
How can I solve this exercise? I probably have to disprove it because it is not true for numbers $> 1$.
All the other examples were with very tiny formulas.
I really tried to find a solution; what should I do?
| By induction we show that:
$$1+2^4+3^4+....+n^4 =\frac 1 {5}n^5+\frac 1 {2}n^4+\frac 1 {3}n^3-\frac 1 {30}n$$
For $n=1$:
$$1=\frac 1 {30}\cdot2\cdot3\cdot5=1$$
Now suppose that is true for $n$, and show that and show that it is also true for $n+1$, that is:
$$1+2^4+3^4+....+n^4+(n+1)^4 =\frac 1 {5}(n+1)^5+\frac 1 {2}(n+1)^4+\frac 1 {3}(n+1)^3-\frac 1 {30}(n+1)$$
This equality is verified if the following equality is verified:
$$\frac 1 {5}(n+1)^5+\frac 1 {2}(n+1)^4+\frac 1 {3}(n+1)^3-\frac 1 {30}(n+1)=$$
$$=\frac 1 {5}n^5+\frac 1 {2}n^4+\frac 1 {3}n^3-\frac 1 {30}n + (n+1)^4$$
, calculating the solution with a few additions!
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Solve $\frac{4}{x+1}+\frac{2}{x-2} \leq 3$ Solve
$$\frac{4}{x+1}+\frac{2}{x-2} \leq 3$$
I must be making a very stupid mistake somewhere ... been stuck on this for 1hr+ or even more ...
| One solution is given below. Another one, very close in spirit to your attempt, is given in a comment at the end.
It turns out that the numbers fit together perfectly! Rewrite our inequality as
$$\frac{4}{x+1}-4 +\frac{2}{x-2}-1 \le 0.$$
This simplifies to
$$-\frac{4x}{x+1}+\frac{x}{x-2} \le 0.$$
The problem now breaks up into two natural cases.
Case (i): $x\ge 0$. Here our inequality is equivalent to
$$\frac{1}{x-2}\le \frac{4}{x+1}.$$
For $x >2$ this can be rewritten as $4(x-2)\ge x+1$, giving $x\ge 3$. For $0\le x<2$, we get in a similar way $x \le 3$, which holds automatically. We have obtained the intervals $[3,\infty)$ and $[0,2)$.
Case (ii): $x<0$. Here our inequality is equivalent to
$$\frac{1}{x-2}\ge \frac{4}{x+1}.$$
This is automatically false if $-1<x<0$, since the left side is negative and the right side is not. When $x<-1$, the inequality is equivalent to $x+1 \ge 4(x-2)$, which holds. This gives us the additional interval $(-\infty, -1)$.
Comment: In order to solve the inequality, you used the nice strategy of multiplying by $(x+1)^2(x-2)^2$. When we do that, we have to note that $x\ne -1, 2$. Minor point. The major point is that you should "multiply out" only when you have to. Multiplying out tends to create a mess. If you keep the common factors of the two sides separated out, your calculation pushes through smoothly. Starting from your
$$2(x-1)(x+1)(x-2)\le (x+1)^2(x-2)^2,$$
we obtain
$$(x+1)(x-2)\left[(x+1)(x-2)-2(x-1)\right]\ge 0,$$
which simplifies to
$$(x+1)(x-2)(x^2-3x) \ge 0.$$
Since $x^2-3x=x(x-3)$, the analysis becomes routine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/100092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Simple Partial Fractions Question For practice, I am integrating,
$$\int \frac{x}{3x^2 + 8x -3}dx$$
So, I can then factor it as,
$$\int \frac{x}{(3x-1)(x+3)}dx$$
By partial fractions, I decompose
$$\frac{x}{(3x-1)(x+3)}= \frac{A}{3x-1} + \frac{B}{x+3}$$
For finding $A$, I multiply both sides by $3x-1$, which gives
$$\frac{x(3x-1)}{(3x-1)(x+3)} = \frac{A(3x-1)}{3x-1} + \frac{B(3x-1)}{x+3}$$
So, we have that
$$\frac{x}{(x+3)} = A + \frac{B(3x-1)}{x+3}$$
Letting $3x-1=0$, we have that $x=\frac{1}{3}$, so then
$$\frac{\frac{1}{3}}{(\frac{1}{3}+3)} = A$$
Thus, we have that $A=\frac{1}{10}$. For determining $B$, we then multiply both sides by $x+3$ and receive, as a similar process to the previous,
$$\frac{x(x+3)}{(3x-1)(x+3)} = \frac{A(x+3)}{3x-1} + \frac{B(x+3)}{x+3}$$
Then,
$$\frac{x}{3x-1} = \frac{A(x+3)}{3x-1} + B$$
So, if we let $x+3=0$, we then have that $x=-3$ and so,
$$\frac{-3}{3(-3)-1}=B$$
So, we then have that $B=\frac{3}{10}$. Thus, our original integral can then be written as,
$$\int \frac{x}{(3x-1)(x+3)}dx = \int \frac{1}{10(3x-1)} + \frac{3}{10(x+3)} dx$$
We can, by splitting up the integral find,
$$\int \frac{x}{(3x-1)(x+3)}dx = \frac{1}{10} \int \frac{1}{3x-1} dx + \frac{3}{10} \int \frac{1}{x+3} dx$$
Thus, we conclude that,
$$\int \frac{x}{3x^2 + 8x -3}dx = \frac{\ln|3x-1|}{30} + \frac{3 \ln|x+3|}{10} + C$$
Wolframalpha shows that, the answer is:
$$\frac{1}{30}(\ln(1-3x)+ 9 \ln(3+x)) +C$$
What am I doing wrong, did I miss a negative sign somewhere?
| Wolfram forgot the absolute value signs. Further both answers are the same.
$$\frac{9}{30}=\frac{3}{10}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the recurrence $T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$ I am trying to solve the recurrence below but I find myself stuck.
$T(n) = \frac{n}{2}T(\frac{n}{2}) + \log n$
I have tried coming up with a guess by drawing a recurrence tree. What I have found is
number of nodes at a level: $\frac{n}{2^{i}}$
running time at each node: $\log \frac{n}{2^{i}}$
total running time at each level: $\frac{n}{2^{i}} \log \frac{n}{2^{i}}$
I try to sum this over through n > $\log n$ which is the height of the tree
$$\begin{align}
\sum\limits_{i=0}^n \frac{n}{2^{i}} \log \frac{n}{2^{i}}&=
n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log \frac{n}{2^{i}}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} (\log n - \log 2^{i})\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n - \sum\limits_{i=0}^n \frac{1}{2^{i}} \log 2^{i}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n - \sum\limits_{i=0}^n \frac{ \log 2^{i}}{2^{i}}\\
&=n \sum\limits_{i=0}^n \frac{1}{2^{i}} \log n - d \sum\limits_{i=0}^n \frac{i}{2^{i}}\\
&=2 n \log n - d \sum\limits_{i=0}^n \frac{i}{2^{i}}
\end{align}$$
I am still trying to figure out the sum of the second summation above but I somehow feel that it will be bigger than $2n \log n$ which makes my whole approach wrong.
Is there another way to tackle this problem?
| Consider the equality
$$
1+x+x^2+...+x^n=\frac{x^{n + 1} -1}{x-1}
$$
Differentiate it by $x$, then multiply by $x$:
$$
x+2x^2+3x^3+...+n x^n=\frac{nx^{n+2} - (n + 1)x^{n+1} + x}{(x-1)^2}
$$
Now we can substitute $x=\frac{1}{2}$ and obtain
$$
\sum\limits_{i=0}^n\frac{i}{2^i}=n\left(\frac{1}{2}\right)^{n} - (n + 1)\left(\frac{1}{2}\right)^{n-1} + 2
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to set up the existence condition of an absolute value $$
\frac{\sqrt{4 + \arccos\left|\frac{2-x}{x+3}\right|}}{\sqrt{x^2 - 4x + 5} - 3}
$$
I'm trying to find the natural domain of the function above. I set up this conditions:
$$
\begin{cases}\sqrt{x^2 - 4x + 5} - 3\neq0&(denominator)\\x^2 - 4x + 5\ge0&(root)\\4 + \arccos\left|\frac{2-x}{x+3}\right|\ge0&(root)\\\left|\frac{2-x}{x+3}\right|\ge-1\cup\left|\frac{2-x}{x+3}\right|\le1&(arccos)\end{cases}
$$
Now I know that is necessary to set up two more conditions: the existence condition of the absolute value, and the existence condition of the fraction denominator. But I don't know how to deal with absolute values. Could someone explain it to me maybe using this example?
Note: These are not homework, but an exercise chose, because has an absolute value inside, to understand the theory.
| We first do it in a slow and tedious way, and then a quick way. For the $\arccos$ to be defined, we need $\frac{|2-x|}{|3+x|} \le 1$ (it is naturally non-negative).
Thus we need the inequality $|2-x|\le |x+3|$. First suppose that $x \ge 2$. Then $|2-x|=x-2$, so we want $x-2 \le x+3$, which is true.
For $-3 <x<2$, we have $|2-x|=2-x$ and $|x+3|=x+3$. So we want $2-x\le x+3$, that is, $x \ge -1/2$.
At $x=-3$ things are clearly bad. But also if $x<-3$, then $|2-x|=2-x$ and $|x+3|=-(3+x)$. Thus our inequality becomes $2-x \le -(3+x)$, which is false.
To sum up, the $\arccos$ part is defined if $x\ge -1/2$.
Now we worry about the bottom. The function $x^2-4x+5$ is always positive, no trouble there. We want to make sure that $x^2-4x+5\ne 9$. We have equality at $-1$ and $5$, but $-1$ has already been ruled out.
Conclusion: Our function is defined for all $x\ge -1/2$ except $x=5$.
A quick way: The inequality $|2-x|\le |3+x|$ is equivalent to $(2-x)^2\le (3+x)^2$. Expand. The $x^2$'s cancel, and we get $x\ge -\frac{5}{10}$.
The tedious breaking up into cases of the first solution is unfortunately a useful tool in dealing with absolute value problems. The slick procedure of the "quick" solution is often much messier than doing things by cases.
| {
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"timestamp": "2023-03-29T00:00:00",
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differential equation nondevelopable I try to solve this differential equation whose solution seems not to be constructable in power series
$y''+(x+a/x^2+b)y=0$, where $a$ and $b$ are some positive real numbers.
If one can help me please?
| Hint:
$y''+\left(x+\dfrac{a}{x^2}+b\right)y=0$
$x^2y''+(x^3+bx^2+a)y=0$
Let $y=x^ku$ ,
Then $y'=x^ku'+kx^{k-1}u$
$y''=x^ku''+kx^{k-1}u'+kx^{k-1}u'+k(k-1)x^{k-2}u=x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u$
$\therefore x^2(x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u)+(x^3+bx^2+a)x^ku=0$
$x^{k+2}u''+2kx^{k+1}u'+(x^3+bx^2+k(k-1)+a)x^ku=0$
$x^2u''+2kxu'+(x^3+bx^2+k(k-1)+a)u=0$
Choose $k(k-1)+a=0$ , i.e. $k=\dfrac{1\pm\sqrt{1-4a}}{2}$ , the ODE becomes
$x^2u''+(1\pm\sqrt{1-4a})xu'+(x^3+bx^2)u=0$
$xu''+(1\pm\sqrt{1-4a})u'+(x^2+bx)u=0$
$4x\dfrac{d^2u}{dx^2}+2\dfrac{du}{dx}+(2\pm4\sqrt{1-4a})\dfrac{du}{dx}+(4x^2+4bx)u=0$
Let $x=r^2$ ,
Then $\dfrac{du}{dr}=\dfrac{du}{dx}\dfrac{dx}{dr}=2r\dfrac{du}{dx}$
$\dfrac{d^2u}{dr^2}=\dfrac{d}{dr}\left(2r\dfrac{du}{dx}\right)=2r\dfrac{d}{dr}\left(\dfrac{du}{dx}\right)+2\dfrac{du}{dx}=2r\dfrac{d}{dx}\left(\dfrac{du}{dx}\right)\dfrac{dx}{dr}+2\dfrac{du}{dx}=2r\dfrac{d^2u}{dx^2}2r+2\dfrac{du}{dx}=4r^2\dfrac{d^2u}{dx^2}+2\dfrac{du}{dx}=4x\dfrac{d^2u}{dx^2}+2\dfrac{du}{dx}$
$\therefore\dfrac{d^2u}{dr^2}+\dfrac{1\pm2\sqrt{1-4a}}{r}\dfrac{du}{dr}+(4r^4+4br^2)u=0$
Let $u=e^{mr^3}v$ ,
Then $\dfrac{du}{dr}=e^{mr^3}\dfrac{dv}{dr}+3mr^2e^{mr^3}v$
$\dfrac{d^2u}{dr^2}=e^{mr^3}\dfrac{d^2v}{dr^2}+3mr^2e^{mr^3}\dfrac{dv}{dr}+3mr^2e^{mr^3}\dfrac{dv}{dr}+(9m^2r^4+6mr)e^{mr^3}v=e^{mr^3}\dfrac{d^2v}{dr^2}+6mr^2e^{mr^3}\dfrac{dv}{dr}+(9m^2r^4+6mr)e^{mr^3}v$
$\therefore e^{mr^3}\dfrac{d^2v}{dr^2}+6mr^2e^{mr^3}\dfrac{dv}{dr}+(9m^2r^4+6mr)e^{mr^3}v+\dfrac{1\pm2\sqrt{1-4a}}{r}\left(e^{mr^3}\dfrac{dv}{dr}+3mr^2e^{mr^3}v\right)+(4r^4+4br^2)e^{mr^3}v=0$
$\dfrac{d^2v}{dr^2}+\left(6mr^2+\dfrac{1\pm2\sqrt{1-4a}}{r}\right)\dfrac{dv}{dr}+((9m^2+4)r^4+4br^2+3mr(3\pm2\sqrt{1-4a}))v=0$
Choose $9m^2+4=0$ , i.e. $m=\pm\dfrac{2i}{3}$ , the ODE becomes
$\dfrac{d^2v}{dr^2}+\left(\pm4ir^2+\dfrac{1\pm2\sqrt{1-4a}}{r}\right)\dfrac{dv}{dr}+(4br^2\pm2ir(3\pm2\sqrt{1-4a}))v=0$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Finding Limit $\lim_{x \to \infty} (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$ Finding Limit
$$\lim_{x \to \infty} (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$$
So I let
$$y = (2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)^{\frac{1}{x}}$$
$\ln$ both sides:
$$\ln{y} = \frac{1}{x} \ln {(2^x + 3^x + 5^x + 7 ^x + 11 ^x +13^x)}$$
Now what?
| From the idea in the question here, for $x>0$:
$$
13^x<2^x+3^x+5^x+7^x+11^x+13^x <6\cdot 13^x;
$$
whence
$$
13 <(2^x+3^x+5^x+7^x+11^x+13^x )^{1/x}<6^{1/x}\cdot13.
$$
Now use the squeeze theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/112698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
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Integration of rational functions.. Some rational function is giving me some trouble...
\begin{aligned}
\ \int \frac {x^2-9x+16}{(x-1)(x^2+6x-7)} dx
\end{aligned}
I simplified it like so:
\begin{aligned}
\ \int \frac {x^2-9x+16}{(x-1)^2(x+7)} dx
\end{aligned}
If I get it right I can transform it into this:
\begin{aligned}
\ \int \frac {A}{(x-1)} + \frac {B}{(x-1)^2} + \frac {C}{(x+7)} dx
\end{aligned}
And then I get this:
\begin{aligned}
\ Ax^2 +6Ax-7A+Bx+Cx^2-2Cx+C = x^2-9x+16
\end{aligned}
So I forgot how to get ABC values. Could somebody remind me this?
I am showing all the steps because I am not 100% confident that with have not made any mistakes.
| You have:
$$
{x^2-9x+16\over(x-1)^2(x+7)}={A\over x-1\vphantom{(^2}}+{B\over (x-1)^2}+{C\over x+7\vphantom{(^2}}.
$$
Clearing denominators in the above equality gives:
$$\tag{1}
x^2-9x+16=A(x-1)(x+7)+B(x+7)+C(x-1)^2.
$$
This holds for all $x$. We can be clever and take advantage of this. Set $x=1$ in $(1)$ to obtain
$$
8=B\cdot 8.
$$
So $B=1$.
Set $x=-7$ in $(1)$ to obtain
$$
128=C\cdot64.
$$
So $C=2$.
Then, substituting the known values Of $B$ and $C$ into $(1)$, we have
$$\tag{2}
x^2-9x+16=A(x-1)(x+7)+ (x+7)+2(x-1)^2.
$$
Putting the right hand side of $(2)$ in standard form gives:
$$
x^2-9x+16=(A+2)x^2 +(6A-3)x +(9-7A);
$$
and we see that $A=-1$. (All we really had to do here, is note that the $x^2$ term on the right hand side of $(2)$ is $Ax^2+2x^2$. Then, since the $x^2$ term on the left hand side of $(2)$ is $x^2$, we'd know $A+2=1$; whence $A=-1$.)
Alternatively, as I suggested in the comments, expand the right hand side of $(1)$ and write the resulting quadratic in standard form:
$$
x^2-9x+16=(A+C)x^2+(6A+B-2C)x+(-7A+7B+C ).
$$
Then equate the $x^2$, $x$, and constant coefficients of the right and left hand sides of the above equality:
$$
\eqalign{
1&=A+C\cr
-9&=6A+B-2C\cr
16&=-7A+7B+C.
}
$$
This gives you three linear equations in the unknowns $A$, $B$, and $C$. Solve the system for the unknowns. (I prefer the first method as it reduces the chances (big for me) of making the types of errors that the second method is prone to.)
| {
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Evaluating $\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$ Evaluating
$$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}$$
Its $\frac{\infty}{\infty}$ so I should use L Hospital rule, but the terms are exponential and differentiation won't do much good? I am thinking maybe I somehow use $\ln$ both sides? But how? Or perhaps I should do something else?
UPDATE
Following @Paul's answer: Since $|r| < 1$ so sequence is convergent. So I use the formula
$$\sum_{n=1}^\infty ar^{n-1} = \frac{a}{1-r}$$
$$\frac{1}{10} \cdot \frac{1}{1-\frac{4}{5}} = \frac{1}{10}\cdot 5 = \frac{1}{2}$$
But answer is $\frac{2}{5}$
| You can rewrite it as
$$\sum_{n=1}^\infty \frac{2^{2n-1}}{5^{n+1}}=\sum_{n=1}^\infty \frac{\frac{1}{2}\cdot2^{2n}}{5\cdot 5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{2^{2n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \frac{4^{n}}{5^n}=\frac{1}{10}\sum_{n=1}^\infty \left(\frac{4}{5}\right)^n$$
which is a geometric series with ratio $r=\displaystyle\frac{4}{5}$. I think you can finish it from here.
| {
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Try to prove $\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$ $$A_n=\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}$$
Try to prove $$\lim_{n \to \infty}n(\ln 2-A_n) = \frac{1}{4}$$
I try to decompose $\ln 2$ as $$\ln(2n)-\ln(n)=\ln\left(1+\frac{1}{2n-1}\right)+\dots+\ln\left(1+\frac{1}{n}\right)\;,$$ but I can't continue, is that right?
| Let's cheat and use one of Euler's many results:
$$\sum_{i=1}^{n} \frac{1}{i} = \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$
Note that:
$$A_n + \sum_{i=1}^{n} \frac{1}{i} = \sum_{i=1}^{2n} \frac{1}{i}$$
Substituting Euler's result for both summations, we get:
$$A_n + \ln n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right) = \ln 2n + \gamma + \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$
Rearranging, and using $\ln 2n = \ln 2 + \ln n$, we get
$$A_n = \ln 2 - \frac{1}{4n} + O\left(\frac{1}{n^2}\right)$$
Thus the requested limit becomes
$$\lim_{n \to \infty} n (\ln 2 - A_n) = \lim_{n \to \infty} n \left(\frac{1}{4n} - O\left(\frac{1}{n^2}\right)\right) = \frac{1}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem:
Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$
I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer.
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.
| Joe, since your original question was to prove that it is an integer:
An extension of Sivaram Ambikasaran's answer:
First claim that
$\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} = 1^4 + 2^4 +\cdots + n^4$
And show your claim by induction. (This should be straightforward)
Once your claim is proven, the given expression $\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30} $ is the sum $1^4 + 2^4 +\cdots + n^4$ for every $n$, which means it is an integer. Because $1^4 + 2^4 +\cdots + n^4$ for every $n$ is an integer.
| {
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If $f(x)f(y)=f(\sqrt{x^2+y^2})$ how to find $f(x)$ As we know, for the $$f(x)f(y)=f(x+y)$$ $f(x)=\mathrm e^{\alpha x}$ is a solution.
What about
$f(x)f(y)=f(\sqrt{x^2+y^2})$?
Does anybody know about the solution of the function equation?
I tried to find $f(x)$.
See my attempts below to find $f(x)$.
$$f(x)=a_0+a_1x+\frac{a_2x^2}{2!}+\frac{a_3x^3}{3!}+\cdots$$
$$f(y)=a_0+a_1y+\frac{a_2y^2}{2!}+\frac{a_3y^3}{3!}+\cdots$$
$$f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
$$f(\sqrt{x^2+y^2})=a_0+a_1\sqrt{x^2+y^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3(x^2+y^2)^{3/2}}{3!}+\cdots=$$
$$f(\sqrt{x^2+y^2})=a_0+a_1y\sqrt{1+(x/y)^2}+\frac{a_2(x^2+y^2)}{2!}+\frac{a_3y^2(1+(x/y)^2)^{3/2}}{3!}+\cdots=f(x)f(y)=a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots$$
if we use binom expansion for $(1+(x/y)^2)^{m}$
$$(1+(x/y)^2)^{m}=1+\frac{mx^2}{y^2}+\frac{m(m-1)x^4}{2!y^4}+\frac{m(m-1)(m-2)x^6}{3!y^6}+\cdots$$
Let's put the expansion to the equation $f(\sqrt{x^2+y^2})$
$$
\begin{align}
& f(\sqrt{x^2+y^2}) =a_0 + a_1 y \left( 1 + \frac{(1/2)x^2}{y^2} + \frac{(1/2)((1/2)-1)x^4}{2!y^4} \right. \\ \\
& \left. {} + \frac{(1/2)((1/2)-1)((1/2)-2)x^6}{3!y^6} + \cdots\right) + \frac{ a_2 (x^2+y^2)}{2!} \\ \\
& + \frac{a_3y^2 \left(1+\frac{(3/2)x^2}{y^2}+\frac{(3/2)((3/2)-1)x^4}{2!y^4}+\frac{(3/2)((3/2)-1)((3/2)-2)x^6}{3!y^6}+\cdots\right)}{3!} +\cdots \\ \\
& = a_0f(y)+a_1f(y)x+\frac{a_2f(y)x^2}{2!}+\frac{a_3f(y)x^3}{3!}+\cdots
\end{align}
$$
If we equal for all $x^n$ terms in both sides
we can see $a_{2n-1}=0$, but to find $a_{2n}$ seems hard for me.
Any idea to find $a_{2n}$
Thanks in advice.
| If you set $$g(x) := f(\sqrt{x})$$ for $x \in [0, \infty)$ then you get $$g(x)g(y) = f(\sqrt{x})f(\sqrt{y}) = f(\sqrt{x+y}) = g(x+y)$$
You see that $g(x) \geq 0$, and if $g(x) = 0$ for some $x > 0$ then $g \equiv 0$. Thus, you can look at $$h(x) := \log(g(x))$$ It satisfies $$h(x) + h(y) = \log(g(x)g(y)) = \log(g(x+y)) = h(x+y)$$ If you impose any reqularity condition on $f$ you can think of, you will get $h(x) = \alpha x$, and consequently $$f(x) = \exp(\alpha x^2)$$ for $x > 0$. You can generalise this result to $x < 0$ using the fact that from the initial equation it follows that $f$ is even.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/115784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 4,
"answer_id": 1
} |
how to find out remainder of $3^{256}$ divided by $13$ here is a question that about finding the remainder when dividing $3^{256}$ divided by $13$. Can anyone suggest how to find the solution
| In general when the modulus $m$ is small, you can check the successive powers $x^r$ modulo $m$, the remainders will cycle quickly.
For instance let's examine $4^r\pmod 7$
$\begin{cases}
4^0\equiv 1\pmod 7\\
4^1\equiv 4\pmod 7\\
4^2\equiv 16\equiv 2\pmod 7\\
4^3\equiv 4^2\times 4\equiv 2\times 4\equiv 8\equiv 1\pmod 7\\
4^4\equiv 4^3\times 4\equiv 1\times 4\equiv 4\pmod 7\\
\cdots
\end{cases}$
You can see the remainders will always be $1,2,4,1,2,4,1,\cdots$
Thus the result depends of the divisibility of $r$ by $3$.
*
*If $r=3k$ then $4^r$ will be the same as $4^0\equiv 1\pmod 7$
*If $r=3k+1$ then $4^r$ will be the same as $4^1\equiv 4\pmod 7$
*If $r=3k+2$ then $4^r$ will be the same as $4^2\equiv 2\pmod 7$
For instance $4^{857}\equiv 4^{(857\bmod 3)}\equiv 4^2\equiv 2\pmod 7$
Of course with another base, the cycle may be shorter or longer (e.g. $3^r\pmod 7$ leads to remainders $1,3,2,6,5,4,1,\cdots$
And this time $3^r\equiv 3^{(r\bmod 6)}\pmod 7$
This technique works very well for small modulus, however when $m$ becomes larger the cycle length may grow as well. In some cases, depending on the base, the cycle can still be short, but in general the method become less applicable.
So for bigger modulus, you need to uses more advanced stuff, like little Fermat theorem or Euler's totient function, both allow to reduce the exponent $r$ when $m$ is a prime number (or a power of a prime).
When $m$ is a composed number (of $n$ primes), Chinese remainder theorem, allows to transform the problem into $n$ smaller problems where only primes modulus are involved.
Applying to your problem $3^r\equiv 1,3,9,1,\cdots \pmod{13}$ the cycle is only $3$ steps short.
Thus $3^{256}\equiv 3^{(256\bmod 3)}\equiv 3^1\equiv 3\pmod{13}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 7
} |
Help to evaluate determinant I want to evaluate the determinant of the $n \times n$ matrix
$\left|\begin{array}{ccccc} 1 & 0 & \ldots & 0& 0 \\
0 & 0 & \ldots & 0 & -a\\
0 & 0 & \ldots & -a & 0\\
&&&\vdots \\
0 & -a & 0 &\dots & 0
\end{array}\right|.$
So I try to say that it is $(-1)^{ n + (n-1) + \ldots n-(n-2)}(-a)^{n-1}$. So power of -1 should be $\frac{(n-1)(n+2)}{2} + n-1$. However answer given is $n(n-1)/2$. Where is wrong?
| In fact, your answer is correct and agrees with the given answer. Here is the reason:$$\frac{(n-1)(n+2)}{2} + n-1=\frac{n(n-1)+2(n-1)}{2}+n-1=\frac{n(n-1)}{2}+2(n-1),$$
which implies that
$$(-1)^{\frac{(n-1)(n+2)}{2} + n-1}=(-1)^{\frac{n(n-1)}{2}+2(n-1)}=(-1)^{\frac{n(n-1)}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/116240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the value of equation? Suppose
\begin{align*}
a+b+c &= 20\\
\frac 1a + \frac 1b + \frac 1c &= 30
\end{align*}
Then find the value of
$$
\frac ab + \frac ba + \frac ac + \frac ca + \frac bc + \frac cb
$$
How i can apply A.M. $\ge$ G.M $\ge$ H.M. in this?
How i can achieve this?
| $$\frac ab + \frac ba + \frac ac + \frac ca + \frac bc + \frac cb $$ can be simplified as
$$ \frac{a^2c+b^2c+a^2b+bc^2+ab^2+ac^2}{abc} $$
which is
$$ \frac{\left[ac(a+c)+bc(b+c)+ab(a+b)\right]}{abc}$$
and that is
$$ \frac{\left[ac(20-b)+bc(20-a)+ab(20-c)\right]}{abc}$$
$$ = \frac{\left[20(ab+bc+ca)-3abc\right]}{abc} $$
But
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=30$$ implies
$$ ab+bc+ca=30abc$$
So you can do further simplification to get $597$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/117004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ In Spivak's Calculus 3rd Edition, there is an exercise to prove the following:
$$x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$$
I can't seem to get the answer. Either I've gone wrong somewhere, I'm overlooking something, or both. Here's my (non) proof:
$$\begin{align*}
x^n - y^n &= (x - y)(x^{n-1} + x^{n-2}y +\cdots+ xy^{n-2} + y^{n-1}) \\
&= x \cdot x^{n-1} + x \cdot x^{n-2} \cdot y + \cdots + x \cdot x \cdot y^{n-2} + x \cdot y^{n-1}\\
&\qquad + (-y) \cdot x^{n-1} + (-y) \cdot x^{n-2} \cdot y + \cdots + (-y) \cdot x \cdot y^{n-2} + (-y) \cdot y^{n-1}\\
&= x^n + x^{n-1} y + \cdots + x^2 y^{n-2} + x y^{n-1} - x^{n-1}y - y^2 x^{n-2} - \cdots- x y^{n-1} - y^n \\
&= x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n \\
&\neq x^n - y^n
\end{align*}$$
Is there something I can do with $x^n + x^2 y^{n-2} - x^{n-2} y^2 - y^n$ that I'm not seeing, or did I make a mistake early on?
EDIT:
I should have pointed out that this exercise is meant to be done using nine of the twelve basic properties of numbers that Spivak outlines in his book:
*
*Associate law for addition
*Existence of an additive identity
*Existence of additive inverses
*Commutative law for additions
*Associative law for multiplication
*Existence of a multiplicative identity
*Existence of multiplicative inverses
*Commutative law for multiplication
*Distibutive law
| I think it would be easier for you to recall
$$\left(1+x+x^2+\cdots+x^{n-1}\right)(x-1) = x^n-1$$
and put $x=\dfrac{b}{a}$
$$\eqalign{
& \left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right)\left( {\frac{b}{a} - 1} \right) = \frac{{{b^n}}}{{{a^n}}} - 1 \cr
& \left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right)\left( {\frac{{b - a}}{a}} \right) = \frac{{{b^n} - {a^n}}}{{{a^n}}} \cr
& {a^{n - 1}}\left( {1 + \frac{b}{a} + \frac{{{b^2}}}{{{a^2}}} + \cdots + \frac{{{b^{n - 1}}}}{{{a^{n - 1}}}}} \right)\left( {b - a} \right) = {b^n} - {a^n} \cr
& \left( {{a^{n - 1}} + b{a^{n - 2}} + {b^2}{a^{n - 3}} + \cdots + {b^{n - 1}}} \right)\left( {b - a} \right) = {b^n} - {a^n} \cr} $$
A little bit "tidier", so that we know what happens in between the dots...
$$\eqalign{
& {x^n} - 1 = \left( {x - 1} \right)\sum\limits_{k = 0}^{n - 1} {{x^k}} \cr
& \frac{{{b^n}}}{{{a^n}}} - 1 = \left( {\frac{b}{a} - 1} \right)\sum\limits_{k = 0}^{n - 1} {\frac{{{b^k}}}{{{a^k}}}} \cr
& \frac{{{b^n} - {a^n}}}{{{a^n}}} = \left( {\frac{{b - a}}{a}} \right)\sum\limits_{k = 0}^{n - 1} {\frac{{{b^k}}}{{{a^k}}}} \cr
& {b^n} - {a^n} = \left( {b - a} \right)\sum\limits_{k = 0}^{n - 1} {{b^k}{a^{n - k - 1}}} \cr} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 9,
"answer_id": 5
} |
Limit of a Sequence involving $\frac {n^2} {\sqrt{n^{6}+k}}$ I am stuck on this problem:
Compute the limit of the sequence $(a_{n})_{n=1}^{\infty}$ defined by
$$a_{n}:=\frac {n^2} {\sqrt{n^{6}+1}}+\frac {n^2} {\sqrt{n^{6}+2}}+\cdot \cdot \cdot + \frac {n^2} {\sqrt{n^{6}+n}}=\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$
So I am trying to find:
$$\lim_{n \to \infty}\sum_{k=1}^{n} \frac {n^2} {\sqrt{n^{6}+k}}$$
In a situation like this should I be noting that the denominator is increasing in value faster than the numerator?
My first thought was to do some manipulation. I may have done something incorrectly. I began with the following.
$$\frac {n^2} {\sqrt{n^6+k}}=\frac {n^2} {\sqrt{n^6(1+k/n^6)}}=\frac {n^2} {\sqrt{n^6}\sqrt{1+k/n^6}}=\frac {1} {n} \cdot \frac {1} {\sqrt {1+k/n^6}}$$
So I now have
$$\lim_{n \to \infty} \frac {1} {n} \sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}$$
Now, I am unsure about the following. It looks to me as if $k/n^6$ goes to zero as $n \to \infty $. That would result in $\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}=n$. So I would be left with
$$\lim_{n \to \infty} \frac {n} {n}=1$$
| The conclusion is correct, but the step $$\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}=n$$ is clearly wrong, a fact that the other answers unfortunately failed to mention: each of the $n$ denominators is greater than $1$, so each term of the sum is less than $1$, and the sum itself is less than $n$.
The largest term in the sum is $$\frac1{\sqrt{1+1/n^6}}\;,$$ and the smallest is $$\frac1{\sqrt{1+k/n^6}}\;,$$ so you do know that
$$\frac{n}{\sqrt{1+k/n^6}}\le\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}\le\frac{n}{\sqrt{1+1/n^6}}$$
and hence that
$$\frac1{\sqrt{1+k/n^6}}\le\frac1n\sum_{k=1}^{n} \frac {1} {\sqrt {1+k/n^6}}\le\frac1{\sqrt{1+1/n^6}}\;.$$
With this in hand you’re in business, since the limit as $n\to\infty$ of each of the bounds is $1$.
(Note that all of the answers work in essentially the same way, by trapping the expression between two simpler ones with the same limit.)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Check my solution to $x^2 + x + 1 > 0$ I spent an hour or so yesterday trying to solve the inequality $x^2 + x + 1 > 0$. Since I'd spent so long on a problem didn't seem like it should be that difficult, I decided I'd call it a day and try it again later.
I just had another look at it and this solution became immediately obvious:
$$x^2 + x + 1 > 0 \ \ \forall \ \ x \in \mathbb{R}$$
I'd justify this by stating that $x^2 > x \ \ \forall \ \ x \in \mathbb{R}$. Because of this, even if $x < 0$, the right hand side of the inequality will always be positive. Am I correct?
| First I will show that with small modifications, your approach can be made to work. (It is, however, simpler to use one of the methods in other answers.) At the end, I suggest an additional simple approach.
It is clear that $x^2+x+1 \ge 1$ if $x \ge 0$. There is potential trouble only when $x<$, so suppose that $x<0$.
Because it is all too easy to make errors when handling negative numbers, let $w=-x$. Then $x^2+x+1=w^2-w+1$, and $w$ is positive.
If $w \ge 1$ (that is, if $x\le -1$), then $w^2 \ge w$, so $w^2 -w+1\ge 0$. (Here we used the method that you proposed.)
That method doesn't work when $0<w<1$, for then $w^2<w$. But $w^2>0$ and $w<1$, so $w^2-w>-1$, and therefore $w^2-w+1>0$.
Another way: Note that $(x-1)(x^2+x+1)=x^3-1$, and that $x-1$ and $x^3-1$ are positive if and only if $x>1$, and are negative if and only if $x<1$.
If $x>1$, then $x-1$ and $x^3-1$ are both positive, so the ratio $\frac{x^3-1}{x-1}$, that is, $x^2+x+1$, is positive.
If $x-1$ and $x^3-1$ are both negative, then again their ratio $x^2+x+1$ is positive.
And finally if $x=1$, then $x^2+x+1$ is positive.
Note that in exactly the same way, we can show that $x^n+x^{n-1}+\cdots +x+1$ is always positive if $n$ is even.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
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For complex $z$, $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$
If $|z|=1$, show that: $$\mathrm{Re}\left(\frac{1 - z}{1 + z}\right) = 0$$
I reasoned that for $z = x + iy$, $\sqrt{x^2 + y^2} = 1\implies x^2 + y ^2 = 1$ and figured the real part would be:
$$\frac{1 - x}{1 + x}$$
I tried a number of manipulations of the equation but couldn't seem to arrive at any point where I could link the two to show that the real part was = 0.
| Yet another approach, is to note that $Re(\overline z_1z_2)=z_1 \cdot z_2$, where $\cdot$ denotes the euclidean dot product in $R^2$.
Let $z=x+iy$ for $x,y$ in $R$, then we have $Re(\frac{1-z}{1+z})=\overline{1-z}{\frac{1}{1+z}}=(1-x,y)\cdot (\frac{1+x}{1+2x+x^2+y^2},\frac{-y}{1+2x+x^2+y^2})=\frac{1-(x^2+y^2)}{2+2x}=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Sums of Double series where $a_{m,n}=\frac {m-n} {2^{m+n}}\frac {\left( m+n-1\right) !} {m!n!}$ I am trying to show if $$a_{m,n}=\dfrac {m-n} {2^{m+n}}\dfrac {\left( m+n-1\right) !} {m!n!}$$ such that $$\left( m,n>0\right) $$ that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right) =-1$, $\sum _{n=0}^{\infty }\left(\sum _{m=0}^{\infty } a_{m,n}\right) =1$.
Now we observe that $a_{m,0}=2^{-m}$,$a_{0,n}=-2^{-n}$ and $a_{0,0}=0$.
We can rewrite the first series as
that $\sum _{m=0}^{\infty }\left( \sum _{n=0}^{\infty }a_{m,n}\right)$.
$\Rightarrow \sum _{m=0}^{\infty }\left(\dfrac {1} {2^{m}}+\dfrac {\left( m-1\right) } {2^{m+1}}+\dfrac {\left( m-2\right) \left( m+1\right) } {2^{m+2}2!}+\dfrac {\left( m-3\right) \left( m+1\right) \left( m+2\right) } {2^{m+3}3!}+\ldots\right)$
$\Rightarrow \sum _{m=0}^{\infty }\dfrac {1} {2^{m}}\left(1+\dfrac {\left( m-1\right) } {2^{1}}+\dfrac {\left( m-2\right) \left( m+1\right) } {2^{2}2!}+\dfrac {\left( m-3\right) \left( m+1\right) \left( m+2\right) } {2^{3}3!}+\ldots\right)$
I can n't recognize the expression inside the brackets, any help of how to proceed forward would be much appreciated.
| It's not clear how to define $a_{0,0}$, since $(-1)!$ is undefined. But if you take $a_{0,0}$ to be $0$ you should have $\sum_{n=1}^\infty a_{0,n} = -1$ while $\sum_{n=0}^\infty a_{m,n} = 0$ otherwise. Note that $$a_{m,n} = 2^{-m-n} \left({m+n-1 \choose n} - {m+n-1 \choose m} \right)$$
and
$$ { m+n-1\choose n} = (-1)^n {-m \choose n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/119063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how many zeroes does 2012! have at the end?
Possible Duplicate:
How come the number $N!$ can terminate in exactly $1,2,3,4,$ or $6$ zeroes but never $5$ zeroes?
How many zeroes does $2012!$ end with?
My idea is:
402 zeroes come from $2\times 5$, 80 from $2\times 25$, 16 from $2\times 125$ and 3 from $2\times 625$
How can we "show" that this is true?
| The correct answer is 501.
In order to find the number of zeros is same as finding the number of factors of powers of $5$. There are more factors of powers of $2$ than the factors of powers of $5$.
For instance $10! = 3628800 = \hspace{3pt}2^8 \hspace{3pt} 3^4 \hspace{3pt}5^2\hspace{3pt} 7$
$$\left \lfloor \frac{n}{p} \right \rfloor +\left \lfloor \frac{n}{p^2} \right \rfloor +\left \lfloor \frac{n}{p^3}\right \rfloor + \cdots \left \lfloor \frac{n}{p^{k-1}} \right \rfloor$$
where $\left \lfloor \frac{n}{p^k} \right \rfloor=0$.
In this case $k=5$ because $\left \lfloor \frac{2012}{5^5} \right \rfloor=0$
$$\left \lfloor \frac{2012}{5} \right \rfloor =402, \hspace{6pt} \left \lfloor \frac{2012}{5^2} \right \rfloor = \left \lfloor \frac{402}{5} \right \rfloor =80$$
$$\left \lfloor \frac{2012}{5^3} \right \rfloor = \left \lfloor \frac{80}{5} \right \rfloor =16, \hspace{6pt} \left \lfloor \frac{2012}{5^4} \right \rfloor = \left \lfloor \frac{16}{5} \right \rfloor =3$$
$$\left \lfloor \frac{2012}{5} \right \rfloor+ \left \lfloor \frac{2012}{5^2}\right \rfloor +\left \lfloor \frac{2012}{5^3}\right \rfloor +\left \lfloor \frac{2012}{5^4} \right \rfloor = 402+80+16+3=501$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Please help me prove by induction that $n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)$ Please help me prove by induction that
$$
n^n>1\cdot3\cdot5\cdot\ldots\cdot(2n-1)
$$
| This is proof without induction.
$$
1\cdot 3\cdot 5\cdot\ldots\cdot(2n-1)=
\frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{2\cdot 4\cdot 6\cdot\ldots\cdot 2n}=
\frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{(2\cdot 1)\cdot (2\cdot 2)\cdot (2\cdot 3)\cdot\ldots\cdot (2\cdot n)}=
$$
$$
\frac{1\cdot 2\cdot 3\cdot\ldots\cdot 2n}{2^n\cdot 1\cdot 2\cdot 3\cdot\ldots\cdot n}=
\frac{(n+1)\cdot (n+2)\cdot\ldots\cdot 2n}{2^n}
$$
So we had to prove that
$$
n^n>\frac{(n+1)\cdot (n+2)\cdot\ldots\cdot 2n}{2^n}
$$
But the last inequality is obvious, since
$$
\frac{(n+1)\cdot (n+2)\cdot\ldots\cdot 2n}{2^n}=
\frac{n+1}{2}\frac{n+2}{2}\cdots\frac{n+n}{2}<n\cdot n\cdot\ldots \cdot n=n^n
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find curve equation from data? How do I find the formula when I only know some data points ?
Usually I just use the Trendline option for diagrams in Excel, but this one eludes me.
I expect it to be something like : Ax^2 + or - Bx + or - C.
Sample data:
X Y
1 4
2 8
3 13
4 18
5 24
6 30
7 37
8 44
9 51
10 60
11 68
12 78
13 88
14 99
15 110
16 122
17 136
18 150
19 166
20 180
21 197
22 216
23 235
24 255
25 277
26 300
27 325
28 351
29 378
30 408
| From plotting you data (Wolfram|Alpha link), it does not look linear. So it better be fit by a polynomial. I assume you want to fit the data:
X Y
1 4
2 8
3 13
4 18
5 24
..
using a quadratic polynomial $y = ax^2 + bx + c.$ If so, then put your data in a matrix form (note that $x^0, x^1, x^2, y$ below are not actually in the matrix. They're just comments for your understanding):
$$
\begin{pmatrix}
\color{red}{x^0} & {\color{red} x} & \color{red}{x^2} \\
1 & 1 & 1 \\
1 & 2 & 4 \\
1 & 3 & 9 \\
& \ldots &
\end{pmatrix}
\begin{pmatrix}
c \\ b \\ a
\end{pmatrix} =
\begin{pmatrix}
{\color{red} y} \\ 4 \\ 8 \\ 13 \\ \ldots
\end{pmatrix} \tag{1}
$$
Now, equation $(1)$ is really of the following form:
$$ Xv = y \tag{2}$$
where each row encodes $cx^0 + bx+ax^2 = y$ for a particular pair of $(x,y)$ values. And we're looking for a solution vector $v^{T} = \begin{pmatrix}
c & b & a
\end{pmatrix}$ that gives the coefficients of that best fitting polynomial $ax^2 + bx+c$.
To solve for $v$, multiply $(2)$ both sides by $X^{T},$ we have
$X^{T}Xv= X^{T}y,$ or
$$ v = (X^{T}X)^{-1} X^{T}y.$$
This is a called linear least squares method because best means minimize the squared error. Several software packages can handle that for you. Luckily, Wolfram|Alpha can do.
To repeat for a polynomial of degree, say 4, construct
$$
\begin{pmatrix}
\color{red}{x^0} & {\color{red} x} & \color{red}{x^2} & \color{red}{x^3} & \color{red}{x^4} \\
1 & 1 & 1 & 1 & 1\\
1 & 2 & 4 & 8 & 16 \\
1 & 3 & 9 & 27 & 81 \\
& \ldots &
\end{pmatrix}
\begin{pmatrix}
e \\ d\\ c \\ b \\ a
\end{pmatrix} =
\begin{pmatrix}
{\color{red} y} \\ 4 \\ 8 \\ 13 \\ \ldots
\end{pmatrix} \tag{1}
$$
and solve for $v$, this should give you the parameters $a,b,c,d,e$ s.t. $$ax^4+bx^3+cx^2+dx+e$$ best fits your data.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 1,
"answer_id": 0
} |
$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$ is a $C^1$-function I would like to show that the function:
$$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$$
is a $C^1$-function.
$$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$$
$$ \frac{\partial f}{\partial y}(x,y)=-\frac{\partial f}{\partial y}(y,x)=... $$
So I just have to show that:
$$ \frac{\partial f}{\partial x}(x,y)\rightarrow_{(0,0)}0$$
When $y\geq0$ :
$$ -\frac{y^3}{6(x^2+y^2)}+\frac{x^2y}{x^2+y^2}-\frac{x^4y}{4!(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}$$
When $y<0$ :
$$ -\frac{y^3}{6(x^2+y^2)}+\frac{y^5}{5!(x^2+y^2)}+\frac{x^2y}{2(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}-\frac{yx^4}{4!(x^2+y^2)}$$
So $$ \frac{\sin(y)-y\cos(x)}{x^2+y^2}\rightarrow_{(0,0)}0 $$
How can I directly find an upper bound of $$ \left| \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2} \right|$$ that tends to 0 ?
| Because of the Taylor expansion $\sin(x) = x - {x^3 /6} + ...$, if $x$ and $y$ are small enough you can write $\sin(x) = x + E(x)$ and $\sin(y) = y + E(y)$, where $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$. So you have
$${2x(y\sin(x) - x\sin(y)) \over (x^2 + y^2)^2} = {2x(yx + yE(x) - xy - xE(y)) \over (x^2 + y^2)^2}$$
$$= {2x(yE(x) - xE(y)) \over (x^2 + y^2)^2}$$
Taking absolute values and bounding, this is at most
$$= 2|x|{|yE(x)| + |xE(y)| \over (x^2 + y^2)^2}$$
Inserting $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$ this is bounded by
$$= 2|x|{|x^3y| + |xy^3| \over (x^2 + y^2)^2}$$
$$= 2|x|{|xy|(x^2 + y^2) \over (x^2 + y^2)^2}$$
$$= 2|x| {|xy| \over x^2 + y^2}$$
I think you can take it from there...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Let $n$ such that $\displaystyle{2^{n-2005}} | n!$ Let $n$ such that $\displaystyle{2^{n-2005}} | n!$
Prove that this number has at most
$2005$ non-zero digits when written in base $2$.
| there are $\sum_{k\ge 1} \lfloor \frac{n}{2^k}\rfloor$ 2s in $n!$. So we must have $n - 2005 \le \sum_{k\ge 1} \lfloor \frac{n}{2^k}\rfloor$. Now write $n = \sum_{i} 2^{\ell_i}$ in base $2$ with $\ell_1 < \ldots < \ell_m$. We have
\begin{align*}
\left \lfloor \frac{n}{2^k}\right\rfloor &=
\sum_{i:\ell_i \ge k} 2^{\ell_i - k}\\
\sum_{k\ge 1} \left \lfloor \frac{n}{2^k}\right\rfloor
&= \sum_{k\ge 1} \sum_{i:\ell_i \ge k} 2^{\ell_i - k}\\
&= \sum_{i=1}^m \sum_{k = 1}^{\ell_i} 2^{\ell_i - k}\\
&= \sum_{i=1}^m 2^{\ell_i} \sum_{k=1}^{\ell_i}2^{-k}\\
&= \sum_{i=1}^m 2^{\ell_i} \left(\frac{1 - 2^{-1-\ell_i}}{1-2^{-1}} - 1\right)\\
&= \sum_{i=1}^m 2^{\ell_i} \left(1 - 2^{-\ell_i}\right)\\
&= \sum_{i=1}^m 2^{\ell_i} - m\\
&= n - m
\end{align*}
So $n - 2005 \le n - m$, i. e. $m \le 2005$ as to be proved.
AB,
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/123597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$ Let $a,b$ be positive real numbers. Prove $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}} \geq \frac{2}{\sqrt{1+ab}}$
if either
$(1) 0 \leq a,b \leq 1$
OR
$(2) ab \geq 3$
Since this question was under Trigonometry, I assumed the following.
Since $a,b$ are positive real numbers with $0 \leq a,b \leq 1$, I can assume that for some $x,y, a=\tan(x), b=\tan(y)$ and therefore it is to be shown that
$$\frac{1}{\sec x} + \frac{1}{\sec y} = \cos x+ \cos y \geq \frac{2\cos x \cos y}{\sqrt{cos(x-y)}}$$
(Originally posted without that $2$ on the right - Sorry!)
I do know that
$$\cos x + \cos y \geq 2 \sqrt{\cos x \cos y}$$
Now how to proceed? Just give me hints !
| Nice approach, you are very very close but this is how you finish it off: let $u^2=\cos x$ and $t^2=\cos y$ . Substituting, you have $u^2+t^2 \ge 2ut$, or $u^2-2ut+t^2 \ge 0$ , and this can be factored into $(u-t)^2 \ge 0$ , which is always true of course.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Plot $|z - i| + |z + i| = 16$ on the complex plane
Plot $|z - i| + |z + i| = 16$ on the complex plane
Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.
I get to the point:
$x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$
$2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$
It seems like I'm and doing something the hard way.
| Don't we all love algebraic solutions...
The ellipse has an equation of the form
$$(\frac {x} {a})^2 + (\frac {y} {b})^2 = 1$$
The focals are aligned on the y axis, therefore
*
*a is the side of a rectangle triangle, the other site being 1 and the hypothenuse 16/2:
$$a = \sqrt{8^2-1}$$
*
*b is 16/2.
The equation of the ellipse is
$$(\frac {x} {\sqrt{63}})^2 + (\frac {y} {8})^2 = 1$$
Or, to write it as above
$$\frac {x^2} {63} + \frac {y^2} {64} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/126518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 2
} |
Solving $\lim\limits_{(x,y)\to(0,0)}\;\frac{x^5 + \,y^5}{x^3+\,y^3}$ How do I solve the limit $$\lim_{(x,y)\to(0,0)}\;\frac{x^5+y^5}{x^3+y^3}\quad ?$$ I have tried using polar coordinates, but I don't think an answer would be valid because theta is not fixed. What else can I do?
| Hint
$$ \frac{x^5+y^5}{x^3+y^3} =\frac{x^5+x^2y^3}{x^3+y^3}+ \frac{x^3y^2+y^5}{x^3+y^3} - \frac{x^2y^2(x+y)}{x^3+y^3} $$
And
$$x^2-xy+y^2 \geq |xy| \,.$$
OK, to make it more clear. If you combine the two hints, you get:
$$\left|\frac{x^5+y^5}{x^3+y^3} \right| \leq x^2+y^2+|xy| \,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
how to find x in the following formula For a software project, I need to calculate some things. on of the formulas looks like:
26280 = 2 * SQRT((149,598,000,000 - x) / 1.4) + x / 10,217,588,910,387,300,000
My colegue says you can't solve the above equation (you won't be able to find x) I quite convinced you should be able to find x, the problem is, I don't know how :(
I came as far as rewriting the above into:
((26280 - x) / 20,435,177,820,774,600,000)^2 = (149,598,000,000 - x) / 1.4
but now I'm stuck.
Could anyone explain to me how to move on, in order to find x?
| First, it's useful to write the expression in the standard quadratic form. For ease of reading/writing, I'm going to set $A = 20,435,177,820,744,600,00$ and $B = 149,598,000,000$.
$$
\begin{align*}
\left(\frac{26280 - x}{A}\right)^2 &= \frac{B - x}{1.4}\\
\frac{(26280 - x)^2}{A^2} &= \frac{B}{1.4} - \frac{x}{1.4}\\
\frac{26280^2 - 52560x - x^2}{A^2} &= \frac{5}{7}B - \frac{5}{7}x\\
\frac{26280^2}{A^2} - \frac{52560x}{A^2} - \frac{x^2}{A^2} &= \frac{5}{7}B - \frac{5}{7}x\\
0 &= \frac{x^2}{A^2} + \left(\frac{52560}{A^2} - \frac{5}{7}\right)x + \frac{5}{7}B - \frac{26280^2}{A^2}.
\end{align*}
$$
You can finish this up with the quadratic formula.
EDIT: Alternatively, there is WolframAlpha.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Differentiate respect to $x$ $(x^2+2x+1)^3$
let u=$x^2+2x+1$ $\frac{du}{dx} = 2x$ or $2x+2$ $\frac{dy}{dx}=3u^2 $
if $\frac{du}{dx} = 2x$ then
$3(x^2+2x+1)^2 (2x)$
answer is $6x(x^2+2x+1) $
Or if $\frac{du}{dx} = 2x+2$ then
$3(x^2+2x+1)^2 (2x+2)$
$6x(x^2+2x+1)+2$
however the right answer is $6(x+1)^5$
can please help me out?
thanks in advance!
| Hint
$$\begin{align}\dfrac{du}{dx}&=2x+2=2(x+1)\\x^2+2x+1 &=(x+1)^2\\(x^a)^b&=x^{ab}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/128624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Finding $\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$ Finding $$\int \frac{1}{x}\sqrt{1+\frac{1}{x^4}}\,dx$$
I suppose I need integration by parts and trigo substitution
Let $u=\frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2}$, $\frac{1}{x^2} = \tan{\theta}$. Is my substitution OK?
So $x=\frac{1}{\sqrt{\tan{\theta}}} \Rightarrow dx = -\frac{\sec^2{\theta}}{2\sqrt{\tan{\theta}}}\,d\theta$. But this will be very complicated to integrate later?
Am I supposed to be trying something else?
UPDATE: An attempt
$$\int \frac{1}{x} \sqrt{1+\frac{1}{x^4}} dx$$
Let $u = \frac{1}{x} \Rightarrow du = -\frac{1}{x^2} dx$
Let $dv = \sqrt{1+(\frac{1}{x^2})^2} dx$
Let $\alpha = \frac{1}{x^2} \Rightarrow d\alpha = -\frac{1}{2x^3}$
$dv=\sqrt{1+\alpha^2} d\alpha$
Let $\alpha = \tan{\theta} \Rightarrow d\alpha = \sec^2{\theta} d\theta$
$dv = \sqrt{1+\tan^2{\theta}} \sec^2{\theta} d\theta = \sec^3{\theta}$. Looks wrong here ?
| $$I = \int \frac{\sqrt{x^4+1}}{x^3} \mathrm{d}x$$
Substitute $u=\sqrt{x^4+1}, \mathrm{d}u=\frac{2x^3}{\sqrt{x^4+1}}\mathrm{d}x$
$$\int \frac{u^2}{2(u^2-1)^{\frac{3}{2}}} \mathrm{d}u$$
Now substitute $u=\sec \theta$ $\mathrm{du} = \sec \theta \tan \theta \mathrm{d}\theta$
Then $(u^2-1)^{\frac{3}{2}} = \tan^3 \theta$
$$I = \frac{1}{2}\int \csc^2 \theta \sec \theta \mathrm{d}\theta = \frac{1}{2}\int (\cot^2 \theta+1) \sec \theta \mathrm{d}\theta$$
And take it from there
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ ?. How do we prove $\cos(\pi/5) - \cos(2\pi/5) = 0.5$ without using a calculator.
Related question: how do we prove that $\cos(\pi/5)\cos(2\pi/5) = 0.25$, also without using a calculator
| The imaginary parts of $\left(\cos\left(\frac{k\pi}5\right)+i\sin\left(\frac{k\pi}5\right)\right)^5=(-1)^k$ give
$$
\sin^5\left(\frac{k\pi}5\right)-10\sin^3\left(\frac{k\pi}5\right)\cos^2\left(\frac{k\pi}5\right)+5\sin\left(\frac{k\pi}5\right)\cos^4\left(\frac{k\pi}5\right)=0
$$
Dividing by $\sin\left(\frac{k\pi}5\right)$ and substituting $\sin^2\left(\frac{k\pi}5\right)=1-\cos^2\left(\frac{k\pi}5\right)$, we get
$$
16\cos^4\left(\frac{k\pi}5\right)-12\cos^2\left(\frac{k\pi}5\right)+1=0
$$
which, using the quadratic formula, has the solutions
$$
\pm\frac{1\pm\sqrt5}4
$$
Thus, since $\cos(x)$ is decreasing from $1$ to $-1$ for $x\in\left[0,\pi\right]$,
$$
\begin{align}
\cos\left(\frac\pi5\right)&=\frac{1+\sqrt5}4\\
\cos\left(\frac{2\pi}5\right)&=\frac{-1+\sqrt5}4\\
\cos\left(\frac{3\pi}5\right)&=\frac{1-\sqrt5}4\\
\cos\left(\frac{4\pi}5\right)&=\frac{-1-\sqrt5}4\\
\end{align}
$$
From which we get
$$
\cos\left(\frac\pi5\right)-\cos\left(\frac{2\pi}5\right)=\frac12
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/130817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 18,
"answer_id": 15
} |
Proving $\sqrt 3$ is irrational. There is a very simple proof by means of divisibility that $\sqrt 2$ is irrational. I have to prove that $\sqrt 3$ is irrational too, as a homework. I have done it as follows, ad absurdum:
Suppose
$$\sqrt 3= \frac p q$$
with $p/q$ irreducible, then
$$\begin{align}
& 3q^2=p^2 \\
& 2q^2=p^2-q^2 \\
&2q^2=(p+q)(p-q) \\
\end{align}$$
Now I exploit the fact that $p$ and $q$ can't be both even, so it is the case that they are either both odd, or have different parity. Suppose then that $p=2n+1$ and $q=2m+1$
Then it is the case that
$$\begin{align}
&p-q=2(n-m) \\
&p+q=2(m+n+1) \\
\end{align}$$
Which means that
$$\begin{align}
&2q^2=4(m-n)(m+n+1) \\
&q^2=2(m-n)(m+n+1) \\
\end{align}$$
Then $q^2$ is even, and so is then $q$, which is absurd. Similarly, suppose
$q=2n$ and $p=2m+1$.
Then $p+q=2(m+n)+1$ and $p-q=2(m-n)+1$. So it is the case that
$$\begin{align}
&2q^2=(2(m-n)+1)(2(m+n)+1)\\
&2q^2=4(m^2+m-n^2)+1 \\
\end{align}$$
So $2q^2$ is odd, which is then absurd.
Is this valid?
| Here are some proofs I've found (link at bottom):
If $\sqrt 3 = m/n$:
$$ \frac{m}{n} = \sqrt 3 \frac{\sqrt 3 - 1}{\sqrt 3 - 1} = \frac{3-\sqrt 3}{\sqrt 3 - 1}
= \frac{3-m/n}{m/n-1} = \frac{3 n - m}{m-n}$$
and the right side has a smaller denominator, since $m < 2n$ (i.e., $\sqrt 3 < 2$).
$x = \sqrt{3} - 1$ is a root of the equation $x^2 + 2x - 2 = 0$, thus:
$$x(3+x) = 2+x$$
$$x = \frac{2+x}{3+x} = \cfrac{1}{1 + \cfrac{1}{2 + x}}$$
And thus
$$x = [1,2,1,2,\dots]$$
So $\sqrt{3}$ os irrational.
These proofs and others: How to prove that $\sqrt 3$ is an irrational number?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 10,
"answer_id": 7
} |
How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.
| $n^5 -n = n^5 + 6n -n = n(n^(4) - 1) + 6n
=n(n+1)(n^2 + 1)(n-1) + 6n
6(n^2 +1) m + 6n
6[{n^2+1+n}]$
by fermat 's theorem** $n^5$ is congruent to $n\mod 5$
$5|n^5 -n$
hence,
$30|n^5-n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 21,
"answer_id": 20
} |
How do I solve this equation? $\displaystyle \large \cos 2x + 1 - \sin 2x=\frac{2 \cos 2x \cos x}{\cos x + \sin x}$
I've been trying for a long time but I can't get it.
| Using the identities
$$
{\cos 2x=\cos^2 x -\sin^2 x},\quad \cos 2x=2\cos^2 x-1,\quad \sin 2x=2\sin x\cos x
$$
We have, if $\cos x+\sin x\ne0$
$$\eqalign{
{2 {\cos 2x} \cos x\over \cos x+\sin x}
&={2( {\cos^2 x-\sin^2 x}) \cos x\over \cos x+\sin x}\cr
&={2(\cos x+\sin x)(\cos x-\sin x) \cos x\over \cos x+\sin x}\cr
&={2 (\cos x-\sin x) \cos x }\cr
&=2\cos^2 x -2\sin x\cos x\cr
&=\cos 2x +1 -\sin 2x.
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/133217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate the improper integral $\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx$ How to Prove: $$\int_{0}^{\infty} \log\biggl(x+\frac{1}{x}\biggr) \cdot \frac{1}{1+x^{2}} \ dx = \pi \: \log{2}$$
| Here I give another way to solve the problem. We rewrite the integral as
\begin{eqnarray}
I&=&\int_0^\infty\log(x+\frac{1}{x})\frac{1}{1+x^2}dx\\
&=&\int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx-\int_0^\infty\frac{\log x}{1+x^2}dx
\end{eqnarray}
Note
$$ \int_0^\infty\frac{\log(1+x^2)}{1+x^2}dx=\pi\log 2$$
from Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$
and
$$ \int_0^\infty\frac{\log x}{1+x^2}dx=0. $$
Thus we have
$$ I=\pi \log 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/134459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Solving the recursion $3a_{n+1}=2(n+1)a_n+5(n+1)!$ via generating functions I have been trying to solve the recurrence:
\begin{align*}
a_{n+1}=\frac{2(n+1)a_n+5((n+1)!)}{3},
\end{align*}
where $a_0=5$, via generating functions with little success. My progress until now is this:
Let $A(x)=\sum_{n=0} ^{\infty} a_nx^n$. By multiplying both sides of our recurrence relation by $x^n$ and summing over $n$ from $0$ to $\infty$, we see that
\begin{align}
\sum_{n=0} ^{\infty} a_{n+1} x^n = \frac{2}{3}\sum_{n=0} ^{\infty} (n+1)a_nx^n + \sum_{n=0} ^{\infty} (n+1)!x^n.
\end{align}
Using our definition of $A(x)$ we can rewrite the left hand side as
\begin{align*}
\sum_{n=0} ^{\infty} a_{n+1} x^n=\frac{A(x)-a_0}{x}.
\end{align*}
Such manipulations of the right hand side have been difficult because of the coefficients of the power series.
Is there anyway to proceed from here, or are generating functions not suited to solve such a recurrence?
| A first order linear non-homogeneous recurrence:
$$
a_{n + 1} - c_n a_n = f_n
$$
can be reduced to a telescoping sum by dividing by the summing factor $s_n = \prod_{0 \le k \le n} c_n$:
$$
\begin{align*}
\frac{a_{n + 1}}{s_n} - \frac{a_n}{s_{n - 1}}
&= \frac{f_n}{s_n} \\
\sum_{0 \le n \le m - 1} \frac{a_{n + 1}}{s_n} - \frac{a_n}{s_{n - 1}}
&= \sum_{0 \le n \le m - 1} \frac{f_n}{s_n} \\
\frac{a_m}{s_{m - 1}} - \frac{a_0}{1}
&= \sum_{0 \le n \le m - 1} \frac{f_n}{s_n}
\end{align*}
$$
It is easier to go through this dance each time. Here the summing factor is:
$$
\prod_{0 \le k \le n} \frac{2}{3}(n + 1) = \left( \frac {2}{3} \right)^{n + 1} (n + 1)!
$$
Dividing through by this gives:
$$
\begin{align*}
\frac{a_{n + 1}}{(2 / 3)^{n + 1} (n + 1)!} - \frac{a_n}{(2/3)^n n!}
&= \frac{5}{3 (2 / 3)^{n + 1}} \\
\frac{a_n}{(2/3)^n n!} - \frac{a_0}{1}
&= \frac{5}{3} \sum_{0 \le k \le n - 1} (3/2)^{k + 1} \\
\frac{a_n}{(2/3)^n n!}
&= 5 + \frac{5}{3} \cdot \frac{3}{2} \cdot \frac{(3/2)^n - 1}{3/2 - 1} \\
&= 5 + 5 \left( (3/2)^n - 1 \right) \\
a_n
&= 5 n!
\end{align*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/135803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 0
} |
Finding $\int e^{2x} \sin{4x} \, dx$ Finding $$\int e^{2x} \sin 4x \, dx$$
I think I should be doing integration by parts...
If I let $u=e^{2x} \Rightarrow du = 2e^{2x}$,
$dv = \sin{4x} \Rightarrow v = -\frac{1}{4} \cos{4x}$
$\int{ e^{2x} \sin{4x}} dx = e^{2x}(-\frac{1}{4}\cos 4x) - \color{red}{\int (-\frac{1}{4} \cos 4x) 2e^{2x} \, dx}$
Highlighted in red, I seem to integrating an exponential times trig expression again ... doesn't seem like te right way to go
So I let $u=\sin{4x} \Rightarrow du = 4\cos 4x \, dx$
$dv = e^{2x} dx \Rightarrow v = \frac{1}{2} e^{2x}$
$\sin{4x}(\frac{1}{2} e^{2x}) - \int (\frac{1}{2} e^{2x})(4\cos4x) \, dx$
Again, its an exponential times a trig function?
Am I using the wrong substitution?
| To find $\int e^{2x}\sin4x\,dx,$ we can let $u=e^{2x}$ and $dv=\sin4x\,dx,$ which implies $du=2e^{2x}\,dx$ and $v=-\frac{1}{4}\cos4x.$ Therefore, $$\begin{align}\int e^{2x}\sin4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(-\frac{1}{4}\cos4x\right)-\int-\frac{1}{4}\cos4x\left(2e^{2x}\right)\,dx\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\int e^{2x}\cos4x\,dx.\end{align}$$ Apply another integration for $\int e^{2x}\cos4x\,dx,$ this time letting $u=e^{2x}$ and $dv=\cos4x\,dx,$ so that $du=2e^{2x}\,dx$ and $v=\frac{1}{4}\sin4x.$ Hence, $$\begin{align}\int e^{2x}\cos4x\,dx &= uv-\int v\,du\\ &= e^{2x}\left(\frac{1}{4}\sin4x\right)-\int\frac{1}{4}\sin4x\left(2e^{2x}\right)\,dx\\ &= \frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx.\end{align}$$
Combining both formulas, we get $$\begin{align}\int e^{2x}\sin4x\,dx &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{2}\left(\frac{1}{4}e^{2x}\sin4x-\frac{1}{2}\int e^{2x}\sin4x\,dx\right)\\ &= -\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x-\frac{1}{4}\int e^{2x}\sin4x\,dx.\end{align}$$ So, $$\frac{5}{4}\int e^{2x}\sin4x\,dx=-\frac{1}{4}e^{2x}\cos4x+\frac{1}{8}e^{2x}\sin4x.$$ Therefore, multiplying by $\frac{4}{5}$ and adding in the constant of integration, we see that $$\int e^{2x}\sin4x\,dx=-\frac{1}{5}e^{2x}\cos4x+\frac{1}{10}e^{2x}\sin4x+C,$$ which we can rewrite as $$\int e^{2x}\sin4x\,dx=\frac{1}{10}e^{2x}\left(-2\cos4x+\sin4x\right)+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/136595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Showing $\int_{0}^{\infty} \frac{1}{(x^2+1)^2(x^2+4)}=\frac{\pi}{18}$ via contour integration I want to show that:
$$\int_{0}^{\infty} \frac{1}{(x^2+1)^2(x^2+4)}=\frac{\pi}{18}$$
so considering:
$$\int_{\gamma} \frac{1}{(z^2+1)^2(z^2+4)}$$ where gamma is the curve going from $0$ to $-R$ along the real axis, from $-R$ to R via a semi-circle in the upper plane and then from $R$ to 0 along the real axis.
Using the residue theorem we have that:
$$\int_{\gamma} \frac{1}{(z^2+1)^2(z^2+4)}=2\pi i \sum Res$$
so re-writing the integrand as $\displaystyle\frac{1}{(z-2i)(z+2i)(z+i)^2(z-i)^2}$
we can see that there is two simple poles at $2i$,$-2i$ and two poles of order 2 at $i$,$-i$.
Calculating the residues:
$$Res_{z=2i}=\lim_{z\rightarrow 2i} \displaystyle\frac{1}{(z+2i)(z+i)^2(z-i)^2}=\frac{1}{36i}$$
$$Res_{z=-2i}=\lim_{z\rightarrow 2i} \displaystyle\frac{1}{(z-2i)(z+i)^2(z-i)^2}=\frac{-1}{36i}$$
$$Res_{z=i}\lim_{z\rightarrow i} \frac{d}{dz} \frac{1}{(z-2i)(z+2i)(z+i)^2}=\frac{2i}{36}+\frac{2}{24i}$$
$$Res_{z=-i}\lim_{z\rightarrow -i} \frac{d}{dz} \frac{1}{(z-2i)(z+2i)(z-i)^2}=\frac{-2i}{36}+\frac{-2}{24i}$$
But now the sum of the residues is 0 and so when I integrate over my curve letting R go to $\infty$ (and the integral over top semi-circle goes to 0) I will just get 0?
Not sure what I've done wrong?
Thanks very much for any help
| Consider the contour $C$ that spans along $-R$ to $R$ and around the arc $Re^{i\theta}$ for $0\le\theta\le \pi$.
Letting
$$f(z):=\frac{1}{(z^2+1)^2(z^2+4)}=\frac{1}{(z+i)^2(z-i)^2(z+2i)(z-2i)}$$
and we see the poles are located at $\pm i$ and $\pm 2i$. Letting $R \to \infty$, it is very clear that the denominator explodes, causing the integral around the arc to disappear. Then
$$\oint_C f(z)\, dz = 2\pi i(\operatorname*{Res}_{z = i}f(z) + \operatorname*{Res}_{z = 2i}f(z))$$
because $2i$ and $i$ are the only poles in $C$.
The pole of $i$ is of order 2:
$$
\operatorname*{Res}_{z = i}f(z) =
\lim_{z \to i} \frac{1}{1!}\frac{d}{dz} (z-i)^2 f(z)=
\lim_{z \to i} \frac{d}{dz}\frac{1}{(z+i)^2(z^2+4)}=
\lim_{z \to i} \frac{2(2z^2 +iz+4)}{(i+z)^3(4+z^2)^2}=-\frac{i}{36}
$$
The pole of $2i$ is simple:
$$
\operatorname*{Res}_{z = 2i}f(z) =
\lim_{z \to 2i} (z-2i)f(z) = \frac{1}{(-4+1)^2(2i+2i)}=-\frac{i}{36}
$$
So finally
$$
\int_0^\infty f(x)\, dx = \frac{1}{2}\int_{-\infty}^\infty f(x)\, dx = \pi i\left(-\frac{i}{36}-\frac{i}{36}\right) = \frac{\pi}{18}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/137167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How do I show that the sum $(a+\frac12)^n+(b+\frac12)^n$ is an integer for only finitely many $n$?
Show that if $a$ and $b$ are positive integers, then $$\left(a +\frac12\right)^n + \left(b+\frac{1}{2}\right)^n$$is an integer for only finitely many positive integers $n$.
I tried hard but nothing seems to work. :(
| First of all, rewrite the equation as others have, to yield $(2a+1)^n+(2b+1)^n = m\cdot 2^n$, or in other words $(2a+1)^n+(2b+1)^n \equiv 0\pmod {2^n}$. Now, if $n=2k$ is even then for the equation to hold $\bmod 2^{2k}$ it must certainly hold $\bmod 4$; i.e., $\bigl((2a+1)^k\bigr)^2 + \bigl((2b+1)^k\bigr)^2\equiv 0\pmod 4$. But this can't work, because both of the squares on the left must be congruent to $1 \bmod 4$ and so their sum is congruent to $2$. Therefore, $n$ must be odd, say $n=2k+1$.
Now, the left side can be factored using the classic formula for $\frac{x^n-y^n}{x-y}$ (substitute $x=2a+1, y=-(2b+1)$), yielding
$$(2a+1)^{2k+1}+(2b+1)^{2k+1} = (2a+2b+2)\cdot(x^{2k}+x^{2k-1}y+x^{2k-2}y^2+\cdots+y^{2k})$$
But the factor on the right is odd (it's the sum of $2k+1$ terms each of which is odd), so for the LHS to be $0\bmod 2^n$, we must have $2a+2b+2\equiv 0 \pmod {2^n}$, and this can only be true for finitely many $n$; it becomes impossible as soon as $2^n\gt 2a+2b+2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
Absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ I need to show absolute max for $f(x,y,z)=x^ay^bz^c$, with constraint $g(x,y,z)=x+y+z-1$ is $$\frac{a^ab^bc^c}{(a+b+c)^{a+b+c}}$$
So I I do have $$ax^{a-1}y^bz^c = bx^ay^{b-1}z^c = cx^ay^bz^{c-1} =\lambda$$
then I went on to equating each of the 2 equations giving
$$y=\frac{b}{a}x, z=\frac{c}{a}x$$
So I have $x+\frac{b}{a}x+\frac{c}{a}x=1$ but I am not so sure how to continue
The answer instead had
$$\lambda x = ax^ay^bz^c, \lambda y = bx^ay^bz^c, \lambda z = cx^ay^bz^c$$, then making observation that $x:y:z=a:b:c$, which means
$$x=\frac{a}{a+b+c}, y=\frac{b}{a+b+c}, z=\frac{c}{a+b+c}$$
I don't quite get this ... how do I derive this?
The rest ...
$$(\frac{a}{a+b+c})^a(\frac{b}{a+b+c})^b(\frac{c}{a+b+c})^c=1$$
...
| You were doing fine, the argument from the textbook that you are quoting is more symmetrical, that's all.
You obtained $x+\frac{b}{a}x+\frac{c}{a}x=1$. That almost finishes things! Multiply through by $a$. You get $(a+b+c)x=a$ and therefore
$$x=\frac{a}{a+b+c}.$$
From your $y=\frac{b}{a}x$ you can then get $y=\dfrac{b}{a+b+c}$, and similarly $z=\dfrac{c}{a+b+c}$. That gets you to exactly the same place as the solution you quoted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\int_0^1\left(\sum\limits_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx$ American Mathematical Monthly problem 11611 essentially asks you to show that
$$\lim_n\ n \int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx=2\log(2).\tag1$$
This would follow easily from (2) below, which is true for small values of $n$
according to Maple.
But I couldn't prove equation (2) in general, so I found a direct solution for (1) instead.
$$\int_0^1\left(\sum_{k=n}^\infty {x^k\over k}\right)^2\,dx = \int_0^1 2x^{n-1}\log\left(1+{1\over\sqrt{x}}\right)\,dx.\tag2$$
But I'm still curious about (2). What am I missing? How can equation (2) be proven?
| Let $\mathcal{R}_n$ denote the integral on the right-hand-side of eq. (2):
$$
\mathcal{R}_n = \int_0^1 2 x^{n-1} \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x
$$
Consider
$$
\begin{eqnarray}
(n+1) \mathcal{R}_{n+1} - n \mathcal{R}_n &=& \int_0^1 2 \left( (n+1) x^{n} - n x^{n-1} \right) \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d}x \\
&=& \int_0^1 2 \log\left(1+\frac{1}{\sqrt{x}}\right) \mathrm{d} \left( -x^n \left(1-x\right)\right) \\
&=& \int_0^1 \left(\sqrt{x}-1\right)x^{n-1} \mathrm{d} x = \frac{2}{2n+1} - \frac{1}{n}
\end{eqnarray}
$$
Therefore:
$$
n \mathcal{R}_n = \mathcal{R}_1 + \sum_{m=1}^{n-1} \left( \frac{2}{2m+1} - \frac{1}{m} \right) = \psi\left(n+\frac{1}{2}\right) - \psi(n) + 2 \left(\log(2)-1\right) + \mathcal{R}_1
$$
Integral $\mathcal{R}_1$ can be easily integrated by parts:
$$
\mathcal{R}_1 = 2 \int_0^1 \log\left(1+\frac{1}{\sqrt{x}}\right)\mathrm{d} x\stackrel{\text{by parts}}{=} \left. 2\left( x \log\left( 1 + \frac{1}{\sqrt{x}}\right) + \sqrt{x} - \log\left(1+\sqrt{x}\right) \right) \right|_{x \downarrow 0}^{x = 1} = 2
$$
Thus
$$
n \mathcal{R}_n = 2 \log(2) + \psi\left(n+\frac{1}{2}\right) - \psi(n)
$$
Similarly, denoting $\mathcal{L}_n = \int_0^1 f_n(x)^2 \mathrm{d} x$ the integral on the left-hand-side of eq. (2):
$$ \begin{eqnarray}
n \left(\mathcal{L}_{n+1} - \mathcal{L}_n\right) &=& \int_0^1 \left( n f_{n+1}(x)^2 - n \left( \frac{x^n}{n} + f_{n+1}(x) \right)^2 \right)\mathrm{d} x \\
&=& \int_0^1 \left( -2 x^n f_{n+1}(x) - \frac{x^{2n}}{n}\right)\mathrm{d} x \\
&=& \color\maroon{2 \int_0^1 x^n \log(1-x) \mathrm{d} x} + {\color\blue{2 \int_0^1 x^n \sum_{k=1}^{n} \frac{x^k}{k} \mathrm{d} x}} - \frac{1}{n(2n+1)} \\
&=& \color\maroon{-\frac{2}{n+1} H_{n+1}} + 2 \sum_{k=1}^n \frac{1}{k(k+n+1)} - \frac{1}{n(2n+1)} \\
&=& \frac{2}{n+1} \left( \psi(n+1) - \psi(2+2n) \right) - \frac{1}{n(2n+1)} \\
&=& \frac{1}{n (n+1)} \left( \psi(n+1)-\psi\left(n+\frac{3}{2}\right)-2 \log (2) \right) -\frac{1}{n^2 (2 n+1)}
\end{eqnarray}
$$
where $f_n(x) = \sum_{k=n}^\infty \frac{x^k}{k}$. Now since $\mathcal{L}_1 = \int_0^1 \log^2(1-x)\mathrm{d} x = 2$, and since $\mathcal{R}_{n+1} - \mathcal{R}_n$ equals to
$\mathcal{L}_{n+1} -\mathcal{L}_n$, we establish that $\mathcal{L}_n = \mathcal{R}_n$:
$$
\mathcal{R}_{n+1} - \mathcal{R}_n = 2\log(2) \left( \frac{1}{n+1}- \frac{1}{n} \right)
+ \frac{1}{n+1} \left( \psi\left(n+\frac{3}{2}\right) - \psi (n+1)\right) -
\frac{1}{n} \left( \psi\left(n+\frac{1}{2}\right) - \psi (n)\right) \stackrel{\color\maroon{\text{use recurrence equation for } \psi}}{=} \mathcal{L}_{n+1} - \mathcal{L}_n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/139729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
} |
A trigonometric identity: $(\sin x)^{-2}+(\cos x)^{-2}=(\tan x+\cot x)^2$ I've been trying to prove it for a while, but can't seem to get anywhere.
$$\frac{1}{\sin^2\theta} + \frac{1}{\cos^2\theta} = (\tan \theta + \cot \theta)^2$$
Could someone please provide a valid proof?
I am not allowed to work on both sides of the equation.
Work so far:
RS:
$$
\begin{align}
& \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} + 2 \\[10pt]
& = \frac{\sin^4\theta}{(\cos^2\theta)(\sin^2\theta)} + \frac{\cos^4\theta}{(\sin^2\theta) (\cos^2\theta)} + \frac{(\sin^4\theta)(\cos^2\theta)}{(\sin^2\theta)(\cos^2\theta)} + \frac{(\sin^2\theta)(\cos^4\theta)}{(\sin^2\theta)(\cos^2\theta)} \\[10pt]
& = \frac{\sin^4\theta + \cos^4\theta + (\sin^4\theta)(\cos^2\theta) + (\sin^2\theta)(\cos^4\theta)}{(\cos^2\theta)(\sin^2\theta)}
\end{align}
$$
I am completely lost after this.
| Look at the largest triangle.
(There's a reason my avatar --the logo of my software company-- is a stylized version of this figure.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/142252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Solutions to $z^3 - (b+6) z^2 + 8 b^2 z - 7+b^2 = 0, b\in \mathbb R, z \in \mathbb C$ $z_1 = 1+i$ is a given solution.
I guess what I have to find is $z_2$ and $z_3$ in
$(z - (1 + i))(z - z_2)(z-z_3) = z^3 - (b+6) z^2 + 8 b^2 z - 7+b^2$.
I tried to divide the polynomial by $(z - (1 + i))$, but that didn’t seem to work because of the $b$. According to the Complex conjugate root theorem $z_2 = \overline{z_1} = 1 - i$ is a solution too and somebody mentioned that it’s a hint that all coefficients are real. But I still don’t know how to proceed. What am I missing?
| Consider the cubic polynomial
$$\begin{equation*}
P(z)=z^{3}+Az^{2}+Bz+C
\end{equation*}\tag{1},$$
where the coefficients $A,B$ and $C$ are real numbers. If we denote its
roots by $z_{1},z_{2}$ and $z_{3}$, then it factors as
$$\begin{eqnarray*}
P(z) &=&\left( z-z_{1}\right) \left( z-z_{2}\right) (z-z_{3}) \\
&=&z^{3}-\left( z_{1}+z_{3}+z_{2}\right) z^{2}+\left(
z_{1}z_{2}+z_{2}z_{3}+z_{1}z_{3}\right) z-z_{1}z_{2}z_{3}.
\end{eqnarray*}\tag{2}$$
The constant term is
$$\begin{equation*}
P(0)=C=-z_{1}z_{2}z_{3}
\end{equation*}.$$
In the present case $A=-(b+6)$, $B=8b^{2}$ and $C=-7+b^{2}$. Since $z_{1}=1+i$ is a given solution, then $z_{2}=\overline{z}_{1}=\overline{1+i}=1-i$ is another solution, as you concluded. We thus have $z_{1}z_{2}=\left( 1+i\right) \left( 1-i\right)=2$ and
$$\begin{equation*}
-7+b^{2}=-2z_{3}
\end{equation*}\tag{3},$$
whose solution is
$$\begin{equation*}
z_{3}=\frac{7-b^{2}}{2}.
\end{equation*}\tag{4}$$
Since $P(z_1)=P(z_2)=0$, we have
$$\begin{eqnarray*}
&&\left( 1+i\right) ^{3}-(b+6)\left( 1+i\right) ^{2}+8b^{2}\left( 1+i\right)
-7+b^{2} \\
&=&-9+9b^{2}+i\left( -10-2b+8b^{2}\right)=0,
\end{eqnarray*}\tag{5}$$
$$\begin{eqnarray*}
&&\left( 1-i\right) ^{3}-(b+6)\left( 1-i\right) ^{2}+8b^{2}\left( 1-i\right)
-7+b^{2} \\
&=&-9+9b^{2}+i\left( 10+2b-8b^{2}\right)=0,
\end{eqnarray*}\tag{6}$$
which means that $b$ satisfies the system
$$\begin{equation*}
\left\{
\begin{array}{c}
-9+9b^{2}=0 \\
10+2b-8b^{2}=0.
\end{array}
\right.
\end{equation*}\tag{7}$$
The solution of $(7)$ is $b=-1$. Using $(4)$ we find $$z_{3}=3.\tag{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/144512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Using contour integration, or other means, is there a way to find a general form for $\int_{0}^{\infty}\frac{\sin^{n}(x)}{x^{n}} \, dx$? While studying some CA, I ran across methods of evaluating $$\int_0^\infty \frac{\sin x}{x} \, dx, \;\ \int_0^\infty \frac{\sin^2 x}{x^2} \, dx, \;\ \text{and} \ \int_0^\infty \frac{\sin^3 x}{x^{3}} \, dx.$$
Is there a way to find a closed form for $$\int_0^\infty \frac{\sin^n x}{x^n} \, dx, \ n \in \mathbb{N}_{>0} ?$$
Rather it be contour integration or some clever method using real analysis.
| I have a generalized elementary method for this problem，If f (x) is an even function, and the period is $\pi$,we have:
$$\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx=\int_{0}^\frac{\pi}{2}f(x)g_n(x)\sin^nxdx \qquad (1)$$
Where the $g_n(x)$ in (1) is as follows
$$g_n(x)=\begin{cases}\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\csc x\right),& \text{for n is odd $n\in\Bbb N$ and}\\[2ex]
\frac{(-1)^{n-1}}{(n-1)!}\frac{d^{n-1}}{dx^{n-1}}\left(\cot x\right),& \text{ for n is even .}
\end{cases}$$
——————————————————————————————————————————————————
Proof:
\begin{align}
\int_{0}^\infty f(x)\frac{\sin^nx}{x^n}dx&=\sum_{k=0}^\infty\int_{k\pi}^{(2k+1)\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x}\right)^ndx+\sum_{k=1}^\infty\int_{(2k-1)\frac{\pi}{2}}^{k\pi}f(x)\left(\frac{\sin x}{x}\right)^ndx\\
&=\sum_{k=0}^\infty\int_{0}^{\frac{\pi}{2}}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty\int_{-\frac{\pi}{2}}^{0}f(x+k\pi)\left(\frac{\sin (x+k\pi)}{x+k\pi}\right)^ndx\\
&=\sum_{k=0}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(x)\left(\frac{\sin x}{x+k\pi}\right)^ndx+\sum_{k=1}^\infty(-1)^{nk}\int_{0}^{\frac{\pi}{2}}f(-x)\left(\frac{\sin x}{x-k\pi}\right)^ndx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nx\left(\frac{1}{x^n}+\sum_{k=1}^\infty(-1)^{nk}\left[\frac{1}{(x+k\pi)^n}+\frac{1}{(x-k\pi)^n}\right]\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}f(x)\sin^nxg_n(x)dx
\end{align}
We know by the Fourier series
\begin{align}
\csc x&=\frac{1}{x}+\sum_{k=1}^\infty(-1)^k\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)\\
\end{align}
and
\begin{align}
\cot x&=\frac{1}{x}+\sum_{k=1}^\infty\left(\frac{1}{x+k\pi}+\frac{1}{x-k\pi}\right)
\end{align}
Take the n-1 order derivative,thus we obtain $g_n(x)$.
——————————————————————————————————————————————————
Example:
\begin{align}
(1.)\qquad\int_{0}^{\infty}\frac{\sin^3x}{x}dx&=\int_{0}^{\frac{\pi}{2}}\sin^2xg_1(x)\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2x\frac{1}{\sin x}\sin xdx\\
&=\int_{0}^{\frac{\pi}{2}}\sin^2xdx\\
&=\frac{\pi}{4}\\
\end{align}
\begin{align}
(2.)
\int_{0}^{\infty}(1+\cos^2x)\frac{\sin^2x}{x^2}dx
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)g_2(x)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(-\frac{d}{dx}\cot x\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)\left(\frac{1}{\sin^2x}\right)\sin^2xdx\\
&=\int_{0}^{\frac{\pi}{2}}(1+\cos^2x)dx\\
&=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}\\
\end{align}
\begin{align}
(3.)
\int_{0}^{\infty}\frac{1}{(1+\cos^2x)}\frac{\sin^3x}{x^3}dx
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}g_3(x)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\left(\frac{1}{2}\frac{d^2}{dx^2}(\csc x)\right)dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{\sin^3x}{(1+\cos^2x)}\frac{(1+\cos^2x)}{2\sin^3x}dx\\
&=\int_{0}^{\frac{\pi}{2}}\frac{1}{2}dx=\frac{\pi}{4}\\
(4.)
\int_{0}^{\infty}\cos 2xy\frac{\sin^2x}{x^2}dx
&=\int_{0}^{\frac{\pi}{2}}\cos 2xydx
=\frac{\sin\pi y}{2y}\\
\end{align}
\begin{align}
\end{align}
| {
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} |
General expression for determinant of a block-diagonal matrix Consider having a matrix whose structure is the following:
$$
A =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{2,1} & a_{2,2} & a_{2,3} & 0 & 0 & 0 & 0 & 0 & 0\\
a_{3,1} & a_{3,2} & a_{3,3} & 0 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & a_{4,4} & a_{4,5} & a_{4,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{5,4} & a_{5,5} & a_{5,6} & 0 & 0 & 0\\
0 & 0 & 0 & a_{6,4} & a_{6,5} & a_{6,6} & 0 & 0 & 0\\
0 & 0 & 0 & 0 & 0 & 0 & a_{7,7} & a_{7,8} & a_{7,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{8,7} & a_{8,8} & a_{8,9}\\
0 & 0 & 0 & 0 & 0 & 0 & a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
Question.
What about its determinant $|A|$?.
Another question
I was wondering that maybe matrix $A$ can be expressed as a product of particular matrices to have such a structure... maybe using these matrices:
$$
A_1 =
\begin{pmatrix}
a_{1,1} & a_{1,2} & a_{1,3}\\
a_{2,1} & a_{2,2} & a_{2,3}\\
a_{3,1} & a_{3,2} & a_{3,3}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{4,4} & a_{4,5} & a_{4,6}\\
a_{5,4} & a_{5,5} & a_{5,6}\\
a_{6,4} & a_{6,5} & a_{6,6}\\
\end{pmatrix}
$$
$$
A_2 =
\begin{pmatrix}
a_{7,7} & a_{7,8} & a_{7,9}\\
a_{8,7} & a_{8,8} & a_{8,9}\\
a_{9,7} & a_{9,8} & a_{9,9}\\
\end{pmatrix}
$$
I can arrange $A$ as a compination of those: $A = f(A_1,A_2,A_3)$
Kronecker product
One possibility can be the Kronecker product:
$$
A=
\begin{pmatrix}
1 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_1 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 0\\
\end{pmatrix} \otimes A_2 +
\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 1\\
\end{pmatrix} \cdot A_3
$$
But what about the determinant??? There are sums in this case which is not good...
| A "functorial approach" using the exterior product: If $\phi: V \rightarrow V$ is an endomorphism of a vector space, you may calculate the determinant of the endomorphism $\phi$ as the induced map
$$\wedge^n (\phi): \wedge^n V \rightarrow \wedge^n V$$
where $n:=\dim(V)$. Since $\wedge^n$ is a functor you get a canonical map $\wedge^l (\phi)$ for any integer $l \geq 1$.
It follows $\wedge^n (\phi)$ is an endomorphism of a one dimensional vector space $\wedge^n V$ and hence it is given as multiplication with a number $a$. The number $a$ is the determinant: $a=\det(\phi)$ of the map $\phi$. If you choose a basis $B$ of $V$ and the matrix of $\phi$ in this basis is a matrix $A$, it follows $a=det(A)$ is the determinant of the matrix $A$.
There is a formula:
$$\wedge^{n_1+n_2}(V_1\oplus V_2) \cong \wedge^{n_1}V_1 \otimes \wedge^{n_2}V_2,$$
where $n_i:=\dim(V_i)$. Let
\begin{align*} \phi= \begin{pmatrix} \phi_1 & 0 \\ 0 & \phi_2 \end{pmatrix} \end{align*}
where $\phi_i$ is an endomorphism of $V_i$. It "follows"
$$\det(\phi)=\wedge^{n_1+n_2}(\phi) \cong \wedge^{n_1}(\phi_1) \otimes \wedge^{n_2}(\phi_2)$$
But the tensor product $\wedge^{n_1}V_1 \otimes \wedge^{n_2}V_2$ is a one dimensional vector space and any linear endomorphism of such a space is given (in a basis) as multiplication with a number. Choosing a basis it follows the endomorphism $\wedge^{n_1}(\phi_1) \otimes \wedge^{n_2}(\phi_2)$ is multiplication with the number
$$\det(A_1)\det(A_2),$$
where $A_i$ is a matrix of $\phi_i$ in a basis $B_i$ for $V_i$.
Question: "But what about the determinant???"
By induction it follows that if $M$ is a matrix with square matrices $A_i$ along the diagonal, you get the formula
$$\det(M)=\det(A_1) \cdots \det(A_n).$$
| {
"language": "en",
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"answer_count": 4,
"answer_id": 3
} |
How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?
| You've already solved the case $n=1$, so I'll not repeat that there.
Assuming as the induction hypothesis that $n$ has the property that $3|n(n^2+8)$, we can rewrite $(n+1)((n+1)^2+8)$ to obtain
$$
\begin{split}
(n+1)((n+1)^2+8) & = (n+1)(n^2+2n+1+8)\\
& =n(n^2+8)+n(2n+1)+((n+1)^2+8)\\
& = n(n^2+8)+3n^2+3n+9
\end{split}
$$
In the latter, all the terms are divisible by $3$, hence it follows that $(n+1)((n+1)^2+8)$ is also divisible by $3$. This finishes the induction proof, so we may conclude that $n(n^2+8)$ is divisible by $3$ for all $n\geq 1$.
| {
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"answer_count": 9,
"answer_id": 0
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Is $n = k \cdot p^2 + 1$ necessarily prime if $2^k \not\equiv 1 \pmod{n}$ and $2^{n-1} \equiv 1 \pmod{n}$? $p$ is an odd prime and $k$ is a positive integer.
Let $n=k \cdot p^2+1$.
If $2^k \not\equiv 1 \pmod n$ and $2^{n-1} \equiv 1 \pmod n$, is $n$ prime? If it is, why?
| Not necessarily. Let $p = 3$ and $k = 154$. Then $n = 3^2 * 154 + 1 = 1387$, $2^{154} \equiv 1024 \not\equiv 1 \pmod{1387}$, and $2^{1386} \equiv 1 \pmod{1387}$. But $n = 1387 = 19 * 73$ is not prime.
| {
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"answer_count": 2,
"answer_id": 0
} |
Integral of $\int_0^{\pi/2} \ (\sin x)^7\ (\cos x)^5 \mathrm{d} x$ I am trying to find this by using integration by parts but I am not sure how to do it.
$$\int_0^{\pi/2} (\sin x)^7 (\cos x)^5 \mathrm{d} x$$
I tried rewriting as
$$\int_0^{\pi/2} \sin x\cdot\ (\sin x)^6\cdot\ (\cos x)^5 \mathrm{d} x = \int_0^{\pi/2}\sin x(1-\ (\cos x)^3)\cdot\ (\cos x)^5 \mathrm{d} x$$
but that seems to only give me a very, very long loop that doesn't help me at all. How do I proceed?
$$\int_0^{\pi/2} \sin x\cdot (\sin x)^6\cdot (\cos x)^5 \mathrm{d} x = \sin x(1- (\cos x)^3)\cdot (\cos x)^5 \mathrm{d} x$$
$u = \cos x$, then $du = -\sin xdx$
$\int \frac{-u^6}{6} \mathrm{d} u - \int \frac{-u^9}{9} \mathrm{d} u$
From here it looks like I have an incredibly long string of $u$ substitutions to make to get to something I can find an antiderivative for.
| Beta version:
$$\begin{aligned}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{7}\left(\cos x\right)^{5}dx & =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\sin x\right)^{6}\left(\cos x\right)^{4}2\sin x\cos xdx\\
& =\frac{1}{2}\int_{0}^{\frac{\pi}{2}}\left(\left(\sin x\right)^{2}\right)^{3}\left(1-\left(\sin x\right)^{2}\right)^{2}d\left(\left(\sin x\right)^{2}\right)\\
& =\frac{1}{2}\int_{0}^{1}t^{3}\left(1-t\right)^{2}dt=\frac{1}{2}\int_{0}^{1}t^{4-1}\left(1-t\right)^{3-1}dt\\
& =\frac{1}{2}B\left(4,3\right)=\frac{1}{2}\frac{\Gamma\left(4\right)\Gamma\left(3\right)}{\Gamma\left(7\right)}=\frac{1}{2}\frac{6\cdot2}{720}=\frac{1}{120}
\end{aligned}$$
| {
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"answer_count": 7,
"answer_id": 6
} |
Proving $\sin x + \sin x \cdot \cot^2 x = \csc x $ The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process:
$$\begin{align*}
\tan \theta &= \frac{\sin \theta}{\cos \theta}\\
(\sin \theta)^2 + (\cos \theta)^2 &= 1\\
\csc \theta &= \frac{1}{\sin \theta}\\
\sec \theta &= \frac{1}{\cos \theta}\\
\cot \theta &= \frac{1}{\tan \theta}.
\end{align*}$$
The trig identity given is:
$$\sin x + \sin x \cdot \cot^2 x = \csc x $$
I simplified it to:
$$\sin x + \sin x \cdot \left(\frac{\cos x}{\sin x}\right)^2 = \frac{1}{\sin x} $$
After which I get lost in a mess of term rewriting that never seems to lead anywhere fruitful. Any hints would be greatly appreciated.
| Taking the left hand side, try cancelling the $\sin^2{x}$ with the $\sin{x}$ in the numerator
(i.e. $\sin{x} .(\frac{\cos^2{x}}{\sin^2{x}}) = \frac{\cos^2{x}}{\sin{x}}$)
Then add the remaining two terms ($\sin{x} + \frac{\cos^2{x}}{\sin{x}}$) and use the identity $\sin^2{x}+\cos^2{x}=1$. That should do the trick :)
| {
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"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
The calculation of a series Calculate the series
\begin{equation}
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.
\end{equation}
| We let
$$f(z)=\dfrac{1}{(4z+1)(4z+3)}$$
There are two poles of $f(z)$. They are $4z_0+3=0 \implies z_0=-\frac{3}{4}$ and $4z_1+1=0 \implies z_1=-\frac{1}{4}$.
Residue calculus tells us that
$$\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(\text{sum of residues of }\pi\cot(\pi z)f(z))$$
We calculate the residues of the poles ($b_0$ and $b_1$ for $z_0$ and $z_1$ respectively):
$$b_0=\operatorname {Res}_{z=z_0}=\lim_{z \to z_0}\frac{(z-z_0)\pi\cot (\pi z)}{(4z+1)(4z+3)}$$
Using L'Hopital's rule we have
$$b_0=\lim_{z \to z_0} \frac{\pi\cot (\pi z)-(z-z_0)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-3\pi/4)}{4(4)}=-\frac{\pi}{8}$$
Similarly, we have
$$b_1=\lim_{z \to z_1} \frac{\pi\cot (\pi z)-(z-z_1)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-\pi/4)}{4(4)}=-\frac{\pi}{8}$$
So
$$
\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(-\frac{\pi}{8}-\frac{\pi}{8})=\frac{\pi}{4}\implies
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=\frac{1}{2}\frac{\pi}{4}=\frac{\pi}{8}
$$
QED
| {
"language": "en",
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"source": "stackexchange",
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"answer_id": 2
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Using the roots of polynomial finding the value of sum. If $a,b$ and $c$ are the roots of $x^{3}+px^{2}+qx+r$, then how can we find the value of $\displaystyle \sum \frac{b^{2}+c^{2}}{bc}$.
| I think you are asking for
$$\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca},\tag{$1$}$$
or perhaps twice this quantity.
If you bring Expression $(1)$ to a common denominator, you will get
$$\frac{a^2c+b^2c+b^2 a+c^2 a+c^2b+a^2b}{abc}.$$
Note that
$$(a+b+c)(ab+bc+ca)=a^2c+b^2c+b^2 a+c^2 a+c^2b+a^2b+3abc.$$
Thus
$$\frac{a^2+b^2}{ab}+\frac{b^2+c^2}{bc}+\frac{c^2+a^2}{ca}=\frac{(a+b+c)(ab+bc+ca)-3abc}{abc}.$$
Everything term on the right-hand side is expressible simply in terms of the coefficients.
| {
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"answer_count": 1,
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Calculating $\int \frac{dx}{1+3\sin x+\cos x}$ Can I get a detailed answer on integrating this? I am currently stumped.
$$
\int \frac{dx}{1+3\sin x+\cos x}
$$
Thanks.
| Let $y=x+\arctan(1/3)$, then because $\sin(x+\arctan(1/3))=\frac{3}{\sqrt{10}}\sin(x)+\frac{1}{\sqrt{10}}\cos(x)$
$$
\int\frac{\mathrm{d}x}{1+3\sin(x)+\cos(x)}=\int\frac{\mathrm{d}y}{1+\sqrt{10}\sin(y)}\tag{1}
$$
The standard substitution $z=\tan(y/2)$ yields $\sin(y)=\dfrac{2z}{1+z^2}$ and $\mathrm{d}y=\dfrac{2\mathrm{d}z}{1+z^2}$. Therefore,
$$
\begin{align}
\int\frac{\mathrm{d}y}{1+\sqrt{10}\sin(y)}
&=\int\frac{2\mathrm{d}z}{1+z^2+2\sqrt{10}\,z}\\
&=\int\frac{2\mathrm{d}z}{(z+\sqrt{10})^2-9}\\
&=\frac13\int\frac{\mathrm{d}z}{z+\sqrt{10}-3}-\frac13\int\frac{\mathrm{d}z}{z+\sqrt{10}+3}\\
&=\frac13\log\left(\frac{z+\sqrt{10}-3}{z+\sqrt{10}+3}\right)+C\tag{2}
\end{align}
$$
Putting $(1)$ and $(2)$ together and reversing the substitutions, we get
$$
\int\frac{\mathrm{d}x}{1+3\sin(x)+\cos(x)}=\frac13\log\left(\frac{\tan(x/2+\arctan(1/3)/2)+\sqrt{10}-3}{\tan(x/2+\arctan(1/3)/2)+\sqrt{10}+3}\right)+C\tag{3}
$$
| {
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Integrate a complex function Suppose $\gamma$ is the circle $|z|=1$ traversed counterclockwise. Evaluate
$$
\int_\gamma\frac{\cot(z)}{z^2}dz
$$
I was able to show $0$ is a pole of order $3$ for the integrand, and I think I have to use residues, but I am stuck. Is this right so far?
| Yes, you are correct. The only pole inside the contour for the integrand is at $0$ with a multiplicity of $3$. You can make use of the Laurent series for $\cot(z)$ at $z=0$. $$\cot(z) = \dfrac1z - \dfrac{z}{3} + \mathcal{O}(z^3)$$
Hence, $$\dfrac{\cot(z)}{z^2} = \dfrac1{z^3} - \dfrac1{3z} + \mathcal{O}(z)$$
Hence, $$\oint_{\gamma} \dfrac{\cot(z)}{z^2} dz = 2 \pi i \times \text{Res} \left( \dfrac{\cot(z)}{z^2}\right)_{z=0} = - \dfrac{2 \pi i}{3}$$
EDIT
Below is a way to compute the residue easily. Note that if $n \in \mathbb{Z}^+$,
$$\lim_{z \to 0} z^n \cot(z) = \lim_{z \to 0} z^{n-1} \cdot \left(z \dfrac{\cos(z)}{\sin(z)} \right) = \lim_{z \to 0} z^{n-1} \cdot \lim_{z \to 0} \left(z \dfrac{\cos(z)}{\sin(z)} \right) = \lim_{z \to 0} z^{n-1} \cdot 1$$
Hence, $$\lim_{z \to 0} z^n \cot(z) = \begin{cases} 1 & \text{ if }n =1\\ 0 & \text{ if }n \in \mathbb{Z}^+\backslash\{1\}\end{cases}
$$
This means the Laurent series for $\cot(z)$ is of the form $\dfrac1z + a_0 + a_1 z + a_2 z^2 + a_3 z^3 + \cdots$.
Further, since $\cot(z)$ is an odd function, we have that $$\cot(z) = \dfrac1z + a_1 z + a_3 z^3 + a_5 z^5 + \cdots$$
Hence, the residue of $ \dfrac{\cot(z)}{z^2}$ is the coefficient $a_1$. From, the Laurent series, we can see that $a_1$ is nothing but $$a_1 = \lim_{z \to 0} \dfrac{z \cot(z) - 1}{z^2} = \lim_{z \to 0} \dfrac{z \cos(z) - \sin(z)}{z^2 \sin(z)} = \lim_{z \to 0} \dfrac{z (1-z^2/2+O(z^4)) - (z-z^3/6 + O(z^5))}{z^2 (z + O(z^3))}\\ =\lim_{z \to 0} \dfrac{-z^3/2 + O(z^5) +z^3/6 + O(z^5)}{z^3 + O(z^5)} =\lim_{z \to 0} \dfrac{-1/3 + O(z^2)}{1 + O(z^2)} = - \dfrac13$$
Hence, $$\text{Res} \left( \dfrac{\cot(z)}{z^2}\right)_{z=0} = -\dfrac13$$
| {
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} |
How to calculate these summations? How to find the values of these kind of summations:
$$\large\sum_{i=0}^6(6-i)\;\ast\;\sum_{j=1}^6(7-j)\;\ast\;\sum_{k=2}^7(8-k)\;\ast\;\sum_{\ell=3}^8(9-\ell)$$
| Use that $$\begin{align}\sum_{t=a}^b(c-t)&=\left(\sum_{t=a}^bc\right)-\left(\sum_{t=a}^b t\right)\\\\&=\left(\sum_{t=a}^bc\right)-\left(\sum_{s=0}^{b-a} (s+a)\right)\\\\ &=\left(\sum_{t=a}^bc\right)-\left(\sum_{s=0}^{b-a} s\right)-\left(\sum_{s=0}^{b-a}a\right)\\\\ &=(b-a+1)c-(b-a+1)a-\left(\sum_{s=0}^{b-a} s\right)\\\\&=(b-a+1)(c-a)-\left(\sum_{s=0}^{b-a} s\right)\\\\&=(b-a+1)(c-a)-\frac{(b-a)(b-a+1)}{2}\\\\&=(b-a+1)(c-a-\tfrac{b}{2}+\tfrac{a}{2})\\\\&=(b-a+1)(c-\tfrac{b}{2}-\tfrac{a}{2})\end{align}$$
for each term. For example,
$$\sum_{i=0}^6(6-i)=6+5+4+3+2+1=\fbox{21}=7\cdot 3=(6-0+1)(6-\tfrac{6}{2}-\tfrac{0}{2})\qquad\checkmark$$
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Reducing a fraction in algebra? How would I reduce the following fraction?
$$\frac{4A^2-B^2}{4A^2-4AB+B^2}$$
I am not sure how I would reduce it.
| The numerator is a difference of squares, so it factors:
$$4A^2 - B^2 = (2A)^2 - B^2= (2A-B)(2A+B).$$
The denominator is a perfect square:
$$4A^2 - 4AB + B^2 = (2A)^2 - 2(2A)B + B^2 = (2A-B)^2.$$
Then you can cancel one factor:
$$\frac{4A^2 - B^2}{4A^2 - 4AB+B^2} = \frac{(2A-B)(2A+B)}{(2A-B)^2} = \frac{2A+B}{2A-B}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving a biquadratic $x^{4}-2x^3 + x^2 - 2x +1 =0$ How do I find the roots of $$x^{4}-2x^3 + x^2 - 2x +1 =0$$
I am not able to find any roots by trial and error.
| *
*Divide throughout by $x^2$.
*Then you have $x^{2}-2x + 1 - \frac{2}{x} + \frac{1}{x^2} = 0$
*You can re-write as $x^{2}+\frac{1}{x^2} - 2(x+\frac{1}{x}) + 1 =0$
*Then use $x^{2}+\frac{1}{x^2} = (x+\frac{1}{x})^{2} -2$ and then reduce it to a quadratic by putting $y=x+\frac{1}{x}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Multiplication in the field $F = \mathbb{Z}_2[x]/f(x)$ Let $f(x) = x^6 + x + 1$ and define the field $F = \mathbb{Z}_2[x]/f(x)$
Compute the following in this field:
1. $(x^5 + x + 1)(x^3 + x^2 +1)$
I start by multiplying (in $\mathbb{Z}_2[x]$):
$(x^5 + x + 1)(x^3 + x^2 +1)$ = $(x^8 + x^7 + x^5 +x^4 + x^2 + x + 1)$
Then dividing the result with $f(x)$:
$(x^2 + x)$ and the remainder $(x^5 + x^4 + x^3 + x^2 + 1)$
Is this the right approach for solving this problem? Do I understand it correct that I want the result of my multiplication mod $f(x)$? Can I think of it as a simple modulus calculation:
$11$ mod $7 = a$
$7*1 + 4$ mod $7 = 4$
In my case I have
$(x^6 + x +1)*(x^2 + x) + \mbox{remainder}$ mod $(x^6 +x + 1) = \mbox{remainder}$
So my answer to the question would be the remainder, $(x^5 + x^4 + x^3 + x^2 + 1)$?
2. $(x + 1)^{-1}$
I read (wiki) that the inverse to $(x + 1)$ could be found by using the extended euclidean alg. for $a = (x^6 + x +1)$, $b = (x+1)$ but I don't really get it since $a$ is irreducible.
Any hints in the right direction would be appreciated!
| Your first idea is right, and I think you need to read the extended Euclidean algorithm over again (because $x^6+x+1$ being irreducible should not deter you.)
Dividing $\frac{x^6+x+1}{x+1}$ you'll find $(x+1)(x^5+x^4+x^3+x^2+x)+1=x^6+x+1$, and so $(x+1)(x^5+x^4+x^3+x^2+x)=1$ in the quotient.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplifying a radical? How would I simplify the following two radicals.
$$\sqrt{\frac{X^3}{50}}$$
For my answer I got $\frac{X^2}{50X}$ but I am not sure if this is correct.
My second question is how would I simplify $\sqrt{\frac{1}{12}}$
| No, your simplification is incorrect.
The first radical only makes sense if $x\geq 0$. Assuming this is the case, remember that:
*
*If $a$ and $b$ are both nonnegative, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$, and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$.
*$(\sqrt{a})^2 = a$ for any $a\geq 0$.
So:
$$\sqrt{\frac{x^3}{50}} = \frac{\sqrt{x^3}}{\sqrt{50}} = \frac{\sqrt{x^2}\sqrt{x}}{\sqrt{25}\sqrt{2}} = \frac{x\sqrt{x}}{5\sqrt{2}} = \frac{x\sqrt{x}\sqrt{2}}{5\sqrt{2}\sqrt{2}} = \frac{x\sqrt{2x}}{10}.$$
There are other ways of expressing it as well.
For the second, $12 = 4\times 3$, so
$$\sqrt{12}=\sqrt{4}\sqrt{3} = 2\sqrt{3}.$$
Can you take it from there?
| {
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Solution to a system of quadratics I am learning about a Bell State, and am trying to show that they are entangled. I believe that the required proof is to show that the system
$$\alpha_0^2+\alpha_1^2=1$$
$$\beta_0^2+\beta_1^2=1$$
$$\alpha_0\beta_0=1/\sqrt{2}$$
$$\alpha_1\beta_1=1/\sqrt{2}$$
has no solutions. I have tried various ways of rewriting each variable in terms of others etc. without success. Any hints?
EDIT: I apologize, I should have made clear. In QM probabilities can be complex - what exactly this means intuitively is unclear to me, but algebraically it means that $\alpha_i,\beta_i$ can be complex.
| From (3) and (4), we have:
\begin{align*}
\beta_0^2 &= \frac{1}{2\alpha_0^2} \\
\beta_1^2 &= \frac{1}{2\alpha_1^2}
\end{align*}
Plug into (2):
\begin{align*}
\frac{1}{2\alpha_0^2} + \frac{1}{2\alpha_1^2} &= 1 \\
\frac{\alpha_1^2}{\alpha_0^2\alpha_1^2} + \frac{\alpha_0^2}{\alpha_0^2\alpha_1^2} &= 2 \\
\frac{\alpha_0^2 + \alpha_1^2}{\alpha_0^2\alpha_1^2} &= 2
\end{align*}
Use (1) in the numerator to get:
$$
\alpha_0^2 \alpha_1^2 = \frac{1}{2}
$$
Therefore:
$$
\alpha_1^2 = \frac{1}{2\alpha_0^2}
$$
Plug into (1) and multiply both sides by $\alpha_0^2$:
$$
\alpha_0^4 - \alpha_0^2 + \frac{1}{2} = 0
$$
This is a quadratic equation for $\alpha_0^2$ with no real solutions, as $\Delta = -1 \lt 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\
\end{pmatrix}$$
It suffices to show that $\operatorname{det}M > 0$, then the claim follows.
Any hints how to show the positivity of this determinant?
| The eigenvalues of an $n$ by $n$ matrix consisting entirely of 1's are $n$ and $0.$ An eigenvector for $n$ can be all entries 1. $0$ has multiplicity $n-1,$ for $1 \leq i \leq n-1$ take an eigenvector to have mostly 0's, but $-1$ at position $i$ and 1 at position $n.$ Adding the identity matrix makes the eigenvalues $n+1$ and $1.$ Dividing by 2 makes the eigenvalues $\frac{n+1}{2}$ and $\frac{1}{2}.$
It took me a few minutes to check, but the doubled version of this, with all 2's on the main diagonal and all 1's elsewhere, is isometric to the root lattice $A_n$ which comes from a Lie algebra. See A3, A4, A5, A6, A7 and so on.
| {
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prove that $\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ Assume: $a,b,c >0$ prove that :
$$\frac{b+c}{a^2}+\frac{c+a}{b^2}+\frac{a+b}{c^2}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$
| Using the AM-GM inequality we obtain:
$$\frac{b+c}{a^{2}}+\frac{c+a}{b^{2}}+\frac{a+b}{c^{2}}=\frac{b^{3}c^{2}+b^{2}c^{3}+a^{2}c^{3}+a^{3}c^{2}+a^{3}b^{2}+a^{2}b^{3}}{a^{2}b^{2}c^{2}}\ge2\frac{a^{3}bc+b^{3}ac+c^{3}ab}{a^{2}b^{2}c^{2}}=2\frac{a^{2}+b^{2}+c^{2}}{abc}$$
Now a bit of juggling around proves an even stronger result:
$$2\frac{a^{2}+b^{2}+c^{2}}{abc}=\frac{(a-b)^{2}+(a-c)^{2}+(b-c)^{2}+2(ab+bc+ac)}{abc}\ge2\frac{ab+bc+ac}{abc}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
In both estimates above equality is attained when $a=b=c$ which can be checked directly.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding all integer solutions for $x^2 - 2y^2 =2 $ I'd love your help with finding all the integer solutions to the following equation:
$x^2 - 2y^2 =2 $. I want to use Pell's theorem so I changed the equation to $-\frac{1}{2}x^2+ y^2 =-1$, Can I use Pell's Theorem now? I got a private solution for $-\frac{1}{2}x^2+ y^2 =1$ $y=3, x=4$, so form Pell I get that $\alpha= (4+3\sqrt{2})^n$ for every integer $n$, and a private solution for $-\frac{1}{2}x^2+ y^2 =-1$ is $y=1, x=2$, so the total solution is $\alpha= (1+\sqrt{2}) \cdot (+/- (4+3\sqrt{2})^n)$. Are all these steps correct? and if not- how should I solve this one?
Thank you!
| Let's say $\alpha_n$ and $\beta_n$ the $n$-solution of the equation $x^2 - 2y^2 = 2$. We have: $$\left\{\begin{matrix}
\alpha_0 = 2
\\\beta_0 = 1
\end{matrix}\right.
\land \left\{\begin{matrix}
\alpha_1 = 10
\\\beta_1 = 7
\end{matrix}\right.
\land \left\{\begin{matrix}
\alpha_2 = 58
\\\beta_2 = 41
\end{matrix}\right.$$
From here, you can deduce the recoursive relation:
$$\left\{\begin{matrix}
\alpha_{n} = 3\alpha_{n-1}+4\beta_{n-1}
\\\beta_{n} = 2\alpha_{n-1}+3\beta_{n-1}
\end{matrix}\right.$$
This is the list of the values of $\alpha_n$ and $\beta_n$:
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral of determinant Good evening. I need help with this task
$$
\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi\int\limits_{-\pi}^\pi{\det}^2\begin{Vmatrix}\sin \alpha x&\sin \alpha y&\sin \alpha z\\\sin \beta x&\sin \beta y&\sin \beta z\\\sin \gamma x&\sin \gamma y&\sin \gamma z\end{Vmatrix} \text{d}x\,\text{d}y\,\text{d}z
$$
where $\alpha,\beta,\gamma$ are integers.
Computations are horrible, I gave up.
| This is a bit too long to comment. A change in notation $(\alpha, \beta, \gamma) \to (a,b,c)$.
If $a=0$ or $b=0$ or $c=0$ or $a^2 = b^2$ or $b^2=c^2$ or $c^2 = a^2$, the integral is zero.
Let's assume $a,b,c \in \mathbb{Z} \backslash \{0\}$ and are distinct. $$
D(x,y,z,a,b,c) = {\det}^2\begin{Vmatrix}\sin a x&\sin a y&\sin a z\\\sin b x&\sin b y&\sin b z\\\sin c x&\sin c y&\sin c z\end{Vmatrix}\\ = (d_1(b,c,y,z) \sin(ax) + d_2(c,a,y,z) \sin(bx) + d_3(a,b,y,z) \sin(cx))^2
$$
Hence, $$\int_{-\pi}^{\pi}D(x,y,z,a,b,c) dx= \int_{-\pi}^{\pi} \left(d_1^2(b,c,y,z) \sin^2(ax) + d_2^2(c,a,y,z) \sin^2(bx) + d_3^2(a,b,y,z) \sin^2(cx) \right) dx$$ (The cross-terms integrals vanish by our initial assumption)
Hence, $$\int_{-\pi}^{\pi}D(x,y,z,a,b,c) dx= \left(d_1^2(b,c,y,z) + d_2^2(c,a,y,z) + d_3^2(a,b,y,z) \right) \pi$$
Lets now consider $ \displaystyle \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz$. The other two can be computed using symmetry arguments.
$$ \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz =\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left(\sin(by) \sin(cz) - \sin(cy) \sin(bz) \right)^2 dy dz $$
Again by our assumption, the integrals of the cross-terms vanish and hence we get
$$ \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz =\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} \left(\sin^2(by) \sin^2(cz) + \sin^2(cy) \sin^2(bz) \right) dy dz = \pi^2 + \pi^2 = 2\pi^2$$
Hence, $$\int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_1^2(b,c,y,z) dy dz = \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_2^2(c,a,y,z) dy dz = \int_{-\pi}^{\pi} \int_{-\pi}^{\pi} d_3^2(a,b,y,z) dy dz = 2\pi^2$$
Hence, $$\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}D(x,y,z,a,b,c) dx dy dz= \left(2 \pi^2 + 2\pi^2 + 2\pi^2 \right) \pi = 6 \pi^3$$
To summarize, $$I(a,b,c) = \begin{cases} 0 & \text{ if $a=0$ or $b=0$ or $c=0$ or $a^2 = b^2$ or $b^2=c^2$ or $c^2 = a^2$}\\ 6 \pi^3 & \text{ if $a,b,c \in \mathbb{Z} \backslash \{0\}$ and are distinct}\\ ? & \text{else}\end{cases}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $3$ distinct prime numbers $a| (bc+b+c)$, $b|(ac+a+c)$, $c|(ab+a+b)$ How to find $3$ prime numbers $a,b$ and $c$ such that:
$$a| (bc+b+c)$$
$$b|(ac+a+c)$$
$$c|(ab+a+b)$$
| There are no such primes:
Without loss of generality assume $a<b<c$. Clearly 2 cannot be one of the primes, so $a\geq 3$ , $b\geq 5$ and $c\geq 7$.
Now $abc+ab+bc+ca+a+b+c$ is divisible by $a$, by $b$ and by $c$ so it is also divisible by $abc$,
but this is impossible since
$1<\frac{abc+ab+bc+ca+a+b+c}{abc}<\frac{a+1}{a}\frac{b+1}{b}\frac{c+1}{c}\leq\frac{4}{3}\frac{6}{5}\frac{8}{7}<2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Show that $6m \mid (2m+3)^n + 1$ if and only if $4m \mid 3^n + 1$ Let $m$ and $n$ be positive integers. How to prove that $$6m \mid (2m+3)^n + 1$$ if and only if $$4m \mid 3^n + 1$$
| $\newcommand{\jaco}[2]{{\left(\frac{#1}{#2}\right)}}$I have tried to solve this - I hope I did not make too many mistakes there. Maybe there is much easier solution than this one.
Let us start by a few easy observations.
Since $2m+3 \equiv 3 \pmod m$, we see that
$$m\mid (2m+3)^n+1 \Leftrightarrow m\mid 3^n+1. \tag{1}$$
It is obvious that
$$3\mid (2m+3)^n+1 \Rightarrow 3\nmid m. \tag{2}$$
Since $3\nmid x$ implies $x^2\equiv 1 \pmod 3$ (Little Fermat's theorem), we get (using this fact for $x=2m+3$) that
$$3\mid (2m+3)^n+1 \Rightarrow n\text{ is odd.}\tag{3}$$
Now we get
$$3\mid (2m+3)^{n}+1 \Rightarrow m\equiv1\pmod3, \tag{4}$$
since $m\equiv2\pmod3$ $\Rightarrow$ $2m+3\equiv1\pmod3$ $\Rightarrow$ $(2m+3)^n\equiv1\pmod3$ $\Rightarrow$ $(2m+3)^n+1\equiv2\pmod3$.
It is easy to see that
$$4\mid 3^n+1 \Leftrightarrow n\text{ is odd}\tag{5}$$
and
$$m\mid 3^n+1 \Rightarrow 3\nmid m.\tag{6}$$
For odd $n=2k+1$ we have $3^{2k+1}+1=(3+1)(3^{2k}-3^{2k-1}+\dots+3^2-3+1)$, which shows that $$\frac{3^{2k+1}+1}4\equiv 1 \pmod 2.$$ Hence
$$4m\mid 3^n+1 \Rightarrow m\text{ is odd.}\tag{7}$$
Now a few more facts with a little less elementary proof.
We will use Legendre symbol and quadratic reciprocity.
We first show:
$$4m\mid 3^n+1 \Rightarrow m\equiv1 \pmod 3.\tag{8}$$
We know that $n$ is odd, let $n=2k+1$. It suffices to show that each odd prime factor $p$ of $3^{2k+1}+1$ fulfills $p\equiv1\pmod 3$.
Suppose that $p\mid 3^{2k+1}+1$ and, consequently $p\mid 3^{2(k+1)}+3$. Therefore $-3$ is a quadratic residue modulo $p$ and $\jaco{-3}p=1$.
We get:
$$\jaco{-3}p=\jaco{-1}p\jaco3p$$
$$1=(-1)^{\frac{p-1}2}(-1)^{\frac{p-1}2}\jaco{p}3.$$
Hence $\jaco{p}3=1$, which implies $p\equiv1\pmod{3}.$
Now we are ready to show that $6m\mid (2m+3)^n+1$ $\Leftrightarrow$ $4m\mid 3^n+1$.
$\boxed{\Rightarrow}$ If $6m\mid (2m+3)^n+1$ then:
*
*$\gcd(m,4)=1$ by (7)
*$n$ is odd by (3)
*$4\mid 3^n+1$ by (5)
*$m\mid 3^n+1$ by (1)
Together we have $4m\mid 3^n+1$.
$\boxed{\Leftarrow}$ If $4m\mid 3^n+1$ then:
*
*$n$ is odd by (5)
*$\gcd(m,6)=1$ by (6), (7)
*$m\mid (2m+3)^n+1$ by (1)
*$2\mid (2m+3)^n+1$ since $2m+3$ is odd
*$3\mid (2m+3)^n+1$ since $m\equiv1 \pmod 3$ by (8), which yields $2m+3\equiv 2\pmod 3$ and $(2m+3)^n\equiv2\pmod3$ (using the fact that $n$ is odd).
Together we have $2\cdot 3 \cdot m \mid (2m+3)^n+1$.
| {
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Value of $n$ for which an improper integral is convergent. A question from the Calculus book that I'm self-studying is asking me to determine the value of $n$ for which the improper integral below is convergent:
$$\int_1^{+\infty}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$$
My attempt is below:
Using the definition of improper integral:
$$\lim_{b\to+\infty} \int_1^{b}\left( \frac{n}{x+1} - \frac{3x}{2x^2 + n} \right ) dx$$
I will first solve the indefinite integral:
$$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = \int \frac{n}{x+1}dx-\int \frac{3x}{2x^2+n}dx$$
For the first integral, $\int \frac{n}{x+1}dx=n\ln (x+1) + \text{constant}$. For the second integral, substituting $u = 2x^2+n$ (and therefore $dx=\frac{du}{4x}$), $\int \frac{3x}{2x^2+n}dx=\frac{3}{4}\int \frac{1}{u}du=\frac{3}{4}\ln(2x^2+n) + \text{constant}$. So, the result for the whole integral is:
$$\int\left( \frac{n}{x+1} - \frac{3x}{2x^2+n} \right )dx = n\ln(x+1) - \frac{3}{4}\ln(2x^2+n) + \text{constant} = \ln\left(\frac{(x+1)^n}{(2x^2+n)^{3/4}}\right) + \text{constant}$$
Thus, the value of the definite integral from $x = 1$ to $x = b$ is:
$$\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{(2)^n}{(2+n)^{3/4}}\right)$$
Thus, the value of the improper integral is:
$$\lim_{b\to+\infty} \left[\ln\left(\frac{(b+1)^n}{(2b^2+n)^{3/4}}\right)-\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)\right]$$
For this limit to exist, it only depends on the limit of the term $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$, because the term $\ln\left(\frac{2^n}{(2+n)^{3/4}}\right)$ is a constant. To calculate the limit of $\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right)$ as $b\to +\infty$, I suppose we can neglect the terms $1$ in $(b+1)^n$ and $n$ in $(2b^2+n)^{3/4}$, because they become negligible as $b$ gets larger. Thus we get, for the limit of this term:
$$\lim_{b\to+\infty} \ln\left(\frac{(b)^n}{(2b^2)^{3/4}}\right)$$
For the above limit to exist, it seems that $n = 3/2$ is the only possible value, because $\lim_{b\to+\infty} \ln\left(\dfrac{(b)^{3/2}}{(2b^2)^{3/4}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{b^{3/2}}{2^{3/4}b^{3/2}}\right) = \lim_{b\to+\infty} \ln\left(\dfrac{1}{2^{3/4}}\right)$. If any different value for $n$ were chosen, the logarithm would approach $-\infty$ or $+\infty$, and therefore the limit would not exist.
For $n = 3/2$, the value of the limit is:
$$\lim_{b\to+\infty} \left[\ln\left(\frac{(b)^{3/2}}{(2b^2)^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(2+3/2)^{3/4}}\right)\right]=\lim_{b\to+\infty} \left[\ln\left(\frac{1}{2^{3/4}}\right)-\ln\left(\frac{2^{3/2}}{(7/2)^{3/4}}\right)\right]=\frac{3}{4}\ln\frac{7}{16}$$
This is equal to the answer given by the book. So, this informal argument of neglecting the terms $1$ and $n$ worked.
But I would like to ask if the reasoning above is correct, and if this informal way of finding the value of $n$ is good. Is there a more formal way?
Edit: A more formal way of showing that $n$ should be $3/2$ is suggested by David Mitra in the comments to this question below:
$$\lim_{b\to+\infty} \ln \left( {(b+1)^n\over (2b^2+n)^{3/4} } \right) = \lim_{b\to+\infty} \ln \left( {b^n\over b^{3/2}} \cdot {\bigl(1+{1\over b}\bigr)^n\over \bigl(2 +{n\over b^2}\bigr)^{3/4}} \right)$$
It is clear from above that $n$ should equal $3/2$. If it were a different value, the limit would not exist, because the expression inside the logarithm would approach either $0$ (causing the logarithm to approach $-\infty$) or $+\infty$ (causing the logarithm to tend to $+\infty$).
| Your method is correct and it is easy to formalize your neglecting of minor terms.
We can see that $$\ln\left(\dfrac{(b+1)^n}{(2b^2+n)^{3/4}}\right) - \ln\left( \frac{b^n}{(2b^2)^{3/4}}\right)= \ln \left( (1+1/b)^n (1+n/(2b^2))^{-3/4} \right) \to \ln 1=0$$
as $b\to \infty.$ So the limit of the first term equals the limit of the second term if either limit exists.
However, this is probably not the best way to do this problem. The point of these types of exercises is to get you used to estimating the value of an integral, by doing something to get a feel for the growth of the integrand. Often we can't explicitly integrate the term.
Here, if you put the terms on a common denominator you get $$ \frac{n^2 + 2nx^2 - 3x(x+1)}{(x+1)(2x^2+n)}.$$
The critical fact used often is: $\displaystyle \int^{\infty}_1 \frac{1}{x^a} dx $ converges only for $a>1.$
So the $\frac{n^2}{(x+1)(2x^2+n)}$ part clearly converges. So we only need to worry about $$\frac{2nx^2 - 3x(x+1) }{(x+1)(2x^2+n)}.$$ If $n=3/2$ then $2nx^2=3x^2$ is able to cancel with the $-3x^2$ term in the next term, so we have a linear term in the top, cubic denominator, so behaviour like $1/x^2$ which is convergent. But if not, then we always get some quadratic term in the numerator, so the behaviour is like some constant term times $1/x$, which diverges. It is instructive to formalize these ideas with some inequalities, which I will leave you to try.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\cdots+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$ Let:$a,b,c,d>0$ be real numbers ,how to prove that :
$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}+\frac{b^2+c^2+d^2}{b^5+c^5+d^5}+\frac{c^2+d^2+a^2}{c^5+d^5+a^5}+\frac{d^2+a^2+b^2}{d ^5+a^5+b^5}\le\frac{a+b+c+d}{abcd}$$.
Edit : I think I proved it. From Cauchy inequality we have
$$
\left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)\geq \left(\sum\limits_{cyc}x^3\right)^2
$$
From Chebyshev inequality it follows
$$
\left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)\leq 3\left(\sum\limits_{cyc}x^3\right)^2
$$
hence
$$
\frac{\left(\sum\limits_{cyc}x^5\right)}{\left(\sum\limits_{cyc}x^2\right)}=
\frac{\left(\sum\limits_{cyc}x^5\right)\left(\sum\limits_{cyc}x\right)}{\left(\sum\limits_{cyc}x\right)\left(\sum\limits_{cyc}x^2\right)}\leq
\frac{3}{\left(\sum\limits_{cyc}x^3\right)}\leq
\frac{1}{xyz}
$$
In the last step I used AM-GM inequality. The rest is clear.
Is there a different way to prove it ?
| This question can be solved only using the AM-GM inequality. Building on an idea that Pan Yang suggests in his comment, it suffices to show the following.
$$\frac{a^2+b^2+c^2}{a^5+b^5+c^5}\le\frac{d}{abcd}=\frac{1}{abc}$$
This is equivalent to showing that
$$a^3 bc+b^3ca+c^3ab\le a^5+b^5+c^5 $$
However, this is true by taking a weighted AM-GM as such (I'm writing it in full):
$$\frac{1}{5}a^5+\frac{1}{5}a^5+\frac{1}{5}a^5+\frac{1}{5}b^5+\frac{1}{5}c^5\ge(a^{15}b^{5}c^{5})^{\frac{1}{5}}=a^3bc$$
and similarly for the terms $b^3ca, c^3ab$. This completes the proof.
| {
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"source": "stackexchange",
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Analytic expression for the primitive of square root of a quadratic Can an analytic expression be given for
$$\int \sqrt{ax^2 + bx +c} \, dx$$
I think substitution doesn't work in this case (I need to compute the integral $\int_0^t \ldots$).
| When you see
$$
ax^2 + \underbrace{{}\quad bx\quad{}}_\text{1st-degree term} + c,
$$
it may help to remember that there is a standard technique in algebra for reducing problems involving quadratic polynomials with a first-degree term to problems involving quadratic polynomials with no first-degree term. It's called "completing the square". You write
$$
ax^2+bx+c = a\left( x^2 + \frac b a x\right) + c.
$$
Then you need to work on $\displaystyle x^2 + \frac b a x$.
Half of the coefficient of the first-degree term is $\dfrac{b}{2a}$. If you square that and add it to this expression you're working on, you get a perfect square—i.e. something squared:
$$
\underbrace{x^2 + \frac b a x} \quad +\quad \frac{b^2}{4a^2} = \left( x + \frac{b}{2a} \right)^2.
$$
So
\begin{align}
ax^2+bx+c = a\left( x^2 + \frac b a x\right) + c & = a\left( x^2 + \frac b a x + \frac{b^2}{4a^2} \right) - a\left( \frac{b^2}{4a^2} \right) + c \\[12pt]
& = a\left( x+ \frac{b}{2a} \right)^2 + \frac{4ac-b^2}{4a} \\[12pt]
& au^2 + \text{constant}.
\end{align}
Let's call that last constant capital $C$, and later we'll recall that it's $\dfrac{4ac-b^2}{4a}$.
Then since $u= x + \dfrac{b}{2a}$, we have $du = dx$, and the integral becomes
$$
\int \sqrt{au^2+C}\,du.
$$
Now we'd like a "$1$" where $C$ is, so that we can apply trigonometric identities. So do a bit of algebra:
$$
\int \sqrt{au^2+C}\,du = \int \sqrt{\frac{a}{C} u^2 + 1} \, du.
$$
We also need $(\text{something})^2+1$, in order to apply the identity involving $\tan^2\theta+1$. So we write:
$$
\int\sqrt{\left(u\sqrt{\frac{a}{C}}\right)^2+1}\ du.
$$
Then we have
$$
\int \sqrt{w^2 + 1}\ du.
$$
Since $w=u\sqrt{\dfrac{a}{C}}$, we have $dw = du\sqrt{\dfrac{a}{C}}$, so $du = dw\sqrt{\dfrac{C}{a}}$.
Now we have
$$
\sqrt{\dfrac{C}{a}} \int \sqrt{w^2+1}\ dw.
$$
This is
$$
\sqrt{\dfrac{C}{a}} \int \sqrt{\tan^2\theta+1}\ \sec^2\theta\,d\theta.
$$
$$
= \sqrt{\dfrac{C}{a}} \int \sec^3\theta\,d\theta.
$$
In April 2007, I wrote this Wikipedia article, which has since been edited by a number of others, and by me, explaining how to treat that integral and why it matters.
Later note: The above works if $a$ and (capital) $C$ are positive. This implies (among other things) that $b^2-4ac<0$, so the quadratic polynomial cannot be factored using real numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
inequality with sum of powers How to prove the following inequality:
$$\forall n\geqslant 4:\dfrac {3^{n}+4^{n}+\cdots +\left( n+2\right) ^{n}} {\left( n+3\right) ^{n}} < 1$$
| Induction base: verify for $n=4$:
$$
\underbrace{3^4 + 4^4 + 5^4 + 6^4}_{2258} < 7^4 = 2401
$$
Induction step: Assuming inequality true for $n$, prove it true for $n+1$:
$$ \begin{eqnarray}
\sum _{k=3}^{n+3} k^{n+1} &=& \sum _{k=3}^{n+2} k^{n+1}+(n+3)^{n+1} \leqslant (n+2)\sum _{k=3}^{n+2} k^{n}+(n+3)^{n+1} \\
&\stackrel{\text{by assumption}}{\leqslant}& (n+2)
(n+3)^n+(n+3)^{n+1} \\ & < & 2 (n+3)^{n+1}<(n+4)^{n+1}
\end{eqnarray}
$$
The last inequality follows because the sequence $a_n = \left( 1 + \frac{1}{n+3} \right)^{n+1}$ is increasing (proof of Brian M. Scott, given in comments):
$$\begin{eqnarray}
\frac{a_{n+1}}{a_n} &=& \left(1+\frac{1}{n+4}\right) \left( 1 - \frac{1}{(n+4)^2} \right)^{n+1} \\ &\stackrel{\text{binomial theorem}}{>}& \left(1+\frac{1}{n+4}\right) \left( 1 - \frac{n+1}{(n+4)^2} \right) = 1 + \frac{2n+11}{(n+4)^3} > 1
\end{eqnarray}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality for cosines Is the following inequality in a triangle known?
$$4(\cos A + \cos B + \cos C) \le 3 + \cos \left(\frac{B-C}{2}\right) + \cos \left(\frac{C-A}{2}\right) + \cos \left(\frac{A-B}{2}\right)$$
It looks correct to me but I would appreciate if someone confirm it.
| WLOG let $C$ be the largest angle, let
$x=\frac{A-B}{2}, y=\frac{A+B}{2}$ then $-\frac{\pi}{4}<x<\frac{\pi}{4}, 0<y<\frac{\pi}{2}$.
$$
A = x+y,~~ B=y-x,~~ C = \pi-2y \\
\cos\left(\frac{B-C}{2}\right) = \sin\left(\frac{3y-x}{2}\right), ~~
\cos\left(\frac{C-A}{2}\right) = \sin\left(\frac{3y+x}{2}\right) \\
\sin\left(\frac{3y-x}{2}\right)+\sin\left(\frac{3y+x}{2}\right) = 2\cos(x/2)\sin(3y/2) \\
\cos(x/2)\ge \cos x>\frac{1}{\sqrt{2}}, ~\sin(3y/2)>0 \Rightarrow 2\cos(x/2)\sin(3y/2)\ge2\cos x\sin(3y/2)
$$
Then we can write
$$
\begin{align}
3 \!&+\cos\left(\frac{A-B}{2}\right)+\cos\left(\frac{B-C}{2}\right)+\cos\left(\frac{C-A}{2}\right) -4(\cos A + \cos B + \cos C) \\
& = 3 + \cos x + 2\cos(x/2)\sin(3y/2)-4(\cos(x+y)+\cos(y-x)-\cos2y) \\
& = -1+\cos x+2\cos(x/2)\sin(3y/2)-8\cos x\cos y+8\cos^2y \\
& \ge 8\cos^2y-1+\cos x(1+2\sin(3y/2)-8\cos y) \\
& = 8\cos^2y-1+\cos x \cdot f(y)
\end{align}
$$
where we define $f(y)=1+2\sin(3y/2)-8\cos y$.
Writing $v=y/2$ we can work through
$$
\begin{align}
8\cos^2y-1+f(y) & = 7-8\sin^22v+1+2\sin3v-8\cos 2v\\
& = 8-16\sin^2v(1-2\sin^2v)+6\sin v-8\sin^3v-8(1-2\sin^2v) \\
& = 32\sin^4v-8\sin^3v-16\sin^2v+6\sin v \\
& = 2\sin v(1-2\sin v)^2(3+4\sin v)
\end{align}
$$
This last expression is clearly $\ge 0$ when $\sin v\ge 0$, but to answer the original question we now need to consider cases.
If $f(y)\le 0$, which corresponds to $0<y\le R=1.192797\cdots$ then
$$
8\cos^2y-1+\cos x\cdot f(y)\ge 8\cos^2y-1+f(y) \ge 0
$$
as we just described.
On the other hand if $R<y<\pi/2, ~f(y)>0$ then since $|x|<\pi/4$
$$
\begin{align}
8\cos^2y-1+\cos x\cdot f(y) & \ge 8\cos^2y-1+\frac{1}{\sqrt{2}}f(y)\\
& = 8\cos^2y-1+f(y)-f(y)\left(1-\frac{1}{\sqrt{2}}\right) \\
& = 2\sin v(1-2\sin v)^2(3+4\sin v)-f(y)\left(1-\frac{1}{\sqrt{2}}\right)
\end{align}
$$
I don't have an algebraic way to proceed from here, but it is easy to check numerically that the last expression is positive for $y$ in the desired range, which establishes the original inequality in $A,B,C$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding a pair of elements to satisfy an inequation Let $F$ be a field of characteristic 2 with more than 2 elements. Show that there are elements $a$ and $b$ in $F$ such that $(a+b)^3 \not= a^3 + b^3$.
$F$ couldn't possibly have less than 2 elements, and if it had exactly 2 — that is, $F = \mathbb Z_2$ —, $(a+b)^3$ would actually always be $a^3+b^3$. With $\#F>2$, how do I find $a$ and $b$ to violate that?
All I could do so far is simplify the inequation, using the characteristic and the commutative prroperty:
$$\begin{align}
(a+b)^3 &\not= a^3 + b^3\\
a^3 + 3a^2b + 3ab^2 + b^3 &\not= a^3 + b^3\\
a^3 + a^2b + ab^2 + b^3 &\not= a^3 + b^3\\
a^2b + ab^2 &\not= 0\\
ab(a + b) &\not= 0\quad.
\end{align}$$
| It might help to know the addition and multiplication tables of $F_4$, the smallest extension of the simplest Galois field of characteristic 2.
Suppose $F_4 = \{0, 1, \alpha, \alpha+1\}$, then addition works as you might expect, and multiplication is such that $\alpha^2 = \alpha + 1$. This lets you work out the full addition and multiplication tables, and now it's child's play to pick $a, b$ such that $ab(a+b) \neq 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Determining and (dis)proving if $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $ converges I am trying to determine if $ \sum_{n = 1}^{\infty} (-1)^{n + 1} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) $ converges using an alternating series test. The test in question requires me to prove $ 1 - n \log \left( \frac{n + 1}{n} \right) $ is decreasing and that $ \lim_{n \to \infty} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) = 0 $ to prove this series is convergent.
My intutition says that this series is convergent because $ n \log \left( \frac{n + 1}{n} \right) $ will tend towards 1 as n goes to infinity (due to the definition of e). I am having trouble proving the sequence $ 1 - n \log \left( \frac{n + 1}{n} \right) $ is decreasing.
I set up the inequality $ 1 - (n + 1) \log \left( \frac{n + 2}{n + 1} \right) \leq 1 - n \log \left( \frac{n + 1}{n} \right) $ but I feel like I am stuck on some simple algebra. Any hints for proving the sequence is nonincreasing? Or am I just wrong?
EDIT: Could this be done using absolute convergence?
EDIT2: I am seeing some really great answers, but I am trying to prove this without calculus. (No derivatives or Taylor series expansions.)
| Notice that $$\lim_{n\to\infty} n \ln\left( \frac{n + 1}{n} \right)=\lim_{n\to\infty} \ln\left( 1+ \frac{1}{n} \right)^n=\ln e=1.$$
So $\lim_{n \to \infty} \left( 1 - n \log \left( \frac{n + 1}{n} \right) \right) = 0$ as required and we are left with monotonicity:
$$
1- (n + 1) \log \left( \frac{n + 2}{n + 1} \right) \leq 1- n \log \left( \frac{n + 1}{n} \right)\\
\log \left( \frac{n + 2}{n + 1} \right)^{n+1} \geq \log \left( \frac{n + 1}{n} \right)^n\\
\left( \frac{n + 2}{n + 1} \right)^{n+1} \geq \left( \frac{n + 1}{n} \right)^n\\
\left( 1+ \frac1{n+1} \right)^{n+1} \Biggr/ \left(1+ \frac1n \right)^n \geq 1\\
\left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)^n \geq \frac{n+1}{n+2}\\
$$
Now use $ \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)^n \geq \left(\left( 1+ \frac1{n+1} \right) \Biggr/ \left(1+ \frac1n \right) \right)=\frac{n(n+2)}{(n+1)^2}$ to conclude:
$$
\frac{n(n+2)}{(n+1)^2} \geq \frac{n+1}{n+2}\\
\frac{n(n+2)^2}{(n+1)^3}=\frac{n^3+4n^2+4n}{n^3+3n^2+3n+1} \geq 1\; ,\\
$$
for $n\geq 1$, which is not elegant, I know.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $ How would I verify the following double angle identity.
$$
\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B
$$
So far I have done this.
$$
(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)
$$But I am not sure how to proceed.
| I will prove the result by starting with the right hand side of the identity:
$$\begin{align}\sin^2A-\sin^2B&=(\sin A+\sin B )(\sin A-\sin B)\\
&=(2\sin\frac{A+B}{2}\cos\frac{A-B}{2})(2\sin\frac{A-B}{2}\cos\frac{A+B}{2})\\
&=(2\sin\frac{A+B}{2}\cos\frac{A+B}{2})(2\sin\frac{A-B}{2}\cos\frac{A-B}{2})\\
&=\sin(A+B)\sin(A-B)
\end{align}$$
as required, on using the sum to product formulae in the second line of working and the double angle formulae in the final line. I hope you found that interesting :)
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}$ I have to prove this most difficult trigonometric identity.
$$\tan(A+B+Y)=\frac{\tan A+\tan B+\tan Y-\tan A\tan B\tan Y}{1-\tan A \tan B-\tan B\tan Y-\tan Y\tan A}.$$
I know
$$\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$$
My problem is with the extra $Y$ in this problem. What can I do about I think I know a solution which is to do $\tan(A+B)$ then $\tan(B+Y)$ but I am not sure how to apply it.
| Just use the sum formula twice:
$$
\begin{align*}
\tan(A+B+Y)=\tan((A+B)+Y) &= \frac{\tan(A+B)+\tan Y}{1-\tan(A+B)\tan Y}\\
&=\frac{\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)+\tan Y}{1-\left(\frac{\tan A +\tan B}{1-\tan A\tan B}\right)\tan Y}\\
&= \frac{(\tan A + \tan B)+ \tan Y(1-\tan A\tan B)}{(1-\tan A\tan B)-(\tan A + \tan B)\tan Y}\\
&=\frac{\tan A + \tan B + \tan Y - \tan A\tan B\tan Y}{1-\tan A\tan B-\tan A\tan Y-\tan B\tan Y}
\end{align*}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Cauchy principal value integral Doing numerical integration I encountered the following calculation of the Cauchy principal value of the following integral:
$$\int_0^{\pi/2}d\varphi\frac{1}{\cos(\varphi)-c},$$
with $0<c<1$.
What is the strategy to solve it?
| The indefinite integral is elementary, can be found using half-angle substitution:
$$
\int \frac{\mathrm{d} \varphi}{\cos(\varphi)-c} = \frac{2}{\sqrt{1-c^2}} \operatorname{arctanh}\left(\frac{1+c}{\sqrt{1-c^2}} \tan\frac{\varphi}{2} \right)+C =: F(\varphi)
$$
The Cauchy principal value is found as a symmetric limit:
$$
\operatorname{P.V.} \int_0^\frac{\pi}{2} \frac{\mathrm{d} \varphi}{\cos(\varphi)-c} = \lim_{\epsilon\to 0} \left(F\left(\frac{\pi}{2}\right) - F\left( \arccos(c)+\epsilon\right) + F\left( \arccos(c)-\epsilon\right) - F(0)\right)
$$
Easily
$$
F\left(\frac{\pi}{2}\right) - F(0) = \frac{2}{\sqrt{1-c^2}} \operatorname{arctanh}\left(\sqrt{\frac{1+c}{1-c}}\right) = \frac{2}{\sqrt{1-c^2}} \operatorname{arctanh}\left(\sqrt{\frac{1-c}{1+c}}\right) - \frac{i \pi}{\sqrt{1-c^2}}
$$
and, using $\tan\left(\frac{1}{2} \arccos(c)\right) = \sqrt{\frac{1-c}{1+c}}$
$$
\lim_{\epsilon\to 0} \left( F\left( \arccos(c)-\epsilon\right) - F\left( \arccos(c)+\epsilon\right) \right) = \frac{2}{\sqrt{1-c^2}} \lim_{\epsilon \to 0} \left( \operatorname{arctanh}\left(1+\frac{\epsilon}{\sqrt{1-c^2}}\right) - \operatorname{arctanh}\left(1-\frac{\epsilon}{\sqrt{1-c^2}}\right) \right) = \frac{i \pi}{\sqrt{1-c^2}}
$$
Combining,
$$
\operatorname{P.V.} \int_0^\frac{\pi}{2} \frac{\mathrm{d} \varphi}{\cos(\varphi)-c} = \frac{2}{\sqrt{1-c^2}} \operatorname{arctanh}\left(\sqrt{\frac{1-c}{1+c}}\right) = \frac{1}{\sqrt{1-c^2}} \log \frac{1+\sqrt{1-c^2}}{c}
$$
| {
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$\sqrt{x} +y = 4$, $x+ \sqrt{y}= 6$, find the solution $(x,y)$ $\sqrt{x} +y = 4$, $\sqrt{y} +x= 6$, find the solution (x,y). $NOTE$ : $\sqrt{4}+1= 4-1$, $\sqrt{1} +4 =1+4$
| Let, $\sqrt{x}=s$, $\sqrt{y}=t$
we have, $s^4 -12s^2+s+32=0$, which is a 'biquadratic' equation of the form,
$$(s^2+ks+l)(s^2-ks+m)=0$$
i.e. $$s^4 -12s^2+s+32=(s^2+ks+l)(s^2-ks+m)$$
now by equating coefficients, we have
$$l+m-k^2 = -12, k(m-l) = 1, lm = 32$$
from the first two of these equations, we obtain
$$2m=k^2-12+(1/k), 2l=k^2-12-(1/k)$$
hence substituting in the third equation, the values of l,m,
$$(4)(32)=(k^2-12-1/k)(k^2-12+1/k)$$
$$128=(k^2-12)^2-1/k^2$$
$$128=k^4+144-24k^2-1/k^2$$
$$k^6-24k^4+16k^2-1=0$$
this is a cubic in $k^2$ which always has one real positive solution and we can find $k^{2}$,$l$,$m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/180520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_0^\infty (u\log(a^2+u^2))/(e^u-1)\, du$ Does the following definite integral have a known "closed form" value?
$$\int_0^\infty\frac{u\log(a^2+u^2)}{e^u-1}~du,$$
or can anyone see a way to integrate it?
| I was only able to find the following formula: for $a = 2\pi\alpha$,
$$\begin{align*}
\int_{0}^{\infty} \frac{u \log(4\pi^2\alpha^2+u^2)}{e^u - 1} \; du
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\
&\quad + \pi^2 \alpha \left(2\alpha \log\alpha - \alpha + 2 - 4 \log\Gamma(\alpha+1)\right) \\
&\quad + 4\pi^2 \int_{0}^{\alpha} \log\Gamma(u+1)\;du.
\end{align*}$$
Indeed, let
$$I(a) = \int_{0}^{\infty} \frac{u \log(a^2+u^2)}{e^u - 1} \; du.$$
Then we have
$$\begin{align*}I(0)
& = 2 \int_{0}^{\infty} \frac{u \log u}{e^u - 1} \; du = 2 \left. \frac{d}{ds}\zeta(s)\Gamma(s) \right|_{s=2} \\
& = 2(\zeta'(2) + \zeta(2)\psi_0(2)) = 2\left(\zeta'(2) + \frac{\pi^2}{6}(1-\gamma)\right).
\end{align*}$$
Also,
$$\begin{align*}
I'(a)
&= \int_{0}^{\infty} \frac{2au}{u^2+a^2} \frac{du}{e^u-1} \\
&= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} e^{-nu} \left(\int_{0}^{\infty} \sin(ux)e^{-ax}\;dx\right)\;du \\
&= 2a \int_{0}^{\infty} \sum_{n=1}^{\infty} \left(\int_{0}^{\infty} \sin(xu)e^{-nu}\;du\right)e^{-ax}\;dx \\
&= 2a \int_{0}^{\infty} \left(\sum_{n=1}^{\infty} \frac{x}{x^2+n^2}\right)e^{-ax}\;dx \\
&= a \int_{0}^{\infty} \left(\pi\coth(\pi x) - \frac{1}{x}\right)e^{-ax}\;dx
\end{align*}$$
Proceeding,
$$\begin{align*}
I'(a)
&= a \left[ \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \right]_{0}^{\infty} + a^2 \int_{0}^{\infty} \left(\log\sinh(\pi x)-\log x \right)e^{-ax} \; dx \\
&= -a\log\pi + a^2 \int_{0}^{\infty} \left(\pi x + \log\left(1 - e^{-2\pi x}\right)-\log2 + \log a -\log ax \right)e^{-ax} \; dx \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right)+a^2 \int_{0}^{\infty} e^{-ax}\log\left(1 - e^{-2\pi x}\right) \; dx \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a^2 \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n} e^{-(a+2n\pi)x} \; dx \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \sum_{n=1}^{\infty} \frac{\frac{a}{2\pi}}{n\left(n + \frac{a}{2\pi}\right)} \\
&= \pi + a\left(\gamma + \log\left(\frac{a}{2\pi}\right)\right) - a \left(\gamma + \psi_0 \left( \frac{a}{2\pi} +1 \right)\right) \\
&= \pi + a\left(\log\left(\frac{a}{2\pi}\right) - \psi_0 \left( \frac{a}{2\pi} +1 \right)\right).
\end{align*}$$
Thus integrating,
$$\begin{align*}
I(a)
&= I(0) + \int_{0}^{a} I'(t) \; dt \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + \pi a + \int_{0}^{a} t \left(\log\left(\frac{t}{2\pi}\right) - \psi_0 \left( \frac{t}{2\pi} +1 \right)\right) \; dt \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) + 2 \pi^2 \alpha + 4\pi^2 \int_{0}^{\alpha} u \left(\log u - \psi_0 (u+1)\right) \; du \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\
&\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \int_{0}^{\alpha} u \psi_0 (u+1) \; du \\
&= 2\zeta'(2) + \frac{\pi^2}{3}(1-\gamma) \\
&\quad + \pi^2 \alpha \left(2 \alpha \log \alpha - \alpha + 2 \right) - 4\pi^2 \left[ u \log\Gamma(u+1) \right]_{0}^{\alpha} \\
&\quad + 4\pi^2 \int_{0}^{\alpha} \log\Gamma(u+1) \; du,
\end{align*}$$
which gives the desired result. You may verify this formula numerically with the following Mathematica code
a = 8;
NIntegrate[(u Log[4 Pi^2 a^2 + u^2])/(Exp[u] - 1), {u, 0, Infinity}]
2 Zeta'[2] + Pi^2/3 (1 - EulerGamma) +
Pi^2 a (2 a Log[a] - a + 2 - 4 LogGamma[1 + a]) +
4 Pi^2 NIntegrate[LogGamma[1 + u], {u, 0, a}]
Clear[a];
In fact, we can give a neater form as follows:
$$ \begin{align*}
\int_{0}^{\infty} \frac{u \log(a^2+u^2)}{e^{2\pi u} - 1} \; du
&= \zeta'(-1) + \frac{a}{2}\left(1 - \frac{a}{2}\right) + \frac{a^2}{2}\log a - a \log\Gamma(a+1)\\
&\quad + \int_{0}^{a} \log\Gamma(t+1) \; dt.
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Inequality. $a^2+b^2+c^2 \geq a+b+c$ Let $a,b,c$ be positive real numbers such that $abc=1$. Prove that
$a^2+b^2+c^2 \geq a+b+c$.
Thanks
| We have:
$$\sqrt{\frac{a^2+b^2+c^2}{3}}\underset{Q\geq G}{\geq} \sqrt[3]{abc} = 1$$
But $\sqrt{x} \geq 1 \iff x \geq 1 \iff x\geq \sqrt{x}$ so:
$$\frac{a^2+b^2+c^2}{3} \geq\sqrt{\frac{a^2+b^2+c^2}{3}}\underset{Q\geq A}{\geq}\frac{a+b+c}{3}$$
$$a^2+b^2+c^2 \geq a+b+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 8,
"answer_id": 6
} |
Solve / Simplify for x $$\frac{1}{\sqrt{a^{2}-x^{2}}}+\frac{1}{\sqrt{b^{2}-x^{2}}}-\frac{1}{c}=0$$
Hello
I'm wondering whether anyone can help me rearrange this to solve for $x$, where $a$, $b$, $c$ are constants. I initially thought about a trig substitution $x=b\cdot \cos(z)$ but that just lead to another brick wall. Any insights gratefully appreciated!
Thanks
Guy
| Put $1/c$ on one side of the equation, square both sides, separate out the square root and do again so that
$x$ is a root of $$\begin{multline}x^8+(-2b^2-2a^2+2c^2+2a^2c^2)x^6\\+(-4b^2a^2c^2+a^4-2a^4c^2-4a^2c^2-2b^2c^2+a^4c^4\\+b^4+4b^2a^2-2a^2c^4+c^4)x^4\\+(4b^2a^2c^2+2b^2a^2c^4-2a^4c^4b^2+4a^4c^2b^2+2b^4a^2c^2-2b^4a^2\\-2a^4b^2-2a^2c^4+2a^4c^2+2a^4c^4)x^2\\-2a^4c^4b^2+b^4a^4c^4-2b^4a^4c^2-2a^4c^2b^2+b^4a^4+a^4c^4\end{multline}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/182788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Using the sum of squares formula to solve more complex sums. I'm studying integration and trying to figure out how to use the sum of squares formula to solve more complicated sums.
For example: knowing that $$\sum_{i=1}^n i^2 = \frac{n (n+1) (2 n+1)}{6}$$
how can we simplify
$$\sum_{i=1}^n (i/n)^2$$
| The key here is what anon wrote in his comment. It's helpful to expand out the terms of the sum:
$$ \begin{eqnarray}
\sum_{i=1}^n (i/n)^2 & = & \sum_{i=1}^n \frac{i^2}{n^2} \\
&=& \frac{1^2}{n^2} + \frac{2^2}{n^2} + \dots + \frac{n^2}{n^2} \\
&=& \Big( \frac{{1^2} + {2^2} + \dots + {n^2}} {n^2} \Big) \\
& = & \frac{1}{n^2} \sum_{i=1}^n i^2 \\
& = & \frac{1}{n^2} \frac{n (n+1) (2 n+1)}{6}
\end{eqnarray} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that: $\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $ If $a\geq0$, $b\geq 0$ then the following inequality holds:
$$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1 $$
There are at least three things to try here:
a). Use AM-GM for the denominator of the second fraction, $(a+b)\le \frac{(a+b+1)^2}{4}$
b). Use the fact that $\frac{b}{(a+b+1)^2}\le\frac{b}{(a+b+1)(a+b)}$
c). Consider that $a\geq b$ or $b\geq a$ simply or combined with a). and\or b).
No one of these attempts has brought me to a nice, simple solution and I'm trying to see if there is something around that missed me. This is an inequality given at a high school competition math.
I'd appreciate to receive your valuable feedback in terms of this question. Thanks.
| To prove:$$\frac{a}{a+1}+\frac{b}{(a+b+1)^2}+\frac{1}{a+b+1}\le1$$
Prove:$$\frac{b}{(a+b+1)^2}+\frac{a+b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$
$$\frac{a+2b+1}{(a+b+1)^2}\le\frac{1}{a+1}$$
$$\color{green}{a^2}+\color{blue}{2ab}+\color{red}{a+a+2b+1}\le\color{green}{a^2}+b^2+\color{blue}{2ab}+\color{red}{2a+2b+1}$$
$$0\le b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
How to construct minimal polynomial? This is an exam question from last semester.
We have the finite field
$$ \mathbb F_{81} = \mathbb Z_3 [x]/(x^4+x^2+x+1)$$
(a) Prove that the polynomial $$ x^4+x^2+x+1 $$
is irreducible
(b) Construct the minimal polynomial of the element $$ x^3+x^2+x+1 \space\epsilon\space Z_3 [x]/(x^4+x^2+x+1)$$
Use y as a formal variable in this polynomial. Hint: using $$ x^3+x^2+x+1 = (x^2+1)(x+1) $$ should help with the calculations.
(c) Construct the subfield F9 in $$ Z_3 [x]/(x^4+x^2+x+1)$$
I tried a and I think you can prove it by showing the polynomial has no Zeros? So assuming we call the polynomial g(x). I just filled in {0,1,2} and none of them gave 0 --> You can't split up the polynomial in polynomials of lower orders -> it's irreducible?
I don't know how to do b and c though.
Can someone please tell me how to do it in general and what the solution is here? Really need the answer.
| I will answer only (b).
By abuse of notation, we write $x$ for the image of $x$ in $\mathbb{Z}_3[x]/(x^4 + x^2 + x + 1)$.
Let $y = x^3 + x^2 + x + 1$.
Since $y = (x^2 + 1)(x + 1)$, $y^2 = (x^2 + 1)^2(x + 1)^2$.
Hence $y^2 = (x^4 + 2x^2 + 1)^2(x + 1)^2 = (-x + x^2)(x + 1)^2 = x(x - 1)(x + 1)^2 = x(x^2 - 1)(x + 1) = x(x^3 + x^2 - x - 1) = x^4 + x^3 -x^2 - x$
$= x^3 - 2x^2 -2x - 1 = x^3 + x^2 + x - 1$.
Hence $y^2 + 2 = y$.
Since $y$ is not contained in $\mathbb{F}_3$, $Y^2 - Y - 1$ is the minimal polynomial of $y$ over $\mathbb{F}_3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/186056",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.
| The claim is false, for example
$$n=2\Longrightarrow n^4+4n^2+11=16+16+11=43$$
which is not a multiple of 16. Check your expression.
Now, if $\,n=2k+1\,$ is odd, then the claim is true, since then
$$n^4+4n^2+11=8k(k+1)(2k^2+2k+3)+16$$
and since $\,8k(k+1)=0\pmod {16}\,$ no matter what parity $\,k\,$ has, we're done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
How to compute this limit related to series? How to compute $\lim_{N\rightarrow+\infty}\frac{\ln^2N}{N}\sum_{k=1}^{N-1}\left(\frac{k}{N}\right)^{2\ln N-2}$? thank you.
| Because $f(x) =x^{2 \ln(n) - 2}$ is increasing function of $x$ for $n \geqslant 3$, we have
$$
\int_0^n \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} =
\sum_{m=0}^{n-1} \int_{m}^{m+1} \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} \lt \frac{1}{n} \sum_{m=0}^{n-1} \left( \frac{m+1}{n} \right)^{2 \ln(n) - 2}
$$
Thus:
$$
\frac{\ln^2(n)}{n} \sum_{k=1}^{n-1} \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \gt \ln^2(n) \left( \int_0^n \left( \frac{k}{n} \right)^{2 \ln(n) - 2} \frac{\mathrm{d} k}{n} - 1\right) = \ln^2(n) \left( \frac{1}{2 \ln(n) - 1} -\frac{1}{n} \right)
$$
Now,
$$
\lim_{n \to \infty } \frac{\ln^2(n)}{n} \sum_{k=1}^{n-1} \left( \frac{k}{n} \right)^{2 \ln(n) - 2} > \lim_{n \to \infty} \ln^2(n) \left( \frac{1}{2 \ln(n) - 1} -\frac{1}{n} \right) = \infty
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Prove that $\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$ let $ABC$ be an acute triangle with all angles greater than $45^o$
Prove that
$$\frac{2}{1+\tan A}+\frac{2}{1+\tan B}+\frac{2}{1+\tan C} \le 3(\sqrt{3}-1)$$
I let $\tan A=a$, $\tan B=b$, $\tan C=c$ with $a+b+c=abc$ then the inequality equivalent to $\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c} \le 3(\sqrt{3}-1)$ but it's not work.
I don't known what to do. Algebra or trigonometric?
| Let's use Lagrange multiplier method to maximize
$$
f(a,b,c) = \frac{2}{1+a} + \frac{2}{1+b} + \frac{2}{1+c}
$$
subject to the constraint $0 = g(a,b,c) = a+b+c-a b c$. Derivatives of $f - \lambda g$ read:
$$ \begin{eqnarray}
\frac{\partial(f- \lambda g)}{\partial a} &=& -\frac{2}{(1+a)^2} + \lambda (b c -1) = \frac{-2+ \lambda (b c-1)(1+a)^2}{(1+a)^2} = 0\\
\frac{\partial(f- \lambda g)}{\partial b} &=& -\frac{2}{(1+b)^2} + \lambda (a c -1) = \frac{-2+ \lambda (a c-1)(1+b)^2}{(1+b)^2} = 0\\
\frac{\partial(f- \lambda g)}{\partial c} &=& -\frac{2}{(1+c)^2} + \lambda (a b -1) = \frac{-2+ \lambda (a b-1)(1+c)^2}{(1+c)^2} = 0
\end{eqnarray}
$$
Eliminating out $\lambda$ this results into two equation:
$$
(1+a)^2 (b c-1) = (1+b)^2 (a c-1) , \quad (1+b)^2 (a c-1) = (1+c)^2 (a b-1)
$$
substituting $c = \frac{a+b}{a b-1}$ we get:
$$
\frac{(1+a)^2 (1+b^2}{a b -1} = \frac{(1+a^2)(1+b)^2}{a b -1}, \quad
\frac{(1+a^2)(1+b)^2}{a b -1} = \frac{\left(a b + a +b -1\right)^2}{a b -1}
$$
Subtracting:
$$
\frac{2 (b-a)(a b -1)}{a b-1} = 0, \quad \frac{4 b - 2 a(b^2-1)}{a b -1} = 0
$$
Giving $a=b$ and $a^2 = 3$. Thus $a=b=c=\sqrt{3}$ is the extremum. It is easy to see that this is a maximum. The inequality, we see, is saturated at $a=b=c=\sqrt{3}$, corresponding to $A=B=C=\frac{\pi}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 1
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I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
| Here is a closed form solution to your recurrence relation obtained by Maple,
$$ s(n)={n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}s \left( 1
\right) +\frac{{n}^{3}}{3}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}
\left( \left( -1 \right)^{{\frac {\ln \left( n \right) }{\ln
\left( 2 \right) }}} \right)^{-1}+\frac{{n}^{2}}{2}{n}^{{\frac {i\pi }{
\ln \left( 2 \right) }}} \left( \left( -1 \right) ^{{\frac {\ln
\left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1}+\frac{1}{6}\,n{n}^
{{\frac {i\pi }{\ln \left( 2 \right) }}} \left( \left( -1 \right) ^{
{\frac {\ln \left( n \right) }{\ln \left( 2 \right) }}} \right) ^{-1
}-{n}^{2}{n}^{{\frac {i\pi }{\ln \left( 2 \right) }}}
\,$$
Here is a more compact form
$$ s(n) = \left( {n}^{2}\cos \left( {\frac {\pi \,\ln \left( n \right) }{\ln
\left( 2 \right) }} \right) +i{n}^{2}\sin \left( {\frac {\pi \,\ln
\left( n \right) }{\ln \left( 2 \right) }} \right) \right) s
\left( 1 \right) -{n}^{2}\cos \left( {\frac {\pi \,\ln \left( n
\right) }{\ln \left( 2 \right) }} \right) +\frac{{n}^{3}}{3}+\frac{{n}^{2}}{2}
+\frac{n}{6}-i{n}^{2}\sin \left( {\frac {\pi \,\ln \left( n \right) }{\ln
\left( 2 \right) }} \right) \,.$$
where $s(1)$ is your initial condition. If you plug in $s(1)=1$ in the above formula you get the simple formula, just as it has been mentioned in the comments,
$$ \frac{n}{6} \left( n+1 \right) \left( 2\,n+1 \right) \,,$$
which is equal to $ \sum_{i=1}^{n} i^2 $.
Note
If you are interested only in finding sums of the form $ \sum_{i=1}^{n} i^m \,, m=1,2,3,\dots $, then they are simple techniques to find them. See here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of equation as x tends to -1 I was given the following expression and had to find the limit as: $$ x \rightarrow 1, x \rightarrow - 1, x \rightarrow \infty $$
$$ \lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1} = \lim_{x \to -1} \frac{\frac{x^2}{x^2} + \frac{3x}{x^2} + \frac{2}{x^2}}{\frac{x^2}{x^2} - \frac{1}{x^2}} = \lim_{x \to -1} \frac{\frac{1}{1} + 0 + 0}{\frac{1}{1} - 0} = \lim_{x \to -1} \frac{1}{1} = 1
$$
So for $-1$, I got 1. However the text book says it's $1/2$. I tried pluging in -1 but I don't get $1/2$, no matter how I shift this.
| $$ \lim_{x \to -1} \frac{x^2 +3x +2}{x^2 -1}$$
$$=\lim_{x \to -1} \frac{(x+1)(x+2)}{(x+1)(x-1)} $$
$$=\lim_{x \to -1} \frac{(x+2)}{(x-1)} $$ $$\text{as}:x \to -1, x≠-1$$
$$=\frac{-1+2}{-1-1}=-\frac{1}{2}$$
Now $ \lim_{x \to 1} \frac{x^2 +3x +2}{x^2 -1}$
$=\lim_{x \to 1} \frac{(x+2)}{(x-1)} $ as $\lim_{x \to 1}, x≠-1$
$\lim_{x \to 1^{+}} \frac{x^2 +3x +2}{x^2 -1}=\infty$
$\lim_{x \to 1^{-}} \frac{x^2 +3x +2}{x^2 -1}=-\infty$
SO, the limit does not exist at $x=1$(as identified by Quintofron )
Now $ \lim_{x \to \infty} \frac{x^2 +3x +2}{x^2 -1}$
$=\lim_{x \to \infty} \frac{(x+2)}{(x-1)} $ as $\lim_{x \to \infty}, x≠-1$
$=\lim_{x \to \infty}\frac{1+\frac{2}{x}}{1-\frac{1}{x}}=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Proving an equality involving compositions of an integer Let's consider various representations of a natural number $n \geq 4$ as a sum of positive integers, in which the order of summands is important (i.e. compositions). The task is to prove the number $3$ appears altogether $n2^{n-5}$ times in all of them.
I know there're $2^{n-1}$ compositions of $n$. However, I have no clue as to how to count only those involving the number(s) $3$. I can't think of any sensible generating function for this. Maybe there's a nice combinatorial interpretation of the given formula, which I can't figure out? Could anyone lend me a hand with handling this problem?
| It appears that the generating function approach is quite simple here. We have by inspection that the bivariate generating function of compositions with the number three marked is
$$M(z, u) = \sum_{q\ge 1}\left(\frac{z}{1-z} - z^3 + uz^3\right)^q.$$
Therefore the generating function of the total number of ocurrences is
$$\left.\frac{d}{du} M(z, u)\right|_{u=1}
= \left. \sum_{q\ge 1} q \left(\frac{z}{1-z} - z^3 + uz^3\right)^{q-1} \times z^3
\right|_{u=1}
\\= z^3 \sum_{q\ge 1} q \left(\frac{z}{1-z}\right)^{q-1}
= z^3 \frac{1}{(1-z/(1-z))^2}
= z^3 \frac{(1-z)^2}{(1-2z)^2}
\\ = z^3 \left(\frac{1}{1-2z} + \frac{z^2}{(1-2z)^2}\right).$$
To conclude extract coefficients, getting
$$[z^{n-3}] \left(\frac{1}{1-2z} + \frac{z^2}{(1-2z)^2}\right)
= 2^{n-3} + [z^{n-5}] \frac{1}{(1-2z)^2}
\\ = 2^{n-3} + (n-4) 2^{n-5} = 4 \times 2^{n-5} + (n-4) 2^{n-5} = n 2^{n-5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/191165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
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