integration - Evaluating a trigonometric integral by means of contour $int_0^{pi} frac{cos(4theta)}{1+cos^2(theta)} dtheta$

I am studying for a qualifying exam, and this contour integral is getting pretty messy:

$\displaystyle I = \int_0^{\pi} \dfrac{\cos(4\theta)}{1+\cos^2(\theta)} d\theta $

I first notice that the integrand is an even function hence

$\displaystyle I = \dfrac{1}{2} \int_{-\pi}^{\pi} \dfrac{\cos(4\theta)}{1+\cos^2(\theta)} d\theta $

Then make the substitutions $\cos(n\theta) = \dfrac{e^{in\theta}+e^{-in\theta}}{2}$, and $z=e^{i\theta}$ to obtain:

$\displaystyle I = \dfrac{1}{2} \int_{||z||=1} \dfrac{\dfrac{z^4+z^{-4}}{2}}{1+\left(\dfrac{z+z^{-1}}{2}\right)^2} \dfrac{-i}{z}dz = -i \int_{||z||=1} \dfrac{z^8+1}{z^3(z^4+6z^2+1)} dz$

Now, assuming this is right so far, this seems "straight-forward" in the sense that I know what to do. However, it gets ugly. I was hoping there might be a better method to evaluate this integral.

Thanks

Answer

Note that $\sin 4\theta$ is an odd function, so we can simplify to

$$\begin{align}
I &= \frac12 \int_{-\pi}^\pi \frac{e^{4i\theta}}{1 + \cos^2\theta}\,d\theta\\
&= \frac{1}{2i}\int_{\lvert z\rvert = 1} \frac{z^4}{1 + \left(\frac{z+z^{-1}}{2}\right)^2}\,\frac{dz}{z}\\
&= \frac{1}{2i}\int_{\lvert z\rvert = 1} \frac{z^5}{z^2+\frac14(z^2+1)^2}\,dz\\
&= -2i \int_{\lvert z\rvert = 1} \frac{z^5}{z^4 + 6z^2+1}\,dz.
\end{align}$$

That looks a little simpler to my untrained eye.

Then we need the zeros of the denominator, which are $\pm \sqrt{-3\pm\sqrt{8}}$, where the inner square root shall be the positive, and the outer can be either square root. The zeros inside the unit disk are $\zeta_\pm = \pm i\sqrt{3-\sqrt{8}}$, both are simple, so

$$\operatorname{Res}\left(\frac{z^5}{z^4+6z^2+1};\,\zeta_\pm\right) = \frac{\zeta_\pm^5}{4\zeta_\pm^3 + 12\zeta_\pm} = \frac{\zeta_\pm^4}{4\zeta_\pm^2 + 12} = \frac{(\sqrt{8}-3)^2}{4(\sqrt{8}-3)+12}=\frac{17-12\sqrt{2}}{8\sqrt{2}},$$

both residues are the same, and

$$I = 8\pi \frac{17-12\sqrt{2}}{8\sqrt{2}} = \pi\frac{17-12\sqrt{2}}{\sqrt{2}} = \frac{\pi}{24+17\sqrt{2}},$$

if I haven't miscalculated.

linear algebra - How to calculate this determinant?

How to calculate this determinant?

$$A=\begin{bmatrix}n-1&k&k&k&\ldots& k\\k&n-1&k&k&\ldots &k\\\ldots&\ldots&\ldots &&\ldots\\\\k&k&k&k&\ldots &n-1\\
\end{bmatrix}_{n\times n}$$

where $n,k\in \Bbb N$ are fixed.

I tried for $n=3$ and got the characteristic polynomial as $(x-2-k)^2(x-2+2k).$

How to find it for general $n\in \Bbb N$?

Answer

Here I've followed the same initial step as K. Miller. Instead of using a determinant identity I examine the eigenvalues $A$ and consider their product.

If $J$ denotes the $n\times n$ matrix of all $1$'s, then then eigenvalues of $J$ are $0$ with multiplicity $n-1$ and $n$ with multiplicity $1$. This can be seen by noting that $J$ has $n-1$ dimensional kernel and trace $n$.

Your matrix $A$ is exactly $kJ+(n-k-1)I$ where $I$ denotes the $n\times n$ identity matrix. The eigenvalues of $A$ are therefore $n-k-1$ with multiplicity $n-1$ and $nk+n-k-1$ with multiplicity $1$. The determinant of $A$ is then $(nk+n-k-1)(n-k-1)^{n-1}$.

number theory - Prove that $sqrt{5n+2}$ is irrational

I'm trying to follow this answer to prove that $\sqrt{5n+2}$ is irrational. So far I understand that the whole proof relies on being able to prove that $(5n+2)|x^2 \implies (5n+2)|x$ (which is why $\sqrt{4}$ doesn't fit, but $\sqrt{7}$ etc. does), this is where I got stuck. Maybe I'm overcomplicating it, so if you have a simpler approach, I'd like to know about it. :)

A related problem I'm trying to wrap my head around is: Prove that $\frac{5n+7}{3n+4}$ is irreducible, i.e. $(5n+7)\wedge(3n+4) = 1$.

Answer

Well, one way to say it is irrational is to see that $5n+2$ isn't an integer square for any $n\in\mathbb{Z}$ (it only finish in $2$ or $7$). The other way you're trying lets you the same ending $(p,q\in \mathbb{Z}, q\neq 0)$:
\begin{align*}
\sqrt{5n+2}=\frac{p}{q}&& \\
5n+2=\frac{p^2}{q^2} &&(1)\\
q^2(5n+2)=p^2 && (2)

\end{align*}

Let
$$p=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_t^{\alpha_t}$$ $$q=q_1^{\beta_1}q_2^{\beta_2}\dots q_s^{\beta_s}$$

where $p_i,q_j$ are primes and $\alpha_i,\beta_j$ are positive integers.

From $(1)$, $p^2/q^2$ is an integer (is equal to $5n+2$), so you have that the $q_i$ are certain primes $p_j$. Renaming the prime factors in a way such that $q_i=p_i$, you can let you the fraction (considering that $t>s$)

$$\frac{p^2}{q^2}=\frac{p_1^{2\alpha_1}p_2^{2\alpha_2}\dots p_t^{2\alpha_t}}{q_1^{2\beta_1}q_2^{2\beta_2}\dots q_s^{2\beta_s}}=p_1^{2(\alpha_1-\beta_1)}p_2^{2(\alpha_2-\beta_2)}\dots p_t^{2(\alpha_s-\beta_s)}\dots p_t^{2\alpha_t}=5n+2$$

then this implies that $5n+2$ is a square, but by the original statement, it can't be.

functions - $f:Xrightarrow X$ such that $f(f(x))=x$

Let $X$ be a metric space and $f:X\rightarrow X$ be such that $f(f(x))=x$, for all $x\in X$.

Then $f$

1. is one-one and onto;


2. is one-one but not onto;


3. is onto but not one-one;


4. need not be either.

From the given condition I have that $f^2=i$ where $i$ is the identity function. If $f$ itself is the identity function then the conditions are satisfied as well as $f$ is bijection. Is that the only such function or are there other possibilities ?

My guess is that it will be bijection i.e. option $1$ will be correct .

For see, if $$f(x_1)=y \ \text{and}\ f(x_2)=y \ \text{then} \ f(y)=x_1\ \text{and}\ f(y)=x_2$$ will be possible iff $x_1=x_2$. So this is injective.

Now an injection from a set to itself is trivially surjective so it is bijective. Is my proof correct?

calculus - Does an integral converge/diverge if its sum converges/diverges

I'm sure this has been asked many times before, sorry if it's a duplicate, but when googling this I mostly found instructions on how to do the integral test.

So I was given an integral $$\int_{1}^{\infty} f(x) dx$$
and asked to find whether it converges/diverges. I figured out that it diverges, and the next task was to find the sum

$$\sum_{n=1}^{\infty} f(n) $$ where $f(x)$ was the same expression in both tasks. So I concluded that since the integral diverges, the sum also diverges by the integral test. But assume I was given the sum of the series first, and let's say I were to use a limit-comparison test to figure out that the series diverges. If the next task was to calculate the integral, would it still hold to conclude that since the sum of the series $a_n$ diverges, then so does the integral, or does the implication not go both ways? I hope my question was clear, thanks in advance.

edit: Original $f(x)$ was $arctan(\frac{1}{x})$ and $f(n) = arctan(\frac{1}{n})$

Answer

If you let $f(x) = \sin(\pi x)$ you have a divergent integral but a convergent sum.

convergence divergence - How to prove an increasing sequence that converges is bounded above by its limit

I am trying to prove that an increasing sequence that converges to $ L$ is bounded above by its limit.
By using $a_n \le a_{n+1}$ and the definition of limit of a sequence, I can prove that for $\epsilon > 0$ , $ a_n \lt {L + \epsilon} $ for all $a_n$.
But is there a way to proceed to $ a_n \le L $ ? because I can't think of a case in which the former is true but the latter isn't.

Answer

HINT

You can easily show that if for some n $a_n>L$ then by definition of limit $a_n$ must decrease which is impossible.

You only need to formalize this idea by setting “assume exists n such that ...then by definition of limit...contradiction”.

Notably

• suppose $\exists n_1$ such that $a_{n_1}>L$ with $d=a_{n_1}-L>0$


• set $\epsilon=d$ by definition of limit must exists $n_2>n_1$ such that $|a_{n_2} -L|<\epsilon \implies a_{n_2}
  
  

elementary number theory - Proof of Fermat's Little Theorem using Primitive Roots

I just learned about primitive roots today, and then I thought of this proof of Fermat's Little Theorem. Seeing that most proofs of this theorem aren't simple, I think I'm either completely wrong in my application of primitive roots (must have missed something fundamental, having just learned about them), or primitive roots are extremely powerful. Which one is it?

Proof: We want to prove that if $\gcd(a,p)=1$, with $p$ a prime, then $a^{p-1}\equiv 1\pmod p$. Since $p$ is prime there exists a primitive root $\operatorname{mod}\, p$, say $j$. It is well known that we can write every least residue $\operatorname{mod}\, p$ as a power of $j$, so e.g. we can write $a\equiv j^k$. Thus, it suffices to prove that $a^{p-1}\equiv j^{k(p-1)}\equiv 1\pmod p$. But this is obvious because $j^{k(p-1)}\equiv (j^{p-1})^k\equiv (1)^k\equiv 1\pmod p$ since $\operatorname{ord}_p(j)=p-1$ by definition.

QED

Thanks for your help guys!