linear algebra - Can two matrices with the same characteristic polynomial have different eigenvalues?

The polynomial is $-\lambda^3+3\lambda -2$

which factorizes into ($\lambda-1$)($\lambda +1$)($\lambda -2$)

A matrix A has the above characteristic polynomial, and so its eigenvalues are 1, -1, and 2.

However, another matrix B, already in diagonal form, has the same characteristic polynomial, but with eigenvalues 1,1,-2, i.e., diagonal entries 1,1,-2.

Is this possible? Or have I gone wrong in my computations?

The problem statement does ask to show that the characteristic polynomials are the same but that the matrices A and B are not similar. So, perhaps I have found exactly what I needed, but it just seems weird...

Thanks,

Answer

$-\lambda^3+3\lambda - 2 = -(\lambda-1)^2(\lambda+2) \neq -(\lambda-1)(\lambda+1)(\lambda-2)$.

Power series convergence in boundary problem

Say I have a power series $\sum_{k=0}^\infty a_k x^k $ which converge uniformly on $\left[0, 1\right)$ . Now I need to prove that series $\sum_{k=0}^\infty a_k $ are convergent.

My idea is to use equivalent Cauchy form $
\forall \epsilon\ \exists N \text{ such that }\sup_\left[0, 1\right) |S_n(x)-S_m(x)|<\epsilon\quad \forall m,n\ge N\
$ where $S_n = \sum_{k=0}^n a_k x^k$. Because of continuity of $P(x) = |S_n(x)-S_m(x)|$ we see that $ \sup_\left[0, 1\right) P(x) = \sup_\left[0, 1\right] P(x) $ and this way it can be proved that $\sum_{k=0}^\infty a_k $ convergent as Cauchy sequence.

inequality - Prove through induction that $3^n > n^3$ for $n geq 4$

I'm new to induction and have not done induction with inequalities before, so I get stuck at proving after the 3rd step.

The question is:

Use induction to show that $3^n > n^3$ for $n \geq 4$.

I have so far:

Step 1:
Prove for $n=4$ (since question states this)
$3^4 > 4^3$

$81 > 64 $

which is true

Step 2:
Assume true for $n=k$

$3^k > k^3$

Step 3:
Prove for $n = k+1$

$3^{k+1} > (k+1)^3$

Here I expand to:

$3^k \cdot 3 > k^3 + 3k^2 + 3k + 1$

However I have no idea how to prove this.

Thanks for any help given

Answer

$3^{k+1}>3k^3$

We need $3^{k+1}>(k+1)^3 $

So, it sufficient to prove $3k^3>(k+1)^3\iff \left(1+\frac1k\right)^3<3$

For $k=3,\left(1+\frac1k\right)^3=\frac{64}{27}<3$

and $\left(1+\frac1{k+1}\right)^3<\left(1+\frac1k\right)^3$

$\implies \left(1+\frac1k\right)^3<3$ for $k\ge3$

limits - What is the breakpoint of a piecewise function?

For

$$
f(x)=\left\{\begin{matrix}
0, & x\leq -1\\
\sqrt{1-x^2},& -1 < x < 1 \\
x, & x\geq 1
\end{matrix}\right.
$$

My book says that the breakpoints are x = -1 and x = 1. How breakpoints are defined so that -1 and 1 is chosen?

gamma function - Sum of series $frac{ 4}{10}+frac{4cdot 7}{10cdot 20}+frac{4cdot 7 cdot 10}{10cdot 20 cdot 30}+cdots cdots$

Sum of the series

$$\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots $$

Sum of series $\frac {4}{10}+\frac {4\cdot7}{10\cdot20}+ \frac {4\cdot7\cdot10}{10\cdot20\cdot30}+\cdots$

But i want to solve it using Beta ganmma function.

My Process follows

Let $$S = \frac{1}{10}\bigg[\frac{ 4}{1}+\frac{4\cdot 7}{1\cdot 2}+\frac{4\cdot 7 \cdot 10}{1\cdot 2 \cdot 3}+\cdots \cdots \bigg]$$

Let $\displaystyle a_{n} = \prod^{n}_{k=1}(3k+1)=3^n \prod^{n}_{k=1}\bigg(k+\frac{1}{3}\bigg)$

$\displaystyle =3^n\Gamma\bigg(n+\frac{4}{3}\bigg)\cdot \frac{1}{\Gamma \bigg(\frac{4}{3}\bigg)}$

And Let $\displaystyle b_{n} = \prod^{n}_{k=1}k = \Gamma (n+1)$

So $\displaystyle S =\frac{1}{10} \sum^{\infty}_{n=1}3^n \frac{\Gamma \bigg(n+\frac{4}{3}\bigg)\cdot \Gamma \bigg(\frac{2}{3}\bigg)}{\Gamma \bigg(\frac{2}{3}\bigg)\Gamma \bigg(\frac{4}{3}\bigg)\cdot \Gamma(n+1)}$

using $\displaystyle \Gamma(x)\cdot \Gamma(1-x) = \frac{\pi}{\sin (\pi x)}$ and

$\displaystyle \frac{\Gamma (a) \Gamma(b)}{\Gamma(a+b)} =B(a,b)= \int^{1}_{0}x^{a-1}(1-x)^{b-1}dx$

So $\displaystyle S = \frac{1}{10}\sum^{\infty}_{n=1}3^n\int^{1}_{0}x^{n+\frac{1}{3}}(1-x)^{-\frac{1}{3}}dx$

$\displaystyle =\frac{1}{10}\int^{1}_{0}\frac{3x}{1-3x}\cdot \bigg(\frac{x}{1-x}\bigg)^{\frac{1}{3}}dx$

Now put $\displaystyle \frac{x}{1-x} = t\Rightarrow x = \frac{t}{1+t} = 1-\frac{1}{1+t}$

So $\displaystyle S = \frac{3}{10}\int^{\infty}_{0}\frac{t^{\frac{4}{3}}}{(2t-1)(1+t)^2}dt$

Could some Help me to solve it, Thanks

although it has solved Here

Answer

For a complex number $z$ such that $|z|<1$, let
$$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k+\frac{1}{3}}{k}\right)\,z^n\,.$$

Then, the required sum $S:=\displaystyle\sum_{n=1}^\infty\,\prod_{k=1}^n\,\left(\frac{3k+1}{10k}\right)$ satisfies
$$S=f\left(\frac{3}{10}\right)-1\,.$$
We claim that $f(z)=(1-z)^{-\frac43}$ for all complex numbers $z$ such that $|z|<1$. As a consequence,
$$S=\left(\frac{7}{10}\right)^{-\frac43}-1=\left(\frac{10}{7}\right)^{\frac43}-1\approx0.608926\,.$$

Note that, for $n=0,1,2,3,\ldots$,
$$\begin{align}t_n&:=\prod_{k=1}^n\,\left(\frac{k+\frac{1}{3}}{k}\right)\\&=\frac{\Gamma\left(n+\frac43\right)}{\Gamma\left(\frac43\right)\,\Gamma(n+1)}\\&=\frac{(n+1)\,\Gamma\left(n+\frac43\right)}{\Gamma\left(\frac43\right)\,\Gamma(n+2)}\,,\end{align}$$
where $\Gamma$ is the usual gamma function.
That is,
$$t_n=\frac{n+1}{\Gamma\left(\frac43\right)\,\Gamma\left(\frac23\right)}\,\text{B}\left(n+\frac43,\frac23\right)\,,$$

where $\text{B}$ is the usual beta function.
From the identity
$$\Gamma(x)\,\Gamma(1-x)=\frac{\pi}{\sin(\pi\,x)}\,,$$
we conclude that
$$\Gamma\left(\frac43\right)\,\Gamma\left(-\frac13\right)=\frac{\pi}{\sin\left(\frac{4\pi}{3}\right)}=-\frac{2\pi}{\sqrt{3}}\,.$$
Ergo,
$$\Gamma\left(\frac43\right)\,\Gamma\left(\frac23\right)=\left(-\frac13\right)\,\left(-\frac{2\pi}{\sqrt{3}}\right)=\frac{2\pi}{3\sqrt{3}}\,.$$
In other words,
$$t_n=\frac{3\sqrt{3}}{2\pi}\,(n+1)\,\int_0^1\,u^{n+\frac13}\,(1-u)^{-\frac13}\,\text{d}u\,.$$

We now obtain
$$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^1\,\left(\sum_{n=0}^\infty\,(n+1)\,u^{n+\frac13}\,z^n\right)\,(1-u)^{-\frac13}\,\text{d}u\,.$$
This gives
$$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^1\,\frac{u^{\frac13}\,(1-u)^{-\frac13}}{(1-zu)^2}\,\text{d}u\,.$$
Let $v:=\dfrac{u}{1-u}$. Then,
$$f(z)=\frac{3\sqrt{3}}{2\pi}\,\int_0^\infty\,\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\,\text{d}v\,.$$

The integral
$$J:=\int_0^\infty\,\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\,\text{d}v$$
can be evaluated using the same keyhole contour as in this thread. It is not difficult to see that

$$\begin{align}J&=\left(\frac{2\pi\text{i}}{1-\exp\left(\frac{2\pi\text{i}}{3}\right)}\right)\,\text{Res}_{v=-\frac{1}{1-z}}\left(\frac{v^{\frac13}}{\big(1+(1-z)v\big)^2}\right)
\\&=-\frac{\pi\,\exp\left(-\frac{\pi\text{i}}{3}\right)}{\sin\left(\frac{\pi}{3}\right)}\,\Biggl(-\frac{\exp\left(\frac{\pi\text{i}}{3}\right)}{3}\,(1-z)^{-\frac43}\Biggr)
\\&=\frac{2\pi}{3\sqrt{3}}\,(1-z)^{-\frac{4}{3}}\,.\end{align}$$
That is,
$$f(z)=(1-z)^{-\frac43}\text{ for all }z\in\mathbb{C}\text{ such that }|z|<1\,.$$

Well, the OP requested a long and combersome solution. The result can be easily proven via the binomial theorem:
$$(1-z)^{-\frac43}=\sum_{n=0}^\infty\,\binom{-\frac43}{n}\,(-1)^n\,z^n=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k+\frac13}{k}\right)\,z^n$$
for every complex number $z$ with $|z|<1$.

analysis - Characterization of Almost-Everywhere convergence

Given a $\sigma$-finite measure $\mu$ on a set $X$ is it possible to formulate a topology on the space of functions $f:X \rightarrow \mathbb{R}$ that gives convergence $\mu$-almost everywhere?

I can't seem to find any way to write this and am suspecting that no such topology exists! Is this true? If so, is there some generalisation of a topological space where one can make sense of convergence without having open sets?

Any comments, references or tips would be greatly appreciated.

Answer

Given I understand you correctly (topologize almost everywhere convergence), we can show that this is not possible. If we have a topological space, then we have convergence of a sequence if and only if every subsequence has a further convergent subsequence and so on.

So pick a sequence that converges in measure but not almost everywhere. There is a theorem that states that every subsequence of this also converges in the same measure. So it has a subsequence that converges almost everywhere (by another theorem). But the original sequence does not converge almost everywhere so we cannot have convergence in a topology.

Should I add more detail?

elementary set theory - How do I prove the following: $f(Scup T) = f(S) cup f(T)$

I have a question.

How do I prove the following identity?
$$

f(S\cup T) = f(S) \cup f(T)
$$

Answer

Element chasing is a promising method here.

$y\in f(s\cup t)$ if and only if there is some $x\in s\cup t$ such that $f(x)=y$. If $x\in s$ then $y\in f(s)$, if $x\in t$ then $y\in f(t)$. Therefore $y\in f(s)\cup f(t)$.

I leave the second inclusion to you.