soft question - Most useful heuristic?

As opposed to the most harmful heuristics, what are the most useful heuristics which

• are hand-waving,




• are conducive to proper mathematical education, and




• you have seen taught or taught yourself?

In this context:

• Hand-waving means imprecise, intuitive, ambiguous, with a purpose of impressing or convincing.





• Proper mathematical education means that a person can understand, use, discuss, and derive the learnt mathematical claims after finishing the education process to the levels (a) advertised by goals of the education process and at the same time (b) having, up to some allowed degree of ambiguity, the same, widely accepted meaning in the community. Example: "Real Calculus" could mean "basics of differentiation and integration over the functions $\mathbb{R}\to\mathbb{R}$".




• Seen taught means you closely observed or participated as a learner in the educational process.




• Taught yourself means you were a lecturer or an author of used educational material.

What's the formula to solve summation of logarithms?

I'm studying summation. Everything I know so far is that:

$\sum_{i=1}^n\ k = \frac{n(n+1)}{2}\ $

$\sum_{i=1}^{n}\ k^2 = \frac{n(n+1)(2n+1)}{6}\ $

$\sum_{i=1}^{n}\ k^3 = \frac{n^2(n+1)^2}{4}\ $

Unfortunately, I can't find neither on my book nor on the internet what the result of:

$\sum_{i=1}^n\log i$.

$\sum_{i=1}^n\ln i$.

is.

Can you help me out?

Answer

By using the fact that $$\log a + \log b = \log ab $$ then

$$ \sum^n \log i = \log (n!) $$

$$ \sum^n \ln i = \ln (n!) $$

summation - Prove the identity $binom{2n+1}{0} + binom{2n+1}{1} + cdots + binom{2n+1}{n} = 4^n$

I've worked out a proof, but I was wondering about alternate, possibly more elegant ways to prove the statement. This is my (hopefully correct) proof:

Starting from the identity $2^m = \sum_{k=0}^m \binom{m}{k}$ (easily derived from the binomial theorem), with $m = 2n$:

$2^{2n} = 4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{2n-1} + \binom{2n}{2n}$

Applying the property $\binom{m}{k} = \binom{m}{m-k}$ to the second half of the list of summands in RHS above:

$4^n = \binom{2n}{0} + \binom{2n}{1} + \cdots + \binom{2n}{n-1} + \binom{2n}{n} + \underbrace{\binom{2n}{n-1} +\cdots \binom{2n}{1} + \binom{2n}{0}}_{\binom{m}{k} = \binom{m}{m-k} \text{ has been applied}}$

Rearranging the above sum by alternately taking terms from the front and end of the summand list in RHS above (and introducing the term $\binom{2n}{-1} = 0$ at the beginning just to make explicit the pattern being developed):

$4^n = (\binom{2n}{-1} + \binom{2n}{0}) + (\binom{2n}{0} + \binom{2n}{1}) + \cdots + (\binom{2n}{n-1} + \binom{2n}{n})$

Finally, using the property $\binom{m}{k} + \binom{m}{k-1} = \binom{m+1}{k}$ on the paired summands, we get the desired result:

$4^n = \binom{2n+1}{0} + \binom{2n+1}{1} + \cdots + \binom{2n+1}{n}$

Answer

Why not just
$$ \begin{align} 2^{2n+1} &=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n+1}+\cdots+\binom{2n+1}{2n+1} \\
&=\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}+\binom{2n+1}{n}+\cdots+\binom{2n+1}{0} \\
&=2\left[\binom{2n+1}{0}+\cdots+\binom{2n+1}{n}\right] \end{align} $$
Then divide each extremity by 2.

summation - Evaluate $sumlimits_{k=1}^{n} frac{k}{2^k}$

Evaluate $$\sum\limits_{k=1}^{n} \frac{k}{2^k}$$

sequences and series - Concerning the sum $sum_{n = 1}^infty sin nx$

I recently came across this question and I posted an answer. It has been pointed out that my answer is incorrect. I cannot work out what is wrong with my reasoning. The answer I gave corresponds with the Abel and Cesaro sum, so perhaps $\sum$ is not the usual summation operator? Am I correct in asserting that if $x$ is in the upper half-plane, i.e., $\textbf{I}[x] > 0$, then $|e^{ix}| < 1$ and consequently
$$\sum_{n = 1}^\infty e^{inx} = \frac{e^{ix}}{1 - e^{ix}},$$
or is my argument flawed? Any help would be appreciated.

Answer

Because you assume that $x$ is not real, the imaginary part of $e^{inx}$ is not $\sin nx$.

Also, when computing the conjugate if $1-e^{ix}$, you don't get $1-e^{-ix}$ when $x$ is non-real, but rather $$1-e^{-i\bar x}$$

real analysis - How do i evaluate this sum $sumlimits_{n=1}^{infty} frac{(-1)^{n+1}}{n^2n!}$?

How do I evaluate this sum:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2n!}$$

Note: The series converges by the ratio test. I have tried to use this sum:$$ \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}= \ln (2) $$ but I didn't succeed. Might there be others techniques which I don't know?

Thank you for any help

Vector $p$-norm for square matrices is submultiplicative for $1 le p le 2$

I'm trying to prove that the vector $p$-norm for square matrices is submultiplicative for values of $p$ between $1$ and $2$. The vector $p$-norm for a square matrix $A$ is defined as

$\displaystyle \|A\|_p:=\left(\sum_{i=1}^n \sum_{j=1}^n |a_{ij}|^p \right)^{\frac{1}{p}}$

For $p=1$ and $p=2$ the result follows easily from the Cauchy-Schwarz inequality. Now for $1

My approach is as follows:

Let $A=(a_{ij})$ and $B=(b_{ij})$ be two square $n \times n$ matrices and let $q=\frac{p}{p-1}$. Then

$\displaystyle \left(\|AB\|_p\right)^p=\sum_{i=1}^n \sum_{j=1}^n \left| \sum_{k=1}^n a_{ik}b_{kj} \right|^p \le \sum_{i=1}^n \sum_{j=1}^n \left( \sum_{k=1}^n \left| a_{ik}\right|^p\right) \left( \sum_{k=1}^n \left| b_{ik}\right|^q\right)^\frac{p}{q}$

Now, since $p<2$, we have that $q>2$ and as such $\frac{p}{q}<1$. Here comes my doubt:
Can I assure that $\left( \sum_{k=1}^n \left| b_{ik}\right|^q\right)^\frac{p}{q} \le \sum_{k=1}^n \left| b_{ik}\right|^p$ ? Because if so, then the result would follow immediately.

Any help would be greatly appreciated!

Answer

The answer to your question follows from this: let $\|x\|_p = \left(\sum_{i=1}^n |x_i|^p \right)^{1/p}$. Then is $\|x\|_q \le \|x\|_p$ if $q \ge p$?

This is a very well known result, But here is a proof.

Answer: yes. Let $M = \|x\|_p$. So $\|\frac1Mx\|_p = 1$. Therefore $|\frac{x_i}M|^p \le 1 \Rightarrow |\frac{x_i}M| \le 1 \Rightarrow |\frac{x_i}M|^q \le |\frac{x_i}M|^p$. Hence
$$ \|\frac1M x\|_q^q = \sum_{i=1}^n |\frac{x_i}M|^q \le \sum_{i=1}^n |\frac{x_i}M|^p = \|\frac1Mx\|_p^p = 1.$$
Therefore $\|x\|_q \le M = \|x\|_p$.