calculus - Gaussian Integral using single integration

So the Gaussian integral basically states that:

$$ I = \int_{-\infty}^{\infty} e^{-x^2} \ dx =\sqrt{\pi}$$

So the way to solve this is by converting to polar co-ordinates and doing a double integration.

Since I haven’t learnt double integration, I have searched a lot to solve this kind of integral using single integration. But to no avail. I have even tried it myself a couple of times but have been unsuccessful.

So here’s my question, is it possible to integrate the above using only single integration and if so how? If this is not possible to integrate using single integration then what is the reason behind it.

optimization - How to simplify the summation of log

I have a summation that involve log. I don't know how to solve this summation. I want to find an expression (even a good approximation is enough) for this summation.

$\sum_{k=0}^{n}{log(a_k)}$

or
$log(\prod_{k=0}^{n}{a_k})$

I only know that
$\sum_{k=0}^{n}{a_k}= N$

Any help Please?

Edit

Let $a_k$ is a random variable which can take its value according to binomial (or normal) distribution. Then how to solve the above log summation problem.

trigonometry - Find t for $Asin(w_1t) = -Bsin(w_2t)$

I am trying to solve t for the next equation:

$$A\sin(\omega_{1}t)+B\sin(\omega_{2}t)=0$$

Where:

$A > B$

$\omega_{1}$ AND $\omega_{2}$ are constants

Reading in wikipedia about trigonometric identities and following this post: Identity for a weighted sum of sines / sines with different amplitudes

I tried:

$$A\sin(\omega_{1}t)+B\sin(\omega_{2}t)=C\sin(\omega_{1}t+\varkappa)=0$$

Where:

$$C^2=A^2+B^2+2AB\cos(\omega_{2}t-\omega_{1}t)$$

And

$$\varkappa=\arcsin\frac{B\sin(\omega_{2}t-\omega_{1}t)}{C}$$

if $C\sin(\omega_{1}t+\varkappa)=0$, means that $C$ and/or $sin(\omega_{1}t+\varkappa)$ are equal to $0$, so I can get a partial solution by making $C=0$:

$$t=\frac{\arccos-\frac{A^2+B^2}{2AB}}{\omega_{2}-\omega_{1}}$$

But for the expression $sin(\omega_{1}t+\varkappa)=0$ I have not been able to solve it.

Someone can help me with this please?

Thanks

Help debunk a proof that zero equals one (no division)?

Unlike the more common variant of proof that 0=1, this does not use division.

So, the reasoning goes like this:

\begin{align}

0 &= 0 + 0 + 0 + \ldots && \text{not too controversial} \\
&= (1-1) + (1-1) + (1-1) + \ldots && \text{by algebra}\\
&= 1 + (-1 + 1) + (-1 + 1) \ldots && \text{by associative property}\\
&= 1\\
\\
&\therefore 0 =1
\end{align}

I can't help but feel that something went wrong here, specifically with the use of the associative property. However, I can't come up with a mathematically compelling reason.

Where's the error?

Answer

The error is that the "..." denotes an infinite sum, and such a thing does not exist in the algebraic sense. The usual way to make sense of adding infinitely many numbers is to use the notion of an infinite series: We define the sum of an infinite series to be the limit of the partial sums. (So the notion of convergence from analysis is involved in addition to algebra.)

Not all algebraic rules generalize to infinite series in the way that one might hope. When they fail, it is because something fails to converge. In this case, what fails to converge is the series that should appear between the two lines in the middle of the "proof":
$$1-1+1-1+1 \cdots.$$
Indeed, this series fails to converge because the
sequence of partial sums $\{1, 1-1, 1-1+1,\ldots\}$ oscillates between $1$ and $0$ and does not converge to any value.

linear algebra - Eigenvectors times diagonal matrix, still eigenvectors?

Suppose we have a $n\times n$ real symmetric positive definite matrix $\Sigma$, and $V=(v_1,...,v_n)$ whose columns are the eigenvectors corresponding to the $n$ eigenvalues $\lambda_1\geq \lambda_2 ...\geq \lambda_n$. Let $\Delta$ be a diagonal matrix $diag(\delta_1,...,\delta_n)$, where $\delta_i>0$ for $i=1,...,n$.
If $M=\Delta V$, then I want to ask that whether the columns of $M$ correspond to the eigenvectors of some matrix, saying $\tilde{\Sigma}$?

Many thanks!

Answer

$V$ is the eigenvector matrix, $D$ is the diagonal matrix.

$VD$ is still an eigenvector matrix. Multiplying a diagonal matrix from the right, scales the colums. Columns are eigenvectors. Scaled eigenvectors are still eigenvectors (of the same matrix that $V$ belongs to).

$DV$, is not. Multiplying a diagonal matrix from the left, scales the rows. This destroys the proportions of all the eigenvectors.

Regarding your last sentence, all vectors are the eigenvectors of the identity matrix, so that statement means very little.

discrete mathematics - Proof of the Hockey-Stick Identity: $sumlimits_{t=0}^n binom tk = binom{n+1}{k+1}$

After reading this question, the most popular answer use the identity
$$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1}.$$

What's the name of this identity? Is it the identity of the Pascal's triangle modified.

How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?

Thanks for your help.

---

EDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.

Answer

This is purely algebraic. First of all, since $\dbinom{t}{k} =0$ when $k>t$ we can rewrite the identity in question as
$$\binom{n+1}{k+1} = \sum_{t=0}^{n} \binom{t}{k}=\sum_{t=k}^{n} \binom{t}{k}$$

Recall that (by the Pascal's Triangle),
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$

Hence

$$\binom{t+1}{k+1} = \binom{t}{k} + \binom{t}{k+1} \implies \binom{t}{k} = \binom{t+1}{k+1} - \binom{t}{k+1}$$

Let's get this summed by $t$:
$$\sum_{t=k}^{n} \binom{t}{k} = \sum_{t=k}^{n} \binom{t+1}{k+1} - \sum_{t=k}^{n} \binom{t}{k+1}$$

Let's factor out the last member of the first sum and the first member of the second sum:
$$\sum _{t=k}^{n} \binom{t}{k}
=\left( \sum_{t=k}^{n-1} \binom{t+1}{k+1} + \binom{n+1}{k+1} \right)
-\left( \sum_{t=k+1}^{n} \binom{t}{k+1} + \binom{k}{k+1} \right)$$

Obviously $\dbinom{k}{k+1} = 0$, hence we get
$$\sum _{t=k}^{n} \binom{t}{k}
=\binom{n+1}{k+1}
+\sum_{t=k}^{n-1} \binom{t+1}{k+1}
-\sum_{t=k+1}^{n} \binom{t}{k+1}$$

Let's introduce $t'=t-1$, then if $t=k+1 \dots n, t'=k \dots n-1$, hence
$$\sum_{t=k}^{n} \binom{t}{k}
= \binom{n+1}{k+1}
+\sum_{t=k}^{n-1} \binom{t+1}{k+1}

-\sum_{t'=k}^{n-1} \binom{t'+1}{k+1}$$

The latter two arguments eliminate each other and you get the desired formulation
$$\binom{n+1}{k+1}
= \sum_{t=k}^{n} \binom{t}{k}
= \sum_{t=0}^{n} \binom{t}{k}$$

Sum of the series $frac{2}{5cdot10}+frac{2cdot6}{5cdot10cdot15}+frac{2cdot6cdot10}{5cdot10cdot15cdot20 }+cdots$

How do I find the sum of the following infinite series:
$$\frac{2}{5\cdot10}+\frac{2\cdot6}{5\cdot10\cdot15}+\frac{2\cdot6\cdot10}{5\cdot10\cdot15\cdot20 }+\cdots$$

I think the sum can be converted to definite integral and calculated but I don't know how to proceed from there.