Recent content by SatyaDas

Great. Thanks. How did you find the value of \( S_{n} \)?

How to prove that \[ \sum_{i=1}^{\infty}\frac{1}{2^{3i}}\left(\csc^{2}\left(\frac{\pi x}{2^{i}}\right)+1\right)\sec^{2}\left(\frac{\pi x}{2^{i}}\right)\sin^{2}\left(\pi x\right)=1 \] for all \( x\in\mathbb{R} \). Using graph, we can see that the value of this series is 1 for all values of x...

Here is my solution. Since $F(x;n)$ is periodic with period $n$, we can assume that the function can be expressed as $$ F(x;n)=\sum_{i=0}^{n-1}\left(a_i\cos\left(\frac{2\pi i x}{n}\right)+b_i\sin\left(\frac{2\pi i x}{n}\right)\right). $$ There are total of $2n$ unknowns. We get $n$ equations by...

Real solutions have already been given so I am focusing on complex solutions. To find solutions in complex plane we can rewrite the equation as $(x^2-7x+11)^{x^2-13x+42}=\exp \left({2k\pi i}\right)$ where $k \in Z$. Now, we can rearrange the terms to write $$ x=\frac{\exp \left(\frac{2k \pi...

I wonder about existence of complex solutions. Can't it happen that x is a complex number such that $(x^2-7x + 11)^n = 1$ and $x^2-13x + 42= n$ where $n\in Z^+$? It appears there are other alternatives as well if we allow complex solutions.

Indeed, I meant smooth function and your finding of constant term matches with mine. So, high five. Below is my output for F(x;4). I guess it will help to get the idea that there is a pattern.

Nice attempt. Below is my output for F(x;3).

Define a continuous function $$F(x;n)$$ that interpolates points (x, x mod n) for a given integer n and all integer x. For example $$F(x;2)=\frac{1}{2}-\frac{1}{2}\cos\left(\pi x\right)$$ interpolates all points (x, x mod 2) when x is an integer. Similarly $$F(x;3)$$ should interpolate points...

You missed taking the summation into account. The lower case 'n' is the index for summation and the expression is summed till n=N. We need to find the limit of the sum as the upper case 'N' tends to infinity. And certainly the limit exists and is non zero that is demonstrated by the graph also...

I forgot to mention that the above graph is plotted as function of x and value of N can be changed by using the slider and that shows that the graph stabilizes pretty fast if N is increased.

I wonder if the limit of the following can be converted into integral or some elegant form as N tends to infinity: \[ \sum_{n=0}^{N}\frac{a}{2^{n}}\sin^{2}\left(\frac{a}{2^{n}}\right) \] If we plot or evaluate the value then it does appear that the series converges very fast...

This problem just needs straight forward application of Rolle's Mean Value Theorem as already mentioned by the problem poster.

This solution in my view is the simplest and still rigorous.

Wow. I had struggled to formalize my solution. You solved this problem in the way it should be solved. Great.