# Thread: How to calculate the maximum for a Fourier series?

1. ## How to calculate the maximum for a Fourier series?

In

$$\frac{Gt}{j} = \frac{Gj}{R}\Big ( \frac{1}{6}+\frac{2}{\pi^2} \sum_{n=1}^{\infty}\frac{(-1)^2}{n^2} \exp(\frac{-n^2\pi^2 Rt}{j^2}) \Big )$$

How to find the range of $Rt/j^2$ for which the equation is valid?

2. ## Re: How to calculate the maximum for a Fourier series?

does $j = \sqrt{-1}$ ?

I'm assuming it does. Thus we have

$b_n = \dfrac{e^{n^2\pi^2 R t}}{n^2}$

is the underlying sequence in the alternating series.

In order for the series to converge this sequence must have two properties

i) $b_n$ is ultimately a decreasing sequence

ii) $\lim \limits_{n \to \infty}~b_n = 0$

we need the sign of $R t$

if $R t > 0$ then this sequence tends to infinity

if $R t \leq 0$ then it tends to 0.

additionally if $R t \leq 0$ then the sequence is decreasing.

So it looks like your answer is the formula is valid for $R t \leq 0$

3. ## Re: How to calculate the maximum for a Fourier series?

Originally Posted by romsek
does $j = \sqrt{-1}$ ?
no, it is a positive veriable

Originally Posted by romsek
does $j = \sqrt{-1}$ ?

is the underlying sequence in the alternating series.
yes,

The entire solution to a partial differential equation is

$$y(x,t) = \frac{Gt}{j}+\frac{Gj}{D}\Bigg(\frac{3x^2-j^2}{6j^2}-\frac{2}{\pi^2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}\exp\Big(\frac{-n^2\pi^2 Rt}{j^2}\Big) \cos\frac{n\pi x}{j}\Bigg)$$

The above equation is for the case at $x=0$. At a constant $R/j^2$, $y$ increases with $t$. The purpose is to find up to what value of $Rt/j^2$, $y$ remains zero.