Why do we have fast convergence?

[evinda]

Hello! (Wave)

Suppose that we have $u(x,t)= \frac{80}{\pi} \sum_{n=1,3,5, \dots}^{\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2 a^2 t}{2500}} \sin{\frac{n \pi x}{50}}$.

According to my notes, the negative exponential factor at each term of the series has as a result the fast convergence of the series except for small values of $t$ or of $a^2$. So we can have exact results using usually only some of the first terms of the series.

Could you explain me the above? Why does it hold?

I thought that we have that if $\sum a_n$ converges then $a_n \to 0$ but the converse does not hold...

 

[I like Serena]

Hello! (Wave)

Suppose that we have $u(x,t)= \frac{80}{\pi} \sum_{n=1,3,5, \dots}^{\infty} \frac{1}{n} e^{-\frac{n^2 \pi^2 a^2 t}{2500}} \sin{\frac{n \pi x}{50}}$.

According to my notes, the negative exponential factor at each term of the series has as a result the fast convergence of the series except for small values of $t$ or of $a^2$. So we can have exact results using usually only some of the first terms of the series.

Could you explain me the above? Why does it hold?

I thought that we have that if $\sum a_n$ converges then $a_n \to 0$ but the converse does not hold...

Hey evinda! (Smile)

The exponential function with a negative exponent is known to converge extremely fast.
And indeed, just because $a_n\to 0$, does not mean that $\sum a_n$ converges.

More to the point, we have:
$$
\left|\sum_{n=1}^\infty \frac 1n e^{-n^2}\sin(nx)\right| \le \sum_{n=1}^\infty \frac 1n e^{-n^2}|\sin(nx)| < \sum_{n=1}^\infty e^{-n^2} < \sum_{n=1}^\infty e^{-n}
= \frac{e^{-1}}{1-e^{-1}} = \frac{1}{e-1}
$$
So we can see that the series converges.We also have:
$$\sum_{n=1}^\infty e^{-n} = \sum_{n=1}^k e^{-n} + R_k
$$
where the remainder $R_k$ is:
$$R_k=\sum_{n=k+1}^k e^{-n} = \frac{e^{-(k+1)}}{1-e^{-1}}
$$
So the remainder for this upper bound is reduced by a factor of $e \approx 2.7$ with every additional term.
And since we actually have $e^{-n^2}$, the remainder is reduced faster.

 

[Euge]

$$
\sum_{n=1}^\infty \frac 1n e^{-n^2}\sin(nx) < \sum_{n=1}^\infty e^{-n^2} < \sum_{n=1}^\infty e^{-n}
= \frac{e^{-1}}{1-e^{-1}} = \frac{1}{e-1}
$$

Shouldn't there be absolute value bars around $\sin(nx)$?

 

[I like Serena]

Shouldn't there be absolute value bars around $\sin(nx)$?

Yep. Added.