3
$\begingroup$

Any one know how to graph a function defined as an infinite series? I need to graph the function $f(x)=\sum_{n=1}^{\infty}\bigg(n^{2}\tan^{-1}(x- n^{2}) +n^{2}\tan^{-1}(x+ n^{2}) \bigg) $ $x\in \mathbb R$.

EDIT: So, as Peter suggested, we can simplify the function using $\tan^{-1} a + \tan^{-1} b = \tan^{-1}((a + b) / (1 - ab))$ to be $f(x)=\sum_{n=1}^{\infty} n^{2} \tan^{-1}\bigg(\frac{2x}{ 1-(x^{2}-n^{4})}\bigg)$

  • 0
    ok, then I see no problem. Choose $k$ and plot $f_k(x)=\sum_{n=1}^{k}\bigg(n^{2}\tan^{-1}(x- n^{2}) +n^{2}\tan^{-1}(x+ n^{2}) \bigg)$ then the error is proportional to the first term that is being omitted.2012-12-08

3 Answers 3

2

Here you can see a plot for $n=6,10,20$. It seems that when we sum up to $n$, we obtain a set of $2n$ vertical sigmoids bounded by the lines $y= \pi/2\cdot x$ and $y=-\pi/2\cdot x$. You can open the image to see it larger.

$\hspace{0.5 cm}$enter image description here

1

Here is a graph of the series by maple 12. To see that there are jumps at some point (look where the vertical segments are), just notice that, the value of the series at $x=4.12$ is positive $9.2885852$ and at the point $x=4.13$ is negative $-3.2263024$.

enter image description here

This is the plot of $f'(x)$

enter image description here

  • 0
    @Mhenni Benghorbal: Something wrong in the graph of $f$ above: since $f'$ is strictly positive then the function $f$ must be strictly increasing, but from the values of $f$ at $x=4.12$ and $x=4.13$ it is not! Did I miss anything?! (I don't know why I couldn't add this as a comment, so I just used the answer box!)2012-12-09
1

If you are able to find an explicit formula for the sum of the series, you use that formula to plot the graph of the sum. If not, you choose $N$ large and plot the sum of the first $N$ terms of the series.

In your example I doubt that there is a simpler expression for the sum. $f$ is odd, so that it is enough to consider $x\ge0$. Since $|\tan^{-1}x|\le\max(|x|,\pi/2)$, we see that the series is uniformly convergent. The sum is discontinuous at the points $x=\pm\sqrt{1+n^4}$, $n\in\mathbb{N}$. Below are the graphs of $8$, $16$ and $32$ terms of the sum, drawn with Mathematica 9.

enter image description here