# Is it possible to graph a function using its taylor series?

1. Jan 26, 2017

### Vitani11

1. The problem statement, all variables and given/known data
For example

cosh(x) = 1+x2/2!+x4/4!+x6/6!+....

2. Relevant equations

3. The attempt at a solution
So plugging in x=0 you get that coshx = 1 at the origin. The approximate graph for the coshx function up to the second order looks like a 1+x2/2! graph, but what about graphing coshx to the term afterwards? and so on.

2. Jan 26, 2017

### Vitani11

Also, what about its negative values? What about the point where cosh slope changes direction? Etc.

3. Jan 26, 2017

### Staff: Mentor

The more terms you use in the Taylor series (really, what you have is the Maclaurin series), the closer your Taylor/Maclaurin polynomial will approximate the graph of y = cosh(x).

4. Jan 26, 2017

### Staff: Mentor

The graph of this function has a sort of parabola shape, opening upward. The slope changes direction at x = 0, the low point on the graph.

Due to the fact that there are only even-degree terms in the Maclaurin series, the graph is symmetric about the y-axis. IOW cosh(-x) = cosh(x), for all real x.

5. Jan 26, 2017

### Ray Vickson

I suggest you answer your question yourself, by trying different numbers of terms and comparing the results with the exact graphs.

However, no finite number of terms can give the correct behavior for $\cosh(x)$ when $|x|$ is very large; can you see why (theoretically)?