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

[Vitani11]

Homework Statement

For example

cosh(x) = 1+x²/2!+x⁴/4!+x⁶/6!+...

Homework Equations

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.

 

[Vitani11]

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

 

[Mark44]

Homework Statement

For example

cosh(x) = 1+x²/2!+x⁴/4!+x⁶/6!+...

Homework Equations

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.

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).

 

[Mark44]

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

See the graphs on this page: https://en.m.wikipedia.org/wiki/Hyperbolic_function
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.

 

[Ray Vickson]

Homework Statement

For example

cosh(x) = 1+x²/2!+x⁴/4!+x⁶/6!+...

Homework Equations

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.

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)?