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

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Homework Help Overview

The discussion revolves around the graphing of the hyperbolic cosine function, cosh(x), using its Taylor series expansion. Participants explore how the approximation improves with additional terms and consider the implications of using the series for both positive and negative values of x.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the approximation of cosh(x) using its Taylor series, particularly focusing on the behavior at the origin and how the graph changes with more terms. Questions arise regarding the function's behavior for negative values and the point where the slope changes direction.

Discussion Status

The discussion is active, with participants sharing insights about the symmetry of the function and the limitations of finite Taylor series in capturing the behavior of cosh(x) for large values of x. Some guidance is offered regarding experimentation with different numbers of terms in the series.

Contextual Notes

There is an emphasis on the nature of the Maclaurin series and its even-degree terms, which leads to the symmetry of the graph about the y-axis. Participants are also considering the theoretical aspects of approximation limits.

Vitani11
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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.
 
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Also, what about its negative values? What about the point where cosh slope changes direction? Etc.
 
Vitani11 said:

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).
 
Vitani11 said:
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.
 
Vitani11 said:

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)?
 

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