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

In summary, the coshx function has an approximate graph that looks like a 1+x2/2! graph up to the second order, but beyond that it becomes more difficult to approximate. Additionally, it is negative for some x values and its slope changes direction at x = 0.
  • #1
Vitani11
275
3

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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  • #2
Also, what about its negative values? What about the point where cosh slope changes direction? Etc.
 
  • #3
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).
 
  • #4
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.
 
  • #5
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)?
 

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

1. Can any function be represented by its Taylor series?

Yes, any function which is infinitely differentiable can be represented by its Taylor series. This is known as Taylor's theorem.

2. How accurate is the Taylor series approximation of a function?

The accuracy of the Taylor series approximation depends on the number of terms used in the series. The more terms that are included, the closer the approximation will be to the original function.

3. Can a Taylor series be used to graph a function?

Yes, a Taylor series can be used to graph a function. By adding more terms to the series, the graph will become a better approximation of the function.

4. How do you find the coefficients of a Taylor series?

The coefficients of a Taylor series can be calculated using the formula: cn = f(n)(a)/n!, where cn is the coefficient of the xn term, f(n)(a) is the nth derivative of the function at the point a, and n! is the factorial of n.

5. Are there any limitations to using a Taylor series to graph a function?

Yes, the Taylor series can only accurately represent a function within a certain interval around the point a. Outside of this interval, the series may not converge and will not accurately represent the function.

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