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: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.
See the graphs on this page: https://en.m.wikipedia.org/wiki/Hyperbolic_functionVitani11 said: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.
Yes, any function which is infinitely differentiable can be represented by its Taylor series. This is known as Taylor's theorem.
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.
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.
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.
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.