What did I do wrong in my Taylor Series Expansion for y=kcosh(x/k)?

In summary, the conversation discusses an equation involving kcosh(x/k) and its simplification using the definition of cosh(x). The individual provides a summary of their thought process and identifies where their error lies.
  • #1
Nusc
760
2
y=kcosh(x/k) = k(e^(x/w) + e^(-x/k))

y ~ k[1 + x/k + (1/2!)(x^2/k^2) + (1/3!)(x^3/k^3)] +...+ k[1 - x/k +(1/2!)(x^2/k^2) - (1/3!)(x^3/k^3) +...]

All odd terms except 1 cancel out.

So we are left with

y = k [2 + (2/2!)(x^2/k^2) + (2/4!)(x^4/k^4) + (2/6!)(x^6/k^6) +...]

I've been staring at this for hours and I'm supposed to show that:

y= k + (1/2k)(x^2) for k>>x


What did I do wrong? My error lies somewhere where the twos are bolded.
 
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  • #2
you lost a 1/2 in your definition of cosh:

[tex]cosh(x)=\frac{e^x+e^{-x}}{2}[/tex]
 
  • #3
Well, you obviously are missing a factor of two somewhere... but I suppose you figured that out already. The answer lies here:

[tex]\cosh x = \frac{e^x + e^{-x}}{2}[/tex]
 

1. What is a Taylor Series Expansion?

A Taylor Series Expansion is a mathematical technique that allows us to approximate any function as a polynomial with an infinite number of terms. It is named after the mathematician Brook Taylor, who first described the concept in the 18th century.

2. How is a Taylor Series Expansion calculated?

A Taylor Series Expansion is calculated using the derivatives of a function at a given point. The formula for the Taylor Series Expansion is f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)^2/2! + f'''(a)(x-a)^3/3! + ..., where f'(a) is the first derivative of the function at the point a, f''(a) is the second derivative, and so on.

3. What is the purpose of a Taylor Series Expansion?

The purpose of a Taylor Series Expansion is to approximate a complicated function with a simpler polynomial function. This allows us to make calculations and predictions about the behavior of the function without having to use complex methods or tools.

4. Can a Taylor Series Expansion be used for any function?

In theory, yes, a Taylor Series Expansion can be used for any function. However, the accuracy of the approximation depends on the smoothness of the function and the number of terms included in the expansion. For some functions, a large number of terms may be needed for a good approximation.

5. What are some applications of Taylor Series Expansion?

Taylor Series Expansion has numerous applications in mathematics, physics, and engineering. It is commonly used in calculus to solve complex problems, in numerical analysis to approximate solutions to differential equations, and in computer graphics to create smooth curves and surfaces. It is also used in fields such as finance and economics to model and predict the behavior of functions.

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