Does the Taylor series expansion for e^x converge quickly?

In summary, the conversation discusses the Taylor series expansion of f(x)=e^x and the calculation of its value at e^(-2) using different numbers of terms. The calculated relative errors are 2609%, 2905%, and 2952% for the first 4, 6, and 8 terms respectively. It is concluded that the error decreases with each successive term and converges quickly. A mistake is also acknowledged and corrected in the calculation process.
  • #1
sandy.bridge
798
1
Hello all,
My question is in regards to the Taylor series expansion of
[tex]f(x)=e^x=1+x+x^2/(2!)+x^3/(3!)...[/tex]
I calculated the value of
[tex]e^(-2)[/tex]
using the first 4 terms, 6 terms, and then the first 8 terms. I then calculated the relative error to compare it to the true value, depcited by my calculator to 6 significant figures. Using the first four terms, I found an error of 2609%. Using the first 6 terms I found an error of 2905%, and lastly using the first 8 terms I found an error of 2952%.

What can I conclude from this? Does the error increase (at a decreasing rate) until it begins decreasing (at an increasing rate)?
 
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  • #2
How did you get to 2609%?
If I take 4 terms I get an error of around 46%.

Are you sure you entered the even powers correctly? One of the most common mistakes is to enter (-2)2 as -22 = -(22).
 
  • #3
Oops, I had a mistake.n However, upon doing it again, I get an error of 346.3.
relative error=100((e^(-2)-(sum of first four terms))/e^(-2))=100((e^(-2)-(-0.333333))/e^(-2))=346%
 
  • #4
Yes the error after 4 terms is 346%. But in the next 4 terms the errors are 146%, -51%, 15% and 4%; after another 4 terms the error is only 5 parts in a thousand and another four, -2 in 1 million. The error decreases in magnitude with each successive term (except for the second term, x) and converges fast enough for me.
 
Last edited:
  • #5


Based on your calculations, it appears that the error does decrease as more terms are added to the Taylor series expansion. However, it does not decrease at a constant rate and may even initially increase before decreasing. This is a common characteristic of Taylor series expansions, as they are only approximations and become more accurate as more terms are added. In general, the convergence rate of a Taylor series expansion depends on the function being approximated, and some may converge faster than others. It is important to use a sufficient number of terms to ensure accuracy, especially for functions with more complex behavior.
 

What is a Taylor series expansion?

A Taylor series expansion is a mathematical representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point.

Why is Taylor series expansion important?

Taylor series expansion is important because it allows us to approximate complex functions with simpler polynomial functions, making it easier to calculate values and perform mathematical operations.

What is the purpose of using a Taylor series expansion?

The purpose of using a Taylor series expansion is to approximate the behavior of a function near a specific point, by using polynomial functions that are easier to work with. This allows us to better understand the behavior of the function and make predictions.

What is the difference between a Taylor series expansion and a Maclaurin series?

A Taylor series expansion is an infinite series that represents a function near a specific point, while a Maclaurin series is a special case of a Taylor series where the expansion is centered at 0 (i.e. the specific point is 0).

What are some applications of Taylor series expansion?

Taylor series expansion has many applications in physics, engineering, and other scientific fields. It is commonly used in numerical analysis, optimization, and to solve differential equations. It is also used in the development of algorithms for scientific computing and in signal processing.

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