Taylor series expansion question

In summary, the conversation discusses the use of Taylor series to expand a function about a point x=a and explores the possibility of expanding about the translated point x+a. It is suggested to simply plug x+a into the original expansion, with a note to be mindful of the radii of convergence.
  • #1
Hi, I have a question about Taylor series:

I know that for a function f(x), you can expand it about a point x=a, which is given by:
[tex] f(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + ... [/tex]

but I would like to do it for f(x+a) instead of f(x), and expand it about the very same point on the graph (I don't know if it is still x=a or something else after this translation).

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  • #2
why not just plug x+a into the expansion (you should be a little careful with radii of convergence - but this can be bypassed)?

[tex]f(x) = f(a) + f^\prime(a)(x-a) + \cdots \Longrightarrow f(x+a) = f(a) +f^\prime(a)x + f^{\prime \prime}(a)x^2 + \cdots.[/tex]

Take a look and see if this satisfies what you want for being about "the same point on the graph" (I suspect you will find it does).
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  • #3

Hi there,

That's a great question! When expanding a function about a point x=a, we are essentially looking at the behavior of the function at that specific point. So, if we want to expand f(x+a) about the point x=a, we would still use the same formula as before, but with the variable x replaced by (x+a). This would give us the Taylor series expansion for f(x+a) about the point x=a:

f(x+a) = f(a) + f'(a)(x+a - a) + \frac{f''(a)}{2!}(x+a - a)^2 + ...

Simplifying this, we get:

f(x+a) = f(a) + f'(a)x + \frac{f''(a)}{2!}x^2 + ...

So, the point on the graph is still x=a, but we are now expanding the function about the translated point (x+a). I hope this helps clarify things for you! Let me know if you have any other questions.

1. What is a Taylor series expansion?

A Taylor series expansion is a mathematical method for representing a function as an infinite sum of terms, with each term becoming increasingly complex. It is used to approximate a function around a specific point by evaluating its derivatives at that point.

2. How is a Taylor series expansion calculated?

The Taylor series expansion is calculated by taking the derivatives of a function at a specific point and plugging them into a formula that involves raising the difference between the point and the center of expansion to various powers. This creates a series of terms that, when summed, approximate the original function.

3. What is the purpose of a Taylor series expansion?

The purpose of a Taylor series expansion is to approximate a function that may be difficult or impossible to solve directly. It is particularly useful for estimating the behavior of a function near a specific point, and can also be used for numerical analysis and optimization.

4. What is the difference between a Taylor series and a Maclaurin series?

A Taylor series is a generalization of a Maclaurin series, which is a special case where the center of expansion is at 0. In other words, a Maclaurin series is a Taylor series expansion of a function around x=0. They both use the same formula, but a Maclaurin series has simpler coefficients since some terms are eliminated when x=0.

5. How accurate is a Taylor series expansion?

The accuracy of a Taylor series expansion depends on the number of terms used in the approximation. The more terms included, the closer the approximation will be to the original function. However, since it is an infinite series, it can never be an exact representation of the function. The accuracy can also be affected by the behavior of the function at the point of expansion.

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