Taylor series expansion question

[meteorologist1]

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

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).

Thanks.

 

[Data]

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

$$f(x) = f(a) + f^\prime(a)(x-a) + \cdots \Longrightarrow f(x+a) = f(a) +f^\prime(a)x + f^{\prime \prime}(a)x^2 + \cdots.$$

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).

 

[M98Ranger]

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.