how do I change the centre of a taylor series from 0 to another values say -5?

Question

Taylor Series and Calculus

Given a function like 5/sqrt(1-x)

Answer 1

Use the general form of the Taylor Series. Evaluate the derivatives at -5 instead of 0.
f(x) ≈
∞
∑ ((f(n)(a))/n!)(x - a)^n
n=0

Answer 2

Generally speaking you have to just recalculate the series at the new point, but for some functions that have nice translation properties you can use short cuts. For example, let's calculate the Taylor series for e^x based at -5. The trick is to write:

e^x = e^((x+5)-5) = e^(-5) e^(x+5)

This last step is what I mean by a "nice translation property". Anyway, now you can just plug (x+5) into the series for e^x to get:

e^x = e^(-5) * sum_(k=0)^infty (x+5)^k/(k!)

In general, this trick works as follows. If you want to expand f(x) about -5, then you write f(x) = f((x+5)-5) and use some sort of special property of f to express f((x+5) - 5) in a nice way.

Edit:

For f(x) = 5/sqrt(1-x) you would write:

f(x) = 5/sqrt(1 - ((x+5)-5)) = 5/sqrt(6-(x+5)) = 5/(sqrt(6)*sqrt(1 -(x+5)/6))

Thus if you know the series for f expanded around 0, you multiply this series by 1/sqrt(6) and plug in (x+5)/6 to get the series expanded around -5.

Answer 3

This is done by translating the initial function, hence rather than expanding f(x), use the taylor series to expand f(x+5), in this case that means rather than using taylor/ maclauring series to expand:

5/sqrt(1-x)
use it to expand:
5/sqrt(1-[x+5]) i.e. expand 5/sqrt(-4-x) with taylor

this expansion will be valid for values between 4 and 6 for f(x), or -1 and 1 for f(x+5)

This works because you translate the initial function so that the required part is centered over "0" and then expand the translated function.

I hope that helped, have fun :)