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In our maths course we are using Maclaurin Series to represent functions, but I do not understand how we can do this. In our textbook (below) it says that you can represent that function by a power series expansion since it is the sum to infinity of a geometric series. However I cannot see how is a sum as well as a simple function. Would anybody be able to explain this? Many thanks :)

I do not understand what they are trying to say here

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You're probably used to numbers having more than one representation:

$$2 = 1+1 = \frac{14}{7} = 1 + \frac{1}{2} + \frac{1}{4} + \cdots = \sum_{n=0}^{\infty} \frac{1}{2^n}.$$

The next step is to express functions in more than one way:

$$e^x = \cosh x + \sinh x = 2^{x\log_2 e}.$$

We know from the formula for geometric series that (for $-1<r<1$)

$$1+r+r^2+\cdots = \frac{1}{1-r}$$

and replacing $r$ by $x$, we have two different expressions for the same function. Your text is replacing $r$ by $-3x$ (and insisting that $-1<-3x<1$.)

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You can view this two ways.

A series of functions of increasing complexity that approximate the function around $0$ better and better. The first and crudest approximation is just its value at $0$ so $1$ in this case. The next best approximation is to use its derivative at $0$ to give it the correct slope at $0$ so $1 -3x$ in this case. Next we can add an $x^2$ term so that the second derivative matches so $1 -3x + 9x^2$ etc.

The other way is consider a specific value of $x$. Then you have an infinite series that might sum to the value $f(x)$.

I just said "might", so does it always work? No, there is no guarantee that these series will converge to the correct value or at all. There will be a radius of convergence in which they work, in this case you are told that it is $|x| < \frac{1}{3}$. The radius of convergence could be from $0$ (never) to $\infty$ (always).

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