For each function, find it’s Taylor Series Expansion. Only consider its valid ranges as indicated in the notes when you are computing the Taylor Series expansion.
1:
\(f(x) = 1/(1-x)\)
A Taylor Series is the expansion of a function into the infinite sum of it’s derrived terms. We can represent this as:
\(\sum _{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n\)
where \(f^{(n)}(a)\) is the function evaluated at the \(n\)’th derivitiave, and \(a\) is the point at which they are evaluated. We’ll set \(a\) to 0; now, for example, the function \(f(x) = 1/(1-x)\) becomes \(f(0) = 1/(1-0)\).
What we really want is a repeating pattern. We can try to find one by solving for the first few terms of the Taylor Series, which we do below.
Starting at n = 0 (the first term), we have \((x-0)^{-1} *(x-0)^0/0! = 1\)
at n = 1 (second term) we have \((1-0)^{-2} *(x-0)^1/1! = x\)
at n = 2 we have\(2(1-0)^{-3} *(x-0)^2/2! = x^2\)
at n = 3 we have\(3(1-0)^{-4} *(x-0)^3/3! = x^3\)
At this point, we’re starting to see that the each expansion of n towards \(\infty\) yields us another term of \(x^n\). Thus we can define this Taylor Series expansion as: \(\sum_{n=0}^{\infty}x^n\) for the function \(f(x) = \frac{1}{x-1}\)
It should be noted that this is only valid when the functions are evaluated at points that are not 0.
2:
\(f(x) = e^x\)
Similar to before, let’s try to find the first few terms of the Taylor Series and see if a pattern emerges. We’ll evaluate each term at \(a = 0\).
at n = 0 we have \(e^0 *(x-0)^0/0! = 1\)
at n = 1 we have \(e^0 *(x-0)^1/1! = x\)
at n = 2 we have \(e^0 *(x-0)^2/2! = x^2/2\)
at n = 3 we have \(e^0 *(x-0)^3/3! = x^3/6\)
And we get the picture. Each term is \(x^n/n!\). The derivitave part of the function oesn’t change. We can write this Taylor Series Expansion as: \(\sum_{n=0}^{\infty}x^n/n!\). The function can be evaluated at any point
3:
\(f(x)=ln(1+x)\)
Same as before:
at n = 0 we have \(ln(1+0) * (x-0)^0/0! = 0\)
at n = 1 we have \(\frac{1}{1+0} * (x-0)^1/1! = x\)
at n = 2 we have \(-\frac{1}{(1+0)^2} * (x-0)^2/2! = -x^2/2\)
at n = 3 we have \(\frac{2}{(1+0)^3} * (x-0)^3/3! = x^3/3\)
at n = 4 we have \(-\frac{6}{(1+0)^4} * (x-0)^4/4! = -x^4/4\)
at n = 5 we have \(\frac{24}{(1+0)^5} * (x-0)^5/5! = x^5/5\)
Here it seems the term at n is \(-1^{n-1}*x^n/n\), but only when n does not equal 0. We can write this expansion as \(\sum_{n=1}^{\infty}-1^{n-1}*x^n/n\)
We can only evaluate the function at points greater than -1, because \(ln(\leq0)\) is undefined.