Gods Gift To Calculators: The Taylor Series

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polynomial, f0(x)=a0, that passes
through the y-intercept of the graph (0,f(0)). So f(0)=ao.

Next, we see that the graph of f1(x)= a0 + a1x will also pass through x=
0, and will have the same slope as f(x) if we let a0=f1(0).

Now, if we want to get a better polynomial approximation for this
function, which we do of course, we must make a few generalizations. First, we
let the polynomial fn(x)= a0 + a1x + a2x2 + ... + anxn approximate f(x) near x=0,
and let this functions first n derivatives match the the derivatives of f(x) at
x=0. So if we want to make the derivatives of fn(x) equal to f(x) at x=0, we
have to chose the coefficients a0 through an properly. How do we do this?
We'll write down the polynomial and its derivatives as follows.

fn(x)= a0 + a1x + a2x2 + a3x3 + ... + anxn

f1n(x)= a1 + 2a2x + 3a3x2 +... + nanxn-1

f2n(x)= 2a2 + 6a3x +... +n(n-1)anxn-2

.
.
f(n)n(x)= ...

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