# B Multiterm Taylor expansion

1. Jul 15, 2017

### knockout_artist

So in the book it says expend function ƒ in ε to get following.

ƒ=√ (1 + (α + βε)2) = √ (1 + α2) + (αβε)/√ (1 + α2) + (β2ε2)/2 (1 + α2)3/2 + O(e3)

When I expend I get(keeping ε = 0)
ƒ(0) = √ (1 + α2) -->first term
ƒ'(0) = (αβ)/√ (1 + α2) --> sec term with gets multiplied by ε

for third term
I used
d/de (αβ + εβ2 ) * √ (1 + (α + βε)2) + d/de √ (1 + (α + βε)2) * (αβ + εβ2)

at most I got

ƒ''(0) = β2 / √ (1 + (α + βε)2) + -1/2 (α + βε) β(αβ + εβ2) /3√ (1 + ( α + βε)2)

= β2 / √ (1 + (α )2) + -1/2 (α + β0) β(αβ + 0β2) /3√ (1 + α 2)

Why I am not getting the term as in the book which is
2ε2)/2 (1 + α2)3/2

2. Jul 15, 2017

### andrewkirk

The error is in the differentiation to get the third term.

The function being expanded is
$$f(x)=\sqrt{1+x^2}$$
and it is being expanded at $x=\alpha$.

The first derivative is correct:
$$\frac d{dx}\sqrt{1+x^2} = x(1+x^2)^\frac12$$

The second derivative is
$$\frac d{dx}\left(x(1+x^2)^\frac12\right) = (1+x^2)^{-\frac12} - x^2(1+x^2)^{-\frac32}$$