- #1
knockout_artist
- 70
- 2
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
(β2ε2)/2 (1 + α2)3/2
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
(β2ε2)/2 (1 + α2)3/2