Why Is My Second Derivative Calculation for Taylor Expansion Incorrect?

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SUMMARY

The forum discussion centers on the calculation of the second derivative for the Taylor expansion of the function \( f(x) = \sqrt{1+x^2} \) at the point \( x = \alpha \). The user attempts to derive the second derivative but encounters discrepancies when comparing their results to those presented in a reference book. The correct second derivative is confirmed as \( f''(x) = (1+x^2)^{-\frac{1}{2}} - x^2(1+x^2)^{-\frac{3}{2}} \), which leads to the expected term \( \frac{\beta^2 \epsilon^2}{2(1+\alpha^2)^{\frac{3}{2}}} \) when evaluated at \( \epsilon = 0 \). The error in the user's calculation is attributed to incorrect differentiation steps.

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