How do I calculate approximately the value of $\frac{1}{\sqrt{1.1}}$ with Linear approximation of the function $\frac{1}{\sqrt{1+x}}$ around point $0$.
And here is a follow-up question: Show that the error in the approximation of $\frac{1}{\sqrt{1.1}}$ in previous question is between $0.001$ and $0.01$
Here is what I have got after trying to find the Mclaurin expansion:
$$ f(x) = 1 – \frac{x}{2} +\frac{3x^2}{8} + R_n(x)$$
im not sure whether if the derivatives were correct, but how do I continue from here? can someone give a clear explanation?
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Best Answer
You have written a quadratic expansion, and your derivatives are not correct.
The linear expansion of $f(x) = \dfrac{1}{\sqrt{1 + x}}$ is just
$$ f(x) \sim f(0) + f'(0)x $$
so
$$ f(0.1) \sim 1 -\dfrac{1}{2}0.1 = 0.95 $$
The error can be estimated by
$$ E \leq \dfrac{M}{2!}|0.1|^2 $$
where
$$ M \geq |f''(x)| \,\, \forall x \in (0.9,1.1) $$
is an upper bound for the second derivative.
Compute
$$ f''(x) = \left( -\dfrac12 (1 + x)^{-3/2} \right)' = \dfrac{3}{4} (1 + x)^{-5/2} = \dfrac{3}{4} \dfrac{1}{(1 + x)^{5/2}}; $$
this gets larger as the denominator gets smaller, so plugging in $x = 0.9$ gives you an upper bound.
Can you take it from here?