So I have got the expansion of $$ (4+5x)^.5 = 2 + (5/4)x - (25/64)x^2 + \dots $$ I am told to use $ x = 1/10 $ to find an approximation of $ \sqrt2 $. I can do this, giving $ 181/128 $, however the last part asks: "Explain why substituting $ x = 1/10 $ into this binomial expansion leads to a valid approximation. "
The answer is said to be because $ |x| < 4/5 $. Why?
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$$(4-5x)^.5 = 2 + (5/4)x + (25/64)x^2+....$$
You are using a few terms of an infinite series to approximate $\sqrt 2$
$$(4-5x)^.5 =2(1-5/4 x)^{.5}$$
The Newton binomial series $$(1+x)^{\alpha}$$ converges where $|x|<1$ which in this case translates in $|5/4 x|<1$ which is satisfied with $x=1/10.$
The actual explanation has something to do with the convergence of Taylor (or Maclaurin) series but it involves quite a high-level calculus.
A not-so-satisfying answer is that you have to put the LHS into the form $k(1+ax)$ and the approximation is valid for $\lvert ax \rvert <1$