Question: Find an approximation of $\sqrt(11)$ using some quadratic Taylor polynomial. Estimate the error using Lagrange's Form of Remainder.

Attempt: Consider $f(x)=\sqrt(1+x)$. Taylor Series for this function around x=0 is:$1+1/2x-1/8x^2+R_2(x)$.

$R_2(x)=1/16$* $(1+c)^-5/2 x^3$ $\leq 1/16x^3$, as $c\in[0,x]$.

Thus, plugging in x=10, to find the approximation of $\sqrt(11)$ , I get: $1+(1/2*10)-1/8*(10^2)$=$-6.5$, which is clearly wrong.

Error$\leq$ $(1/16)*(10^3)$=$62.5$

Obviously,I am doing something wrong as I'm getting numbers that don't make any sense, so I could really use some help. Thank you.

  • $\begingroup$ perhaps I should a function like sqrt(10+x) instead? $\endgroup$
    – kemb
    Nov 7 '17 at 23:43

$x=10$ is too far away to expect good results out of a series expansion of $\sqrt{1+x}$ about $x=0$. But you could take an expansion of $\sqrt{9+x}$ about $x=0$; then you know the first term of the expansion so you can continue.

Also, there is a way to reuse your work by writing $\sqrt{9+x}=a\sqrt{1+bx}$ for appropriate constants $a,b$.

  • $\begingroup$ Great just figured out the same thing. $\endgroup$
    – kemb
    Nov 7 '17 at 23:54
  • $\begingroup$ Would it even be possible to use sqrt(11+x)? $\endgroup$
    – kemb
    Nov 7 '17 at 23:55
  • $\begingroup$ @kemb Strictly speaking yes but all that that gives you is the tautology $\sqrt{11}=\sqrt{11}$. This does not help you to evaluate the square root. You want to expand around a point where you know the answer. The two nearby points where you know the answer are $9$ and $16$; either of these will do the job (though you will get different errors). $\endgroup$
    – Ian
    Nov 7 '17 at 23:57

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