I'm trying to approximate $\sqrt{101}$ using the Taylor series for the function $f(x)=\sqrt{x}$ centered at the point $x=100$. I need to obtain an approximation that is within $0.01$ of the correct answer. The Taylor series is given by

$$ f(x) = \sum_{k=0}^{n-1} \frac{f^{(k)}(100)}{k!}(x-100)^k + R_n(x) $$

where $R_n(x)$ is the remainder term. For the case $n=2$, we have

$$ R_2(x) = \frac{f''(c)}{2!}(101-100)^2 = \frac{f''(c)}{2}. $$

where $c \in [100, \, 101]$. Since $f''(x) = -1/4x^{3/2}$, we have

$$ \left|R_2(101)\right| = \left| \frac{1}{2} \cdot \frac{-1}{4c^{3/2}} \right| = \left| \frac{1}{8c^{3/2}} \right| $$

Since we are considering the interval $[100, \, 101]$, we can bound this by

$$ \left|R_2(101)\right| \leq \frac{1}{8\cdot 100^{3/2}} = \frac{1}{8000} = 0.000125. $$

Thus, the approximation using the first two terms of the Taylor expansion should be sufficiently accurate. Those two terms are given by

$$ f(101) = f(100) + f'(100)/2 = 10 + 1/40 = 401/40. $$

However, using numerical software to confirm the approximation, we see that

$$ \left| (401/40) - \sqrt{101} \right| \approx 0.024875621, $$

which is not sufficiently accurate. Can anyone tell me why I am not obtaining a sufficiently accurate approximation? Thanks in advance!

  • 2
    $\begingroup$ Check your formula. You should have $f(101) \approx f(100) + f'(100)/1! = \dfrac{201}{20}$. $\endgroup$
    – Macavity
    Apr 28 '14 at 5:14
  • $\begingroup$ I think what you are referring to as $R_2$ is more commonly referred to as $R_1$, because it is the error from approximating the function with a degree $1$ polynomial. Perhaps that misindexing is somehow connected to the mistake that @Macavity noticed. $\endgroup$ Apr 28 '14 at 5:24

I also think that most of the problem is related to the mistake @Macavity noticed and, accordingly, $$\left| (201/20) - \sqrt{101} \right| \approx 0.000124379$$ But what I would like to suggest for this kind of problem is to write $$\sqrt x=\sqrt {x_0+(x-x_0)}=\sqrt {x_0} \sqrt {1+\frac{x-x_0}{x_0}}=\sqrt {x_0} \sqrt {1+y}$$ and consider the Taylor expansion of the last term around $y=0$. I suppose that this could make life slightly easier.

This would make $$\sqrt x=x_0\left(1+\frac{k}{2 x_0^2}-\frac{k^2}{8 x_0^4}+O\left(\frac{1}{x^8}\right)\right)$$ So, in your case $$\sqrt{101}=10\left(1+\frac{1}{200}-\frac{1}{80000}+\cdots\right)\approx \frac{80399}{8000} =10.04987500$$ while the "exact" value would be $\approx 10.04987562$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.