Application of the Taylor Polynomial

Question

I was tasked to find, using the Taylor polynomial or any other way a rational number such that subtraction between it and $\sqrt{10}$ is smaller in it's absolute value than $\frac{1}{100}$. I tried using the Maclaurin series for $\sqrt{x}$, but didn't find it to be particularly useful, or practical , in this case.

Answer 1

Peter gives you one way in the comments. Here is another:

The Taylor series of $\sqrt x$ about $x = 9$ is $$ \sqrt x = \sqrt9 + \frac{1}{2\sqrt9}(x-9) - \frac{1}{4\sqrt{9^3}\cdot 2!}(x-9)^2 + \frac{3}{8\sqrt{9^5}\cdot 6!}(x-9)^3-\cdots $$ Taylor's theorem tells you an upper bound for the error at $x = 10$ at any given number of terms, so you just include enough terms to make that error smaller than $\frac1{100}$, and you're good to go.

---

A third, more elementary way is the following: Start with $3$ as an initial guess for $\sqrt{10}$. Now clearly $\sqrt{10}$ is somewhere between $3$ and $\frac{10}3$, so as a next approximation we take the average of $3$ and $\frac{10}3$. You would really want to take the geometric average $\sqrt{3\cdot \frac{10}3}$, because that's exactly what you're after, but that's what we are trying to in the first place. So we take the arithmetic average instead. That gives us $$ \frac{3 + \frac{10}3}{2} = \frac{19}{6} $$ as a second approximation. Is this good enough?

How can we even tell that we are close enough? Well, if $x$ is our approximation so far, then $\sqrt{10}$ must lie between $x$ and $\frac{10}x$, so when $|x-\frac{10}{x}|<\frac1{100}$, you're within the bounds you want.

---

A fourth way: continued fractions. We have $$ \sqrt{10} = 3 + \frac{1}{3+\sqrt{10}}\\ = 3 + \cfrac1{6 + \frac{1}{3+\sqrt{10}}}\\ = 3 + \cfrac1{6+\cfrac1{6 + \frac{1}{3+\sqrt{10}}}}\\ \vdots $$ Now, we clearly have $3<\sqrt{10}<4$, so it's straight-forward to insert $3$ and $4$ instead of the lowermost $\sqrt{10}$ at any step and evaluate the entire fraction to get an upper and a lower bound on $\sqrt{10}$. Once the bounds are closer than $\frac1{100}$ to one another, then you know that they're both closer to $\sqrt{10}$ than that, so pick one of them, and you're done.

Answer 2

Here, the most obvious and easiest method - bisection:

• set $x_{-1}=3, x_0 = 4$
• $3 < \sqrt{10} < 4 \rightarrow x_1 := \frac{x_{-1}+x_0}{2}= \frac{7}{2} \rightarrow |\sqrt{10} - x_1| < \frac{1}{2}$
• $3 < \sqrt{10} < \frac{7}{2} \rightarrow x_2 := \frac{x_0+x_1}{2}= \frac{13}{4} \rightarrow |\sqrt{10} - x_2| < \frac{1}{2^2}$
• ... go on till $\frac{1}{2^n} < \frac{1}{100}$
• $\rightarrow x_7$ satisfies your condition

:-)