# Having trouble calculating approximations using Taylor polynomials

I have a problem to approximate $\sqrt{1.06}$ using a third degree Taylor polynomial. The way I learned was to pick a center that we would know the answer to that is close to the value we're trying to find. So, I chose the center to be 1.

Next I took the derivatives to find the coefficients. This could be where I made my mistake, because I'm not sure what function to use when a function is not given, but I chose my function to be $f(x)=\sqrt{1+x}$. I found the derivatives to get the coefficients as follows:

$f'(x)=\dfrac{1}{2\sqrt{1+x}}$ so $f'(1)=\dfrac{1}{2\sqrt2}$

$f''(x)=-\dfrac{1}{4\cdot(x+1)^{3/2}}$ so $f''(1)=-\dfrac{1}{4\cdot(2)^{3/2}}$

$f''(x)=\dfrac{3}{8\cdot(x+1)^{5/2}}$ so $f''(1)=\dfrac{3}{8\cdot(2)^{5/2}}$

Then I wrote the 3rd degree taylor polynomial as:

$p_3(x)=\sqrt2+\dfrac{(x-1)}{2\sqrt2}-\dfrac{(x-1)^2}{8(2)^{3/2}}+\dfrac{3(x-1)^3}{40(2)^{5/2}}$

So then I plugged in .06 for $x$ and got $1.031811$.

The actual approximated value from my calculator is $1.029563$ so somewhere I did something wrong. Yes, $1.031811$ is close to $1.029563$, but my book says the calculated value should have been $1.029564$, so there is a very small difference between the expected calculated estimate and the actual estimate (much smaller than the result I got).

I suspect it's either with the function I used or the value for $x$, but I'm not sure what to do instead of what I did. Can anyone help?

• A small comment, since there have been answers but they may not have left it clear: if your $f(x)=\sqrt{1+x}$ then it's the expansion about $0$, not $1$, that you want; expanding $\sqrt{1+x}$ around $x=1$ will give good approximations to numbers near $\sqrt{2}$ (as you can see by plugging $x=1$ into $f(x)=\sqrt{x+1}$). Your choice of function was excellent, though. – Steven Stadnicki Apr 28 '14 at 5:42
• So when choosing a center, you want that center value plugged into the function to be close to the approximation you're trying to find...I think I get it now. Thank you. – Sabien Apr 28 '14 at 7:04

Given that $f(x) = \sqrt{1+x}$, we know that $\sqrt{1.06} = f(0.06)$. Since $0.06$ is close to $x_0 = 0$, this suggests that we center our approximation at $0$. [Alternatively, you could have defined $g(x) = \sqrt{x}$, in which case $\sqrt{1.06} = g(1.06)$, so you could center it at $x_0 = 1$]. Hence, we obtain: \begin{align*} f(x) &\approx p_3(x) \\ &= f(0) + f'(0)(x - 0) + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 \\ &= \sqrt{1+0} + \frac{1}{2\sqrt{1+0}}x + \frac{\frac{-1}{4(0+1)^{3/2}}}{2}x^2 + \frac{\frac{3}{8(0+1)^{5/2}}}{6}x^3 \\ &= 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 \\ \end{align*} Thus, substituting $x = 0.06$ yields: \begin{align*} \sqrt{1.06} &= f(0.06) \\ &\approx p_3(0.06) \\ &= 1 + \frac{1}{2}(0.06) - \frac{1}{8}(0.06)^2 + \frac{1}{16}(0.06)^3 \\ &= 1.0295635 \end{align*}

• I see, that makes sense. I guess I don't fully understand how to choose the center. Thank you. – Sabien Apr 28 '14 at 4:49

You should derive the series at the point $x=0$ as

$$\sqrt{1+x} \approx 1+\frac{1}{2}\,x-\frac{1}{8}\,{x}^{2}+\frac{1}{16}x^3 \implies \sqrt{1+0.06}\approx 1.029563500.$$

You can use $\ a \ = \ 1 \$ as the center, but you must then choose $\ f(x) \ = \ \sqrt{x} \ = \ x^{1/2} \$ as your function. The first three derivatives are then

$$f'(x) \ = \ \frac{1}{2}x^{-1/2} \ \ , \ \ f''(x) \ = \ -\frac{1}{4}x^{-3/2} \ \ \ , \ \ f'''(x) \ = \ \frac{3}{8}x^{-5/2} \ \ .$$

This makes your third-degree Taylor polynomial

$$T_3(x) \ = \ 1^{1/2} \ + \ \frac{1/2}{1^{1/2} \cdot 1!} \ (x - 1) \ + \ \frac{-1/4}{1^{3/2} \cdot 2!} \ (x - 1)^2 \ + \ \frac{3/8}{1^{5/2} \cdot 3!} \ (x - 1)^3 \ \ ,$$

from which you would compute

$$\sqrt{1.06} \ \approx \ 1^ \ + \ \frac{1}{2} \ (0.06) \ - \ \frac{1}{8} \ (0.06)^2 \ + \ \frac{1}{16} \ (0.06)^3 \ \ ,$$

leading to the same result as Adriano shows. (In fact, the alternative here must produce the same Taylor coefficients as the series for $\ (1+x)^{1/2} \$ at $\ a \ = \ 0 \$ .)

The use of $\ \sqrt{1 + x} \$ allows us to use $\ x \ = \ 0 \$ as center to write a Maclaurin series (you'll find that it is a frequently-tabulated function). It can be viewed as making a variable-substitution. It was fine for use to use the center you did, but then the function must be adapted appropriately.

• How do you adapt the function based on the center choice? I'm not quite sure about that part. Originally, I thought you just choose a number that you know the answer to ($\sqrt1$ is 1) that is close to the approximation you're looking for, but obviously that's not right. – Sabien Apr 28 '14 at 7:01
• You are looking to set up a series where the terms raised to the $\ k$th power will be numbers much smaller than 1; that is what makes the Maclaurin or Taylor series useful approximations of the value of a function close to the "center" $\ x = a \$ . Using $\ \sqrt{x} \$ with a center $\ a = 1 \$ will have terms in the Taylor series $\ (x - 1)^k \$ ; using $\ \sqrt{1+x} \$ with center $\ a = 0 \$ will have "Maclaurin" terms $\ x^k \$ . The Taylor series for $\ \sqrt{1.06} \$ will thus have terms $\ (1.06 - 1)^k \$ , [continued] – colormegone Apr 28 '14 at 15:33
• while the Maclaurin series will use $\ (0.06)^k \$ because we are working with $\ 1 + x \ = 1.06 \$ . So the function $\ \sqrt{1+x} \$ is just being used as a version of $\ \sqrt{x} \$ "shifted" 1 unit "to the left". (And, yes, you do want to choose a center with a known value of the function and derivatives close to the number for which you plan to make the evaluation of $\ f(x) \$ . That is what keeps the terms $\ (x-a)^k \$ small.) – colormegone Apr 28 '14 at 15:35