0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Steven Stadnicki Apr 28 '14 at 5:42
  • $\begingroup$ 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. $\endgroup$ – Sabien Apr 28 '14 at 7:04
1
$\begingroup$

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*}

$\endgroup$
  • $\begingroup$ I see, that makes sense. I guess I don't fully understand how to choose the center. Thank you. $\endgroup$ – Sabien Apr 28 '14 at 4:49
1
$\begingroup$

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

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Sabien Apr 28 '14 at 7:01
  • 1
    $\begingroup$ 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] $\endgroup$ – colormegone Apr 28 '14 at 15:33
  • 1
    $\begingroup$ 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.) $\endgroup$ – colormegone Apr 28 '14 at 15:35

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.