1
$\begingroup$

Use the first order approximation to determine $\sqrt{3.9}$. Use your calculator to determine the error in your approximation.

I am really confused by this question. There is no function so I don't know how can I use the Taylor approximation? Can I assume the $f(x) = \sqrt{x}$ ? If I can assume that I can use the first order approximation but I am not sure whether I can do that assumption.

Note: Thank you all for your help. As it turns out I can assume $f(x) = \sqrt{x}$. I also found this YouTube video which I believe will help anyone who has the same problem: Approximation of Square Roots by Taylor Approximation

$\endgroup$
  • 1
    $\begingroup$ Yes, you want to use $f(x) = \sqrt{x}$ and expand around $x=4$ $\endgroup$ – Moya Jul 17 at 15:19
  • 3
    $\begingroup$ The question is a bit ambiguous; you could expand a Taylor series for $f(x) = \sqrt x$ around any point $a \in (0,\infty)$ and git a different "first order approximation." But because the value that you want to compute is $\sqrt {3.9}$ it seems likely that are asking you to compute the first order Taylor approximation for $f(x) = \sqrt x$ centered at the point $a = 4$. $\endgroup$ – User8128 Jul 17 at 15:19
  • $\begingroup$ Hint: "First Order Approximation" means: In the case of a first-order approximation, at least one number given is exact - See: en.wikipedia.org/wiki/Order_of_approximation $\endgroup$ – NoChance Jul 17 at 15:24
9
$\begingroup$

The first order approximation means you use the Taylor series of $f(x) = \sqrt{x}$ up to the term of degree 1 (=first order). In this case it is very convenient to develop it around $x_0=4$, as it is close to $3.9$ and we already know that $\sqrt{4}=2$:

$$f(x_0+h)\approx f(x_0) + hf'(x_0)$$

In our case $x_0=4$ and $h=-0.1$ and with $f(x) = \sqrt{x}$ we get $f'(x) = \frac{1}{2\sqrt{x}}$ so

$$f(3.9) = f(x_0+h) \approx f(4) -0.1f'(4) = 2 -0.1\frac14 = 2-0.025 = 1.975$$

Actually $\sqrt{3.9} = 1.974841...$ (computed using a calculator) so our error is at about $1.6\cdot 10^{-4}$.

$\endgroup$
  • $\begingroup$ Thank you so much. I will accept this as an answer as soon as I can do - system doesn't let me at the moment. $\endgroup$ – curiouseng Jul 17 at 15:30
0
$\begingroup$

Since $\sqrt{1+x} \approx 1+\frac{x}{2} $ for small $x$,

$\begin{array}\\ \sqrt{a^2+b} &=a\sqrt{1+\frac{b}{a^2}}\\ &\approx a(1+\frac{b}{2a^2})\\ & a+\frac{b}{2a}\\ \end{array} $

If $a=2, b=-.1$, $\sqrt{3.9} \approx 2-\frac{.1}{2\cdot 2} = 2-\frac1{40} =2-.025 =1.975 $.

Note that $(2-\frac1{40})^2 =4-4\cdot\frac1{40}+\frac1{1600} =3.9+\frac1{1600} $ and, in general, $(a+\frac{b}{2a})^2 =a^2+2a\frac{b}{2a}+\frac{b^2}{4a^2} =a^2+b+\frac{b^2}{4a^2} $.

$\endgroup$

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.