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

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    $\begingroup$ Yes, you want to use $f(x) = \sqrt{x}$ and expand around $x=4$ $\endgroup$ – Moya Jul 17 at 15:19
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    $\begingroup$ 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

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 $1.6\cdot 10^{-4}$.

  • $\begingroup$ 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

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} $.


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