# First Order Approximation of a $\sqrt{3.9}$

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

• Yes, you want to use $f(x) = \sqrt{x}$ and expand around $x=4$ – Moya Jul 17 at 15:19
• 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 $a = 4$. – User8128 Jul 17 at 15:19
• 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 – NoChance Jul 17 at 15:24

## 2 Answers

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

• 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. – 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, $$(a+\frac{b}{2a})^2 =a^2+2a\frac{b}{2a}+\frac{b^2}{4a^2} =a^2+b+\frac{b^2}{4a^2}$$.