Try to approximate this function at these three points for a deviation Δx =0.1

Question

Take the function y=x^2 . Take three points, say x=0, x=1, x=3. Try to approximate this function at these three points for a deviation Δx =0.1. For which of the three points the approximation works best? For which point it works worst? Can you explain why, in a qualitative way, just by looking at the graph of the function y=x2. Are my answers correct?

Here's what I have so far dy=f'(x)dx

Find dy if y=x^2 dy=2x

Find dy when x=0, x=1, x=3 and dx=.01 dy = 2(0)(.01)=0 dy = 2(1)(.01)=.02 dy = 2(3)(.01)=.06

Answer 1

Well, if we use Euler's Method as an approximation, then we can find approximations easily as follows: $$f'(x) = 2x$$ Now, we apply this with a step size $\Delta x = 0.1$ to solve the problem. Notice that for this specific problem, we can simply write the approximation of $f(1)$ as a finite series: $$f(1) \approx \sum_{k = 0}^{9} 0.2k,$$ and the series can be evaluated simply: $$\sum_{k = 0}^{9} 0.2k = \frac{(0 + 1.8)(10)}{2} = \boxed{0.9}.$$

The approximation is low. This makes sense because the curve is always concave up. I leave you to evaluate the approximations of $f(2)$ and $f(3).$

Answer 2

If you are required to use a linear function, you can use least squares techniques to find the best fit to your three points. The fact that you are evaluating $\Delta x$ suggests that you should fit $x=\sqrt y$ instead of the other way around. When I plug the data into Excel it finds $x=1.6888-0.2046y$ as the best linear approximation. Now evaluate the error in $x$. I find they all exceed $0.1$