Euler method without defined function

Question

Before I begin, I must say that reading and quickly understanding the definitions of mathematical theorems is not my strongest suit.

We had a test where one of the subjects was Eulers methods. The problems we have in our study book (Adams and Essex) are problems that I felt went relatively easy, but on this test we encountered an even worse problem. The problem is:

The function f(x) is differentiable with a continuous derivative f'(x).

Some of the values of this derivative is listed in the table below:

(I will not list them now, but there is a row for values for x and what the value of f'(x) will be if you substitute this x with the value above. I'm not quite sure if I explained this well, but I hope someone gets the picture).

Furthermore, the problem says:

Suppose that f(1.4) = -0.9. Use Eulers method with stepsize h = 0.2 to find an approximate value for f(2.0).

This is the task. I'm not looking to have the problem solved for me, I only wish to understand the method of how I can solve this task. In my calculus book, the problems tied to Eulers method has contained a function, while in this problem it seems we have to figure out what the function is.

Thank you so much in advance if anyone replies.

Answer 1

But you only use the function to find the y-value at the beginning and the value of its derivative at the various steps along the way.

Since the starting point is $x=1.4$, the ending point is $x=2.0$, and the step size is $h=0.2$, you know the x-values you will be stepping through.

You are given the y-value at the beginning. Euler's method uses that and the value of the derivative there to produce an approximation of the y-value at the next step.

Presumably your table has the values of the derivatives at all of the steps along the way.

So start computing the successive y-value approximations until you get the one at the end of the stepping.

Answer 2

The standard forward Euler method is based on the approximation $$ f(x + h) \approx f(x) + f'(x) \cdot h $$ In your case, this means that with $h = 0.2$, you have the following approximations:

$f(1.4)$	$f(1.6)$	$f(1.8)$	$f(2.0)$
$-0.9$ (given)	$f(1.4) + f'(1.4)\times 0.2$	$f(1.6) + f'(1.6)\times 0.2$	$f(1.8) + f'(1.8)\times 0.2$

(Sorry about the formatting; not sure why MathJax plays oddly with Markdown Tables)

Based on the table of values of $f'$ that is given to you, you can fill in this table starting from the left successively.