Euler method without defined function 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.
 A: 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.
A: 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.
