The book here (sorry not in English) on page 676:

$$\begin{cases}y'=-y^{2} \\ y(0)=1\end{cases}$$

when $x_{k}=\frac{k}{5}$ which means $h=0.2$ i.e. $\Delta x=x_{k+1}-x_{k}=0.2:=h$. The task is to calculate 5 steps with the logic on pages 670-671:

$$\begin{align*} y_{k+1}&=y_{k}+hf(x_{k},y_{k}) \\ h&=x_{k+1}-x_{k} \\ k&=0,1,...\end{align*}$$ where $f(x_{k},y_{k})=y'$.

My solution

I calculated the Euler values on the left and the right having the exact values. For example, $y_{1}=y_{0}+h \left|y'(y_{0})\right|=1-0.2*(1)^{2}=\frac{4}{5}$. I got the same as the R verified values below but my exact values are $0.5, 0.66, 0.6, 0.625, 0.61...$ which I think are wrong. The way I calculated the exact values is this:

  1. separate-integrate $y'=-y^{2}$
  2. you get $y(x)=\frac{1}{x+1}$ because $y(0)=1$
  3. Then $y(y_{0})=y(1)=0.5, y(y_{1})=y(0.5)=2/3,...$ until recursion ends to the fifth.

Are my results correct?

R -verification for the Euler -method

I was eventually able to get the same values as here by hand.

> euler<-function(y){a<-y; for(i in 1:5){b<-a-0.2*a**2; print(b); a<-b}}
> euler(1)
[1] 0.8
[1] 0.672
[1] 0.5816832
[1] 0.5140121
[1] 0.4611704

Thanks to Julian, I had misunderstanding in the parameters. The below should be correct.

> 1/((1:5/5)+1)
[1] 0.8333333 0.7142857 0.6250000 0.5555556 0.5000000

Let's compare the values. The exact values are always greater than the Euler values because of the derivative $y'(x)=\frac{-1}{(x-1)^{2}}$, I am not yet sure how to vizualize this in R, basically the above values into data.frame and then matplot(df), investigating...

plot(1/(1:100+1), type='l' )

enter image description here

Perhaps related

  1. Euler's Method for slope fields
  • $\begingroup$ ...found instructions here how to verify this kind of things on Maple, unfortunately I have no Maple. Perhaps, there is some other way to check whether I am doing things correctly, investigating... $\endgroup$ – hhh Feb 25 '12 at 21:20
  • $\begingroup$ You need to explain what is "the same book as earlier", what you meant by h, $y_0$ and $y_1$, where k has gone, how you came up with "recursion" (instead of iteration), ... lots of things $\endgroup$ – nodakai Feb 25 '12 at 21:22
  • $\begingroup$ @nodakai: Wow, I just realized the book has become online! Cool -- now you can see it in the updated question. If I can understand the Euler method correctly, it is a recursive equation. Please, see the bottom: given $y_{0}$, then calculate $y_{1}$, then $y_{2}$ and so on. $\endgroup$ – hhh Feb 25 '12 at 21:44

You are confusing $x_k$ and $y_k$ in the exact values. Expresions like $y(y_1)$ make no sense.

$y_k$ is an approximation of the exact value $y(x_k)$. The exact values are $$ y(0)=1,\quad y(0.2)=\frac1{1+0.2}=0.833,\quad y(0.4)=\frac1{1+0.4}=0.714, $$ $$ y(0.6)=\frac1{1+0.6}=0.625,\quad y(0.8)=\frac1{1+0.8}=0.555,\quad y(1)=\frac1{1+1}=0.5. $$

  • $\begingroup$ You apparently approximate values with $y_{k+1}\approx y(x_{k+1})$ (the page 670) where $x_{k+1}=h+x_{k}$? Yes that is clear now, no recursion in this function. $\endgroup$ – hhh Feb 25 '12 at 23:14

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