MHB How do we get the approximation?

[evinda]

Hello! (Wave)

Approximating $y'(t^n)$ at the relation $y'(t^n)=f(t^n,y(t^n))$ with the difference quotient $\left[\frac{y(t^{n+1})-y(t^n)}{h} \right]$ we get to the Euler method.

Approximating the same derivative with the quotient $\left[\frac{y(t^{n})-y(t^{n-1})}{h} \right]$ we get to the backward Euler method

$$y^{n+1}=y^n+hf(t^{n+1},y^{n+1}), n=0, \dots, N-1$$

where $y^0:=y_0$.
In order to find the formula for the forward Euler method, we use the limit $\lim_{h \to 0} \frac{y(x_0+h)-y(x_0)}{h}$ for $x_0=t^n, h=t^{n+1}-t^n$.
In order to find the formula for the backward Euler method, could we pick $h=t^{n-1}-t^n$ although it is negative?Or how do we get otherwise to the approximation:$$y'(t^n) \approx \frac{y(t^n)-y(t^{n-1})}{h}$$

?

 

[I like Serena]

In order to find the formula for the backward Euler method, could we pick $h=t^{n-1}-t^n$ although it is negative?

Or how do we get otherwise to the approximation:

$$y'(t^n) \approx \frac{y(t^n)-y(t^{n-1})}{h}$$

?

Hi! (Wave)

We should still pick $h$ positive.
I previously explained to http://mathhelpboards.com/members/mathmari/ how to find that derivative in this thread. (Thinking)

 

[evinda]

Hi! (Wave)

We should still pick $h$ positive.
I previously explained to http://mathhelpboards.com/members/mathmari/ how to find that derivative in this thread. (Thinking)

Could you explain it further to me? (Thinking)

 

[I like Serena]

Or how do we get otherwise to the approximation:

$$y'(t^n) \approx \frac{y(t^n)-y(t^{n-1})}{h}$$

?

Could you explain it further to me? (Thinking)

Since we don't know $y(t^n)$ yet, we need to iterate to find it. (Thinking)

We start with an initial estimate:
$$y_n^{[0]} = y(t^{n-1})$$
And then we can iterate with:
$$y_{n}^{[i+1]} = y(t^{n-1}) + h f(t_{n}, y_{n}^{})$$
(Wasntme)