Talk:Backward Euler method

Mistake in formula for fixed-point iteration

$y_{k+1}^{[0]}=y_{k},\quad y_{k+1}^{[i+1]}=y_{k}+hf(t_{k+1}^{[i]},y_{k+1}^{[i]}).$

in this formula, i guess it should read:

$y_{k+1}^{[0]}=y_{k},\quad y_{k+1}^{[i+1]}=y_{k}+hf(t_{k+1},y_{k+1}^{[i]}).$

as we don't improve the value of $t_{k+1}$ during iterations. wrong?

Thanks for the correction. I changed the article accordingly. -- Jitse Niesen (talk) 15:10, 24 April 2012 (UTC)[reply]

thanks — Preceding unsigned comment added by 182.73.186.2 (talk) 05:22, 28 March 2013 (UTC)[reply]

Local truncation error

I'm fairly certain that the 'Local truncation error' is of size $-{\frac {1}{2}}h^{2}y''(\xi )$ , thus $O(h^{2})$ . And, assuming the function and method don't 'amplify' the errors, the error at time $t_{n}$ is given by $n\times \tau _{n}={\frac {1}{h}}\times O(h^{2})=O(h)$ . So, it is only the 'global truncation error' which is of order $O(h)$ . — Preceding unsigned comment added by 213.124.211.182 (talk) 11:35, 18 January 2014 (UTC)[reply]

This does depend on the definition of 'local truncation error'. But I agree, using the definition in our article (which I think is the most common one), it is

O(h^{2})

, so I'll change it. -- Jitse Niesen (talk) 16:56, 18 January 2014 (UTC)[reply]

Region of absolute stability

The article says: "The region of absolute stability for the backward Euler method is the complement in the complex plane of the disk with radius 1 centered at 1, depicted in the figure.", and then provides a nice figure of a circle on a plane. It's probably evident to whomever constructed this figure, but it isn't to the reader: What are the axes of this plane? Could the figure provide a title for the axes? — Preceding unsigned comment added by 2001:480:20:22:0:0:0:10 (talk) 14:35, 3 June 2014 (UTC)[reply]