4 ms·
Lost me a bit at Im(f(x+ih))/h near the end (that's a fancy I that android FF won't paste). Can anyone explain where 'm' came from? Or is 'Im' just a fn
by edoloughlin 11y ago
Lost me a bit at
Im(f(x+ih))/h
near the end (that's a fancy I that android FF won't paste). Can anyone explain where 'm' came from?
Or is 'Im' just a fn returning the imaginary part of its argument?
EDIT: reading the code following, it's clear that Im is just that.
- czzarr 11y agoIm is for imaginary part of its argument as you say
- deleted 11y ago[deleted]
- mbroshi 11y agoI was also lost at that part, although for a different reason. The function f was assumed to be from R to R, so it does not make formal sense to plug in f(x+ih). I guess the author has the unstated assumption that f is the restriction to R of a function on C.
- eriktaubeneck 11y agoGreat point, and I apologize for the confusion. The function should be C to C, with the added restriction of R to R.
- eriktaubeneck 11y agoThanks, I just realized that when I introduced the notation above, I just used I(f), but then when I used it below I wrote Im(f). I will clear that up.