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Re: **trend function**
Posted:
Nov 11, 2017 4:27 AM


k(x+n,xm) is always parallel to f'(x) is anyway not possible, there are differentiable functions
and points x, where this is not possible. Take this function f:[1,1]>[0,1]:
/ 1 x>=0 f(x) = < \ sqrt(1x) x<0
And the point x=0. There are no parallel secants to f'(x) at the point x=0.
Am Samstag, 11. November 2017 10:13:51 UTC+1 schrieb burs...@gmail.com: > For any n+m<>0. But I did not check whether k(x+n,xm) is always > parallel to f'(x), I guess its a little weaker. It is kind of > a mean slope between the two secants (xm,x) and (x,x+n). > > But at least we have k(x,x)=f'(x).



