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Topic: Re: **trend function**
Replies: 1   Last Post: Nov 11, 2017 4:32 AM

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

k(x+n,x-m) 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(1-x) 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,x-m) is always
> parallel to f'(x), I guess its a little weaker. It is kind of
> a mean slope between the two secants (x-m,x) and (x,x+n).
>
> But at least we have k(x,x)=f'(x).

Date Subject Author
11/11/17 Guest
11/11/17 bursejan@gmail.com