The Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

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

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Guest

Watch Watch this User
Re: **trend function**
Posted: Nov 11, 2017 4:27 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply

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
Read Re: **trend function**
Guest
11/11/17
Read Re: **trend function**
bursejan@gmail.com

Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© The Math Forum at NCTM 1994-2017. All Rights Reserved.