Re: to first order ??


"David C. Ullrich" <ullrich@xxxxxxxxxxxxxxxx> wrote in message news:42vsj158fj647e0dh96kb3fkrssg7hk6dd@xxxxxxxxxx
> On Fri, 30 Sep 2005 20:02:46 GMT, "Dirk Van de moortel"
> <dirkvandemoortel@xxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
>
> >
> >"Zinc Potterman" <zincnews@xxxxxxxxxxxxxxxxx (delete 123's to reply)> wrote in message
> >news:dhk4s7$h7e$1@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
> >> My physics book says "
> >> to first order in delta x,
> >>
> >> delta f = (df/dx) delta x
> >>
> >> What does this mean please?
> >
> >Abbreviating
> > delta x to dx
> >and
> > delta f to df
> >
> >See
> > http://mathworld.wolfram.com/TaylorSeries.html
> > x --> x + dx
> > a --> x
> > x-a --> dx
> >
> >df = f(x + dx) - f(x) by definition
> > = f(x) + f'(x) dx + 1/2 f''(x) dx^2 + 1/6 f'''(x) dx^3 + ....
> > - f(x) by Taylor expansion
> > = f'(x) dx + 1/2 f''(x) dx^2 + 1/6 f'''(x) dx^3 + .... by simplification
> > = f'(x) dx by throwing away higher powers of dx
> > because too small.
>
> No, that's really not quite what it means. If f is any
> differentiable function then it's true that delta f = (df/dx) delta x
> "to first order in delta x", but it's not necessarily true that
> f equals its Taylor series, in fact f need not even _have_ a
> Taylor series.
>
> What it does mean is what Wade said: delta f = (df/dx) delta x
> plus an error term which is asyptotically smaller than delta x
> as delta x -> 0.

I bet it's quite what it means. The guy was reading a physics
book.
On the next page he'll read "to second order in delta x".
So with what I gave him, he is ready for that.
What WWW wrote can be correct, but is useless for a
physics student.

Dirk Vdm


.

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