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Re: Turning Derivative into Function (Newbie Question)


Thanks guys! That pure function approach looks the most concise...that would
be the best approach with regards to Mathematica's design philosophy?

I'll sink my teeth into the documentation. :)

On Tue, Jan 11, 2011 at 7:10 AM, Murray Eisenberg <murray at math.umass.edu>wrote:

> Because g is NOT a "function" when you set:
>
>  g := f'[x]
>
> (No more than f would be had you started with f := x^2.)
>
> Whenever you now call g, you get the value of f[x], namely, 2x.
>
> If you really want to define g that way, then you can use a rule and
> replacement:
>
>  g /. x-> 2
>
> But it's probably better to do things like this:
>
>  Clear[g]
>  g[x_] = f'[x]   (* note Set, =, not SetDelayed, := *)
>  g[2]
> 4
>
> Or, if you prefer a more functional approach:
>
>  Clear[g]
>  g = f'
>  g[2]
> 4
>
>
> On 1/11/2011 6:58 AM, Just A Stranger wrote:
>
>> When I apply a function that outputs another function, how come I can not
>> pass arguments to the new function? Am I missing some principle about how
>> Mathematica functions work? Could someone please point me to the relevant
>> documentation?
>>
>>
>> A simple example to illustrate my confusion:
>>
>> In[1]: f[x_] := x^2
>> In[2]: g := f'[x]
>>
>> So now I have:
>>
>> In[3]: g
>> out[3]: 2x
>>
>> as expected
>>
>> So how come I get the following when trying to pass an argument to g:
>>
>> In[4]: g[2]
>> Out[4]: 2x[2]
>>
>> Instead of the output I want:
>>
>> *Ou[4]: 4
>>
>>
>> I tried
>>
>> In: g[x_] := f'[x]
>>
>> But it seems to think I'm trying to assign a function to Times.
>> "SetDelayed::write: Tag Times in (2 x)[y_] is Protected.>>"
>>
>> Thank you very much for any help :)
>>
>>
>>
> --
> Murray Eisenberg                     murray at math.umass.edu
> Mathematics & Statistics Dept.
> Lederle Graduate Research Tower      phone 413 549-1020 (H)
> University of Massachusetts                413 545-2859 (W)
> 710 North Pleasant Street            fax   413 545-1801
> Amherst, MA 01003-9305
>


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