Re: Variant of inner Product ...

• To: mathgroup at smc.vnet.net
• Subject: [mg56738] Re: [mg56683] Variant of inner Product ...
• From: Bob Hanlon <hanlonr at cox.net>
• Date: Thu, 5 May 2005 06:01:21 -0400 (EDT)
• Reply-to: hanlonr at cox.net
• Sender: owner-wri-mathgroup at wolfram.com

A sum of equal length Lists is the List of sums. Hence your requested output
is

{f[1,a],f[1,b]}+{f[2,c],f[2,d]}+{f[3,r],f[3,s]}

{f(1,a)+f(2,c)+f(3,r),f(1,b)+f(2,d)+f(3,s)}

which is to say

A={1,2,3};B={{a,b},{c,d},{r,s}};

{f[1,a],f[1,b]}+{f[2,c],f[2,d]}+{f[3,r],f[3,s]} == Inner[f,A,B]

True

Bob Hanlon

>
> From: Detlef Müller at smc.vnet.net
To: mathgroup at smc.vnet.net
> Date: 2005/05/04 Wed AM 12:32:49 EDT
> Subject: [mg56738] [mg56683] Variant of inner Product ...
>
> Hello,
>
> I have the following to do:
>
> Given
>
> In[1]:= A={1,2,3}; B={{a,b},{c,d},{r,s}};
>
> And a Function f, I like to have
>
> Out[2] = {f[1,a],f[1,b]}+{f[2,c],f[2,d]}+{f[3,r],f[3,s]}
>
> The trial
>
> In[8]:=A={1,2,3}; B={{a,b},{c,d,e},{r,s}};
> In[9]:= Inner[f,A,B]
> Out[9]= f[1,{a,b}]+f[2,{c,d,e}]+f[3,{r,s}]
>
> looks promising,
> but if the Lists in B have the same length, "Inner"
> makes something different:
>
> In[15]:=
> A={1,2,3}; B={{a,b},{c,d},{r,s}}; Inner[f,A,B]
>
> Out[16]= {f[1,a]+f[2,c]+f[3,r],f[1,b]+f[2,d]+f[3,s]}
>
> So for now I have an ugly Table-Construction doing the job,
> but I can't imagine there is no elegant and clear solution
> for this ... any suggestions?
>
> Greetings,
>    Detlef
>
>

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