Functional Equations PSet : From Very Basic

For discussing Olympiad Level Algebra (and Inequality) problems
Corei13
Posts:153
Joined:Tue Dec 07, 2010 9:10 pm
Location:Chittagong
Functional Equations PSet : From Very Basic

Unread post by Corei13 » Tue Mar 27, 2012 9:39 pm

Here are 50 Functional Equations ( and Inequalities ) from Mathlinks, various contests, own and including basic equations like Cauchy's, Jensen's and D-Alembert's.

( There may be some confusion with the definition of range and co-domain, Wiki says Range $\subseteq$ Co-domain, but somehow not only me, but also a lot of Mathlinker knows the opposite. Anyway, definition is what we define :) )
Attachments
fe_camp.pdf
(72.66KiB)Downloaded 284 times
ধনঞ্জয় বিশ্বাস

User avatar
Nadim Ul Abrar
Posts:244
Joined:Sat May 07, 2011 12:36 pm
Location:B.A.R.D , kotbari , Comilla

Re: Functional Equations PSet : From Very Basic

Unread post by Nadim Ul Abrar » Tue Mar 27, 2012 9:56 pm

অনেকগুলা ধন্যবাদ
$\frac{1}{0}$

User avatar
Phlembac Adib Hasan
Posts:1016
Joined:Tue Nov 22, 2011 7:49 pm
Location:127.0.0.1
Contact:

Re: Functional Equations PSet : From Very Basic

Unread post by Phlembac Adib Hasan » Tue Mar 27, 2012 9:59 pm

Again I am requesting you to use ''image'' instead of ''co-domain" and write image$\subseteq $Range.Bacause somewhere co-domain refers to range (According to our text book) and somewhere to image (According to mathlinks).That's why Saumitra Vaia advised me not to use "co-domain".And a lot of thanks for the note.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules

User avatar
nayel
Posts:268
Joined:Tue Dec 07, 2010 7:38 pm
Location:Dhaka, Bangladesh or Cambridge, UK

Re: Functional Equations PSet : From Very Basic

Unread post by nayel » Tue Mar 27, 2012 10:29 pm

Adib is right, your use of codomain is wrong and should be replaced by image/range (as should be done here). It is always best not to use "range" because it can mean either codomain or image.

If $f:A\to B$, then
the codomain of $f$ is $B$
the image of $f$ is $\{f(x):x\in A\}$

So, for example, $f:\mathbb R\to\mathbb R$, $f(x)=x^2$ has codomain $\mathbb R$, and image $\mathbb R_{\ge 0}$. The domain and codomain must be specified when defining a function.
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

Post Reply