say you have the set of all functions defined at 1 with f(1)=0 or a homogeneous differential equation y''+ay'+by=0, these are both linear spaces because they satisfy closure under addition/multiplication but why isn't it closed if you change 0 to a number c?
I see the sum must be 0 but how exactly does this follow from addition closure which says that there is a unique sum x+y, isn't 2c unique? is there a deeper insight for differential equations and a relation to one to one injectiveness?
Yes. The differential equations of the form some left hand side = 0, are known as the homogeneous solutions to that differential equation. Because the right hand side is zero if you have two solutions to that equation then their sum is also a solution and in general the homogeneous solutions form a space and finding various bases for that space is important.
Some of the harder core math guys on here can expand on this I'm sure.
I remember that we'd use a series of solutions to the homogeneous equation to solve for boundary conditions using Fourier analysis.
this isn't what closure says
closure says if you have a set of elements with an addition operation then if you add two elements of that set the sum is also in that set.
so if my set of functions if specified as
$$f_i(1)=c$$
then if closure is to be satisfied
$$(f_i+f_j)(1)=c\;\;\forall i,j$$
which is violated if $$c \neq 0$$
Thanks very much!!! Apostol defines addition closure as- For every pair of elements x and y in V there corresponds a unique elemnt in V called the sum of x and y, denoted by x+y.
Yes that is so beautiful that the solutions to homogeneous equations form a space and superposition can be used. there is a really nice Braun Springer text on DE
I wonder why Apostol adds bijection...there's a really nice online latex editor- I'm trying to get the type on this site to load with mathjax
Online LaTeX Equation Editor - create, integrate and download
take this proof from Apostol on the uniqueness of 0. if there existed two 0s 01=x, 02=0 then 01+02=01. then define now 02=x, 01=0, 02+01=02. by commutativity 01+02=02+01, but I don't see how this proves the hypothesis. haven't you proven only that x=x? maybe I'm missing something in the proof