Question

what does the notation (f+g)(x)=f(x)+g(x)

Answer 1

It means exactly what you have written down. In words one could say that it's the same thing to add the functions \(f\) and \(g\) and calculate their sums value at \(x\) (\((f+g)(x)\)) or to calculate each functions value at \(x\) and then add these numbers \(f(x)+g(x)\).

Answer 2

You could also think of it defining how to sum two functions \(f,g\) to get a new function \(f+g\).

Answer 3

That's better actually, since the thing you wrote down actually is a definition, not some deep fact about functions.

Answer 4

could you please give an example..

Answer 5

yep you are right i encountered this fact while reading facts about vector spaces

Answer 6

Let's say \(f(x)=\sin^2x\) and \(g(x)=\cos^2(x)\), then to get the sum of these functions you can do
\(f(x)+g(x)=\sin^2x+\cos^2x=1=(f+g)(x)\).
So the sum of the function \(f\) and \(g\) is the constant function \((f+g)(x)=1\)

Answer 7

awesum..!!could you give an example when this property will not be satisfied..
and thanks in advance

Answer 8

That's right, with this addition rule and just the usual scalar multiplication (evaluate the function at x and multiply this number by a scalar). Most spaces of functions become a vector space, with the additive 0 just being the zero function.

Answer 9

It will always hold under certain restrictions since it's a definition. But the addition rule breaks down for example when \(f\) and \(g\) have different domains.

Answer 10

finally got it .. thanks a lot..

if i am not wasting your time can a subspace of a vector space(I know that V.S. may have infinite dim) have infinite dimension?

Answer 11

Take for example \(f(x)=x\) and \(g(x)=1/x\), then the addition rule breaks down since \(g\) does not have 0 in it's domain.

Answer 12

Yes, subspaces of infinite dimensional vector spaces can have infinite dimension as well.

Answer 13

then in that case is not that subspace "equal to" the vector space..??

Answer 14

as both are infinitely huge..we cannot visualize how subset falls inside the vector space

Answer 15

The subspace and the originial space would both be equally big (namely infinitely big), but one would be contained in the other. It's the same situation as this, imagine the set of all the integers, so \(\mathbb{Z}=\{0,1,-1,2,-2,...\}\). Now let's create a subset, the positive integers so that \(\mathbb{Z^+}=\{1,2,3,...\}\subset\mathbb{Z}\). They are both infinitely big, but one is contained in the other. We can also show that they are equally big, since we can pair up all the elements of the two sets without getting any elements left.

Answer 16

thanx a lot ..your help is very much appreciated

Answer 17

No problem.