anonymous
  • anonymous
what does the notation (f+g)(x)=f(x)+g(x)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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dape
  • dape
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 \(f(x)+g(x)\).
dape
  • dape
You could also think of it defining how to sum two functions \(f,g\) to get a new function \(f+g\).
dape
  • dape
That's better actually, since the thing you wrote down actually is a definition, not some deep fact about functions.

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anonymous
  • anonymous
could you please give an example..
anonymous
  • anonymous
yep you are right i encountered this fact while reading facts about vector spaces
dape
  • dape
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 \((f+g)(x)=1\)
anonymous
  • anonymous
awesum..!!could you give an example when this property will not be satisfied.. and thanks in advance
dape
  • dape
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.
dape
  • dape
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.
anonymous
  • anonymous
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?
dape
  • dape
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.
dape
  • dape
Yes, subspaces of infinite dimensional vector spaces can have infinite dimension as well.
anonymous
  • anonymous
then in that case is not that subspace "equal to" the vector space..??
anonymous
  • anonymous
as both are infinitely huge..we cannot visualize how subset falls inside the vector space
dape
  • dape
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.
anonymous
  • anonymous
thanx a lot ..your help is very much appreciated
dape
  • dape
No problem.

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