OpenStudy (anonymous):

Let C1 be the set of all differentiable functions. Show that C1 is a vector space.

5 years ago
OpenStudy (anonymous):

Help Please

5 years ago
OpenStudy (anonymous):

ok lets go slow, one at a time

5 years ago
OpenStudy (anonymous):

ok

5 years ago
OpenStudy (anonymous):

starting with number one on the list here http://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpaces.aspx

5 years ago
OpenStudy (anonymous):

first one says if \(u, v\in V\) then \(u+b\in V\) so what do we have to check in this case? we need to check that if \(f,g\) are differentiable functions, then so is \(f+g\) because that will mean it is closed under addition

5 years ago
OpenStudy (anonymous):

but this is straightforward because we know that the derivative of a sum is the sum of the derivative, so we know that \((f+g)'=f'+g'\) and so if \(f,g\) are differentiable, then so is \(f+g\)

5 years ago
OpenStudy (anonymous):

that takes care of axiom a from the list

5 years ago
OpenStudy (anonymous):

ok I understand that

5 years ago
OpenStudy (anonymous):

what does the next one say?

5 years ago
OpenStudy (anonymous):

if it's multuple?

5 years ago
OpenStudy (anonymous):

yes, you have to check that if \( c\) is any number, and \(f \) is a differentiable function, then \(cf\) is differentiable as well

5 years ago
OpenStudy (anonymous):

I think it's right

5 years ago
OpenStudy (anonymous):

because if I have f is differentiable then cf is differentiable .So, cf prime is differentiable

5 years ago
OpenStudy (anonymous):

Yes^^

5 years ago
OpenStudy (anonymous):

ok but I need to proof it

5 years ago
OpenStudy (anonymous):

So: \[(cf)' = c(f)' \] Which is definitely a differentiable function.

5 years ago
OpenStudy (anonymous):

I want to proof the question not this

5 years ago
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