Linear Dependence and Span Question

[Nexttime35]

Homework Statement

Is the following set linearly dependent or independent? And does this set span the given space?
{e^X, e^-x}\inC^∞(R)

Homework Equations

The Attempt at a Solution

So, if it's linearly independent, then:
k1e^x +k2e^-x = 0 where k1,k2=0 and only 0. But if you let k1= -1/(e^x) and k2 = 1, then you also get 0, so this set is linearly dependent. I believe I am correct with this logic.

Now how do I go about checking whether the set is in the span of the vector space?

 

[hilbert2]

But if you let k1= -1/(e^x) and k2 = 1, then you also get 0, so this set is linearly dependent. I believe I am correct with this logic.

The coefficients in the linear combination must be numbers, not functions of $x$. You should approach the problem by making the assumption that for some numbers $a,b$, the function $f(x) = ae^{x}+be^{-x}$ is zero for any value of $x$. Then derive a contradiction, the function must be nonzero at some point.

The set $C^{\infty}(\mathbb{R})$ is infinite dimensional. You can't span it with only two functions.

 

[Nexttime35]

Ahh, ok, that makes sense. Thank you for the help. I'll see what I can do!

 

[Nexttime35]

So, deriving f(x)=ae^x+be^-x I get f'(x) = ae^x -be^-x, and f''(x) = ae^x+b^-x, so I'm back to the original function with the second derivative, and so forth. When proving linear dependence with functions, I understand that I need to show f(x) = 0 for any x, and where a,b≠0. Is that correct?

 

[hilbert2]

If $a$ and $b$ are nonzero numbers, one of the terms $ae^{x}$, $be^{-x}$ grows without bound in absolute value and the other one approaches zero when $x \rightarrow \infty$. From this it follows that the function $f(x)$ must be nonzero for some $x$. You just have to express this in a more exact way to make it a real mathematical proof.

 

[Nexttime35]

Gotcha. I follow. Thanks for your help, hilbert2.

 

[Nexttime35]

One more question: is C^∞(R) represents the vector space of functions with infinitely continuous derivatives, then why wouldn't the above functions in that set span that vector space? Don't those functions have infinitely continuous derivatives?

 

[hilbert2]

One more question: is C^∞(R) represents the vector space of functions with infinitely continuous derivatives, then why wouldn't the above functions in that set span that vector space? Don't those functions have infinitely continuous derivatives?

Saying that a set of functions {$f_{1},f_{2},f_{3},\dots$} spans the space $A$ means that any function in $A$ can be represented as a linear combination of functions $f_{i}$. This is clearly not the case here. You need a set of infinitely many functions to span $C^{\infty}(\mathbb{R})$.