[sin(t) , cos(t), e^t] are these linearly dependent?

[Cecile]

[sin(t) , cos(t), e^t] are these linearly dependent?


can someone solve this q?
I write

Asint+BcosT+C(e^t) = 0

but then i cannot proceed...

 

[dextercioby]

Yes,they are independent.What are u trying to do...?Post the text of the problem...

Daniel.

 

[Cecile]

I'm trying to prove that they are whether dependent or independent. how can I show iy?

 

[dextercioby]

Are u trying to prove that they can form a basis in some vector space...?Define the vector space.

Daniel.

 

[Cecile]

No it just asks whether they are dep or in dependent.
for ex if it would have given me sets like v=(0,2,3) , u=(1,2,3,) & z=(1,1,0) then i could write
Av+Bu+Cz=0 and I could show that they are indep of not just by assuming one of the coef. is not zero. but i cannot show this with cos sin & e

 

[arildno]

Cecile: What is the DEFINITION of linear independence?

 

[dextercioby]

Here's a nice way to do it.Use differential equations.Consider the homogenous,linear,constant coefficient ODE

\frac{d^{4}y(x)}{dx^{4}}-y(x)=0

Daniel.

 

[Cecile]

if we don't have nontrivial solution for the combination of n vectors then these n vectors are said to be linearly dependent.

I know the DEFINITION just i cannot show that rule for sin cos & e
?!? am I not clear yet? :(

 

[arildno]

Remember the vector is the WHOLE function here; that is:
If you can show that the following equation,

a_{1}\sin(t)+a_{2}\cos(t)+a_{3}e^{t}=0

in order to be valid (that is, holds) for ALL values of "t" implies that a_{1}=a_{2}=a_{3}=0
then you have concluded that the three functions are linearly independent.

 

[dextercioby]

Or you could just compute the wronskian...

Daniel.

 

[Cecile]

thank you daniel:)

 

[snoble]

less machinery

Can I suggest an idea that doesn't involve the machinery of the Wronskian.
First consider the periodicity of the three functions. What can you conclude about the coefficient of e^x.
Now consider a root of cos. What does this tell you about the coefficient of sin. Now what must the coefficient of cos be?

 

[Data]

Or a (very tiny little) bit cleaner, note that we just need

A\sin{x} + B\cos{x} + Ce^x \equiv 0 \Longrightarrow A=B=C=0

setting x=0 immediately gives B+C = 0 \Longrightarrow B=-C. But as you noted the limiting behaviour of e^x at infinity implies C=0 so B=-C=0. But then A\sin{x}\equiv 0 obviously implies A=0 so we're done.

 

[matt grime]

Yes, thank goodness someone came up with the sensible and obvious approach. No nonn-trivial combination can be the zero *function*, and that can be gotten just from putting some values of x in.

 

[HallsofIvy]

Saying that sin(x), cos(x), and e^x are independent means that

In order for C₁sin(x)+ C₂cos(x)+ C₃e<sup>x[/sub]= 0 for all x, we must have C₁= C₂= C₃.

Take 3 different values for x:

x= 0 is especially easy: if x= 0, C₁sin(x)+ C₂cos(x)+ C₃ex[/sub]= 0 becomes C₂+ C₃= 0.
If x= \frac{\pi}{2}[/b], C_1+ e^{\frac{\pi}{2}}C_3= 0.<br /> <br /> Okay, C_2= -C_3 and C_1= -e^{\frac{\pi}{2}}C_3.<br /> Now put those into the original equation and take x to be some third number. Solve that for C₃. If C₃= 0 then so must C₁= 0 and <br /> C₂= 0 and the functions are independent.<br /> <br /> Yes, using the Wronskian is simpler.</sup>