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

1. Apr 4, 2005

### 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...

2. Apr 4, 2005

### dextercioby

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

Daniel.

3. Apr 4, 2005

### Cecile

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

4. Apr 4, 2005

### dextercioby

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

Daniel.

5. Apr 4, 2005

### 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

6. Apr 4, 2005

### arildno

Cecile: What is the DEFINITION of linear independence?

7. Apr 4, 2005

### 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.

8. Apr 4, 2005

### Cecile

if we dont 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? :(

9. Apr 4, 2005

### 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.

10. Apr 4, 2005

### dextercioby

Or you could just compute the wronskian...:uhh:

Daniel.

11. Apr 4, 2005

### Cecile

thank you daniel:)

12. Apr 4, 2005

### 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?

13. Apr 4, 2005

### 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.

Last edited: Apr 4, 2005
14. Apr 4, 2005

### 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.

15. Apr 4, 2005

### HallsofIvy

Staff Emeritus
Saying that sin(x), cos(x), and ex are independent means that

In order for C1sin(x)+ C2cos(x)+ C3ex[/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 $$x= \frac{\pi}{2}[/b], [tex]C_1+ e^{\frac{\pi}{2}}C_3= 0$$.

Okay, $$C_2= -C_3$$ and $$C_1= -e^{\frac{\pi}{2}}C_3$$.
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
C2= 0 and the functions are independent.

Yes, using the Wronskian is simpler.