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

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The discussion centers on determining whether the functions sin(t), cos(t), and e^t are linearly dependent or independent. Participants suggest using the definition of linear independence, which states that if a linear combination of the functions equals zero only when all coefficients are zero, they are independent. A proposed method involves evaluating the functions at specific values to derive relationships between the coefficients, ultimately showing that all must equal zero. The conversation highlights that while the Wronskian can simplify the proof, a more intuitive approach using properties of the functions can also demonstrate their independence. The conclusion is that sin(t), cos(t), and e^t are indeed linearly independent functions.
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[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...
 
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Yes,they are independent.What are u trying to do...?Post the text of the problem...

Daniel.
 
I'm trying to prove that they are whether dependent or independent. how can I show iy?
 
Are u trying to prove that they can form a basis in some vector space...?Define the vector space.

Daniel.
 
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
 
Cecile: What is the DEFINITION of linear independence?
 
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.
 
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? :(
 
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
Or you could just compute the wronskian...:rolleyes:

Daniel.
 
  • #11
thank you daniel:)
 
  • #12
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
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.
 
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  • #14
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
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], 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 C<sub>3</sub>. If C<sub>3</sub>= 0 then so must C<sub>1</sub>= 0 and <br /> C<sub>2</sub>= 0 and the functions are independent.<br /> <br /> Yes, using the Wronskian is simpler.
 

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