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

by Cecile
Tags: dependent, linearly, sint
 P: 5 [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...
 HW Helper Sci Advisor P: 11,722 Yes,they are independent.What are u trying to do...?Post the text of the problem... Daniel.
 P: 5 I'm trying to prove that they are whether dependent or independent. how can I show iy?
HW Helper
P: 11,722

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

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

Daniel.
 P: 5 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
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 Cecile: What is the DEFINITION of linear independence?
 HW Helper Sci Advisor P: 11,722 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.
 P: 5 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? :(
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 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.
 HW Helper Sci Advisor P: 11,722 Or you could just compute the wronskian... Daniel.
 P: 5 thank you daniel:)
 P: 127 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?
 P: 998 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.
 HW Helper Sci Advisor P: 9,395 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.
 PF Patron Sci Advisor Thanks Emeritus P: 38,429 Saying that sin(x), cos(x), and ex 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)+ C3e[sup]x[/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.

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