
#1
Apr405, 01:50 PM

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... 



#2
Apr405, 01:54 PM

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P: 11,863

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



#3
Apr405, 01:57 PM

P: 5

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




#4
Apr405, 02:01 PM

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P: 11,863

[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. 



#5
Apr405, 02:06 PM

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 



#6
Apr405, 02:08 PM

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PF Gold
P: 12,016

Cecile: What is the DEFINITION of linear independence?




#7
Apr405, 02:09 PM

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P: 11,863

Here's a nice way to do it.Use differential equations.Consider the homogenous,linear,constant coefficient ODE
[tex] \frac{d^{4}y(x)}{dx^{4}}y(x)=0 [/tex] Daniel. 



#8
Apr405, 02:12 PM

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? :( 



#9
Apr405, 02:19 PM

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PF Gold
P: 12,016

Remember the vector is the WHOLE function here; that is:
If you can show that the following equation, [tex]a_{1}\sin(t)+a_{2}\cos(t)+a_{3}e^{t}=0[/tex] in order to be valid (that is, holds) for ALL values of "t" implies that [tex]a_{1}=a_{2}=a_{3}=0[/tex] then you have concluded that the three functions are linearly independent. 



#10
Apr405, 02:22 PM

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Or you could just compute the wronskian...
Daniel. 



#11
Apr405, 02:25 PM

P: 5

thank you daniel:)




#12
Apr405, 02:44 PM

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 [tex]e^x[/tex]. 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
Apr405, 03:01 PM

P: 998

Or a (very tiny little) bit cleaner, note that we just need
[tex]A\sin{x} + B\cos{x} + Ce^x \equiv 0 \Longrightarrow A=B=C=0[/tex] setting [itex]x=0[/itex] immediately gives [itex]B+C = 0 \Longrightarrow B=C[/itex]. But as you noted the limiting behaviour of [itex]e^x[/itex] at infinity implies [itex]C=0[/itex] so [itex]B=C=0[/itex]. But then [itex]A\sin{x}\equiv 0[/itex] obviously implies [itex]A=0[/itex] so we're done. 



#14
Apr405, 03:22 PM

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P: 9,398

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




#15
Apr405, 04:24 PM

Math
Emeritus
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Thanks
PF Gold
P: 38,877

Saying that sin(x), cos(x), and e^{x} are independent means that
In order for C_{1}sin(x)+ C_{2}cos(x)+ C_{3}e[sup]x[/sub]= 0 for all x, we must have C_{1}= C_{2}= C_{3}. Take 3 different values for x: x= 0 is especially easy: if x= 0, C_{1}sin(x)+ C_{2}cos(x)+ C_{3}e[sup]x[/sub]= 0 becomes C_{2}+ C_{3}= 0. If [tex]x= \frac{\pi}{2}[/b], [tex]C_1+ e^{\frac{\pi}{2}}C_3= 0[/tex]. Okay, [tex]C_2= C_3[/tex] and [tex]C_1= e^{\frac{\pi}{2}}C_3[/tex]. Now put those into the original equation and take x to be some third number. Solve that for C_{3}. If C_{3}= 0 then so must C_{1}= 0 and C_{2}= 0 and the functions are independent. Yes, using the Wronskian is simpler. 


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