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

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But it isn't the only way.In summary, the conversation discusses the linear dependence of the functions sin(t), cos(t), and e^t. The participants suggest using differential equations or considering the periodicity of the functions to prove their independence. Ultimately, it is determined that the functions are independent if their coefficients are all equal to zero.
  • #1
Cecile
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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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  • #2
Yes,they are independent.What are u trying to do...?Post the text of the problem...

Daniel.
 
  • #3
I'm trying to prove that they are whether dependent or independent. how can I show iy?
 
  • #4
Are u trying to prove that they can form a basis in some vector space...?Define the vector space.

Daniel.
 
  • #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
Cecile: What is the DEFINITION of linear independence?
 
  • #7
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
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? :(
 
  • #9
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
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 [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
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.
 
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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 [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 C3. If C3= 0 then so must C1= 0 and
C2= 0 and the functions are independent.

Yes, using the Wronskian is simpler.
 

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

What does it mean for [sin(t), cos(t), e^t] to be linearly dependent?

Linear dependence refers to the relationship between vectors in a vector space. If [sin(t), cos(t), e^t] are linearly dependent, it means that at least one of these vectors can be expressed as a linear combination of the others. In other words, one vector can be written as a multiple of another vector.

How can I determine if [sin(t), cos(t), e^t] are linearly dependent?

To determine if [sin(t), cos(t), e^t] are linearly dependent, we can use the concept of a determinant. The determinant of a matrix containing these vectors will be equal to zero if they are linearly dependent. Alternatively, we can also check if the rank of the matrix is less than the number of vectors (in this case, 3).

Why is it important to know if [sin(t), cos(t), e^t] are linearly dependent?

Knowing if [sin(t), cos(t), e^t] are linearly dependent can help us understand the relationships between these functions. It can also be useful in solving systems of equations and determining the span of a set of vectors.

What does it mean for [sin(t), cos(t), e^t] to be linearly independent?

If [sin(t), cos(t), e^t] are linearly independent, it means that none of these vectors can be expressed as a linear combination of the others. Each vector is unique and cannot be written as a multiple of another vector.

Can [sin(t), cos(t), e^t] be linearly dependent in certain cases and linearly independent in others?

Yes, it is possible for [sin(t), cos(t), e^t] to be linearly dependent in certain cases and linearly independent in others. For example, if we are working with a specific range of values for t, these vectors may be linearly dependent. However, if we consider a larger range of values for t, they may become linearly independent. It depends on the specific values and relationships between the vectors.

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