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May 15th, 2018, 01:25 AM   #1
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Linear dependence

Why the functions cos(3x) and cos(3x+pi/2) are linearly dependent functions?
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May 15th, 2018, 02:50 AM   #2
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They aren't!

Cos(a+ b)= cos(a)cos(b)- sin(a)sin(b) so cos(3x+ pi/2)= cos(3x)cos(pi/2)- sin(3x)sin(pi/2)= -sin(3x) which is NOT a multiple of cos(3x).
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May 15th, 2018, 10:55 AM   #3
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Math Focus: Algebraic Number Theory, Arithmetic Geometry
Perhaps you mean algebraically dependent? In that case, it follows from the standard trig identity $\sin^2\left(x\right) + \cos^2\left(x\right) \equiv 1$.

To fill in the details, let $f(X,Y)$ be the polynomial $X^2 + Y^2 - 1$. Then, as Country Boy shows, $\cos\left(3x + \frac{\pi}{2}\right) = -\sin\left(3x\right)$, so

$\begin{align*}
f\left(\cos\left(3x\right), \cos\left(3x + \frac{\pi}{2}\right) \right) &= f(\cos\left(3x\right), -\sin\left(3x\right)) \\
&= (\cos\left(3x\right))^2 + (-\sin\left(3x\right))^2 - 1 \\
&= \cos^2\left(3x\right) + \sin^2\left(3x\right) - 1 \\
&= 1 - 1 \\
&= 0
\end{align*}$
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May 15th, 2018, 12:52 PM   #4
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By definition, f(x) and g(x) are linearly independent over an interval if no constants a and b exist st af(x) + bg(x) =0 on the interval.

If you're familiar with Wronskians, if W(f,g) $\displaystyle \neq$ 0 for some x on the interval, the functions are LI on the interval. W(f,g)=1.
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May 15th, 2018, 02:59 PM   #5
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I really appreciate the response.Then how we will judge for the linear independency of the functions e^x and xe^(x)?
The wronskian comes out to be e^(2x)
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May 16th, 2018, 01:04 PM   #6
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Which is never 0!

From the definition, let for all x. Taking x= 0, we have . Then, taklng x= 1, so B= 0. The two functions are linearly independent.
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