10.1 are linear independent ....

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SUMMARY

The discussion confirms that the functions \(e^{2x}\) and \(\sin 2x\) are linearly independent over the interval \(\left\{-\infty,\infty\right\}\). This is established by demonstrating that the only solution to the equation \(ae^{2x} + b\sin(2x) = 0\) for all \(x\) is \(a = 0\) and \(b = 0\). The analysis utilizes specific values of \(x\) to validate the independence, particularly at \(x = 0\) and \(x = \frac{\pi}{2}\). The discussion emphasizes that amplitude is not a factor in determining linear independence in vector spaces.

PREREQUISITES
  • Understanding of linear independence in vector spaces
  • Knowledge of exponential functions and trigonometric functions
  • Familiarity with scalar multiplication and vector addition
  • Basic calculus concepts, including evaluation of functions at specific points
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Mathematicians, students of linear algebra, and anyone studying vector spaces and their properties will benefit from this discussion.

karush
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show that
$e^{2x},\quad \sin 2x$
are linear independent on
$\left\{-\infty,\infty\right\}$

new concept to me
but
$\sin 2x$
has an amplitude
$e^{2x}$
doesn't
 
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Whether or not a "vector" has an "amplitude" is not relevant. All we require in a vector space is that we be able to add vectors and multiply vectors by numbers (more generally, "scalars").

Two vectors, u and v, in a vector space are "independent" if the only values or a and b such that au+ bv= 0 are a= b= 0. Here that gives us the equation ae^{2x}+ bsin(2x)= 0 for all x. In particular, if x= 0 that becomes a(1)+ b(0)= a= 0 and, if x= \pi/2, ae^{\pi/2}+ bsin(\pi/2)= 0 which, since a= 0, gives b= 0.
 
SSCwt.png
 
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?? That appears to be impossible to read!
 
.
 
Last edited:

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