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If I have two functions $$ f_0(x) = \sin(\pi a_0 x ) $$ and $$ f_1(x) = \sin(\pi a_1 x ) $$ How do I determine which one is increasing more rapidly with respect to itself over a closed interval (say 0 to $\pi/2$ or 0 to $\pi/4$). Not sure if I am saying this correctly, one will double more quickly than the other depending on the values of $a_0$ and $a_1$. I believe this is equivalent to asking how to determine if $$ \frac{\sin(\pi a_0 x)}{\sin(\pi a_1 x)} $$ is monotonically increasing (or decreasing) with repsect to $x$ over a specified closed interval. Any suggestions on how this might be accomplished?

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A function $f$ is increasing more rapidly than another function $g$ if $f' > g'$. – Antonio Vargas Jan 21 '13 at 0:06
up vote 1 down vote accepted

There are some various ways to determine which one increases the fastest, for example :

Determine if the difference $f_1 - f_2$ is heading towards + inf, - inf, or a limit (then you need to know the sign). You just need to find a value of $x$ past which the sign of the sum is constant.

You can do the same with the quotient $f_1/f_2$. (+ inf, 0, limit, - inf, etc).

In both cases you can derive and see.

In your case (neither functions are converging) you are acting on a limited intervall, so just derive both functions and substract/build the quotient. Here a substraction might be better because of the constantly changing signs.

$f_0' - f_1' = \pi*a_0* cos(\pi*a_0*x) - \pi*a_1* cos(\pi*a_1*x)$

Now you can have fun with the cos addition formulas to get something you can actually analyze lightly. The sign will give you the fastest growing function.

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