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Thread: multiple solutions to equations involving sine

  1. #1
    Junior Member pinion's Avatar
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    Lightbulb multiple solutions to equations involving sine

    hello
    i'll get straight to the problem

    i'm looking for non-numerical, non-iterative methods for finding the solutions to equations such as this:

    sin(pi*2^x) = sin(pi*2^(1-x)) = 0, over the range [-1 < x < 2]

    in words, i'm looking for where the given functions simultaneously intersect each other and the x-axis
    so,
    sin(pi * 2^x) = 0
    sin(pi * 2^(1-x)) = 0

    pi * 2^x = c1 * pi
    pi * 2^(1-x) = c2 * pi
    where c1 and c2 are integers >= 0

    2^x = c1
    2^(1-x) = c2

    ..ok so clearly i'm just transforming this a little to see if i jog someone's memory

    i vaguely remember something in calculus that would lead to multiple solutions..however it somehow involved integrals..does anyone know what those types of calculus equations are called? how to solve them?

    *edit*
    oh can't calculus find the points of intersection of a parabola and a line? can i use a similar process on my equations?
    Last edited by pinion; Jan 31st 2008 at 01:10 PM.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    $\displaystyle sin(\pi2^x)=sin(\pi2^{1-x})=0$

    As you pointed out, you are looking for instances where:
    $\displaystyle 2^x=c_1$
    $\displaystyle 2^{1-x}=c_2$
    Where $\displaystyle c_1, c_2$ are natural numbers.

    So, you need $\displaystyle x=log_2{c_1}$ AND $\displaystyle x=1-log_2{c_2}$ (for the same values of x), which implies that:
    $\displaystyle log_2{c_1}=1-log_2{c_2}$
    $\displaystyle c_1=2^{1-log_2{c_2}}$
    $\displaystyle c_1=\frac{2}{c_2}$

    Since $\displaystyle c_1$ and $\displaystyle c_2$ must be natural numbers but are related such that $\displaystyle c_1=\frac{2}{c_2}$, $\displaystyle c_2=2,1$ where $\displaystyle c_1=1,2$. In either case, $\displaystyle x=log_2{c_1}=0,1$ are the only solutions.
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  3. #3
    Junior Member pinion's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Since $\displaystyle c_1$ and $\displaystyle c_2$ must be natural numbers but are related such that $\displaystyle c_1=\frac{2}{c_2}$, $\displaystyle c_2=2,1$ where $\displaystyle c_1=1,2$. In either case, $\displaystyle x=log_2{c_1}=0,1$ are the only solutions.
    ok, but is there a way of finding $\displaystyle x=0,1$ without looking at a graph or checking $\displaystyle c_1=0$ $\displaystyle c_1=1$ $\displaystyle c_1=2$ $\displaystyle c_1=3$ ... ?

    for instance, how is it you find the points where $\displaystyle y=3^x$ and $\displaystyle y=x+1.2$ intersect? i know i've done that before, so i think i can do something similar to find where two functions intersect each other and the x axis.

    *edit*
    well the intersection of a parabola and a line is fairly simple and straightforward..just use the quadratic formula. i'm positive i've had calculus problems which would lead to multiple solutions, however as i cannot remember their form i don't know if i could make use of that method.

    now, in the case of finding the intersection of $\displaystyle y=3^x$ and $\displaystyle y=x+1.2$, if one could somehow rotate $\displaystyle y=3^x$ so that $\displaystyle y=x+1.2$ became the x-axis, then you would simply solve the rotated function for $\displaystyle y=0$, then un-rotate the solutions. in the case of my two functions involving sine it might be slightly more difficult to adjust one of them so that the other became the x-axis. perhaps a rather off-the-wall idea, but we'll see if it goes anywhere. any ideas?

    i guess this doesn't have so much to do with calculus, i shall start a different thread.
    Last edited by pinion; Feb 1st 2008 at 11:14 AM. Reason: clarification
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