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Math Help - mx=sin x

  1. #1
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    mx=sin x

    Find roots of the equation mx=sin x considering different values of m

    I have figured out that 0<m<1 or else the graph wont intersect other than at x=0. But I cant calculate the roots generally, unless i plug in different values of m.
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  2. #2
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    What more do you want? There is no algebraic method to solve mx= sin x.

    In general, there is no algebraic method to solve x= f(x) where f is any transcendental function.
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    What more do you want? There is no algebraic method to solve mx= sin x.

    In general, there is no algebraic method to solve x= f(x) where f is any transcendental function.
    That's all the question said, so I am guessing that we are just meant to consider that there are 3 solutions if the gradient is less than 1, or else there will be just 1. I wasn't sure what else we could do with it.

    There was also another question on the sheet which was similar that I thought couldn't be solved:

    Solve a^b=b^a for all real a and b

    but it might be solvable through logs?
    Last edited by Aquafina; November 2nd 2009 at 06:14 AM.
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  4. #4
    Super Member dhiab's Avatar
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    Hello
    You have 3 cases :
    1 )  -1\leq(mx)\leq1
    2 ) (mx)<-1
    3 ) (mx)>1
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  5. #5
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    Hello Aquafina
    Quote Originally Posted by Aquafina View Post
    That's all the question said, so I am guessing that we are just meant to consider that there are 3 solutions if the gradient is less than 1, or else there will be just 1...
    Are you sure about that? See the attached diagram.

    There was also another question on the sheet which was similar that I thought couldn't be solved:

    Solve ab=ba for all real a and b

    but it might be solvable through logs?
    You obviously can't mean ab = ba. Do you mean a^b = b^a?

    Grandad
    Attached Thumbnails Attached Thumbnails mx=sin x-untitled.jpg  
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  6. #6
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    Quote Originally Posted by Grandad View Post
    Hello AquafinaAre you sure about that? See the attached diagram.
    Hi Grandad! How does that help? Both the gradients are less than 1, so it doesn't disprove what I said?

    Quote Originally Posted by Grandad View Post
    You obviously can't mean ab = ba. Do you mean a^b = b^a?
    Yes sorry typo, I have corrected it now to a^b = b^a. Any ideas with this?
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  7. #7
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    Hello Aquafina

    y = x/10 clearly intersects y = \sin x at 7 points. So how can you say there are only 3 solutions?

    Grandad
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  8. #8
    Super Member dhiab's Avatar
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    Quote Originally Posted by Grandad View Post
    Hello Aquafina

    y = x/10 clearly intersects y = \sin x at 7 points. So how can you say there are only 3 solutions?

    Grandad
    Hello Aquafina: y = x/100 clearly intersects y=sinx at 63 points conclusion 63 solutions
    look here:
    Attached Files Attached Files
    • File Type: doc 3.doc (38.0 KB, 24 views)
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