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Math Help - How to show graphically that sinx=mx has no non-zero solutions...

  1. #1
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    How to show graphically that sinx=mx has no non-zero solutions...

    The question is:
    show graphically that sinx=mx has no non-zero solutions for -pie<x<pie when m<0.

    Solution:
    I drew the two graphs just by estimates,and they dint intercept.As the m is always going to be steeper,but I wanted to know if there is a concrete way to prove this.You know like doing a table or something.

    I know it is a very nooby question,and I apologize for the inconvenience.
    But honestly,with my Uni exams in a month and me missing most of the classes,this forum is my only source of getting ANY problems sorted out and hence I came here.

    Thanks in advance for the help.
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  2. #2
    Junior Member ignite's Avatar
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    Re: How to show graphically that sinx=mx has no non-zero solutions...

    You are given m<0.
    Consider two regions:
    0<x<\pi:Here sin(x) is positive and mx is negative,hence they don't intersect.
    -\pi<x<0:Here sin(x) is negative and mx is positive,hence they don't intersect.
    Only intersection point is 0.
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  3. #3
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    Re: How to show graphically that sinx=mx has no non-zero solutions...

    Quote Originally Posted by ignite View Post
    You are given m<0.
    Consider two regions:
    0<x<\pi:Here sin(x) is positive and mx is negative,hence they don't intersect.
    -\pi<x<0:Here sin(x) is negative and mx is positive,hence they don't intersect.
    Only intersection point is 0.

    thanks,could you solve one more problem for me?
    i don't know if I should put up another thread for this,so posting it here anyways.
    Here is the question:
    by considering the case x>=0,-1<x<0 and x<=-1 prove that 1+x+x^2+x^3+x^4 is always positive.

    Attempt at solution:
    I did,f(0),F(1),F(-1),F(-0.5) and F(-2) and it all came positive,and so I said it is proven,am I right or is there something more I should do?
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    Re: How to show graphically that sinx=mx has no non-zero solutions...

    Also they had asked me before to show that 1+x+x^2 is always positive,I did that by completing square method.

    The part c asks me to generalize the answers.What Should I write?

    Edit:
    I wrote:
    All Polynomials with all coefficients equal to 1 and the constant equal to 1,will always be positive,regardless of their degree.
    Last edited by syedmaheen; April 26th 2012 at 01:14 PM.
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  5. #5
    Junior Member ignite's Avatar
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    Re: How to show graphically that sinx=mx has no non-zero solutions...

    Quote Originally Posted by syedmaheen View Post
    thanks,could you solve one more problem for me?
    i don't know if I should put up another thread for this,so posting it here anyways.
    Here is the question:
    by considering the case x>=0,-1<x<0 and x<=-1 prove that 1+x+x^2+x^3+x^4 is always positive.

    Attempt at solution:
    I did,f(0),F(1),F(-1),F(-0.5) and F(-2) and it all came positive,and so I said it is proven,am I right or is there something more I should do?
    No!

    What you did was just to check functions value at particular points.You have to prove in general.
    Case I: x \ge 0 You can easily see that as all 1,x,x^2,x^3,x^4 \ge 0 \Rightarrow F(x) > 0
    Case II: x \le -1 In this case  x^2 and x^4 are positive while  x and  x^3 are negative.But  x^2+x>0 and  x^4 + x^3 >0 \Rightarrow F(x) > 0
    Case III: -1 < x < 0 Here  1+x > 0 ,  x^2 + x^3 >0 ,  x^4 > 0 \Rightarrow F(x) > 0

    Your generalisation is almost correct,except for the degree part.Think about it again.
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  6. #6
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    Re: How to show graphically that sinx=mx has no non-zero solutions...

    Quote Originally Posted by ignite View Post
    No!

    What you did was just to check functions value at particular points.You have to prove in general.
    Case I: x \ge 0 You can easily see that as all 1,x,x^2,x^3,x^4 \ge 0 \Rightarrow F(x) > 0
    Case II: x \le -1 In this case  x^2 and x^4 are positive while  x and  x^3 are negative.But  x^2+x>0 and  x^4 + x^3 >0 \Rightarrow F(x) > 0
    Case III: -1 < x < 0 Here  1+x > 0 ,  x^2 + x^3 >0 ,  x^4 > 0 \Rightarrow F(x) > 0

    Your generalisation is almost correct,except for the degree part.Think about it again.
    Yes i know now its only that way when the polynomial has even powers.

    If you or anyone else can have a look at this,how do you sketch the curve: y= ((e^x)+x)/((e^-x)-x) ?

    I found the intercept at x=0 but I don't know how to solve equations in the form of e^x=x.

    And how do you solve the question uploaded in the picture?How to show graphically that sinx=mx has no non-zero solutions...-27-04-12-7-35-04-pm.jpg
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