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Math Help - Number of roots

  1. #1
    Senior Member pankaj's Avatar
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    Number of roots

    Find the number of roots of the equation
     <br />
2^x=x^2+1<br />
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  2. #2
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    Quote Originally Posted by pankaj View Post
    Find the number of roots of the equation
     <br />
2^x=x^2+1<br />
    2^x=x^2+1

     e^{\ln(2^x)} = x^2+1

     e^{x\ln(2)} = x^2+1

     e^{x\ln(2)} = x^2+1

    x = \pm(e^{x\ln(2)} -1)

    Now. The exponential function has only 1 value at any point. There is no x value for which the exponential function has more than 1 value.

    So, the RHS, inside the brackets, as x varies, there will only ever be 1 unique solution inside the brackets. However outside the brackets we have written  \pm so... can you draw a conclusion from there?
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  3. #3
    Senior Member pankaj's Avatar
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    Let f(x)=2^x-x^2-1

    Now, f(0)=0;f(1)=0 and f(4)f(5)<0

    Therefore the equation has clearly atleast 3 roots.

    How to prove or disprove that the equation has exactly 3 roots???
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  4. #4
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    Quote Originally Posted by pankaj View Post
    Let f(x)=2^x-x^2-1

    Now, f(0)=0;f(1)=0 and f(4)f(5)<0

    Therefore the equation has clearly atleast 3 roots.

    How to prove or disprove that the equation has exactly 3 roots???
    so you only need to show that the equation has at most 3 (real) roots. here's a general fact:

    Fact: suppose a>0 and let p(x) be a polynomial of degree n with real coefficients. then the equation a^x=p(x) has at most n+1 (distinct) real roots.

    Proof: if a=1, there's nothing to prove by the fundamental theorem of algebra. for 0 < a \neq 1 let f(x)=a^x - p(x). suppose f(x)=0 has at least n+2 roots. then, by Rolle's theorem, the

    equation f^{(n+1)}(x)=0 must have at least one real root. but f^{(n+1)}(x)=a^x(\ln a)^{n+1} \neq 0, \ \forall x \in \mathbb{R}. contradiction! \Box
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