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    given that n (is a member) of +integers, x is a member of REal numbers, and M= cosh^2 x cosh^2 x
    sinh^2 x sinh^2 x (2x2 matrix)

    use induction to show that M^n=M
    I can show that M^2=m but not sure how to use the let n=k method!
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    (Original post by ogs)
    given that n (is a member) of +integers, x is a member of REal numbers, and M= cosh^2 x cosh^2 x
    sinh^2 x sinh^2 x (2x2 matrix)

    use induction to show that M^n=M
    I can show that M^2=m but not sure how to use the let n=k method!
    You have to use "let n=k+1", throw into the equation and see if your result would be as expected.
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    M^3
    = M^2 M
    = M^2 . . . because M^2 = M
    = M . . . again because M^2 = M

    M^4
    = M^3 M
    = M^2 . . . because M^3 = M
    = M . . . because M^2 = M

    M^5
    = M^4 M
    = M^2 . . . because M^4 = M
    = M . . . because M^2 = M
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    that won't work as a proof by induction though! as you have to consider a general value...
    or will it? do you think that it would get the marks?
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    (Original post by ogs)
    that won't work as a proof by induction though! as you have to consider a general value...
    or will it? do you think that it would get the marks?
    Since M^k.M = M.M = M^2 by induction hypothesis the inductive proof is equivalent to showing M^2 = M which can be done simply by multiply the matrices and seeing what you get (hopefully M!) remembering that cosh^2x - sinh^2x = 1
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    Here is a more formal presentation of the inductive step. Suppose that k >= 2 is an integer and M^k = M. Then

    M^(k + 1)
    = M^k M
    = M^2 . . . because M^k = M
    = M . . . proved in the first part of the question

    So M^(k + 1) = M.
 
 
 
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Updated: June 28, 2004
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