Question #144384
You have given a function λ : R → R with the following properties (x ∈ R, n ∈ N):

λ(n) = 0 , λ(x + 1) = λ(x) , λ(n+1/2)=1
Find two functions p, q : R → R with q(x)6=0 for all x such that λ(x) = q(x)(p(x) + 1).
1
Expert's answer
2020-11-17T17:21:34-0500

The required function is,

λ(x)=1cos(2πx)2\lambda(x) = \dfrac{1-cos(2πx)}{2}



Comparing the function to λ(x)=q(x)(p(x)+1)\lambda(x) = q(x)(p(x) + 1)


p(x)=cos(2πx),and q(x)=12\therefore p(x) = -cos(2πx),\\ and\ q(x) =\frac{1}{2}



For x = 1,

λ(1)=1cos(2π(1))2\lambda(1) = \dfrac{1-cos(2π(1))}{2}


λ(1)=1cos(2π)2=0\lambda(1) = \dfrac{1-cos(2π)}{2} = 0


For x= 2,

λ(2)=1cos(2π(2))2\lambda(2) = \dfrac{1-cos(2π(2))}{2}


λ(2)=1cos(4π)2=0\lambda(2) = \dfrac{1-cos(4π)}{2} = 0



\therefore when x = n,

λ(x) = λ(n) = 0


and λ(n) derived is;


λ(n)=1cos(2π(n))2\lambda(n) = \dfrac{1-cos(2π(n))}{2}


λ(n)=1cos2nπ2\lambda(n) = \dfrac{1-cos2nπ}{2}


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