Question

A program devise a recursive algorithm to find a2n, where a is a real number and n is a positive integer (hint: use the equality a2n+1 = (a^2n)^2

Answer 1

In this question, we have to find the recursive algorithm to find a^{2^n}, where a is a real number and n is a positive integer.

Also we know the equality that a^{2^n+1} = (a^{2^n})².

This implies that a^{2^n+1} = (a^{2^n})² = a² . a^{2^n} , for n>0.

For initial condition i.e n=0 , a^{2^1} = a² = a*a . So to find recursive algorithm for a^{2^n}, successively reduce it by using the recursive step .

The recursive algorithm to find a^{2^n} is given below :

procedure power (a:real number, n:positive number)

if(n=1) then power(a,n) = a*a

else pow=power(a,n-1)

power(a,n) = pow * pow

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