David your method is perfect but i just rearranged the formula and value of P to arrive at the answer mine formula is also working fine enough.please take a look.
P(x)=100/(1+s1/m)^x[ my P(x) is different then yours this saved me from embedding extra m in power)
P(.5)=100/(1+.02020/2)^.5=99.498 [m=2 for semiannual compounding]
P(1)=100/(1+.030692/2)^1=98.488
f(.5)=m * {[P(s1)/P(s2)]^(1/n) - 1}=2*([99.498/98.488]^(1/.5)-1]=2(1.0206-1)=2*.0206~.0412 which is coming out as correct .this saved me from including extra m as mn.
thanks
P(x)=100/(1+s1/m)^x[ my P(x) is different then yours this saved me from embedding extra m in power)
P(.5)=100/(1+.02020/2)^.5=99.498 [m=2 for semiannual compounding]
P(1)=100/(1+.030692/2)^1=98.488
f(.5)=m * {[P(s1)/P(s2)]^(1/n) - 1}=2*([99.498/98.488]^(1/.5)-1]=2(1.0206-1)=2*.0206~.0412 which is coming out as correct .this saved me from including extra m as mn.
thanks