F fonixbob New member Joined Sep 6, 2010 Messages 3 Sep 7, 2010 #1 Given that, C(n+1) = (1+r)[C(n) - (1+i)^n * S] How can I prove that: C(n) = (1+r)^n [ Co - ((1+r)/(r-i))* S] + ((1+r)/(r-i))*(1+i)^n * S I get really confused with the exponents
Given that, C(n+1) = (1+r)[C(n) - (1+i)^n * S] How can I prove that: C(n) = (1+r)^n [ Co - ((1+r)/(r-i))* S] + ((1+r)/(r-i))*(1+i)^n * S I get really confused with the exponents
D Denis Senior Member Joined Feb 17, 2004 Messages 1,700 Sep 7, 2010 #2 fonixbob said: Given that, C(n+1) = (1+r)[C(n) - (1+i)^n * S] How can I prove that: C(n) = (1+r)^n [Co - ((1+r)/(r-i))* S] + ((1+r)/(r-i))*(1+i)^n * S Click to expand... Looks to me like that comes from some financial formula; C = Coupons?...do you know which? Anyway, what is "Co" ????? May be less confusing/unwieldy if you let a = 1+r, b = 1+i : C(n+1) = a[C(n) - b^n * S] C(n) = a^n [Co - a/(r-i) * S] + a/(r-i) * b^n * S That's the extent of what I can do, as I'm not sure what you're doing :shock:
fonixbob said: Given that, C(n+1) = (1+r)[C(n) - (1+i)^n * S] How can I prove that: C(n) = (1+r)^n [Co - ((1+r)/(r-i))* S] + ((1+r)/(r-i))*(1+i)^n * S Click to expand... Looks to me like that comes from some financial formula; C = Coupons?...do you know which? Anyway, what is "Co" ????? May be less confusing/unwieldy if you let a = 1+r, b = 1+i : C(n+1) = a[C(n) - b^n * S] C(n) = a^n [Co - a/(r-i) * S] + a/(r-i) * b^n * S That's the extent of what I can do, as I'm not sure what you're doing :shock:
