Prove the identity: 1 / (i + s' / ((1+s')ⁿ − 1)) = (1-PV)/i, assuming s' = i and PV = (1+i)^-n.

Question

ScanSolve AI Tutor · Accepted Answer

To prove the given identity, we will simplify both sides of the equation. The identity is:

\frac{1}{i + \frac{s'}{(1+s')^n - 1}} = \frac{1-PV}{i}

Identify the left-hand side (LHS) and the right-hand side (RHS).
LHS: $\frac{1}{i + \frac{s'}{(1+s')^n - 1}}$
RHS: $\frac{1-PV}{i}$
Interpret $PV$ and assume $s'=i$ .
In financial mathematics, $PV$ typically represents the present value factor, which is $(1+i)^{-n}$ . The RHS, $\frac{1-(1+i)^{-n}}{i}$ , is the formula for the present value of an ordinary annuity, $a_{\overline{n}|i}$ . For the given identity to hold true, the interest rate $s'$ on the LHS must be equal to the interest rate $i$ on the RHS. Therefore, we assume $s' = i$ .
Substitute $s'=i$ into the LHS.
$\text{LHS} = \frac{1}{i + \frac{i}{(1+i)^n - 1}}$
Simplify the denominator of the LHS.
$i + \frac{i}{(1+i)^n - 1} = i \left( 1 + \frac{1}{(1+i)^n - 1} \right)$ $= i \left( \frac{(1+i)^n - 1 + 1}{(1+i)^n - 1} \right)$ $= i \left( \frac{(1+i)^n}{(1+i)^n - 1} \right)$
Substitute this simplified denominator back into the LHS expression.
$\text{LHS} = \frac{1}{i \left( \frac{(1+i)^n}{(1+i)^n - 1} \right)} = \frac{(1+i)^n - 1}{i (1+i)^n}$
Simplify the RHS using $PV = (1+i)^{-n}$ .
$\text{RHS} = \frac{1-(1+i)^{-n}}{i} = \frac{1 - \frac{1}{(1+i)^n}}{i}$ $= \frac{\frac{(1+i)^n - 1}{(1+i)^n}}{i} = \frac{(1+i)^n - 1}{i (1+i)^n}$
Compare the simplified LHS and RHS.
Since both the simplified LHS and RHS are equal to $\frac{(1+i)^n - 1}{i (1+i)^n}$ , the identity is proven.
$\text{LHS} = \frac{(1+i)^n - 1}{i (1+i)^n} = \text{RHS}$
The identity is proven.

Prove the identity: 1 / (i + s' / ((1+s')ⁿ − 1)) = (1−PV)/i, assuming s' = i and PV = (1+i)^-n.