Where is the error in my attempted derivation of the result $i^{(p)} = p[(1+i)^frac{1}{p}-1]$?
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Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.
Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that
$$
frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
$$
Then, by using the rule
$$
sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
$$
and taking
$$
a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
$$
we can express $(*)$ as
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
$$
Thus, by dividing through by $i$ we have
$$
frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
$$
Giving the result
$$
i^{(p)} = p[1 - (1+i)^frac{1}{p}]
$$
This result is incorrect, since the correct result should be
$$
i^{(p)} = p[(1+i)^frac{1}{p}-1]
$$
Where in my attempted derivation did I go wrong?
finance
add a comment |
up vote
2
down vote
favorite
Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.
Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that
$$
frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
$$
Then, by using the rule
$$
sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
$$
and taking
$$
a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
$$
we can express $(*)$ as
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
$$
Thus, by dividing through by $i$ we have
$$
frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
$$
Giving the result
$$
i^{(p)} = p[1 - (1+i)^frac{1}{p}]
$$
This result is incorrect, since the correct result should be
$$
i^{(p)} = p[(1+i)^frac{1}{p}-1]
$$
Where in my attempted derivation did I go wrong?
finance
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.
Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that
$$
frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
$$
Then, by using the rule
$$
sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
$$
and taking
$$
a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
$$
we can express $(*)$ as
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
$$
Thus, by dividing through by $i$ we have
$$
frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
$$
Giving the result
$$
i^{(p)} = p[1 - (1+i)^frac{1}{p}]
$$
This result is incorrect, since the correct result should be
$$
i^{(p)} = p[(1+i)^frac{1}{p}-1]
$$
Where in my attempted derivation did I go wrong?
finance
Suppose $i^{(p)}$ denotes the nominal interest rate, payable as $frac{i^{(p)}}{p}$, at a rate of $p$ times per period, and assume that $i$ is the effective interest rate (payable once per period) that is equivalent.
Then, by considering the accumulated value of the interest payments made under the nominal rate, we deduce that
$$
frac{i^{(p)}}{p} + frac{i^{(p)}}{p}(1+i) + frac{i^{(p)}}{p}(1+i)^2 + dots + frac{i^{(p)}}{p}(1+i)^frac{p-1}{p} = i hspace{10mm} (*)
$$
Then, by using the rule
$$
sum_{k=0}^{n-1}ar^k = a cdot frac{(1-r^n)}{1-r}
$$
and taking
$$
a = frac{i^{(p)}}{p}, hspace{10mm} r = (1+i)^frac{1}{p}, hspace{10mm} n = p
$$
we can express $(*)$ as
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{i}{1 - (1+i)^frac{1}{p}} = i
$$
Thus, by dividing through by $i$ we have
$$
frac{i^{(p)}}{p[1 - (1+i)^frac{1}{p}]} = 1
$$
Giving the result
$$
i^{(p)} = p[1 - (1+i)^frac{1}{p}]
$$
This result is incorrect, since the correct result should be
$$
i^{(p)} = p[(1+i)^frac{1}{p}-1]
$$
Where in my attempted derivation did I go wrong?
finance
finance
asked Nov 23 at 15:33
J. Chapman
132
132
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1 Answer
1
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up vote
2
down vote
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You've forgotten a minus sign:
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
$$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You've forgotten a minus sign:
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
$$
add a comment |
up vote
2
down vote
accepted
You've forgotten a minus sign:
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
$$
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You've forgotten a minus sign:
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
$$
You've forgotten a minus sign:
$$
frac{i^{(p)}}{p} cdot frac{1 - left( (1+i)^frac{1}{p} right)^p}{1 - (1+i)^frac{1}{p}} = frac{i^{(p)}}{p} cdot frac{-i}{1 - (1+i)^frac{1}{p}} = i
$$
answered Nov 23 at 20:25
alexjo
12.2k1329
12.2k1329
add a comment |
add a comment |
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