How to calculate compounding frequency for an loan (Using Lambert-W Function)
$begingroup$
A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.
I tried to solve the question with trial and error, but is there a better way to solve it?
The function looks like this:
$$500 = 450(1 +{0.0213over x})^{5x}$$
exponential-function lambert-w
asked Jan 14 at 1:59
KevinKevin
496
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add a comment |
$begingroup$
A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.
I tried to solve the question with trial and error, but is there a better way to solve it?
The function looks like this:
$$500 = 450(1 +{0.0213over x})^{5x}$$
exponential-function lambert-w
asked Jan 14 at 1:59
KevinKevin
496
$endgroup$
add a comment |
$begingroup$
A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.
I tried to solve the question with trial and error, but is there a better way to solve it?
The function looks like this:
$$500 = 450(1 +{0.0213over x})^{5x}$$
exponential-function lambert-w
asked Jan 14 at 1:59
KevinKevin
496
$endgroup$
A bond will become worth 500 dollars when it becomes due in 5 years. If the bond was purchased today for 450 dollars at 2.13% per year, determine how frequently the interest was compounded.
I tried to solve the question with trial and error, but is there a better way to solve it?
The function looks like this:
$$500 = 450(1 +{0.0213over x})^{5x}$$
exponential-function lambert-w
exponential-function lambert-w
asked Jan 14 at 1:59
KevinKevin
496
asked Jan 14 at 1:59
KevinKevin
496
edited Jan 15 at 11:58
Kevin
asked Jan 14 at 1:59
KevinKevin
496
asked Jan 14 at 1:59
KevinKevin
496
asked Jan 14 at 1:59
KevinKevin
496
496
add a comment |
add a comment |
1 Answer
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Consider the more general case of the equation
$$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
$$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
Just remember than any equation which can write or rewrite as
$$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.
For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
$$left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)+(x-1) left(5 log
left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$ and ignoring the higher order terms, this would give, as an approximation,
$$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)}{5 left(10213 log
left(frac{10213}{10000}right)-213right)}approx 0.98086$$ while the exact solution is $0.98121$.
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
The explanation is really good, wish I could give you an upvote if I am reputation 15
$endgroup$
– Kevin
Jan 14 at 20:29
$begingroup$
@KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
$endgroup$
– Claude Leibovici
Jan 15 at 3:36
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the more general case of the equation
$$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
$$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
Just remember than any equation which can write or rewrite as
$$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.
For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
$$left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)+(x-1) left(5 log
left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$ and ignoring the higher order terms, this would give, as an approximation,
$$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)}{5 left(10213 log
left(frac{10213}{10000}right)-213right)}approx 0.98086$$ while the exact solution is $0.98121$.
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
The explanation is really good, wish I could give you an upvote if I am reputation 15
$endgroup$
– Kevin
Jan 14 at 20:29
$begingroup$
@KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
$endgroup$
– Claude Leibovici
Jan 15 at 3:36
add a comment |
$begingroup$
Consider the more general case of the equation
$$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
$$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
Just remember than any equation which can write or rewrite as
$$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.
For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
$$left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)+(x-1) left(5 log
left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$ and ignoring the higher order terms, this would give, as an approximation,
$$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)}{5 left(10213 log
left(frac{10213}{10000}right)-213right)}approx 0.98086$$ while the exact solution is $0.98121$.
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
$begingroup$
The explanation is really good, wish I could give you an upvote if I am reputation 15
$endgroup$
– Kevin
Jan 14 at 20:29
$begingroup$
@KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
$endgroup$
– Claude Leibovici
Jan 15 at 3:36
add a comment |
$begingroup$
Consider the more general case of the equation
$$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
$$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
Just remember than any equation which can write or rewrite as
$$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.
For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
$$left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)+(x-1) left(5 log
left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$ and ignoring the higher order terms, this would give, as an approximation,
$$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)}{5 left(10213 log
left(frac{10213}{10000}right)-213right)}approx 0.98086$$ while the exact solution is $0.98121$.
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
$endgroup$
Consider the more general case of the equation
$$a=left(1+frac{b}{x}right)^{c x}$$ The solution of it is given in terms of Lambert function. I shall skip all intermediate steps and just provide the result; it write
$$x=-frac{b log (a)}{log (a)+b, c, W_{-1}(-t)}qquad text{where} qquad t=frac{ a^{-frac{1}{b, c}}}{b ,c}log (a)$$
Just remember than any equation which can write or rewrite as
$$A+B x+C log(D+Ex)=0$$ has solution(s) i terms of Lambert function.
For your specific problem, since graphing, you see that the solution is close to $x=1$, you could have done the following : take logarithms of both sides and use Taylor expansion around $x=1$. Working with whole numbers you would get
$$left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)+(x-1) left(5 log
left(frac{10213}{10000}right)-frac{1065}{10213}right)+Oleft((x-1)^2right)$$ and ignoring the higher order terms, this would give, as an approximation,
$$x=1-frac{10213 left(5 log left(frac{10213}{10000}right)-log
left(frac{10}{9}right)right)}{5 left(10213 log
left(frac{10213}{10000}right)-213right)}approx 0.98086$$ while the exact solution is $0.98121$.
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
edited Jan 14 at 6:32
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
answered Jan 14 at 6:19
Claude LeiboviciClaude Leibovici
124k1158135
124k1158135
$begingroup$
The explanation is really good, wish I could give you an upvote if I am reputation 15
$endgroup$
– Kevin
Jan 14 at 20:29
$begingroup$
@KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
$endgroup$
– Claude Leibovici
Jan 15 at 3:36
add a comment |
$begingroup$
The explanation is really good, wish I could give you an upvote if I am reputation 15
$endgroup$
– Kevin
Jan 14 at 20:29
$begingroup$
@KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
$endgroup$
– Claude Leibovici
Jan 15 at 3:36
$begingroup$
The explanation is really good, wish I could give you an upvote if I am reputation 15
$endgroup$
– Kevin
Jan 14 at 20:29
$begingroup$
The explanation is really good, wish I could give you an upvote if I am reputation 15
$endgroup$
– Kevin
Jan 14 at 20:29
$begingroup$
@KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
$endgroup$
– Claude Leibovici
Jan 15 at 3:36
$begingroup$
@KevinXu. You are very welcome ! I upvoted you question because it reveals an intelligent attitude (curiosity). Cheers :-)
$endgroup$
– Claude Leibovici
Jan 15 at 3:36
add a comment |
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