Mixing
How do I calculate the average interest rate for a dollar cost averaging investment?
$begingroup$
Say I invested $100 every month into the S&P 500 since January 2004. If I sold out on 2019-01-11 (not considering fees or dividends), I will have invested $18,100 and my investment would be worth $32,331.15.
How do I calculate my average yearly return (interest rate) on this investment?
What I've tried so far
My starting equation is
$$
totalValue = monthlyAmount cdot interest^{numYears} + monthlyAmount cdot interest^{numYears - 1/12} + dots + monthlyAmount cdot interest^{1/12}
$$
I can then extract $monthlyAmount$ to simplify the expression:
$$
totalValue = monthlyAmount cdot (interest^{numYears} + interest^{numYears - 1/12} + dots + interest^{1/12})
$$
The part in the parentheses, if I'm not mistaken, seems to be the sum of a geometric sequence - where both the first term and the common ratio is $interest^{1/12}$. If so, then I can use the formula for calculating the sum of the first $n$ terms of a geometric sequence:
$$
totalValue = monthlyAmount cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{numYears}}
$$
Alright, given the values in the introduction, I have an equation with a single unknown - the $interest$:
$$
32331.15 = 100 cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{15}}
$$
I gave this to WolframAlpha, and it gave me a solution of 1.074 (i.e. 7.4%). That's nice, but how do I arrive at the solution?
I'm writing a program that calculates this in the general case, so I'm ideally looking for some sort of algorithm that can be written in code.
Source of financial information: https://finance.yahoo.com/quote/%5EGSPC/history?p=%5EGSPC
linear-algebra
asked Jan 22 at 22:57
ZoltánZoltán
1156
$endgroup$
|
show 6 more comments
$begingroup$
Say I invested $100 every month into the S&P 500 since January 2004. If I sold out on 2019-01-11 (not considering fees or dividends), I will have invested $18,100 and my investment would be worth $32,331.15.
How do I calculate my average yearly return (interest rate) on this investment?
What I've tried so far
My starting equation is
$$
totalValue = monthlyAmount cdot interest^{numYears} + monthlyAmount cdot interest^{numYears - 1/12} + dots + monthlyAmount cdot interest^{1/12}
$$
I can then extract $monthlyAmount$ to simplify the expression:
$$
totalValue = monthlyAmount cdot (interest^{numYears} + interest^{numYears - 1/12} + dots + interest^{1/12})
$$
The part in the parentheses, if I'm not mistaken, seems to be the sum of a geometric sequence - where both the first term and the common ratio is $interest^{1/12}$. If so, then I can use the formula for calculating the sum of the first $n$ terms of a geometric sequence:
$$
totalValue = monthlyAmount cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{numYears}}
$$
Alright, given the values in the introduction, I have an equation with a single unknown - the $interest$:
$$
32331.15 = 100 cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{15}}
$$
I gave this to WolframAlpha, and it gave me a solution of 1.074 (i.e. 7.4%). That's nice, but how do I arrive at the solution?
I'm writing a program that calculates this in the general case, so I'm ideally looking for some sort of algorithm that can be written in code.
Source of financial information: https://finance.yahoo.com/quote/%5EGSPC/history?p=%5EGSPC
linear-algebra
asked Jan 22 at 22:57
ZoltánZoltán
1156
$endgroup$
$begingroup$
Hint: write it out as a sum involving the rate and note that you have a geometric series.
$endgroup$
– lulu
Jan 22 at 23:04
$begingroup$
@lulu No, because the stock price changes.
$endgroup$
– Acccumulation
Jan 22 at 23:07
$begingroup$
@Acccumulation Why does that matter? You are looking for a single rate $r$ which would match the return over the period.
$endgroup$
– lulu
Jan 22 at 23:09
$begingroup$
I'ver misread the problem a couple of times now, so let me state how I am currently reading it. I believe the OP is looking for the rate, $r$, such that a monthly investment of $100$ over $181$ periods would result in a total value of $32,331.15$ Thus, the first $100$ will have become $100times (1+frac r{12})^{181}$, the second will have grown to $100times (1+frac r{12})^{180}$ and so on. Is this what was intended?
$endgroup$
– lulu
Jan 22 at 23:14
$begingroup$
@lulu yes, that's what I meant. See my update for more information
$endgroup$
– Zoltán
Jan 22 at 23:15
|
show 6 more comments
$begingroup$
Say I invested $100 every month into the S&P 500 since January 2004. If I sold out on 2019-01-11 (not considering fees or dividends), I will have invested $18,100 and my investment would be worth $32,331.15.
How do I calculate my average yearly return (interest rate) on this investment?
What I've tried so far
My starting equation is
$$
totalValue = monthlyAmount cdot interest^{numYears} + monthlyAmount cdot interest^{numYears - 1/12} + dots + monthlyAmount cdot interest^{1/12}
$$
I can then extract $monthlyAmount$ to simplify the expression:
$$
totalValue = monthlyAmount cdot (interest^{numYears} + interest^{numYears - 1/12} + dots + interest^{1/12})
$$
The part in the parentheses, if I'm not mistaken, seems to be the sum of a geometric sequence - where both the first term and the common ratio is $interest^{1/12}$. If so, then I can use the formula for calculating the sum of the first $n$ terms of a geometric sequence:
$$
totalValue = monthlyAmount cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{numYears}}
$$
Alright, given the values in the introduction, I have an equation with a single unknown - the $interest$:
$$
32331.15 = 100 cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{15}}
$$
I gave this to WolframAlpha, and it gave me a solution of 1.074 (i.e. 7.4%). That's nice, but how do I arrive at the solution?
I'm writing a program that calculates this in the general case, so I'm ideally looking for some sort of algorithm that can be written in code.
Source of financial information: https://finance.yahoo.com/quote/%5EGSPC/history?p=%5EGSPC
linear-algebra
asked Jan 22 at 22:57
ZoltánZoltán
1156
$endgroup$
Say I invested $100 every month into the S&P 500 since January 2004. If I sold out on 2019-01-11 (not considering fees or dividends), I will have invested $18,100 and my investment would be worth $32,331.15.
How do I calculate my average yearly return (interest rate) on this investment?
What I've tried so far
My starting equation is
$$
totalValue = monthlyAmount cdot interest^{numYears} + monthlyAmount cdot interest^{numYears - 1/12} + dots + monthlyAmount cdot interest^{1/12}
$$
I can then extract $monthlyAmount$ to simplify the expression:
$$
totalValue = monthlyAmount cdot (interest^{numYears} + interest^{numYears - 1/12} + dots + interest^{1/12})
$$
The part in the parentheses, if I'm not mistaken, seems to be the sum of a geometric sequence - where both the first term and the common ratio is $interest^{1/12}$. If so, then I can use the formula for calculating the sum of the first $n$ terms of a geometric sequence:
$$
totalValue = monthlyAmount cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{numYears}}
$$
Alright, given the values in the introduction, I have an equation with a single unknown - the $interest$:
$$
32331.15 = 100 cdot frac{interest^{1/12} cdot (1-interest^{1/12})}{1-interest^{15}}
$$
I gave this to WolframAlpha, and it gave me a solution of 1.074 (i.e. 7.4%). That's nice, but how do I arrive at the solution?
I'm writing a program that calculates this in the general case, so I'm ideally looking for some sort of algorithm that can be written in code.
Source of financial information: https://finance.yahoo.com/quote/%5EGSPC/history?p=%5EGSPC
linear-algebra
linear-algebra
asked Jan 22 at 22:57
ZoltánZoltán
1156
asked Jan 22 at 22:57
ZoltánZoltán
1156
edited Jan 22 at 23:29
Zoltán
asked Jan 22 at 22:57
ZoltánZoltán
1156
asked Jan 22 at 22:57
ZoltánZoltán
1156
asked Jan 22 at 22:57
ZoltánZoltán
1156
1156
$begingroup$
Hint: write it out as a sum involving the rate and note that you have a geometric series.
$endgroup$
– lulu
Jan 22 at 23:04
$begingroup$
@lulu No, because the stock price changes.
$endgroup$
– Acccumulation
Jan 22 at 23:07
$begingroup$
@Acccumulation Why does that matter? You are looking for a single rate $r$ which would match the return over the period.
$endgroup$
– lulu
Jan 22 at 23:09
$begingroup$
I'ver misread the problem a couple of times now, so let me state how I am currently reading it. I believe the OP is looking for the rate, $r$, such that a monthly investment of $100$ over $181$ periods would result in a total value of $32,331.15$ Thus, the first $100$ will have become $100times (1+frac r{12})^{181}$, the second will have grown to $100times (1+frac r{12})^{180}$ and so on. Is this what was intended?
$endgroup$
– lulu
Jan 22 at 23:14
$begingroup$
@lulu yes, that's what I meant. See my update for more information
$endgroup$
– Zoltán
Jan 22 at 23:15
|
show 6 more comments
$begingroup$
Hint: write it out as a sum involving the rate and note that you have a geometric series.
$endgroup$
– lulu
Jan 22 at 23:04
$begingroup$
@lulu No, because the stock price changes.
$endgroup$
– Acccumulation
Jan 22 at 23:07
$begingroup$
@Acccumulation Why does that matter? You are looking for a single rate $r$ which would match the return over the period.
$endgroup$
– lulu
Jan 22 at 23:09
$begingroup$
I'ver misread the problem a couple of times now, so let me state how I am currently reading it. I believe the OP is looking for the rate, $r$, such that a monthly investment of $100$ over $181$ periods would result in a total value of $32,331.15$ Thus, the first $100$ will have become $100times (1+frac r{12})^{181}$, the second will have grown to $100times (1+frac r{12})^{180}$ and so on. Is this what was intended?
$endgroup$
– lulu
Jan 22 at 23:14
$begingroup$
@lulu yes, that's what I meant. See my update for more information
$endgroup$
– Zoltán
Jan 22 at 23:15
$begingroup$
Hint: write it out as a sum involving the rate and note that you have a geometric series.
$endgroup$
– lulu
Jan 22 at 23:04
$begingroup$
Hint: write it out as a sum involving the rate and note that you have a geometric series.
$endgroup$
– lulu
Jan 22 at 23:04
$begingroup$
@lulu No, because the stock price changes.
$endgroup$
– Acccumulation
Jan 22 at 23:07
$begingroup$
@lulu No, because the stock price changes.
$endgroup$
– Acccumulation
Jan 22 at 23:07
$begingroup$
@Acccumulation Why does that matter? You are looking for a single rate $r$ which would match the return over the period.
$endgroup$
– lulu
Jan 22 at 23:09
$begingroup$
@Acccumulation Why does that matter? You are looking for a single rate $r$ which would match the return over the period.
$endgroup$
– lulu
Jan 22 at 23:09
$begingroup$
I'ver misread the problem a couple of times now, so let me state how I am currently reading it. I believe the OP is looking for the rate, $r$, such that a monthly investment of $100$ over $181$ periods would result in a total value of $32,331.15$ Thus, the first $100$ will have become $100times (1+frac r{12})^{181}$, the second will have grown to $100times (1+frac r{12})^{180}$ and so on. Is this what was intended?
$endgroup$
– lulu
Jan 22 at 23:14
$begingroup$
I'ver misread the problem a couple of times now, so let me state how I am currently reading it. I believe the OP is looking for the rate, $r$, such that a monthly investment of $100$ over $181$ periods would result in a total value of $32,331.15$ Thus, the first $100$ will have become $100times (1+frac r{12})^{181}$, the second will have grown to $100times (1+frac r{12})^{180}$ and so on. Is this what was intended?
$endgroup$
– lulu
Jan 22 at 23:14
$begingroup$
@lulu yes, that's what I meant. See my update for more information
$endgroup$
– Zoltán
Jan 22 at 23:15
$begingroup$
@lulu yes, that's what I meant. See my update for more information
$endgroup$
– Zoltán
Jan 22 at 23:15
|
show 6 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Trusting that the schedule of payments I outlined in the comments are correct, the problem comes down to the following: $$100times sum_{i=1}^{181}left(1+frac r{12}right)^i=32,311.15$$
Now, it's possible to evaluate that sum as a rational function, but as that leads to a polynomial of degree $182$ in $r$ you'll be stuck with numerical methods anyway. Instead, I'd leave the function as it stands and work numerically. A simple binary search converges rapidly and we get $$rapprox 7.056%$$
answered Jan 23 at 1:03
lulululu
42.9k25080
$endgroup$
add a comment |
$begingroup$
Considering the general case, you want to solve for $x$ the equation
$$k=frac TM=sum_{i=1}^n(1+x)^i=frac{(x+1) left((x+1)^n-1right)}{x}$$ where $T$ is the total value, $M$ the monthly amount, $n$ the number of months and $x$ the monthly interest.
As lulu explained, you need some numerical method (Newton method would be the easiest).
You can have good estimates of the solution considering that $x ll 1$. So, let us expand the summation as a Taylor series (or use the binomial theorem) and get
$$k=n+frac{n (n+1)}{2} x+frac{(n-1)n(n+1)}{6} n x^2+Oleft(x^3right)$$
Neglecting the higher order terms, you just face a quadratic equation in $x$ anf then
$$x_0=frac{sqrt{24(n-1)n(n+1)k-3n^2(5 n^2-6 n-11) }-3n(n+1) } {2(n-1)n(n+1) }$$
For your example, this would give as an estimate $x=0.00626955$ corresponding to $7.52$%. Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.0062695483 \
1 & 0.0058852014 \
2 & 0.0058753116 \
3 & 0.0058753053
end{array}
right)$$ and then the answer $7.050366$%.
We could have a better estimate (still at the price of a quadratic equation in $x$) using, instead of Taylor series, a simple Padé approximant. This would give
$$k=frac{n+frac{1}{10} left(n^2+13 nright) x+frac{1}{60} left(n^3+3 n^2+20 nright)
x^2 } {1-frac{2}{5} (n-2) x+frac{1}{20} left(n^2-3 n+2right) x^2 }$$ Applied to the example, this would give as an estimate $x=0.00587785$ corresponding to $7.053$%.
Edit
We can even avoid solving quadratic equations building at $x=0$ the $[1,p]$ Padé approximant of
$$frac{(x+1) left((x+1)^n-1right)}{x}-k=frac{(k-n)+a_{(p)} x}{1+sum_{j=1}^p b_j x^j}implies x_{(p)}=frac{n-k}{a_{(p)} }$$
For the simplest , the required coefficient is given by
$$a_{(1)}=frac{1}{6} n (n+5)+frac{1}{3} k (n-1)$$
$$a_{(2)}=frac{3 n^2 (n+3)+2 (n-1) n (n+4)k+ (n-2) (n-1)k^2 } {2 n (n+5)+4 (n-1)k }$$ The formula becomes really long for $p geq 3$ but they will be very easy to code.
Applied to the example, this would give
$$x_{(1)}approx 0.00568499implies 6.82198text{ %}$$
$$x_{(2)}approx 0.00585822implies 7.02987text{ %}$$
$$x_{(3)}approx 0.00587585implies 7.05102text{ %}$$
$$x_{(4)}approx 0.00587561implies 7.05073text{ %}$$
$$x_{(5)}approx 0.00587532implies 7.05039text{ %}$$
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
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2 Answers
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2 Answers
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$begingroup$
Trusting that the schedule of payments I outlined in the comments are correct, the problem comes down to the following: $$100times sum_{i=1}^{181}left(1+frac r{12}right)^i=32,311.15$$
Now, it's possible to evaluate that sum as a rational function, but as that leads to a polynomial of degree $182$ in $r$ you'll be stuck with numerical methods anyway. Instead, I'd leave the function as it stands and work numerically. A simple binary search converges rapidly and we get $$rapprox 7.056%$$
answered Jan 23 at 1:03
lulululu
42.9k25080
$endgroup$
add a comment |
$begingroup$
Trusting that the schedule of payments I outlined in the comments are correct, the problem comes down to the following: $$100times sum_{i=1}^{181}left(1+frac r{12}right)^i=32,311.15$$
Now, it's possible to evaluate that sum as a rational function, but as that leads to a polynomial of degree $182$ in $r$ you'll be stuck with numerical methods anyway. Instead, I'd leave the function as it stands and work numerically. A simple binary search converges rapidly and we get $$rapprox 7.056%$$
answered Jan 23 at 1:03
lulululu
42.9k25080
$endgroup$
add a comment |
$begingroup$
Trusting that the schedule of payments I outlined in the comments are correct, the problem comes down to the following: $$100times sum_{i=1}^{181}left(1+frac r{12}right)^i=32,311.15$$
Now, it's possible to evaluate that sum as a rational function, but as that leads to a polynomial of degree $182$ in $r$ you'll be stuck with numerical methods anyway. Instead, I'd leave the function as it stands and work numerically. A simple binary search converges rapidly and we get $$rapprox 7.056%$$
answered Jan 23 at 1:03
lulululu
42.9k25080
$endgroup$
Trusting that the schedule of payments I outlined in the comments are correct, the problem comes down to the following: $$100times sum_{i=1}^{181}left(1+frac r{12}right)^i=32,311.15$$
Now, it's possible to evaluate that sum as a rational function, but as that leads to a polynomial of degree $182$ in $r$ you'll be stuck with numerical methods anyway. Instead, I'd leave the function as it stands and work numerically. A simple binary search converges rapidly and we get $$rapprox 7.056%$$
answered Jan 23 at 1:03
lulululu
42.9k25080
answered Jan 23 at 1:03
lulululu
42.9k25080
answered Jan 23 at 1:03
lulululu
42.9k25080
answered Jan 23 at 1:03
lulululu
42.9k25080
42.9k25080
add a comment |
add a comment |
$begingroup$
Considering the general case, you want to solve for $x$ the equation
$$k=frac TM=sum_{i=1}^n(1+x)^i=frac{(x+1) left((x+1)^n-1right)}{x}$$ where $T$ is the total value, $M$ the monthly amount, $n$ the number of months and $x$ the monthly interest.
As lulu explained, you need some numerical method (Newton method would be the easiest).
You can have good estimates of the solution considering that $x ll 1$. So, let us expand the summation as a Taylor series (or use the binomial theorem) and get
$$k=n+frac{n (n+1)}{2} x+frac{(n-1)n(n+1)}{6} n x^2+Oleft(x^3right)$$
Neglecting the higher order terms, you just face a quadratic equation in $x$ anf then
$$x_0=frac{sqrt{24(n-1)n(n+1)k-3n^2(5 n^2-6 n-11) }-3n(n+1) } {2(n-1)n(n+1) }$$
For your example, this would give as an estimate $x=0.00626955$ corresponding to $7.52$%. Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.0062695483 \
1 & 0.0058852014 \
2 & 0.0058753116 \
3 & 0.0058753053
end{array}
right)$$ and then the answer $7.050366$%.
We could have a better estimate (still at the price of a quadratic equation in $x$) using, instead of Taylor series, a simple Padé approximant. This would give
$$k=frac{n+frac{1}{10} left(n^2+13 nright) x+frac{1}{60} left(n^3+3 n^2+20 nright)
x^2 } {1-frac{2}{5} (n-2) x+frac{1}{20} left(n^2-3 n+2right) x^2 }$$ Applied to the example, this would give as an estimate $x=0.00587785$ corresponding to $7.053$%.
Edit
We can even avoid solving quadratic equations building at $x=0$ the $[1,p]$ Padé approximant of
$$frac{(x+1) left((x+1)^n-1right)}{x}-k=frac{(k-n)+a_{(p)} x}{1+sum_{j=1}^p b_j x^j}implies x_{(p)}=frac{n-k}{a_{(p)} }$$
For the simplest , the required coefficient is given by
$$a_{(1)}=frac{1}{6} n (n+5)+frac{1}{3} k (n-1)$$
$$a_{(2)}=frac{3 n^2 (n+3)+2 (n-1) n (n+4)k+ (n-2) (n-1)k^2 } {2 n (n+5)+4 (n-1)k }$$ The formula becomes really long for $p geq 3$ but they will be very easy to code.
Applied to the example, this would give
$$x_{(1)}approx 0.00568499implies 6.82198text{ %}$$
$$x_{(2)}approx 0.00585822implies 7.02987text{ %}$$
$$x_{(3)}approx 0.00587585implies 7.05102text{ %}$$
$$x_{(4)}approx 0.00587561implies 7.05073text{ %}$$
$$x_{(5)}approx 0.00587532implies 7.05039text{ %}$$
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Considering the general case, you want to solve for $x$ the equation
$$k=frac TM=sum_{i=1}^n(1+x)^i=frac{(x+1) left((x+1)^n-1right)}{x}$$ where $T$ is the total value, $M$ the monthly amount, $n$ the number of months and $x$ the monthly interest.
As lulu explained, you need some numerical method (Newton method would be the easiest).
You can have good estimates of the solution considering that $x ll 1$. So, let us expand the summation as a Taylor series (or use the binomial theorem) and get
$$k=n+frac{n (n+1)}{2} x+frac{(n-1)n(n+1)}{6} n x^2+Oleft(x^3right)$$
Neglecting the higher order terms, you just face a quadratic equation in $x$ anf then
$$x_0=frac{sqrt{24(n-1)n(n+1)k-3n^2(5 n^2-6 n-11) }-3n(n+1) } {2(n-1)n(n+1) }$$
For your example, this would give as an estimate $x=0.00626955$ corresponding to $7.52$%. Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.0062695483 \
1 & 0.0058852014 \
2 & 0.0058753116 \
3 & 0.0058753053
end{array}
right)$$ and then the answer $7.050366$%.
We could have a better estimate (still at the price of a quadratic equation in $x$) using, instead of Taylor series, a simple Padé approximant. This would give
$$k=frac{n+frac{1}{10} left(n^2+13 nright) x+frac{1}{60} left(n^3+3 n^2+20 nright)
x^2 } {1-frac{2}{5} (n-2) x+frac{1}{20} left(n^2-3 n+2right) x^2 }$$ Applied to the example, this would give as an estimate $x=0.00587785$ corresponding to $7.053$%.
Edit
We can even avoid solving quadratic equations building at $x=0$ the $[1,p]$ Padé approximant of
$$frac{(x+1) left((x+1)^n-1right)}{x}-k=frac{(k-n)+a_{(p)} x}{1+sum_{j=1}^p b_j x^j}implies x_{(p)}=frac{n-k}{a_{(p)} }$$
For the simplest , the required coefficient is given by
$$a_{(1)}=frac{1}{6} n (n+5)+frac{1}{3} k (n-1)$$
$$a_{(2)}=frac{3 n^2 (n+3)+2 (n-1) n (n+4)k+ (n-2) (n-1)k^2 } {2 n (n+5)+4 (n-1)k }$$ The formula becomes really long for $p geq 3$ but they will be very easy to code.
Applied to the example, this would give
$$x_{(1)}approx 0.00568499implies 6.82198text{ %}$$
$$x_{(2)}approx 0.00585822implies 7.02987text{ %}$$
$$x_{(3)}approx 0.00587585implies 7.05102text{ %}$$
$$x_{(4)}approx 0.00587561implies 7.05073text{ %}$$
$$x_{(5)}approx 0.00587532implies 7.05039text{ %}$$
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
add a comment |
$begingroup$
Considering the general case, you want to solve for $x$ the equation
$$k=frac TM=sum_{i=1}^n(1+x)^i=frac{(x+1) left((x+1)^n-1right)}{x}$$ where $T$ is the total value, $M$ the monthly amount, $n$ the number of months and $x$ the monthly interest.
As lulu explained, you need some numerical method (Newton method would be the easiest).
You can have good estimates of the solution considering that $x ll 1$. So, let us expand the summation as a Taylor series (or use the binomial theorem) and get
$$k=n+frac{n (n+1)}{2} x+frac{(n-1)n(n+1)}{6} n x^2+Oleft(x^3right)$$
Neglecting the higher order terms, you just face a quadratic equation in $x$ anf then
$$x_0=frac{sqrt{24(n-1)n(n+1)k-3n^2(5 n^2-6 n-11) }-3n(n+1) } {2(n-1)n(n+1) }$$
For your example, this would give as an estimate $x=0.00626955$ corresponding to $7.52$%. Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.0062695483 \
1 & 0.0058852014 \
2 & 0.0058753116 \
3 & 0.0058753053
end{array}
right)$$ and then the answer $7.050366$%.
We could have a better estimate (still at the price of a quadratic equation in $x$) using, instead of Taylor series, a simple Padé approximant. This would give
$$k=frac{n+frac{1}{10} left(n^2+13 nright) x+frac{1}{60} left(n^3+3 n^2+20 nright)
x^2 } {1-frac{2}{5} (n-2) x+frac{1}{20} left(n^2-3 n+2right) x^2 }$$ Applied to the example, this would give as an estimate $x=0.00587785$ corresponding to $7.053$%.
Edit
We can even avoid solving quadratic equations building at $x=0$ the $[1,p]$ Padé approximant of
$$frac{(x+1) left((x+1)^n-1right)}{x}-k=frac{(k-n)+a_{(p)} x}{1+sum_{j=1}^p b_j x^j}implies x_{(p)}=frac{n-k}{a_{(p)} }$$
For the simplest , the required coefficient is given by
$$a_{(1)}=frac{1}{6} n (n+5)+frac{1}{3} k (n-1)$$
$$a_{(2)}=frac{3 n^2 (n+3)+2 (n-1) n (n+4)k+ (n-2) (n-1)k^2 } {2 n (n+5)+4 (n-1)k }$$ The formula becomes really long for $p geq 3$ but they will be very easy to code.
Applied to the example, this would give
$$x_{(1)}approx 0.00568499implies 6.82198text{ %}$$
$$x_{(2)}approx 0.00585822implies 7.02987text{ %}$$
$$x_{(3)}approx 0.00587585implies 7.05102text{ %}$$
$$x_{(4)}approx 0.00587561implies 7.05073text{ %}$$
$$x_{(5)}approx 0.00587532implies 7.05039text{ %}$$
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
$endgroup$
Considering the general case, you want to solve for $x$ the equation
$$k=frac TM=sum_{i=1}^n(1+x)^i=frac{(x+1) left((x+1)^n-1right)}{x}$$ where $T$ is the total value, $M$ the monthly amount, $n$ the number of months and $x$ the monthly interest.
As lulu explained, you need some numerical method (Newton method would be the easiest).
You can have good estimates of the solution considering that $x ll 1$. So, let us expand the summation as a Taylor series (or use the binomial theorem) and get
$$k=n+frac{n (n+1)}{2} x+frac{(n-1)n(n+1)}{6} n x^2+Oleft(x^3right)$$
Neglecting the higher order terms, you just face a quadratic equation in $x$ anf then
$$x_0=frac{sqrt{24(n-1)n(n+1)k-3n^2(5 n^2-6 n-11) }-3n(n+1) } {2(n-1)n(n+1) }$$
For your example, this would give as an estimate $x=0.00626955$ corresponding to $7.52$%. Newton iterates would be
$$left(
begin{array}{cc}
n & x_n \
0 & 0.0062695483 \
1 & 0.0058852014 \
2 & 0.0058753116 \
3 & 0.0058753053
end{array}
right)$$ and then the answer $7.050366$%.
We could have a better estimate (still at the price of a quadratic equation in $x$) using, instead of Taylor series, a simple Padé approximant. This would give
$$k=frac{n+frac{1}{10} left(n^2+13 nright) x+frac{1}{60} left(n^3+3 n^2+20 nright)
x^2 } {1-frac{2}{5} (n-2) x+frac{1}{20} left(n^2-3 n+2right) x^2 }$$ Applied to the example, this would give as an estimate $x=0.00587785$ corresponding to $7.053$%.
Edit
We can even avoid solving quadratic equations building at $x=0$ the $[1,p]$ Padé approximant of
$$frac{(x+1) left((x+1)^n-1right)}{x}-k=frac{(k-n)+a_{(p)} x}{1+sum_{j=1}^p b_j x^j}implies x_{(p)}=frac{n-k}{a_{(p)} }$$
For the simplest , the required coefficient is given by
$$a_{(1)}=frac{1}{6} n (n+5)+frac{1}{3} k (n-1)$$
$$a_{(2)}=frac{3 n^2 (n+3)+2 (n-1) n (n+4)k+ (n-2) (n-1)k^2 } {2 n (n+5)+4 (n-1)k }$$ The formula becomes really long for $p geq 3$ but they will be very easy to code.
Applied to the example, this would give
$$x_{(1)}approx 0.00568499implies 6.82198text{ %}$$
$$x_{(2)}approx 0.00585822implies 7.02987text{ %}$$
$$x_{(3)}approx 0.00587585implies 7.05102text{ %}$$
$$x_{(4)}approx 0.00587561implies 7.05073text{ %}$$
$$x_{(5)}approx 0.00587532implies 7.05039text{ %}$$
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
edited Jan 24 at 7:33
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
answered Jan 23 at 6:24
Claude LeiboviciClaude Leibovici
124k1157135
124k1157135
add a comment |
add a comment |
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$begingroup$
Hint: write it out as a sum involving the rate and note that you have a geometric series.
$endgroup$
– lulu
Jan 22 at 23:04
$begingroup$
@lulu No, because the stock price changes.
$endgroup$
– Acccumulation
Jan 22 at 23:07
$begingroup$
@Acccumulation Why does that matter? You are looking for a single rate $r$ which would match the return over the period.
$endgroup$
– lulu
Jan 22 at 23:09
$begingroup$
I'ver misread the problem a couple of times now, so let me state how I am currently reading it. I believe the OP is looking for the rate, $r$, such that a monthly investment of $100$ over $181$ periods would result in a total value of $32,331.15$ Thus, the first $100$ will have become $100times (1+frac r{12})^{181}$, the second will have grown to $100times (1+frac r{12})^{180}$ and so on. Is this what was intended?
$endgroup$
– lulu
Jan 22 at 23:14
$begingroup$
@lulu yes, that's what I meant. See my update for more information
$endgroup$
– Zoltán
Jan 22 at 23:15