I ran into this problem at the end of a differential equation problem on interest. If I ever learned to solve such a problem I am drawing a complete blank.

Specifically the part I'm stuck at is:

500r + 1 = e^(40r), solve for r

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- Jan 23rd 2010, 08:23 PM #1

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- Jan 23rd 2010, 08:46 PM #2

- Jan 24th 2010, 06:08 AM #3

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It has an infinite number of solutions since $\displaystyle f(z)=e^{40 z}-500z-1$ is a non-polynomial "entire" complex function and the zeros can be written in terms of the Lambert-W function. Consider:

$\displaystyle ax+b=e^{cx}$

$\displaystyle (x+b/a)=1/ae^{cx}$

$\displaystyle e^{b/a c}\left(x+b/a=1/a e^{cx}\right)$

$\displaystyle e^{b/a c}(x+b/a)=1/ae^{c(x+b/a)}$

$\displaystyle -c(x+b/a)e^{-c(x+b/a)}=-c/a e^{-b/ac}$

and at this point we can take the W-function of both sides and obtain:

$\displaystyle -c(x+b/a)=\text{W}\left(-c/ae^{-b/a c}\right)$

or:

$\displaystyle x=-b/a-1/c \text{W}\left(-c/a e^{-b/a c}\right)$

but the W-function is infinitely-valued and to distinguish each value, we can write it as:

$\displaystyle x=-b/a-1/c \text{W}\left(n,-c/a e^{-b/a c}\right),\quad n=0,\pm 1,\pm 2,\cdots$

where $\displaystyle n$ is the sheet number of the infinitely twisted (real and complex) sheets in the complex plane. Now in the specific case above:

$\displaystyle x_0=-1/500-1/40 \text{W}\left(0,-40/500 e^{-40/500}\right)=0$

$\displaystyle x_{-1}=-1/500-1/40 \text{W}\left(-1,-40/500 e^{-40/500}\right)\approx 0.098$

$\displaystyle x_1=-1/500-1/40 \text{W}\left(1,-40/500 e^{-40/500}\right)\approx 0.012-0.18i$

and so on where I'm using ProductLog[n,val] in Mathematica to compute those values.

- Jan 24th 2010, 01:24 PM #4

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Ok, all of that is far and above what I'm supposed to be learning here, so maybe I missed something.

The premise verbatim is as follows:

A young person with no capital invests k dollars per year at an annual rate of return r. Assume that investments are made continuously and that the return is compounded continuously.

a)Determine the sum S(t) accumulated at any time t

[Answer: S(t) = k(e^rt-1)/r]

b)If r = 7.5%, determine K so that $1 million will be available for retirement in 40 years.

[Answer: k= about $3930]

c)If k = $2000/year, determine the return rate r that must be obtained to have $1 million available in 40 years

Part c is where I'm stuck and the back of the book gives the answer r = 9.77%.

S(40) = 2,000(e^40r-1)/r = 1,000,000

500r + 1 = e^40r

Infinite solutions and a rate of 0% won't make $1 million, though I see both points. I guess r > 0 would be important for me to mention as well.

- Jan 25th 2010, 03:47 AM #5

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Well since we're only talkin' about money here then we're only interested in the bottom-line: the answer. Ok then, just plot it (Mathematica code shown below):

Plot[{500 r + 1, Exp[40 r]}, {r, 0, .2}]

which is shown below. So they cross at about 0.1. Next do a find root to get a closer answer, remember, we're only interested in the value and not some elegant irrelevant math equation:

In[3]:= FindRoot[Exp[40 r] == 500 r + 1, {r, 0.1}]

Out[3]= {r -> 0.097734}

and that's 9.8 % right. Bingo-bango. I wanna' cut.