April 3rd, 2014, 05:30 PM  #1 
Newbie Joined: Oct 2011 Posts: 3 Thanks: 0  Finance & Compounding Interest
It's been years since I studied math. I've forgotten the basics. I've come across this question, and I can't solve it (at least I'm not getting the same solution as a book). Here's the problem: How much money do you have to deposit into an account monthly, for 30 years, at an interest rate of 12%, compounded monthly, in order to have $1.5 million at the end of the 30 years? Here's what I have: y = ca^x, where c is a constant, a is the growth factor, and x is the number of months c E (1.12)^n (i = 1, n = 360) Therefore, 1.5x10^6 = c[1.12^360 + 1.12^359 + 1.12^358 + ... + 1.12] Solving for c (after computing the summation with a calculator) yields a very small number. I'm not sure my Sigma formula is correct, and I don't know how to calculate it by hand. I'd like to know how to manually do it. I totally forgot how to compute this like a human. 
April 3rd, 2014, 07:45 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 
I would model this using a linear difference equation. Let $A_t$ be the amount in the account at time $t$ measured in months and $D$ be the monthly deposit. We will let the clock begin at the moment the first deposit is made. Hence, our difference equation is: $\displaystyle A_{t+1}1.01A_{t}=D$ The homogeneous solution is: $\displaystyle h_t=c_1(1.01)^t=c_1\left(\frac{101}{100} \right)^t$ and the particular solution is: $\displaystyle p_t=B$ Using the method of undetermined coefficients, we find: $\displaystyle B=100D$ Thus, by the principle of superposition, the general solution is: $\displaystyle A_t=c_1\left(\frac{101}{100} \right)^t100D$ Using the initial value $A_0=D$, we may determine the value of the parameter: $\displaystyle A_0=c_1100D=D\implies c_1=101D$ And so the solution satisfying the model is: $\displaystyle A_t=D\left(101\left(\frac{101}{100} \right)^t100 \right)$ Now, to find find the value of $D$ such that: $\displaystyle A_{360}=1500000$ we may state: $\displaystyle D\left(101\left(\frac{101}{100} \right)^{360}100 \right)=1500000$ Solving for $D$, we find: $\displaystyle D=\frac{1500000}{101\left(\frac{101}{100} \right)^{360}100}\approx424.82$ 
April 4th, 2014, 12:11 AM  #3  
Newbie Joined: Oct 2011 Posts: 3 Thanks: 0  Quote:
I am not familiar with your method (despite what I read on Google a few minutes ago). As for my Summation Notation Equation, was my equation incorrect? That is: cE1.12^n (i=1, n=360)  
April 4th, 2014, 12:31 AM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 463 Math Focus: Calculus/ODEs 
It could be that they intended an effective rate of 12% which means we would get the following monthly interest rate instead: $\displaystyle r=0.94888\overline{3}\%$ And this would change the deposit amount slightly...but I don't want to go through the derivation again. You need to use a monthly rate instead of the annual rate in your summation, that's why you are getting a small value. 

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