Any help on this is much appreciated!
i'll use $\displaystyle s_n$ to denote the accumulated value of the annuity since i dont know the code in LaTex..
so, $\displaystyle s_{10} = \frac{(1+i)^{10} - 1}{i}$
and assuming that this is a compound interest, then we can use
$\displaystyle a(t) = \underbrace{e^{\int_0^t \delta_s \, ds}}_{in \, terms \, of \, force \, of \, interest} = \underbrace{(1+i)^t}_{basic \, formula}$
therefore, the middle part can be simplified (evaluated at $\displaystyle t = 10$) as $\displaystyle \ln 2$ (verify) and together with the right hand side, we have $\displaystyle (1+i)^{10} = \ln 2$ or $\displaystyle i = \sqrt[10]{\ln 2} - 1$
therefore, $\displaystyle s_{10} = \frac{(1+i)^{10} - 1}{i} = \frac{\ln 2 - 1}{\sqrt[10]{\ln 2} - 1}$
Please do not double post. You question was first answered in the Business Math forum: http://www.mathhelpforum.com/math-he...-function.html
You should have these:
$\displaystyle a(t) = e^{\int_{0}^{t}\delta_{r}dr}$
$\displaystyle a_{angle-n} = \sum_{t=1}^{n}\left(\frac{1}{a(t)}\right)$
$\displaystyle s_{angle-n} = a(n)*a_{angle-n}$
Let's see what you get.
Note: I'm not sure what Kalagota is missing. It doesn't make sense for 10 payments to be accumulated to only 8½, so you get to tell us what you manage to calulate.
now i wonder what i missed?
i think, my mistake was that $\displaystyle \ln 2$ thing.. i forgot that the integral was a power of $\displaystyle e$ and thus, $\displaystyle a(t)$ should be equal to $\displaystyle 2$.. this should make sense...
now, let me correct my mistake..
$\displaystyle (1+i)^{10} = 2$ or $\displaystyle i = \sqrt[10]{2} - 1$
thus, $\displaystyle s_{10} = \frac{(1+i)^{10} - 1}{i} = \frac{1}{\sqrt[10]{2} - 1} \approx 13.9327262$
Formulas shown above. I think yours is missing some moving parts.
Start from $\displaystyle s_{10} = (1+i)^{10} + (1+i)^{9} + ... + (1+i)$, rather than the summed version that assumes regularity of interest for each payment.
With variable interest, $\displaystyle a(10) \neq \left(a(5)\right)^{2}$.
did you notice notice that your summation is just the sum of the first n geometric sequence.. i think, you must be the one to check your solution.. or the best way is to post your whole solution in here..
besides, if you will push your 14.5, try to work it backward, and you will derive at $\displaystyle a(t) = 1.71...$ and not 2..
maybe i'll put the derivation here..
$\displaystyle a(t) = (1+i)^t$
thus, $\displaystyle a_{angle-n} = \sum_{t=1}^n \frac{1}{a(t)}|_{n=10} = \frac{1}{a(1)} + \frac{1}{a(2)} + ... + \frac{1}{a(10)} $
or $\displaystyle a_{angle-10} = \frac{1}{(1+i)} + \frac{1}{(1+i)^2} + ... + \frac{1}{(1+i)^{10}} $
and $\displaystyle s_{angle-10} = a(10) \, a_{angle-10} = (1+i)^{10}\, a_{angle-10} = (1+i)^9 + (1+i)^8 + ... + (1+i) + 1 $
$\displaystyle = \frac{(1+i)^{10} - 1}{(1+i) - 1}$ and just do your substitution..
Thanks, I've got the answer. Do you mind if I ask a question though?
Why can I not just use this? Or at least, the answer doesn't turn out right, but I don't understand why.
$\displaystyle
\int _1 ^{10} e^{\delta t} dt
$
exp(-ln(20 - 10)+ ln(20-1)) = exp (ln(19/10)) = 19/10
Did I make a mistake somewhere?
Perfect.
This is where it falls apart. Your interest is static.or $\displaystyle a_{angle-10} = \frac{1}{(1+i)} + \frac{1}{(1+i)^2} + ... + \frac{1}{(1+i)^{10}} $
Calculate from the integrals:
a(1) = 20/19
a(2) = 10/9
a(3) = 20/17
a(4) = 5/4
It is not a Geometric Sequence.
Another Possible Answer:
As it stands, we have 't' defined as the time from EACH deposit. That may not seem quite right. If we use 't' to mean the time from the beginning of the financial structure, we get:
$\displaystyle s(t) = e^{\int_{t}^{10}\delta^{t}\;dr}$
Then:
$\displaystyle s_{angle-10} = \sum_{0}^{9}s(t) = 15.5$
Very interestingly, we get:
s(0) = 2.00
s(1) = 1.90
s(2) = 1.80
etc...
It IS an arithmetic sequence.