+0

# Find the accumulated value of an investment of $15,000 for 5 years at an interest rate of 7% if the money is compounded continuously. 0 1360 8 Find the accumulated value of an investment of$15,000 for 5 years at an interest rate of 7% if the money is compounded continuously.

Guest Feb 16, 2015

#6
+80956
+10

Why does Pe^(rt) work??

Look at

lim n →∞   (1 + r/n)^n

Let  m = n/r    So we have

lim m →∞ ( 1 + 1/m)^(r*m)  =

lim m →∞ [(1 + 1/m)^m ] ^r       but ...    lim m →∞ (1 + 1/m)^m   = e

So we have

e^r

And

Pe^[r(1)]  = P(e^r) = would represent continuous compounding for one year

So....

Pe^(rt)  =

P(e^r)(e^r)(e^r)...........   where (e^r) is multiplied "t" times ....{representing "t" years (or periods)......}

And that's it.......!!!

CPhill  Feb 16, 2015
Sort:

#1
+10

Let there be n divisions of time in 1 year.

formula after 1 year 15000*(1+7/n)^n

after 5 years 15000*(1+7/n)^(n*5)

n approaches infinity:

lim n->infinity [15000*(1+7/n)^(n*5)

=15000*e^7*5

=15000*e^35

Guest Feb 16, 2015
#2
+91435
0

I'd like to see a proof of this :))

Melody  Feb 16, 2015
#3
+80956
+10

The proof is this, Melody.....it's in the way that "e" is defined...

e = lim as n →∞  (1 + 1/n)^n

Notice, that Anonymous is letting n, the number of compounding periods per year, grow larger and larger......if we theoretically make "n" "huge," then the number of compoundings per year ≈ "continuous" because the time intervals of compounding → 0........thus, we have an "infinite" number of compoundings per year of extremely short duration each....!!!!

CPhill  Feb 16, 2015
#4
+26399
+10

This graph doesn't constitute a proof, but simply suggests that it's likely to be true (though you should have put 0.07 rather than 7 Melody!):

.

Alan  Feb 16, 2015
#5
+91435
0

It wasn't me Alan, I didn't put anything but I also noticed that error, I suppose I should have said so :))

Melody  Feb 16, 2015
#6
+80956
+10

Why does Pe^(rt) work??

Look at

lim n →∞   (1 + r/n)^n

Let  m = n/r    So we have

lim m →∞ ( 1 + 1/m)^(r*m)  =

lim m →∞ [(1 + 1/m)^m ] ^r       but ...    lim m →∞ (1 + 1/m)^m   = e

So we have

e^r

And

Pe^[r(1)]  = P(e^r) = would represent continuous compounding for one year

So....

Pe^(rt)  =

P(e^r)(e^r)(e^r)...........   where (e^r) is multiplied "t" times ....{representing "t" years (or periods)......}

And that's it.......!!!

CPhill  Feb 16, 2015
#7
+26399
+5

"It wasn't me Alan, I didn't put anything but I also noticed that error, I suppose I should have said so :))

Apologies Melody, I should have looked more carefully!

Alan  Feb 16, 2015
#8
+91435
+5

Thank you Chris and Alan,

I have finally looked properly at what you are telling me, and hopefully I will remember.

I have always had problems with this because I could never see why it was true and I have a reall problem reproducing things when I don't really understand.

I did not know that e could be defined that way

I will try and remember.

$$\boxed{e=\displaystyle \lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n}$$

Melody  Feb 19, 2015

### 27 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details