Continuous Compounding Interest

  • Thread starter ecoo
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  • #1
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Hello

So from what I understand, the continuous compound formula finds out the most you can get from interest no matter how many times you compound the interest in a set amount of time. So how come when I plugin in a big number into the regular compounding formula for the rate, the end amount is more than the amount I get when calculating with the continuous compounding formula? I think that it's a calculator inaccuracy, or is my view incorrect?

Thanks for the help!
 

Answers and Replies

  • #2
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Hello

So from what I understand, the continuous compound formula finds out the most you can get from interest no matter how many times you compound the interest in a set amount of time. So how come when I plugin in a big number into the regular compounding formula for the rate, the end amount is more than the amount I get when calculating with the continuous compounding formula? I think that it's a calculator inaccuracy, or is my view incorrect?

Thanks for the help!
Interest compounded continuously should give the larger amount. Can you show us an example where the normal compounding formula seems to give a larger interest amount?

Are you forgetting to divide the annual interest rate by the number of compounding periods per year?
 
  • #3
87
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Interest compounded continuously should give the larger amount. Can you show us an example where the normal compounding formula seems to give a larger interest amount?

Are you forgetting to divide the annual interest rate by the number of compounding periods per year?
So for example, if I use the regular compounding interest for 6% for 10 years, and the rate is 999999999999 (add more if you want), then the answer is more than the continuous equation result. Besides the answer, is my thinking correct?
 
  • #4
TeethWhitener
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Noncontinuous: [tex]A_{noncont}=P(1+\frac{0.06}{999999999999})^{10\times999999999999}=1.82124518238P[/tex]
Continuous: [tex]A_{cont}=Pe^{(0.06\times10)}=1.82211880039P[/tex]
[tex]A_{noncont}<A_{cont}[/tex]
 
  • #5
35,231
7,053
So for example, if I use the regular compounding interest for 6% for 10 years, and the rate is 999999999999 (add more if you want), then the answer is more than the continuous equation result. Besides the answer, is my thinking correct?
The rate refers to the interest rate, which in your example is 6%. The periodic interest rate would be .06/999999999999.
 

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