Number e and compound interest - Revision history

Alexandk at 20:03, 3 April 2022

2022-04-03T20:03:55Z

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 Suppose, for simplicity, that one period of investment is one year. Suppose further that you invest <math> P = 1 </math> for the period of one year, and the rate of interest is 100% or, equivalently, <math> i = 1 </math> (this means that the amount you invested will double at the end of the year, which, of course, rarely occurs in practice). Alternatively, you may consider investing the same amount, <math> P =1 </math>, but so that the effective interest period is a fraction of a year, say, 1 month. In this case, the stated rate of interest, <math> i = 1 </math> becomes nominal, in the sense that the effective monthly rate of compound interest is computed as <math> j= i/12 = 1/12 </math>. Thus, at the end of one year you will have
-<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^12 </math>
+<math> (1+j)^{12} = \left (1+\frac{1}{12}\right)^{12} </math>
 in your account. Using the Binomial formula

← Older revision		Revision as of 20:03, 3 April 2022
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	which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>) This observation leads to the definition of famous mathematical constant - number <math> e</math>:		which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>) This observation leads to the definition of famous mathematical constant - number <math> e</math>:

−	<math> e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>	+	<math> e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n.</math>

← Older revision		Revision as of 20:02, 3 April 2022
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	<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>		<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

−	which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>)	+	which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>) This observation leads to the definition of famous mathematical constant - number <math> e</math>:
		+
		+	<math> e = \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

← Older revision		Revision as of 20:00, 3 April 2022
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	<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>		<math> \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>

−	which corresponds to the interest compounded "continuously". It can be shown that the above limit exists and is approximately equal to <math> 2.7.</math>	+	which corresponds to the interest compounded "continuously". It can be shown that the above limit exists (and is approximately equal to <math> 2.7.</math>)

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 we see that
-<math> \left(1+\frac{1}{12}\right)^{12|} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>
+<math> \left(1+\frac{1}{12}\right)^{12} \ge 1 + 12\cdot \frac{1}{12} = 2 </math>
 and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.
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 <math>  \lim _{n\to\infty} \left(1+\frac{1}{n}\right)^n </math>
-which corresponds to the interest compounded "continuously".
+which corresponds to the interest compounded "continuously". It can be shown that the above limit exists and is approximately equal to <math> 2.7.</math>

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 Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be
-<math>  (1+\frac{1}{n}\right)^n. </math>
+<math>  \left(1+\frac{1}{n}\right)^n. </math>
 For very large values of <math> n </math> we therefore can consider the limit
-<math>  \lim _{n\to\infty} (1+\frac{1}{n}\right)^n </math>
+<math>  \lim _{n\to\infty \left(1+\frac{1}{n}\right)^n </math>
-which corresponds to the interest compounded `` continuously".
+which corresponds to the interest compounded "continuously".

← Older revision		Revision as of 19:57, 3 April 2022
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	we see that		we see that

−	<math> (1+\frac{1}{12}\right)^12	+	<math> (1+\frac{1}{12}\right)^12 \ge 1 + 12\cdot \frac{1}{12} = 2 </math>
		+
		+	and thus investing at nominal rate of interest 1 is more profitable than at effective annual rate equal to 1.
		+
		+	Further, we may consider scenarios in which the interest is compounded at higher frequencies, e.g. weekly or daily. In general, if <math> n </math> is the number of times compounding occurs per year, and 1 is the corresponding nominal rate of interest, the value accumulated after one year will be
		+
		+	<math> (1+\frac{1}{n}\right)^n. </math>
		+
		+	For very large values of <math> n </math> we therefore can consider the limit
		+
		+	<math> \lim _{n\to\infty} (1+\frac{1}{n}\right)^n </math>
		+
		+	which corresponds to the interest compounded `` continuously".

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 in your account. Using the Binomial formula
-<math> (1+x)^n = \sum_{k=0}^n n\choose k x^k </math>
+<math> (1+x)^n = \sum_{k=0}^n {n\choose k} x^k </math>
 we see that
 <math> (1+\frac{1}{12}\right)^12