Question

Why is the value of e = 2.7183(approx.) ?

Answer 1

interesting question

Answer 2

it is constant

Answer 3

Why has it been given that value ?

Answer 4

no idea lol

Answer 5

me neither....

Answer 6

It is the limit of (1 + 1/n)n as n approaches infinity.

Answer 7

\[(1+1/n)^{n}\]

Answer 8

not got the appropriate answer.

Answer 9

smart walking stick....

Answer 10

\[\LARGE \lim_{n \rightarrow ∞}~~ 1+\frac{1}{n}\] I haven't ever thought f ...

Answer 11

Can we get to the answer, by using Exponential expansion series ?

Answer 12

The number e is an important mathematical constant that is the base of the natural logarithm. It is approximately equal to 2.71828,and is the limit of (1 + 1/n)n as n approaches infinity. It can also be calculated as the sum of the infinite series.

e = \displaystyle\sum\limits_{n = 0}^{ \infty} \dfrac{1}{n!} = 1 + \frac{1}{1} + \frac{1}{1\cdot 2} + \frac{1}{1\cdot 2\cdot 3} + \cdots

Answer 13

put your latex in the equation editor before posting.

Answer 14

\[\lim_{n \rightarrow \infty}(1+1/n)^{n}\]Thank you SolomonZelman. It is often used in measuring things with instantaneous growth, like compound interest.

Answer 15

\[e=1+\frac{ 1 }{ 1! }+\frac{ 1 }{ 2! }+\frac{ 1 }{ 3! }+\frac{ 1 }{ 4! }+\frac{ 1 }{ 5! }+\frac{ 1 }{ 6! }+\frac{ 1 }{ 7! }+\frac{ 1 }{ 8! }+....\]
\[=1+\frac{ 1 }{ 1 }+\frac{ 1 }{ 2*1 }+\frac{ 1 }{ 3*2*1 }+\frac{ 1 }{ 4*3*2*1 }+\frac{ 1 }{ 5*4*3*2*1 }\]
\[+\frac{ 1 }{ 6*5*4*3*2*1 }+\frac{ 1 }{ 7*6*5*4*3*2*1 }+\frac{ 1 }{ 8*7*6*5*4*3*2*1 }+....\]
\[=1+1+\frac{ 1 }{ 2 }+\left( \frac{ 1 }{ 6 }+\frac{ 1 }{ 24 } \right)+\left( \frac{ 1 }{ 120 }+\frac{ 1 }{ 720 } \right)+\left( \frac{ 1 }{ 5040 }+\frac{ 1 }{ 40320 } \right)+....\]
\[=2.5+0.20833333...+0.00972222...+0.00022321..+..=2.71827872...=2.7183 (\approx.)\]

Answer 16

wow!

Answer 17

Holy math....