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arindameducationusc
 one year ago
Why is e^x = 1 + x + (1/2!) x^2 + (1/3!)x^3 + ....
and how is this equal to 2.718
arindameducationusc
 one year ago
Why is e^x = 1 + x + (1/2!) x^2 + (1/3!)x^3 + .... and how is this equal to 2.718

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Empty
 one year ago
Best ResponseYou've already chosen the best response.2Well essentially it's because if you take the derivative of both sides, they're the same.

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2\[e^x = \sum_{n=0}^{\infty} \frac{ x^n }{ n! } = 1+x+\frac{ x^2 }{ 2! }+\frac{ x^3 }{ 3! }+...\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Use Maclaurin series

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0@Astrophysics ofcourse I know the series but why? Is it a defination?

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0What is a Maclaurin series? Please send me a link

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2If \[f(x) = e^x\] we can then say \[f^{n+1}(x) = e^x\] for all n. And if d is any positive integer number and \[x \le d\], then \[f^{n+1}(x) = e^x \le e^d\] so we can say the taylor inequality, with a = 0 and \[M=e^d\] tells us \[R_n(x) \le \frac{ e^d }{ (n+1)! }x^{n+1}~~~~~\text{for}~`x \le d\] see how \[M=e^d\] we can then say \[\lim_{n \rightarrow \infty} \frac{ e^d }{ (n+1)! }x^{n+1} = e^d \lim_{n \rightarrow \infty} \frac{ x^{n+1} }{ (n+1) }=0\] we can use the squeeze theorem to conclude then \[e^x = \sum_{n=0}^{\infty} \frac{ x^n }{ n! }\] for all x.

Empty
 one year ago
Best ResponseYou've already chosen the best response.2A maclaurin series just is a polynomial that has all the same derivatives as a function at a single point. So if we only made this approximation with 3 terms, \[e^x \approx 1+x+\frac{x^2}{2}\] then that means if we evaluate them at x=0 we get the same values for the function, its first, and second derivatives. After you differentiate that polynomial a 3rd time, then it's no longer approximately equal. So if you make a polynomial that has infinitely many of the same derivatives at a point as another function, then you basically can say they're the same function.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Thank you, @Astrophysics and @Empty
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