## 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

1. Empty

Well essentially it's because if you take the derivative of both sides, they're the same.

2. Astrophysics

$e^x = \sum_{n=0}^{\infty} \frac{ x^n }{ n! } = 1+x+\frac{ x^2 }{ 2! }+\frac{ x^3 }{ 3! }+...$

3. Astrophysics

Use Maclaurin series

4. arindameducationusc

@Astrophysics ofcourse I know the series but why? Is it a defination?

5. arindameducationusc

What is a Maclaurin series? Please send me a link

6. Astrophysics

If $f(x) = e^x$ 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.

7. Empty

A 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.

8. arindameducationusc

Thank you, @Astrophysics and @Empty

9. Astrophysics

Np :)