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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

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  1. Empty
    • one year ago
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    Well essentially it's because if you take the derivative of both sides, they're the same.

  2. Astrophysics
    • one year ago
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    \[e^x = \sum_{n=0}^{\infty} \frac{ x^n }{ n! } = 1+x+\frac{ x^2 }{ 2! }+\frac{ x^3 }{ 3! }+...\]

  3. Astrophysics
    • one year ago
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    Use Maclaurin series

  4. arindameducationusc
    • one year ago
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    @Astrophysics ofcourse I know the series but why? Is it a defination?

  5. arindameducationusc
    • one year ago
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    What is a Maclaurin series? Please send me a link

  6. Astrophysics
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Thank you, @Astrophysics and @Empty

  9. Astrophysics
    • one year ago
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    Np :)

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