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DANIEL93

  • 2 years ago

single variable calculus . help me friends find the maclaurine series for (i) f(x)=1/1−x

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  1. klimenkov
    • 2 years ago
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    \(\frac1{1-x}=1+x+x^2+x^3+..., |x|<1\)

  2. DANIEL93
    • 2 years ago
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    solution?

  3. klimenkov
    • 2 years ago
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    Yes.

  4. DANIEL93
    • 2 years ago
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    can you explain how to get it?

  5. klimenkov
    • 2 years ago
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    It is the sum of the geometric sequence.

  6. DANIEL93
    • 2 years ago
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    oh.. okay2. how to find taylor series then?

  7. DANIEL93
    • 2 years ago
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    do you know?

  8. klimenkov
    • 2 years ago
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    \(f(x)=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x-x_0)}{2!}(x-x_0)^2+\ldots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+\ldots\)

  9. sirm3d
    • 2 years ago
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    \[\huge f(x)=\sum_{n=0}^{+\infty} \frac{ f^n(a) }{ n! }(x-a)^n\] compute the value of \[\huge f^{(n)}(a)\]

  10. sirm3d
    • 2 years ago
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    where f^0(a) is the value of the function, f^1(a) is the value of the first derivative. For the maclaurin series, use a = 0.

  11. DANIEL93
    • 2 years ago
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    @klimenkov : thank you :D @sirm3d : tHANK yOU :D

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