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Find the maclaurin series for f(x) using the definition of a maclaurin series. [assume that f has a power series expansion. do not show that rn(x) → 0. ] f(x) = e−6x

Sagot :

The equation of [tex]f(x) = e^{-6x}[/tex] by maclaurin series is [tex]f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i} }{i!}[/tex].

The maclaurin series for f(x) is defined by the following formula:

[tex]f(x) = \sum_{i=0}^{\infty} \frac{f^{(i)} (0)}{i!} .x^{i}[/tex]--------------(1)

Where [tex]f^{i}[/tex] is the i - th derivative of the function

If f(x) = [tex]e^{-6x}[/tex], then the formula of the i - th derivative of the function is:

[tex]f^{i} =(-6)^{i} .e^{-6x}[/tex]----------------------(2)

Then,

[tex]f^{i}(0) = (-6)^{i}[/tex]

Lastly, the equation of the trascendental function by Maclaurin series is: [tex]f(x)=\sum_{i=0}^{\infty} \frac{(-6)^{i}.x^{i} }{i!} \\f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i} }{i!}[/tex]---------------(3)

Hence,

The equation of [tex]f(x) = e^{-6x}[/tex] by maclaurin series is [tex]f(x)=\sum_{i=0}^{\infty} \frac{(-6.x)^{i} }{i!}[/tex].

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