Question

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

Answer 1

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

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

$f(x) = \sum_{i=0}^{\infty} \frac{f^{(i)} (0)}{i!} .x^{i}$--------------(1)

Where $f^{i}$ is the i - th derivative of the function

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

$f^{i} =(-6)^{i} .e^{-6x}$----------------------(2)

Then,

$f^{i}(0) = (-6)^{i}$

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

Hence,

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

Find out more information about maclaurin series here

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