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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−3x

Sagot :

xero099

Answer:

The equation of [tex]f(x) = e^{-3\cdot x}[/tex] by Maclaurin series is [tex]f(x) = \Sigma\limits_{i=0}^{\infty} \frac{(-3\cdot x)^{i}}{i!}[/tex].

Step-by-step explanation:

The Maclaurin series for [tex]f(x)[/tex] is defined by the following formula:

[tex]f(x) = \Sigma\limits_{i = 0}^{\infty} \frac{f^{(i)}(0)}{i!} \cdot x^{i}[/tex] (1)

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

If [tex]f(x) = e^{-3\cdot x}[/tex], then the formula of the i-th derivative of the function is:

[tex]f^{(i)} = (-3)^{i}\cdot e^{-3\cdot x}[/tex] (2)

Then,

[tex]f^{(i)}(0) = (-3)^{i}[/tex] (2b)

Lastly, the equation of the trascendental function by Maclaurin series is:

[tex]f(x) = \Sigma\limits_{i=0}^{\infty} \frac{(-3)^{i}\cdot x^{i}}{i!}[/tex]

[tex]f(x) = \Sigma\limits_{i=0}^{\infty} \frac{(-3\cdot x)^{i}}{i!}[/tex] (3)

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