SOLUTION: 1.Find the Maclaurin series for 𝑒 exponential kx, 𝑘 is a real number?

Question 1137179: 1.Find the Maclaurin series for 𝑒 exponential kx, 𝑘 is a real number?

Found 2 solutions by MathLover1, rothauserc:
Answer by MathLover1(20850) (Show Source): You can put this solution on YOUR website!
Find the Maclaurin series for e^(kx), k is a real number.
Calculate the derivatives:
′′=, ′′′=,…....,

Then, at we have
, ′, ′′,…

Hence, the Maclaurin expansion for the given function is
=+…=

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
A Maclaurin series is an infinite series that is a method to calculate an approximation of a function, f(x), for x values close to zero, given that the successive derivatives of f(x) at zero can be calculated.
:
For this problem, f(x) = e^(kx), where k is a real number
:
1) Calculate the derivatives for f(x)
:
f'(x) = ke^(kx), f''(x) = k^2e^(kx), ,,,,, f^n(x) = k^n * e^(kx)
:
2) At x = 0,
:
f(0) = 1, f'(0) = ke^0 = k, f''(0) = k^2, .... f^n(0) = k^n
:
Using the definition of the Maclaurin for e^(kx) is
:
e^(kx) = summation from n=0 to +infinity of f^n (0) (x^n/n!) =
:
1 +kx +(k^2 * x^2 / 2!) +(k^3 * x^3 / 3!) +.... = summation n=0 to +infinity of (k^n * x^n / n!)
:

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