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

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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) About Me  (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:
f%28x%29=%28e%5E%28kx%29%29′=ke%5E%28kx%29, f′′%28x%29=%28ke%5E%28kx%29%29′=k%5E2%2Ae%5E%28kx%29,…....,f%5E%28n%29%28x%29=k%5En%2Ae%5E%28kx%29

Then, at x=0 we have
f%280%29=e%5E0=1, f%280%29=ke%5E0=k, f′′%280%29=k%5E2%2Ae%5E0=k%5E2,…f%5E%28n%29%280%29=k%5En%2Ae%5E0=k%5En

Hence, the Maclaurin expansion for the given function is
e%5E%28kx%29=sum%28f%5E%28n%29%280%29%28x%5En%2Fn%21%29%2Cn=0%2Cinfinity%29=1%2Bkx%2B%28k%5E2x%5E2%29%2F2%21%2B%28k%5E3x%5E3%29%2F3%21+…=sum%28%28k%5En%2Ax%5En%29%2Fn%21%2Cn=0%2Cinfinity%29

Answer by rothauserc(4718) About Me  (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.
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For this problem, f(x) = e^(kx), where k is a real number
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1) Calculate the derivatives for f(x)
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f'(x) = ke^(kx), f''(x) = k^2e^(kx), ,,,,, f^n(x) = k^n * e^(kx)
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2) At x = 0,
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f(0) = 1, f'(0) = ke^0 = k, f''(0) = k^2, .... f^n(0) = k^n
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Using the definition of the Maclaurin for e^(kx) is
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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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