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): 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.
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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!) =
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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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