# Taylor and MacLaurin Series: Consider the approximation of the exponential by its third degree Taylor Polynomial:?

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• 2 months ago

I don't see where's the problem. Just plug in the numbers and calculate.

e^0 - (1+0+0²/2+0³/6) = 0

e^0.1 - (1+0.1+0.1²/2+0.1³/6) = 4.25 × 10^-6

e^0.5 - (1+0.5+0.5²/2+0.5³/6) = √e - 79/48 ≈ 2.89 × 10^-3

e^1 - (1+1+1²/2+1³/6) = e - 8/3 ≈ 0.0516

e^2 - (1+2+2²/2+2³/6) = e² - 19/3 ≈ 1.056

e^-1 - (1-1+1²/2-1³/6) = 1/e - 1/3 ≈ 0.0345