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Aer wi rié error and percent relative error. Add terms until the absolute value of the error estimate falls below an error cr
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Answer #1

3) a) f(1) =e, a = 0

Then f(1) = f(x) = ... f(n) (2) =

Which means f(a) = f(a) = ... f(n)(a) = e = 1

Thus the Taylor series about a=0 is

(ir - a) f(n)(a) n! + +

That is, (ix - a) f(n)(a) n! + T + NEO

b) The Taylor series is

for (x – a)f(n)(a) = -0.2 - 2 -0.2 (2 – 0.2) + (2 n! e-02--0.26.-0.2) - (x -0.2)e-0.2 - + ..

0th order approximate is e^{-0.2}

First order approximate is e^{-0.2}-e^{-0.2}(1-0.2)\approx 0.163

Second order approximate is e-0.2(1 -0.2) e-0.2 - e-0.2 (1 -0.2) + 0.426

And third order approximate is e^{-0.2}-e^{-0.2}(1-0.2)+\frac{e^{-0.2}(1-0.2)^2}{2!}-\frac{e^{-0.2}(1-0.2)^3}{3!}\approx 0.355

The actual value is e- 10.368

\blacksquare

Hope this was helpful. Please do leave a positive rating if you liked this answer. Thanks and have a good day!

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