Answer to Question #218962 in Calculus for nicimo

Question #218962

find the taylor series expansion of

f(x)=(9-e-x)1/2 about x=0 hence determine (9-e-0.03) to 3 decimal places using the first three terms


1
Expert's answer
2021-08-03T13:09:24-0400

Find the Taylor series expansion of f(x)=(9ex)f(x)=(9-e^{-x}) about x=0x=0

A Maclaurin series is given by 

f(x)=k=0f(k)(a)k!xkf(x)=\sum_{k=0}^{\infin } {f^{(k)}(a)\over k!} x^k

In our case, f(x)P(x)=k=0nf(k)(a)k!xk=k=03f(k)(a)k!xkf(x)≈P(x)=\sum_{k=0}^{n} {f^{(k)}(a)\over k!} x^k=\sum_{k=0}^{3 } {f^{(k)}(a)\over k!} x^k

So, what we need to do to get the desired polynomial is to calculate the derivatives, evaluate them at the given point, and plug the results into the given formula.

f(0)(x)=f(x)=9exf^{(0)}(x)=f(x)=\sqrt{9−e^{−x}}

Evaluate the function at the point: f(0)=9e0=22f(0)=\sqrt{9−e^{−0}}=2 \sqrt 2

Find the 1st derivative: 

f(x)=12(9ex)12exf'(x)={1\over 2}(9-e^{-x})^{-{1\over 2}}e^{-x}

When x=0x=0

f(0)=12(9e(0))12e(0)=12(91)12.(1)=12(8)12.(1)=12(8)12=28f'(0)={1\over 2}(9-e^{-(0)})^{-{1\over 2}}e^{-(0)}={1\over 2}(9-1)^{-{1\over 2}}.(1)={1\over 2}(8)^{-{1\over 2}}.(1)={1\over 2}(8)^{-{1\over 2}}={ \sqrt 2 \over 8}

Find the 2nd derivative:

f(x)=(12)(12)(9ex)32.ex.ex12(9ex)12.exf''(x)=(-{1\over 2})({1\over 2})(9-e^{-x})^{-{3\over 2}}.e^{-x}.e^{-x}-{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}

=(14)(9ex)32.e2x12(9ex)12.ex=(-{1\over 4})(9-e^{-x})^{-{3\over 2}}.e^{-2x}-{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}

When x=0x=0

f(0)=(14)(9e(0))32.e2(0)12(9e(0))12.e(0)f''(0)=(-{1\over 4})(9-e^{-(0)})^{-{3\over 2}}.e^{-2(0)}-{1\over 2}(9-e^{-(0)})^{-{1\over 2}}.e^{-(0)}

=(14)(91)32.112(91)12.1=(-{1\over 4})(9-1)^{-{3\over 2}}.1-{1\over 2}(9-1)^{-{1\over 2}}.1

=(14)(8)3212(8)12=172128=(-{1\over 4})(8)^{-{3\over 2}}-{1\over 2}(8)^{-{1\over 2}}=-{17 \sqrt 2 \over 128}

Find the 3rd derivative:

f(x)=(32)(14)(9ex)52.ex.e2x+24(9ex)32.e2x[(12)(12)(9ex)32.ex.ex12(9ex)12.ex]f'''(x)=(-{3\over 2})(-{1\over 4})(9-e^{-x})^{-{5\over 2}}.e^{-x}.{e^{-2x}}+{2\over 4}(9-e^{-x})^{-{3\over 2}}.e^{-2x}-[(-{1\over 2})({1\over 2})(9-e^{-x})^{-{3\over 2}}.e^{-x}.{e^{-x}}-{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}]

=(38)(9ex)52.e3x+24(9ex)32.e2x+(14)(9ex)32.e2x+12(9ex)12.ex=({3\over 8})(9-e^{-x})^{-{5\over 2}}.e^{-3x}+{2\over 4}(9-e^{-x})^{-{3\over 2}}.e^{-2x}+({1\over 4})(9-e^{-x})^{-{3\over 2}}.e^{-2x}+{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}

When x=0x=0

f(0)=(38)(9e(0))52.e3(0)+24(9e(0))32.e2(0)+(14)(9e(0))32.e2(0)+12(9e(0))12.e(0)f'''(0)=({3\over 8})(9-e^{-(0)})^{-{5\over 2}}.e^{-3(0)}+{2\over 4}(9-e^{-(0)})^{-{3\over 2}}.e^{-2(0)}+({1\over 4})(9-e^{-(0)})^{-{3\over 2}}.e^{-2(0)}+{1\over 2}(9-e^{-(0)})^{-{1\over 2}}.e^{-(0)}

=(38)(91)52.1+24(91)32.1+(14)(91)32.1+12(91)12.1=({3\over 8})(9-1)^{-{5\over 2}}.1+{2\over 4}(9-1)^{-{3\over 2}}.1+({1\over 4})(9-1)^{-{3\over 2}}.1+{1\over 2}(9-1)^{-{1\over 2}}.1

=(38)(8)52+24(8)32+(14)(8)32+12(8)12=30722048=({3\over 8})(8)^{-{5\over 2}}+{2\over 4}(8)^{-{3\over 2}}+({1\over 4})(8)^{-{3\over 2}}+{1\over 2}(8)^{-{1\over 2}}={307 \sqrt 2 \over 2048}

From Taylor series,

f(x)=f(0)0!x0+f(0)1!x1+f(0)2!x2+f(0)3!x3+...f(x)={f(0)\over 0!}x^0+{f'(0)\over 1!}x^1+{f''(0)\over 2!}x^2+{f'''(0)\over 3!}x^3+...

Now, use the calculated values to get a polynomial:

    f(x)=220!x0+281!x1+1721282!x2+307220483!x3+...\implies f(x)={2 \sqrt 2\over 0!}x^0+{{ \sqrt 2 \over 8}\over 1!}x^1+{-{17 \sqrt 2 \over 128}\over 2!}x^2+{{307 \sqrt 2 \over 2048}\over 3!}x^3+...

=221.1+281x+1721282x2+307220486x3+...={2 \sqrt 2\over 1}.1+{{ \sqrt 2 \over 8}\over 1}x+{-{17 \sqrt 2 \over 128}\over 2}x^2+{{307 \sqrt 2 \over 2048}\over 6}x^3+...

=22+28x172256x2+307212288x3+...=2 \sqrt 2+{ \sqrt 2 \over 8}x-{17 \sqrt 2 \over 256}x^2+{307 \sqrt 2 \over 12288}x^3+...

Determine (9e0.03)(9-e^{-0.03}) to 3 decimal places using the first three terms

(9e0.03)1222+28(0.03)172256(0.03)2+307212288(0.03)3(9-e^{-0.03})^{1\over 2} \approx 2 \sqrt 2+{ \sqrt 2 \over 8}(-0.03)-{17 \sqrt 2 \over 256}(-0.03)^2+{307 \sqrt 2 \over 12288}(-0.03)^3

2.8284271250.0053033008590.000084521357440.000000953972674\approx 2.828427125-0.005303300859-0.00008452135744-0.000000953972674

2.823038349\approx 2.823038349

    (9e0.03)=((9e0.03)12)2(2.823038349)27.970\implies (9-e^{-0.03})=((9-e^{-0.03})^{1\over 2})^2 \approx(2.823038349)^2 \approx7.970 correct to 3 decimal places



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