Question

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

Answer 1

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

A Maclaurin series is given by 

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

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

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.

$f^{(0)}(x)=f(x)=\sqrt{9−e^{−x}}$

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

Find the 1st derivative: 

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

When $x=0$

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

Find the 2nd derivative:

$f''(x)=(-{1\over 2})({1\over 2})(9-e^{-x})^{-{3\over 2}}.e^{-x}.e^{-x}-{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}$

$=(-{1\over 4})(9-e^{-x})^{-{3\over 2}}.e^{-2x}-{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}$

When $x=0$

$f''(0)=(-{1\over 4})(9-e^{-(0)})^{-{3\over 2}}.e^{-2(0)}-{1\over 2}(9-e^{-(0)})^{-{1\over 2}}.e^{-(0)}$

$=(-{1\over 4})(9-1)^{-{3\over 2}}.1-{1\over 2}(9-1)^{-{1\over 2}}.1$

$=(-{1\over 4})(8)^{-{3\over 2}}-{1\over 2}(8)^{-{1\over 2}}=-{17 \sqrt 2 \over 128}$

Find the 3rd derivative:

$f'''(x)=(-{3\over 2})(-{1\over 4})(9-e^{-x})^{-{5\over 2}}.e^{-x}.{e^{-2x}}+{2\over 4}(9-e^{-x})^{-{3\over 2}}.e^{-2x}-[(-{1\over 2})({1\over 2})(9-e^{-x})^{-{3\over 2}}.e^{-x}.{e^{-x}}-{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}]$

$=({3\over 8})(9-e^{-x})^{-{5\over 2}}.e^{-3x}+{2\over 4}(9-e^{-x})^{-{3\over 2}}.e^{-2x}+({1\over 4})(9-e^{-x})^{-{3\over 2}}.e^{-2x}+{1\over 2}(9-e^{-x})^{-{1\over 2}}.e^{-x}$

When $x=0$

$f'''(0)=({3\over 8})(9-e^{-(0)})^{-{5\over 2}}.e^{-3(0)}+{2\over 4}(9-e^{-(0)})^{-{3\over 2}}.e^{-2(0)}+({1\over 4})(9-e^{-(0)})^{-{3\over 2}}.e^{-2(0)}+{1\over 2}(9-e^{-(0)})^{-{1\over 2}}.e^{-(0)}$

$=({3\over 8})(9-1)^{-{5\over 2}}.1+{2\over 4}(9-1)^{-{3\over 2}}.1+({1\over 4})(9-1)^{-{3\over 2}}.1+{1\over 2}(9-1)^{-{1\over 2}}.1$

$=({3\over 8})(8)^{-{5\over 2}}+{2\over 4}(8)^{-{3\over 2}}+({1\over 4})(8)^{-{3\over 2}}+{1\over 2}(8)^{-{1\over 2}}={307 \sqrt 2 \over 2048}$

From Taylor series,

$f(x)={f(0)\over 0!}x^0+{f'(0)\over 1!}x^1+{f''(0)\over 2!}x^2+{f'''(0)\over 3!}x^3+...$

Now, use the calculated values to get a polynomial:

$\implies f(x)={2 \sqrt 2\over 0!}x^0+{{ \sqrt 2 \over 8}\over 1!}x^1+{-{17 \sqrt 2 \over 128}\over 2!}x^2+{{307 \sqrt 2 \over 2048}\over 3!}x^3+...$

$={2 \sqrt 2\over 1}.1+{{ \sqrt 2 \over 8}\over 1}x+{-{17 \sqrt 2 \over 128}\over 2}x^2+{{307 \sqrt 2 \over 2048}\over 6}x^3+...$

$=2 \sqrt 2+{ \sqrt 2 \over 8}x-{17 \sqrt 2 \over 256}x^2+{307 \sqrt 2 \over 12288}x^3+...$

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

$(9-e^{-0.03})^{1\over 2} \approx 2 \sqrt 2+{ \sqrt 2 \over 8}(-0.03)-{17 \sqrt 2 \over 256}(-0.03)^2+{307 \sqrt 2 \over 12288}(-0.03)^3$

$\approx 2.828427125-0.005303300859-0.00008452135744-0.000000953972674$

$\approx 2.823038349$

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