Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation. f(x)=cosh(2x-2),a=1

Question
Series
asked 2020-12-29
Taylor series Write out the first three nonzero terms of the Taylor series for the following functions centered at the given point a. Then write the series using summation notation.
\(\displaystyle{f{{\left({x}\right)}}}={\text{cosh}{{\left({2}{x}-{2}\right)}}},{a}={1}\)

Answers (1)

2020-12-30
We have to write first three nonzero terms of the taylor series for the function
\(\displaystyle{f{{\left({x}\right)}}}={\text{cosh}{{\left({2}{x}-{2}\right)}}},{a}={1}\)
\(\displaystyle{f{{\left({x}\right)}}}={\text{cosh}{{\left({2}{x}-{2}\right)}}},{f{{\left({1}\right)}}}={\text{cosh}{{\left({0}\right)}}}={1}\)
\(\displaystyle{f}'{\left({x}\right)}={2}{\text{sinh}{{\left({2}{x}-{2}\right)}}},{f}'{\left({1}\right)}={2}{\text{sinh}{{\left({0}\right)}}}={0}\)
\(\displaystyle{f}{''}{\left({x}\right)}={4}{\text{cosh}{{\left({2}{x}-{2}\right)}}},{f}{''}{\left({1}\right)}={4}{\text{cosh}{{\left({0}\right)}}}={4}\)
\(\displaystyle{f}{'''}{\left({x}\right)}={8}{\text{sinh}{{\left({2}{x}-{2}\right)}}},{f}{'''}{\left({1}\right)}={8}{\text{sinh}{{\left({0}\right)}}}={0}\)
\(\displaystyle{{f}^{{4}}{\left({x}\right)}}={16}{\text{cosh}{{\left({2}{x}-{2}\right)}}},{{f}^{{4}}{\left({1}\right)}}={16}{\text{cosh}{{\left({0}\right)}}}={16}\)
\(\displaystyle{{f}^{{5}}{\left({x}\right)}}={32}{\text{sinh}{{\left({2}{x}-{2}\right)}}},{{f}^{{5}}{\left({1}\right)}}={32}{\text{sinh}{{\left({0}\right)}}}={32}\)
\(\displaystyle{{f}^{{6}}{\left({x}\right)}}={64}{\text{cosh}{{\left({2}{x}-{2}\right)}}},{{f}^{{6}}{\left({1}\right)}}={64}{\text{cosh}{{\left({0}\right)}}}={64}\)
First term \(\displaystyle{f{{\left({1}\right)}}}={1}\)
Second term \(\displaystyle{f}'{\left({1}\right)}{\left({x}-{1}\right)}={0},{\left({x}-{1}\right)}={0}\)
Third term \(\displaystyle{\frac{{{f}{''}{\left({1}\right)}}}{{{2}!}}}{\left({x}-{1}\right)}^{{2}}={\frac{{{4}}}{{{2}!}}}{\left({x}-{1}\right)}^{{2}}={2}{\left({x}-{1}\right)}^{{2}}\)
Third term \(\displaystyle{\frac{{{f}{'''}{\left({1}\right)}}}{{{3}!}}}{\left({x}-{1}\right)}^{{3}}={\frac{{{0}}}{{{3}!}}}{\left({x}-{1}\right)}^{{3}}={0}\)
Fourth term \(\displaystyle{\frac{{{{f}^{{4}}{\left({1}\right)}}}}{{{4}!}}}{\left({x}-{1}\right)}^{{4}}={\frac{{{16}}}{{{4}!}}}{\left({x}-{1}\right)}^{{4}}={\frac{{{2}}}{{{3}}}}{\left({x}-{1}\right)}^{{4}}\)
Fifth term \(\displaystyle{\frac{{{{f}^{{5}}{\left({1}\right)}}}}{{{5}!}}}{\left({x}-{1}\right)}^{{5}}={\frac{{{0}}}{{{5}!}}}{\left({x}-{1}\right)}^{{5}}={0}\)
Sixth term \(\displaystyle{\frac{{{{f}^{{6}}{\left({1}\right)}}}}{{{6}!}}}{\left({x}-{1}\right)}^{{6}}={\frac{{{64}}}{{{6}!}}}{\left({x}-{1}\right)}^{{6}}\)
So, first three non-zero term of taylor series are
\(\displaystyle{1},{2}{\left({x}-{1}\right)}^{{2}},{\frac{{{2}}}{{{3}}}}{\left({x}-{1}\right)}^{{4}}\)
Now, we will write series summation notation
\(\displaystyle{f{{\left({x}\right)}}}={\sum_{{{n}={0}}}^{\infty}}{\frac{{{{f}^{{{\left({n}\right)}}}{\left({1}\right)}}}}{{{n}!}}}{\left({x}-{1}\right)}^{{n}}\)
\(\displaystyle={f{{\left({1}\right)}}}+{\frac{{{{f}^{{{\left({1}\right)}}}{\left({1}\right)}}}}{{{1}!}}}{\left({x}-{1}\right)}+{\frac{{{{f}^{{{\left({2}\right)}}}{\left({1}\right)}}}}{{{2}!}}}{\left({x}-{1}\right)}^{{2}}+{\frac{{{{f}^{{{\left({3}\right)}}}{\left({1}\right)}}}}{{{3}!}}}{\left({x}-{1}\right)}^{{3}}+{\frac{{{{f}^{{{\left({4}\right)}}}{\left({1}\right)}}}}{{{4}!}}}{\left({x}-{1}\right)}^{{4}}+{\frac{{{{f}^{{{\left({5}\right)}}}{\left({1}\right)}}}}{{{5}!}}}{\left({x}-{1}\right)}^{{5}}+{\frac{{{{f}^{{{\left({6}\right)}}}{\left({1}\right)}}}}{{{6}!}}}{\left({x}-{1}\right)}^{{6}}+\ldots\)
\(\displaystyle={1}+{0}+{\frac{{{4}}}{{{2}!}}}{\left({x}-{1}\right)}^{{2}}+{0}+{\frac{{{16}}}{{{4}!}}}{\left({x}-{1}\right)}^{{4}}+{0}+{\frac{{{64}}}{{{6}!}}}{\left({x}-{1}\right)}^{{6}}+\ldots\)
\(\displaystyle={1}+{\frac{{{4}}}{{{2}!}}}{\left({x}-{1}\right)}^{{2}}+{\frac{{{16}}}{{{4}!}}}{\left({x}-{1}\right)}^{{4}}+{\frac{{{64}}}{{{6}!}}}{\left({x}-{1}\right)}^{{6}}+\ldots\)
\(\displaystyle={\sum_{{{n}={0}}}^{\infty}}{\frac{{{2}^{{{2}{n}}}}}{{{\left({2}{n}\right)}!}}}{\left({x}-{1}\right)}^{{{2}{n}}}\)
0

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