What is maclaurin (e^x+e^{-x})/2 ?

The answer to maclaurin (e^x+e^{-x})/2 is 1+1/2 x^2+1/24 x^4+1/720 x^6+1/40320 x^8+\ldots

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maclaurin (e^x+e^{-x})/2

maclaurin

\frac{e ^{x} + e ^{- x}}{2}

1 + \frac{1}{2} x^{2} + \frac{1}{24} x^{4} + \frac{1}{720} x^{6} + \frac{1}{40320} x^{8} + \dots

Solution steps

Apply the Maclaurin Formula

= 1 + \frac{\frac{d}{d x} ( \frac{e ^{x} + e ^{- x}}{2} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( \frac{e ^{x} + e ^{- x}}{2} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( \frac{e ^{x} + e ^{- x}}{2} ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{1}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{1}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{1}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{1}{8 !} x^{8} + \dots

Refine

= 1 + \frac{1}{2} x^{2} + \frac{1}{24} x^{4} + \frac{1}{720} x^{6} + \frac{1}{40320} x^{8} + \dots

\int 2 x (x^{2} + 7)^{9} d x

n = 1 \sum \infty \frac{1}{5 n ^{2} - n}

\int (x + 4) d x

\int \frac{9}{x ( x ^{4} + 5 )} d x

f (t) = 2 cos (t) + sin (2 t)

What is maclaurin (e^x+e^{-x})/2 ?
The answer to maclaurin (e^x+e^{-x})/2 is 1+1/2 x^2+1/24 x^4+1/720 x^6+1/40320 x^8+\ldots