What is taylor f(x)=sqrt(1+x^2) ?

The answer to taylor f(x)=sqrt(1+x^2) is 1+1/2 x^2-1/8 x^4+1/16 x^6-5/128 x^8+\ldots

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taylor f(x)=sqrt(1+x^2)

taylor

f (x) = 1 + x^{2}

1 + \frac{1}{2} x^{2} - \frac{1}{8} x^{4} + \frac{1}{16} x^{6} - \frac{5}{128} x^{8} + \dots

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( 1 + x ^{2} ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( 1 + x ^{2} ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( 1 + 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{- 3}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{45}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{- 1575}{8 !} x^{8} + \dots

Refine

= 1 + \frac{1}{2} x^{2} - \frac{1}{8} x^{4} + \frac{1}{16} x^{6} - \frac{5}{128} x^{8} + \dots

\frac{\partial}{\partial x} (x y (\frac{8 - x y}{x + y}))

\int_{0}^{\frac{π}{7}} cos (4 x) d x

\int x^{2} + 1 d x

\frac{d}{d x} (e^{2 x - x^{2}})

n = 0 \sum \infty \frac{6 ^{n}}{n + 1}

What is taylor f(x)=sqrt(1+x^2) ?
The answer to taylor f(x)=sqrt(1+x^2) is 1+1/2 x^2-1/8 x^4+1/16 x^6-5/128 x^8+\ldots