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

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

Popular >

taylor sqrt(x^2+1)

taylor

x^{2} + 1

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} ( x ^{2} + 1 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{2} + 1 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{2} + 1 ) ( 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

y^{^{'}} = 2 y + 39

\frac{d}{d x} (\frac{x ^{\frac{3}{5}} x ^{3} + 2 x - 6}{( 4 x - 3 ) ^{8}})

f (x) = \frac{4}{x - 2}

x \to 4 - lim (e^{\frac{3}{4 - x}})

\int x^{5} (x - 9) d x

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