What is taylor x^4-3x^2+1 ?

The answer to taylor x^4-3x^2+1 is 1-3x^2+x^4

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taylor x^4-3x^2+1

taylor

x^{4} - 3 x^{2} + 1

1 - 3 x^{2} + x^{4}

Solution steps

Apply the Taylor Formula

= 1 + \frac{\frac{d}{d x} ( x ^{4} - 3 x ^{2} + 1 ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( x ^{4} - 3 x ^{2} + 1 ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( x ^{4} - 3 x ^{2} + 1 ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 1 + \frac{0}{1 !} x + \frac{- 6}{2 !} x^{2} + \frac{0}{3 !} x^{3} + \frac{24}{4 !} x^{4} + \frac{0}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{0}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{0}{9 !} x^{9} + \frac{0}{10 !} x^{10}

Refine

= 1 - 3 x^{2} + x^{4}

n = 0 \sum \infty \frac{1}{n ^{3} + 1}

7 x^{\frac{1}{5}} (x^{3} - 2)

\frac{d}{d x} (y^{2} (8 - x))

x^{2} y^{^{'}} = 4 x^{2} y^{3} - 2 x y

n = 1 \sum \infty \frac{( 3 ^{n} )}{n !}