What is taylor sqrt(1-3x) ?

The answer to taylor sqrt(1-3x) is 1-3/2 x-9/8 x^2-27/16 x^3-405/128 x^4+\ldots

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taylor sqrt(1-3x)

taylor

1 - 3 x

1 - \frac{3}{2} x - \frac{9}{8} x^{2} - \frac{27}{16} x^{3} - \frac{405}{128} x^{4} + \dots

Solution steps

Apply the Taylor Formula

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

Evaluate Derivatives

= 1 + \frac{- \frac{3}{2}}{1 !} x + \frac{- \frac{9}{4}}{2 !} x^{2} + \frac{- \frac{81}{8}}{3 !} x^{3} + \frac{- \frac{1215}{16}}{4 !} x^{4} + \dots

Refine

= 1 - \frac{3}{2} x - \frac{9}{8} x^{2} - \frac{27}{16} x^{3} - \frac{405}{128} x^{4} + \dots

\int sin^{2} (r) d r

\int (6 csc (x) cot (x)) d x

x^{2} \cdot sin (x)

f (x) = 3 x^{2} - 2 x

- 9 x^{4}

What is taylor sqrt(1-3x) ?
The answer to taylor sqrt(1-3x) is 1-3/2 x-9/8 x^2-27/16 x^3-405/128 x^4+\ldots