What is taylor sin(9x) ?

The answer to taylor sin(9x) is 9x-243/2 x^3+19683/40 x^5-531441/560 x^7+4782969/4480 x^9+\ldots

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taylor sin(9x)

taylor

sin (9 x)

9 x - \frac{243}{2} x^{3} + \frac{19683}{40} x^{5} - \frac{531441}{560} x^{7} + \frac{4782969}{4480} x^{9} + \dots

Solution steps

Apply the Taylor Formula

= 0 + \frac{\frac{d}{d x} ( sin ( 9 x ) ) ( 0 )}{1 !} x + \frac{\frac{d ^{2}}{d x ^{2}} ( sin ( 9 x ) ) ( 0 )}{2 !} x^{2} + \frac{\frac{d ^{3}}{d x ^{3}} ( sin ( 9 x ) ) ( 0 )}{3 !} x^{3} + \dots

Evaluate Derivatives

= 0 + \frac{9}{1 !} x + \frac{0}{2 !} x^{2} + \frac{- 729}{3 !} x^{3} + \frac{0}{4 !} x^{4} + \frac{59049}{5 !} x^{5} + \frac{0}{6 !} x^{6} + \frac{- 4782969}{7 !} x^{7} + \frac{0}{8 !} x^{8} + \frac{387420489}{9 !} x^{9} + \dots

Refine

= 9 x - \frac{243}{2} x^{3} + \frac{19683}{40} x^{5} - \frac{531441}{560} x^{7} + \frac{4782969}{4480} x^{9} + \dots

\frac{d}{d x} ((a + \frac{b}{x ^{2}})^{3})

sin (x) ln (x)

f (x) = 2 sin^{5} (x)

x sin (x^{2})

y^{^{''}} - 4 y^{^{'}} + 3 y = - e^{- s}, y (0) = 0, y^{^{'}} (0) = 0

What is taylor sin(9x) ?
The answer to taylor sin(9x) is 9x-243/2 x^3+19683/40 x^5-531441/560 x^7+4782969/4480 x^9+\ldots