The Maclaurin’s series expansion of esin x is

Maclaurin’s Series Expansion of e^{sin x}

We know that,

• f(x) = e^{sin x} ⇒ f(0) = e = 1
• f'(x) = e^{sin x} (cos x) ⇒ f'(0) = 1
• f”(x) = e^{sin x} cos2 x + esin x (-sin x)
• f”(0) = 1 = e^{sin x} [cos2 x – sin x]
• f”'(x) = e^{sin x} [-sin 2x – cos x] + e^{sin x} [cos3 x – sin x cos x]
• f”'(x) = -1 + 1 = 0
• f (x) = e^{sin x} [-2 cos 2x + sin x] + e^{sin x} cos x [-sin 2x – cos x] + e^{sin x} [3 cos2 x(−sin x) − cos 2x/2 × 2] + e^{sin x} [cos4 x – sin x cos2 x]
• f (0) = -2 -1 -1 + 1 = -3

Substitue in Maclaurin Series

e^{sin x} = 1 + x + x2/2 + x3/3! (0) + x4/4! X (-3) + ……

= 1 + x + x2/2 – x4/8 + …..

Therefore, the Maclaurin’s series expansion of e^{sin x} is 1 + x + x2/2 – x4/8 + …..

Summary:

The Maclaurin’s series expansion of e^{sin x} is? (a) 1 + x – x2/2 + x4/12 – ….. (b) 1 – x +x2/2 – x4/8 + ….. (c) 1 + x + x2/2 – x4/8 + ….. (d) 1 + x + x2/2 – x4/12 + …..

1 + x + x2/2 – x4/8 + ….. is the Maclaurin’s series expansion of e^sinx. A Taylor series expansion of a function about 0 is called Maclaurin’s series. It is renamed after the Scottish mathematician Colin Maclaurin.