Maclaurinandlsquo;s series expansion off(x) = ex2+sinxhas the form f(x) = A0+ A1x + A2x2+ A3x3+ andhellip;.The coefficient A4(correct to two decimal place) is equal to _____.Correct answer is '0.875'. Can you explain this answer?

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Maclaurin‘s series expansion of f(x) = e^x2+sinx has the form f(x) = A₀ + A₁x + A₂x² + A₃x³ + ….

The coefficient A₄ (correct to two decimal place) is equal to _____.

Correct answer is '0.875'. Can you explain this answer?

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Maclaurin‘s series expansion off(x) = ex2+sinxhas the form f(x) ...

So, Maclaurin’s series expansion of

Also, Maclaurin’s series expansion of e^sinx

g(x) = e^sinx (say)

g′(x) = e^sinx.cosx = f(x).cosx

g′′(x) = g′(x)cosx − g(x)sinx

g′′′(x) = g′′(x)cosx − g′sinx − g′sinx − g(x)cosx

= g′′(x)cosx − 2g′sinx − g(x)cosx

g′′′′(x) = g′′′(x)cosx − 3g′′(x)sinx − 3g′cosx + g(x)sinx

g(x = 0) = e^{sin 0} = e⁰ = 1

g'(x = 0) = g(x = 0) × cos (0) = 1

g''(x = 0) = g'(x = 0) cos (0) – g(0) sin (0)

= 1 × 1 – 1 × 0

= 1

g'''(x = 0) = g''(x = 0) cos (0) – 2g'(x = 0) sin (0) – g(x = 0) × cos (0)

= 1 × 1 – 2 × 1 × 0 – 1 × 1

= 0

g′′′′(x = 0) = g′′′(x = 0) cos0 − 3g′′(x = 0) sin0−3g′(x = 0) cos0 + g(x = 0) sin0

g''''(x = 0) = -3

Hence,

e^sinx = 1 + (1 × x) +

Coefficient of

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Maclaurin‘s series expansion off(x) = ex2+sinxhas the form f(x) = A0+ A1x + A2x2+ A3x3+ ….The coefficient A4(correct to two decimal place) is equal to _____.Correct answer is '0.875'. Can you explain this answer?

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