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प्रश्न
Write the Maclaurin series expansion of the following functions:
log(1 – x); – 1 ≤ x ≤ 1
उत्तर
f(x) = log(1 – x), f(0) = 0
f'(x) = `- 1/(1 - x)`, f'(0) = – 1
f''(x) = `- 1/(1 - x)^2`, f''(0) = – 1
f'''(x) = `- 2/(1 - x)^3`, f'''(0) = – 2
fIV(x) = `- 6/(1 - x)^4`, fIV(0) = – 6
Maclaurin ‘s expansion is
f(x) = `sum_("n" = 0)^oo ("f"^(("n"))(0) x^"n")/("n"!)`
= `"f"(0) + ("f'"(0))/(1!) x + ("f''"(0)x^2)/(2!) + ... + ("f"^(("n"))(0)x^"n")/("n"!) + ...`
`log(1 - x) = 0 - 1/(1!) x - 1/(2!) x^2 - 2/(3!) x^3 + ... + 6/(4!) x^4 + ...`
= `- [x + x^2/2 + x^3/3 + x^4/4 + ...]`
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