Lf fx-x-x2-x3-x4- for -x-1 then f-0-

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Higher Order Derivatives

lf fx=x-x2+...

Question

lf $f (x) = x - x^{2} + x^{3} - x^{4} + \dots .$ . for $| x | < 1$ then $f^{^{''}} (0) =$

\frac{- 2}{(1 - x)^{2}}

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Similar questions

f (x) = \frac{x^{2}}{2}

0 \leq x \leq 1

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

1 \leq x \leq 2

f^{''} (x)

f (x) = \frac{x^{2}}{2}

0 \leq x \leq 1

f (x) = 2 x^{2} - 3 x + \frac{3}{2}

1 \leq x \leq 2

f^{''}

f (x) = x - x^{2} + x^{3} - x^{4} + \dots . . \infty

| x | < 1

f (1 / 2)

f^{- 1} (1 / 2)

f (- 1 / 2)

f^{- 1} (- 1 / 2)

x \neq 0

x + \frac{1}{x} = 2

x^{2} + \frac{1}{x^{2}} = x^{3} + \frac{1}{x^{3}} = x^{4} + \frac{1}{x^{4}}

f (x) = \frac{x^{3}}{x^{2} - 1}

f^{^{'''}} (0) =

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