Cubic function and Alternating Series 2

Let $(0 \leq x <1)$ .

$\sum_{n = 1}^{\infty} n x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} x^n = \dfrac{1}{1 - x}\sum_{j = 1}^{\infty} x^{j} = \dfrac{1}{1 - x}(x)\dfrac{1}{1 - x} = \dfrac{x}{(1 - x)^2}$

$(\dfrac{x}{x - 1})(\dfrac{1}{x})(\dfrac{x}{x - 1}) = \dfrac{x}{(x - 1)^2}.$

$\sum_{n = 1}^{\infty} n^2 x^n = \sum_{j = 1}^{\infty} (\sum_{n = j}^{\infty} n x^n) =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} (j x^j) + \dfrac{x}{(1 - x)^2}\sum_{j = 1}^{\infty} x^j =$

$(\dfrac{1}{1 - x})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x}{(1 - x)^2})(\dfrac{x}{1 - x}) =$ $\dfrac{x^2 + x}{(1 - x)^3}$ .

$\sum_{n = 1}^{\infty} n^3 x^n = \sum_{j = 1}^{\infty} \sum_{n = j}^{\infty} n^2 x^{n} =$ $\dfrac{1}{1 - x}\sum_{j = 1}^{\infty} j^2 x^j + \dfrac{2x}{(1 - x)^2}\sum_{j = 1}^{\infty} j x^j + \dfrac{x^2 + x}{(1 - x)^3}\sum_{j = 1}^{\infty} x^j = (\dfrac{1}{1 - x})(\dfrac{x^2 + x}{(1 - x)^3}) +$ $(\dfrac{2x}{(1 - x)^2})(\dfrac{x}{(1 - x)^2}) + (\dfrac{x^2 + x}{(1 - x)^3})(\dfrac{x}{1 - x}) =$ $\dfrac{x^3 + 4x^2 + x}{(1 - x)^4}$

$\implies g(x) = \sum_{n = 1}^{\infty} (-1)^{n + 1} n^3 x^n = -\sum_{n = 1}^{\infty} n^3 (-x)^n = -(\dfrac{-x^3 + 4x^2 - x}{(1 + x)^4}) = \dfrac{x^3 - 4x^2 + x}{(1 + x)^4}$

Let $f(x) = x^3 + ax^2 + bx + c$ .

$f(x) = g(x) = 0 \implies c = 0$ and $f(\dfrac{1}{2}) = g(\dfrac{1}{2}) = \dfrac{-2}{27} \implies$ $\boxed{54a + 108b = -43}$ .

Let $u = 1+ x \implies du = dx \implies \int_{0}^{\frac{1}{2}} g(x) dx = \int_{1}^{\frac{3}{2}} \dfrac{u^3 - 7u^2 + 12u - 6}{u^4} du = \int_{1}^{\frac{3}{2}} \dfrac{1}{u} - 7u^{-2} + 12u^{-3} - 6u^{-4} du =$

$(\ln(u) + \dfrac{7}{u} - \dfrac{6}{u^2} + \dfrac{2}{u^3})|_{1}^{\frac{3}{2}} =$ $\ln(\dfrac{3}{2}) + \dfrac{14}{3} - \dfrac{24}{9} + \dfrac{16}{27} - 3 = \ln(\dfrac{3}{2}) - \dfrac{11}{27}$

$\implies \ln(\dfrac{3}{2}) - \dfrac{11}{27} = \int_{0}^{\frac{1}{2}} f(x) dx = (\dfrac{x^4}{4} + \dfrac{ax^3}{3} + \dfrac{bx^2}{2})|_{0}^{\frac{1}{2}} = \dfrac{24 + 64a + 192b}{1536} \implies$

$\boxed{1728a + 5184b = 41472\ln(\dfrac{3}{2}) - 17546}$

$\boxed{54a + 108b = -43}$

$\implies a =\dfrac{15482 - 41472\ln(\dfrac{3}{2})}{864}$ and $b = \dfrac{20736\ln(\dfrac{3}{2}) - 8085}{864} \implies f(x) = x^3 + \dfrac{15482 - 41472\ln(\dfrac{3}{2})}{864}x^2 + \dfrac{20736\ln(\dfrac{3}{2}) - 8085}{864}x$

Letting $u = 1+ x \implies du = dx \implies$

$\int_{0}^{\frac{1}{10}} g(x) dx = \int_{1}^{\frac{11}{10}} \dfrac{u^3 - 7u^2 + 12u - 6}{u^4} du = \int_{1}^{\frac{11}{10}} \dfrac{1}{u} - 7u^{-2} + 12u^{-3} - 6u^{-4} du =$ $(\ln(u) + \dfrac{7}{u} - \dfrac{6}{u^2} + \dfrac{2}{u^3})|_{1}^{\frac{11}{10}} =$

$\ln(\dfrac{11}{10}) + \dfrac{70}{11} - \dfrac{600}{11^2} + \dfrac{2000}{11^3} - 3 = \ln(\dfrac{11}{10}) - \dfrac{123}{1331} \approx \boxed{0.00289846}$

and,

$\int_{0}^{\frac{1}{10}} f(x) dx = \dfrac{1}{4 * 10^4} + \dfrac{15482 - 41472\ln(\dfrac{3}{2})}{864}(\dfrac{1}{3 * 10^3}) + \dfrac{20736\ln(\dfrac{3}{2}) - 8085}{864}(\dfrac{1}{2 * 10^2}) =$

$\dfrac{2592 + 40 * (15482 - 41472\ln(\dfrac{3}{2})) + 600 * (20736\ln(\dfrac{3}{2}) - 8085)}{103680000} =$ $\dfrac{10782720\ln(\dfrac{3}{2}) - 4229128}{103680000} \approx \boxed{0.00137817}$

$\implies \int_{0}^{\frac{1}{10}} g(x) - f(x) dx \approx \boxed{0.00152029}$