Let [y(x) = (1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 + x^16)].

Question

Let $$y(x) = (1 + x)(1 + {x^2})(1 + {x^4})(1 + {x^8})(1 + {x^{16}})$$. Then $y' - y''$ at $x =  - 1$ is equal to

Accepted Answer

$$
\begin{aligned}
& y=\frac{1-x^{32}}{1-x}=1+x+x^2+x^3+\ldots+x^{31} \\
& y^{\prime}=1+2 x+3 x^2+\ldots+31 x^{30} \\
& y^{\prime}(-1)=1-2+3-4+\ldots+31=16 \\
& y^{\prime \prime}(x)=2+6 x+12 x^2+\ldots+31.30 x^{29} \\
& y^{\prime \prime}(-1)=2-6+12 \ldots 31.30=480 \\
& y^{\prime \prime}(-1)-y^{\prime}(-1)=-496
\end{aligned}
$$

Solutions · 1