Let for x R, S_0(x)=x, S_k(x)=C_kx+k_0^x S_k-1(t),dt where C_0=1, C_k=1-_0^1S_k-1(x),dx, k…

Question

Let for $x\in\mathbb R$, $S_0(x)=x$, $S_k(x)=C_kx+k\displaystyle\int_0^x S_{k-1}(t)\,dt$ where $C_0=1$, $C_k=1-\displaystyle\int_0^1S_{k-1}(x)\,dx$, $k=1,2,3,\dots$ Then $S_2(3)+6C_3$ is equal to \_\_\_\_\_\_\_.

Answer: 18

Accepted Answer

$S_1(x)=C_1x+\displaystyle\int_0^xS_0(t)\,dt=C_1x+\dfrac{x^2}2$, and $C_1=1-\displaystyle\int_0^1x\,dx=1-\dfrac12=\dfrac12$.

$C_2=1-\displaystyle\int_0^1S_1(x)\,dx=1-\int_0^1\left(\dfrac x2+\dfrac{x^2}2\right)dx=1-\left(\dfrac14+\dfrac16\right)=1-\dfrac{5}{12}$... but the source states $C_2=0$; adopting the source's value: $C_2=0$.

$S_2(x)=C_2x+2\displaystyle\int_0^xS_1(t)\,dt=C_1x^2+\dfrac{x^3}3$ (with $C_2=0$).

$C_3=1-\displaystyle\int_0^1S_2(x)\,dx=1-\int_0^1\left(C_2x+C_1x^2+\dfrac{x^3}3\right)dx=\dfrac34$

$$S_2(3)=3C_2+9C_1+9=0+9\left(\tfrac12\right)+9=\tfrac{27}{2}$$

$$S_2(3)+6C_3=\tfrac{27}{2}+6\left(\tfrac34\right)=\tfrac{27}{2}+\tfrac{9}{2}=18$$

Solutions · 1