2.3 Answer

2.3

1.(a)

$\displaystyle (\sin{x})'$ $\displaystyle =$ $\displaystyle \cos{x} = \sin{(x + \frac{\pi}{2})}$  
$\displaystyle (\sin{x})^{\prime\prime}$ $\displaystyle =$ $\displaystyle \cos{(x + \frac{\pi}{2})} = \sin{(x + \frac{\pi}{2} + \frac{\pi}{2}})$  

Using the mathematical induction, we prove this.For $n = 1$, this is ture. Now assume that

$\displaystyle (\sin{x})^{(k)} = \sin{(x + \frac{k\pi}{2})} $

Then we show it is true for $n = k+1$
$\displaystyle (\sin{x})^{(k+1)}$ $\displaystyle =$ $\displaystyle ((\sin{x})^{(k)})^{\prime} = (\sin{(x + \frac{k\pi}{2})})^{\prime}$  
  $\displaystyle =$ $\displaystyle \cos{(x + \frac{k\pi}{2})} = \sin{(x + \frac{k\pi}{2} + \frac{\pi}{2})}$  
  $\displaystyle =$ $\displaystyle \sin{(x + \frac{(k+1)\pi}{2}}$  

1.(b)

$\displaystyle (\cos{x})'$ $\displaystyle =$ $\displaystyle -\sin{x} = \cos{(x + \frac{\pi}{2})}$  
$\displaystyle (\sin{x})^{\prime\prime}$ $\displaystyle =$ $\displaystyle -\sin{(x + \frac{\pi}{2})} = \cos{(x + \frac{\pi}{2} + \frac{\pi}{2}})$  

Using the mathematical induction, we prove this.For $n = 1$, this is true. Now assume that

$\displaystyle (\cos{x})^{(k)} = \cos{(x + \frac{k\pi}{2})} $

Then we show it is true for $n = k+1$
$\displaystyle (\cos{x})^{(k+1)}$ $\displaystyle =$ $\displaystyle ((\cos{x})^{k})^{\prime} = (\cos{(x + \frac{k\pi}{2})})^{\prime}$  
  $\displaystyle =$ $\displaystyle -\sin{(x + \frac{k\pi}{2})} = \cos{(x + \frac{k\pi}{2} + \frac{\pi}{2})}$  
  $\displaystyle =$ $\displaystyle \cos{(x + \frac{(k+1)\pi}{2}}$  

1.(c) Using the mathematical induction, we prove this. For $n = 1$

$\displaystyle ((1 + x)^{\alpha})' = \alpha(1 + x)^{\alpha - 1} $

Next we assume that,

$\displaystyle ((1 + x)^{\alpha})^{(k)} = \alpha(\alpha - 1)\cdots(\alpha - k + 1)((1 + x)^{\alpha - k} $

Then we show it is true for $n = k+1$
$\displaystyle ((1+x)^{\alpha})^{(k+1)}$ $\displaystyle =$ $\displaystyle ((1+x)^{\alpha})^{(k))^{\prime}}$  
  $\displaystyle =$ $\displaystyle (\alpha(\alpha - 1)\cdots(\alpha - k + 1)((1 + x)^{\alpha - k})^{\prime}$  
  $\displaystyle =$ $\displaystyle \alpha(\alpha - 1)\cdots(\alpha - k + 1)(\alpha - k)((1 + x)^{\alpha - k -1}$  
  $\displaystyle =$ $\displaystyle \alpha(\alpha - 1)\cdots(\alpha - k + 1)(\alpha - k)((1 + x)^{\alpha - (k +1)}$  

2.

(a)

$\displaystyle{\frac{x^3}{1 - x} = -x^2 - x - 1 + \frac{1}{1-x}}$ また $\displaystyle{(\frac{1}{1-x})^{(n)} = n!(1 - x)^{-(n+1)}}$ より

$\displaystyle (\frac{x^3}{1 - x})^{(n)} = \begin{array}{ll}
-2x - 1 + (1-x)^{-...
...\\
-2 + 2(1-x)^{-3}, & n = 2\\
n!(1 - x)^{-(n+1)}, & n \geq 3
\end{array} $

(b) By Leibniz's theorem, we have

$\displaystyle (x^{2}\sin{x})^{(n)} = \sum_{j=0}^{n} \binom{n}{j}(x^{2})^{j}(\sin{x})^{n-j} $

Note that $(x^{2})^{j} = 0 (n \geq 3)$. Then
$\displaystyle \sum_{j=0}^{n}\binom{n}{j}(x^{2})^{j}(\sin{x})^{n-j}$ $\displaystyle =$ $\displaystyle \underbrace{x^2 \sin{\left(x + \frac{n\pi}{2}\right)}}_{j=0} + \underbrace{2xn\sin{\left(x + \frac{(n-1)\pi}{2}\right)}}_{j=1}$  
  $\displaystyle +$ $\displaystyle \underbrace{n(n-1)\sin{\left(x + \frac{(n-2)\pi}{2}\right)}}_{j=2}$  

(c) By Leibniz's theorem, we have

$\displaystyle (e^{x}\sin{x})^{(n)}$ $\displaystyle =$ $\displaystyle \sum_{j=0}^{n}\binom{n}{j}(e^{x})^{n-j}(\sin{x})^{j}$  
  $\displaystyle =$ $\displaystyle e^{x}\sum_{j=0}^{n}\binom{n}{j}\sin{( x + \frac{j\pi}{2})}$  

Unfortunately it can't be easier.But by letting $y = e^{x}\sin{x}$ , we can write $y' = e^{x}\sin{x} + e^{x}\cos{x} = \sqrt{2}e^{x}\sin{(x + \frac{\pi}{4})}$.So, we suppose that $y^{(k)} = (\sqrt{2})^{k}e^{x}\sin{(x + \frac{k\pi}{4})}$. Then
$\displaystyle y^{(k+1)}$ $\displaystyle =$ $\displaystyle (\sqrt{2})^{k}e^{x}\{\sin{x + \frac{k\pi}{4})} + \cos{x + \frac{k\pi}{4})}\}$  
  $\displaystyle =$ $\displaystyle (\sqrt{2})^{k+1}e^{x}\{\sin{x + \frac{(k+1)\pi}{4})}\}$  

Thus by the mathematical induction, we have

$\displaystyle (\sqrt{2})^{n}e^{x}\sin{(x + (n\pi/4))}$