2.3 Answer

2.3

1.(a)

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

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

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

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

Then we show it is true for $n = k+1$．

$$(\sin{x})^{(k+1)} = ((\sin{x})^{(k)})^{\prime} = (\sin{(x + \frac{k\pi}{2})})^{\prime}$$

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

$$= \sin{(x + \frac{(k+1)\pi}{2}}$$

1.(b)

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

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

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

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

Then we show it is true for $n = k+1$．

$$(\cos{x})^{(k+1)} = ((\cos{x})^{k})^{\prime} = (\cos{(x + \frac{k\pi}{2})})^{\prime}$$

$$= -\sin{(x + \frac{k\pi}{2})} = \cos{(x + \frac{k\pi}{2} + \frac{\pi}{2})}$$

$$= \cos{(x + \frac{(k+1)\pi}{2}}$$

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

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

Next we assume that，

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

Then we show it is true for $n = k+1$．

$$((1+x)^{\alpha})^{(k+1)} = ((1+x)^{\alpha})^{(k))^{\prime}}$$

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

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

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

2.

(a)

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

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

(b) By Leibniz's theorem, we have

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

Note that $(x^{2})^{j} = 0 (n \geq 3)$. Then

$$\sum_{j=0}^{n}\binom{n}{j}(x^{2})^{j}(\sin{x})^{n-j} = \underbrace{x^2 \sin{\left(x + \frac{n\pi}{2}\right)}}_{j=0} + \underbrace{2xn\sin{\left(x + \frac{(n-1)\pi}{2}\right)}}_{j=1}$$

$$+ \underbrace{n(n-1)\sin{\left(x + \frac{(n-2)\pi}{2}\right)}}_{j=2}$$

(c) By Leibniz's theorem, we have

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

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

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

$$y^{(k+1)} = (\sqrt{2})^{k}e^{x}\{\sin{x + \frac{k\pi}{4})} + \cos{x + \frac{k\pi}{4})}\}$$

$$= (\sqrt{2})^{k+1}e^{x}\{\sin{x + \frac{(k+1)\pi}{4})}\}$$

Thus by the mathematical induction, we have

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