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早稲田大学 創造理工学研究科 経営システム工学専攻 2021年7月実施 数理基礎 問題A

标签：

Author​

思齐塾, 祭音Myyura

Description​

日本語​

小問A2​

マクローリン展開 (Maclaurin series)

f(x)=f(0)+f′(0)1!x+f′′(0)2!x2+f′′′(0)3!x3+…f(x) = f(0) + \frac{f'(0)}{1!}x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots

を用いて、次の極限 (limit) を求めよ。

lim⁡x→01x3{log⁡e(1+sin⁡x)−x+x22}\lim_{x \to 0} \frac{1}{x^3} \left\{\log_e(1 + \sin x) - x + \frac{x^2}{2}\right\}

小問A3​

z=f(r),r=x2+y2z = f(r), r = \sqrt{x^2 + y^2} とするとき,

∂2z∂x2+∂2z∂y2=0\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0

を満たすような関数 (function) f(r)f(r) を求めよ。

题目描述​

小问A2​

使用麦克劳林展开

f(x)=f(0)+f′(0)1!x+f′′(0)2!x2+f′′′(0)3!x3+⋯f(x)=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2 +\frac{f'''(0)}{3!}x^3+\cdots

求极限

lim⁡x→01x3{log⁡e(1+sin⁡x)−x+x22}.\lim_{x\to0}\frac1{x^3} \left\{\log_e(1+\sin x)-x+\frac{x^2}{2}\right\}.

小问A3​

设

z=f(r),r=x2+y2.z=f(r),\qquad r=\sqrt{x^2+y^2}.

求使

∂2z∂x2+∂2z∂y2=0\frac{\partial^2z}{\partial x^2} +\frac{\partial^2z}{\partial y^2}=0

成立的函数 f(r)f(r)。原题未说明区域是否包含 r=0r=0，也未给出原点处的光滑性要求；因此应先求 r>0r>0 上的径向通解，并另行说明若要求 zz 在原点也为 C2C^2 调和函数时所受的限制。

Kai​

小問A2​

Let f(x)=log⁡e(1+sin⁡x)f(x) = \log_e(1 + \sin x) . Using Maclaurin series expansion:

sin⁡x=x−x33!+x55!−…\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots
log⁡e(1+x)=x−x22+x33−x44+…\log_e(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots

Therefore,

log⁡e(1+sin⁡x)=sin⁡x−(sin⁡x)22+(sin⁡x)33−…\log_e(1+\sin x) = \sin x - \frac{(\sin x)^2}{2} + \frac{(\sin x)^3}{3} - \dots
log⁡e(1+sin⁡x)=(x−x36+… )−12(x−x36+… )2+13(x−x36+… )3−…\log_e(1+\sin x) = (x - \frac{x^3}{6} + \dots) - \frac{1}{2}(x - \frac{x^3}{6} + \dots)^2 + \frac{1}{3}(x - \frac{x^3}{6} + \dots)^3 - \dots
log⁡e(1+sin⁡x)=x−x36−12(x2−x43+… )+13(x3−… )−…\log_e(1+\sin x) = x - \frac{x^3}{6} - \frac{1}{2}(x^2 - \frac{x^4}{3} + \dots) + \frac{1}{3}(x^3 - \dots) - \dots
log⁡e(1+sin⁡x)=x−x22−x36+x33+O(x4)=x−x22+x36+O(x4)\log_e(1+\sin x) = x - \frac{x^2}{2} - \frac{x^3}{6} + \frac{x^3}{3} + O(x^4) = x - \frac{x^2}{2} + \frac{x^3}{6} + O(x^4)
log⁡e(1+sin⁡x)−x+x22=x−x22+x36−x+x22+O(x4)=x36+O(x4)\log_e(1 + \sin x) - x + \frac{x^2}{2} = x - \frac{x^2}{2} + \frac{x^3}{6} - x + \frac{x^2}{2} + O(x^4) = \frac{x^3}{6} + O(x^4)
lim⁡x→01x3{log⁡e(1+sin⁡x)−x+x22}=lim⁡x→01x3(x36+O(x4))=16\lim_{x \to 0} \frac{1}{x^3} \left\{\log_e(1 + \sin x) - x + \frac{x^2}{2}\right\} = \lim_{x \to 0} \frac{1}{x^3} \left(\frac{x^3}{6} + O(x^4)\right) = \frac{1}{6}

小問A3​

We are given that z=f(r)z = f(r) and r=x2+y2r = \sqrt{x^2 + y^2} . We want to find f(r)f(r) such that ∂2z∂x2+∂2z∂y2=0\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = 0 .

First, we find the first partial derivatives:

∂z∂x=dfdr∂r∂x=f′(r)xx2+y2=f′(r)xr\frac{\partial z}{\partial x} = \frac{df}{dr} \frac{\partial r}{\partial x} = f'(r) \frac{x}{\sqrt{x^2+y^2}} = f'(r) \frac{x}{r} ∂z∂y=dfdr∂r∂y=f′(r)yx2+y2=f′(r)yr\frac{\partial z}{\partial y} = \frac{df}{dr} \frac{\partial r}{\partial y} = f'(r) \frac{y}{\sqrt{x^2+y^2}} = f'(r) \frac{y}{r}

Next, we find the second partial derivatives:

∂2z∂x2=∂∂x(f′(r)xr)=f′′(r)xrxr+f′(r)r−x(xr)r2=f′′(r)x2r2+f′(r)r2−x2r3=f′′(r)x2r2+f′(r)y2r3\frac{\partial^2 z}{\partial x^2} = \frac{\partial}{\partial x} (f'(r) \frac{x}{r}) = f''(r) \frac{x}{r} \frac{x}{r} + f'(r) \frac{r - x(\frac{x}{r})}{r^2} = f''(r) \frac{x^2}{r^2} + f'(r) \frac{r^2 - x^2}{r^3} = f''(r) \frac{x^2}{r^2} + f'(r) \frac{y^2}{r^3} ∂2z∂y2=∂∂y(f′(r)yr)=f′′(r)yryr+f′(r)r−y(yr)r2=f′′(r)y2r2+f′(r)r2−y2r3=f′′(r)y2r2+f′(r)x2r3\frac{\partial^2 z}{\partial y^2} = \frac{\partial}{\partial y} (f'(r) \frac{y}{r}) = f''(r) \frac{y}{r} \frac{y}{r} + f'(r) \frac{r - y(\frac{y}{r})}{r^2} = f''(r) \frac{y^2}{r^2} + f'(r) \frac{r^2 - y^2}{r^3} = f''(r) \frac{y^2}{r^2} + f'(r) \frac{x^2}{r^3}

Now, we add the second partial derivatives:

∂2z∂x2+∂2z∂y2=f′′(r)x2r2+f′(r)y2r3+f′′(r)y2r2+f′(r)x2r3=f′′(r)x2+y2r2+f′(r)x2+y2r3=f′′(r)r2r2+f′(r)r2r3=f′′(r)+f′(r)r\frac{\partial^2 z}{\partial x^2} + \frac{\partial^2 z}{\partial y^2} = f''(r) \frac{x^2}{r^2} + f'(r) \frac{y^2}{r^3} + f''(r) \frac{y^2}{r^2} + f'(r) \frac{x^2}{r^3} = f''(r) \frac{x^2 + y^2}{r^2} + f'(r) \frac{x^2 + y^2}{r^3} = f''(r) \frac{r^2}{r^2} + f'(r) \frac{r^2}{r^3} = f''(r) + \frac{f'(r)}{r}

We are given that this sum equals 0, so we have:

f′′(r)+f′(r)r=0f''(r) + \frac{f'(r)}{r} = 0

Let g(r)=f′(r)g(r) = f'(r) . Then g′(r)=f′′(r)g'(r) = f''(r) , so we have:

g′(r)+g(r)r=0g'(r) + \frac{g(r)}{r} = 0 dgdr=−gr\frac{dg}{dr} = -\frac{g}{r} dgg=−drr\frac{dg}{g} = -\frac{dr}{r}

Integrating both sides, we get:

∫dgg=−∫drr\int \frac{dg}{g} = -\int \frac{dr}{r} ln⁡∣g∣=−ln⁡∣r∣+C1\ln|g| = -\ln|r| + C_1 g(r)=e−ln⁡∣r∣+C1=eC1e−ln⁡∣r∣=Crg(r) = e^{-\ln|r| + C_1} = e^{C_1} e^{-\ln|r|} = \frac{C}{r}

So f′(r)=Crf'(r) = \frac{C}{r} . Integrating again, we get:

f(r)=∫Crdr=Cln⁡∣r∣+C2f(r) = \int \frac{C}{r} dr = C \ln|r| + C_2

Thus, r>0r>0 では

f(r)=Aln⁡r+Bf(r)=A\ln r+B

where AA and BB are constants. ただし原点 r=0r=0 まで含む領域で z=f(r)z=f(r) が C2C^2 級かつ調和であることを要求する場合、 ln⁡r\ln r は原点で発散するため A=0A=0 でなければならず、解は定数関数 f(r)=Bf(r)=B に限られる。

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