式展開: 10月 2023

n乗の3つの数の和を基本対称式で表す

$\alpha, \beta, \gamma$を解とする3次方程式を考える． $$\begin{eqnarray} \left(x-\alpha\right)\left(x-\beta\right)\left(x-\gamma\right)&=&0 \\x^3-\left(\alpha+\beta+\gamma\right)x^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)x-\alpha\beta\gamma&=&0 \end{eqnarray}$$ $\alpha,\beta,\gamma$は式の解なので，$x$に代入しても式は成り立つ．代入した2つの式を得る． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^3-\left(\alpha+\beta+\gamma\right)\alpha^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha-\alpha\beta\gamma&=&0 \\\beta^3-\left(\alpha+\beta+\gamma\right)\beta^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta-\alpha\beta\gamma&=&0 \\\gamma^3-\left(\alpha+\beta+\gamma\right)\gamma^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma-\alpha\beta\gamma&=&0 \end{array} \right. \end{eqnarray} $$ 第一式には$\alpha^{n-3}$，第二式には$\beta^{n-3}$，第三式には$\gamma^{n-3}$を両辺に掛ける． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^{n-3}\left\{\alpha^3-\left(\alpha+\beta+\gamma\right)\alpha^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha-\alpha\beta\gamma\right\}&=&\alpha^{n-3}\cdot0 \\\beta^{n-3}\left\{\beta^3-\left(\alpha+\beta+\gamma\right)\beta^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta-\alpha\beta\gamma\right\}&=&\beta^{n-3}\cdot0 \\\gamma^{n-3}\left\{\gamma^3-\left(\alpha+\beta+\gamma\right)\gamma^2+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma-\alpha\beta\gamma\right\}&=&\gamma^{n-3}\cdot0 \end{array} \right. \\ \\\left\{ \begin{array}{l} \alpha^n&-\left(\alpha+\beta+\gamma\right)\alpha^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha^{n-2}&-\alpha\beta\gamma\alpha^{n-3}&=&0 \\\beta^n&-\left(\alpha+\beta+\gamma\right)\beta^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta^{n-2}&-\alpha\beta\gamma\beta^{n-3}&=&0 \\\gamma^n&-\left(\alpha+\beta+\gamma\right)\gamma^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma^{n-2}&-\alpha\beta\gamma\gamma^{n-3}&=&0 \end{array} \right. \end{eqnarray} $$ 3式を足し合わせ，n乗の和について解くことで．n乗の和を基本対称式で求める式を得る． $$\begin{eqnarray} &\alpha^n&-\left(\alpha+\beta+\gamma\right)\alpha^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\alpha^{n-2}&-\alpha\beta\gamma\alpha^{n-3}&=&0 \\&\beta^n&-\left(\alpha+\beta+\gamma\right)\beta^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\beta^{n-2}&-\alpha\beta\gamma\beta^{n-3}&=&0 \\+)&\gamma^n&-\left(\alpha+\beta+\gamma\right)\gamma^{n-1}&+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\gamma^{n-2}&-\alpha\beta\gamma\gamma^{n-3}&=&0 \\\hline &\alpha^{n}+\beta^{n}+\gamma^{n} &-\left(\alpha+\beta+\gamma\right)\left(\alpha^{n-1}+\beta^{n-1}+\gamma^{n-1}\right) &+\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\left(\alpha^{n-2}+\beta^{n-2}+\gamma^{n-2}\right) &-\alpha\beta\gamma\left(\alpha^{n-3}+\beta^{n-3}+\gamma^{n-3}\right)&=&0 \end{eqnarray}$$ $$\begin{eqnarray} \alpha^{n}+\beta^{n}+\gamma^{n}&=& \left(\alpha+\beta+\gamma\right)\left(\alpha^{n-1}+\beta^{n-1}+\gamma^{n-1}\right) -\left(\alpha\beta+\beta\gamma+\alpha\gamma\right)\left(\alpha^{n-2}+\beta^{n-2}+\gamma^{n-2}\right) +\alpha\beta\gamma\left(\alpha^{n-3}+\beta^{n-3}+\gamma^{n-3}\right) \end{eqnarray}$$

$\alpha, \beta, \gamma$は任意の数でよいのでそれを$x, y, z$とすれば以下の式を得る． $$\begin{eqnarray} x^{n}+y^{n}+z^{n}&=&\left(x+y+z\right)\left(x^{n-1}+y^{n-1}+z^{n-1}\right)-\left(xy+yz+xz\right)\left(x^{n-2}+y^{n-2}+z^{n-2}\right)+xyz\left(x^{n-3}+y^{n-3}+z^{n-3}\right) \end{eqnarray}$$

n乗された2つの数の和を基本対称式で表す

$\alpha, \beta$を解とする2次方程式を考える． $$\begin{eqnarray} \left(x-\alpha\right)\left(x-\beta\right)&=&0 \\x^2-\left(\alpha+\beta\right)x+\alpha\beta&=&0 \end{eqnarray}$$ $\alpha,\beta$は式の解なので，$x$に代入しても式は成り立つ．代入した2つの式を得る． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^2-\left(\alpha+\beta\right)\alpha+\alpha\beta&=&0 \\\beta^2-\left(\alpha+\beta\right)\beta+\alpha\beta&=&0 \end{array} \right. \end{eqnarray} $$ 第一式には$\alpha^{n-2}$，第二式には$\beta^{n-2}$を両辺に掛ける． $$ \begin{eqnarray} \left\{ \begin{array}{l} \alpha^{n-2} \cdot\left\{\alpha^2-\left(\alpha+\beta\right)\alpha+\alpha\beta\right\}&=&\alpha^{n-2}\cdot 0 \\\beta^{n-2}\cdot\left\{\beta^2 -\left(\alpha+\beta\right)\beta +\alpha\beta\right\}&=&\beta^{n-2} \cdot 0 \end{array} \right. \\ \\\left\{ \begin{array}{l} \alpha^{n}-\left(\alpha+\beta\right)\alpha^{n-1}+\alpha\beta\alpha^{n-2}&=&0 \\\beta^{n}-\left(\alpha+\beta\right)\beta^{n-1}+\alpha\beta\beta^{n-2}&=&0 \end{array} \right. \end{eqnarray} $$ 両式を足し合わせ，n乗の和について解くことで．n乗の和を基本対称式で求める式を得る． $$\begin{eqnarray} &\alpha^{n}&-\left(\alpha+\beta\right)\alpha^{n-1}&+\alpha\beta\alpha^{n-2}&=&0 \\+)&\beta^{n}&-\left(\alpha+\beta\right)\beta^{n-1}&+\alpha\beta\beta^{n-2}&=&0 \\\hline &\alpha^{n}+\beta^{n}&-\left(\alpha+\beta\right)\left(\alpha^{n-1}+\beta^{n-1}\right)&+\alpha\beta\left(\alpha^{n-2}+\beta^{n-2}\right)&=&0 \end{eqnarray}$$ $$\begin{eqnarray} \alpha^{n}+\beta^{n}&=&\left(\alpha+\beta\right)\left(\alpha^{n-1}+\beta^{n-1}\right)-\alpha\beta\left(\alpha^{n-2}+\beta^{n-2}\right) \end{eqnarray}$$

$\alpha, \beta$は任意の数でよいのでそれを$x, y$とすれば以下の式を得る． $$\begin{eqnarray} x^{n}+y^{n}&=&\left(x+y\right)\left(x^{n-1}+y^{n-1}\right)-xy\left(x^{n-2}+y^{n-2}\right) \end{eqnarray}$$

問

original:https://www.youtube.com/watch?v=KPT862KhxRM $$\begin{eqnarray} \left(\frac{1+\sqrt{13}}{2}\right)^7+\left(\frac{1-\sqrt{13}}{2}\right)^7の値を求めよ． \end{eqnarray}$$ 第一項,第二項の括弧内をそれぞれ$\alpha, \beta$とする． $$\begin{eqnarray} \alpha^7+\beta^7\;\cdots\;\alpha=\frac{1+\sqrt{13}}{2},\;\beta=\frac{1-\sqrt{13}}{2} \end{eqnarray}$$ n乗の和を基本対称式で求める上記式を用い，2乗から7乗まで順次求めていく． $$\begin{eqnarray} \alpha^2+\beta^2&=&\left(\alpha+\beta\right)\left(\alpha^{2-1}+\beta^{2-1}\right)-\alpha\beta\left(\alpha^{2-2}+\beta^{2-2}\right) \\&=&\left(\alpha+\beta\right)\left(\alpha+\beta\right)-\alpha\beta\left(\alpha^0+\beta^0\right) \\&=&1\cdot1-(-3)2=1+6=7\;\cdots\;\alpha+\beta=1,\;\alpha\beta=-3,\;\alpha^0+\beta^0=1+1=2 \\ \\\alpha^3+\beta^3&=&\left(\alpha+\beta\right)\left(\alpha^{3-1}+\beta^{3-1}\right)-\alpha\beta\left(\alpha^{3-2}+\beta^{3-2}\right) \\&=&\left(\alpha+\beta\right)\left(\alpha^{2}+\beta^{2}\right)-\alpha\beta\left(\alpha+\beta\right) \\&=&1\cdot7-(-3)1=7+3=10 \\ \\\alpha^4+\beta^4&=&\left(\alpha+\beta\right)\left(\alpha^{4-1}+\beta^{4-1}\right)-\alpha\beta\left(\alpha^{4-2}+\beta^{4-2}\right) \\&=&\left(\alpha+\beta\right)\left(\alpha^{3}+\beta^{3}\right)-\alpha\beta\left(\alpha^{2}+\beta^{2}\right) \\&=&1\cdot10-(-3)7=10+21=31 \\ \\\alpha^5+\beta^5&=&\left(\alpha+\beta\right)\left(\alpha^{5-1}+\beta^{5-1}\right)-\alpha\beta\left(\alpha^{5-2}+\beta^{5-2}\right) \\&=&\left(\alpha+\beta\right)\left(\alpha^{4}+\beta^{4}\right)-\alpha\beta\left(\alpha^{3}+\beta^{3}\right) \\&=&1\cdot31-(-3)10=31+30=61 \\ \\\alpha^6+\beta^6&=&\left(\alpha+\beta\right)\left(\alpha^{6-1}+\beta^{6-1}\right)-\alpha\beta\left(\alpha^{6-2}+\beta^{6-2}\right) \\&=&\left(\alpha+\beta\right)\left(\alpha^{5}+\beta^{5}\right)-\alpha\beta\left(\alpha^{4}+\beta^{4}\right) \\&=&1\cdot61-(-3)31=61+93=154 \\ \\\alpha^7+\beta^7&=&\left(\alpha+\beta\right)\left(\alpha^{7-1}+\beta^{7-1}\right)-\alpha\beta\left(\alpha^{7-2}+\beta^{7-2}\right) \\&=&\left(\alpha+\beta\right)\left(\alpha^{6}+\beta^{6}\right)-\alpha\beta\left(\alpha^{5}+\beta^{5}\right) \\&=&1\cdot154-(-3)61=154+183=337 \end{eqnarray}$$

2を底とする指数凾数の微分

与式

$$\begin{eqnarray} y&=&2^x \end{eqnarray}$$

両辺とも自然対数をとってネイピア数を底とする指数凾数にする(変形例1)

$$\begin{eqnarray} \ln{\left(y\right)}&=&\ln{\left(2^x\right)} \\e^{\ln{\left(y\right)}}&=&e^{\ln{\left(2^x\right)}} \\y&=&e^{x\ln{\left(2\right)}}\;\cdots\;\ln{\left(A^B\right)}=B\ln{\left(A\right)} \end{eqnarray}$$

逆凾数を底の変換及び分母をはらった後，ネイピア数を底とする指数凾数にする(変形例2)

$$\begin{eqnarray} x&=&\log_2{\left(y\right)} \\&=&\frac{\ln{\left(y\right)}}{\ln{\left(2\right)}}\;\cdots\;\log_A{\left(B\right)}=\frac{\log_C{\left(B\right)}}{\log_C{\left(A\right)}} \\\ln{\left(y\right)}&=&x\ln{\left(2\right)} \\e^{\ln{\left(y\right)}}&=&e^{x\ln{\left(2\right)}} \\y&=&e^{x\ln{\left(2\right)}} \end{eqnarray}$$

合成凾数の微分

$$\begin{eqnarray} \frac{\mathrm{d}y}{\mathrm{d}x}&=& \frac{\mathrm{d}}{\mathrm{d}x} 2^x \\&=& \frac{\mathrm{d}}{\mathrm{d}x} e^{x\ln{\left(2\right)}} \\&=&\frac{\mathrm{d}e^u}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}\;\cdots\;u=x\ln{\left(2\right)} \\&=&e^u\cdot\ln{\left(2\right)}\;\cdots\;\frac{\mathrm{d}e^x}{\mathrm{d}x}=e^x,\;\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}x\ln{\left(2\right)}=\ln{\left(2\right)} \\&=&e^{x\ln{\left(2\right)}}\ln{\left(2\right)} \\&=&e^{\ln{\left(2^x\right)}}\ln{\left(2\right)} \\&=&2^x\ln{\left(2\right)} \end{eqnarray}$$

2でなく変数aとした場合

$$\begin{eqnarray} y&=&a^x \end{eqnarray}$$

両辺とも自然対数をとってネイピア数を底とする指数凾数にする(変形例1)

$$\begin{eqnarray} \ln{\left(y\right)}&=&\ln{\left(a^x\right)} \\e^{\ln{\left(y\right)}}&=&e^{\ln{\left(a^x\right)}} \\y&=&e^{x\ln{\left(a\right)}}\;\cdots\;\ln{\left(A^B\right)}=B\ln{\left(A\right)} \end{eqnarray}$$

逆凾数を底の変換及び分母をはらった後，ネイピア数を底とする指数凾数にする(変形例2)

$$\begin{eqnarray} x&=&\log_a{\left(y\right)} \\&=&\frac{\ln{\left(y\right)}}{\ln{\left(a\right)}}\;\cdots\;\log_A{\left(B\right)}=\frac{\log_C{\left(B\right)}}{\log_C{\left(A\right)}} \\\ln{\left(y\right)}&=&x\ln{\left(a\right)} \\e^{\ln{\left(y\right)}}&=&e^{x\ln{\left(a\right)}} \\y&=&e^{x\ln{\left(a\right)}} \end{eqnarray}$$

合成凾数の微分

$$\begin{eqnarray} \frac{\mathrm{d}y}{\mathrm{d}x}&=& \frac{\mathrm{d}}{\mathrm{d}x} a^x \\&=& \frac{\mathrm{d}}{\mathrm{d}x} e^{x\ln{\left(a\right)}} \\&=&\frac{\mathrm{d}e^u}{\mathrm{d}u}\frac{\mathrm{d}u}{\mathrm{d}x}\;\cdots\;u=x\ln{\left(a\right)} \\&=&e^u\cdot\ln{\left(a\right)}\;\cdots\;\frac{\mathrm{d}e^x}{\mathrm{d}x}=e^x,\;\frac{\mathrm{d}u}{\mathrm{d}x}=\frac{\mathrm{d}}{\mathrm{d}x}x\ln{\left(a\right)}=\ln{\left(a\right)} \\&=&e^{x\ln{\left(a\right)}}\ln{\left(a\right)} \\&=&e^{\ln{\left(a^x\right)}}\ln{\left(a\right)} \\&=&a^x\ln{\left(a\right)} \end{eqnarray}$$

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