式展開

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

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

1からnまでの三乗の和は，1からnまでの和の二乗

$$\begin{eqnarray} k^2-(k-1)^2&=&k^2-(k^2-2k+1)\;\cdots\;左辺を展開する \\&=&k^2-k^2+2k-1 \\&=&2k-1 \end{eqnarray}$$
$k$に$1\cdots n$を代入した式を作る． $$\begin{eqnarray} 1^2-(1-1)^2&=&2\cdot 1-1 \\2^2-(2-1)^2&=&2\cdot 2-1 \\3^2-(3-1)^2&=&2\cdot 3-1 \\\vdots& \\n^2-(n-1)^2&=&2\cdot n-1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^2}&-0^3&=&2\cdot 1&-1 \\&\cancel{2^2}&\cancel{-1^2}&=&2\cdot 2&-1 \\&\cancel{3^2}&\cancel{-2^2}&=&2\cdot 3&-1 \\&&\vdots& \\+)&n^2&\cancel{-(n-1)^2}&=&2\cdot n&-1 \\\hline \\&n^2&-0^2&=&2\sum_{k=1}^n k&-\sum_{k=1}^n 1 \end{eqnarray}$$
$\sum_{k=1}^n k$について解く． $$\begin{eqnarray} n^2&=&2\sum_{k=1}^n k-\sum_{k=1}^n 1 \\2\sum_{k=1}^n k&=&n^2+\sum_{k=1}^n 1 \\&=&n^2+n\;\cdots\;\sum_{k=1}^n 1=n \\&=&n(n+1) \\\sum_{k=1}^n k&=&\frac{n(n+1)}{2} \end{eqnarray}$$

$$\begin{eqnarray} k^3-(k-1)^3&=&k^3-(k-1)(k-1)^2\;\cdots\;左辺を展開する \\&=&k^3-(k-1)(k^2-2k+1) \\&=&k^3-\left\{k(k^2-2k+1)-(k^2-2k+1)\right\} \\&=&k^3-\left(k^3-2k^2+k-k^2+2k-1\right) \\&=&k^3-\left(k^3-3k^2+3k-1\right) \\&=&k^3-k^3+3k^2-3k+1 \\&=&3k^2-3k+1 \end{eqnarray}$$
$k$に$1\cdots n$を代入した式を作る． $$\begin{eqnarray} 1^3-(1-1)^3&=&3\cdot 1^2-3\cdot 1+1 \\2^3-(2-1)^3&=&3\cdot 2^2-3\cdot 2+1 \\3^3-(3-1)^3&=&3\cdot 3^2-3\cdot 3+1 \\&\vdots& \\n^3-(n-1)^3&=&3\cdot n^2-3\cdot n+1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^3}*&-0^3&=&3\cdot 1^2&-3\cdot 1&+1 \\&\cancel{2^3}&\cancel{-1^3}&=&3\cdot 2^2&-3\cdot 2&+1 \\&\cancel{3^3}&\cancel{-2^3}&=&3\cdot 3^2&-3\cdot 3&+1 \\&&\vdots& \\+)&n^3&\cancel{-(n-1)^3}&=&3\cdot n^2&-3\cdot n&+1 \\\hline \\&n^3&-0^3&=&3\sum_{k=1}^n k^2&-3\sum_{k=1}^n k&+\sum_{k=1}^n 1 \end{eqnarray}$$
$\sum_{k=1}^n k^2$について解く． $$\begin{eqnarray} n^3&=&3\sum_{k=1}^n k^2-3\sum_{k=1}^n k+\sum_{k=1}^n 1 \\3\sum_{k=1}^n k^2&=&n^3+3\sum_{k=1}^n k-\sum_{k=1}^n 1 \\&=&n^3+3\frac{n(n+1)}{2}-n\;\cdots\;\sum_{k=1}^n k=\frac{n(n+1)}{2},\;\sum_{k=1}^n 1=n \\&=&n^3+\frac{3}{2}n^2+\frac{3}{2}n-n \\&=&n^3+\frac{3}{2}n^2+\frac{1}{2}n \\&=&\frac{\left(2n^3+3n^2+n\right)}{2} \\&=&\frac{n\left(2n^2+3n+1\right)}{2} \\&=&\frac{n\left(n+1\right)\left(2n+1\right)}{2} \\\sum_{k=1}^n k^2&=&\frac{n\left(n+1\right)\left(2n+1\right)}{6} \end{eqnarray}$$

$$\begin{eqnarray} k^4-(k-1)^4&=&k^4-(k-1)^2(k-1)^2\;\cdots\;左辺を展開する \\&=&k^4-\left\{(k^2-2k+1)(k^2-2k+1)\right\} \\&=&k^4-\left\{k^2(k^2-2k+1)-2k(k^2-2k+1)+(k^2-2k+1)\right\} \\&=&k^4-\left(k^4-2k^3+k^2-2k^3+4k^2-2k+k^2-2k+1\right) \\&=&k^4-\left(k^4-4k^3+6k^2-4k+1\right) \\&=&k^4-k^4+4k^3-6k^2+4k-1 \\&=&4k^3-6k^2+4k-1 \end{eqnarray}$$
$k$に$1\cdots n$を代入した式を作る． $$\begin{eqnarray} 1^4-(1-1)^4&=&4\cdot1^3&-6\cdot1^2&+4\cdot1&-1 \\2^4-(2-1)^4&=&4\cdot2^3&-6\cdot2^2&+4\cdot2&-1 \\3^4-(3-1)^4&=&4\cdot3^3&-6\cdot3^2&+4\cdot3&-1 \\&\vdots& \\n^4-(n-1)^4&=&4\cdot n^3&-6\cdot n^2&+4\cdot n&-1 \end{eqnarray}$$
左辺，右辺を各項ごとに足し合わせる． $$\begin{eqnarray} &\cancel{1^4}&-0^4&=&4\cdot1^3&-6\cdot1^2&+4\cdot1&-1 \\&\cancel{2^4}&\cancel{-1^4}&=&4\cdot2^3&-6\cdot2^2&+4\cdot2&-1 \\&\cancel{3^4}&\cancel{-2^4}&=&4\cdot3^3&-6\cdot3^2&+4\cdot3&-1 \\&&&\vdots& \\+)&n^4&\cancel{-(n-1)^4}&=&4\cdot n^3&-6\cdot n^2&+4\cdot n&-1 \\\hline \\&n^4&-0^4&=&4\sum_{k=1}^n k^3&-6\sum_{k=1}^n k^2&+4\sum_{k=1}^n k&-\sum_{k=1}^n 1 \end{eqnarray}$$
$\sum_{k=1}^n k^3$について解く． $$\begin{eqnarray} n^4&=&4\sum_{k=1}^n k^3-6\sum_{k=1}^n k^2+4\sum_{k=1}^n k-\sum_{k=1}^n 1 \\4\sum_{k=1}^n k^3&=&n^4+6\sum_{k=1}^n k^2-4\sum_{k=1}^n k+\sum_{k=1}^n 1 \\&=&n^4+6\frac{n(n+1)(2n+1)}{6}-4\frac{n(n+1)}{2}+n\;\cdots\;\sum_{k=1}^n k^2=\frac{n(n+1)(2n+1)}{6},\sum_{k=1}^n k=\frac{n(n+1)}{2},\;\sum_{k=1}^n 1=n \\&=&n^4+n(n+1)(2n+1)-2n(n+1)+n \\&=&n^4+(n^2+n)(2n+1)-2n^2-2n+n \\&=&n^4+n^2(2n+1)+n(2n+1)-2n^2-2n+n \\&=&n^4+2n^3+n^2+2n^2+n-2n^2-2n+n \\&=&n^4+2n^3+n^2 \\&=&n^2(n^2+2n+1) \\&=&n^2(n+1)^2 \\\sum_{k=1}^n k^3&=&\frac{n^2(n+1)^2}{4}=\left(\frac{n(n+1)}{2}\right)^2=\left(\sum_{k=1}^n k\right)^2 \end{eqnarray}$$

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