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公式（指数・対数関数）

指数の定義               自然数mについて am= m個 z }| { a× a × · · · × a 整数に拡張 am−n₌ a m an a 0_{= 1 a}−n₌ 1 an 有理数に拡張（a > 0） amn = n √ am _a1_{n =} √n_a 無理数に拡張（a > 0） 1 < a の場合 ax_{= y⇔ ∀ p, q :有理数（p < x < q → a}p _{< y < a}q_） 0 < a < 1 の場合 ax_{= y⇔ ∀ p, q :有理数（p < x < q → a}p _{> y > a}q_） 指数法則         ax+y_{= a}x_{· a}y _ax−y ₌ a x ay ax·y_{= (a}x₎y = (ay₎x axy _{= (a}x₎ 1 y ₌ ( ay1 )x (a· b)x_{= a}x_{· b}x (a b )x = a x bx 指数関数の大小関係 1 < a の場合 ：  x < y ⇐⇒ ax< ay 0 < a < 1 の場合 ：  x < y ⇐⇒ ax> ay

対数の定義         x = log_aX ⇐⇒ X = ax logaax= x alogaX= X 対数法則        

log_a(X· Y ) = logaX + log_aY log_a X

Y = logaX− logaY log_aYx_{= x log} aY 底の変換公式 log_XY = logaY log_aX 対数関数の大小関係 1 < a の場合 ：  x < y ⇐⇒ logax < log_ay 0 < a < 1 の場合 ：  x < y ⇐⇒ logax > log_ay 常用対数         log102 = 0.30103 . . . log103 = 0.47712 . . .

log₁₀4 = 0.60206 . . . (= 2 log₁₀2) log₁₀5 = 0.69897 . . . (= 1− log₁₀2) log₁₀6 = 0.77815 . . . (= log₁₀2 + log₁₀3) log₁₀7 = 0.84509 . . .

log₁₀8 = 0.90309 . . . (= 3 log₁₀2) log₁₀9 = 0.95424 . . . (= 2 log₁₀3)

桁数と最高位の数字

M がn桁の数 ⇐⇒ 10n−1_{5 M < 10}n

⇐⇒ n = (log10M の整数部分) + 1 最高位の数字がa ⇐⇒ a × 10n−15 M < (a + 1) × 10n−1

Napierの定数eの定義         e = ( lim t→0 at− 1 t = 1となるa ) lim t→0 et− 1 t = 1 eの定義と同値な関係         e = ( lim t→0 log_a(t + 1) t = 1となるa ) lim t→0 log_e(t + 1) t = 1 lim n→+∞ ( 1 + 1 n )n = e lim n→−∞ ( 1 + 1 n )n = e lim x→+∞ ( 1 + 1 x )x = e lim x→−∞ ( 1 + 1 x )x = e 自然対数の定義 log x = log_ex (底eを省略する) ln xと書くこともある 双曲線関数               cosh θ = eθ + e −θ 2 sinh θ = eθ− e−θ 2 tanh θ = eθ− e−θ eθ + e−θ 相互関係               tanh θ = sinh θ cosh θ cosh 2 θ− sinh2θ = 1 1− tanh2θ = 1 cosh2θ 加法定理        

cosh (α + β) = cosh α cosh β + sinh α sinh β cosh (α− β) = cosh α cosh β − sinh α sinh β sinh (α + β) = sinh α cosh β + cosh α sinh β sinh (α− β) = sinh α cosh β − cosh α sinh β tanh (α + β) = tanh α + tanh β

1 + tanh α tanh β tanh (α− β) =

tanh α− tanh β 1− tanh α tanh β

逆双曲線関数

θ = cosh−1x ⇐⇒ cosh θ = xかつθ= 0    cosh−1x = log(x +√x2− 1)

θ = sinh−1y ⇐⇒ sinh θ = y     sinh−1y = log

(

y +√y2_{+ 1})

θ = tanh−1m ⇐⇒ tanh θ = m     tanh−1m = 1

2 log 1 + m 1− m

証明

対数法則

log_aX = x，log_aY = y  とおく

X = ax，Y = ay

X· Y = ax· ay = ax+y

∴ loga(X· Y ) = x + y = logaX + logaY X Y = ax ay = a x_−y ∴ loga X Y = x− y = logaX− logaY Yx= (ay)x= axy ∴ logaY x_{= xy = x log} aY 底の変換 Y = ay= ax·yx = (ax) y x = X y x ∴ logXY = y x = logaY log_aX

• lim x_→0 ex_{− 1} x = 1 ⇒ tlim_→0 log_e(t + 1) t = 1 ( 逆も同様) x = log_e(t + 1)とおく t = ex− 1 t→ 0 のとき x→ loge1 = 0 ∴ lim t→0 loge(t + 1) t = limx→0 x ex− 1 = 1 lim x→0 ex_{− 1} x = 1 1 = 1 • lim x→∞ ( 1 + 1 x )x = e ⇒ lim t→−∞ ( 1 + 1 t )t = e (逆も同様) x =−t − 1とおく t =−x − 1 t→ −∞ のとき x→ ∞ ∴ lim t_→−∞ ( 1 + 1 t )t = lim x_→∞ ( 1− 1 x + 1 )−x−1 = lim x_→∞ ( x x + 1 )−x−1 = lim x_→∞ ( x + 1 x )x+1 = lim x_→∞ ( 1 + 1 x )x( 1 + 1 x ) = e· 1 = 1 • lim x_→±∞ ( 1 + 1 x )x = e ⇒ lim t→0 log_e(t + 1) t = 1 ( 逆も同様) x = 1 t とおく t = 1 x t→ 0 のとき x→ ±∞ ∴ lim t→0 log_e(t + 1) t =x_→±∞lim log_e ( 1 x + 1 ) 1 x = lim x_→±∞x loge ( 1 + 1 x ) = lim x→±∞loge ( 1 + 1 x )x = log_ee = 1

• lim n→∞ ( 1 + 1 n )n = e ⇒ lim x→∞ ( 1 + 1 x )x = e (逆は明らか) n = [x] = (xの整数部分)，m = n + 1とする n5 x < m 1 + 1 m < 1 + 1 x 5 1 + 1 n ( 1 + 1 m )n 5 ( 1 + 1 m )x < ( 1 + 1 x )x 5 ( 1 + 1 n )x < ( 1 + 1 n )m x→ ∞のときn→ ∞，m→ ∞ lim m→∞ ( 1 + 1 m )n = lim m→∞ ( 1 + 1 m )m−1 = lim m→∞ ( 1 + 1 m )m ( 1 + 1 m ) = e 1 = e lim n→∞ ( 1 + 1 n )m = lim n→∞ ( 1 + 1 n )n+1 = lim n→∞ ( 1 + 1 n )n( 1 + 1 n ) = e· 1 = e ∴ lim x_→∞ ( 1 + 1 x )x = e 双曲線関数の相互関係 tanh θ = eθ− e −θ eθ + e−θ = 2 sinh θ 2 cosh θ = sinh θ cosh θ

cosh2θ− sinh2θ = (cosh θ + sinh θ)(cosh θ− sinh θ) = eθe−θ = 1

1− tanh2θ = 1− sinh 2 θ cosh2θ = cosh2θ− sinh2θ cosh2θ = 1 cosh2θ

cosh α cosh β = eα + e −α 2 eβ + e−β 2 = eα+β + eα−β + e−α+β + e−α−β 4 = eα +β + e−α−β 4 + eα−β + e−α+β 4 = cosh (α + β) 2 + cosh (α− β) 2 sinh α sinh β = eα− e

−α 2 eβ− e−β 2 = eα+β − eα−β − e−α+β + e−α−β 4 = eα +β + e−α−β 4 − eα−β + e−α+β 4 = cosh (α + β) 2 − cosh (α− β) 2 ゆえに

cosh (α + β) = cosh α cosh β + sinh α sinh β cosh (α− β) = cosh α cosh β − sinh α sinh β

sinh α cosh β = eα− e

−α 2 eβ + e−β 2 = eα+β + eα−β − e−α+β − e−α−β 4 = eα +β − e−α−β 4 + eα−β − e−α+β 4 = sinh (α + β) 2 + sinh (α− β) 2 cosh α sinh β = eα + e

−α 2 eβ− e−β 2 = eα+β − eα−β + e−α+β − e−α−β 4 = eα +β − e₋α₋β 4 − eα−β − e−α+β 4 = sinh (α + β) 2 − sinh (α− β) 2 ゆえに

sinh (α + β) = sinh α cosh β + cosh α sinh β sinh (α− β) = sinh α cosh β − cosh α sinh β

tanh (α + β) = sinh(α + β) cosh (α + β) =

sinh α cosh β + cosh α sinh β cosh α cosh β + sinh α sinh β

= sinh α cosh α + sinh β cosh β 1 + sinh α cosh α sinh β cosh β = tanh α + tanh β 1 + tanh α tanh β tanh (α− β) = sinh(α− β) cosh (α− β) =

sinh α cosh β− cosh α sinh β cosh α cosh β− sinh α sinh β

= sinh α cosh α − sinh β cosh β 1− sinh α cosh α sinh β cosh β = tanh α− tanh β 1− tanh α tanh β

逆双曲線関数 θ = cosh−1x ⇐⇒ x = cosh θ かつ θ= 0 ゆえに x = eθ + e −θ 2 = ( eθ )2 + 1 2eθ ( eθ )2 − 2xeθ + 1 = 0 θ= 0よりeθ= 1だから eθ = x +√x2_{− 1} ∴ cosh−1_{x = θ = log}(_{x +}√_x2− 1) θ = sinh−1y ⇐⇒ y = sinh θ ゆえに y = eθ− e −θ 2 = ( eθ )2 − 1 2eθ ( eθ )2 − 2yeθ − 1 = 0 eθ > 0だから eθ = y +√y2_{+ 1} ∴ sinh−1y = θ = log ( y +√y2_{+ 1}) θ = tanh−1m ⇐⇒ m = tanh θ ゆえに m = eθ− e −θ eθ + e−θ = e2θ − 1 e2θ + 1 m(e2θ + 1) = e2θ − 1 1 + m = (1− m)e2θ e2θ = 1 + m 1− m 2θ = log 1 + m 1− m ∴ tanh−1m = θ = 1 2 log 1 + m 1− m