p 181 n°2
règle du cours : \(\exp(x+y) = \exp(x) \times \exp(y)\)
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\(\exp(2) = \exp(1 + 1)\)
- \(\exp(2) = \exp(1) \times \exp(1)\)
- \(\exp(2) = (\exp(1))^2\)
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\(\exp(3) = \exp(1 + 2)\)
- \(\exp(3) = \exp(1) \times \exp(2)\)
- \(\exp(3) = \exp(1) \times (\exp(1))^2\)
- \(\exp(3) = (\exp(1))^1 \times (\exp(1))^2\)
- \(\exp(3) = (\exp(1))^{1+2} = (\exp(1))^{3}\)
Remarque : \(\exp(3) = \exp(1,5 + 1,5) = \exp(1,5) \times \exp(1,5) = (\exp(1,5))^2\)
donc \((\exp(1))^{3} = (\exp(1,5))^2\)
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\(\exp(10) = \exp( 5 + 5 )\)
- \(\exp(10) = \exp( 5 ) \times \exp(5)\)
- \(\exp(10) = (\exp( 5 ))^2\)
- \(\exp(10) = (\exp( 2+3 ))^2\)
- \(\exp(10) = \left(\exp( 2) \times \exp(3 )\right)^2\)
- \(\exp(10) = \left(\exp( 1)^2 \times \exp(1 )^3\right)^2\)
- \(\exp(10) = \left(\exp( 1)^{2+3}\right)^2\)
- \(\exp(10) = \left(\exp( 1)^{5}\right)^2\)
- \(\exp(10) = \exp( 1)^{5 \times 2} = \exp( 1)^{10}\)
Conjecture : \(\exp(n) = \exp(1)^n\)
Exercice du cours
Rappels sur les puissances
- \(a^0 = 1\) (avec \(a \neq 0\))
- \(a^{-n} = \dfrac{1}{a^n}\)
- \(a^{n-p} = \dfrac{a^n}{a^p}\)
- \(2^3 \times 2^5 =2^8\)
- \(10^5 \times 10^{-5} =10^0 = 1\)
- \((2^3)^4 = 2^{12}\)
- \(\dfrac{8^9}{8^8} =8^1 = 8\)
- \(\dfrac{3^{10}}{3^6} =3^4\)
- \(\dfrac{4^2}{4^8} =4^{-6}\)
histoire de la racine carrée
Supposons qu'on puisse écrire la racine carrée sous forme d'une puissance : \(\sqrt{x} = x^p\) (avec \(x \geqslant 0\))
on sait \(\left(\sqrt{x}\right)^2 = x\)
donc il faut \(\left(x^p\right)^2 = x\)
c'est à dire \(x^{2p} = x\)
comparons des éléments de même nature : \(x^{2p} = x^1 \Leftrightarrow 2p = 1 \Leftrightarrow p = \frac{1}{2}\)
notation e
\(\exp(-x) = \dfrac{1}{\exp(x)}\) et \(e^{-x} = \dfrac{1}{e^x}\)
\(\exp(x + y) = \exp(x) \times \exp(y)\) et \(e^{x + y} = e^x \times e^y\)
au lieu d'écrire \(f(x)= \exp(x)\) , on écrira \(f(x) = e^x\)
p 191 n° 42
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\(A = 3 \times \exp(x)^2 \times \exp(x)\)
- \(A = 3 \times \exp(x)^3\)
- \(A = 3 \times (e^x)^3\)
(règle : \((a^x)^y = a^{x \times y}\))
- \(A = 3 \times e^{x \times 3}\)
- \(A = 3 e^{3x}\)
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\(B = \dfrac{\exp{(3x+1)}}{\exp{(x^2)}}\)
- \(B = \dfrac{e^{3x+1}}{e^{x^2}}\)
(règle : \(\dfrac{a^x}{a^y} = a^{x-y}\))
- \(B = e^{(3x+1) - x^2} = e^{-x^2 + 3x+1}\)
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\(C= \exp(-x) \times \exp(x)\)
- \(C= e^{-x} \times e^{x} = e^{-x +x} = e^0 = 1\)