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MagicAlex
@MagicAlex
October 2021
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Доказать справедливость неравенства ln(1+x)<x при x>0.
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Jamilua
Ln(1+x)<x
ln(1+x)<ln(e^x)
1+x<e^x
Разложим в ряд e^x
e^x = 1+x + x^2/2! + o(x^n)
1+x<1+x + x^2/2! + o(x^n)
x^2/2! + o(x^n)>0 для любого x>0
1 votes
Thanks 7
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Answers & Comments
ln(1+x)<ln(e^x)
1+x<e^x
Разложим в ряд e^x
e^x = 1+x + x^2/2! + o(x^n)
1+x<1+x + x^2/2! + o(x^n)
x^2/2! + o(x^n)>0 для любого x>0