已知X属于R,求证:COSX≥1-X^2/2 已知X属于R,求证:COSX≥1-X^2/2
\u8bc1\u660e:1/2*(e^x e^(-x))>=x^2 cosx,x\u5c5e\u4e8er(x1^2-E(x))^2\u00d7p1+(x2^2-E(x))^2\u00d7p2+\u2026\u2026+(xn^2-E(x))^2\u00d7pn =x1^2 \u00d7p1-2x1p1E(x)+(E(x))^2\u00d7p1+x2^2 \u00d7p2-2x2p2E(x)+(E(x))^2\u00d7p2+\u2026\u2026+xn^2 \u00d7pn-2xnpnE(x)+(E(x))^2\u00d7pn =(x1^2 \u00d7p1+x2^2 \u00d7p2+\u2026\u2026+xn^2 \u00d7pn)-\uff082x1p1E(x)+2x2p2E(x)+\u2026\u2026+2xnpnE(x)\uff09+((E(x))^2\u00d7p1+(E(x))^2\u00d7p2+\u2026\u2026+(E(x))^2\u00d7pn)\u8003\u8651\u5230x1p1+x2p2+\u2026\u2026+xnpn=E(x) p1+p2+\u2026\u2026+pn=1\u6240\u4ee5\u4e0a\u5f0fD(x)=(x1^2 \u00d7p1+x2^2 \u00d7p2+\u2026\u2026+xn^2 \u00d7pn)-2(E(x))^2+(E(x))^2 =E(X^2)-(E(x))^2
\u4e0d\u77e5\u9053\u4f60\u662f\u54ea\u4e2a\u5e74\u7ea7\u7684\u54ce \u5176\u5b9e\u6700\u76f4\u63a5\u5c31\u662f\u6c42\u5bfc\u6765\u505a \u521d\u7b49\u505a\u6cd5\u7684\u8bdd
\u5373\u8bc1x^2/2>1-COSX=2(SIN(X/2))^2 \u518d\u7528\u4e00\u4e0bx\u5927\u4e8e\u96f6\u65f6sin(x\uff0f2)<x\uff0f2\u548cx\u5c0f\u4e8e\u96f6\u65f6x\uff0f2<sin\uff08x\uff0f2\uff09\u8fd9\u4e24\u4e2a\u7ed3\u8bba
因为x∈R，所以f(X)∈[-2，0],求得g'(x)=x，x≥0时，g'(x)≥0,x≤0时，g'(x)≤0，所以g(X)在R上先递减再递增，最小值为g(0)=0，所以g(x)≥f(x),所以cosX≥1-X^2/2
cosx=-sinx
(1-x^2/2)'=-x
因为-sinx和-x在趋于0时是等价无穷小
并且设cosx=f(x) 1-x^2/2=g(x)
则f(x)=g(x)
而在不接近0的时候(-sinx)'=-cosx
|-cosx|<=1
而(-x)'=-1所以g'(x)的斜率大单调性快
所以f(x)>=g(x)在R上恒成立
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