无穷小问题:设f(x)有连续的导数,f(0)=0,f'(0)不等于0,F(x)=(x到0)(x^2-t^2)f(t)dt,且当x趋于0时,... 设f(x)有连续导数,f(0)=0,f′(0)≠0,且F(x...
\u9ad8\u6570\u5fae\u79ef\u5206\u95ee\u9898\u8bbef(x)\u6709\u8fde\u7eed\u7684\u5bfc\u6570,f(a)=0,f \u2019(a)\u4e0d\u7b49\u4e8e0,F\uff08x\uff09=[\u5b9a\u79ef\u5206a\u5230x\u4f60\u786e\u5b9a\u9898\u76ee\u7684f(a)=0\u4e48\uff1f\u8fd9\u6837\u7b97\u4e0d\u51fa\u6765k\u5416
\u2235F(x)\uff1dx2\u222bx0f(t)dt?\u222bx0t2f(t)dt\u2234F\u2032\uff08x\uff09=2x\u222bx0f(t)dt+x2f\uff08x\uff09-x2f\uff08x\uff09=2x\u222bx0f(t)dt\u2234\u7531\u5df2\u77e5\u6761\u4ef6F\u2032\uff08x\uff09\u4e0exk\u662f\u540c\u9636\u65e0\u7a77\u5c0f\uff0c\u4e14f\uff080\uff09=0\uff0cf\u2032\uff080\uff09\u22600\uff0c\u6709limx\u21920F\u2032(x)xk\uff1dlimx\u219202x\u222bx0f(t)dtxk\uff1dlimx\u219202\u222bx0f(t)dtkxk?1\u2550limx\u219202f(x)k(k?1)xk?2=limx\u219202f\u2032(x)k(k?1)(k?2)xk?3=2f\u2032(0)limx\u219201k(k?1)(k?2)xk?3\u22600\u2234k=3\u6545\u9009\uff1aC\uff0e
简单分析一下,答案如图所示
F(x)=(x到0)(x^2-t^2)f(t)dt
=x^2*(x到0)f(t)dt-(x到0)t^2f(t)dt
F'(x)=2x*(x到0)f(t)dt+x^2*f(x)-x^2*f(x)
=2x*(x到0)f(t)dt
第一次求导:F''(x)=2*(x到0)f(t)dt+2xf(x)
把2消掉,第二次求导:F‘’‘(x)=2f(x)+xf'(x)
F'''(x)/x=2*f(x)/x+f’(x)
以为f(x)/x 在x趋近于零时,根据洛必达法则,等于f'(0)
所以F'''(x)/x在x趋近于零时,等于3f'(0)
因为f'(0)不为零
所以F'''(x)与x同阶
所以F'(x)与x^3同阶
k=3
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