高等数学B：CH2习题解答.pdf

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1、P52SK18.dúª1+x+x2+


xn=xn+11(n•ê§x,1)§

2、x1^¦꧉ÑeÚª¦Úúª(1)1+2x+3x2+


+nxn1;(2)12
x+22
x2+33x3+


+n2xn;)))(1)f(x)def=1+2x+3x2+


+nxn10xn+110nxn+1(n+1)xn+1=1+x+x2+


xn==x1(x1)2(2)g(x)def=12
x+22
x2+33x3+


+n2xn0=x(xf(x))=


P67SKo1117.y²Ô‚x2+y2=a2(a>0)þ?˜:?ƒ‚3ü‹I¶þ

3、åƒÚua.1.00.80.60.40.20.20.40.60.81.0111Figure:x2+y2=12ã”111)))3x2+y2=a2(a>0)ü>éx¦§r1111dydyyp+p=0;=2x2ydxdxxÔ‚þ?¿˜:(x0;y0)?ƒ‚:ry0yy0=(xx0)x03x;y¶þå©Oµpp(0;y0+x0y0);(x0+x0y0;y)åÚpppp2p2y0+x0y0+x0+x0y0=x0+y0=a=aP67SKo17.®•4‹I•‚r=asin2(a>0•~ê)§

4、=žƒ4A:?ƒ‚†{‚•§")))r4‹I•§=z•ëê

5、•§8>>>:y=asin2sinpp22ƒ:(a;a)).ƒ‚Ç22dy2acos2sin+asin2cos==1dx=2acos2cosasin2sin=44pƒ‚•§x+y2a=0;{‚•§xy=0:4‹I•‚r=asin2(a>0•~ê)¡•o“p4‚§Ùã”1.00.5-0.50.51.0-0.5Figure:r=sin2ã”