为方便,令x+3=t原式=1/(2x+3)(x-1)=1/5*[1/(x-1)-2/(2x+3)]=1/5[1/(x+3-2)-1/(x+3-1.5)]=1/5[ 1/(t-2)-1/(t-1.5)]=1/5[-0.5/(1-t/2)+1/(1-t/1.5)]=1/5[ -0.5(1+t/2+t^2/2^2+.)+(1+t/1.5+t^2/1.5^2+...)]
幂级数 求教 求1/(2x^2+x-3)在x=3处的泰勒级数
为方便,令x+3=t原式=1/(2x+3)(x-1)=1/5*[1/(x-1)-2/(2x+3)]=1/5[1/(x+3-2)-1/(x+3-1.5)]=1/5[ 1/(t-2)-1/(t-1.5)]=1/5[-0.5/(1-t/2)+1/(1-t/1.5)]=1/5[ -0.5(1+t/2+t^2/2^2+.)+(1+t/1.5+t^2/1.5^2+...)]