Question

求幂函数∑(n=1,到无穷)(2n+3)x^(2n)在收敛域内的和函数

Answer 1

令f(x)=∑(n=1,到无穷)(2n+3)x^(2n),则x²f(x)=∑(n=1,到无穷)(2n+3)x^(2n+2)以0为下限,x为上限对两边取定积分得

∫x²f(x)dx=∑(n=1,到无穷)x^(2n+3)=x5+x7+x9+……=x^5(1+x²+x^4+……)=x^5*[1/(1-x²)]=x^5/(1-x²).再对两边求导得

x²f(x)=[4x^4(1-x²)-(-2x)*x^5]/(1-x²)²=[16x^4(1-x²)+2x^6]/(1-x²)²,因此f(x)=[16x²(1-x²)+2x^4]/(1-x²)²