作业帮数列求和极限问题已知f[1] = 1/3; f[2] = 2/27f[n] = 2/3*(f[1]*f[n-1] + f[2]*f[n-2] + ...+ f[n-1]*f[1])求(f[1]+f[2]+...+f[n])当n趋于无穷大的极限.请问能否求出f[n]的通项公式或者s[n]的呢?f[n] = 2*(f[1]*f[n-1] + f[2]*f[n-2
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![数列求和极限问题已知f[1] = 1/3; f[2] = 2/27f[n] = 2/3*(f[1]*f[n-1] + f[2]*f[n-2] + ...+ f[n-1]*f[1])求(f[1]+f[2]+...+f[n])当n趋于无穷大的极限.请问能否求出f[n]的通项公式或者s[n]的呢?f[n] = 2*(f[1]*f[n-1] + f[2]*f[n-2](/uploads/image/z/14277263-23-3.jpg?t=%E6%95%B0%E5%88%97%E6%B1%82%E5%92%8C%E6%9E%81%E9%99%90%E9%97%AE%E9%A2%98%E5%B7%B2%E7%9F%A5f%5B1%5D+%3D+1%2F3%3B+f%5B2%5D+%3D+2%2F27f%5Bn%5D+%3D+2%2F3%2A%28f%5B1%5D%2Af%5Bn-1%5D+%2B+f%5B2%5D%2Af%5Bn-2%5D+%2B+...%2B+f%5Bn-1%5D%2Af%5B1%5D%29%E6%B1%82%28f%5B1%5D%2Bf%5B2%5D%2B...%2Bf%5Bn%5D%29%E5%BD%93n%E8%B6%8B%E4%BA%8E%E6%97%A0%E7%A9%B7%E5%A4%A7%E7%9A%84%E6%9E%81%E9%99%90.%E8%AF%B7%E9%97%AE%E8%83%BD%E5%90%A6%E6%B1%82%E5%87%BAf%5Bn%5D%E7%9A%84%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%E6%88%96%E8%80%85s%5Bn%5D%E7%9A%84%E5%91%A2%3Ff%5Bn%5D+%3D+2%2A%28f%5B1%5D%2Af%5Bn-1%5D+%2B+f%5B2%5D%2Af%5Bn-2)
数列求和极限问题已知f[1] = 1/3; f[2] = 2/27f[n] = 2/3*(f[1]*f[n-1] + f[2]*f[n-2] + ...+ f[n-1]*f[1])求(f[1]+f[2]+...+f[n])当n趋于无穷大的极限.请问能否求出f[n]的通项公式或者s[n]的呢?f[n] = 2*(f[1]*f[n-1] + f[2]*f[n-2
数列求和极限问题
已知f[1] = 1/3; f[2] = 2/27
f[n] = 2/3*(f[1]*f[n-1] + f[2]*f[n-2] + ...+ f[n-1]*f[1])
求(f[1]+f[2]+...+f[n])当n趋于无穷大的极限.
请问能否求出f[n]的通项公式或者s[n]的呢?
f[n] = 2*(f[1]*f[n-1] + f[2]*f[n-2] + ...+ f[n-1]*f[1])/3
数列求和极限问题已知f[1] = 1/3; f[2] = 2/27f[n] = 2/3*(f[1]*f[n-1] + f[2]*f[n-2] + ...+ f[n-1]*f[1])求(f[1]+f[2]+...+f[n])当n趋于无穷大的极限.请问能否求出f[n]的通项公式或者s[n]的呢?f[n] = 2*(f[1]*f[n-1] + f[2]*f[n-2
你只要能看到卷积形式应该对应于多项式乘法就能做出来.
考察母函数g(x)=f[1]x+f[2]x^2+...f[n]x^n+...,条件相当于2/3*g^2=g-x/3,解出g(x)=3/4*(1-sqrt(1-8/9*x)),幂级数的收敛半径为9/8,所以1在收敛圆内,直接得到s[n]的极限就是g(1)=1/2.s[n]的通项可能没什么很简洁的形式.
电灯剑客的卷积用的好
但是g(1)=1/2或1。选哪个好呢？
这题有难度...
题没写清楚
你那是三分之二乘(f[1]*f[n-1] + f[2]*f[n-2] + ... + f[n-1]*f[1]) 还是2除以3*(f[1]*f[n-1] + f[2]*f[n-2] + ... + f[n-1]*f[1])