讨论1(1 x) (1 x的2n次方)的连续性
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/22 01:49:26
n超过20,开始不稳定.用数值解法比较好.%解析方法clc;clearsymsxn=1:50f=[x.^n/(x+5)]'I=int(f,'x',0,1)I=eval(I)plot(n,I,'o-')
x的n次方乘x的n+1次方+x的n+1次方乘x的n-2的次方+(-x)的3次方乘(-x)的2n-4次方=x^(2n+1)+x^(2n-1)-x((2n-1)=x^(2n+1)-2x^(2n-1)
x^(3n+1)*x+3x^(n+1)*x^(2n+1)=x^(3n+2)+3x^(n+1+2n+1)=x^(3n+2)+3x^(3n+2)=4x^(3n+2)
【(x的n+1次方)的四次方*x²】/【(x的n+2次方)的³/(x²)的n次方=x的4n+4×x²÷(x的3n+6次方)×x的2n次方=x的4n+6次方÷(x
x^3n*x+3x^(n+1)*x^(2n+1)=x^(3n+1)+3*x^(2n+2)=x^(3n+1)*(1+3x)再问:问题补充!!!再答:x^(3n+1)*x+3x^(n+1)*x^(2n+1
当|x|1时,f(x)=-x,作图知为跳跃间断点
1.首先他是关于n的偶函数,所以分析一边的情况就可以了.2.关于x^2n,(n→+∞),分界点是1,所以当x>1时【也即x→(1+0)】,x^2n=+∞,lim(n→+∞)f(x)=-1;当x
F(x)=lim(n→∞)x*(1+x^2n)/(1-x^2n)whenx=1or-1F(x)isundefinedF(x)在x=1or-1不连续if|x|1lim(n→∞)x*(1+x^2n)/(1
x的n+1次方-1即x^(n+1)-1[x^n+x^(n-1)+...+x+1](x-1)=[x^n+x^(n-1)+...+x+1]x-[x^n+x^(n-1)+...+x+1]=[x^(n+1)+
n为偶数时:2*x^(n+2)n为奇数时:0
原式=-x^(3+2n-1)-x^(2n+2)=-x^(2n+2)-x^(2n+2)=-2x^(2n+2)
2x*(x^n+2)=2x^(n+1)-42x^(n+1)+4x=2x^(n+1)-44x=-4x=-1
(-x)^3*x^(n-1)+x^2n*(-x)^2=-x^3*x^(n-1)+x^2n*x^2=-x^(3+n-1)+x^(2n+2)=-x^(n+2)+x^(2n+2)=-x^(n+2)+x^n*
可以用等比公式啊公比q=x利用公式Sn=a1(1-q^n)/(1-q)x的1次方+x的2次方+.+x的n次方=x(1-x^n)/(1-x)同时,注意,当x=1时,x的1次方+x的2次方+.+x的n次方
因为x的m-1次方+1>0,3x的2n次方-2x
2x^n-6x^(n-1)-8x^(n-2)=2x^(n-2)(x^2-3x-4)=2x^(n-2)(x-4)(x+1)
∵x的3n次方-x的2n次方+1=0∴x的3n次方=x的2n次方-1∴x的5n次方+x的n次方+2013=x的3n次方*x的2n次方+x的n次方+2013=(x的2n次方-1)*x的2n次方+x的n次
3x^(2n-1)-12x^(2n+1)=3x^(2n-1)(1-4x^2)=3x^(2n-1)(1+2x)(1-2x)