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(1%1/2)x(1+1/3)x(1%1/4)x(1+1/5)x......x(1%1/1998)

(1+1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)x(1/2+1/3+1/4) =(1/2+1/3+1/4+1/5)+(1/2+1/3+1/4)×(1/2+1/3+1/4+1/5)-(1/2+1/3+1/4)-(1/2+1/3+1/4+1/5)×(1/2+1/3+1/4) =(1/2+1/3+1/4+1/5)-(1/2+1/3+1/4) =1/5

设 (1+1/2+1/3+1/4)=a (1/2+1/3+1/4)=b 则a-b=1 把a,b代入上面式子得 a(b+1/5)-(a+1/5)b =ab+1/5a-ab-1/5b =1/5(a-b) =1/5 (上面的1/6打错,应该是1/5)

(1+2/3)x(1-2/3)x(1+2/5)x(1-2/5)x...x(1+2/99)x(1-2/99) =(1-2/3)x(1+2/3)x(1-2/5)x(1+2/5)x(1-2/7)x(1+2/7)...x(1-2/97)x(1+2/97)x(1-2/99)x(1+2/99) =1/3 x 5/3 x 3/5 x 7/5 x 5/7 x 9/7 ... x 95/97 x 99/97 x 97/99 x 101/99 = 1/3 x 1 x 1...

(x+3)/(x+2)+(x+4)/(x+3)=(x+5)/(x+4)+(x+2)/(x+1) 1+1/(x+2)+1+1/(x+3)=1+1/(x+4)+1+1/(x+1) x+3+x+2/(x+2)(x+3)=x+1+x+4/(x+4)(x+1) 2x+5/(x+2)(x+3)=2x+5/(x+4)(x+1) (1)2x+5不为0 (x+2)(x+3)=(x+4)(x+1) x^2+5x+...

1/(x+2)-1/(x+3)-1/(x+4)+1/(x+5)=0 1/(x+2)-1/(x+3)=1/(x+4)-1/(x+5) 1/(x+2)(x+3)=1/(x+4)(x+5) (x+2)(x+3)=(x+4)(x+5) x²+5x+6=x²+9x+20 4x=-14 x=-3.5

分析: 用循环来实现累加,循环变量取值:1,2,3,...每次循环加1 循环结束条件=当前项绝对值=1e-5 每项的分子:1,2,3,...就是循环变量取值 每项的分母:从第2项开始是1,x,x*x,x*x*x,...是前一项的分母再乘以x得到 每项的符号:1,-1,1...

程序如下,运行如果在最后。 clear,close all,x=-10:0.02:10; % x=-10:0.5:10 N=300; % N=input('please input N='),f=1/2;for n=1:N f=f+2/pi/(2*n-1)*sin((2*n-1)*x);end % f,plot(f)

设(1-1/2-1/3-1/4-1/5)=X;(1/2+1/3+1/4+1/5)=Y; 原式=X*(Y+1/6)-(X-1/6)Y=XY+X*1/6-XY+Y*1/6=1/6*(X+Y)=1/6*1=1/6

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