[求助]怎样求和含有未知数的表达式？

[zhyanchao]

以m=1:5，求和的值为1为例：
syms x;
m=1:5;
s=sum((x-m)/(x-0.5))
solve(f-1)
结果
s=(x-1)/(x-1/2)+(x-2)/(x-1/2)+(x-3)/(x-1/2)+(x-4)/(x-1/2)+(x-5)/(x-1/2)
ans =

29/8

[evaline.duan]

Thanks a lot. 自己试着编写了类似程序：
syms x;
y=0;
p=0;
syms x;
while (p<10)
p=p+1;
y=y+trimf1(x,[p p+1])
end;
solve(y-0.25)，显示出每一步y的表达式为：
y =

(x-2)/(x-1)

y =

(x-2)/(x-1)+(x-3)/(x-2)

y =

(x-2)/(x-1)+(x-3)/(x-2)+(x-4)/(x-3)

y =

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

y =

(x-2)/(x-1)+(x-3)/(x-2)+(x-4)/(x-3)+(x-5)/(x-4)+(x-6)/(x-5)

y =

(x-2)/(x-1)+(x-3)/(x-2)+(x-4)/(x-3)+(x-5)/(x-4)+(x-6)/(x-5)+(x-7)/(x-6)

y =

(x-2)/(x-1)+(x-3)/(x-2)+(x-4)/(x-3)+(x-5)/(x-4)+(x-6)/(x-5)+(x-7)/(x-6)+(x-8)/(x-7)

y =

(x-2)/(x-1)+(x-3)/(x-2)+(x-4)/(x-3)+(x-5)/(x-4)+(x-6)/(x-5)+(x-7)/(x-6)+(x-8)/(x-7)+(x-9)/(x-8)

y =

(x-2)/(x-1)+(x-3)/(x-2)+(x-4)/(x-3)+(x-5)/(x-4)+(x-6)/(x-5)+(x-7)/(x-6)+(x-8)/(x-7)+(x-9)/(x-8)+(x-10)/(x-9)

y =

(x-2)/(x-1)+(x-3)/(x-2)+(x-4)/(x-3)+(x-5)/(x-4)+(x-6)/(x-5)+(x-7)/(x-6)+(x-8)/(x-7)+(x-9)/(x-8)+(x-10)/(x-9)+(x-11)/(x-10)
输入solve命令：solve(y-0.25)后，结果却显示：
x =

1.0786898967617613085887798657648
2.0855661586263124742606885245834
3.0901392962871015446663250353338
4.0940124115549277509671954064637
5.0977113712866084211968191987618
6.1015696954689081199203797592360
7.1059563518414547306012042485634
8.1115164312857313681455235886801
9.1199593222558947264347081304284
10.140520090272325196244017267826
明显和方程式x的求解值不符么，按计算，y的求解值x不会这么多的吧，应该只有一个的么，怎么会有这么多呢？不知matlab是怎么计算的，为什么计算出来不是我想要的答案呢？
恳请高手指点！:lovely:

[evaline.duan]

忘记说明了,trimf1(x,[a b])=(x-b)./(x-a)
谢谢！

[zhyanchao]

上面的求解结果是正确的。你把y的表达式通分以后，分子就是一个十次方程，所以它的解是应该有十个的。

[evaline.duan]

Make sense! Thanks a lot!

[xiezhh]

我认为没有必要用循环，还可以这样
syms x m
f=symsum((x-m)/(x-0.5),m,1,5);
x=solve(f-1,x)

结果：
x = 29/8

[evaline.duan]

果然是高手，谢谢提示！