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汕头大学海洋科学接受调剂
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chaofan1231

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[求助] matlab解代数方程的问题

哪位大侠帮忙看下这是怎么回事啊?谢谢啊
y=2;
>> subs(solve('y=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1))','x'))
Warning: Explicit solution could not be found.
我把y=2直接代进去的话solve('2=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1))','x')
ans =matrix([[14197.808632879686391435779145547]]),但是我在origin中将这个方程画成曲线的话,y=2时x的两个值分别在1100和1900附近,这是怎么回事啊?
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csgt0

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感谢参与,应助指数 +1
三个都是啊,我画出来3个点附近都接近2
showmethemoney
2楼2012-05-14 10:01:04
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csgt0

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忘传图乐


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3楼2012-05-14 10:01:30
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chaofan1231

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引用回帖:
3楼: Originally posted by csgt0 at 2012-05-14 10:01:30:
忘传图乐
5d/87/449046_1336960887_743.jpg

是这样的,但是为什么matlab中解出来只有一个解?我想要的是前两个。
4楼2012-05-15 08:05:50
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csgt0

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chaofan1231: 金币+5, ★★★★★最佳答案 2012-05-15 14:53:43
chaofan1231: 回帖置顶 2012-05-15 14:54:44
>> evalin(symengine, 'numeric::solve(2=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1)), x = 1000..2000)')

ans =

1092.7957902606343947864233870012

>> evalin(symengine, 'numeric::solve(2=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1)), x = 1500..2000)')

ans =

1826.2296709384982930177909403237

Solve the following equation:

syms x;
solve('sin(x) = x^2 - 1')

The symbolic solver cannot find an exact symbolic solution for this equation, and therefore, it calls the numeric solver. Because the equation is not polynomial, an attempt to find all possible solutions can take a long time. The numeric solver does not try to find all numeric solutions for this equation. Instead, it returns only the first solution that it finds:

ans =
-0.63673265080528201088799090383828

Plotting the left and the right sides of the equation in one graph shows that the equation also has a positive solution:

ezplot(sin(x), -2, 2);
hold on;
ezplot(x^2 - 1, -2, 2)
hold off

You can find this solution by calling the MuPAD numeric solver directly and specifying the interval where this solution can be found. To call MuPAD commands from the MATLAB Command Window, use the evalin or feval function:

evalin(symengine, 'numeric::solve(sin(x) = x^2 - 1, x = 0..2)')

ans =
1.4096240040025962492355939705895

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5楼2012-05-15 09:18:57
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chaofan1231

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引用回帖:
5楼: Originally posted by csgt0 at 2012-05-15 09:18:57:
>> evalin(symengine, 'numeric::solve(2=exp((1/0.0000005461-1/0.000000700)*0.014388/x+ln((x-993.74731)/2869.03078)-5*ln(700/546.1)), x = 1000..2000)')

ans =

1092.7957902606343947864233 ...

6楼2012-05-15 14:54:04
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