Solving for a single unknown (non-linear equation)
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I apologize for my naivety, I am new to MatLab. I have this very long equation typed out and the unknown I need to solve for is "P"
6.2==1.414*(1-(0.548+0.71*(3.83/49.5)^4)*(1-(1-(P/194166))))*sqrt(P/1.184)*(((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(6/49.5)^4))^0.5))^0.5)*0.018241)+(3*((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(4/49.5)^4))^0.5))^0.5)*0.008107))+((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(3/49.5)^4))^0.5))^0.5)*0.00456)+((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(2/49.5)^4))^0.5))^0.5)*0.002027))
very messy, but it should be typed out correctly. The hard part has been done..
How can I get MatLab to solve P for me? I've tried
solve('that equation', P)
But it gives me an error, saying P is an unknown variable. Any help would be appreciated. I'm sure this is easy, I just have no idea what the command is to solve an equation. Or what must be done to any variables you plan on using (like P, here)
Thank you!
回答（4 个）
G A
2012-3-12
syms P
solve('6.2=1.414*(1-(0.548+0.71*(3.83/49.5)^4)*(1-(1- (P/194166))))*sqrt(P/1.184)*(((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(6/49.5)^4))^0.5))^0.5)*0.018241)+(3*((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(4/49.5)^4))^0.5))^0.5)*0.008107))+((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(3/49.5)^4))^0.5))^0.5)*0.00456)+((0.9975-0.00653*(10^6/(70900*49.5*((1.184*P)/(1-(2/49.5)^4))^0.5))^0.5)*0.002027))',P)
ans =
408270.34774603219832835714795887
290297.52311957774144902012949622
290318.2813236540219441455483253
10023.461083668597932042362965426
408256.9318415621052355263787515
10037.722237649978451175667443743
10030.586416547686816457207201435 + 7.1305756978312245604932980434246*i
10030.586416547686816457207201435 - 7.1305756978312245604932980434246*i
290307.90584288889227473668488892 + 10.379101358088288664349740315312*i
290307.90584288889227473668488892 - 10.379101358088288664349740315312*i
408263.63817095856098494380133019 - 6.7079520439268252489438119583435*i
408263.63817095856098494380133019 + 6.7079520439268252489438119583435*i
3 个评论
Walter Roberson
2012-3-13
The computation I am doing is taking a long time, but the structure of it is suggesting to me that the solution will involve polynomials of at least degree 16 and possibly higher. Could be multiple roots though.
Walter Roberson
2012-3-13
G A, when I plot the equations, I show no sign of double roots near 10000 or 290000 or 408250 -- only single roots.
G A
2012-3-13
Walter, I have just demonstrated to Anthony that function solve() works giving some result - no matter is solution meaningless or not. However, your are right. I have not checked equation graphically or algebraically as you have done.
Walter Roberson
2012-3-13
1 个投票
Anthony, your equation is numerically fairly meaningless. You are creating polynomials of at least degree 16 because of the ^4 expressions. The roots (especially the transition between real and complex) of such polynomials are usually quite sensitive to the exact coefficients of the polynomials. You do not, however, have meaningful exact coefficients: you use a mix of floating point values varying from 1 significant place to 5 significant places. Logically you are getting major round-off in that calculation, so the high-degree polynomials that result are near meaningless except as examples of a wide range of coefficient values that are within the round-off of the figures you give.
In some places in your equation, you use sqrt(), but in other places you use ^0.5 . The two are not equivalent in algebra. In algebra, it is not known whether the 0.5 is the representation of 1/2 or of 5/10 or of 10/20 and so on -- interpretations that can affect the sign of the result and can affect which complex root is chosen. Using floating point values as a power causes problems for algebraic solutions, especially when negative values are being taken to the power (which can happen incidentally due to a rewriting of the expression to make it easier to calculate.)
1 个评论
Bjorn Gustavsson
2012-3-13
In addition to that: 1.414...
...might be a very short version of sqrt(2)...
Honglei Chen
2012-3-12
Move 6.2 to the right side of the equation and then use fzero.
doc fzero
for example, to solve x+1 = 2
fzero(@(x)x-1,0)
Anthony
2012-3-12
4 个评论
Sean de Wolski
2012-3-12
Hope isn't lost. You have to figure out what youw ant to do with the square root of a negative number. Do you want to keep just the real part (doc real) or the magnitude (doc abs)?
Sean de Wolski
2012-3-12
Or, does it mean you have an error since you shouldn't be able to have a negative?
Anthony
2012-3-12
Honglei Chen
2012-3-12
Try change the initial guess. I'm using 0 for the initial guess but you can use a different number so it is not easy to fall into negative P region.
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