Asked by xueqi
on 6 Jan 2013

Hi,

I am using the code below so solve function

syms x solve(1-x-x^0.3 == 0,x)

and MATLAB gives me the answer

2.2806043918081745336771958436901 0.301860807970099685692066514691 0.29115437161383716945904165867324 + 0.29287833041472217013239499780749*i 0.42827262660009503368869244743538 - 0.77711337147723605404039493650918*i 0.42827262660009503368869244743538 + 0.77711337147723605404039493650918*i 1.9090820154790144088601993088112 - 0.85475656146972727566334457636303*i 0.29115437161383716945904165867324 - 0.29287833041472217013239499780749*i 1.0802583864179162783074354058896 + 1.1426229105821451060150636481001*i 1.9090820154790144088601993088112 + 0.85475656146972727566334457636303*i 1.0802583864179162783074354058896 - 1.1426229105821451060150636481001*i

Clearly matlab sovle this problem numerically. The problem is that I find the first answer is wrong since I type

1-2.2806043918081745336771958436901-2.2806043918081745336771958436901^0.3

in the command code and get the result is

ans =

-2.5612

This is really bizzar! Is there anyone could tell me why? Really appreciated!

*No products are associated with this question.*

Answer by Roger Stafford
on 6 Jan 2013

That first answer is not as wrong (or bizarre) as you might think. When you present 'solve' with fractional powers you must expect some unusual behavior. In the case of that first value, 2.28060439...., if you interpret its 0.3 power as the cube of its tenth root, one of those tenth roots is -1.085937913... whose cube is -1.280604392... and that will satisfy your equation.

In order to achieve a solution to your original equation, very likely 'solve' converted your problem to the polynomial equation 1 - y^10 - y^3 = 0 with the substitution y = x^(1/10) and got ten y roots. Then it took each of these to the tenth power to obtain the x values and those ten values you received are the result. You can check that by giving that equation in y to 'solve' and then taking each solution to the tenth power.

Show 1 older comment

xueqi
on 8 Jan 2013

ha! Thanks! It clear things up! Do you also have any suggestions that how to avoid this kind of solution?

Walter Roberson
on 8 Jan 2013

Yes: do not use floating point exponents when you solve(); it is just going to confuse you and confuse the readers.

When x and a are specific floating point numbers, x^a is calculated as exp(ln(x)*a) . When x is negative and a is real, then this produces a complex-valued result.

When you use symbolic forms, x symbolic and a real, then unless you are careful about what you are doing, then for x^a, the a is converted to the rational fraction in lowest terms, c/d, leading to x^(c/d) . (c, d integer). This is then calculated algebraically (x^c)/(x^d) which is going to come out real-valued for negative x if "c" is even, a distinctly different value than if x^a had been calculated numerically.

You need to decide which of the several possible meanings you want for x^a when a is floating point.

Roger Stafford
on 8 Jan 2013

I would advise that rather than avoiding such an outcome as you received, Xueqi, the important thing is to understand it as a natural consequence of fractional exponents. As I think you might agree, the mystery you faced is not so mysterious when a logical explanation of it is provided. Often an equation or set of equations have more solutions than the user expects and the 'solve' function tries to find as many of them as it knows how, even though some of them may seem strange. Perhaps the real problem is that equations can turn out to be too difficult for 'solve' to find solutions for. After all, it is no "smarter" than the humans that created it.

Log in to comment.

Related Content

MATLAB and Simulink resources for Arduino, LEGO, and Raspberry Pi

Learn moreOpportunities for recent engineering grads.

Apply Today
## 0 Comments

Log in to comment.