I Rearranging equation with more than variable instance

[Jehannum]

This question is no doubt absurdly simple to many here, but the answer will help me immensely.

Say that:

c = a ^ 4

Then it's simple to rearrange for a:

a = c ^ 0.25

But what if there is more than one term, for example:

c = a ^ 4 + a ^ 3

Is it possible to rearrange this to the form:

a = something?

 

[BvU]

No it is not. At least not in general. There may be special cases, but they are rare.

Note that in your first example, $a = -c^{0.25}$ also satisfies the equation !

 

[mfb]

Up to quartic polynomials, it is possible, but only with messy formulas. Beyond that, there is no proper solution in general.

 

[Jehannum]

Thank you for your answers.

Now, may I generalise my question so that it touches on engineering and physics?

If a relationship between physical quantities exists and can be expressed as a function:

a = f1 (x)

and we wish to obtain x in terms of a, i.e.:

x = f2 (a)

what do we do if f1 (x) cannot be rearranged algebraically to return x, as the answers above have indicated?

Is there a general method used by engineers and physicists to obtain a serviceable approximation of f2 so that x can be determined in terms of a?

It seems to me that this must be a common problem, since it is one that I have come across in a particular application I am trying to develop.

​

 

[BvU]

It happens all the time and is called equation solving ...
If you reveal a little more about your specific case perhaps we can say something about usable methods ?
(for example: every control problem involves a form of inverting a transfer function)

 

[Jehannum]

Yes, I can certainly give details of the problem I'm working on. It's to do with pressure loss in low pressure gas pipes.

The equation I have to work with is:

Q = K1 . [p (d ^ 5) / (s L f)] ^ 0.5

where Q = flow rate, K1 = a constant, p = pressure loss, d = pipe diameter, s = relative density, L = pipe length and f = friction factor​

I want to rearrange this equation for d. This would of course be simple were it not for the fact that:

f = K2 . (1 + K3 / d)

where K2 and K3 are constants
​

So, you see, d is actually in the numerator and denominator of the function for Q. This means it is beyond my mathematical ability to rearrange into the form d = function (Q). As far as I know, it may not be possible to do so anyway.

I have constructed an iterative algorithm which returns d, but it is not suitable for human computation. It works by trying successively larger values for d until the flow rate is sufficient with an acceptable pressure loss.

I would like to know if there is a way I can present the relationship in a human-computable (preferably non-iterative) form.

 

[mfb]

There is no solution in closed form. No finite number of calculations will give you an exact result, so iterations are the best you can do to find an approximate solution.

The best approach will depend on typical values of K3/d (only this ratio matters). Is it small? Is it large? Is it close to 1 (worst case)?

Newton's method can be useful.

 

[pasmith]

Setting x = d/K_3 and squaring both sides yields <br /> \frac{Q^2 sLK_2}{K_1^2 pK_3^5} = \frac{x^6}{1 + x}. Thus you need to solve <br /> y = \frac{x^6}{1 + x} for x. That in principle must be done numerically, but you need only tabulate the value of x for sufficient values of y and x could then be found for other values of y by interpolation - and then d = K_3 x.

Alternatively you can graph \frac{x^6}{1 + x} against x and for a given value of y = \frac{Q^2 sLK_2}{K_1^2 pK_3^5} on the vertical axis one can move across to the curve and then down to the horizontal axis to find x = d/K_3.

 

[BvU]

Note that this Spitzglass equation is one of many design equations which have limitations . A high degree of accuracy in doing the calculations may not be justified. I agree with mfb that a doing few Newton steps would be the a very sensible approach, but if your trial method works fast enough, then why bother.

 

[Jehannum]

Thank you, guys.

BvU, you are probably right about the degree of accuracy. In any case, commercially-available steel pipework comes in well-separated discrete sizes, not a continuum, so the ideal diameter d 'snaps' to the next available size up anyway.

The information in this thread has been useful as I am trying to increase my knowledge of applied maths in general.

 

[Jehannum]

There is no solution in closed form. No finite number of calculations will give you an exact result, so iterations are the best you can do to find an approximate solution.

The best approach will depend on typical values of K3/d (only this ratio matters). Is it small? Is it large? Is it close to 1 (worst case)?

Newton's method can be useful.

Just one more thing ... in general, how does one determine whether there is a solution in closed form?

 

[mfb]

Just one more thing ... in general, how does one determine whether there is a solution in closed form?

This can be a complicated problem. Proving that quintic equations don't have a general solution was an open problem until 1824, while solutions for lower powers were known in the 16th century already. There are two typical cases indicating that no solution in closed form exists:

- polynomials with a degree of at least 5
- everything like $x e^x$ or $x \log (x)$