# Finding roots of 4 order poynomial equation for PT100 transfer function

#### aamirali

Joined Feb 2, 2012
412
PTD100 has following equation:

RTD(t) = R0 [1 + A*t + B*t*t + C (t  100°C)t*t*t ] t<= 0
RTD(t) = R0 [1 + A*t + B*t*t ] t>= 0

I have to calculate T from known value of R. For that I have to find inverse transfer function.

Roots of second equation are easy to find as it is quadratic.
But can we find roots of first equation also as it is of fourth order.

However for project, I have used best fit polynomial method & can calculate value of T from known R.

This is just for knowledge that if exact inverse transfer can be calculated for 4th order polynomial.

#### WBahn

Joined Mar 31, 2012
30,303
1) How do you figure that the first equation is fourth order?

2) In general you aren't guaranteed that even a second order polynomial has an inverse function, let alone a higher-order polynomial.

3) The point made above can be ignored if you are confident that the forward function is strictly monotonic, which is probably a reasonable assumption in this case.

4) Just as you use the Quadratic Formula to find the roots of a quadratic equation, you might see if there is such a thing as the Cubic Formula to find the roots of cubic equation.

#### MrChips

Joined Oct 2, 2009
31,093
What temperature range are you interested in?

#### The Electrician

Joined Oct 9, 2007
2,973
PTD100 has following equation:

RTD(t) = R0 [1 + A*t + B*t*t + C (t  100°C)t*t*t ] t<= 0
RTD(t) = R0 [1 + A*t + B*t*t ] t>= 0

I have to calculate T from known value of R. For that I have to find inverse transfer function.

Roots of second equation are easy to find as it is quadratic.
But can we find roots of first equation also as it is of fourth order.

However for project, I have used best fit polynomial method & can calculate value of T from known R.

This is just for knowledge that if exact inverse transfer can be calculated for 4th order polynomial.
If by "exact inverse transfer", you mean something analogous to the quadratic formula for a 2nd order polynomial equation, the answer is yes, such a formula exists, but you wouldn't want to use it:

http://mathworld.wolfram.com/QuarticEquation.html

#### Papabravo

Joined Feb 24, 2006
21,322
Generally speaking you will want to use numerical methods for polynomial of higher order than two. Your chances of manually executing the complex steps of Cardano's method for cubics, or The Electricians link for the quartic are dim. The good news is that Matlab and the freeware clones Scilab or Octave will find the roots for you.

If you insist on doing it the manual way I can recommend Newton's Method as a first order approach. For a cubic equation you can take the derivative, solve that for the local minimum and local maximum points. The value of the original at the points of zero slope will tell you where to look for the one root that must be real. The other two roots must also be real or be a complex conjugate pair.

#### amilton542

Joined Nov 13, 2010
497
Intuition can solve a lot of polynomial problems. If I was to say, find the pure values of x that makes y = (4x^4 - 4x^2 + 1) equate to zero, how will you do it?

You've avoided the calculus if you figure it out.

#### Papabravo

Joined Feb 24, 2006
21,322
Could it be?
Perhaps it is...substitution....
x^2 = u

#### amilton542

Joined Nov 13, 2010
497
You could, but I said intuition and you've avoided the calculus if you figure it out.

#### WBahn

Joined Mar 31, 2012
30,303
Intuition can solve a lot of polynomial problems. If I was to say, find the pure values of x that makes y = (4x^4 - 4x^2 + 1) equate to zero, how will you do it?

You've avoided the calculus if you figure it out.
Could it be?
Perhaps it is...substitution....
x^2 = u
You could, but I said intuition and you've avoided the calculus if you figure it out.
What is a "pure" value of x? What makes a value "impure"?

As Papabravo suggested, the "intuition" is recognizing that x always appears in the form (x^2)^n and so using the substitution u = x^2 makes it obvious that you have a simple quadratic. After that you can simply use the quadradic equation to solve for them or recognize that you have an easily factored quadratic polynomial that is, in fact, a perfect square. So you have a single repeated root. Thus, for the problem given you have two repeated roots. No calculus anywhere in site. But also an example of a contrived situation.

#### amilton542

Joined Nov 13, 2010
497
The sub' is a generic protocol. I hint not using calculus because you actually compute the location of the turning points without taking the derivative with respect to x hence the pure standpoint and not a numerical one.

#### WBahn

Joined Mar 31, 2012
30,303
The sub' is a generic protocol. I hint not using calculus because you actually compute the location of the turning points without taking the derivative with respect to x hence the pure standpoint and not a numerical one.
I'm not following. Maybe it's just terminology. What do you mean by "turning points"? How would taking a derivative, even if you chose to, make it a numerical approach? Yes, using Newton-Raphson would be a numerical approach, but the use of a derivative isn't what makes it so.

#### amilton542

Joined Nov 13, 2010
497
O.K, maybe I'm not being too clear.

The quadratic I've posted is a nit-pick that's remained with me ever since.

I was asked to compute the values of x that make it equate to zero. For sure, I could have used Pappa B's u substitution, but I decided to do something different.

In doing so, I actually found the co-ordinates where maxima and minima occur, without the calculus! But, a pure coincidence.

(I do apologise, "turning points" is a bad habit of a nomenclature I've picked up for critical points.)

Let,

$$4x^4 - 4x^2 + 1 = 0$$

$$4x^2(x^{2} - 1) = -1$$

Now,

"something" x "something else" = -1

Either,

1 x -1 = -1

Or,

-1 x 1 = -1

If,

$$4x^2 = -1$$

Then,

$$x = \frac {(-1)^\frac{1}{2}}{4^\frac{1}{2}}$$

But,

x is defined over the real domain R.

So,

$$x = 4x^2 = 1$$

$$x = \pm \frac{1}{\sqrt 2}$$

From which,

$$x^2 -1 = -1$$

Which leaves,

$$x = \pm 0 = 0$$

Then,

$$f(\pm \frac{1}{\sqrt 2}) = 0$$

and,

$$f(0) = 1$$

#### WBahn

Joined Mar 31, 2012
30,303
Okay.... so f(0) = 1. So what? That's trivially obvious from the original equation.

What is your basis for claiming that you have found the minima and maxima?

That f(0) is one or the other is required by the fact that x only appears a x^2, but do you see why that is the case?

Why do you claim that the zeroes are extrema of the function?

What would your "method" have concluded about the minima and maxima for the following just slightly different equations:

y2 = 4X^4 - 2X^2 + 1 = 0

y3 = 4X^4 - 8X^2 + 1 = 0