# Homework Help: Polynomial approximation to find function values

1. Apr 1, 2014

### Maxo

1. The problem statement, all variables and given/known data
If we have the following data

Code (Text):
T = [296 301 306 309 320 333 341 349 353];
R = [143.1 116.3 98.5 88.9 62.5 43.7 35.1 29.2 27.2];
(where T = Temperature (K) and R = Reistance (Ω) and each temperature value corresponds to the resistor value at the same position)

2. Relevant equations
We know that
$$R=a \cdot e^{b/T}$$
The question of the problem is the following: How can we, from these data and the given equation, find the values of the constants a and b?

3. The attempt at a solution
In the assignment it is suggested one draws a diagram that shows ln(R/Ω) as a function of 1/T and then use MATLAB "polyfit" function which gives a polynom approximation of this line and then use that polynom to find the values of a and b.

I have used the polyfit function in MATLAB to get a polynom approximation, but how does that polynom give me the constants a and b?

Last edited: Apr 1, 2014
2. Apr 1, 2014

### jaytech

Polyfit outputs polynomial coefficients, so I think what you'd have to do is plot ln(R) = ln(a) + b/T, and look for b = 0. Now you will know what value a is. From there, divide your polyfit matrix by a, and you will have the coefficients to the power series of eb/T. With the power series representation, you may have to do some approximating to find a value for b, but I would imagine the coefficients won't change much after the second order term.

3. Apr 2, 2014

### Maxo

So MATLAB gives me this polynomial:

Code (Text):
1.0e+03 *

3.0412   -0.0053
which I assume means the same as a polynomial like
$$y=10^{3}\cdot 3.0412x-0.0053 = 3041.2x - 5.3$$
I assume then, that -5.3 is the value of ln(a)?

If I divide then the polynomial with -5.3 I get the following matrix:
Code (Text):
[-0.0018    1.0000]
Correct so far?

If so I wonder what does it means that these are the coefficients for the power series of e^(b/T)? I guess that should be a Maclaurin or Taylor series? How can I determine b from such a series?

4. Apr 2, 2014

### Staff: Mentor

If I were in your position, I would take logarithms of both sides of that equation, and plot a graph of your data. The y-intercept and the slope lead you to values for a and b. You will then know the approximate values for those constants, and can recognize when MATLAB is giving you nonsense.

5. Apr 2, 2014

### milesyoung

Right (I'd go with at least 3 significant figures).

You'd like to fit your data to:
$$\ln(R) = b \frac{1}{T} + \ln(a)$$
and MATLAB has provided you with:
$$\ln(R_\mathrm{fit}) = p_1 \frac{1}{T} + p_2$$
You're basically done at this point. How does $p_1$ and $p_2$ relate to $b$ and $a$, respectively?

6. Apr 2, 2014

### Maxo

Thanks, that's how I figured also. Onwards