\section*{Exercise 2 - 031212}

During the calibration of a dynamometer, the load \(P\) required to deflect the pointer to certain divisions on the scale was measured. The following values were obtained:

\begin{tabular}{cc} 
Number of divisions & Load in kp \\
\(N\) & \(P\) \\
\hline 0 & 0 \\
5 & 4.87 \\
10 & 10.52 \\
15 & 17.24 \\
20 & 25.34
\end{tabular}

The load \(P\) can be represented by the following rational function of \(N\):

\[
P(N)=a_{1} N+a_{2} N^{2}+a_{3} N^{3}+a_{4} N^{4}
\]

a) The coefficients \(a_{1}, a_{2}, a_{3}, a_{4}\) are to be calculated!

b) What is the value of the function for \(N=25\)? Compare it with the measured value \(P=35.16 \, \text{kp}\)!
\footnotetext{
*) A dynamometer is a device for measuring forces, where the elastic deformation of a spring is transmitted via a lever mechanism to a (usually circular) scale (spring balance).
}
}
a) It can be observed that the polynomial relationship set between the number of segments \(N\) and the load \(P(N)\) simplifies when one takes the differences between successive values in the value table. Specifically, the polynomial degree and thus the number of undetermined parameters are reduced by one.

The differences between these differences reduce the polynomial degree by another one. This process of forming differences can be continued until only one coefficient remains.

To make the calculations easier, the values are normalized so that only decimal fractions without periods are obtained.

\[
\begin{aligned}
& f_{0}(x):=\frac{3}{5 x} \cdot P(5 x)=3 \cdot\left(a_{1}+5 a_{2} x+25 a_{3} x^{2}+125 a_{4} x^{3}\right)=: b_{0}+b_{1} x+b_{2} x^{2}+b_{3} x^{3} \\
& f_{1}(x) \quad:=f_{0}(x+1)-f_{0}(x)=b_{1}+b_{2}(2 x+1)+b_{3}\left(3 x^{2}+3 x+1\right) \\
& =b_{1}+b_{2}+b_{3}+\left(2 b_{2}+3 b_{3}\right) x+3 b_{3} x^{2}=: c_{0}+c_{1} x+c_{2} x^{2} \\
& f_{2}(x):=f_{1}(x+1)-f_{1}(x)=c_{1}+c_{2}(2 x+1)=c_{1}+c_{2}+2 c_{2} x=: d_{0}+d_{1} x \\
& f_{3}(x):=f_{2}(x+1)-f_{2}(x)=: d_{1} \\
& \begin{array}{lllll}
x & f_{0}(x) & f_{1}(x) & f_{2}(x) & f_{3}(x)
\end{array} \\
& \begin{array}{llllll}
1 & 2.922 & 0.234 & 0.058 & 0.003 &
\end{array} \\
& \begin{array}{llllll}
2 & 3.156 & 0.292 & 0.061
\end{array} \\
& \begin{array}{lllll}
3 & 3.448 & 0.353
\end{array} \\
& \begin{array}{llll}
4 & 3.801 & 1-10 \\
1
\end{array}
\end{aligned}
\]

From this, one can step by step obtain the coefficients of all polynomials:

\[
\begin{aligned}
& d_{1}=f_{3}(1) \\
& =0.003 \\
& d_{0}=f_{2}(1)-1 \cdot d_{1} \\
& =0.055 \\
& c_{2}=d_{1} / 2 \\
& =0.0015 \\
& c_{1}=d_{0}-c_{2} \\
& =0.0535 \\
& c_{0}=f_{1}(1)-1 \cdot c_{1}-1^{2} \cdot c_{2} \\
& =f_{1}(1)-d_{0}=0.179 \\
& b_{3}=c_{2} / 3 \\
& =0.0005 \\
& b_{2}=\left(c_{1}-3 b_{3}\right) / 2 \\
& =\left(c_{1}-c_{2}\right) / 2=0.026 \\
& b_{1}=c_{0}-b_{2}-b_{3} \\
& =0.1525 \\
& b_{0}=f_{0}(1)-1 \cdot b_{1}-1^{2} \cdot b_{2}-1^{3} \cdot b_{3}=f_{1}(1)-c_{0}=2.743
\end{aligned}
\]

![](https://cdn.mathpix.com/cropped/2024_06_06_b85a894aa8dbf722a7b3g-1761.jpg?height=458&width=807&top_left_y=228&top_left_x=670)
b) \(\quad P(25)=\frac{25}{3} \cdot f_{0}(5)=\frac{25}{3} \cdot(((0.0005 \cdot 5+0.026) \cdot 5+0.1525) \cdot 5+2.743)\)

\[
\begin{aligned}
& =\frac{25}{3} \cdot((0.0285 \cdot 5+0.1525) \cdot 5+2.743)=\frac{25}{3} \cdot(0.295 \cdot 5+2.743)=\frac{25}{3} \cdot 4.218 \\
& =25 \cdot 1.406=35.15
\end{aligned}
\]