Question

Some modeling considerations have mandated a search for a function $$ u(x)=\gamma_{0} e^{\gamma_{1} x+\gamma_{2} x^{2}} $$ where the unknown coefficients \(\gamma_{1}\) and \(\gamma_{2}\) are expected to be nonpositive. Given are data pairs to be interpolated, \(\left(x_{0}, z_{0}\right),\left(x_{1}, z_{1}\right)\), and \(\left(x_{2}, z_{2}\right)\), where \(z_{i}>0, i=0,1,2 .\) Thus, we require \(u\left(x_{i}\right)=z_{i}\) The function \(u(x)\) is not linear in its coefficients, but \(v(x)=\ln (u(x))\) is linear in its. (a) Find a quadratic polynomial \(v(x)\) that interpolates appropriately defined three data pairs, and then give a formula for \(u(x)\) in terms of the original data. [This is a pen-and-paper item; the following one should consume much less of your time.] (b) Write a script to find \(u\) for the data \((0,1),(1, .9),(3, .5)\). Give the coefficients \(\gamma_{i}\) plot the resulting interpolant over the interval \([0,6] .\) In what way does the curve beh qualitatively differently from a quadratic?

Short answer

Question: Determine the function \(u(x)\) that has the form \(u(x) = \gamma_0 e^{\gamma_1x + \gamma_2x^2}\), given the interpolation problem: \(u(0)=1, u(1)=0.9, u(3)=0.5\), and discuss the qualitative difference between the resulting curve and a quadratic curve. Answer: The function \(u(x)\) is given by $$u(x) = e^{a + bx + cx^2}$$ where a, b, and c are the coefficients found using the steps provided. The interpolant yields a qualitative difference compared to a quadratic curve, as it has an overall exponential decay shape due to the exponential term, which a quadratic curve does not exhibit.

Step-by-step solution

Convert the interpolation problem

As mentioned, \(v(x)=\ln(u(x))\) is linear in \(\gamma_1\) and \(\gamma_2\). Therefore, we should first convert the interpolation problem from \(u(x)\) to \(v(x)\) by taking the natural logarithm of the known data points: \(y_{i}=\ln (z_{i}), i=0, 1, 2\). This will give us three new data pairs \((x_0, y_0), (x_1, y_1)\), and \((x_2, y_2)\).

Determine the coefficients

Using the three new data points, we can find a quadratic polynomial for \(v(x)\): \(v(x) = a + bx + cx^2\), where a, b, and c are unknown coefficients. We need to determine these coefficients by solving the linear system derived from the requirement \(v(x_i) = y_i\).

Convert back to u(x)

Once we have the coefficients a, b, and c for the quadratic polynomial \(v(x)\), we can convert it back to \(u(x)\) by taking the exponential of \(v(x)\):$$ u(x) = e^{v(x)} = e^{a + bx + cx^2} $$The resulting function \(u(x)\) will have the desired form with the coefficients \(\gamma_0=e^a, \gamma_1=b\), and \(\gamma_2=c\). ## Part B ##

Calculate y_i

For the given data points \((0, 1), (1, 0.9), (3, 0.5)\), calculate \(y_{i}\) for \(i=0,1,2.\) $$ y_0=\ln(1)=0, $$ $$ y_1=\ln(0.9)\approx -0.105, $$ $$ y_2=\ln(0.5)\approx -0.693. $$

Determine coefficients for v(x)

Using the new data pairs \((0, 0), (1, -0.105), (3, -0.693)\), find the coefficients a, b, and c for the quadratic polynomial v(x): $$ \begin{bmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 3 & 9 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix}= \begin{bmatrix} 0 \\ -0.105 \\ -0.693 \\ \end{bmatrix} $$Solve the linear system to find the coefficients a, b, and c.

Obtain the u(x) function

With the coefficients a, b, and c for the quadratic polynomial v(x), find the function u(x): $$ u(x) = e^{v(x)} = e^{a + bx + cx^2} $$

Plot the interpolant

Plot the resulting interpolant \(u(x)\) over the interval \([0, 6]\) and analyze the qualitative difference between the curve and a quadratic. In conclusion, after solving the problem step by step, we found a formula for \(u(x)\) by interpolating the given data points using a quadratic polynomial v(x). Then, we wrote a script to calculate the coefficients \(\gamma_i\) and plotted it to visually compare the curve to a quadratic one.

Key concepts

Exponential Interpolation

Exponential interpolation is a mathematical technique used to estimate unknown values that lie between two known points on an exponentially shaped curve. Unlike linear interpolation, which assumes a constant rate of change between points, exponential interpolation accounts for changes that increase or decrease at a rate proportional to their current value.

To perform exponential interpolation, we often transform the original exponential model into a linear one. This simplifies the process of determining unknown coefficients, which can be quite challenging directly with an exponential model. The linearized equation can then be solved using established methods for linear interpolation. After the coefficients are found, they are translated back into the exponential model to provide the desired interpolation function, such as the function u(x) from our example.

In practical terms, exponential interpolation is useful in a variety of fields, from finance in compounding interest calculations to science in modeling population growth or radioactive decay, where growth rates are proportional to the current amount.

Quadratic Polynomial

Understanding Quadratic Polynomials

Quadratic polynomials are equations of the second degree, generally given in the form f(x) = ax^2 + bx + c, where a, b, and c are coefficients, and a ≠ 0. These polynomials are known for their characteristic 'U' shaped curve, called a parabola.

When tasked with finding a quadratic polynomial that interpolates data, we usually derive a system of equations from the given points. By solving this system, we can determine the values of the coefficients that create the polynomial which best fits our data. In the context of the provided exercise, it's this quadratic polynomial form that allows for the interpolation of the logarithmically transformed data.

Application in Interpolation

In interpolation tasks, these simple quadratic forms can effectively fit a set of three points on a plane, allowing us to create a model that approximates the given data. The robust nature of quadratic polynomials makes them a valuable tool in not just interpolation but also in broader applications such as physics, economics, and any area where modeling complex systems is necessary.

Logarithmic Transformation

Logarithmic transformation is a power tool for converting a nonlinear relationship into a linear one. It involves applying the natural logarithm to both sides of an equation or to a set of data points. This transformation is particularly beneficial when dealing with exponential growth or decay processes as in our exercise.

The usefulness of logarithmic transformation in the context of interpolation is demonstrated in the exercise, where the function u(x) is not linear in its coefficients. By taking the natural logarithm of u(x), resulting in v(x), the problem becomes a linear one in terms of gamma_{1} and gamma_{2}, simplifying the calculation process. The transformed data can then be fitted with a quadratic polynomial, and the coefficients found can be used to reconstruct the original function. This demonstrates how logarithms bridge the gap between linear and nonlinear modeling, broadening the analytical tools we can use to understand and predict complex patterns in data.