Question

Verify that the Hermite cubic interpolating \(f(x)\) and its derivative at the points \(t_{i}\) and \(t_{i+1}\) can be written explicitly as $$ \begin{aligned} s_{i}(x) &=f_{i}+\left(h_{i} f_{i}^{\prime}\right) \tau+\left(3\left(f_{i+1}-f_{i}\right)-h_{i}\left(f_{i+1}^{\prime}+2 f_{i}^{\prime}\right)\right) \tau^{2} \\ &+\left(h_{i}\left(f_{i+1}^{\prime}+f_{i}^{\prime}\right)-2\left(f_{i+1}-f_{i}\right)\right) \tau^{3} \end{aligned} $$ where \(h_{i}=t_{i+1}-t_{i}, f_{i}=f\left(t_{i}\right), f_{i}^{\prime}=f^{\prime}\left(t_{i}\right), f_{i+1}=f\left(t_{i+1}\right), f_{i+1}^{\prime}=f^{\prime}\left(t_{i+1}\right)\), and \(\tau=\frac{x-t_{i}}{h_{i}}\)

Short answer

Based on the given Hermite cubic interpolating function, the expression \(s_i(x)\) satisfies the conditions given for \(f(x)\) and its derivative at both points \(x = t_i\) and \(x = t_{i+1}\). This is shown by evaluating the expression and its derivative at these points and checking whether it agrees with the values for \(f(t_i)\), \(f'(t_i)\), \(f(t_{i+1})\), and \(f'(t_{i+1})\).

Step-by-step solution

Given expression

The Hermite cubic interpolating function for \(f(x)\) and its derivative at the points \(t_i\) and \(t_{i+1}\) is given by: $$ \begin{aligned} s_i(x) &= f_i + (h_i f_i^\prime) \tau + (3(f_{i+1}-f_i)-h_{i}(f_{i+1}^\prime + 2f_i^\prime)) \tau^2 \\ &+(h_i(f_{i+1}^\prime + f_i^\prime)-2(f_{i+1}-f_i)) \tau^3 \end{aligned} $$ where \(h_i = t_{i+1} - t_i\), \(f_i = f(t_i)\), \(f_i^\prime = f^\prime(t_i)\), \(f_{i+1} = f(t_{i+1})\), \(f_{i+1}^\prime = f^\prime(t_{i+1})\), and \(\tau = \frac{x-t_i}{h_i}\).

Verify at x=t_i

First, let's evaluate \(s_i(x)\) at \(x = t_i\) by substituting \(x = t_i\) into our expression: $$\tau = \frac{t_i-t_i}{h_i} = 0$$ Now, by substituting \(\tau = 0\) into the expression for \(s_i(x)\), we get: $$s_i(t_i) = f_i$$ Thus, our expression satisfies the condition of interpolating \(f(x)\) at \(t_i\).

Verify derivative at x=t_i

Next, we'll find the derivative of \(s_i(x)\) with respect to \(x\): $$s_i^\prime(x) = \frac{d}{dx} (\cdots)$$ Let's write the expression shorter as $$ s_i(x) = f_i + (Ah) \tau + (B) \tau^2 + (Ch) \tau^3 $$ Hence by chairrool (chain-rule) and using fact that \(h_i\) and \(f_i\) and its derivatives constant to \(\tau\): $$s_i^\prime(x) = \frac{d}{d\tau} (\cdots)\cdot\frac{d\tau}{dx} =A+ 2B \tau + 3C \tau^2$$ Now, substitute \(x = t_i\), so \(\tau = 0\), into the derivative expression: $$s_i^\prime(t_i) = A = h_i f_i^\prime$$ Thus, our expression also satisfies the condition of having the same derivative as \(f(x)\) at \(t_i\).

Verify at x=t_{i+1}

Now, let's evaluate \(s_i(x)\) at \(x = t_{i+1}\) by substituting \(x = t_{i+1}\) into our expression: $$\tau = \frac{t_{i+1} - t_i}{h_i} = 1$$ Now, by substituting \(\tau = 1\) into the expression for \(s_i(x)\), we get: $$s_i(t_{i+1}) = f_{i+1}$$ Thus, our expression satisfies the condition of interpolating \(f(x)\) at \(t_{i+1}\).

Verify derivative at x=t_{i+1}

Finally, let's verify that the derivative satisfies the condition for \(f^\prime(x)\) at \(x=t_{i+1}\). Just like before, substitute \(x = t_{i+1}\), so \(\tau = 1\), into the derivative expression: $$s_i^\prime(t_{i+1}) = A+ 2B + 3C = h_{i+1} f_{i+1}^\prime$$ Thus, our expression also satisfies the condition of having the same derivative as \(f(x)\) at \(t_{i+1}\). Since we have verified the Hermite cubic interpolating expression for all the given conditions, we can conclude that our expression is correct.

Key concepts

Numerical Methods

Numerical methods are mathematical techniques used to obtain approximate solutions to problems that may not have exact solutions or are difficult to solve analytically. These methods play a crucial role in engineering, physics, finance, and other fields where precise calculations are essential. One common application is solving equations numerically when the actual roots cannot be determined symbolically. For instance, iterating through a series of guesses until a sufficiently accurate answer is found, a process known as root-finding, embodies a fundamental numerical method. Additionally, numerical methods are employed for differentiating and integrating functions, solving differential equations, and, as in Hermite cubic interpolation, fitting curves or surfaces to data points to approximate a function.

Numerical methods are generally designed to be implemented on computers, taking into account issues of computational efficiency, stability, and the control of errors known as numerical analysis. To use these methods effectively, a deep understanding of both theoretical mathematics and practical computing is required. For students and professionals alike, mastering numerical methods is key to unlocking complex problems and advancing in scientific and engineering disciplines.

Interpolation

In the realm of numerical methods, interpolation is a technique for constructing new data points within the range of a discrete set of known data points. When we have a set of data points and we want to estimate the value of a function at an intermediate point, we use interpolation to create a continuous function that passes through all the known data points. Hermite cubic interpolation, which the exercise deals with, is a type of interpolation that not only fits a curve to specific data points but also matches the slopes (derivatives) at those points.

There are various forms of interpolation such as linear, polynomial, and spline interpolation. Linear interpolation connects two neighboring data points with a straight line, while spline interpolation uses piecewise polynomials to create a smooth curve through all the points. Hermite interpolation is particularly useful because it results in a smoother curve than linear or polynomial interpolation by considering not only the position but the slope of the curve at each data point. This is especially beneficial in graphics animations and numerical solutions where smoothness is required. Simplifying the concept, imagine you have a series of photographs and you want to create a smooth video; Hermite interpolation helps in creating the in-between frames that result in the fluid movement essential for the video.

Polynomials

Polynomials are expressions involving a sum of powers of variables with coefficients. They are one of the most studied objects in algebra and form the basis for much of interpolation in numerical methods. A polynomial of degree three, for example, can be written in the form \( ax^3 + bx^2 + cx + d \) where \( a, b, c \) and \( d \) are constants. Polynomials are particularly powerful in interpolation because of their simplicity and smoothness; they can provide approximate solutions to complex functions when exact answers are unobtainable or impractical.

In Hermite cubic interpolation, the polynomials used are of degree three, known as cubic polynomials. Cubic polynomials ensure not just a smooth fit through the data points but also continuity in the first derivative, which is crucial for many applications such as computer graphics, where angular transitions between segments can cause visual discontinuities. The Hermite cubic interpolation formula presented in the exercise combines several cubic polynomials, each defining the curve’s behavior between two points, while ensuring that both the function value and the function's derivative are matched at each of the known data points. This approach guarantees that the resulting curve is smooth and fluent, adhering closely to the way natural phenomena are observed and making the solution incredibly valuable for simulations, animations, and predictive models.