Q1Advanced Engineering Mathematics

In engineering applications, we frequently encounter experimental data presented as discrete sets of points rather than continuous analytical functions. When the objective is to analyze the rate of change (velocity) or acceleration of the underlying physical process, we must perform numerical differentiation. The choice of mathematical algorithm is strictly governed by the location of the point of interest relative to the entire dataset and the spacing of the independent variables. When the data is equally spaced and the point of interest lies near the beginning (upper region) of the tabulated data, Newton's Forward Difference Interpolation formula provides the most robust and theoretically sound basis for differentiation.

1. Theoretical Foundation of the Forward Formula

Consider a rigorous dataset of $n+1$ distinct points: $(x_0,y_0),(x_1,y_1),\dots ,(x_n,y_n)$, where the independent variable $x$ increments by a strictly constant step size $h$. Therefore, $x_i=x_0+i\cdot h$ for any integer index $i$. Newton's Forward Difference formula constructs an interpolating polynomial $P(x)$ that perfectly passes through all these discrete points.

To formulate this polynomial, we introduce a dimensionless parameter $p$, defined as the normalized distance from the starting node: $p=\frac{x-x_0}{h}$. The continuous continuous polynomial is mathematically expressed in terms of forward differences ($\Delta$):

2. Deriving the First Derivative (Velocity)

To extract the derivative $\frac{dy}{dx}$, we must apply the chain rule of differential calculus. Since our polynomial is expressed entirely in terms of the parameter $p$, the chain rule dictates that $\frac{dy}{dx}=\frac{dy}{dp}\cdot \frac{dp}{dx}$.

From our precise definition $p=\frac{x-x_0}{h}$, differentiating with respect to $x$ yields the constant scaling factor: $\frac{dp}{dx}=\frac{1}{h}$.

Next, we meticulously differentiate the interpolating polynomial $y(p)$ with respect to $p$ term by term:

• The derivative of the constant $y_0$ is exactly 0.
• The derivative of $p\Delta y_0$ is simply $\Delta y_0$.
• The term $\frac{p(p-1)}{2!}=\frac{p^2-p}{2}$. Its derivative is $\frac{2p-1}{2}$.
• The term $\frac{p(p-1)(p-2)}{3!}=\frac{p^3-3p^2+2p}{6}$. Its derivative is $\frac{3p^2-6p+2}{6}$.

Combining these derived components via the chain rule yields the general, continuous numerical first derivative formula valid for any point $x$ within the interpolation interval:

Crucially, if we need to evaluate the derivative exactly at the starting data point $x=x_0$, the parameter $p$ becomes exactly zero. Substituting $p=0$ into the general equation simplifies it remarkably into the standard pointwise formula:

3. Deriving the Second Derivative (Acceleration)

The second derivative $\frac{d^{2}y}{d{x}^2}$ is obtained by strictly applying the chain rule a second time to the first derivative equation: $\frac{d^{2}y}{d{x}^2}=\frac{d}{dp}\left(\frac{dy}{dx}\right)\cdot \frac{dp}{dx}$.

We differentiate the bracketed expression of the first derivative with respect to $p$:

• The term $\Delta y_0$ is constant with respect to $p$, so its derivative is 0.
• The term $\frac{2p-1}{2}{\Delta }^2{y}_0$ differentiates to exactly ${\Delta }^2{y}_0$.
• The term $\frac{3p^2-6p+2}{6}{\Delta }^3{y}_0$ differentiates to $\frac{6p-6}{6}{\Delta }^3{y}_0=(p-1){\Delta }^3{y}_0$.

Multiplying by the second $\frac{1}{h}$ factor from the chain rule yields the general second derivative:

Evaluating this explicitly at the base point $x=x_0$ (where $p=0$) results in the final formula:

4. Methodological Execution

To solve the specific problem requiring the derivatives at $x=1.2$, the procedure is as follows: First, mathematically confirm the uniform step size $h=0.2$. Second, construct a comprehensive forward difference table calculating all successive differences ${\Delta }^{n}y$. Third, select $x_0=1.2$ as the base node, identifying the specific diagonal row of forward differences ($\Delta y_0,{\Delta }^2{y}_0,\dots$) corresponding to this exact starting point. Finally, substitute these precise numerical values into the derived theoretical formulas to compute both the gradient and concavity at $x=1.2$.