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How is the finite difference formulation for the first derivative related to the Taylor series expansion of the solution function?

Short Answer

Expert verified
Answer: The finite difference formulation for the first derivative is derived from the Taylor series expansion of the solution function by isolating the first derivative term. In particular, it is derived by expanding the function at the point x + h and canceling out the function terms, leading to the approximation f'(x) ≈ (f(x + h) - f(x)) / h.

Step by step solution

01

Define finite difference formulation for the first derivative

The finite difference formulation for the first derivative approximates the derivative of a function using the difference between function values at adjacent points. For a function f(x), the forward finite difference approximation of f'(x) can be expressed as: f'(x) ≈ (f(x + h) - f(x)) / h where h is a small increment in the x-value.
02

Express Taylor series expansion for a function

The Taylor series is an infinite series representation of a function in the form of a sum of its polynomial derivatives evaluated at a specific point. For a function f(x) expanded around the point x = a, the Taylor series can be expressed as: f(x) = f(a) + f'(a)(x - a) + (f''(a)/2!)(x - a)^2 + (f'''(a)/3!)(x - a)^3 + ...
03

Apply Taylor series expansion to f(x + h)

To relate the forward finite difference approximation to the Taylor series expansion, we can expand the function f(x) at the point x + h using Taylor series: f(x + h) = f(x) + f'(x)h + (f''(x)/2!)h^2 + (f'''(x)/3!)h^3 + ...
04

Derive the finite difference from Taylor series expansion

Now, we want to isolate the first derivative f'(x) by rearranging the expansion of f(x + h): f'(x) ≈ (f(x + h) - f(x)) / h From the Taylor expansion, the left-hand side can be approximated as: (f(x) + f'(x)h + (f''(x)/2!)h^2 - f(x)) / h ≈ f'(x) Simplifying and canceling out the f(x) terms, we get: (f'(x)h + (f''(x)/2!)h^2) / h ≈ f'(x) Finally, dividing both sides by h: f'(x) + (f''(x)/2!)h ≈ f'(x) Since h is a small increment, (f''(x)/2!)h is much smaller in magnitude, the approximation holds: f'(x) ≈ (f(x + h) - f(x)) / h Hence, we have shown that the finite difference formulation for the first derivative is related to the Taylor series expansion of the solution function.

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Most popular questions from this chapter

A 1-m-long and \(0.1\)-m-thick steel plate of thermal conductivity $35 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$ is well insulated on both sides, while the top surface is exposed to a uniform heat flux of $5500 \mathrm{~W} / \mathrm{m}^{2}$. The bottom surface is convectively cooled by a fluid at \(10^{\circ} \mathrm{C}\) having a convective heat transfer coefficient of $150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. Assuming one-dimensional heat conduction in the lateral direction, find the temperature at the midpoint of the plate. Discretize the plate thickness into four equal parts.

What happens to the discretization and the roundoff errors as the step size is decreased?

A hot surface at \(100^{\circ} \mathrm{C}\) is to be cooled by attaching \(3-\mathrm{cm}-\) long, \(0.25-\mathrm{cm}\)-diameter aluminum pin fins $(k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})$ with a center-to-center distance of \(0.6 \mathrm{~cm}\). The temperature of the surrounding medium is \(30^{\circ} \mathrm{C}\), and the combined heat transfer coefficient on the surfaces is \(35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming steady one-dimensional heat transfer along the fin and taking the nodal spacing to be \(0.5 \mathrm{~cm}\), determine \((a)\) the finite difference formulation of this problem, \((b)\) the nodal temperatures along the fin by solving these equations, \((c)\) the rate of heat transfer from a single fin, and \((d)\) the rate of heat transfer from a \(1-\mathrm{m} \times 1-\mathrm{m}\) section of the plate.

Consider a large uranium plate of thickness \(L=9 \mathrm{~cm}\), thermal conductivity \(k=28 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and thermal diffusivity \(\alpha=12.5 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\) that is initially at a uniform temperature of \(100^{\circ} \mathrm{C}\). Heat is generated uniformly in the plate at a constant rate of $\dot{e}=10^{6} \mathrm{~W} / \mathrm{m}^{3}\(. At time \)t=0$, the left side of the plate is insulated while the other side is subjected to convection with an environment at \(T_{\infty}=20^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=35 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Using the explicit finite difference approach with a uniform nodal spacing of $\Delta x=1.5 \mathrm{~cm}\(, determine \)(a)$ the temperature distribution in the plate after \(5 \mathrm{~min}\) and \((b)\) how long it will take for steady conditions to be reached in the plate.

Using appropriate software, solve these systems of algebraic equations. (a) $$ \begin{aligned} 3 x_{1}+2 x_{2}-x_{3}+x_{4} &=6 \\ x_{1}+2 x_{2}-x_{4} &=-3 \\ -2 x_{1}+x_{2}+3 x_{3}+x_{4} &=2 \\ 3 x_{2}+x_{3}-4 x_{4} &=-6 \end{aligned} $$ (b) $$ \begin{aligned} 3 x_{1}+x_{2}^{2}+2 x_{3} &=8 \\ -x_{1}^{2}+3 x_{2}+2 x_{3} &=-6.293 \\ 2 x_{1}-x_{2}^{4}+4 x_{3} &=-12 \end{aligned} $$

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