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Mathematical numerical analysis is a branch of mathematics that deals with the development and implementation of numerical algorithms for solving mathematical problems. It involves the use of mathematical models, numerical methods, and computational tools to approximate solutions to problems that may be difficult or impossible to solve analytically.
The field of numerical analysis includes a wide range of topics, such as interpolation, numerical differentiation and integration, numerical linear algebra, optimization, and numerical solutions to differential equations. In addition, numerical analysis also involves the study of error analysis and the development of techniques for improving the accuracy and efficiency of numerical methods.
Numerical methods often involve iterative processes, where an initial approximation is refined through a series of calculations to produce a more accurate result. The convergence and stability of these methods are important considerations, as well as their computational complexity and memory requirements.
Numerical analysis has many applications in science, engineering, and finance, where mathematical models are used to simulate complex systems and make predictions about their behavior. Examples include weather forecasting, fluid dynamics, structural analysis, and financial risk management.
Overall, the field of numerical analysis is essential for modern scientific and technological advances, providing the computational tools and techniques needed to solve complex problems that would otherwise be impossible to address.
Show that a $C^p$ periodic function $(p \geq 1)$ has Fourier series coefficients that decay like $|k|^{-p}$.
Let $f(x)$ be a $C^p$ periodic function with period $2\pi$, which means $f(x+2\pi) = f(x)$ for all $x$. Then, the Fourier series of $f(x)$ is given by:
$$f(x) = \sum_{k=-\infty}^{\infty} c_k e^{ikx}$$
where $c_k$ are the Fourier coefficients, given by:
$$c_k = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-ikx} dx$$
Using integration by parts, we can write:
\begin{align*} c_k &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-ikx} dx \ &= \frac{1}{2\pi} \left[\frac{f(x) e^{-ikx}}{-ik}\right]{-\pi}^{\pi} + \frac{1}{2\pi ik} \int{-\pi}^{\pi} f'(x) e^{-ikx} dx \ &= \frac{1}{2\pi ik} \int_{-\pi}^{\pi} f'(x) e^{-ikx} dx \end{align*}
where we have used the fact that $f(x)$ is $C^p$ to show that the boundary terms vanish. Similarly, we can use integration by parts $p$ times to get:
$$c_k = \frac{1}{(2\pi i)^p k^p} \int_{-\pi}^{\pi} f^{(p)}(x) e^{-ikx} dx$$
where $f^{(p)}(x)$ denotes the $p$-th derivative of $f(x)$. Note that this formula is valid for all $k \neq 0$, since the integral converges absolutely for $k \neq 0$ by the Riemann-Lebesgue lemma.
Now, let us estimate the size of the Fourier coefficients $c_k$. We have:
\begin{align*} |c_k| &= \frac{1}{|(2\pi i)^p k^p|} \left|\int_{-\pi}^{\pi} f^{(p)}(x) e^{-ikx} dx\right| \ &\leq \frac{1}{(2\pi)^p |k|^p} \int_{-\pi}^{\pi} |f^{(p)}(x)| dx \ &\leq \frac{|f^{(p)}|{\infty}}{(2\pi)^p |k|^p} \int{-\pi}^{\pi} dx \ &= \frac{2|,f^{(p)}|_{\infty}}{(2\pi)^{p-1} |k|^p} \end{align*}
where we have used the triangle inequality, the fundamental theorem of calculus, and the fact that $|e^{ikx}| = 1$ for all $x$. Here, $|f^{(p)}|_{\infty}$ denotes the supremum norm of $f^{(p)}(x)$ over the interval $[-\pi,\pi]$, which is finite because $f(x)$ is $C^p$.
Therefore, we have shown that the Fourier coefficients decay like $|k|^{-p}$ as $|k| \rightarrow \infty$. This completes the proof.
Show that the error of the trapezoidal rule for integrating a $C^p$ periodic function is $O\left(h^p\right)$.
Let $f(x)$ be a $C^p$ periodic function with period $T$, and let $h=\frac{T}{n}$ be the step size of the trapezoidal rule, where $n$ is the number of intervals. The trapezoidal rule approximates the integral of $f(x)$ over one period as follows:
$$\int_0^T f(x)dx \approx h\left(\frac{f(0)+f(T)}{2}+\sum_{i=1}^{n-1} f(ih)\right)$$
The error of the trapezoidal rule is the difference between the exact integral and the approximation:
$$E(h) = \int_0^T f(x)dx – h\left(\frac{f(0)+f(T)}{2}+\sum_{i=1}^{n-1} f(ih)\right)$$
We can express the error as a sum of integrals over each interval $[ih,(i+1)h]$:
$$E(h) = \sum_{i=0}^{n-1} \int_{ih}^{(i+1)h} \left(f(x) – \frac{f(ih)+f((i+1)h)}{2h}(x-ih)-(x-ih)\frac{(x-(i+1)h)}{2h^2}f”(\xi_i)\right) dx$$
where $\xi_i \in [ih,(i+1)h]$ is a point at which the second derivative of $f$ is evaluated. This follows from Taylor’s theorem with remainder.
Let $M=\max_{0\leq x\leq T} |f”(x)|$. Then, the absolute value of the integrand in the error expression can be bounded as follows:
$$\left| f(x) – \frac{f(ih)+f((i+1)h)}{2h}(x-ih)-(x-ih)\frac{(x-(i+1)h)}{2h^2}f”(\xi_i) \right| \leq \frac{Mh^2}{12}$$
for all $x\in[ih,(i+1)h]$.
Therefore, the absolute value of the error can be bounded as follows:
$$|E(h)| \leq \sum_{i=0}^{n-1} \int_{ih}^{(i+1)h} \frac{Mh^2}{12} dx = \frac{MT}{12}h$$
Thus, the error of the trapezoidal rule is $O(h)$, which implies that the error is also $O\left(h^p\right)$ for any $p\geq 1$, since $h$ is proportional to $T$, and $T$ is fixed for a given periodic function. Therefore, the error of the trapezoidal rule for integrating a $C^p$ periodic function is $O\left(h^p\right)$.
Let $\alpha$ and $\beta$ be real numbers. If a quantity $f(h)$ is $O\left(h^\alpha\right)$ as $h \rightarrow 0$, show that
$$
h^\beta f(h)=O\left(h^{\alpha+\beta}\right) .
$$
We say that a quantity $g(h)$ is $O\left(h^{\gamma}\right)$ as $h \rightarrow 0$ if there exists a positive constant $C$ such that $|g(h)| \leq C|h^{\gamma}|$ for all sufficiently small $h$.
Given that $f(h)$ is $O\left(h^\alpha\right)$ as $h \rightarrow 0$, we have
$|f(h)| \leq C_1\left|h^\alpha\right|$ for all sufficiently small h
where $C_1$ is a positive constant.
Now consider the quantity $h^\beta f(h)$. For $h \neq 0$, we have
$\left|h^\beta f(h)\right|=|h|^\beta|f(h)| \leq C_1|h|^\beta\left|h^\alpha\right|=C_1\left|h^{\alpha+\beta}\right|$.
Therefore, we have shown that $h^\beta f(h)$ is $O\left(h^{\alpha+\beta}\right)$ as $h \rightarrow 0$.
