数学代写|实分析作业代写Real analysis代考|Vector-Valued Partial Derivatives and Riemann Integrals

Doug I. Jones

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数学代写|实分析作业代写Real analysis代考|Vector-Valued Partial Derivatives and Riemann Integrals

It is useful to extend the results of Section 2 so that they become valid for functions $f: E \rightarrow \mathbb{C}^m$, where $E$ is an open set in $\mathbb{R}^n$. Up to the chain rule in Theorem 3.10 , these extensions are consequences of what has been proved in Section 2 if we identify $\mathbb{C}^m$ with $\mathbb{R}^{2 m}$. Achieving the extensions by this identification is preferable to trying to modify the original proofs because of the use of the Mean Value Theorem in the proofs of Theorem 3.7 and Proposition 3.9.

The chain rule extends in the same fashion, once we specify what kinds of functions are to be involved in the composition. We always want the domain to be a subset of some $\mathbb{R}^l$, and thus in a composition $g \circ f$, we can allow $g$ to have values in some $\mathbb{C}^k$, but we insist as in Theorem 3.10 that $f$ have values in $\mathbb{R}^m$.
Now let us turn our attention to Taylor’s Theorem as in Theorem 3.11. The statement of Theorem 3.11 allows $\mathbb{R}^1$ as range but not a general $\mathbb{R}^m$. Thus the above extension procedure is not immediately applicable. However, if we allow the given $F$ to take values in $\mathbb{R}^m$, a vector-valued version of Taylor’s Theorem will be valid if we adapt our definitions so that the formula remains true component by component. For this purpose we need to enlarge two definitions – that of partial derivatives of any order and that of 1-dimensional Riemann integration-so that both can operate on vector-valued functions. There is no difficulty in doing so, and we may take it that our definitions have been extended in this way.

In the case of vector-valued partial derivatives, let $f: E \rightarrow \mathbb{R}^m$ be given. Then $\frac{\partial f}{\partial x_j}$ is now defined without passing to components. The entries of this vectorvalued partial derivative are exactly the entries of the $j^{\text {th }}$ column of the Jacobian matrix of $f$. Thus the Jacobian matrix consists of the various vector-valued partial derivatives of $f$, lined up as the columns of the matrix.

数学代写|实分析作业代写Real analysis代考|Exponential of a Matrix

In Chapter IV, we shall make use of the exponential of a matrix in connection with ordinary differential equations. If $A$ is an $n$-by- $n$ complex matrix, then we define
$$\exp A=e^A=\sum_{N=0}^{\infty} \frac{1}{N !} A^N$$
This definition makes sense, according to the following proposition.
Proposition 3.12. For any $n$-by- $n$ complex matrix $A, e^A$ is given by a convergent series entry by entry. Moreover, the series $X \mapsto e^X$ and every partial derivative of an entry of it is uniformly convergent on any bounded subset of matrix space $\left(=\mathbb{R}^{2 n^2}\right)$, and therefore $X \mapsto e^X$ is a $C^{\infty}$ function.

REMARK. The proof will be tidier if we use derivatives of $n$-by- $n$ matrix-valued functions. If $F$ and $G$ are two such functions, the same argument as for the usual product rule shows that $\frac{d}{d t}(F(t) G(t))=F^{\prime}(t) G(t)+F(t) G^{\prime}(t)$.

PROOF. Let us define $|A|$ for an $n$-by- $n$ matrix $A$ to be the operator norm of the member of $L\left(\mathbb{C}^n, \mathbb{C}^n\right)$ with matrix $A$. Fix $M \geq 1$. On the set where $|A| \leq M$, we have
$$\left|\sum_{N=N_1}^{N_2} \frac{1}{N !} A^N\right| \leq \sum_{N=N_1}^{N_2} \frac{1}{N !}\left|A^N\right| \leq \sum_{N=N_1}^{N_2} \frac{1}{N !}|A|^N \leq \sum_{N=N_1}^{N_2} \frac{1}{N !} M^n,$$
and the right side tends to 0 uniformly for $|A| \leq M$ as $N_1$ and $N_2$ tend to infinity. Hence the series for $e^A$ is uniformly Cauchy in the metric built from the operator norm and therefore, by Proposition 3.5 , uniformly Cauchy in the metric built from the Hilbert-Schmidt norm. Uniformly Cauchy in the latter metric means that the series is uniformly Cauchy entry by entry, and hence it is uniformly convergent.

实分析代写

数学代写|实分析作业代写Real analysis代考|Exponential of a Matrix

$$\exp A=e^A=\sum_{N=0}^{\infty} \frac{1}{N !} A^N$$

$$\left|\sum_{N=N_1}^{N_2} \frac{1}{N !} A^N\right| \leq \sum_{N=N_1}^{N_2} \frac{1}{N !}\left|A^N\right| \leq \sum_{N=N_1}^{N_2} \frac{1}{N !}|A|^N \leq \sum_{N=N_1}^{N_2} \frac{1}{N !} M^n,$$

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