Span of Exponential of a x by Cosine of b x and Exponential of a x by Sine of b x is preserved under Differentiation wrt x
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Theorem
Let $a, b, x, \alpha_1, \alpha_2 \in \R$ be real numbers.
Denote:
- $f_1 := \map \exp {a x} \map \cos {b x}$
- $f_2 := \map \exp {a x} \map \sin {b x}$
Let $\map \CC \R$ be the space of continuous real-valued functions.
Let $\tuple {\map {\CC^1} \R, +, \, \cdot \,}_\R$ be the vector space of continuously differentiable real-valued functions.
Let $S = \set {f_1, f_2} \subset \map {\CC^1} \R$ be a vector space.
Let $\map \span S$ be the span of $S$.
Then differentiation with respect to $x$ preserves $\map \span S$.
Proof
\(\ds \map {\dfrac \d {\d x} } {\alpha_1 f_1 + \alpha_2 f_2}\) | \(=\) | \(\ds \map {\dfrac \d {\d x} } {\alpha_1 \map \exp {a x} \map \cos {b x} + \alpha_2 \map \exp {a x} \map \sin {b x} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \alpha_1 a \map \exp {a x} \map \cos {b x} - \alpha_1 b \map \exp {a x} \map \sin {b x} + \alpha_2 a \map \exp {a x} \map \sin {b x} + \alpha_2 b \map \exp {a x} \map \cos {b x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\alpha_1 a + \alpha_2 b} \map \exp {a x} \map \cos {b x} + \paren {\alpha_2 a - \alpha_1 b} \map \exp {a x} \map \sin {b x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \beta_1 f_1 + \beta_2 f_2\) |
$\blacksquare$
This article, or a section of it, needs explaining. In particular: Why this proves the result Because the basis does not reduce to a single function nor it introduces a new function, while the coefficients are real. That will need to be explained then. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Explain}} from the code. |
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.1$: Continuous and linear maps. Linear transformations