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authorPrefetch2022-12-17 18:19:26 +0100
committerPrefetch2022-12-17 18:20:50 +0100
commita39bb3b8aab1aeb4fceaedc54c756703819776c3 (patch)
treeb21ecb4677745fb8c275e54f2ad9d4c2e775a3d8 /source/know/concept/convolution-theorem
parent49cc36648b489f7d1c75e1fde79f0990e08dd514 (diff)
Rewrite "Lagrange multiplier", various improvements
Diffstat (limited to 'source/know/concept/convolution-theorem')
-rw-r--r--source/know/concept/convolution-theorem/index.md18
1 files changed, 9 insertions, 9 deletions
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
index 510417a..d10d85d 100644
--- a/source/know/concept/convolution-theorem/index.md
+++ b/source/know/concept/convolution-theorem/index.md
@@ -36,9 +36,9 @@ rearrange the integrals:
$$\begin{aligned}
\hat{\mathcal{F}}{}^{-1}\{\tilde{f}(k) \: \tilde{g}(k)\}
- &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \exp(i s k x') \dd{x'} \Big) \exp(-i s k x) \dd{k}
+ &= B \int_{-\infty}^\infty \tilde{f}(k) \Big( A \int_{-\infty}^\infty g(x') \: e^{i s k x'} \dd{x'} \Big) e^{-i s k x} \dd{k}
\\
- &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \exp(- i s k (x - x')) \dd{k} \Big) \dd{x'}
+ &= A \int_{-\infty}^\infty g(x') \Big( B \int_{-\infty}^\infty \tilde{f}(k) \: e^{-i s k (x - x')} \dd{k} \Big) \dd{x'}
\\
&= A \int_{-\infty}^\infty g(x') \: f(x - x') \dd{x'}
= A \cdot (f * g)(x)
@@ -49,9 +49,9 @@ this time starting from a product in the $$x$$-domain:
$$\begin{aligned}
\hat{\mathcal{F}}\{f(x) \: g(x)\}
- &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \exp(- i s x k') \dd{k'} \Big) \exp(i s k x) \dd{x}
+ &= A \int_{-\infty}^\infty f(x) \Big( B \int_{-\infty}^\infty \tilde{g}(k') \: e^{-i s x k'} \dd{k'} \Big) e^{i s k x} \dd{x}
\\
- &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \exp(i s x (k - k')) \dd{x} \Big) \dd{k'}
+ &= B \int_{-\infty}^\infty \tilde{g}(k') \Big( A \int_{-\infty}^\infty f(x) \: e^{i s x (k - k')} \dd{x} \Big) \dd{k'}
\\
&= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
= B \cdot (\tilde{f} * \tilde{g})(k)
@@ -86,20 +86,20 @@ because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
$$\begin{aligned}
\hat{\mathcal{L}}\{(f * g)(t)\}
- &= \int_0^\infty \Big( \int_0^\infty g(t') f(t - t') \dd{t'} \Big) \exp(- s t) \dd{t}
+ &= \int_0^\infty \Big( \int_0^\infty g(t') \: f(t - t') \dd{t'} \Big) e^{-s t} \dd{t}
\\
- &= \int_0^\infty \Big( \int_0^\infty f(t - t') \exp(- s t) \dd{t} \Big) g(t') \dd{t'}
+ &= \int_0^\infty \Big( \int_0^\infty f(t - t') \: e^{-s t} \dd{t} \Big) g(t') \dd{t'}
\end{aligned}$$
Then we define a new integration variable $$\tau = t - t'$$, yielding:
$$\begin{aligned}
\hat{\mathcal{L}}\{(f * g)(t)\}
- &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s (\tau + t')) \dd{\tau} \Big) g(t') \dd{t'}
+ &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s (\tau + t')} \dd{\tau} \Big) g(t') \dd{t'}
\\
- &= \int_0^\infty \Big( \int_0^\infty f(\tau) \exp(- s \tau) \dd{\tau} \Big) g(t') \exp(- s t') \dd{t'}
+ &= \int_0^\infty \Big( \int_0^\infty f(\tau) \: e^{-s \tau} \dd{\tau} \Big) g(t') \: e^{-s t'} \dd{t'}
\\
- &= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'}
+ &= \int_0^\infty \tilde{f}(s) \: g(t') \: e^{-s t'} \dd{t'}
= \tilde{f}(s) \: \tilde{g}(s)
\end{aligned}$$
{% include proof/end.html id="proof-laplace" %}