Rewrite "Lagrange multiplier", various improvements

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" %}