From 8dbdde31a49deec3f3791cbd41acdb92eaa1ee49 Mon Sep 17 00:00:00 2001
From: Sam Yates <halfflat@gmail.com>
Date: Wed, 26 Oct 2016 15:01:15 +0200
Subject: [PATCH] Remove tabs

---
 docs/passive_cable/cable_computation.tex | 96 ++++++++++++------------
 tests/validation/CMakeLists.txt          |  2 +-
 validation/CMakeLists.txt                |  6 +-
 validation/ref/neuron/CMakeLists.txt     |  2 +-
 4 files changed, 53 insertions(+), 53 deletions(-)

diff --git a/docs/passive_cable/cable_computation.tex b/docs/passive_cable/cable_computation.tex
index 39a37219..888aa8a1 100644
--- a/docs/passive_cable/cable_computation.tex
+++ b/docs/passive_cable/cable_computation.tex
@@ -54,16 +54,16 @@ is the reversal potential.
 \begin{table}[ht]
     \centering
     \begin{tabular}{lSl}
-	\toprule
-	Term & {Value} & Property\\
-	\midrule
-	$d$    & \SI{1.0}{\um}                 & cable diameter \\
-	$L$    & \SI{1.0}{\mm}                 & cable length \\
-	$R_A$  & \SI{1.0}{\ohm\m}              & bulk axial resistivity \\
-	$R_M$  & \SI{4.0}{\ohm\m\squared}      & areal membrane resistivity \\
-	$C_M$  & \SI{0.01}{\F\per\m\squared}   & areal membrane capacitance \\
-	$E_M$  & \SI{-65.0}{\mV}               & membrane reversal potential \\
-	\bottomrule
+        \toprule
+        Term & {Value} & Property\\
+        \midrule
+        $d$    & \SI{1.0}{\um}                 & cable diameter \\
+        $L$    & \SI{1.0}{\mm}                 & cable length \\
+        $R_A$  & \SI{1.0}{\ohm\m}              & bulk axial resistivity \\
+        $R_M$  & \SI{4.0}{\ohm\m\squared}      & areal membrane resistivity \\
+        $C_M$  & \SI{0.01}{\F\per\m\squared}   & areal membrane capacitance \\
+        $E_M$  & \SI{-65.0}{\mV}               & membrane reversal potential \\
+        \bottomrule
     \end{tabular}
     \caption{Cable properties for the Rallpack 1 model.}
     \label{tbl:rallpack1}
@@ -87,9 +87,9 @@ and the linear membrane capacitance $c$. These determine $\lambda$ and $\tau$ by
 With the model boundary conditions,
 \begin{subequations}
     \begin{align}
-	v(x, 0) &= E, \\
-	\left.\frac{\partial v}{\partial x}\right\vert_{x=0} & = -Ir, \\
-	 \left.\frac{\partial v}{\partial x}\right\vert_{x=L} & = 0,
+        v(x, 0) &= E, \\
+        \left.\frac{\partial v}{\partial x}\right\vert_{x=0} & = -Ir, \\
+         \left.\frac{\partial v}{\partial x}\right\vert_{x=L} & = 0,
     \end{align}
 \end{subequations}
 where $I$ is the injected current and $L$ is the cable length.
@@ -98,18 +98,18 @@ The solution $v(x, t)$ can be expressed in terms of the solution $g(x, t; L)$
 to a normalized version of the cable equation,
 \begin{subequations}
     \begin{align}
-	\label{eq:normcable}
-	\frac{\partial^2 g}{\partial x^2} & =
-	\frac{\partial g}{\partial t} + g,
+        \label{eq:normcable}
+        \frac{\partial^2 g}{\partial x^2} & =
+        \frac{\partial g}{\partial t} + g,
         \\
-	\label{eq:normcableinitial}
-	g(x, 0) &= 0,
-	\\
-	\label{eq:normcableleft}
-	\left.\frac{\partial g}{\partial x}\right\vert_{x=0} & = 1,
-	\\
-	\label{eq:normcableright}
-	\left.\frac{\partial g}{\partial x}\right\vert_{x=L} & = 0
+        \label{eq:normcableinitial}
+        g(x, 0) &= 0,
+        \\
+        \label{eq:normcableleft}
+        \left.\frac{\partial g}{\partial x}\right\vert_{x=0} & = 1,
+        \\
+        \label{eq:normcableright}
+        \left.\frac{\partial g}{\partial x}\right\vert_{x=L} & = 0
     \end{align}
 \end{subequations}
 by
@@ -158,7 +158,7 @@ and thus
 Consequently,
 \begin{equation}
     \begin{aligned}
-	G(x, s) &= \frac{1}{ms}\cdot\frac{e^{mx}+e^{2mL-mx}}{1-e^{2mL}}\\
+        G(x, s) &= \frac{1}{ms}\cdot\frac{e^{mx}+e^{2mL-mx}}{1-e^{2mL}}\\
                 &= - \frac{1}{ms}\cdot\frac{\cosh m(L-x)}{\sinh mL}.
     \end{aligned}
 \end{equation}
@@ -199,10 +199,10 @@ arising from $m=\sqrt{1+s}$, as letting $m=-\sqrt{1+s}$ leaves $G$ unchanged.
 For $|s+1|>\epsilon$,
 \begin{equation}
     \begin{aligned}
-	|s^{3/2}G(x,s)|^2
-	    & \leq (1+\epsilon)^{-1} \left| \frac{\cosh m(L-x)}{\sinh mL} \right|^2
-	\\
-	    & \leq (1+\epsilon)^{-1} (1+|\coth mL|)^2
+        |s^{3/2}G(x,s)|^2
+            & \leq (1+\epsilon)^{-1} \left| \frac{\cosh m(L-x)}{\sinh mL} \right|^2
+        \\
+            & \leq (1+\epsilon)^{-1} (1+|\coth mL|)^2
     \label{eq:gbounds}
     \end{aligned}
 \end{equation}
@@ -218,8 +218,8 @@ Recalling $m=\sqrt{1+s}$, $m$ and $\sinh mL$ are non-zero in
 a neighbourhood of $s=0$, and so the pole is simple and
 \begin{equation}
     \begin{aligned}
-	\Res(G; 0) & = - \frac{1}{m}\cdot\left.\frac{\cosh m(L-x)}{\sinh mL}\right|_{s=0}\\
-		   & = - \frac{\cosh (L-x)}{\sinh L}.
+        \Res(G; 0) & = - \frac{1}{m}\cdot\left.\frac{\cosh m(L-x)}{\sinh mL}\right|_{s=0}\\
+                   & = - \frac{\cosh (L-x)}{\sinh L}.
     \end{aligned}
 \end{equation}
 
@@ -235,9 +235,9 @@ Let $G(x,s)=f(x,s)/h(s)$, where
 Noting that $dm/ds = \frac{1}{2}m^{-1}$,
 \begin{equation}
     \begin{aligned}
-	h'(s) &= \frac{1}{2}m^{-1}\sinh mL + \frac{1}{2}L\cosh mL \\
-	      &= \frac{1}{2}L + \frac{1}{2}L + O(m^2) \quad(m\to 0) \\
-	      &= L + O(s+1) \quad(s\to -1).
+        h'(s) &= \frac{1}{2}m^{-1}\sinh mL + \frac{1}{2}L\cosh mL \\
+              &= \frac{1}{2}L + \frac{1}{2}L + O(m^2) \quad(m\to 0) \\
+              &= L + O(s+1) \quad(s\to -1).
     \label{eq:hprime}
     \end{aligned}
 \end{equation}
@@ -255,11 +255,11 @@ $m_k$ is non-zero for $k\geq 1$ and
 Consequently the pole is simple and
 \begin{equation}
     \begin{aligned}
-	\Res(G; s_k)
-	    & = f(x, s_k)/h'(s_k)\\
-	    & =  -\frac{2}{s_k L}\frac{\cosh m_k(L-x)}{\cosh m_kL} \\
-	    & =  -\frac{2}{s_k L}\frac{\cosh m_kL\cosh m_kx-\sinh m_kL\sinh m_kL}{\cosh m_kL} \\
-	    & =  -\frac{2}{s_k L}\cosh m_k x,
+        \Res(G; s_k)
+            & = f(x, s_k)/h'(s_k)\\
+            & =  -\frac{2}{s_k L}\frac{\cosh m_k(L-x)}{\cosh m_kL} \\
+            & =  -\frac{2}{s_k L}\frac{\cosh m_kL\cosh m_kx-\sinh m_kL\sinh m_kL}{\cosh m_kL} \\
+            & =  -\frac{2}{s_k L}\cosh m_k x,
     \end{aligned}
 \end{equation}
 as $\sinh m_k=0$.
@@ -272,7 +272,7 @@ In terms of $a_k$,
 The series exapnsion for $g(x, t)$ therefore is
 \begin{equation}
     g(x, t) = -\frac{\cosh(L-x)}{\sinh L} + \frac{1}{L}e^{-t}\left\{
-	1+2\sum_{k=1}^\infty \frac{e^{-ta_k^2}}{1+a_k^2}\cos a_k x\right\}.
+        1+2\sum_{k=1}^\infty \frac{e^{-ta_k^2}}{1+a_k^2}\cos a_k x\right\}.
     \label{eq:theg}
 \end{equation}
 
@@ -285,16 +285,16 @@ of stopping criteria for a given tolerance.
 Let $g_n$ be the partial sum
 \begin{equation}
     g_n(x, t) = -\frac{\cosh(L-x)}{\sinh L} + \frac{1}{L}e^{-t}\left\{
-	1+2\sum_{k=1}^n \frac{e^{-ta_k^2}}{1+a_k^2}\cos a_k x\right\}.
+        1+2\sum_{k=1}^n \frac{e^{-ta_k^2}}{1+a_k^2}\cos a_k x\right\}.
 \end{equation}
 so that $g(x, t) =\lim_{n\to\infty} g_n(x,t)$. Let $\bar{g}_n = |g-g_n|$ be the
 residual. The $a_k$ form an increasing sequence, so
 \begin{equation}
     \begin{aligned}
-	\bar{g}_n(x,t)
-	    & \leq \frac{2}{L}e^{-t}\sum_{n+1}^\infty\frac{e^{-ta_k^2}}{1+a_k^2}\\
-	    & \leq \frac{2}{L}e^{-t}\int_{a_n}^\infty \frac{e^{-tu^2}}{1+u^2}\,du\\
-	    & < \frac{2}{L}e^{-t}\int_{a_n}^\infty \frac{e^{-tu^2}}{u^2}\,du.
+        \bar{g}_n(x,t)
+            & \leq \frac{2}{L}e^{-t}\sum_{n+1}^\infty\frac{e^{-ta_k^2}}{1+a_k^2}\\
+            & \leq \frac{2}{L}e^{-t}\int_{a_n}^\infty \frac{e^{-tu^2}}{1+u^2}\,du\\
+            & < \frac{2}{L}e^{-t}\int_{a_n}^\infty \frac{e^{-tu^2}}{u^2}\,du.
     \end{aligned}
     \label{eq:gbar}
 \end{equation}
@@ -313,9 +313,9 @@ For real $\alpha<1$ and $z>0$, \textcite[][Theorem 2.3]{borwein2009} give the up
 Substituting into \eqref{eq:gbar} gives
 \begin{equation}
     \begin{aligned}
-	\bar{g}_n(x,t)
-	    & < \frac{1}{L}e^{-t}\sqrt{t}\,\Gamma(-\frac{1}{2},a_n^2 t) \\
-	    & \leq \frac{1}{L}e^{-t}\sqrt{t}\,(a_n^2 t)^{-\frac{3}{2}}e^{-a_n^2t} \\
+        \bar{g}_n(x,t)
+            & < \frac{1}{L}e^{-t}\sqrt{t}\,\Gamma(-\frac{1}{2},a_n^2 t) \\
+            & \leq \frac{1}{L}e^{-t}\sqrt{t}\,(a_n^2 t)^{-\frac{3}{2}}e^{-a_n^2t} \\
             & = \frac{e^{-t(1+a_n^2)}}{L t a_n^3}.
     \end{aligned}
 \end{equation}
diff --git a/tests/validation/CMakeLists.txt b/tests/validation/CMakeLists.txt
index a38ae3e2..783ef003 100644
--- a/tests/validation/CMakeLists.txt
+++ b/tests/validation/CMakeLists.txt
@@ -38,7 +38,7 @@ foreach(target ${TARGETS})
     )
 
     if(BUILD_VALIDATION_DATA)
-	add_dependencies(${target} validation_data)
+        add_dependencies(${target} validation_data)
     endif()
 endforeach()
 
diff --git a/validation/CMakeLists.txt b/validation/CMakeLists.txt
index 009a27aa..99b4bec9 100644
--- a/validation/CMakeLists.txt
+++ b/validation/CMakeLists.txt
@@ -31,10 +31,10 @@ function(add_validation_data)
     set(out "${VALIDATION_DATA_DIR}/${ADD_VALIDATION_DATA_OUTPUT}")
     string(REGEX REPLACE "([^;]+)" "${CMAKE_CURRENT_SOURCE_DIR}/\\1" deps "${ADD_VALIDATION_DATA_DEPENDS}")
     add_custom_command(
-	OUTPUT "${out}"
+        OUTPUT "${out}"
         DEPENDS ${deps}
-	WORKING_DIRECTORY "${CMAKE_CURRENT_SOURCE_DIR}"
-	COMMAND ${ADD_VALIDATION_DATA_COMMAND} > "${out}")
+        WORKING_DIRECTORY "${CMAKE_CURRENT_SOURCE_DIR}"
+        COMMAND ${ADD_VALIDATION_DATA_COMMAND} > "${out}")
 
     # Cmake, why can't we just write add_dependencies(validation_data "${out}")?!
     make_unique_target_name(ffs_cmake "${out}")
diff --git a/validation/ref/neuron/CMakeLists.txt b/validation/ref/neuron/CMakeLists.txt
index 570d3f72..4df17dc3 100644
--- a/validation/ref/neuron/CMakeLists.txt
+++ b/validation/ref/neuron/CMakeLists.txt
@@ -11,7 +11,7 @@ set(models
 foreach(model ${models})
     set(script "${model}.py")
     add_validation_data(
-	OUTPUT "neuron_${model}.json"
+        OUTPUT "neuron_${model}.json"
         DEPENDS "${script}" "nrn_validation.py"
         COMMAND ${NRNIV_BIN} -nobanner -python "${script}")
 endforeach()
-- 
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