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diff --git a/exercise10/main.typ b/exercise10/main.typ
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--- a/exercise10/main.typ
+++ b/exercise10/main.typ
@@ -36,3 +36,5 @@ this document was created using
 
 #include "problem1.typ"
 
+#include "problem2.typ"
+
diff --git a/exercise10/marine.png b/exercise10/marine.png
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diff --git a/exercise10/problem2.typ b/exercise10/problem2.typ
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--- /dev/null
+++ b/exercise10/problem2.typ
@@ -0,0 +1,131 @@
+#import "@preview/physica:0.9.6": *
+#import "lib.typ": ccases
+
+= problem 2
+
+== a)
+
+click the code to obtain the sketches for the expansions of $f(x)$.
+
+#link(
+  "https://uiua.org/pad?src=0_18_0-dev_2__eJxTVnCtKEgtysxNzStJzFHkqlNQSs8ssVJIzyzJKE3SS87P1ffPzcvMTizSL80sTdQtyMkvUVKoU3BJLElUCMjJL-F61LHyUc_8R11Nj7p2HFr_qL3nUdeOR20zDBUeNU08vP3R_DmP2hcamkKUzXvUNV_X8PB0o8PbdQ1B6pbvPrThUVc_SFYDZBrE3Ee9nZpcAB-1SDo=",
+)[```uiua
+# Experimental!
+~ "git: github.com/Omnikar/uiua-plot" ~ Data Plot
+∩⌟⊂⊸¯⇌⊸↘1 ₑ÷⟜⇡15
+∩⌞⊟-1×2÷-1⊸⧻°⊏
+∩(Plot Data ⍉)
+```]
+
+#image("./2a_even_expansion.png")
+
+this is how the even expansion would look.
+
+#image("./2a_odd_expansion.png")
+
+this is how the odd expansion would look. it is a little off because i didn't
+bother to make a duplicate point at -1 to highlight the discrete jump.
+
+these are both created by evaluating $f(x) = exp(x)$ on $[0, 1]$, then taking
+those values in reverse from 0 to -1, one of them negated to obtain the odd
+property.
+
+== b)
+
+it makes sense that the even expansion is more practical in this case, since it
+is smoother and doesn't contain a discrete jump, only a point at which it isn't
+differentiable, but it is continuous.
+
+the gibbs artifacts will be smaller compared to the odd expansion, which will
+have large artifacts at the discrete jump, making for a less accurate
+approximation of $f(x)$ using a truncated fourier series.
+
+
+== c)
+
+let now $f(x) = x - x^2$ on $[0, 1]$.
+
+we only need to change our single function in the code, namely change from `ₑ`
+to `(-⊸°√)`.
+
+#link(
+  "https://uiua.org/pad?src=0_18_0-dev_2__eJxTVnCtKEgtysxNzStJzFHkqlNQSs8ssVJIzyzJKE3SS87P1ffPzcvMTizSL80sTdQtyMkvUVKoU3BJLElUCMjJL-F61LHyUc_8R11Nj7p2HFr_qL3nUdeOR20zDBU0dEEiGx51zNI8vP3R_DmP2hcamkKUz3vUNV_X8PB0o8PbdQ1B6pfvPrThUVc_SFYDZCrE_Ee9nZpcAGxZTF0=",
+)[```uiua
+... ∩⌟⊂⊸¯⇌⊸↘1 (-⊸°√)÷⟜⇡15 ...
+```]
+
+#image("./2c_even_expansion.png")
+
+#table(
+  image("./marine.png"), align(horizon)[this is how the even expansion looks.],
+
+  stroke: none,
+  columns: 2,
+)
+
+#image("./2c_odd_expansion.png")
+
+this is the odd expansion. very nice.
+
+we can see that the odd expansion already resembles a sine-wave. we can then
+guess that this will the more accurate one of the two.
+
+
+== d)
+
+denote the even and odd expansions as functions respectively
+$
+  g(x) = ccases(x - x^2, x >= 0, -x - x^2)space, quad
+  h(x) = ccases(x - x^2, x >= 0, -x + x^2)
+$
+
+we can find the coefficients for these by calculating the integrals from the
+definitions of the coefficients
+
+$
+  a_0 (g) & = integral_(-1)^1 g(x) dd(x) \
+          & = integral_(-1)^0 (-x - x^2) dd(x) + integral_0^1 (x - x^2) dd(x) \
+          & = -1/2 [x^2]_(-1)^0 - 1/3 [x^3]_(-1)^0
+            + 1/2 [x^2]_0^1 - 1/3 [x^3]_0^1 \
+          & = -1/2 - 1/3 + 1/2 - 1/3 = -2/3
+$
+
+$
+  a_n (g) & = integral_(-1)^1 g(x) cos(2 pi n x) dd(x) \
+          & = integral_(-1)^0 (-x - x^2) cos(2 pi n x) dd(x)
+            + integral_0^1 (x - x^2) cos(2 pi n x) dd(x) \
+          & = 2 integral_0^1 (x - x^2) cos(2 pi n x) dd(x) \
+          & = 2 [(x-x^2)/(2 pi n) sin(2 pi n x)
+              + (1 - 2x)/(4 pi^2 n^2) cos(2 pi n x)
+              + 2/(8 pi^3 n^3) sin(2 pi n x)]_0^1 \
+          & = 2 [-1/(4 pi^2 n^2) - 1/(4 pi^2 n^2)] \
+          & = -1/(pi^2 n^2)
+$
+
+$
+  b_n (g) & = integral_(-1)^1 g(x) sin(2 pi n x) dd(x) = 0
+$
+
+$
+  a_0 (h) & = integral_(-1)^1 h(x) dd(x) \
+          & = integral_(-1)^0 (-x + x^2) dd(x) + integral_0^1 (x - x^2) dd(x) \
+          & = -1/2 [x^2]_(-1)^0 + 1/3 [x^3]_(-1)^0
+            + 1/2 [x^2]_0^1 - 1/3 [x^3]_0^1 \
+          & = -1/2 + 1/3 + 1/2 - 1/3 = 0
+$
+
+$
+  a_n (h) & = integral_(-1)^1 h(x) cos(2 pi n x) dd(x) = 0
+$
+
+$
+  b_n (h) & = integral_(-1)^1 h(x) sin(2 pi n x) dd(x) \
+          & = integral_(-1)^0 (-x + x^2) sin(2 pi n x) dd(x)
+            + integral_0^1 (x - x^2) sin(2 pi n x) dd(x) \
+          & = 2 integral_0^1 (x - x^2) sin(2 pi n x) dd(x) \
+          & = 2 [(x^2 - x)/(2 pi n) cos(2 pi n x)
+              + (1 - 2x)/(4 pi^2 n^2) sin(2 pi n x)
+              -2/(8 pi^3 n^3) cos(2 pi n x)]_0^1 \
+          & = 0
+$
+