diff --git a/exercise3/exercise3.pdf b/exercise3/exercise3.pdf
index cc41ee5..2031976 100644
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diff --git a/exercise3/exercise3.typ b/exercise3/exercise3.typ
index f6e14da..98db255 100644
--- a/exercise3/exercise3.typ
+++ b/exercise3/exercise3.typ
@@ -23,16 +23,14 @@
 these are my solutions to the third exercise set of TMA4135.
 
 there should be a python source code file attached to this deliverable.
-this file can be used to generate the images shown in problem 2.
 
 this document was created using
 #link("https://typst.app/")[#text(blue.darken(5%))[typst]].
 
-#v(42pt)
+#v(1fr)
 
 #outline(title: none)
 
-#v(42pt)
 
 #pagebreak()
 
@@ -212,7 +210,7 @@ $
   |E_([a,b])^n [f]| <= (b-a)^3/(12n^2) max_(xi in [a,b]) |f''(xi)|
 $
 
-=== first problem: $|E_([0,1])^n [exp]| <= 10^(-3)$
+*first problem:* $|E_([0,1])^n [exp]| <= 10^(-3)$
 
 for $f(x) = e^x$, we have $f''(x) = e^x$.
 since $e^x$ is increasing, $max_(xi in [0,1]) |f''(xi)| = e$.
@@ -225,14 +223,14 @@ solving: $n^2 >= e/(12 dot 10^(-3)) = 1000e/12$
 
 therefore $n >= sqrt(1000e/12) approx 15.05$, so $n >= 16$.
 
-=== second problem: $|E_([0,1])^n [ln]| <= 10^(-5)$
+*second problem:* $|E_([0,1])^n [ln]| <= 10^(-5)$
 
 for $f(x) = ln(x)$, we have $f''(x) = -1/x^2$.
 since $|f''(x)| = 1/x^2 -> infinity$ as $x -> 0^+$,
 the function is not twice differentiable at $x = 0$.
 the trapezoidal rule error bound does not apply.
 
-=== third problem: $|E_([1,2])^n [e^x/x]| <= 10^(-3)$
+*third problem:* $|E_([1,2])^n [e^x/x]| <= 10^(-3)$
 
 for $f(x) = e^x/x$, using the quotient rule:
 $f''(x) = (e^x (x^2 - 2x + 2))/x^3$
@@ -363,6 +361,94 @@ needed, making it more efficient than uniform refinement.
 
 
 
+
+
+= problem 2
+
+== a)
+
+to transform the quadrature rule from $[-1,1]$ to $(-3,3)$ we use
+$
+  x = (b - a)/2 t + (b + a)/2 = 3 t
+$
+where $t in [-1,1]$, so $d x = 3 d t$. then
+
+$
+  integral_(-3)^3 e^x d x = integral_(-1)^1 e^(3t) dot 3 d t = 3 integral_(-1)^1 e^(3t) d t
+$
+
+the gauß-legendre approximation is given by
+$
+  G_h = 3 sum_(i=0)^3 w_i e^(3 x_i)
+$
+
+so
+
+#[
+  #let x0 = -1 / 35 * calc.sqrt(525 + 70 * calc.sqrt(30))
+  #let x1 = -1 / 35 * calc.sqrt(525 - 70 * calc.sqrt(30))
+  #let x2 = 1 / 35 * calc.sqrt(525 - 70 * calc.sqrt(30))
+  #let x3 = 1 / 35 * calc.sqrt(525 + 70 * calc.sqrt(30))
+
+  #let w0 = 1 / 36 * (18 - calc.sqrt(30))
+  #let w1 = 1 / 36 * (18 + calc.sqrt(30))
+  #let w2 = 1 / 36 * (18 + calc.sqrt(30))
+  #let w3 = 1 / 36 * (18 - calc.sqrt(30))
+
+  $
+    G_h & = 3(w_0 e^(3 x_0) + w_1 e^(3 x_1) + w_2 e^(3 x_2) + w_3 e^(3 x_3)) \
+        & = 3(#(w0 * calc.exp(3 * x0) + w1 * calc.exp(3 * x1) + w2 * calc.exp(3 * x2) + w3 * calc.exp(3 * x3))) \
+        & = #(3 * (w0 * calc.exp(3 * x0) + w1 * calc.exp(3 * x1) + w2 * calc.exp(3 * x2) + w3 * calc.exp(3 * x3)))
+  $
+
+  wheras the exact value is $e^3 - e^(-3) = #(calc.exp(3) - calc.exp(-3))$.
+]
+
+
+
+== b)
+
+from the given error formula, each subinterval of length $h = (b-a)/m$ contributes error
+
+$
+  E_i prop h^(2n+1) = ((b-a)/m)^(2n+1)
+$
+
+total error from m subintervals
+$
+  E_m prop m dot ((b-a)/m)^(2n+1) = (b-a)^(2n+1)/m^(2n)
+$
+
+we expect error to decrease when we subdivide the interval. more subintervals
+means more precision.
+
+
+
+== c)
+
+let $f(x) = x^8/8!$
+
+we can see that $f^((2n))(x) = x^(8-2n)/(8-2n)!$ for $n > 0$ and $<= 8$ so
+$
+  E_1 = (6^(2n+1) (n!)^4)/((2n+1)[(2n)!]^3) dot xi^(8-2n)/((8-2n)!)
+$
+
+then sum the left and right errors to obtain
+$
+  E_2 = (3^(2n+1) (n!)^4)/((2n+1)[(2n)!]^3) dot (xi_1^(8-2n) + xi_2^(8-2n))/((8-2n)!)
+$
+
+which yields
+$
+  E_2/E_1 = (xi_1^(8-2n) + xi_2^(8-2n))/(2^(2n+1) xi^(8-2n))
+$
+
+we can see that increasing n makes it more accurate very fast, as the numerator
+grows smaller exponentially and the denominator bigger.
+
+
+
+
 = problem 3
 
 == a)
@@ -417,6 +503,7 @@ $
 $
 
 
+
 == c)
 
 let
@@ -431,11 +518,72 @@ where $n - 1 = 2$ since $k = 0, 1, 2$.
 
 computing the cardinal polynomials, we obtain
 $
-  ell_0(x) & = (x - x_1) / (x_0 - x_1) dot (x - x_2) / (x_0 - x_2)
-             // = (x^2 - x_1 x - x_2 x + x_1 x_2) / (x_0^2 - x_1 x_0 - x_2 x_0 + x_1 x_2)
-             = (1 - 2 x) / sqrt(3/5) dot (x - 1/2 + 1/2 sqrt(3/5)) / sqrt(3/5) \
-           & = (5 (x - 1/2 + 1/2 sqrt(3/5) - 2x^2 - x + x sqrt(3/5)))/3 \
-           & = 5 / 3 (-2 x^2 + sqrt(3/5) x + 1/2 sqrt(3/5) - 1/2) \
-  ell_1(x) & = (x - x_0) / (x_1 - x_0) dot (x - x_2) / (x_1 - x_2)
-             = (2 x - 1 + sqrt(3/5))/(sqrt(3/5)) dot (2 x - 1 + sqrt(3/5)) / sqrt(3/5) \
+  ell_0(x) & = (x - x_1) / (x_0 - x_1) dot (x - x_2) / (x_0 - x_2) \
+           & = (x - 1/2) / (-1/2 sqrt(3/5)) dot (x - 1/2 - 1/2 sqrt(3/5)) / (-sqrt(3/5)) \
+           & = 5/3 (-2x^2 + sqrt(3/5) x + 1/2 sqrt(3/5) - 1/2) \
+  ell_1(x) & = (x - x_0) / (x_1 - x_0) dot (x - x_2) / (x_1 - x_2) \
+           & = (x - 1/2 + 1/2 sqrt(3/5)) / (1/2 sqrt(3/5)) dot (x - 1/2 - 1/2 sqrt(3/5)) / (1/2 sqrt(3/5)) \
+           & = 4/3 (x^2 - x + 1/10) \
+  ell_2(x) & = (x - x_0) / (x_2 - x_0) dot (x - x_1) / (x_2 - x_1) \
+           & = 5/3 (-2x^2 - sqrt(3/5) x + 1/2 sqrt(3/5) + 1/2)
 $
+
+integrating to find the weights
+$
+  w_0 & = integral_0^1 ell_0(x) dd(x) = 5/18 \
+  w_1 & = integral_0^1 ell_1(x) dd(x) = 4/9 \
+  w_2 & = integral_0^1 ell_2(x) dd(x) = 5/18
+$
+
+by symmetry we have $w_0 = w_2$.
+
+
+== d)
+
+the 3-point gauß-legendre quadrature rule for $[0,1]$ is
+$
+  integral_0^1 f(x) dd(x) approx 5/18 f(x_0) + 4/9 f(1/2) + 5/18 f(x_2)
+$
+where $x_0 = 1/2 - 1/2 sqrt(3/5)$ and $x_2 = 1/2 + 1/2 sqrt(3/5)$.
+
+to verify that the 3-point rule integrates $x^5$ exactly, we compute
+$
+  integral_0^1 x^5 dd(x) = x^6/6 |_0^1 = 1/6
+$
+and using the symmetry property and binomial expansion for $x_0^5 + x_2^5$ where
+$x_0 = 1/2 - 1/2 sqrt(3/5)$ and $x_2 = 1/2 + 1/2 sqrt(3/5)$, we obtain
+
+$
+  5/18 (x_0^5 + x_2^5) + 4/9 dot 1/32 = 1/6
+$
+
+#h(1fr) $qed$
+
+
+== e)
+
+to show equivalence with the standard table, we transform from the reference interval $[-1, 1]$ to $[0, 1]$ using:
+$
+  x = (t + 1)/2 quad "where" t in [-1, 1] "and" x in [0, 1]
+$
+
+so
+$
+  x_0 & = (-sqrt(15)/5 + 1)/2 = 1/2 - sqrt(15)/10 \
+  x_1 & = (0 + 1)/2 = 1/2 \
+  x_2 & = (sqrt(15)/5 + 1)/2 = 1/2 + sqrt(15)/10
+$
+
+using the change of variables $x = (t+1)/2$, we have $dd(x) = 1/2 dd(t)$.
+so
+$
+  w_0 & = 5/9 dot 1/2 = 5/18 \
+  w_1 & = 8/9 dot 1/2 = 4/9 \
+  w_2 & = 5/9 dot 1/2 = 5/18
+$
+
+note that $sqrt(15)/10 = 1/2 sqrt(3/5)$ so the transformed rule matches exactly
+our derived 3-point gauß-legendre quadrature for $[0, 1]$, confirming that both
+approaches yield the same integration rule.
+
+