diff --git a/exercise3/exercise3.pdf b/exercise3/exercise3.pdf
index 48eb1fe..cc41ee5 100644
Binary files a/exercise3/exercise3.pdf and b/exercise3/exercise3.pdf differ
diff --git a/exercise3/exercise3.typ b/exercise3/exercise3.typ
index 9dc09a2..f6e14da 100644
--- a/exercise3/exercise3.typ
+++ b/exercise3/exercise3.typ
@@ -52,52 +52,36 @@ $
 $
 then $(2)$
 $
-  integral_0^1 ln(x) dd(x) & = [x ln(x) - 1/2 dot x^2 / x - 1/6 dot x^3 / x^2
-                               - 1/12 dot x^4 / x^3 - dots.c]_0^1 \
-                           & = -underbracket((1/2 + 1/6 + 1/12 + dots.c), 1)
-                             - cancel(lim_(t -> 0) t ln(t)) \
-                           & = -1,
-$
-because with $p(x) = x^2 + 3 x + 2$ we get
-$
-  2 + 6 + 12 + dots.c = p(0) + p(1) + dots.c
-$
-so we can compute the telescope sum
-$
-  sum_(k=0)^oo 1/p(k) & = sum_(k=0)^oo 1/(k^2 + 3 k + 2) \
-                      & = sum_(k=0)^oo 1/((k+1)(k+2)) \
-                      & = sum_(k=0)^oo (1/(k+1) - 1/(k+2)) \
-                      & = 1 - cancel(1/2 + 1/2) - cancel(1/3 + 1/3) + dots.c \
-                      & = 1.
+  integral_0^1 ln(x) dd(x) & = [x ln(x) - x]_0^1 \
+                           & = (1 ln(1) - 1) - lim_(t -> 0^+) (t ln(t) - t) \
+                           & = -1 - 0 = -1,
 $
+where we used integration by parts and the fact that $lim_(t -> 0^+) t ln(t) = 0$.
 
 now we know what to expect from our estimates.
 
 recall the trapezoidal rule
 $
-  // I_[a, b][f] approx T_[a, b][f] := (b-a)/2 (f(b) - f(a))
-  I_[a, b][f] approx T_[a, b][f] := (b-a)/2 sum_(i=0)^(n-1) (f(x_(i+1)) - f(x_i))
+  I_[a, b][f] approx T_[a, b][f] := (b-a)/2 (f(a) + f(b))
 $
 
-in our case $a = 0$ and $b = 1$. for $(1) med f(x) = e^x$ we obtain
+in our case $a = 0$ and $b = 1$. for $(1)$ with $f(x) = e^x$ we obtain
 $
-  T_[0, 1][exp] = 1/2 (e - 1),
+  T_[0, 1][exp] = 1/2 (e^0 + e^1) = 1/2 (1 + e) = (1 + e)/2,
 $
 which yields an error of
 $
-  e - 1 - 1/2 (e - 1) = 1/2 (e - 1).
+  E_[0,1][exp] = (e - 1) - (1 + e)/2 = (2e - 2 - 1 - e)/2 = (e - 3)/2.
 $
 
-for $(2) med f(x) = ln(x)$ we obtain
+for $(2)$ with $f(x) = ln(x)$ we obtain
 $
-  T_[0, 1][ln] = 1/2 (lim_(t->0) ln(t) - ln(1)) = 1/2 (-oo - 0) = -oo
+  T_[0, 1][ln] = 1/2 (ln(0) + ln(1)) = 1/2 (-oo + 0) = -oo
 $
-which yields an error of $oo$.
+which yields an error of $+oo$.
 
 thus we can see that using the trapezoidal rule for $ln(x)$ diverges and proves
-to be unwieldy. for $exp(x)$ we can see that it was off by half the actual
-value.
-
+to be unwieldy. for $exp(x)$ the actual error is $(e-3)/2 approx #{ calc.round((calc.e - 3) / 2, digits: 3) }$.
 
 == c)
 
@@ -105,20 +89,278 @@ recall the upper bound of the error for the trapezoidal rule
 $
   abs(E_[a,b][f]) <= (b-a)^3 / 12 dot max_(xi in [a,b]) abs(f''(xi))
 $
-which is for $f = exp$
+which is for $f = exp$ on $[0,1]$
 $
-  E_[0, 1][exp] = e/12 approx #{ calc.round(calc.e / 12, digits: 3) }
+  abs(E_[0, 1][exp]) <= 1^3/12 dot max_(xi in [0,1]) abs(e^xi) = e/12 approx #{ calc.round(calc.e / 12, digits: 3) }
 $
-as opposed to our previous $1/2 (e - 1)
-= #{ calc.round((calc.e - 1) / 2, digits: 3) }$.
+Our actual error magnitude is $abs((e-3)/2) = (3-e)/2 approx #{ calc.round((3 - calc.e) / 2, digits: 3) }$.
 
+Since $#{ calc.round((3 - calc.e) / 2, digits: 3) } <= #{ calc.round(calc.e / 12, digits: 3) }$, the error bound is satisfied.
 
+== d)
+
+recall the error bound derived in the lecture,
+$
+  abs(E_[0,1][f]) <= (max_(xi in
+  [a,b]) abs(f^((n+1))(xi))) / ((n+1)!) integral_a^b product_(i=0)^n abs(
+    x
+    - x_i
+  )
+  dd(x),
+$
+where the integral of $f$ is taken over the interval $[a,b]$, and $f in
+C^(n+1)[0,1]$.
+
+for the trapezoidal rule, $n = 1$, so we have $x_0 = 0$ and $x_1 = 1$. for $f(x)
+= e^x$:
+
+- $f^((n+1))(x) = f^((2))(x) = e^x$
+- $max_(xi in [0,1]) abs(e^xi) = e$
+- $(n+1)! = 2! = 2$
+
+the integral becomes:
+$
+  integral_0^1 product_(i=0)^1 abs(x - x_i) dd(x) & = integral_0^1 abs(x - 0) dot abs(x - 1) dd(x) \
+                                                  & = integral_0^1 x dot abs(x - 1) dd(x) \
+                                                  & = integral_0^1 x dot (1 - x) dd(x) quad "since" x - 1 <= 0 "on" [0,1] \
+                                                  & = integral_0^1 (x - x^2) dd(x) \
+                                                  & = [x^2/2 - x^3/3]_0^1 = 1/2 - 1/3 = 1/6
+$
+
+therefore, the error bound is:
+$
+  abs(E_[0,1][exp]) <= e/2 dot 1/6 = e/12 approx #{ calc.round(calc.e / 12, digits: 3) }
+$
+
+this gives the same bound as part c), confirming that both error bound formulas
+are equivalent for the trapezoidal rule. the actual error
+$(3-e)/2 approx #{ calc.round((3 - calc.e) / 2, digits: 3) }$
+is indeed less than this bound.
 
 == e)
 
-telescope sum ish
+to derive the composite trapezoidal rule, we divide the interval $[a,b]$ into $n$ segments of equal length $h = (b-a)/n$, with points $x_i = a + i h$ for $i = 0, 1, ..., n$.
+
+applying the basic trapezoidal rule to each subinterval $[x_i, x_(i+1)]$:
+$
+  T_([x_i, x_(i+1)])[f] = h/2 (f(x_i) + f(x_(i+1)))
+$
+
+the composite rule sums over all subintervals:
+$
+  T_([a,b])^n [f] = sum_(i=0)^(n-1) T_([x_i, x_(i+1)])[f] = sum_(i=0)^(n-1) h/2 (f(x_i) + f(x_(i+1)))
+$
+
+expanding this sum:
+$
+  T_([a,b])^n [f] = h/2 [f(x_0) + f(x_1) + f(x_1) + f(x_2) + ... + f(x_(n-1)) + f(x_n)]
+$
+
+we observe that $f(x_0)$ and $f(x_n)$ appear once, while $f(x_1), f(x_2), ..., f(x_(n-1))$ each appear twice. collecting terms:
+$
+  T_([a,b])^n [f] = h/2 (f(x_0) + f(x_n)) + h sum_(i=1)^(n-1) f(x_i)
+$
+
+since $x_0 = a$ and $x_n = b$, this gives us the desired composite formula.
+
+== f)
+
+to derive the error bound for the composite trapezoidal rule, we
+start with the error bound for each subinterval.
+
+for each subinterval $[x_i, x_(i+1)]$ with length $h$:
+$
+  |E_([x_i, x_(i+1)])[f]| <= h^3/12 max_(xi in [x_i, x_(i+1)]) |f''(xi)|
+$
+
+the total error for the composite rule is:
+$
+  |E_([a,b])^n [f]| = |sum_(i=0)^(n-1) E_([x_i, x_(i+1)])[f]|
+$
+
+using the triangle inequality:
+$
+  |sum_(i=0)^(n-1) E_([x_i, x_(i+1)])[f]| <= sum_(i=0)^(n-1) |E_([x_i, x_(i+1)])[f]|
+$
+
+substituting the error bound for each subinterval:
+$
+  sum_(i=0)^(n-1) |E_([x_i, x_(i+1)])[f]| <= sum_(i=0)^(n-1) h^3/12 max_(xi in [x_i, x_(i+1)]) |f''(xi)|
+$
+
+since $[x_i, x_(i+1)] subset [a,b]$, we have:
+$
+  max_(xi in [x_i, x_(i+1)]) |f''(xi)| <= max_(xi in [a,b]) |f''(xi)|
+$
+
+therefore:
+$
+  sum_(i=0)^(n-1) h^3/12
+  max_(xi in [x_i, x_(i+1)])
+  |f''(xi)| & <= h^3/12
+              sum_(i=0)^(n-1) max_(xi in [a,b]) |f''(xi)| \
+            & = h^3/12 dot n dot max_(xi in [a,b]) |f''(xi)| \
+            & = h^2 dot (b-a)/12 max_(xi in [a,b]) |f''(xi)| \
+            & = (b-a)^3/(12n^2) max_(xi in [a,b]) |f''(xi)|
+$
+
+
+== g)
+
+using the error bound formula:
+$
+  |E_([a,b])^n [f]| <= (b-a)^3/(12n^2) max_(xi in [a,b]) |f''(xi)|
+$
+
+=== first problem: $|E_([0,1])^n [exp]| <= 10^(-3)$
+
+for $f(x) = e^x$, we have $f''(x) = e^x$.
+since $e^x$ is increasing, $max_(xi in [0,1]) |f''(xi)| = e$.
+
+$
+  1^3/(12n^2) dot e <= 10^(-3)
+$
+
+solving: $n^2 >= e/(12 dot 10^(-3)) = 1000e/12$
+
+therefore $n >= sqrt(1000e/12) approx 15.05$, so $n >= 16$.
+
+=== second problem: $|E_([0,1])^n [ln]| <= 10^(-5)$
+
+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: $|E_([1,2])^n [e^x/x]| <= 10^(-3)$
+
+for $f(x) = e^x/x$, using the quotient rule:
+$f''(x) = (e^x (x^2 - 2x + 2))/x^3$
+
+to find the maximum, we check:
+- $f''(1) = e$
+- $f''(1.5) approx 1.66$
+- $f''(2) = e^2/4 approx 1.85$
+
+since $f''(1) > f''(1.5)$ and $f''(1) > f''(2)$,
+we have $max_(xi in [1,2]) |f''(xi)| = e$.
+
+$
+  1^3/(12n^2) dot e <= 10^(-3)
+$
+
+solving: $n >= sqrt(1000e/12) approx 15.05$, so $n >= 16$.
+
+
+== h) & i)
+
+empirically testing shows that $n = 12$ subintervals already
+achieves $|E_([0,1])^12 [exp]| <= 10^(-3)$. furthermore, we need $120$
+subintervals for error $10^(-5)$.
+
+
+== j)
+
+to estimate the convergence rate, we fit the model $E_n approx C h^p$ by taking logarithms:
+$
+  log |E_n| approx log C + p log h
+$
+
+using linear regression on $(log h, log |E_n|)$ data for various values of $n$:
+
+for $exp(x)$ on $[0,1]$:
+- estimated $p_1 approx 2.0$
+
+for $sqrt(x)$ on $[0,1]$:
+- estimated $p_2 approx 1.5$
+
+
+== k)
+
+for exp(x): $p_1 approx 2.0 =>$ quadratic
+
+for sqrt(x): $p_2 approx 1.5$ shows that $f''(x) = -1/4 x^(-3/2)$ is unbounded as $x -> 0^+$.
+
+
+== l)
+
+given
+$
+  E_([a,b])^1 = -M/12 (b-a)^3 quad "and" quad
+  E_([a,b])^2 = -M/48 (b-a)^3
+$
+
+we obtain $E_([a,b])^1 = 4 E_([a,b])^2$.
+
+since
+$
+  E_([a,b])^1 = I_([a,b])[f] - T_([a,b])^1 quad "and" quad
+  E_([a,b])^2 = I_([a,b])[f] - T_([a,b])^2
+$
+we get
+$
+  E_([a,b])^1 - E_([a,b])^2 = T_([a,b])^2 - T_([a,b])^1
+$
+
+thus
+
+$
+  4 E_([a,b])^2 - E_([a,b])^2 & = T_([a,b])^2 - T_([a,b])^1 \
+                3 E_([a,b])^2 & = T_([a,b])^2 - T_([a,b])^1 \
+              ==> E_([a,b])^1 & = 4 E_([a,b])^2
+                                = 4/3 (T_([a,b])^2 - T_([a,b])^1) =: cal(E)_([a,b])^1 [f] \
+                  E_([a,b])^2 & = 1/3 (T_([a,b])^2 - T_([a,b])^1) =: cal(E)_([a,b])^2 [f]
+$
+
+
+== m)
+
+for $f(x) = sqrt(x)$ on $[0,1]$ with 10 uniform intervals and tolerance $10^(-4)$:
+#[
+  #let sqrt_trap1(a, b) = calc.round((b - a) / 2 * (calc.sqrt(a) + calc.sqrt(b)), digits: 6)
+  #let sqrt_trap2(a, b) = {
+    let c = (a + b) / 2
+    calc.round((b - a) / 4 * (calc.sqrt(a) + 2 * calc.sqrt(c) + calc.sqrt(b)), digits: 6)
+  }
+  #let error_est(a, b) = calc.round(calc.abs(4 / 3 * (sqrt_trap2(a, b) - sqrt_trap1(a, b))), digits: 6)
+
+  #show table.cell.where(y: 0): set text(weight: "bold")
+  #table(
+    columns: 4,
+    inset: 5% + 6pt,
+    [Interval], [$T_([a,b])^1$], [$T_([a,b])^2$], [$|cal(E)_([a,b])^1|$],
+
+    [$[0, 0.1]$], [#sqrt_trap1(0, 0.1)], [#sqrt_trap2(0, 0.1)], [#error_est(0, 0.1)],
+    [$[0.1, 0.2]$], [#sqrt_trap1(0.1, 0.2)], [#sqrt_trap2(0.1, 0.2)], [#error_est(0.1, 0.2)],
+    [$[0.2, 0.3]$], [#sqrt_trap1(0.2, 0.3)], [#sqrt_trap2(0.2, 0.3)], [#error_est(0.2, 0.3)],
+    [$[0.3, 0.4]$], [#sqrt_trap1(0.3, 0.4)], [#sqrt_trap2(0.3, 0.4)], [#error_est(0.3, 0.4)],
+    [$[0.4, 0.5]$], [#sqrt_trap1(0.4, 0.5)], [#sqrt_trap2(0.4, 0.5)], [#error_est(0.4, 0.5)],
+    [$[0.5, 0.6]$], [#sqrt_trap1(0.5, 0.6)], [#sqrt_trap2(0.5, 0.6)], [#error_est(0.5, 0.6)],
+    [$[0.6, 0.7]$], [#sqrt_trap1(0.6, 0.7)], [#sqrt_trap2(0.6, 0.7)], [#error_est(0.6, 0.7)],
+    [$[0.7, 0.8]$], [#sqrt_trap1(0.7, 0.8)], [#sqrt_trap2(0.7, 0.8)], [#error_est(0.7, 0.8)],
+    [$[0.8, 0.9]$], [#sqrt_trap1(0.8, 0.9)], [#sqrt_trap2(0.8, 0.9)], [#error_est(0.8, 0.9)],
+    [$[0.9, 1.0]$], [#sqrt_trap1(0.9, 1.0)], [#sqrt_trap2(0.9, 1.0)], [#error_est(0.9, 1.0)],
+  )
+]
+
+intervals needing refinement (error > $10^(-4)$): first 4 intervals
+
+
+== n)
+
+implemented adaptive trapezoidal quadrature with tolerance $"tol" = 10^(-5)$:
+
+#table(
+  columns: 4,
+  [Function], [Adaptive Result], [Intervals Used], [Uniform (part i)],
+
+  [$e^x$], [1.7183], [64], [120],
+  [$sqrt(x)$], [0.6667], [79], [-],
+)
+
+the adaptive algorithm automatically allocates computational effort where
+needed, making it more efficient than uniform refinement.
 
-overlapping edges
 
 
 = problem 3
diff --git a/exercise3/main.py b/exercise3/main.py
new file mode 100644
index 0000000..ea0d1c4
--- /dev/null
+++ b/exercise3/main.py
@@ -0,0 +1,71 @@
+import numpy as np
+
+
+def step(a, b, n):
+    return (b - a) / n
+
+
+# this is basically linspace
+def interpolation_nodes(a, b, n):
+    h = step(a, b, n - 1)
+    return [a + i * h for i in range(n)]
+
+
+def composite_trapezoidal(f, a, b, n):
+    h = step(a, b, n)
+    x = interpolation_nodes(a, b, n + 1)
+    return h / 2 * sum([f(x[i]) + f(x[i + 1]) for i in range(n)])
+
+
+def error_trapezoidal(f, a, b, n, expected):
+    return expected - composite_trapezoidal(f, a, b, n)
+
+
+def format_error(f, a, b, n, expected):
+    return f"error for {f} from {a} to {b} with {n} subintervals is {error_trapezoidal(f, a, b, n, expected)}."
+
+
+def find_subinterval_count(f, a, b, expected, error_threshold):
+    n = 1
+    e = lambda n: error_trapezoidal(f, a, b, n, expected)
+    while np.abs(e(n)) > error_threshold:
+        n += 1
+    return n
+
+
+def estimate_convergence_rate(f, a, b, expected):
+    m = 20
+    errors = [np.log(abs(error_trapezoidal(f, a, b, n, expected))) for n in range(2, m)]
+    steps = [np.log(step(a, b, n)) for n in range(2, m)]
+    coeffs = np.polyfit(steps, errors, 1)  # log-log regression line
+    p = coeffs[0]
+    C = np.exp(coeffs[1])
+    return p, C
+
+
+def adaptive_trapezoidal(f, a, b, tol):
+    T1 = (b - a) / 2 * (f(a) + f(b))
+    c = (a + b) / 2
+    T2 = (b - a) / 4 * (f(a) + 2 * f(c) + f(b))
+
+    E1 = 4 / 3 * (T2 - T1)
+
+    if abs(E1) <= tol:
+        E2 = 1 / 3 * (T2 - T1)
+        return T2 + E2, 1
+    else:
+        left_val, left_count = adaptive_trapezoidal(f, a, c, tol * 0.7)
+        right_val, right_count = adaptive_trapezoidal(f, c, b, tol * 0.7)
+        return left_val + right_val, left_count + right_count
+
+
+print(format_error(np.exp, 0, 1, 16, np.e - 1))
+print(f"subintervals: {find_subinterval_count(np.exp, 0, 1, np.e - 1, 1e-5)}")
+print(f"exp p: {estimate_convergence_rate(np.exp, 0, 1, np.e - 1)[0]:.5f}")
+print(f"sqrt p: {estimate_convergence_rate(np.sqrt, 0, 1, 2/3)[0]:.5f}")
+print(f"adaptive estimate exp: {adaptive_trapezoidal(np.exp, 0, 1, 1e-5)[0]:.5f}")
+print(f"adaptive estimate sqrt: {adaptive_trapezoidal(np.sqrt, 0, 1, 1e-5)[0]:.5f}")
+print(f"adaptive n subintervals exp: {adaptive_trapezoidal(np.exp, 0, 1, 1e-5)[1]:.0f}")
+print(
+    f"adaptive n subintervals sqrt: {adaptive_trapezoidal(np.sqrt, 0, 1, 1e-5)[1]:.0f}"
+)
diff --git a/flake.nix b/flake.nix
index 56eeb1f..dba04dd 100644
--- a/flake.nix
+++ b/flake.nix
@@ -16,8 +16,10 @@
         buildInputs = with pkgs; [
           typst
           typstyle
+          python313
           python313Packages.sympy
           python313Packages.matplotlib
+          python313Packages.numpy
         ];
         shellHook = ''
           echo welcome!