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TMA4135/exercise1/exercise1.typ
fredrikr79 8690eb9655 ex1: errata
2025-08-27 19:28:36 +02:00

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#import "@preview/cetz:0.3.2";
#import "@preview/cetz-plot:0.1.1": plot
#import "@preview/physica:0.9.4": *
#import "@preview/plotsy-3d:0.1.0": plot-3d-parametric-surface
#import "@preview/fletcher:0.5.4" as fletcher: diagram, edge, node
#set page(paper: "a4", margin: (x: 2.6cm, y: 2.8cm), numbering: "1 : 1")
#set par(justify: true, leading: 0.52em)
#let FONT_SIZE = 18pt;
#set text(font: "FreeSerif", size: FONT_SIZE, lang: "us")
#show math.equation: set text(font: "Euler Math", size: (FONT_SIZE * 1.0), lang: "en")
#set heading(numbering: none)
#show heading.where(level: 1): it => {
rect(inset: FONT_SIZE / 2)[#it]
}
// #show heading.where(level: 2): it => [
// #set align(left)
// #set text(size: FONT_SIZE * 1.1, weight: "semibold")
// #(it.body)
// ]
#align(center)[
#text(size: FONT_SIZE * 2, weight: "bold")[#underline[exercise 1]]
]
these are my solutions to the first exercise set of TMA4135.
i recommend using a PDF-reader with document rotation capabilities, like
#link("https://wiki.archlinux.org/title/Zathura")[#text(blue.darken(5%))[zathura]].
this document was created using
#link("https://typst.app/")[#text(blue.darken(5%))[typst]].
#v(42pt)
#outline(title: none)
= problem 1
== a)
#[#show math.equation: set text(size: (FONT_SIZE * 0.56))
#rotate(
-90deg,
reflow: true,
table(
$bold(u(x, y, t))$, $bold(u_y)$, $bold(u_t)$, $bold(u_(x x))$, $bold(u_(x y))$, $bold(u_(y x))$,
$t^4 - cos(x y)$, $x sin(x y)$, $4 t^3$, $y^2 cos(x y)$, $x y cos(x y) + sin(x y)$, $x y cos(x y) + sin(x y)$,
$-sin(t x y)$,
$- t x cos(t x y)$,
$- x y cos(t x y)$,
$t^2 y^2 sin(t x y)$,
$t^2 x y sin(t x y)$,
$t^2 x y cos(t x y)$,
$e^(-t) sin(x) ln(y)$,
$(e^(-t) sin(x)) slash y$,
$- e^(-t) sin(x) ln(y)$,
$-e^(-t) sin(x) ln(y)$,
$(e^(-t) cos(x)) slash y$,
$(e^(-t) cos(x)) slash y$,
$e^(-x) sqrt(x^3 + y^2)$,
$(2 y e^(-x)) slash sqrt(x^3 + y^2)$,
$0$,
$(dagger)$,
$(dagger.double)$,
$(dagger dagger)$,
$(t e^t) sin(x)$, $0$, $e^t (t + 1) sin(x)$, $- t e^t sin(x)$, $0$, $0$,
$sin(t) e^(-x) + cos(t) e^(-y)$, $- cos(t) e^(-y)$, $cos(t) e^(-x) - sin(t) e^(-y)$, $sin(t) e^(-x)$, $0$, $0$,
rows: 7,
columns: 6,
)
+ [#set text(size: FONT_SIZE * 0.6, fill: gray.darken(35%))
#show math.equation: set text(size: FONT_SIZE * 0.5)
some calculations\
#table(
table.cell(
rowspan: 2,
$(dagger)$,
),
table.cell(
rowspan: 2,
$
& quad pdv(, x, 2)e^(-x) sqrt(x^3 + y^2) \
& = pdv(, x) ((3 x^2 e^(-x) ) / (2 sqrt(x^3 + y^2))
- e^(-x) sqrt(x^3 + y^2)) \
& = 3/2 dot ((2 x e^(-x) - x^2 e^(-x)) sqrt(x^3 + y^2)
- (3 x^4 e^(-x)) / (2 sqrt(x^3 + y^2))) / (x^3 + y^2)
- (3 x^2 e^(-x) ) / (2 sqrt(x^3 + y^2))
+ e^(-x) sqrt(x^3 + y^2) \
& = (3 e^(-x) ((2 x - x^2) (x^3 + y^2) - 3 x^4))
/ (4(x^3 + y^2)^(3/2))
- (6 x^2 e^(-x) (x^3 + y^2)) / (4 (x^3 + y^2)^(3/2))
+ (4 e^(-x) (x^3 + y^2)^2) / (4 (x^3 + y^2)^(3/2)) \
& = (3 e^(-x) (2 x y^2 - x^5 - x^2 y^2 - x^4)
- 6 x^2 e^(-x) (x^3 + y^2)
+ 4 e^(-x) (x^6 + 2 x^3 y^2 + y^4))
/ (4 (x^3 + y^2)^(3/2)) \
& = (6 x y^2 e^(-x) - 3 x^5 e^(-x) - 3 x^2 y^2 e^(-x) - 3 x^4 e^(-x)
- 6 x^5 e^(-x) + 6 y^2 e^(-x)
+ 4 x^6 e^(-x) + 8 x^3 y^2 e^(-x) + 8 y^4 e^(-x))
/ (4 (x^3 + y^2)^(3/2)) \
& = (6 x y^2 e^(-x) - 9 x^5 e^(-x) - 3 x^2 y^2 e^(-x) - 3 x^4 e^(-x)
+ 6 y^2 e^(-x)
+ 4 x^6 e^(-x) + 8 x^3 y^2 e^(-x) + 8 y^4 e^(-x))
/ (4 (x^3 + y^2)^(3/2)) \
& = e^(-x) dot (6 x y^2 - 9 x^5 - 3 x^2 y^2 - 3 x^4
+ 6 y^2
+ 4 x^6 + 8 x^3 y^2 + 8 y^4)
/ (4 (x^3 + y^2)^(3/2)) \
& #[a few errors somewhere, but close enough...]
$,
),
$(dagger.double)$,
$
& quad pdv(, y, x) e^(-x) sqrt(x^3 + y^2) \
& = pdv(, y) ((3 x^2 e^(-x) ) / (2 sqrt(x^3 + y^2))
- e^(-x) sqrt(x^3 + y^2)) \
& = (-3 x^2 y e^(-x))/2 dot (x^3 + y^2)^(-3/2)
- (2 y e^(-x))/(2 sqrt(x^3 + y^2))
$,
$(dagger dagger)$,
$
& quad pdv(, x, y) e^(-x) sqrt(x^3 + y^2) \
& = pdv(, x) space (-y e^(-x))/(sqrt(x^3 + y^2)) \
& = (y e^(-x) sqrt(x^3 + y^2) + y e^(-x) dot 1 slash 2 dot (x^3
+ y^2)^(-1/2)) / (x^3 + y^2) \
& = (y e^(-x) (sqrt(x^3 + y^2) + 1 slash 2 dot (x^3
+ y^2)^(-1/2))) / (x^3 + y^2)
$,
columns: 4,
stroke: none,
)
],
)
]
== b)
define $ f^i_k := (partial f^i) / (partial y^k)
quad quad f^i_(k l) := (partial f^i) / (partial y^k partial y^l) $
and let the jacobian matrix $ frak(J) := [ (f_y)_(i j) = f^i_j ] $
where $f^2_y$ is the matrix product and $f_(y y)$ has entries $(f^i_(k l))$.
consider $ f : RR^m -> RR^m,
space y |-> (f^1 (y^1, ..., y^m), ..., f^m (y^1, ..., y^m))^T $
show that $ (f_y f)_y f = f^T f_(y y) f + f^2_y f $
#align(center)[#line(length: 75%)]
$
y |-> #text(red)[$f(y)$] |-> pdv(, y) (pdv(f, y)(f(#text(red)[$x$])))
$ $
=> (f_y f)_y f & = pdv(, y) (pdv(f, y) (f(f(y)))) \
& = pdv(, y) (pdv(f, y) (f^2(y)))
$
$
y |-> #text(red)[$f(y)$] |-> f^T (pdv(f, y, 2)(#text(red)[$x$])) + pdv(f, y)
(pdv(f, y) (#text(red)[$x$]))
$ $
=> f^T f_(y y) f + f^2_y f & = (f(pdv(f, x, 2)(f(y))))^T + pdv(f, x) (pdv(f, x) (f(y)))
$
we can see through $eta$-reduction that we only need to show $ (f_y f)_y = f^T f_(y y) + f^2_y $
and further that $ f_y f = f^T f_y + f_y f $
thus we need to prove that $ f^T f_y = 0 $
but $ f^T f_y = (f(pdv(f, y)))^T = (f^1(f^1_y, ..., f^m_y), ..., f^m (f^1_y, ..., f^m_y)) $
no more, i yield, i yield!!
= problem 2
== a)
let $f(x) := x^4 + 3 x^3 - 2 x + 5$; find all taylor polynomials around
$x_0 := -2$.
#align(center)[#line(length: 75%)]
recall that each term is given by $ P_k (x) := (f^((k)) (x_0)) / k! (x - x_0)^k $
for $k in [0, deg(f)] inter ZZ$.
first compute
$
f'(x) & = 4x^3 + 9x^2 - 2 \
f''(x) & = 12x^2 + 18x \
f^((3))(x) & = 24 x + 18 \
f^((4))(x) & = 24
$
then
$
P_0(x) & = f(-2) = 16 - 24 + 4 + 5 = 1 \
P_1(x) & = f'(-2) dot (x + 2) = 2 (x+2) = 2x + 4 \
P_2(x) & = (f''(-2))/2 dot (x + 2)^2 = 6 x^2 + 24x + 24 \
P_3(x) & = (f^((3)) (-2))/6 dot (x + 2)^3 = -5 (x + 2)^3 \
& = -5x^3 - 30 x^2 - 60 x - 40 \
P_4(x) & = (x + 2)^4 = x^4 + 8 x^3 + 24 x^2 + 32 x + 16
$
note that there are finitely many unique derivatives, as $f(x)$ is a polynomial
of degree 4.
then the $k$-th taylor polynomial can be expressed as
$
T_k := sum_(i=0)^k P_i
$
or recursively
$
T_k = T_(k-1) + P_k quad and quad T_0 = P_0 = 1
$
for $k in NN^+$.
thus the taylor polynomials for $f$ are
$
T_0 & = P_0 = 1 \
T_1 & = P_0 + P_1 = 2x + 5 \
T_2 & = 6x^2 + 26x + 29 \
T_3 & = -5x^3 - 24x^2 - 34x - 11 \
T_4 & = x^4 + 3x^3 - 2x + 5 = f(x)
$
naturally we are able to perfectly describe a fourth-degree polynomial with
a taylor series.
== b)
let $g(x) := ln(1 + x)$; calculate its maclaurin series -- i.e. taylor series at
$x_0 = 0$.
#align(center)[#line(length: 75%)]
first we differentiate
$
g'(x) & = 1/(1+x) \
g''(x) & = -1/(1+x)^2 \
& dots.v \
g^((k)) (x) & = (-1)^(k-1) dot (k-1)! dot (1 + x)^(-k)
$
for $k in NN^+$.
as the function is infinitely differentiable and continuous around $x = 0$, we
can conclude that $g(x)$ is analytic.
recall that the maclaurin series of an analytic function $f$ is
$
f(x) = sum_(i=0)^infinity (f^((i)) (0))/(i!) x^i
$
fortunately we have
$
g^((k)) (0) = (-1)^(k-1) dot (k - 1)!
$
so
$
g(x) & = sum_(i=1)^k (g^((i)) (0)) / (i!) x^i \
& = sum_(i=1)^k (-1)^(i-1) (i-1)! / (i!) x^i \
& = sum_(i=1)^k ((-1)^(i-1) x^i) / i \
& = x - x^2 / 2 + x^3 / 3 - dots.c
$
thus we have the maclaurin series of $g(x)$.
= problem 3
== a)
are $1+x$, $1-x$ and $x-x^2$ linearly independent in $P_2$? what do they span?
#align(center)[#line(length: 75%)]
let us denote these in vector notation as
$
vec(0, 1, 1), vec(0, -1, 1) #[and] vec(-1, 1, 0)
$
respectively. thus we can determine their dependency via gauss-jordan
elimination
$
mat(0, 0, -1; 1, -1, 1; 1, 1, 0) ~
mat(1, 0, 0; 0, 1, 0; 0, 0, 1) = I_3
$
the three vectors are thus linearly independent. then what is their span?
since they are linearly independent in $P_2$, they form a basis for the vector
space and thus span out $P_2$. of note: $P_2$ is itself isomorphic to $RR^3$.
== b)
our affine space has two conditions, $p(1) = 1$ and $p(2) = 2$.
let
$
p(x) := a x^3 + b x^2 + c x + d
$
such that
$
p(1) = a + b + c + d = 1
$
and
$
p(2) = 8 a + 4 b + 2 c + d = 2
$
we can represent this system with a matrix
$
mat(1, 1, 1, 1 | 1; 8, 4, 2, 1 | 2) tilde
mat(1, 1, 1, 1 | 1; 7, 3, 1, 0 | 1)
$
thus the condition vectors are linearly independent and the matrix has rank 2,
so they span out a two-dimensional constraint space. by the rank-nullity theorem
we can conclude that the solution space must have
$
dim(P_3) - "rank" = 4 - 2 = 2
$
dimensions.
== c)
we can create a basis out of the three vectors
$
x-1, x^2 - 1 #[and] x^3 - 1
$
to form a three-dimensional space, since they are linearly independent.
from the results of the last task we can intuitively guess that the three
conditions will lead to a system of equations with three unknowns. thus the
system is solvable and we may indeed choose arbitrary values for our
conditions.
so let
$
p(x) := alpha (x - 1) + beta (x^2 - 1) + gamma (x^3 - 1)
$
such that
$
p(0) & = - alpha - beta - gamma = y_0, \
p(1) & = 0 = y_1, \
p(2) & = alpha + 3 beta + 7 gamma = y_2
$
this is different from my original expectation. we can tell that $y_1$ must be
0, thus cannot be chosen arbitrarily. then we effectively only have two
equations, thus ending up in the same situation as the last subtask, meaning we
will not be able to choose the remaining values arbitrarily, since the system of
equations will be underdetermined.
as such, if we choose for $y_1 = 0$ to hold, it is possible.
= problem 4
== a)
#let inner(f, g) = $angle.l #f, #g angle.r$
prove that ${sin(t), cos(t), 1}$ is orthogonal in the space $C[0, 2 pi]$ with
inner product
$
inner(f, g) := integral_0^(2 pi) f(s) g(s) dd(s)
$
i.e. that it is an orthogonal basis for $C[0, 2 pi]$.
#align(center)[#line(length: 75%)]
we must compute the pair-wise inner product of each base vector
$
// inner(sin, cos) & = integral_0^(2 pi) sin(t) cos(t) dd(t) = \
inner(sin, 1) & = integral_0^(2 pi) sin(t) dd(t) = 0, \
inner(1, cos) & = integral_0^(2 pi) cos(t) dd(t) = 0
$
because $sin(t)$ and $cos(t)$ all have a period of $2 pi$. lastly,
$
inner(sin, cos) & = integral_0^(2 pi) sin(t) cos(t) dd(t) \
& = 1/2 integral_0^(2 pi) sin(2 t) dd(t) \
& = 1/4 integral_0^(4 pi) sin(u) dd(u) = 0.
$
thus they are all orthogonal and they form a basis under this definition of
inner product. to make it an orthonormal basis, we can scale each base component
by its length, such that
$
frak(O) := {sin(t)/a, cos(t)/b, 1/c}
$
forms an orthonormal basis, where
$
a & = sqrt(inner(sin, sin)) = (integral_0^(2 pi) sin^2 t dd(t))^(1 slash 2) = sqrt(pi) \
b & = sqrt(inner(cos, cos)) = (integral_0^(2 pi) cos^2 t dd(t))^(1 slash 2) = sqrt(pi) \
c & = sqrt(inner(1, 1)) = (integral_0^(2 pi) dd(t)) = sqrt(2 pi)
$
== b)
to form an orthonormal basis for the monomials ${1, x, x^2}$, we use the
gram-schmidt method with $vectorbold(v_1) := 1$.
then
$
vectorbold(v_2) & := x - "proj"_(vectorbold(v_1)) (x) \
& = x - (inner(vectorbold(v_1), x) / inner(
vectorbold(v_1),
vectorbold(v_1)
)) vectorbold(v_1) \
& = x - inner(1, x) / inner(1, 1) \
& = x - 1/2 integral_(-1)^1 x dd(x) \
& = x
$
and similarly for the last vector
$
vectorbold(v_3) & := x^2
- "proj"_(vectorbold(v_1))(x^2)
- "proj"_(vectorbold(v_2))(x^2) \
& = x^2
- 1/2 integral_(-1)^1 x^2 dd(x)
- (integral_(-1)^1 x^2 dd(x))^(-1)
dot integral_(-1)^1 x^3 dd(x) \
& = x^2 - 1/3
$
thus we have found an orthonormal base using gram-schmidt
$
frak(O) := {1/a, x/b, (x^2 - 1/3)/c}
$
where $a & = sqrt(2), b & = sqrt(2/3)$ and $c & = sqrt(8/45)^(dagger)$.
#[#show math.equation: set text(size: FONT_SIZE * 0.75, gray.darken(50%))
$
dagger: inner(x^2-1/3, x^2-1/3) & = integral_(-1)^1 (x^2 - 1/3)^2 dd(x) \
& = integral_(-1)^1 (x^4 - 2/3 x^2 + 1/9) dd(x)
= 8/45
$
]
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