I have to evaluate
$$\int^1_{-1} \int^{ \sqrt {1-x^2}}_{ - \sqrt {1-x^2}} \int^1_{-\sqrt{x^2+y^2}}dzdydx$$
using spherical coordinates.
This is what I have come up with
$$\int^1_{0} \int^{ 2\pi}_0 \int^{3\pi/4}_{0}r^2\sin\theta d\phi d\theta dr$$
by a combination of sketching and...
A satellite is in a circular orbit a distance $h$ above the surface of the earth with speed $v_0$. It suffers a head-on collision with some debris which reduces its speed to $kv_0$, where $k$ is a constant in the range $0<k<1$, but does not change its direction. Calculate the eccentricity of the...
Homework Statement
a satellite is in a circular orbit a distance $h$ above the surface of the earth with speed $v_0$, booster rockets are fired which double the speed of the satellite without changing the direction. Find the subsequent orbit.
Homework Equations
The Attempt at a Solution
Before...
I have found via integration that the y coordinate is $$y =h/2 = 120 mm$$. The x coordinate is $$x = \frac{-4r}{3\pi} = -51.9mm$$ and the z coordinate is $$z = r - \frac{4r}{3\pi} = 69.1 mm$$. I have no answers in my textbook so cant confirm whether i am correct or not.
Thanks, Ray. Maybe i should have made what i HAD done already a little more clear.
I have already done what you said, regarding a 4x4 matrix and eliminating the first row and column, what results is another lower triangular matrix. the question asks me to show that that if i > j then ##B_{ij}##...
Homework Statement
Prove that is ##A## is lower triangular and ##B_{ij}## is the matrix that results when the ith row and jth column of A are deleted, then ##B_{ij}## is lower triangular if i > j.
Homework Equations
The Attempt at a Solution
I know that a square matrix is lower triangular...
Thank you ever so much.
The first line
If m = n+1 then g(x) = m+1. I find this a little confusing. if m is some element of ##S_{n+1}## and m = n+1, i get this part.... but then to say that g(x) = m+1.
This may sound really stupid, but to me, i'm not quite seeing it.
Claim: ##g## is a 1-1...
Actually i don't have a teacher/ lecturer, i made the choice to drop out of university to look after a dying grandparent, this resulted in me taking a degree with the open university which does not cover analysis of this kind. so i am self-studying Analysis 1 by terrence tao. When i do come...
I am setting out to prove that g is onto;
let ## m \in S_{n+1} ##, we must show that there exist ## y \in X \cup \{x\}## such that ## g(y) = m ##.
since ##f## is a bijection between X and ##S_n## there is some ##y \in X ## such that ##f(y) = m## and if ##m= n+1 ## then ##g(x) = m ## by the...
I see the difference. to state that for all ## y \in X, f(y) = m ## is a false statement as it implies that the image of every y in X is m. I understand the difference. I do apologise i wrongfully assumed that you were trying to point out some apparent difference between 'there exists' and...
Well.
first, what is the difference?
there is some,
there exists,
i cant quite make the difference, could you give me an example of where one is true and the other is not?
moving on,
there is some ##y \in X##such that## f(y) = m## and if ## m = n +1 ## then g(x) = m by the definition of ##g##...
g(y) = f(y) for all y in X,
g(y) = n + 1 when x = y.
ok, now to prove the 'there exists' part;
let $$ m \in S_{n+1} $$ we must show that there exists $$ y \in X \cup \{x\} such that g(y) = m $$, now $$ g(y) = f(y) = m $$ for all $$ y \in X. $$and $$ g(y) = n+ 1 = m for y = x $$
so $$ g(y) =...
I'm sorry, a little more detail please?
so is it the wording thats the problem in the first issue? so if i was to say 'we have' instead of 'there exists', secondly that was a typo, i meant to type $$ y_1 = y_2 $$ instead of $$ i_1 = i_2 $$.
surely the last one proves it that it is injective...
right for the onto proof, i have said that;
given any $$ i \in \{i \in N : 1 \le i \le n + 1 \} $$
there exists $$ y \in X \cup \{x\} $$ such that $$ g(y) = i $$ it follows that g is onto, this quite simply has to be true!
for the one to one proof;
suppose that $$ g(y_{1}) = g(y_{2})$$
then $$...
g(y) = n+1 ? because thats the element its mapped to when y = x? i know this is true, surely?
ONE- ONE; so each element in the domain of g is mapped to a unique element in the co-domain. so y = x iff f(y) = f(x).
ONTO; the image set of g is equal to the co-domain.
could i use the fact that f...
g(y): = g(x) when y = x?
To prove that g is a bijection , i just have to show that X U {x} has equal cardinality with the set $$ \{i \in N : 1 \le i \le n +1 \} $$ right?
right, here is some work on a function g;
I will now define a function g, such that
$$g: X\cup\{x\} \rightarrow \{ i \in N:1 \le i \le n+1\} $$
by the following rule
for any $$ y \in X \cup \{x\}, $$
we define $$ g(y):=f(y) for y \in X $$ and $$ g(y):=f(y) + 1 when y = x $$
since f is a...
I cant think of anything more basic other than actually letting $$ X = \{x_1, x_2....x_n\}$$ then showing that $$ X \cup {x} = \{x_1, x_2....x_{n+1}\} $$. since x is not an element of X. Other than that LCKurtz...
Let X be a finite set with cardinality n, denoted by #X, then there exists a bijection;
$$f:X \rightarrow \{ i \in N : 1 \le i \le n\} $$. Since x is not an element of X then $$X \cup \{x\} $$ is the set of elements that contains X or {x}, then there must exist a bijection $$ g:X \cup \{x\}...
Oh there is another one;
Let n be a natural number. A set X is said to have cardinality n, iff it has equal cardinality with $$ \{i \in N : 1 \le i \le n\} $$. We also say that X has n elements iff it has cardinality n.
A set is finite iff it has cardinality n for some natural number n; otherwise, the set if called infintie. If X is a finite set, we use #(X) to denote the cardinality of X.
That is what your asking for, i think?
Homework Statement
Let X be a finite set and let x be an object which is not an element of X. Then X U {x} is finite and #(X U {x}) = #(X) + 1
The Attempt at a Solution
Let X be a finite set such that X has cardinality n, denoted by #X.
Suppose that ## x \notin X##, then the set X U {x} has...
Homework Statement
A catapult projects a stone in the normal direction to a playground which slopes at a angle of 10 degrees to the horizontal.
The initial speed of the stone is 18 m/s. calculate the range parallel to the playground.
Homework Equations
the usual ones!
The Attempt at a...