MATH1251 Questions HELP (1 Viewer)

InteGrand · Oct 18, 2016

leehuan said:
Use double integration to find the area of the region bounded by y=x^3 and y=x^2

Just set up the integral please

$\noindent $\int_{x = 0}^{x =1} \int _{y=x^3}^{y=x^2} 1 \, \mathrm{d}y \, \mathrm{d}x$$

(I wrote like "x = " and "y = " in the bounds for the integrals, but you don't need to write them, I just put them in because it may help with visualisation/understanding and keeping track of which variable is integrated when etc.)

InteGrand · Oct 18, 2016

$\noindent Explanation of how to set up the integral (I'm assuming you've sketched the region or have visualised it): recall that the area of th region (call it $\mathcal{R}$) is $\iint _{\mathcal{R}} 1\,\mathrm{d}A$ (i.e. the integral over the region of the constant function $1$. Note that $\mathrm{d}A$ can be $\mathrm{d}x\, \mathrm{d}y$ or $\mathrm{d}y\,\mathrm{d}x$. Whichever order will be more convenient for the problem at hand will generally be be one used.) So all we need to do is set up the bounds. (Also remember, setting up the bounds for a region depends only on the region, it's completely unaffected by what function we're integrating over said region.)$

$\noindent For our given region, we can see that $x$ varies from $0$ to $1$, and for any fixed $x$ in this range, $y$ varies from $x^3$ to $x^2$ (starts at $x^3$ because the lower curve of the region is $x^3$ here). So the inner integral is going to have $y$ go from $x^3$ to $x^2$, and then $x$ will vary over $0$ to $1$ (explaining the outer limits). Note that the outermost differential is the one for the variable that goes between constant values (so in this case the outermost differential is $\mathrm{d}x$). The inner differential is for the one going between bounds that depend on the other variable, so inner differential is $\mathrm{d}y$ (i.e. because for the inner integral, we've fixed $x$ and integrated with respect to $y$).$

$\noindent In general the inner integral in the iterated integral for a double integral will have limits that are functions of the other variable (they could be constant functions, corresponding to when our region has some boundaries that are parallel to a coordinate axis, but in general they need not be constant. On the other hand, the outer limits \emph{must} be constants.$

Note that using these ideas, we could also set up and evaluate the integral in the other order (dx dy). It would be a good exercise to set it up that way too, to practise setting these integrals up. Oftentimes for multiple integrals, setting it up in some order makes the integral easy to evaluate, whereas in another order it may be extremely difficult or impossible to.

leehuan · Oct 18, 2016

I guess what I never understood (probably cause the tutorial problems are out of order) is that why are we integrating the unit constant function and not say 2. Why is it specifically the correct integrand?

seanieg89 · Oct 18, 2016

In a sense that double integral (technically that is an iterated integral, but Tonelli's/Fubini's tells us this is the same thing as the corresponding double integral) is how we define area, at least until one encounters measure theory.

Notice that this definition of area coincides with the usual definition of the area of a rectangle.

Indeed this is all that matters, because Riemann integration of function in R^n amounts to partitioning the domain into tiny boxes and forming a Riemann sum over this partition. In this case, with integrand 1, the contribution from each of the sub-boxes is simply the n-dimensional volume of that box.

Paradoxica · Oct 18, 2016

$$\noindent Let $\displaystyle (a_n)_{n>0}$ be a sequence of real numbers such that $\displaystyle \sum_{n>0} a_n $ converges.$ \\$ Prove or disprove the convergence of $\displaystyle \sum_{n>0} (a_n)^{1-\frac{1}{n+1}}$

Any ideas? I've tried comparison but to no avail. Certain methods also run into the issue of failing if and when all the a_n are less than 1.

seanieg89 · Oct 19, 2016

leehuan · Oct 21, 2016

To the guys in this course reminder that there's a review session next Thursday 3-6

RenegadeMx · Oct 21, 2016

leehuan said:
To the guys in this course reminder that there's a review session next Thursday 1-4

cool see you then

leehuan · Oct 22, 2016

$\text{Let }A\text{ be a fixed }3\times 3\text{ matrix and define a linear map}\\ T:M_{33}\to M_{33}\text{ by }T(X) = AX$

$\text{If }\lambda\text{ is a real eigenvalue of }T\text{ corresponding to an invertible eigenvector }X,\text{ find }\lambda\text{ in terms of }det(A)$

Ok I have a bit of confusion here. X is a matrix here, but they also call it an "invertible eigenvector".

My brain's getting lost in the words. Can an eigenvector be a matrix?

InteGrand · Oct 22, 2016

leehuan said:
$\text{Let }A\text{ be a fixed }3\times 3\text{ matrix and define a linear map}\\ T:M_{33}\to M_{33}\text{ by }T(X) = AX$

$\text{If }\lambda\text{ is a real eigenvalue of }T\text{ corresponding to an invertible eigenvector }X,\text{ find }\lambda\text{ in terms of }det(A)$

Ok I have a bit of confusion here. X is a matrix here, but they also call it an "invertible eigenvector".

My brain's getting lost in the words. Can an eigenvector be a matrix?

An eigenvector is going to be some element from the domain vector space of the linear map (which are generally called "vectors", even if they may be matrices, polynomials, functions, whatever). So in this case, an eigenvector is a matrix, since the domain vector space is a space of matrices (so the vectors here are matrices. Remember the term "vector" in the general setting of vector spaces doesn't only refer to arrows or n-tuples.).

$\noindent So recall the definition of $\lambda \in \mathbb{F}$ being an eigenvalue of a linear map $T\colon V\to V$, where $V$ is an arbitrary vector space over the field $\mathbb{F}$: it is that there exists some non-zero vector $\mathbf{v}\in V$ such that $T\left(\mathbf{v}\right) = \lambda \mathbf{v}$. When this is the case, by definition $\mathbf{v}$ is called an \textbf{eigenvector} of $T$ (with eigenvalue $\lambda$). So from this definition, we see that eigenvectors are just (special) elements from the space the linear map is defined on.$

InteGrand · Oct 22, 2016

leehuan said:
$\text{Let }A\text{ be a fixed }3\times 3\text{ matrix and define a linear map}\\ T:M_{33}\to M_{33}\text{ by }T(X) = AX$

$\text{If }\lambda\text{ is a real eigenvalue of }T\text{ corresponding to an invertible eigenvector }X,\text{ find }\lambda\text{ in terms of }det(A)$

Ok I have a bit of confusion here. X is a matrix here, but they also call it an "invertible eigenvector".

My brain's getting lost in the words. Can an eigenvector be a matrix?

Here is a solution in spoilers:

$\noindent To do this question, let $\lambda$ be a real eigenvalue of $T$ with invertible eigenvector $X$, then we have $T\left(X\right) = \lambda X \Rightarrow AX = \lambda X$. Take the determinant of both sides: $\left(\det A\right) \left(\det X\right) = \lambda^{3}\det X$. Since $X$ is an \emph{invertible} eigenvector, its determinant is non-zero and hence can be cancelled from the previous equation, so $\lambda^{3} = \det A$. Since $\lambda$ is real, we obtain the final answer of $\boxed{\lambda = \sqrt[3]{\det A}}$.$

leehuan · Oct 22, 2016

Trying to switch the order of integration. Please tell me that their answer is wrong.

$\int_0^1\int_x^{2x}x^2e^ydydx=\int_0^2\int_{\frac{y}{2}}^y x^2e^ydxdy$

Cause I think there's a split at y=1

Paradoxica · Oct 22, 2016

leehuan said:
Trying to switch the order of integration. Please tell me that their answer is wrong.

$\int_0^1\int_x^{2x}x^2e^ydydx=\int_0^2\int_{\frac{y}{2}}^y x^2e^ydxdy$

Cause I think there's a split at y=1

Indeed. The corresponding regions of integration don't represent identical bounds in the x-y plane, so the equality you have given above is false.

leehuan · Oct 22, 2016

Paradoxica · Oct 22, 2016

leehuan said:
$\text{Evaluated: }\int\int_\Omega e^{-(x^2+y^2)}dA=\frac{\pi}{4}\text{ where }\Omega\text{ denotes the first quadrant.}$

$\text{Hence deduce that }\int_{-\infty}^\infty e^{-\frac{x^2}2}dx=\sqrt{2\pi}$

The double integral is equal to the iterated integral which is also equal to the product of the integrals...

leehuan · Oct 22, 2016

Paradoxica said:
The double integral is equal to the iterated integral which is also equal to the product of the integrals...

Yeah, I think it was something I had seen before but why is it the product of the integrals again

Paradoxica · Oct 22, 2016

leehuan said:
Yeah, I think it was something I had seen before but why is it the product of the integrals again

Because the domain of integration is not variable dependent, the two integrals can be seperated as they are independent of each other.

leehuan · Oct 22, 2016

<-2, 0, 1>, <1, 1, 0> and <-1, 1, -2> are eigenvectors of

$A=\begin{pmatrix}0&-1&2\\-1&0&-2\\2&-2&3\end{pmatrix}$

with eigenvalues -1, -1 and 5 respectively.

$\text{Question: Find an orthogonal matrix }Q\text{ which diagonalises }A$

seanieg89 · Oct 22, 2016

leehuan said:
<-2, 0, 1>, <1, 1, 0> and <-1, 1, -2> are eigenvectors of

$A=\begin{pmatrix}0&-1&2\\-1&0&-2\\2&-2&3\end{pmatrix}$

with eigenvalues -1, -1 and 5 respectively.

$\text{Question: Find an orthogonal matrix }Q\text{ which diagonalises }A$

I won't do the calculation for you but here is the idea:

The -1-eigenspace and the 5-eigenspace are orthogonal. (This can be seen directly by computing dot products or as a consequence of A being symmetric.)

So being given these three linearly independent eigenvectors v1,v2,v3, we need to find an orthonormal basis of eigenvectors u1,u2,u3.

Normalise the third one, (so u3=v3/|v3|).

Normalise the first one to obtain u1.

To obtain u2, it is probably quickest to just take the cross product u2=u1xu3, because this will certainly satisfy the desired properties. (Note that althought u3 is unique up to sign, our choice of u1 and u2 are far from unique, any orthonormal basis of the 2-dimensional -1-eigenspace will suffice.)

Then the matrix Q with columns (u1|u2|u3) satisfies AQ=Q*diag(-1,-1,5) as required and is orthogonal by construction.

(Try to make sure you have a clear picture of the geometry of the situation here, as that is more important than the actual mechanical manipulations of diagonalisation.)

leehuan · Oct 23, 2016

This obviously converges but the ratio test is inconclusive, so how would you justify it

$\sum_{k=1}^\infty \frac{k!}{k^k}$

MATH1251 Questions HELP (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Kosovo is Serbian

Well-Known Member

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

-insert title here-

Well-Known Member

Well-Known Member

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)