maths 1B last minute questions (1 Viewer)

Drsoccerball · Oct 31, 2016

InteGrand said:
$\noindent We need to integrate something like $\sqrt{1+\cos x}$. To do this, we can recall that $1+\cos x = 2\cos^{2}\frac{x}{2}$. Taking square roots of both sides yields $\sqrt{1+\cos x} = \sqrt{2}\left| \cos \frac{x}{2}\right|$. So the function we'll end up integrating is (a multiple of) $\left|\cos \frac{x}{2}\right|$. Why absolute values? Because $\sqrt{a^2} = |a|$.$

I still don't get the answer and whats wrong with my way of integration?

InteGrand · Oct 31, 2016

Drsoccerball said:
I still don't get the answer and whats wrong with my way of integration?

Drsoccerball said:
$r^2 = 1 + 2\cos{\theta} + \cos^2{\theta}, \frac{dr}{d\theta} = -\sin{\theta}$

$I = \int_0^{2 \pi}{\sqrt{2 + 2\cos{x}}}dx$

$I = \sqrt{2} \int_0^{2 \pi}{\sqrt{1 + \cos{x}}}dx$

$I = \sqrt{2} \int_0^{2 \pi}{\frac{\sin{x}}{\sqrt{1 - \cos{x}}}}dx$

$I = 2\sqrt{2} [-\sqrt{1- \cos{x}}]_0^{2 \pi}$
?

Edit: Imm guessing you mean that |-cosx| but why ????????/

$\noindent At one stage in your working you had a $\sqrt{\sin^{2} x}$ that you simplified to $\sin x$, but it actually is $\left|\sin x\right|$, because $\sqrt{a^2} = |a|$. In our interval of integration, $\sin x$ will be negative sometimes so it is crucial to include the absolute value signs. You'll then need to split the region up according to when $\sin x$ is positive or negative in order to get rid of the absolute value signs. In the parts of the interval where $\sin x$ is positive, the absolute values will become $|\sin x| = \sin x$, and in the parts where $\sin x$ is negative, it'll become $|\sin x | = -\sin x$.$

Drsoccerball · Oct 31, 2016

InteGrand said:
$\noindent At one stage in your working you had a $\sqrt{\sin^{2} x}$ that you simplified to $\sin x$, but it actually is $\left|\sin x\right|$, because $\sqrt{a^2} = |a|$. In our interval of integration, $\sin x$ will be negative sometimes so it is crucial to include the absolute value signs. You'll then need to split the region up according to when $\sin x$ is positive or negative in order to get rid of the absolute value signs. In the parts of the interval where $\sin x$ is positive, the absolute values will become $|\sin x| = \sin x$, and in the parts where $\sin x$ is negative, it'll become $|\sin x | = -\sin x$.$

Makes sense now

so my integral becomes : $\sqrt{2} (\int_0^{\pi}{\frac{\sin{x}}{\sqrt{1-\cos{x}}}} - \int_{\pi}^{2\pi}...$

thanks !!

Drsoccerball · Oct 31, 2016

1) How can you determine the nullity of this transformation without finding the matrix? Also does P_2 -> P_3 mean that we put in a polynomial of degree 2 and we get out one of degree 3 ?

$T(p(x)) = (x^2 + 1)p'(x) -2xp(x) P_2 -> P_3$

2) Let A be a fixed 3 x 3 matrix and define a linear map T: M_{33} -> M_{33} by T(x) = AX. If lambda is a real eigenvalue of T corresponding to an invertible eigenvector X, find lambda in terms of det(A).

3) Let T be a linear map which reflects vectors in R^2 about the line y = x.

a) Explain why <1,1> and <1,-1> are eigenvectors and give their corresponding eigenvalues.

InteGrand · Oct 31, 2016

Drsoccerball said:
1) How can you determine the nullity of this transformation without finding the matrix? Also does P_2 -> P_3 mean that we put in a polynomial of degree 2 and we get out one of degree 3 ?

$T(p(x)) = (x^2 + 1)p'(x) -2xp(x) P_2 -> P_3$

2) Let A be a fixed 3 x 3 matrix and define a linear map T: M_{33} -> M_{33} by T(x) = AX. If lambda is a real eigenvalue of T corresponding to an invertible eigenvector X, find lambda in terms of det(A).

3) Let T be a linear map which reflects vectors in R^2 about the line y = x.

a) Explain why <1,1> and <1,-1> are eigenvectors and give their corresponding eigenvalues.

3) Remember, geometrically, eigenvectors are those special directions (really vectors) that only get scaled by T, without having their direction changed (a typical vector won't have this property as it would normally get rotated a bit too). And the factor an eigenvector gets scaled by (which is negative if the vector gets flipped in direction too) is the eigenvalue for that eigenvector.

Take a look at this Mona Lisa picture to get a feel for this (and read the picture description): https://en.wikipedia.org/wiki/Eigen...rs#/media/File:Mona_Lisa_eigenvector_grid.png . Basically a linear map is applied to the picture, and the blue vector direction turns out to be unchanged (so vectors lying in the line of the blue direction are eigenvectors), whilst this is not the case for the red one (since that changes direction when acted upon by the linear map).

Now to the question at hand. Clearly the vectors <1,1> and <1,-1>, when flipped about the line y = x, don't change their line direction (one of them flips around but this is just scaling by -1). This is because <1,1> stays as is (since it is on the line y = x), whilst the vector <1, -1> just flips about the origin (gets negated). So <1,1> is an eigenvector with eigenvalue 1 (since T(<1,1>) = <1,1>, since it's unchanged), and <1,-1> is an eigenvector with eigenvalue -1 (since it gets flipped, i.e. T(<1,-1>) = –<1,-1>).

Drsoccerball · Oct 31, 2016

InteGrand said:
3) Remember, geometrically, eigenvectors are those special directions (really vectors) that only get scaled by T, without having their direction changed (a typical vector won't have this property as it would normally get rotated a bit too). And the factor an eigenvector gets scaled by (which is negative if the vector gets flipped in direction too) is the eigenvalue for that eigenvector.

Take a look at this Mona Lisa picture to get a feel for this (and read the picture description): https://en.wikipedia.org/wiki/Eigen...rs#/media/File:Mona_Lisa_eigenvector_grid.png . Basically a linear map is applied to the picture, and the blue vector direction turns out to be unchanged (so vectors lying in the line of the blue direction are eigenvectors), whilst this is not the case for the red one (since that changes direction when acted upon by the linear map).

Now to the question at hand. Clearly the vectors <1,1> and <1,-1>, when flipped about the line y = x, don't change their line direction (one of them flips around but this is just scaling by -1). This is because <1,1> stays as is (since it is on the line y = x), whilst the vector <1, -1> just flips about the origin (gets negated). So <1,1> is an eigenvector with eigenvalue 1 (since T(<1,1>) = <1,1>, since it's unchanged), and <1,-1> is an eigenvector with eigenvalue -1 (since it gets flipped, i.e. T(<1,-1>) = –<1,-1>).

I was looking for this everywhere.

InteGrand · Oct 31, 2016

Drsoccerball said:
1) How can you determine the nullity of this transformation without finding the matrix? Also does P_2 -> P_3 mean that we put in a polynomial of degree 2 and we get out one of degree 3 ?

$T(p(x)) = (x^2 + 1)p'(x) -2xp(x) P_2 -> P_3$

2) Let A be a fixed 3 x 3 matrix and define a linear map T: M_{33} -> M_{33} by T(x) = AX. If lambda is a real eigenvalue of T corresponding to an invertible eigenvector X, find lambda in terms of det(A).

3) Let T be a linear map which reflects vectors in R^2 about the line y = x.

a) Explain why <1,1> and <1,-1> are eigenvectors and give their corresponding eigenvalues.

$\noindent 2) Suppose $\lambda$ is a real eigenvalue of $T$ corresponding to invertible eigenvector (which is a matrix here) $X$. Then $AX = \lambda X$. Taking determinants of both sides, we have $\left(\det A\right)\left(\det X\right) = \lambda^{3} \det X$. Since $X$ is invertible, its determinant is non-zero and can be cancelled from both sides to give $\lambda^{3} = \det A \Rightarrow \lambda = \sqrt[3]{\det A}$.$

Drsoccerball · Oct 31, 2016

InteGrand said:
$\noindent 2) Suplose $\lambda$ is a real eigenvalue of $T$ corresponding to invertible eigenvector (which is a matrix here) $X$. Then $AX = \lambda X$. Taking determinants of both sides, we have $\left(\det A\right)\left(\det X\right) = \lambda^{3} \det X$. Since $X$ is invertible, its determinant is non-zero and can be cancelled from both sides to give $\lambda^{3} = \det A \Rightarrow \lambda = \sqrt[3]{\det A}$.$

Where did you get lambda ^3 from ? Is it because its a 3 by 3 matrix?

InteGrand · Oct 31, 2016

Drsoccerball said:
1) How can you determine the nullity of this transformation without finding the matrix? Also does P_2 -> P_3 mean that we put in a polynomial of degree 2 and we get out one of degree 3 ?

$T(p(x)) = (x^2 + 1)p'(x) -2xp(x) P_2 -> P_3$

2) Let A be a fixed 3 x 3 matrix and define a linear map T: M_{33} -> M_{33} by T(x) = AX. If lambda is a real eigenvalue of T corresponding to an invertible eigenvector X, find lambda in terms of det(A).

3) Let T be a linear map which reflects vectors in R^2 about the line y = x.

a) Explain why <1,1> and <1,-1> are eigenvectors and give their corresponding eigenvalues.

$\noindent 1) Yeah that's essentially what the $P_{2}\to P_{3}$ would mean (though technically it means polynomials of degree \emph{at-most} $2$ get inputted and \emph{at-most} $3$ get outputted. We can still input linear polynomials and lower degree in addition to quadratics, and we can get out quadratic, linear, constant, as well as of course cubics).$

$\noindent For a polynomial $p(x) = ax^2 + bx + c \in P_{2}$, we have$

$\begin{align*}T(p(x)) &= \left(x^2 + 1\right)\left(2ax + b\right) - 2x \left(ax^2 + bx + c\right) \\ &= -b\left(x^{2}-1\right)+2(a-c)x.\end{align*}$

$\noindent It follows that $\mathrm{im}(T) = \mathrm{span}\left\{x^2 -1, x\right\}$, which implies $\mathrm{rank}(T) = 2$, so $\mathrm{nullity}(T) = 3-2 = 1$ (since the domain of $T$ is $P_2$, which has dimension $3$ and using Rank-Nullity Theorem).$

InteGrand · Oct 31, 2016

Drsoccerball said:
Where did you get lambda ^3 from ? Is it because its a 3 by 3 matrix?

$\noindent Yep. Recall the general identity $\det \left(cM\right) = c^{n}\det M$, if $M$ is an $n\times n$ matrix and $c$ a scalar.$

Drsoccerball · Oct 31, 2016

All these identities that I don't know :'( rip

Drsoccerball · Oct 31, 2016

seanieg89 · Oct 31, 2016

What's with the dx's? lol.

And are the a_k real numbers? If so, the answer is no (consider the alternating harmonic series).

If the a_k are positive real numbers, the answer is yes.

seanieg89 · Oct 31, 2016

Drsoccerball said:
All these identities that I don't know :'( rip

If you are approaching these kinds of facts as identities to memorise, you are going to make life hard for yourself as there are too many to count!

Instead, when you learn about a new object/operation, make sure that the definition is really ingrained in your head (multiple equivalent definitions if possible, and understand why they are equivalent). In this case, it should be fairly clear from any definition of the determinant that multiplying a row/column of the matrix by a constant multiplies the determinant by that constant. So multiplying the entire matrix by that constant multiplies the determinant by c^n.

This can also be seen as a consequence of det(AB)=det(A)det(B), which is perhaps the most important determinant identity to know. This identity is less obvious using some definitions of det than others.

In general, when you are introduced to a new mathematical object, it is a valuable exercise to try to guess/deduce/prove properties of it yourself before diving into doing computational questions.

Drsoccerball · Oct 31, 2016

seanieg89 said:
What's with the dx's? lol.

LOL I have done way too much integration.

If the a_k are positive real numbers, the answer is yes.

Should've specified they're all positive.

leehuan · Nov 1, 2016

Drsoccerball said:
LOL I have done way too much integration.

Should've specified they're all positive.

Integration isn't even fun anymore.

$\text{To answer that question, note that whilst the converse is false what you gave is true.}\\ \text{The convergence of }\sum_{k=1}^\infty a_k\text{ implies the convergence of }\sum_{k=1}^\infty(-1)^k a_k\\ \text{as we say the latter is 'absolutely convergent', which always implies convergence.}$

davidgoes4wce · Nov 1, 2016

Curious to know is this Maths 1B UWS (Western Sydney) version?

leehuan · Nov 1, 2016

UNSW.

Drsoccerball · Nov 1, 2016

Drsoccerball · Nov 1, 2016

maths 1B last minute questions (1 Viewer)

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