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First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

Thread starter leehuan
Start date Jul 16, 2016

I

InteGrand

Well-Known Member

#21

Re: MATH1231/1241/1251 SOS Thread

When you divided by y in separating variables, you had to assume y =/= 0, making you lose that solution of y = 0.

And what did you want clarification with for the solution from the back of the book? You can check that it is differentiable everywhere and satisfies the given ODE.

leehuan

Well-Known Member

#22

Re: MATH1231/1241/1251 SOS Thread

Whoops... clumsily forgot about the y≠0 problem...

Yeah that makes sense but, I have no idea where they got a≤x≤b from. Can't figure out the significance of using two different constants a and b to replace C

I

InteGrand

Well-Known Member

#23

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
Whoops... clumsily forgot about the y≠0 problem...

Yeah that makes sense but, I have no idea where they got a≤x≤b from. Can't figure out the significance of using two different constants a and b to replace C

The solution they gave is essentially a more 'general' sort of solution than (x – c)^3, which is a shifted cubic. The more 'general' solution given is essentially taking this cubic but 'widening out' the horizontal point of the cubic to any arbitrary length. This preserves differentiability of the function because the derivatives at these 'joining points' x = a and x = b are still 0, as the derivative approaches 0 from either side and the function is still continuous at these points (and the function is differentiable everywhere else too clearly).

Flop21

Well-Known Member

#24

Re: MATH1231/1241/1251 SOS Thread

How did this get simplified to this??

I

InteGrand

Well-Known Member

#25

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
How did this get simplified to this??

Compute the partial derivatives (which they've done) and sub. in the given (x,y) point.

$\noindent E.g. They found $F_x \left(x,y\right) = \pi y^2 \cos \left(\pi xy^2\right)$. So at $(x,y) = (2,-1)$, $F_x (2,-1) = \pi \cdot (-1)^2 \cos \left(\pi \cdot 2\cdot (-1)^2\right) = \pi \cos 2\pi = \pi$, since $\cos 2\pi = 1$.$

Flop21

Well-Known Member

#26

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Compute the partial derivatives (which they've done) and sub. in the given (x,y) point.

$\noindent E.g. They found $F_x \left(x,y\right) = \pi y^2 \cos \left(\pi xy^2\right)$. So at $(x,y) = (2,-1)$, $F_x (2,-1) = \pi \cdot (-1)^2 \cos \left(\pi \cdot 2\cdot (-1)^2\right) = \pi \cos 2\pi = \pi$, since $\cos 2\pi = 1$.$

Ohh okay! They subbed the points in, explains things. Thanks!

Flop21

Well-Known Member

#27

Re: MATH1231/1241/1251 SOS Thread

Another tangent to surface question, how do I do this one: S: z^2+x^2+y^2 = 1., x0 = (1/3, 1/2, root(23)/6)

Find normal vector and equation of the tangent plane to the surface S at the point x0.

What is confusing me is the z. So do I move everything but the z to the RHS? Then solve?

I

InteGrand

Well-Known Member

#28

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Another tangent to surface question, how do I do this one: S: z^2+x^2+y^2 = 1., x0 = (1/3, 1/2, root(23)/6)

Find normal vector and equation of the tangent plane to the surface S at the point x0.

What is confusing me is the z. So do I move everything but the z to the RHS? Then solve?

$\noindent There's an easier way to do this particular one. This surface is a sphere centred at the origin, so the normal to the surface at any point $\left(x,y,z\right)$ on the surface can geometrically be seen to just be that same vector $\left(x,y,z\right)$. This means a normal to $S$ at that point $\mathbf{x}_0$ is $\mathbf{n} = \begin{bmatrix}1/3 \\ 1/2 \\ \sqrt{23}/6 \end{bmatrix}$ (i.e. same vector as $\mathbf{x}_0$). Note that $\mathbf{n}\cdot \mathbf{x}_0 = \mathbf{x}_0 \cdot \mathbf{x}_0 = \left \|\mathbf{x}_0\right \|^2 = 1^2 = 1$ (as $\mathbf{x}_0$ lies on the unit sphere). So an equation of the tangent plane is$

$\begin{align*} \mathbf{n}\cdot \mathbf{x} &= \mathbf{n}\cdot \mathbf{x}_0 \\ \iff \frac{1}{3} x +\frac{1}{2}y +\frac{\sqrt{23}}{6}z &= 1. \end{align*}$

I

InteGrand

Well-Known Member

#29

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
Another tangent to surface question, how do I do this one: S: z^2+x^2+y^2 = 1., x0 = (1/3, 1/2, root(23)/6)

Find normal vector and equation of the tangent plane to the surface S at the point x0.

What is confusing me is the z. So do I move everything but the z to the RHS? Then solve?

$\noindent Here's the other method for doing these types of Q's that'll work even if the surface isn't something like a sphere where we can easily see a normal. We don't need to try and solve for $z$. We just need to make use of the following fact. If a smooth surface $S$ is defined implicitly by the equation $F\left(x,y,z\right) = C$ (where $C$ is a constant), then a normal to $S$ at the point $\mathbf{x}_0$ on the surface is given by $\nabla F$ evaluated at the point $\mathbf{x}_0$. (Recall that $\nabla F$ is the gradient vector $\left(F_x, F_y, F_z\right)$.)$

$\noindent For your particular question, the surface is $F\left(x,y,z\right)=1$, where $F\left(x,y,z\right) = x^2 + y^2 + z^2$. So $\nabla F = \left(2x,2y,2z\right)$ is a normal to the surface at any point $\left(x,y,z\right)$ on the surface. So at the given point $\mathbf{x}_0 =\left(\frac{1}{3}, \frac{1}{2}, \frac{\sqrt{23}}{6}\right)$ (which we can check is indeed on the surface by substituting it into the surface's equation and seeing it holds), a normal to the surface here is $\left(2\times \frac{1}{3}, 2\times \frac{1}{2}, 2\times \frac{\sqrt{23}}{6}\right)$. From this, we can simplify things and since we have a normal and we know the point, we can find an equation for the tangent plane.$

Flop21

Well-Known Member

#30

Re: MATH1231/1241/1251 SOS Thread

how do I do this one

I

InteGrand

Well-Known Member

#31

Re: MATH1231/1241/1251 SOS Thread

Flop21 said:
how do I do this one

Row-reduce that augmented matrix (get it into row-echelon form) and you should find a zero row at the bottom, with some linear expression involving x,y and z in this row in the right-hand augmented part. There will be solutions, i.e. we will have v be in S, if and only if this expression equals 0.

leehuan

Well-Known Member

#32

Re: MATH1231/1241/1251 SOS Thread

A hint on part c) please before I succumb to being 100% stuck. (Aside from conjug(x)=x)

$$For $n>1$, let $\omega_1, \omega_2, \dots, \omega_n$ be the $n$ distinct $n$th roots of 1 and let $A_k$ be the point on the Argand diagram which represents $\omega_k$. Let P represent any point $z$ on the unit circle, and let $PA_k$ denote the distance from $P$ to $A_k$

$$a) Prove that $(PA_k)^2=(z-w_k)(\overline{z}-\overline{w_k})$

$$b) Deduce that $\sum_{k=1}^{n}(PA_k)^2=2n$

$$c) Now let $P$ represent the point $x$ on the real axis$, -1<x<1,$ prove that:$\\ \prod_{k=1}^n PA_k=1-x^n$

Paradoxica

-insert title here-

#33

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
A hint on part c) please before I succumb to being 100% stuck. (Aside from conjug(x)=x)

$$For $n>1$, let $\omega_1, \omega_2, \dots, \omega_n$ be the $n$ distinct $n$th roots of 1 and let $A_k$ be the point on the Argand diagram which represents $\omega_k$. Let P represent any point $z$ on the unit circle, and let $PA_k$ denote the distance from $P$ to $A_k$

$$a) Prove that $(PA_k)^2=(z-w_k)(\overline{z}-\overline{w_k})$

$$b) Deduce that $\sum_{k=1}^{n}(PA_k)^2=2n$

$$c) Now let $P$ represent the point $x$ on the real axis$, -1<x<1,$ prove that:$\\ \prod_{k=1}^n PA_k=1-x^n$

lol steven was doing this the other day

The point P is on the real axis so the conjugate of P is itself.

So, the conjugate distances from above will be (x-ω)(x-ω*) = x² - 2xcosθ +1

On the other hand, the product of all the conjugate pairs form all the irreducible quadratic factors of the degree n polynomial of unity.

Throw in the factor of (x+1) based on the parity of n.

Lastly, chuck in the 1-x factor which appears for all values of n.

This is equal to 1-x^n

leehuan

Well-Known Member

#34

Re: MATH1231/1241/1251 SOS Thread

Paradoxica said:
lol steven was doing this the other day

The point P is on the real axis so the conjugate of P is itself.

So, the conjugate distances from above will be (x-ω)(x-ω*) = x² - 2xcosθ +1

On the other hand, the product of all the conjugate pairs form all the irreducible quadratic factors of the degree n polynomial of unity.

Throw in the factor of (x+1) based on the parity of n.

Lastly, chuck in the 1-x factor which appears for all values of n.

This is equal to 1-x^n

That was a bit too rushed. I had the idea of the quadratic factors but I don't see how they transform into 1-x^n

Paradoxica

-insert title here-

#35

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
That was a bit too rushed. I don't get how the quadratic factors transform into 1-x^n

TL;DR

Factorise the nth polynomial of unity into it's complex factors and use the knowledge that x is inside the unit circle to obtain the distances you want.

leehuan

Well-Known Member

#36

Re: MATH1231/1241/1251 SOS Thread

This isn't helping sorry. Too rushed and you TLDRd it further. I don't see it....

Paradoxica

-insert title here-

#37

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
This isn't helping sorry. Too rushed and you TLDRd it further. I don't see it....

...

x-ω

ω is one of the nth roots of unity

do I have to say more.

I

InteGrand

Well-Known Member

#38

Re: MATH1231/1241/1251 SOS Thread

leehuan said:
This isn't helping sorry. Too rushed and you TLDRd it further. I don't see it....

Essentially, here is a sketch.

$\noindent Let the product in question equal $\rho$. Recall from Paradoxica's earlier remarks that $PA_k ^2 = \left(x -\omega_k \right)\left(x -\overline{\omega}_k\right)$. Take the product of both sides from $k=1$ to $n$: $\prod \limits _{k=1}^{n} PA_k ^2 = \prod \limits _{k=1} ^{n} \left(x- \omega_k\right)\left(x -\overline{\omega_k}\right) \Rightarrow \rho^2 \equiv \left(\prod \limits _{k=1}^{n} PA_k\right)^2 = \prod _{k=1}^{n}\left(x -\omega_k\right) \cdor \prod \limits _{k=1}^{n}\left(x- \overline{\omega_k}\right)$.$

$\noindent You should be able to convince yourself that $\left\{\omega_k \right\}_{k=1}^{n}$ and $\left\{\overline{\omega_k}\right\}_{k=1}^{n}$ are exactly the same sets. This means the two products on the R.H.S. of the last line are the same, so $\rho^2 = \left(\prod \limits _{k=1}^{n}\left(x-\omega_k\right)\right)^2$. Since this right-hand product is $x^n-1$ (factorise the polynomial $p(z) = z^n -1$ and sub. in $z=x$), we have $\rho^2 = \left(x^n -1\right)^2$. Since $\rho$ is a product of distances, it is positive, so $\rho = 1-x^n$ (as $-1 <x < 1$), as required.$

Flop21

Well-Known Member

#39

Re: MATH1231/1241/1251 SOS Thread

InteGrand said:
Row-reduce that augmented matrix (get it into row-echelon form) and you should find a zero row at the bottom, with some linear expression involving x,y and z in this row in the right-hand augmented part. There will be solutions, i.e. we will have v be in S, if and only if this expression equals 0.

I'm stuck on a similar one

How do I find this vector???

turntaker

Well-Known Member

#40

Re: MATH1231/1241/1251 SOS Thread

why are you still doing matricies

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