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

#### InteGrand

##### Well-Known Member

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

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

#### InteGrand

##### Well-Known Member

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

How did this get simplified to this??

#### InteGrand

##### Well-Known Member

How did this get simplified to this??

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

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

#### Flop21

##### Well-Known Member

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

$\bg_white \noindent E.g. They found F_x \left(x,y\right) = \pi y^2 \cos \left(\pi xy^2\right). So at (x,y) = (2,-1), F_x (2,-1) = \pi \cdot (-1)^2 \cos \left(\pi \cdot 2\cdot (-1)^2\right) = \pi \cos 2\pi = \pi, since \cos 2\pi = 1.$
Ohh okay! They subbed the points in, explains things. Thanks!

#### Flop21

##### Well-Known Member

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?

#### InteGrand

##### Well-Known Member

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

\bg_white \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*}

#### InteGrand

##### Well-Known Member

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

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

#### Flop21

##### Well-Known Member

how do I do this one

#### InteGrand

##### Well-Known Member

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

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

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

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

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

$\bg_white c) Now let P represent the point x on the real axis, -1

##### -insert title here-

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

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

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

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

$\bg_white c) Now let P represent the point x on the real axis, -1
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

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

##### -insert title here-

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

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

##### -insert title here-

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.

#### InteGrand

##### Well-Known Member

This isn't helping sorry. Too rushed and you TLDRd it further. I don't see it....
Essentially, here is a sketch.

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

$\bg_white \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

#### Flop21

##### Well-Known Member

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???