Proof by induction divisibility

[austere sleet]

Help question 13 part b

@patent hollow would u be able to help? I’m lost on what to do

Oh wait

If f(n) is divisible by 7 then 4f(n) will also be divisible by 7

And f(n+1) - 5f(n) has been proven to be divisible by 7 in the first part

[patent hollow]

Yes, very good!

[austere sleet]

@patent hollow what do I do for 16? It’s a series but don’t know where to start

Do I just use the (2^n x n)/(n+1)(n+2) part?

[patent hollow]

Well, let's see.
Let a(n) = 2^n*n/((n + 1)(n + 2)). Then:
S(n) = a(1) + ... + a(n)
S(n + 1) = a(1) + ... + a(n) + a(n + 1)
So:
S(n + 1) = S(n) + a(n + 1)
We need to prove that S(n) = 2^(n + 1)/(n + 2) - 1. So, we need to prove the following equality:
2^(n + 2)/(n + 3) - 1 = 2^(n + 1)/(n + 2) - 1 + 2^(n + 1)*(n + 1)/((n + 2)(n + 3))
Start by adding 1 to both sides, then you will be able to multiply by something to simplify.

[austere sleet]

@patent hollow are you any good with vectors?

[patent hollow]

Well, in some capacity.

[austere sleet]

I just had a test on hyperbolic functions proof by induction and vectors

I took a picture of a vectors question

Could you look at it rq

[patent hollow]

Sure, let's see.

[austere sleet]

So what I did was I got the directional vectors of the 2 wires and then put them both in vector form

And the shortest distance between two skew lines is the perpendicular distance

So then I took co ordinates from both of the equations and then got a directional vector for the line that will be the shortest distance

And then obviously if you wanted the distance you’d do
root((…)^2 + (…)^2 + (…)^2)

But I would’ve had to also solve for mu and lambda

From what I’ve just said i only got to the part where I get the directional vector of the line between them both

I stopped working on that question cause it was gonna take too long

[patent hollow]

There is a general way to do this.
Let us have two lines: one with vector s1 passing through r1, the other with vector s2 passing through r2.
First, let's calculate the normal vector to them both: n = s1⨯s2.
Then the distance between them is:
d = |(r2 - r1)·n|/|n|
The meaning of this expression is that if we take two arbitrary points on the lines, make a vector through them (so, r2 - r1) and project it onto a line perpendicular to them both, we will always get a vector that is also perpendicular to them (obviously), but its length will be the distance between the lines.

[austere sleet]

What do u mean by normal vector?

[patent hollow]

A normal vector is a vector perpendicular to something.
While one line doesn't have a unique normal vector, two lines do.

[austere sleet]

Ah

Quick question

Let’s say those 2 points on the perpendicular line have mu and lamba in it how would I solve for those? Cause I need the length of the line/ the distance but I have mu and lambda there

[patent hollow]

What are μ and λ?

[austere sleet]

Wdym

[patent hollow]

You said you have μ and λ that you need to solve for. What are they?

[austere sleet]

Icl idk wym tho I don’t know what they are

[patent hollow]

Well, if you don't know, I also don't know.

Use the formula that I provided to find the distance between the lines.

[austere sleet]

So I had two lines in vector form and then I took the points from each line which would include lambda and mu

Cause I had to find the equation of the line connecting them

I wish I had my working 💀

@patent hollow I’m gonna do the question again to where I got to and I’ll send it here

@patent hollow

[patent hollow]

Oh, they were the parameters!

I see.

Well, maybe you can do that this way, but that is much longer.

[austere sleet]

@patent hollow is there anyway you can solve the question in the way you said cause I don’t really understand how you’re doing it in words
sorry for asking so much

[patent hollow]

Well, let's see.
First line:
r1 = {0, 0, 0}
s1 = {0, 100, -20} - {0, 0, 0} = {0, 100, -20} = 20{0, 5, -1}; we can take s1 = {0, 5, -1}, as we don't care about the magnitude, just the direction.
Second line:
r2 = {10, 0, 20}
s2 = {-10, 100, -5} - {10, 0, 20} = {-20, 100, -25} = 5{-4, 20, -5}; we can take s2 = {-4, 20, -5} for the same reason as above.
We find the normal vector and its magnitude.
n = s1⨯s2 = {0, 5, -1}⨯{-4, 20, -5} = {-5, 4, 20}
|n| = √(25 + 16 + 400) = 21
Then we find the product (r2 - r1)·n and its absolute value.
(r2 - r1)·n = ({10, 0, 20} - {0, 0, 0})·{-5, 4, 20} = {10, 0, 20}·{-5, 4, 20} = 350
|(r2 - r1)·n| = 350
So, the distance is:
d = |(r2 - r1)·n|/|n| = 350/21 = 50/3

[austere sleet]

Wait so why are you allowed to bring out a 20 for s1 and then the same for s2

[patent hollow]

Because s1 and s2 are direction vectors. We don't care about their magnitude, so we can multiply or divide them by any nonzero constant if we want to.

[austere sleet]

Also you put n = s1 x s2
What are you doing there? Cause the answer doesn’t look like you timesed them together

[patent hollow]

Cross product.

[austere sleet]

What’s that

Scalar product? Dot product?

Are they the same?

[patent hollow]

No, dot product is dot product.

[patent hollow]

It's an operation that you can perform on two 3-dimensional vectors, which produces a vector perpendicular to the both.

[austere sleet]

Ah

How does it work

[patent hollow]

Well, there's a formula depicting how to do it, but you need to know how to calculate determinants.
So, in your case, maybe it'll be easier to use the definition on orts.
Let i, j, k be standard orts of ℝ^3. By that, I mean that if we have a vector r = {x, y, z}, then it can be represented as r = xi + yj + zk.
Then:
i⨯i = j⨯j = k⨯k = 0 (vector)
i⨯j = k, j⨯i = -k
j⨯k = i, k⨯j = -i
k⨯i = j, i⨯k = -j
Note the anticommutativity: for any vectors a, b we have a⨯b = -b⨯a.

[austere sleet]

Determinants as in 1/(ad - bc) for matrices?

[patent hollow]

Well, not by that definition.
Here's how you usually calculate the cross product. You expand this "determinant" with the first row, thus getting (...)i + (...)j + (...)k - a vector.

[austere sleet]

Do I need to do all these multiplications with all the values?

But when I do them multiplications I don’t get the corresponding values

So no?

[patent hollow]

Well, yeah. As I said, first row expansion. So, there will be three terms.

If you count individual products, then six terms.

[austere sleet]

I found this

Oh wait

Now I see

[patent hollow]

Well, yes. That's what you get by expanding it.

[austere sleet]

Cause you got 0 - (-20)

I didn’t realise you took away one from the other

Ty

[patent hollow]

Ah, ok, I see.
You're welcome!

[austere sleet]

I’ve managed to find the ms and this is how they did it

[patent hollow]

Well, this is basically doing the same thing, but longer.

[austere sleet]

I’ve found videos covering the vector product and it does turn out I’m meant to learn it I believe I learn it in year 2 tho

[patent hollow]

Well, I don't remember when you're supposed to learn it, but I learned it myself around grade 8-9.

[austere sleet]

How old r u in grade 8-9?

[patent hollow]

14-15 years old, I think.

[austere sleet]

Ah when I say year 2 you’d be 17-18

[patent hollow]

Year 2 of uni? No, considering we have 11 grades, that would be 18-19 years old.

[austere sleet]

No not year 2 of uni in the UK you do GCSEs then A-Levels then uni

Year 2 of a levels

You do 5 years of GCSEs

[patent hollow]

Don't really know what that means, but ok.

[austere sleet]

💀

[austere sleet]

Well ty for all the help

[dreamy relic]

@austere sleet has given 1 rep to @patent hollow

[lime mulch]

lorddarpinger is goated 🙏🏼

[patent hollow]

You're welcome!