help-27

[long trout]

thank you

.close

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[weak cove]

I have a few questions about what this problem is asking for

it is defining a relation on R, but it looks like this only works on integers? How can we consider mod of x,y irrational numbers?

And then when it says if gamma is rational the sequence is periodic, I don't understand that either. Like if gamma is 1/2, then the series is
1/2, 1, 3/2, 2, ....
is it periodic because <1>=<2>=..?

but then the in between parts like I don't see how the relation is even defined that it could be periodic on them since they aren't integers

[restive river]

is this uni

[desert compass]

yes

[restive river]

phew

[weak cove]

<@&286206848099549185>

[eager nova]

The series would be 1/2, 0, 1/2, 0

It is only the fractional part

[weak cove]

How is that?

[eager nova]

The fractional part of any integer is 0

[weak cove]

yes

Why is the sequence only the fractional part?

[eager nova]

Because of the definition given

If x=y…

[weak cove]

x=y mod 1
is only true for integer x and y?

Sorry I'm confused by the definition given

[eager nova]

For any real number, you tale the fractional part for example

3.7 mod 1 = 0.7

If x and h are different but the decimal part is the same

Then that mod 1 will be the same

Not sure if im explaining myself tbh xd

[weak cove]

I don't understand how we are taking mod over non-integers

like x=y mod 1
<=>
1| (x-y)
<=>
(x-y) is an integer

wait no

I see

yeah nevermind

because if their fractional part is the same

then their difference will be an integer

necessarily

[weak cove]

I get this now

ty

[eager nova]

Great i was trying to find another way

[weak cove]

so the sequence
1/2, 1, 3/2, 2, 5/2, ...
becomes taking all their fractional parts
1/2, 0, 1/2, 0, ..

as you said earlier

which is periodic

[eager nova]

Yes

[weak cove]

great atleast now I understand what is is asking

I do not think I can solve it tonight though, I have no idea what to do and it looks time-consuming.

[eager nova]

Well, u got the hardest part, which is knowing what are u being asked xD

[weak cove]

I'm not sure that is the hardest part here XD

unless you see a way?

[eager nova]

I dont remember now sorry

It would consume too much time for me too lol

[weak cove]

Yeah no that's fair hahe

Thank you for the help so far

going to try it later

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[strange arch]

Does gamma = p/q implicitly also assume that gamma is in its simplest fraction form?

otherwise the initial statement wouldn't be true

[weak cove]

the initial statement being what

[strange arch]

well if gamma isn't in its most simplified form then the statement can't be proven

so I presume it's implicitly assumed when saying gamma = p/q e Q

[weak cove]

because q could range

yeah it is likely assuming that

[strange arch]

k, you can then first perceive the ratio of the set to N as a probability

[weak cove]

Do you want me to reopen this?

[strange arch]

iyw, I just wanted to mention an approach

[weak cove]

Sure

.reopen

[devout snow]

✅

[strange arch]

you can also take sample values, e.g. gamma = 3/7, thereby a e {0,1,2}. It'd then ask you what's the probability that n * gamma = a/7 for all n

(or rather 1 <= n <= N for a sufficiently large N)

which due to gamma being in its simplest form, are equal for all a

therefore by showing it for a=0, it must be true for all a e {0, ..., p-1}

and the probability of hitting the case n * gamma = a/7 is 1/7 for N -> inf.

or generalised the probability of hitting the case n * gamma = a/q is 1/q for N -> inf.

lastly you'd need to retranslate, since we don't have N -> inf., but only N >> 0

which is where the additional + O(1/N) comes in for the probability

hope my rambling was partially understandable :D

[weak cove]

Okay, I'm going to try to digest this and see what I can make of it, thank you for the explanation

[strange arch]

np 🦇

[weak cove]

another maybe unclear thing about the problem statement

actually nevermind sorry for ping, I just realized

it is clear

[strange arch]

dw I don't mind pings, at least the implicit simplified form of gamma is what stood out to me

[weak cove]

Alright well Ty again for the help on this I’m going to have to try it again in the morning though as it is too late for me now

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[restive river]

Hey, can I use length of side = a to solve the following?

[long pasture]

sure

[soft umbra]

Ofc

[restive river]

I ask because my professor used an arbitrary length of 1 in the exam correction, but during the exam I used length = a

[long pasture]

which is decent

[soft umbra]

But setting coordinates to the points would be more efficient

[restive river]

And the angles’ expressions are inevitably different

[long pasture]

using a is decent

[winter patrol]

angles aren't affected in similar figures

so doesn't matter

[restive river]

Well I did, for instance I had A(a, 0, 0)

And so on

[soft umbra]

yeah it’s good

[winter patrol]

so nice of them to tell you that a cube has equal side lengths

[soft umbra]

Indeed

Too many helpers here, I’ll leave

[restive river]

My angle was expressed in terms of a though

That isn’t supposed to be the case though, right?

[winter patrol]

a should've end up being cancelled

[restive river]

I messed up somewhere then

🥲

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[sturdy yew]

i know it was ur question but i have a doubt, what exactly is the meaning of angle between two rays which are not in the same plane

[devout snow]

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[outer spade]

Prove inequality

[outer spade]

<@&286206848099549185>

Help pls

[topaz jolt]

Hint: $\frac{n}{(n+1)!}=\frac{1}{n!}-\frac{1}{(n+1)!}$

[woven radish]

Caroline

[outer spade]

Ty

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[livid carbon]

Common ratio is 3 here, right?

[hasty saffron]

Yes

[livid carbon]

Alright thanks

[hasty saffron]

👍

[livid carbon]

That was a very simple question, I thought there was something fishy

But anyway

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[hasty saffron]

Hmm

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[ionic thistle]

I'm trying to derive y=x/4-x using the power rule, by moving the denominator to the top with a -1 exponent and I'm not getting the right answer and I just can't find where my mistake is

oh wait I'm crazy I found my mistake 😭😭

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[restive river]

why does the base of exponential fn must be positive?

[devout snow]

@restive river Has your question been resolved?

[slender stone]

from the way its phrased, the person seems to be using a definition of "exponential function" where this property is given

[restive river]

is it a general property/definition?

of exp fns

[devout snow]

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[hardy crag]

can someone help me with f) please

[vapid nest]

okay okay

does $cos(90 - \theta)$ look familiar to you?

[woven radish]

nosqldb

[hardy crag]

yeah

..?

[vapid nest]

ah it's equal to $sin(\theta)$ with an identity because in a right triangle you already have a right angle, so 90 of the 180 degrees is already gone, so the sum of the other two angles is 90

[woven radish]

nosqldb

[hardy crag]

ohh

okay

so how would i use that to do 13 f)?

[vapid nest]

well if you know $sin(\theta)$, draw a sample triangle that satisfies

[woven radish]

nosqldb

[vapid nest]

$sin(\theta) = \frac{2}{3}$, then solve for the rest of the components (pythogorean thm), and compute $cos(\theta)$ on the triangle

[woven radish]

nosqldb

[hardy crag]

ohhh

alright

thank you

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[broken solar]

Could someone help me with this problem?

[fresh elbow]

Can you find any similar triangles

What angles are equal

[broken solar]

i dont think so

i believe lines ae and ab are equal

[fresh elbow]

Is DC and AB parallel

[broken solar]

it doesn't say that

so i dont think so

[fresh elbow]

They are since the inside angles add to 180

[broken solar]

oh yeah true

[fresh elbow]

Angle C and angle B

So can you find any congruent angles with that info

since you have a parallel line

[broken solar]

i gues you can

[fresh elbow]

Oh I know there is a weird theorem

[broken solar]

what else can you do

[fresh elbow]

Ok wait I think I can do it without the formula

So can you find similar triangles

[broken solar]

i dont think i learned that yet

[fresh elbow]

Huh

[broken solar]

oh wait i might have

similar triangles

[fresh elbow]

Yes

So find similar angles and from those angles

Find a similar triangle using vertical angles as well

[broken solar]

which angles would you use?

[fresh elbow]

Sorry I don’t know anymore, have you learned geometric means

[broken solar]

i might have

[fresh elbow]

ok I searched up the formula

[broken solar]

explain your idea and ill see if i remenber

[fresh elbow]

Basically when you have a right triangle with an altitude from the right angle to the hypotenuse

There is a relationship between the altitude and the two lengths on the hypotenuse that were created by the altitude

[broken solar]

ok

[fresh elbow]

do you know how to solve for x given a and b

[broken solar]

wouldnt u just use pythag. therom

oh wait u cant do that

uhh then no

[fresh elbow]

These are similar triangles

Can you find identical angles? And explain

[broken solar]

ok

so how would you use that for this problem

[fresh elbow]

Wait what is the formula

To find x

[broken solar]

i dunno

i havent learned it yet

[fresh elbow]

Ok well we can learn it now

[fresh elbow]

In the two triangles, where are the right angles

[broken solar]

nvm

[fresh elbow]

Well the right angles are at the bottom of the altitude

Hard to say srry

[broken solar]

ik its so hard to explain ove discord 😅

[fresh elbow]

Because the bottom line is divided by a perpendicular line right?

ok so

the top angle that is divided by the altitude

Let’s represent the left one as d and the right one as e

Can you right equations for what the angle is equal to

E+d=90

[broken solar]

how would find the angles

[fresh elbow]

algebra

Just isolate the e variable

And then another equation where d is isolated

e+d=90

[broken solar]

would you just get 90 = 90

[fresh elbow]

I’ll do the first one

e=90-d

Can you find what d is

[broken solar]

yes

d = 90 -e

[fresh elbow]

Like the other equation

ok

Next

Look at the angles on the far right and far left

The far left will be f

And the far right will be g

can you say what those angles are

all angles in a triangle add to 180

[broken solar]

so what would you do then

[fresh elbow]

well we know that f+d+90=180

for the left triangle

And g+e+90=180

Right

[broken solar]

mhm

[fresh elbow]

Can you simplify the equations a little further

[broken solar]

so g+e+90 = f+d+90

g+e = f+d

[fresh elbow]

no

Yes

yes correct

[broken solar]

and then what

[fresh elbow]

what is e and what is d

[broken solar]

f and d

[fresh elbow]

Can you replace e with 90-d

[broken solar]

ohhhhh

would do you that with all the other angles?

[fresh elbow]

g=90-e

f=90-d

Right?

[broken solar]

yes

[fresh elbow]

so what is 90-e and what is 90-d also

[broken solar]

so what would the next equation look likee

[fresh elbow]

Because we said what 90-e was before

[broken solar]

g is 90 - e and f is 90 - d

[fresh elbow]

No but

90-e is also equal to something else

scroll up again

[broken solar]

d?

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[fresh elbow]

So g=d f=e I hope that helps you solve

[broken solar]

.reopen

[devout snow]

✅

[broken solar]

i think i got it now

thanks for the help!!!!!
\

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[carmine pumice]

Hello, im rlly confused on how to approach this problem

[carmine pumice]

what does 'set the coefficients of y, y', and y" equal to each other to find
g(t)." want me to do

heres the rest of the problem.

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[drowsy quiver]

Does this make sense for simplifying the radical?

[drowsy quiver]

I don’t exactly know how to check my awnser

[knotty sage]

32*sqrt(45)= 32 sqrt(4) *sqrt(5) ?

[drowsy quiver]

Uh… yes?

Just wondering if I simplified correctly, or if I went too far/got a calculation wrong

[knotty sage]

you got it wrong
5*4=20

[drowsy quiver]

But root4 can be simplified to 2 because it’s one of the square numbers tho right?

Oh wait

I see what u mean

[knotty sage]

Yes ,
but you simplified root 45 wrong

[drowsy quiver]

Nvm

[knotty sage]

It’s ok

[drowsy quiver]

So does that make 32 root45 simplified?

[carmine hollow]

use wolframAlpha

[knotty sage]

45=9*5
so it becomes
32 root 9 * root 5

[drowsy quiver]

Wat

[carmine hollow]

https://www.wolframalpha.com/

[drowsy quiver]

Ohhhh thank you

I didn’t notice that

[knotty sage]

any time

[drowsy quiver]

So 96 root5?

[knotty sage]

Yep

[drowsy quiver]

Alright awesome, thanks for your help

[knotty sage]

anytime

[drowsy quiver]

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[fading pivot]

how to solve this

[late orbit]

When you go with the flow, your speed adds up with the flow. When you go against it, your speed subtracts from the flow.

Calculate the total time taken in both cases and add it to 4

[final token]

relative speed

[late orbit]

Not exactly.

[final token]

$V_R = V_1-V_2$ if v1 and v2 have different directions

[woven radish]

dqvidutzul

[late orbit]

It's still your speed in still water that adds up with the speed of water. That, in turn, is the relative speed of you wrt water.

[final token]

in the first case, when the man rows against the water the speed decreases

but yes i agree with you

[devout snow]

@fading pivot Has your question been resolved?

[devout snow]

@fading pivot Has your question been resolved?

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[sharp wagon]

getting stuck on calculating how to do even the first part

[late orbit]

Make 4 cases: 3 used, 2 used 1 not_used, 2 not_used 1 used, and 3 unused.

[sharp wagon]

yeah i did that now

i feel like theres a better solution tho

@late orbit like this just seems like a dumb way to do it

[late orbit]

Probably.

Even my first intuition was to do this.

[sharp wagon]

i guess i write this all down for my work

but i feel rly stupid

oh im slow

i know how to do it faster

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[echo kraken]

Hi could I get some help on q2 please

[weak cove]

, rotate

[tender lotus]

$y=\int \frac{\dd y }{\dd x} \dd x$

[woven radish]

Adam Chebil

[echo kraken]

i think i have to differentiate first

[tender lotus]

differentiate what ?

[echo kraken]

oh wait no, i integrate the 3 root x - x^2

[echo kraken]

honestly that was a bit of a guess, can i ask why i need to integrate that?

is it because its alr in dy/dx form

so i need to anti differentiate?

[tender lotus]

you have f'(x) you need to find f(x)

that's why you do the indefinite integral

[echo kraken]

ok so im just going back a step basically

[tender lotus]

ye

[echo kraken]

sorry this might seem a dumb question but what does it say in the question which makes u know u need to find the integral?

is it find the value of y?

[tender lotus]

y is a function of x

y = f(x)
finding the value of y is finding f(x)

to find f(x) u integrate f'(x) (which is dy/dx)

[echo kraken]

ahh i see thanks

i got 2x^3/2 - 1/3x^3 + c from integrating

would i plug x= 4 into the equation now?

[tender lotus]

u need to find c first

f(2/3) = 1

[echo kraken]

i got c as -1

ah yea then i plug x=4 in

thanks

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[noble skiff]

what is the point of gauss jordan elimination

[noble skiff]

i have a linear algebra exam tomorrow should i learn gauss jordan elimination or just stick to gaussian elimination

is their even a point when one is needed

[stone stump]

I wouldnt even make a distinction between those two

[noble skiff]

ok awesome

so im not gona waist time practicing it

[stone stump]

its the same thing

why would you need to extra practice

[noble skiff]

is their ever a situation where it is nessecary for rref though

[stone stump]

if you want to find the inverse of a matrix by starting with (A| I ) and bringing the left side into rref

then you get (I | A^-1)

(only works if the rref actually ends up being I )

[noble skiff]

okok thak you thats not gona be on the mid term though we just stated that recently

thank you

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[gleaming pivot]

what is "dx" asking for in this problem? i know how to do the integral test but i don't really get it wants

[acoustic leaf]

it is asking for the value of the integral

[gleaming pivot]

oh so the integral evaluated?

[acoustic leaf]

yes

[gleaming pivot]

why is it worded like that

[acoustic leaf]

it's not dx = ___, it's the entire integral = ___

[gleaming pivot]

ohhhhhh lmao

someone else put this same question on piazza i feel dumb now lol

thank you

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[devout snow]

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[mild sand]

I am unsure if this is the right server but I've been working with the schwarzchild radius equation and I would like to use the Kerr Metric along side but I am unsure of the (1+ root (1- a^2)) primarily figuring out the a variable

[mild sand]

For reference I've used Sagittarius A as a control for the schwartzchild radius which is 0.08 AU

[dense jay]

youll probably have a better shot in the physics server

[mild sand]

Oh I did not know there was one

[dense jay]

#old-network

[mild sand]

Thank you

[dense jay]

np

[mild sand]

.close

[devout snow]

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[devout snow]

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[ruby mauve]

how do i do this please help me

[dense jay]

simplify each radical first (do you know how one can do so?)

[ruby mauve]

what

i dont know hpw to do that

[royal kestrel]

the number under the radical can be split into 2 numbers

for example

[ruby mauve]

whats a radical

[royal kestrel]

20 is the same thing as 10 * 2, or 5 * 4, etc

the radical is the square root

[ruby mauve]

oh ok

[royal kestrel]

yup

[royal kestrel]

with this in mind

you're able to split the numbers under the radicals into something that you can take the square root of

so going back to the example

[ruby mauve]

okay

[royal kestrel]

you can't just straight up take the square root of this right

[ruby mauve]

yeah

[royal kestrel]

you are able to transform it into this

from then what do you think is next?

[ruby mauve]

uhh split four into 2x2 and leave five alone

[royal kestrel]

half true

we are able to do something with the 4

what's the square root of 4?

[ruby mauve]

2

[royal kestrel]

mhm

so now we end up with this

[ruby mauve]

okay but wait

[royal kestrel]

and that is fully simplified

ya

[ruby mauve]

if u take it and make a tree and circle the number that are the same is that a way to do it

[royal kestrel]

yes

[ruby mauve]

also does 112 become 4 or 2 with a power of 2

[royal kestrel]

uhh

[royal kestrel]

as a product of primes 112 is this

[ruby mauve]

yes i got that put then when u put the two on the out side of the radical does it become four

[royal kestrel]

yes

[ruby mauve]

okay and what do i do if there is already a number on the outside of the radical

[royal kestrel]

you will multiply that

by the number you pulled out of the radical

[ruby mauve]

so i will do -8x3

[royal kestrel]

yes

[ruby mauve]

okay

so now i have -27radical 7 + 4radical7

[royal kestrel]

should be -24

otherwise that looks right

[ruby mauve]

mb i always do that when i move fast

[royal kestrel]

np

[ruby mauve]

hey uhh quick question

[royal kestrel]

ya

[ruby mauve]

how educated are you on algebra

[royal kestrel]

uhhh

depends

what do you need help with

[ruby mauve]

okay radicals exponetintal functions and equation systems u know those

[royal kestrel]

ya

[ruby mauve]

okay what are u doing this weekend

[royal kestrel]

homework

[ruby mauve]

wanna help me i got alot of it to do and my parents are down my back and i really need to pass because i have a lot to do for algebra

please

also what do i do with the adding symbol in the middle

[royal kestrel]

just keep making help channels in this server

there are lots of ppl that can help

[ruby mauve]

okay thanks for the advice

what do i do now

[royal kestrel]

now you can add the coefficients (numbers infront of the radicals)

[ruby mauve]

-28

[royal kestrel]

so just -27 + 4

which is

[ruby mauve]

-28

what do i do with the numbers inside

[royal kestrel]

naur

[ruby mauve]

ohhhh mb

[royal kestrel]

-27 + 4 = -23

[ruby mauve]

yeah

[royal kestrel]

you don't do anything with the radical

[ruby mauve]

shoot im thinking of subtract

[royal kestrel]

they stay the same

[midnight girder]

u could have just rewrite it into sqrt(7)*(...), no?

[ruby mauve]

what

[midnight girder]

i mean like bc they have the same product so u can just bring it in front

and just subtract/add

[royal kestrel]

thats what he's doing

i think

[ruby mauve]

what am i doing

i have it as -23radical7

[midnight girder]

i dont want to confuse u, the answer is correct.

[ruby mauve]

okay so i write it like that

[midnight girder]

if thats the math question, yh

[ruby mauve]

oh shot it was wrong

[midnight girder]

whats the full question

[midnight girder]

is that just shorten the given sum or what?

[royal kestrel]

just putting it in its simplest form

so yea

[royal kestrel]

this is the final answer

[ruby mauve]

i got it wrong though

[royal kestrel]

???

did it give you feedbkac

[ruby mauve]

not yet i have on left

i have one last chance\

[royal kestrel]

ummm

[midnight girder]

hmm seems weird tbh

i mean it says shorten/simplify it

and thats what u did

[ruby mauve]

wait 3x8 equals

[ruby mauve]

this is where we went wrong

[royal kestrel]

oh

[ruby mauve]

we put 27 instead of 24

[royal kestrel]

cus you kept the 27

ya

so it should be -21

mb for not catching that

[ruby mauve]

no -20

thanks for the help though

help

[midnight girder]

now i am confused 🤔

[ruby mauve]

becuase -24 + 4 = -20

he put -27

which would equal negative 23

[midnight girder]

ouh wait

the question was -24+4, but the screen was showing -27?

[ruby mauve]

no jack put -27 instead of -24 so he put -27+4=-23 but it should have been -24+4=-20

[royal kestrel]

-24 + 4 = -20 yea

[midnight girder]

yh but here it was -27

thats what i mean

[ruby mauve]

because 8x3 = 24 but eight was negative

[royal kestrel]

you had this in the calclator so i forgot it was meant to be -24 💀

that's mb

[midnight girder]

mb

i was reading only the half 🫣

sryyy

[ruby mauve]

what do i do here

[midnight girder]

so what do u think which number is at the front?

[ruby mauve]

5

[midnight girder]

no there

[ruby mauve]

oh 0

[midnight girder]

hm

what does 0*sqrt(7) equal?

[ruby mauve]

7

[midnight girder]

no my friend, 0 times a number equal 0

[ruby mauve]

ohhh

mb mb

[midnight girder]

so at the front, there is a "1"

[ruby mauve]

yes

[midnight girder]

so u can proceed with that

u can just add them

[ruby mauve]

6radical7

[royal kestrel]

yes

[midnight girder]

yh

[ruby mauve]

thank you yall

[midnight girder]

y re welcome 🫡

[ruby mauve]

what do i do here

[livid hamlet]

solve the root of 50

reduce It

[ruby mauve]

okay 25

25 radical2

oh 5 radical 2

[royal kestrel]

yup

so now just add them

[ruby mauve]

14 radical 2

what do i do if its subtraction

like

help

[royal kestrel]

same thing

when they have the same radical, you are able to treat them like normal addition and subtractionn

[ruby mauve]

okay

[royal kestrel]

so make sure you get them to the simplest radical possible

and if they're the same, then you can add/subtract the numbers in the front

if you're adding and subtracting them and they don't have the same radical, then you can't simplify any more

and that's it

[ruby mauve]

okay

what if i have two inside the radical

like 2 and 5

do i add

[royal kestrel]

wdym

[ruby mauve]

when sloving i get something like 90 divide by 2 gets u 2 and 45 45 divide by 5 gets u 9 divide by 3 gets u 3 and 3

so then u put three on the out side but ur left with 2 and five on the inside

[royal kestrel]

you can't take anything else out so you will simplify those and keep it in the radical

so in this case you'll just get 10 in the radical

[ruby mauve]

okay so i just multiply

[midnight girder]

just make the number inside the radical a product and see what happens

[royal kestrel]

from 2 * 5

[ruby mauve]

its wrong

its not 1 radical 10

[midnight girder]

ehm, so i would bring it down to sqrt(16*10) e.g.

[ruby mauve]

so 16 radical 10

[midnight girder]

no

sqrt(16) is?

oh wait

u mean the whole solution

[ruby mauve]

yes

[midnight girder]

i was focusing on one part

mb

yh then it should be 16sqrt(10)

[ruby mauve]

but ur subtracting

[midnight girder]

ik

[ruby mauve]

so it should be 1 radical 16?

[midnight girder]

no, is a product between the number and the sqrt

so if u get a number out of the sqrt, u will have to multiply with the number infront of it

[ruby mauve]

yes

[midnight girder]

so in this case it was 7*sqrt(16x10) => 7x4sqrt(10) if i am not mistaken

[ruby mauve]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

mb mb i get it now im sorry thank you

[midnight girder]

yw

[ruby mauve]

how do i do this

[midnight girder]

this one, u just simplify the sqrt and multiply the numbers, if i got the question right

i think the whole product is like this "6x3sqrt(28)"

[ruby mauve]

how about this one

help me

[midnight girder]

The same my friend

The product looks like this "8sqrt(7)×6sqrt(14)"

[ruby mauve]

what do i do with the 7

[midnight girder]

Hmm

I am not sure what the rule was with havin 2 sqrts

I forgot tbh

I mean as a product

[ruby mauve]

bro youve been so hepfull

dont quit on me now

there multiple ways this can go

[midnight girder]

So iirc

U can just bring the numbers inside the two sqrt into one srqt

Sqrt*

And multiply them under the sqrt

[ruby mauve]

so 2x7x7

[midnight girder]

I mean like sqrt(7×14)

The 8 and the 6 can be multiplied without any consequences

[ruby mauve]

so rn i have 48 radical 2x7x7

[midnight girder]

Ok now concentrate on the product inside the sqrt

Look at the numbers and think which of them can be simplified

[ruby mauve]

none i got 98

[midnight girder]

Look at this again

There is a number that appears 2 times

[ruby mauve]

okay so now i have a 7 on the outside

[midnight girder]

Yeah multiply it with the number infront of the sqrt

That should be the solution if we did everything right

[ruby mauve]

336

[midnight girder]

Yh

Dont forget the sqrt behind that

[ruby mauve]

how about this one and it was right

[midnight girder]

This is pretty simple

U have a small product

That is for later tho

So first simplify the number inside sqrt

[ruby mauve]

should i simplify

[midnight girder]

Yes

[ruby mauve]

6 radical 7

now what

[midnight girder]

If thats the simplest answer then thats it

[ruby mauve]

okay

[midnight girder]

Did u divide it by 3 tho?

[ruby mauve]

no idk how to

[midnight girder]

Just the number u take out of the sqrt

Divided by 3

[ruby mauve]

so 7 divide by 3

[midnight girder]

7?

[ruby mauve]

6 divide by 3

[midnight girder]

Yes

[ruby mauve]

2

[midnight girder]

Yh so 2×sqrt()

[ruby mauve]

2x7

[midnight girder]

Sqrt my friend

Dont forget the sqrt

[ruby mauve]

yeah 7

[midnight girder]

It doesnt disappear

Yh

[ruby mauve]

okay 14

[midnight girder]

No u got me wrong i think

2sqrt(7)

[ruby mauve]

so 2 radical 7 is the answer?

[midnight girder]

Yh

If u simplified it correct then it should be the right answer

[ruby mauve]

yes thank you

now ho do i do this one

[midnight girder]

Multiply under one sqrt and then simplify

[ruby mauve]

so 5x15

[midnight girder]

Yh

[ruby mauve]

75 is the answer?

[midnight girder]

Yes

Now sqrt

[ruby mauve]

what

radical 75?

[midnight girder]

Yes but is that the simplest form

[ruby mauve]

no

15 is

so radical 15

[midnight girder]

Why 15 cant follow there

My brain stopped, its 3 AM over here 🙃

[ruby mauve]

lol its only 8:10 pm here

were u at/

[midnight girder]

Europe

[ruby mauve]

dam\

so are u done for the night?

[midnight girder]

So

75 can be written into

[ruby mauve]

15

[midnight girder]

5×5×3

[ruby mauve]

yes

[midnight girder]

Why 15 tho

How do u get 15

[ruby mauve]

mb i messed up but u just get 75 from 5x5x3

[midnight girder]

And sqrt(5×5×3) can be simplified

[ruby mauve]

i got it

5 radical 3

right?

[midnight girder]

Yes

[ruby mauve]

lets go

so are you going to bed?

[midnight girder]

My friend thats it from me, gonna sleep now 😴

Yh i have to

[ruby mauve]

alright thank you for all the help

your better than the tutors

[midnight girder]

Y are welcome 😄

[ruby mauve]

.close

[devout snow]

Channel closed

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Use .reopen if this was a mistake.

[low wing]

Batman with motion💯🔥

[devout snow]

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[sturdy mirage]

Construct an equation that gives the number of diagonals formed by connecting each vertex to each other non-adjacent vertex.

[sturdy mirage]

I constructed the equation correctly and did an induction proof mostly correctly to show that it works, but I'm just having trouble with the summation notation.

Obviously it wouldn't be the sum of i(i-3)/2, but how would I actually do it?

Someone in class said that it would be $\sum_{i=4}^{n} n$ and it matched all the numbers, but just didn't make sense in my head.

[woven radish]

Narutoes

[jade oak]

Wdym by P(n)=3+…+n and also P(n)=n(n-3)/2. These are not equivalent. Also this is not how you would write P(n) for induction

[sturdy mirage]

Yeah I know they're not equivalent. That's why I need help.

[jade oak]

You would do something like let P(n) be denoted by 3+…+n=n(n-3)/2

[sturdy mirage]

Don't I have to have a summation somewhere?

[jade oak]

Yeah

[sturdy mirage]

The summation is what's confusing me.

I know what I need to fix for the rest, I just don't have the laptop with this LaTex file on me rn.

[jade oak]

Lets look at the pattern. It goes 0,2,5,9,14,… right

[sturdy mirage]

Yes.

[jade oak]

So you first add 2, then 3, then 4. So your starting point should be i=2

[sturdy mirage]

Ah ok.

Trying to remember the notes my teacher said for how to find the summation.

[jade oak]

Also we want this to be 0 when n=3, 2 when n=4, etc.

The summation is 0 when the top is 2 (3-1), the summation is 2 when the top is 3 (4-1), each time the top number follows a certain pattern

[sturdy mirage]

What is the top?

Which top are we referring to here?

[jade oak]

The summation upper limit

[sturdy mirage]

Ah ok.

Wait so I'm starting with n = 2?

So somethin like $\sum_{i=2}^{n} ???$

[woven radish]

Narutoes

[jade oak]

Close

We do want to start with n=3 so just do n-1 instead

[sturdy mirage]

Ahh ok.

$\sum_{i=3}^{n-1}$

[woven radish]

Narutoes

[jade oak]

we still have i=2. The confusing part here is that we add 2 when for n=3, add 3 for n=4, add 4 for n=5. Since we want to start by adding 2 we use i=2

And we go to n-1 because of that offset of 1

Maybe its more intuitive to think of it as $\sum_{i=3}^n(i-1)$

[woven radish]

Sora Harewataru

[sturdy mirage]

Gotcha.

If I start at i=3, wouldn't the first sum be 2?

[jade oak]

Yeah

[sturdy mirage]

But the first one is supposed to be 0

3 sides = 0 diagonals

[jade oak]

Oh I see what u mean. Our first value is n=3 but this gives 0

[sturdy mirage]

Right

Then n=4 gives 2, etc

[jade oak]

$\sum_{i=3}^3(i-1)$ is 0 though

[woven radish]

Sora Harewataru

[sturdy mirage]

Right. I guess I just don't fully understand how summations work

[jade oak]

When the upper bound is less than or equal to starting i value, the sum is 0

[sturdy mirage]

But then when I go to 4, the top stays a 3?

[jade oak]

Nah i stays at 3 and n then equals 4

[sturdy mirage]

How do I set that up?

[jade oak]

$\sum_{i=3}^4(i-1)$

[woven radish]

Sora Harewataru

[jade oak]

Actually I think I lied earlier. This is only true when the upper bound is strictly less than the starting value

$\sum_{i=3}^3(i-1)=3-1=2$

[woven radish]

Sora Harewataru

[sturdy mirage]

I think my teacher gave me notes on how to set it up a long while ago

[jade oak]

Yeah should probably double check that cuz I think im trippin a bit. The logic though is that we always want to start at the same value and have the upper limit be some function of n

Where n is number of vertices

[sturdy mirage]

So the first sum I want is 0, then 2, then 3, then 4? Do I actually need the 0

I essentially want 2 3 4 5 to add to the previous, but I also need the first sum to be 0

[jade oak]

Yeah for that part you want to take advantage of this

$\sum_{i=3}^{n-1}(i-1)$ would work because

[woven radish]

Sora Harewataru

[jade oak]

$\sum_{i=3}^{(3)-1}(i-1)=0$ because of that

[woven radish]

Sora Harewataru

[sturdy mirage]

Gotchaa

[jade oak]

$\sum_{i=3}^{(4)-1}(i-1)=(3-1)$

[sturdy mirage]

Ok cool

[woven radish]

Sora Harewataru

[jade oak]

And now it works out yeah

[sturdy mirage]

Cool.

Someone in my class used $\sum_{i=4}^{n}(n)$ this looks similar with shifted indices.

[woven radish]

Narutoes

[jade oak]

I dont think that works (also that n should be i). The shifted version would be from i=2 to n-2

[sturdy mirage]

How do I set up my induction and stuff? Is my first statement $If 0+2+5+9+14+20+...+n=\frac{n(n-3)}{2}, then 0+2+5+9+14+20+...+n+n+1=\frac{(n+1)(n-2)}{2}$

[woven radish]

Narutoes

[jade oak]

You want to show the sum we found is equal to n(n-3)/2 right

[sturdy mirage]

Right.

[jade oak]

Ok the sum would be 0+2+3+…+(n-2) in expanded form

[sturdy mirage]

Ah ok

[jade oak]

We want to show that is equivalent to n(n-3)/2 for the base case

[sturdy mirage]

Through induction?

[jade oak]

Yeah we do this using induction

[sturdy mirage]

So then I end up with n(n-3)/2 + n-1?

[jade oak]

The induction step would be to show if 0+2+3+…+(n-2)=n(n-3)/2 then 0+2+3+…+(n-2)+((n+1)-2)=(n+1)((n+1)-3)/2

[sturdy mirage]

Which simplifies to mine, no?

Since the first bit is P(n)

[jade oak]

Yeah it does

[sturdy mirage]

Then that's n(n-3)/2+(2n-2)/2, then that's (n^2-n-5)/2

[jade oak]

There is a quicker way. Using the induction assumption you can replace the n(n-3)/2 with the equivalent summation part. That immediately shows the equality

[jade oak]

Using this form since the final term of the summation expansion is (n+1)-2=n-1

[sturdy mirage]

.close

[devout snow]

Channel closed

Closed by @sturdy mirage

Use .reopen if this was a mistake.

[devout snow]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[stark anvil]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[stark anvil]

1

[supple knot]

Use binomial theorem

[stark anvil]

$\sum_{r = 0}^{n} \binom{n}{r} x^{r}$

[stark anvil]

How does one solve something like $\binom{\frac{27}{5}}{r}$

[woven radish]

ColdTee

ColdTee

[merry lily]

how can any term in the expression be negative if x is positive?

[stark anvil]

I am confused on that as well

[radiant dune]

This holds even if n is fractional

[heady tulip]

(n r) = n!/((n-r)!r!)

[merry lily]

none of the terms are negative lol

[stark anvil]

Is there a quick way to solve fractions

[merry lily]

,w graph (1+x)^(27/5)

[woven radish]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by anonymemes

[radiant dune]

I hope you can realise 27/5 is 5..something

And 27/5 -6 will be negative

For example

[merry lily]

yes the power is negative

the term is not

[stark anvil]

I'm confused

How does that make a difference

[radiant dune]

Can you go ahead and apply this?

You'll figure it out

[heady tulip]

wait

what kind of expansion

[merry lily]

he's saying the term will be $x^{negative}$ but that doesnt mean the number is negative

[woven radish]

anonymemes

[heady tulip]

is it taylor expansion

[stark anvil]

I kinda get what you mean but

Let's say we had 7CR

And we applied that

(7 - 8) will never be in that

[radiant dune]

Yeah but we don't have a whole number up there this time

That's the catch

[stark anvil]

But r is always lesser than or equal to n

[radiant dune]

nCr isn't defined when n<r

[stark anvil]

How does it matter for

[radiant dune]

But the expressio n(n-1)..../whatever! can be calculated

[stark anvil]

How is it different for fractions

[stark anvil]

So you are telling me $(1 + x)^7$ will also have negative terms in that sense?

[woven radish]

ColdTee

[radiant dune]

Not until you set r=9 in this expression but that power won't even exist in its normal ncr expansion since we know it cant go beyond x^7, that's why it's absurd

x^9 won't even appear in its binomial expansion technically so there won't be a negative term

[stark anvil]

So how is r = 6 supposed to appear when we have n = 27/5

With that logic

[radiant dune]

It's Taylor series expansion will have negative terms, but normally when we have natural indexes we don't use them

[stark anvil]

This is confusing me even more

[crisp niche]

here you must use generalized newton symbol

not newton from school

but

generalised one

!

it is true

for all

real

x

and natural k

[stark anvil]

I know but why that

Specifically

[crisp niche]

GENERALISATION

sio if you put

x = 27 / 5

and appropriate k

then you find negative number

and this generalisaiotn we sue when you play with :

alpha is any real number

so such expansion

[stark anvil]

I see

[crisp niche]

requires geenralsied newtown symbol

tha is yoru case ) above)

[stark anvil]

I didn't know that

[crisp niche]

ok np

so now you need to calculate

whe first term wil be negative

[stark anvil]

$|x| < 1$ can you sum that up for me

[woven radish]

ColdTee

[crisp niche]

left side is your sum

[stark anvil]

Does it mean x can only exist in fractions?

[crisp niche]

it is series

[radiant dune]

I dont know how the hell I missed this but you'll get a (7-7) before you get a (7-8) in that expression lol

So it'll all be 0

[crisp niche]

and is convegent onyl iff | x | <1

[radiant dune]

There won't be a negative term then

I really need some sleep right now

[stark anvil]

Ofcourse 😭

Im dumb

[radiant dune]

Me too

[crisp niche]

take a peice paper and evalue (27 / 5 , 6) and later (27/5, 7)

just for exercise

and you wil see whcih one becoems negative

yo also can veirfy fwe others for practising

[stark anvil]

Alright

[crisp niche]

k

)

[stark anvil]

So generalized newton method

[crisp niche]

yes 🙂

sicen exponent is a fraction

[stark anvil]

Ohhh

That makes sense

Now that I think

About it

[crisp niche]

if expoent is natural, you play with old newton

[stark anvil]

But they should have added that in my book

[crisp niche]

yes

[stark anvil]

This should be a result explicitly for fractions in exponents

But it's not explained in general

[crisp niche]

i agree

wel it i snot my fault smiles)

ask your instructor

[stark anvil]

Alright

Thanks for the help

Guys

And girls

[crisp niche]

this generlaised newtown you can fidn in basis of discrete maths

books

yw )

[stark anvil]

I see

[crisp niche]

example:

[stark anvil]

@crisp niche

[stark anvil]

This wont expand any more right?

[crisp niche]

it is exact result of this generalised symbol

[stark anvil]

So lets say

In our case

r = 4 and n = 27/5

Then that term wont be negative

[crisp niche]

so you end on 27 / 5 - 3

so still positive

[stark anvil]

I see

Makes a lot of sense

Now that we compared integers and fractions

Thank you

[crisp niche]

yw)

[stark anvil]

I was confused about this since yesterday

[crisp niche]

thta happens )

[stark anvil]

[crisp niche]

patience is the key _)

[stark anvil]

Truly

It has satisfied me now that I have learned the truth

[crisp niche]

one needs to search everywhere

[stark anvil]

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