help-26

[topaz sinew]

@heavy surge Has your question been resolved?

[heavy surge]

<@&286206848099549185>

[night glacier]

Am I missing smth or first term is 0? How is that possible?

[arctic briar]

1st term is 1

k=1,2,3...100

[heavy surge]

How the first term is 0?

[night glacier]

How!?

[heavy surge]

What are you talking about

Which zero

[night glacier]

This, $\frac{k-1}{k!}$

[thorny flame]

Čeva18

[arctic briar]

well yea then its 0/1 = 0

[night glacier]

If you put $k=1$?

[thorny flame]

Čeva18

[heavy surge]

So what's the problen

[night glacier]

But, you have g.s.: $0, 0 \cdot \frac{1}{k}$

[thorny flame]

Čeva18

[night glacier]

But, you have g.s.: $0, 0 \cdot \frac{1}{k}, ... 0 \cdot (\frac{1}{k})^{n}$

[thorny flame]

Čeva18

[heavy surge]

So what take K from 2

Not 1

[night glacier]

But $k\in{1,2,...}$

[thorny flame]

Čeva18

[heavy surge]

Here 1 isn't satisfying so we can take it from 2

As S1 is zero

[sour bloom]

Try calculating it for smaller values of 100 \hj

but honestly, replace the 100 with a smaller number, like 4

[heavy surge]

Why?

[granite hound]

Split the quadratic into
(k-1)^2-k
Make the k part as
k^2/k!
then telescopic series

[heavy surge]

I know and I got the answer but it is wrong

[sour bloom]

then you may be able to spot some ideas about how to compute it for 100

[granite hound]

What's the answer

[heavy surge]

I got it as 4

[granite hound]

What? I got it as 1

[heavy surge]

Answer is 3

[granite hound]

💀💀💀💀

How are you getting 4? Maybe I can find some fix in my answer

[heavy surge]

What is telescopic I did it without telescopic

[granite hound]

It's like making a series where when written, most of the terms get cancelled and only the first and last remain

Typically with a subtraction sign in between

So cancelling out everything
I am getting, actually I just checked my answer again, I'm getting 0

[sudden temple]

how?
I am getting 3 too

[granite hound]

I must have missed some numbers here and there, I'm checking my solution once again

[sudden temple]

okok

[granite hound]

I am not able to find any mistakes, just check it once please

[topaz sinew]

@heavy surge Has your question been resolved?

[topaz sinew]

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[shell orbit]

The solution says I need to do sqrt((x(t)-x(t))^2 + (y(t)-y(t))^2))

But thats for the distance between points right?

Like set points where we know the location

These points could be anywhere depending on the value of t right?

[craggy haven]

yes but the idea is that you use the same t for both P and Q

imagine two planets

eg what's the distance between Earth and Mars

[shell orbit]

It depends right?

[craggy haven]

right, it depends on t

[shell orbit]

Yes agreed

[craggy haven]

but at any given time there is some distance

even though we move around and so does Mars

for example, you could find the distance at t = 0

and that would be a number

[shell orbit]

So which number do they want for t in this question

[craggy haven]

they don't

they want a formula for the distance

for arbitrary t

[shell orbit]

Ok I think I get it

Thank you!

❤️

.close

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[molten wren]

hi

[molten wren]

I think the answer is pi/12 but since its the smallest number in options I have doubts

options are

pi
2pi/3
pi/12
pi/2

is pi/12 correct or I'm making a mistake?

<@&286206848099549185>

[native notch]

Are the a and b in the first part and in the question same?

[molten wren]

no no

A B are points

[native notch]

I see

[molten wren]

mb I should have named them something else

[native notch]

And all f belongs to real numbers right

[molten wren]

yes

[native notch]

Ok wait lemme think

Yea it's probably pi/12

[molten wren]

tysm

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[pseudo creek]

yo

[pseudo creek]

can someone help

[plush tulip]

just ask

[topaz sinew]

@pseudo creek Has your question been resolved?

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[pseudo creek]

.reopen

[topaz sinew]

✅

[pseudo creek]

There's like 10 of these questions and i cant find any answer for this one

[brave coral]

Might want to bust out vertex form of a parabola

[topaz sinew]

@pseudo creek Has your question been resolved?

[pseudo creek]

how do i turn a parabola side ways

like i want it opening lelft

[delicate dragon]

!help

[topaz sinew]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

@pseudo creek Has your question been resolved?

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[prisma mesa]

I kind of intuitively get the idea of why the sum works in here, but I can't see how was the formal statement of inclusion exclusion applied here. What are all the A_i's? What does their intersection represent with respect to the problem?

[prisma mesa]

^ Here is the statement of inclusion exclusion principle

[coarse tusk]

A_i represents the set of n people with i men

or something of that sort

i think that would work

maybe it should be i women actually

[prisma mesa]

but then, A_1 and A_2 should have empty intersection

and actually A_i and A_j in general

wait, no

oh

[coarse tusk]

no wait

I think I got it

A_k is the set of sets of n people with atleast k men

[prisma mesa]

i think you were probably right actually

[coarse tusk]

i think

[prisma mesa]

this wouldnt work, because then |A_1| = |Union over Ai|

[coarse tusk]

can you show the original question once?

[prisma mesa]

sure

[coarse tusk]

oh okay

[prisma mesa]

i decided to not include it, because the solution basically relies on solving a similar problem, which is counting the number of sets of n persons in group of n-1 women and m+1 men

through inclusion exclusion

[coarse tusk]

how?

[prisma mesa]

oh, it's not right actually. |A_i| would be just n

[coarse tusk]

yeah

so they're basically saying it's (atleast 0 men) - (atleast 1 man) + (atleast 2 men) - ... - (-1)^n (atleast n men)

[prisma mesa]

hmm

the first one has to be at least 1 man

[coarse tusk]

this gives you the number of ways with exactly 0 men, which is 0

hence the sum is 0

i think that was the idea

[coarse tusk]

why?

[prisma mesa]

because the sum starts at k = 1

[prisma mesa]

this one

[coarse tusk]

oh yeah, hmm

that's weird

[coarse tusk]

wait but isn't this what we want?

the rest of the terms must then sum to 0, which probably gives you the sum you need

[prisma mesa]

it doesnt

the way they get the og sum is that they count the number of sets of n persons in another way (just m+n choose n), and then they say that Inclusion-exclusion sum = m+n choose n and rearrange it such that they obtain the og sum

this is the full solution

[coarse tusk]

sorry, I can't figure it out

i'll keep thinking but you probably wont get a good answer from me

[prisma mesa]

np, dw about it

[coarse tusk]

also, this approach to proving the identity is not something I would've thought about at all

[prisma mesa]

yeah, it's quite unusual. Which is what i like about it

what i dont like is that i cant comprehend it lol

[coarse tusk]

if you want I might have thought of an algebraic way to prove this

[prisma mesa]

through induction?

[coarse tusk]

no

[prisma mesa]

oh, how then?

[coarse tusk]

I think you can use $\binom{r}{j}=(-1)^j\binom{j-r-1}{j}$

[thorny flame]

kheerii

[coarse tusk]

this does use negative binomial coefficients

if you're comfortable with that

[prisma mesa]

i'm probably not

[coarse tusk]

no worries

I'll just show you my work

[prisma mesa]

alr

[coarse tusk]

$$\begin{aligned}
&\sum_{k=0}^n(-1)^k\binom{m+1}{k}\binom{m+n-k}{n-k}\
=&\sum_{k=0}^n(-1)^k\binom{m+1}{k}\cdot (-1)^{n-k}\binom{-m-1}{n-k}\
=&(-1)^n\sum_{k=0}^n\binom{m+1}{k}\binom{-m-1}{n-k}\
=&(-1)^n\binom{(m+1)+(-m-1)}{n}\
=&(-1)^n\binom{0}{n}=0\text{ for }n\ge 1
\end{aligned}$$

[thorny flame]

kheerii

[coarse tusk]

the second last step comes from the Vandermonde identity

although I'm not sure how valid some of these steps are

[prisma mesa]

which steps wouldn't be valid?

[coarse tusk]

the vandermonde identity mainly

[prisma mesa]

why wouldnt it be valid?

oh, -m-1 is negative

[coarse tusk]

i dont really know how you would prove it for negative things

but it probably does work

maybe the exact same proof works lol

i think you can also use the snake oil method for this

although I'm not very familiar with it

[prisma mesa]

it works from what i've found

[coarse tusk]

you would also need to prove this of course but that shouldn't be too bad

[prisma mesa]

yeah, it doesnt involve any sums, so it should be doable

thanks

[coarse tusk]

by the way, this involves defining a function f(n)=sum of .... and then considering the generating function for f(n)

which would be a double sum

then you can (hopefully) simplify that double sum by exchanging the summations and so on, and land on a (hopefully) nice enough generating function, from which you can find the general formula for f(n)

[prisma mesa]

i'll probably learn about it in next chapter

[coarse tusk]

in your case I think the generating function would need to come out as the function 1

since f(n)=1 for n=0 and f(n)=0 for n>=1

[prisma mesa]

thanks for your help

ill close this now and ponder about the book's solution for few more mins

.close

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[modern sonnet]

I need help

[modern sonnet]

?

plz

[raven sparrow]

Ask your question.

[modern sonnet]

Can you help me summarize the lesson of quadrilaterals in geometry

I don't get it

It's a edtell course

[topaz sinew]

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[tulip grove]

w

[tulip grove]

Hi can someone confirm these awnsers for me

A B C D

i think they got it wrong

^ that is the marking key

please try it and lmk

my values were

a 1749.71

b = 26.25

c = 775.46

.close

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[neon iron]

Hi, can someone help me with this problem?

[neon iron]

Well if you have a right angled triangle

4 ft along the base

And three feet up

The hypotenuse of it must be 5 feet using Pythagoras

Then converting feet to inches by x12

Should be 60

So if it’s 55 what do you think that means ?

What does it mean?

Well if your hypotenuse is smaller than 60

That means the angle between the floor and the wall is smaller than 90

Because Pythagoras doesn’t apply to it

So it is not a right triangle?

Yes

If this helps

The left is what it should be

And the right is what it sort of looks like now

Ok that helps, thank you!

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@neon iron Has your question been resolved?

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[lilac plume]

Can someone just check over my work please?

[sudden parcel]

it would be lot better if you inclusively taken log base 3 of both sides in q1

[lilac plume]

I just realized that the first page is from the wrong assignment

[lilac plume]

I did fix that when i handed that one in that was the wrong page lol

hang on

sorry THIS is the right page 1

[lilac plume]

Also im unsure of what to do for 4

[sudden parcel]

4 will be a x-axial parabola facing opposite to the axis

[lilac plume]

Like this?

[lilac plume]

also is the rest of the stuff good?

[sudden parcel]

yes

[lilac plume]

What do you mean by infeasable?

[sudden parcel]

I looked properly, 5 number seems ok

[lilac plume]

did i do it correctly?

[sudden parcel]

have you marked the black one starting from the +- 3?

[lilac plume]

yeah

hang on ill check my calculations

it starts on 0 -3

hits -3, 0 and 3, 0

hits 5, 1.41 and -5, 1.41

[lilac plume]

does this sound right?

[sudden parcel]

yep

[lilac plume]

sick

[lilac plume]

And page 1?

[lilac plume]

what about question 3?

[topaz sinew]

@lilac plume Has your question been resolved?

[lilac plume]

<@&286206848099549185> could someone help me out?

[pure sandal]

Let me see the problem

[errant jolt]

CB
D
Db
D

[lilac plume]

Question 3 here

And this page

I just want someone to look it over to make sure Im doing it right

[pure sandal]

Okay I haven’t worked with this type of graph but I understand the question

Sure I will look at it

[pure sandal]

Oh what is no. 5? Can you explain it to me? It looks like another type to me

[lilac plume]

Number 5 I already did and got it corrected. Its just square rooting the y values of hte points

[pure sandal]

No.1 is wrong

[lilac plume]

oh?

[pure sandal]

Yes

Will you give me a second?

I can do it on paper and explain to you

Okay let me explain

Okay so do you know perfect squares right

[lilac plume]

yup

[pure sandal]

So you want to find numbers that will make the radical a perfect square

It makes this much easier for y

u

Do you understand so far?

Give me 2 numbers that will make the radical a perfect square

[lilac plume]

3 cuz 3 + 1 = 4 and 8 cuz 8+1 = 9 right?

[pure sandal]

Yep!

You can also do 0

[lilac plume]

oh right

[pure sandal]

okay so you plug that number in

So, you have to solve the radical first

let’s work with no. 0 for now since I saw uuu got the y intercep wrong

so can you plug in 0 and solve for me again,

[lilac plume]

-4?

cuz -2*(1) - 2 is -4

[pure sandal]

Yep

Btw for the radical graph

You need 3 coordinates to make a more accurate graph (from my experience)

Give me 1 more coordinates

I just realized ur graph is quite small so I guess 2 will work, you can’t do 8 since ite not on the graph

[lilac plume]

so whats wrong with it?

[pure sandal]

Wdym?

You didn’t plot them right

[lilac plume]

ohhhh

yeah ill fix that

[pure sandal]

Also

For radical graphs

X cannot be negative

Square Root 0 does not exist

because 0 times anything can be undefined

X can only be negative for example if you’re given a number that you can subtract and still get a positive number

For example if you have sqrt x+9 x can be -5 because that would make sqrt of 4 which is 2

And -8

[pure sandal]

Yep

Looks right to me

[lilac plume]

I realized i put the dot on 2,-6 instead of 3,-6

[pure sandal]

Ahh okay

Anything else?

[lilac plume]

what about no 2?

[pure sandal]

I’m not understanding that one

Because it just says write the equation of the function but there are no numbers I can create it with

Can you tell me where you got the numbers from?

[lilac plume]

I actually do not remember how i got that lol

hang on ill check my notes

ok i have actually 0 clue how i figured that out

i checked on a graphing calc and its right but i do not remember how i got it

[pure sandal]

yeah haha when i have questions like that i I usually am given like the y intercept is this etc

[lilac plume]

You jsut have to look at the poorly made graph

the question is look at this graph and write the equation

[pure sandal]

I can tell you if the equation matches the graph

[lilac plume]

so i had to look at the grainy graph to figure it out

[pure sandal]

but thats it lol

okay

[lilac plume]

cuz the graph was already there

so i used trial and error i guess

[pure sandal]

wait can you resend it

[lilac plume]

Sorry last question. Is no 3 right?

[lilac plume]

sure

the graph is there to begin with

[pure sandal]

uhh

i just switched to my laptop so its a bit small

give me a second

It’s wrong

Given the equation is that, the vertical asymptote of the graph cannot beless then3

Wait nvm that’s what u have

The points look wrong though I’ll check again

Yes it is wrong

Let me explain

You have your vertical asymptote correct, which is 3

But remember when I said x values can’t be negative?

Radical graphs can begin at a negative value but then cannot be negative infinity

[lilac plume]

so what would be the right answer?

[pure sandal]

i cant draw a graph

but according to your equation

it looks like this

do you know what desmos is?

wait

did i type it in right

or is the 1/2 outside the sqrt?

[lilac plume]

inside

[pure sandal]

x can be 5, because 5-3 is 2 and half of 1 which is 1, and then that means the x intercept is 0

using 3 will also give you -1

(11,1)

do you understand why?

[topaz sinew]

@lilac plume Has your question been resolved?

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@lilac plume Has your question been resolved?

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[topaz sinew]

Channel closed

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Use .reopen if this was a mistake.

[heavy surge]

.reopen

[topaz sinew]

✅

[heavy surge]

Anyone

<@&286206848099549185>

[reef crest]

add all the equations

and then compare LHS and RHS

[heavy surge]

Done

But still not getting

[reef crest]

a= sintheta

b = costheta

c =0

so the numerator is 1

[heavy surge]

Got till that

But final answer is not coming

[heavy surge]

Final answer isn't coming

Right

[heavy surge]

<@&286206848099549185>

[rigid cloak]

I came

[rigid cloak]

!help

[topaz sinew]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[heavy surge]

@rigid cloak

[heavy surge]

I know

[rigid cloak]

Find minimum value okok

[heavy surge]

It is asked about maximum value

[rigid cloak]

Okok max

Let me do

[reef crest]

well now you'll get 1 + 3sinthetacostheta + 4cos^2theta = 3+ 3/2(sin2theta) + 2cos2theta

then use c- sqrt(a^2+b^2)

will give you max vaue

[reef crest]

this is denominator btw

[rigid cloak]

So u wanna say

That we directly compare lhs and rjs

Rjs*

Rhs*

And say a=sin theta and b =cos theta and c=0?

[heavy surge]

Yes

[heavy surge]

But why can't we Take a's value as 1 because maximum value of sin theta is 1

[reef crest]

bcuz if you take sintheta as 1 that means costheta = 0

which will not give you max value

[heavy surge]

Ok

Ok

[rigid cloak]

No

See

Bro

Try d/dx vro

[reef crest]

ye you could do that as well

probably easier

[rigid cloak]

@heavy surge find min value of denominator for max value of whole expression

,w min(1+sin2t+4(cost)^2)

[thorny flame]

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[rigid cloak]

What

Bro

He doing what

,w calc 1/0.763932

[thorny flame]

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Upgrade to WolframAlpha Pro!

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[sudden temple]

wth?

[reef crest]

its approximating

ig

[rigid cloak]

See we wanna minimise denominator

So that we maximise the whole expression

[reef crest]

or not

[rigid cloak]

Leave it bruh

[terse mason]

yes

[rigid cloak]

Aah

I found my error

I put 2 instead of 3

,w min(1+3sint.cost+4(cost)^2)

[thorny flame]

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[rigid cloak]

The

Wjat

[heavy surge]

Got till here but what afterwards

[rigid cloak]

What

,w minimise (1+3sint.cost+4(cost)^2)

[thorny flame]

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[rigid cloak]

Aaaah

[heavy surge]

Haha

[reef crest]

then the formula for max value is c - sqrt(a^2+b^2)

[rigid cloak]

,w min value of (1+3sint.cost+4(cost)^2)

[reef crest]

where c is the constant

[thorny flame]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by oppailol.

[reef crest]

a = is the coefficient of sin2theta

[heavy surge]

How did you get that

[reef crest]

and b = the coefficient of cos2theta

[rigid cloak]

,w d/dt (1+3sint.cost+4(cost)^2)

[thorny flame]

Results provided by WolframAlpha

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

Requested by oppailol.

[heavy surge]

What the heck is this

[heavy surge]

Cant understand this one

[rigid cloak]

See bro has compared lhs and rjs

Rhs*

And found that a=sin t
B=cost and c=0 right?

Now we just substituting and finding max / min

[heavy surge]

It is not a equation it's an identity

So we can do that

[rigid cloak]

Yes

Ik

I said that only

[heavy surge]

Have you got it

[rigid cloak]

Cz it says for every alpha gamma beta

See after substituting a b

We get 1/(1+3sint.cost+4(cost)^2)

Now to maximise the whole expression we minimise the denominator

@heavy surge

[reef crest]

https://www.quora.com/How-can-I-show-that-max-acosx+bsinx-+sqrt-a-2+b-2

[heavy surge]

Understood

[terse mason]

im getting 2 theta to be tan inverse 3/4

[terse mason]

after doing this

[rigid cloak]

Yus

That only

[heavy surge]

Oh that I know

[rigid cloak]

So we get t

And substitue

[terse mason]

find theta substitue back in the equation and youll get the min value of denominator

[rigid cloak]

Yus

[terse mason]

1/that gives maximum value of expression'

[rigid cloak]

Yus

[heavy surge]

Oh

[heavy surge]

Have you got final answer

[heavy surge]

Understood

[reef crest]

well the problem with this method is that you might get a value of t for which you dont know sint and cost values

[rigid cloak]

Bro

Itf do u know?

Itf

[reef crest]

yes

[rigid cloak]

Inverse trigonometric functions

I am using that

[reef crest]

this is wrong

it should be sin inverse

[rigid cloak]

I got the values as 2/11

Of the whole expression

This means denominator is max at t=arctan(3/4)/2

Now we need to check at end points

We need minima

[terse mason]

you can do sin inverse as well

[rigid cloak]

Calculus it's coming as 2/11

Which is of course not max

[reef crest]

no like its sin inverse 3/4

[rigid cloak]

Of expressoon

Expression*

[reef crest]

not tan inverse 3/4

[terse mason]

sin inverse 3/5 is it not

[reef crest]

yes

[terse mason]

3/5 sorry

[reef crest]

no

3/4

[rigid cloak]

Yes

[terse mason]

um wait

[rigid cloak]

3/5

Na bro it's tan inverse (3/4)

[terse mason]

let me send working

[rigid cloak]

Na bro

[terse mason]

yea

[rigid cloak]

Leave it

Let's just do the derivative of full

,w d/dt 1/(1+3sint.cost+4(cost)^2

[thorny flame]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by oppailol.

[rigid cloak]

Ok

Na bruh

Aaaah

[terse mason]

im getting 1/5.5

@heavy surge

[rigid cloak]

I got 2/11

[heavy surge]

Answer is 2

[heavy surge]

It's same

[reef crest]

yes

its 2

[rigid cloak]

@terse mason it is maximum.value of denominator

[terse mason]

0.5 nvm

[rigid cloak]

We need minimum

Aah

[terse mason]

im stupid

[rigid cloak]

@heavy surge check simply bro

[terse mason]

made a mistake

[rigid cloak]

See

See I will.help.u

See we have (1+3sint.cost+4cos^2t)

We can write cos^2t=(1+cos2t)/2

[heavy surge]

I want a solution without using derivatives

[terse mason]

wait now im not sure

[rigid cloak]

Same we can write for sint.cost that is 3sin(2t)/2

[terse mason]

the answer is 2?

[heavy surge]

We haven't being taught calculus

[rigid cloak]

Now use what dawg was saying

[heavy surge]

Yes

[rigid cloak]

@heavy surge

I got it

[heavy surge]

I used it

[reef crest]

yes

i did a mistake

sry

[heavy surge]

But still stuck

[rigid cloak]

Convert the denominator as sin2t and cos2t

Then use

The range of it

-√(a^2+b^2) to √(a^2+b^2)

[reef crest]

so 2t = 37 degrees approx

[terse mason]

yeaa

[rigid cloak]

Bro

Listen

It giving min value

Not max

2/11 is min

Fr

[terse mason]

that works too

[rigid cloak]

Yes

Use that

[terse mason]

do you know the range for asint+bcost @heavy surge

[heavy surge]

Yes

[terse mason]

use that

and a couple trig identities

[rigid cloak]

Yea this works out

I got 2

Imma show it

[heavy surge]

Got it

I also got 2

[rigid cloak]

Fr

√(a^2+b^2)

Worked

Lol max is 2

[heavy surge]

Yes

[rigid cloak]

And min is 2/11

Fr

We got both

Enjoy

.close

[heavy surge]

Well is derivate easy sol for this

I just wanna know

[rigid cloak]

No

[sudden temple]

if you can find t after setting this equal to zero then yes

[rigid cloak]

Cz from derivative u can only get minimum value

And from √a^2+b^2 u get both

[terse mason]

it's unecessary really

but on first glance everyone would do that

[heavy surge]

Oh ok

.close

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[indigo estuary]

Can anyone help with a and c

I got a as the 5 ohm one but the answers say 10 ohm

And idk how do even start c

[topaz sinew]

@indigo estuary Has your question been resolved?

[vestal sigil]

how did you choose the 5Ω resistor for (a)? did you calculate the current through it?

if you did so, well, you dont have to actually

[indigo estuary]

I calculated the current lol

[vestal sigil]

just think of it logically, the current is greatest when it leaves the source, before it reaches some node, where it divides

where it divides according to the current divider rule

if it divides, how could it possibly be greater than the source current? 😛

[indigo estuary]

But shouldn’t I get the 10 ohm one even if I calculate the current

I get it divides but still

[vestal sigil]

yes, in theory you should

You made a mistake somewhere

[indigo estuary]

If its not too much trouble can u show me working

[vestal sigil]

you get a system of two linear equations in I1 and I2

i1 being the current through the first mesh and i2 being the current through the second mesh

while for the 15Ω resistor, the current is i1-i2

not i2, as i see you wrote 😛

[indigo estuary]

Ohk

[vestal sigil]

but trust me, this is not required, just logic for (a)

[indigo estuary]

Ohk

Thx

.close

[topaz sinew]

Channel closed

Closed by @indigo estuary

Use .reopen if this was a mistake.

[vestal sigil]

wait, how about (c)? xD

did u figure it out!

[indigo estuary]

Oh yea I figured that out

[vestal sigil]

Awesome

[topaz sinew]

Available help channel!

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Remember:
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• 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.

[inner wren]

How to find the area of a parallelogram given it's coordinates of four points?

[inner wren]

Basically, can two parallelograms be congruent if their sides are given?

[inner wren]

Or their area?

I need a @mystic crystal @cold wagon .

[inner wren]

Like for triangle, the criteria is SAS.

[odd pagoda]

you can for example "cut off" one triangle from the parallelogram and attach it at the other side to get a rectangle. this tells you that the area is just base*height. of course, depending on the four points those might be a bit annoying to compute

a much easier way, tho I am not sure if you are supposed to know it, is to use the determinant

[inner wren]

Yes. If the angle between two adjacent sides is equal, then they are congruent, of course. THank you.

[inner wren]

Nope, what are determinants? Idk about them.

.close

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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.

[inner wren]

.close But i'll learn them within a month.

.close haha

[toxic aspen]

Does someone have a good explanation for injective, surjective, bijective and inverse functions?

[toxic aspen]

Neither the book or videos online explain it very well

[toxic aspen]

Yeah i've seen that many times before but it really doesn't make any sense

[ember lodge]

Injective, surjective and bijective and inverse functions are basically describing the way that the function maps from the domain A to the range B

[ember lodge]

For injection, you can simply understand as there should be one dot to one dot, and not two dot to one dot

such as f(1) = 1, f(2) = 2... and so on

one input only output one, and not f(1) = 1 and f(2) = 1

for unique input, comes out a unique output

Does that make sense for u?

[toxic aspen]

Yeah i guess so

I'm just not sure how that is relevant to discrete math

There aren't any examples of injective/surjective functions?

[ember lodge]

What this says, is if x_1 and x_2 are in X (which is the domain of f) such that x_1 is not equal to x_2, then f(x_1) should not equal to f(x_2) as "One to One"

[toxic aspen]

Because everyone keeps saying what you are saying now, "one to one"

[ember lodge]

Yes

That's injective

[toxic aspen]

so x^2 is surjective

[ember lodge]

Yes

you get it?

[toxic aspen]

So if something isn't injective it's surjective?

[ember lodge]

yes

[toxic aspen]

And if something is surjective, does that also imply it's bijective?

[ember lodge]

Liek this case

The general equation

[ember lodge]

A bijective function is a function that is both surjective and injective

[toxic aspen]

How do I figure that out?

[ember lodge]

y = x | y = f(x) is an example

[toxic aspen]

That’s to complicated

[ember lodge]

Nah, that's the most common example of bijective

[toxic aspen]

It is for me

Since I don’t get it

[ember lodge]

Tell me what you don't get it

[toxic aspen]

How do you know it’s bijective

I don’t see it

[ember lodge]

Ok I'm gonna explain

But first you need to know how to proof a function that is injective and bijective

[toxic aspen]

Like if someone gives me a function im supposed to know if it’s inj, surj, bij

How is that done

[ember lodge]

ye so u want to proof

Ok, for proofing inj. you have two ways, One is assume f(x_1) = f(x_2) which if you do further calculation, you will end up with x_1 = x_2, the second method, is to assume x_1 is not equal to x_2 which if you do further calculation, you will end up with f(x_1) is not equal to f(x_2).

I think the first one will be easier for u

[toxic aspen]

Yes I saw someone do that method. But how is f(x_1) = f(x_2) relevant ?

[ember lodge]

wdym?

[toxic aspen]

Didn’t we say that any x gives a different y?

[ember lodge]

ye but we asuume x_1 is equal to x_2

[toxic aspen]

Yeah but if we prove they are equal, doesn’t that prove that it’s NOT an injective function?

[ember lodge]

wait, sorry its not, surjective is simply like if the whole graph takes up whole space of the diagram

my bad, sorry for looking it wrongly

[ember lodge]

how? If x_1 is equal to x_2, they should be equal cuz they are at the same point, they will end up to the same point too

One to one

[toxic aspen]

$f(x)=2x+3$

[ember lodge]

x_1 and x_2 are same dot

[thorny flame]

Merineth

[toxic aspen]

If i have this

[ember lodge]

ye

[toxic aspen]

$2x_1 + 3 = 2x_2+3 \implies x_1 = x_2$

[thorny flame]

Merineth

[toxic aspen]

What does this prove?

[ember lodge]

ye

u are correct

I know its a bit werid for u since u r new to this

[toxic aspen]

That proves that two different x gives the same y?

[ember lodge]

They are same x

they are same point

[toxic aspen]

How does that make sense? Of course we get the same value if we use the same x?

[ember lodge]

x_1 and x_2

you may want to try with x^2

x^2 is not injective

[toxic aspen]

Yes i've seen that x^2 gives pm x

But it still doesn't make sense

So should i just not learn it and jsut do

x_1 = x_2?

[ember lodge]

nah

you should learn the proof

[toxic aspen]

I'd like to, but it doesn't make any sense

Injective functions is a one to one
Surjective is all from A to some of B?

[ember lodge]

basically is f(x) = B

[toxic aspen]

ye okay but i still dont' get injective

[ember lodge]

lol, I thought the surjective is the hardest one

[ember lodge]

why?

[toxic aspen]

Wdym why

I can't explain why i don't understand something when i don't understand something xD

[ember lodge]

I mean where u stuck?

[toxic aspen]

That's like asking someone what they forgot when they forgot something

xD

[ember lodge]

umm

injective is just one to one

one dot to one dot

like f(1) output 1, f(2) output 2 and so on

[toxic aspen]

Yes i'm aware. That is literally what everyone says but what does that mean

[ember lodge]

it means unique input comes out unique output

[toxic aspen]

Yes

That i also get

[ember lodge]

not just f(1) = 1, f(2) = 1, f(3) = 1 and so on

[toxic aspen]

Is that all there is to it?

If someone asks me

[ember lodge]

that's not injecitve

[toxic aspen]

"what is injective" and i responds "one to one" then i get full points?

[ember lodge]

Somehow yes, but since it's not the rigirous meanning

[ember lodge]

No

Since that is just a concept

@sudden temple would u like to help with my problem pls

[toxic aspen]

$f(x) = e^x \sin (x)$

[thorny flame]

Merineth

[toxic aspen]

Is this injective?

[sudden temple]

I was trying but you explained everything already

[toxic aspen]

Does everyone x give a unique y?

[ember lodge]

lol

[toxic aspen]

How do you know that?

[ember lodge]

ok so now you want to proof

if you want to proof it you have two ways like I said, but this function....

[sudden temple]

oh the goat is here

[ember lodge]

is kinda hard to proof

[toxic aspen]

Well yes, isn't that the whole point of the chapter covering inj, sur, bij functions? To determine what type of function it is?

[ember lodge]

umm ye technically

[cedar wagon]

f(x) = f(y) => x=y

[toxic aspen]

Like if someone asked me to give examples of
a) bijective
b) injective but not surjective
c) surjective but not injective
d) neither injective or surjective

[ember lodge]

oh ye, lol

[toxic aspen]

How would i do that?

[cedar wagon]

For injective but not surjective

Like e^x

[toxic aspen]

But how do you know that?

[cedar wagon]

Well, e^x is injective cuz strictly increasing but not surjective cuz for R->R , e^x = -1 dont have a sol

[toxic aspen]

ok...

and surjective?

[cedar wagon]

x^3 - x

[toxic aspen]

But i'm trying to understand why. I don't get what any of you are talking about

[cedar wagon]

Surjective cuz its continious (from -inf to +inf) but not injective cuz 1,0 and -1 are all sol of
f(x) = 0

[stone rapids]

"Why" is a hard question

[toxic aspen]

I can't distinguish between any of them and i can't come up with my own examples and i can't determine if a function is inj, surj or bij

everyone keeps saying "injective is one to one"

Like it's supposed to trigger an epiphany

[stone rapids]

Begin from basics

Bijectivity is easier depending on what tools you have

Can you repeat the definition of an injective function?

[toxic aspen]

I'm WELL aware that injective is one to one. Where one input gives a unique output. But HOW do i determine if a function is injective?

[stone rapids]

I don't care if you call if injective is one to one or ABCXYZ.
If a function f is injective, then in terms of algebra we have ...?????????

[cedar wagon]

Start from f(x) = f(y)

[ember lodge]

one unique input *

[stone rapids]

Because these are (basic) definitions you should always use the algebraic definition

And not the English (or language) one

[ruby tree]

How has nobody mentioned domain, codomain, and image yet

[cedar wagon]

.

[ember lodge]

I MENTIONED, NEL

[cedar wagon]

Well the question is how

[toxic aspen]

$f(x) = 3x+1$

[thorny flame]

Merineth

[toxic aspen]

Would the domain be all of the possible x?

[stone rapids]

But what is all possible x

Is a platypus inside all possible x?

[ruby tree]

The domain is part of the definition of a function

[cedar wagon]

Is it injective ?

The range for the image of this function

[toxic aspen]

Well x can be literally anything?

[stone rapids]

Okay my questions are not helping, I will begin just stating stuff

[toxic aspen]

Any R number for x

[stone rapids]

We assume that $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$.
We define injective functions as functions that, for every (a,b) in domain of said function, (f(a) = f(b) \implies a = b) for every (a, b) in (\mathrm{dom} f).
Then the proof that (f) is injective starts from (f(a) = f(b))

[thorny flame]

fishwhale

[toxic aspen]

So if my domain has a = 2 and b = 3

you are trying to say that f(2) = f(3) ?

That isn't true at all?

[ruby tree]

A function f is defined along with two sets: its domain X and its codomain Y.
f: X -> Y
Usually, when you define a function like f(x) = 3x+1, you imply that the domain and codomain are both the set of real numbers: X = Y = R.
In this context, f is injective because it maps distinct elements of R to distinct elements of R. In other words, if you pick any two different real numbers x1 and x2, and put them into f, you get two different real numbers y1 and y2. x^2 as a function from R to R is not injective because you can pick x1 = 2 and x2 = -2 and get y1 = y2 = 4. Different inputs led to the same output, so that function is not injective. However, x^2 as a function from R+ to R is injective, because you can't pick two different positive real numbers as input and get the same output.

[toxic aspen]

Yeah so far i'm with you

But i'm having a hard time understanding the whole f(2) = f(3) thing

[stone rapids]

No. It's saying if f(2)=f(b), if f is injective, b=2

[toxic aspen]

Huh

[ruby tree]

If f(2) = f(3) that means the outputs from two different inputs are the same, so f is not injective

[toxic aspen]

$f(x) = 3x+2 \implies f(2) = 3 \cdot 2 + 2 = 8$

[thorny flame]

Merineth

[toxic aspen]

$f(x) = 3x+2 \implies f(3) = 3 \cdot 3 + 2 = 11$

[thorny flame]

Merineth

[toxic aspen]

They are clearly not

[stone rapids]

Definition of injective: f is injective if for every pair of inputs, a, b, [a \neq b \implies f(a) \neq f(b) \iff f(a) = f(b) \implies a = b \iff\text{There is at most one $a$ for every $b$ such that $f(a)=b$ for every $b$}]

[toxic aspen]

$f(a) \ne f(b)$

[thorny flame]

Merineth

[toxic aspen]

YES that is what i mean

they are NOT equal

[ruby tree]

That's incorrect

[toxic aspen]

Ok good

[stone rapids]

I'm quoting definitions, so I am not incorrect.

[ruby tree]

Your definitions are incorrect

[stone rapids]

They are not

[ruby tree]

$f(a) \neq f(b) \implies a \neq b$

[thorny flame]

Nel

[ruby tree]

That's a requirement for f to be a function

[stone rapids]

no

That's a different requirement

[ruby tree]

...

[ruby tree]

Take the contrapositive, a = b => f(a) = f(b)

That is saying that the same input must produce the same output

It's literally the requirement for a function

[toxic aspen]

I've never heard of contrapositive before

[stone rapids]

no

[ruby tree]

One input can't have multiple outputs

[ruby tree]

This is the converse, which is not logically equivalent

[stone rapids]

Okay I am stupid

My definition is wrong

You are right

[toxic aspen]

Maybe it is easier after all if i just understand that that a injective function can be found by just doing f(x_1) = f(x_2). And not go inot detail why it is the case

[ruby tree]

(and here it's correct, injectivity means x1 =/= x2 => f(x1) =/= f(x2))

[thorny flame]

fishwhale

[ruby tree]

If you understand this and a couple of examples you should be set, really

[toxic aspen]

but my proof only proved it for one case, right?

[ruby tree]

Idk what your prof did

[toxic aspen]

When i used a = 2 and b = 3 i only proved it for two cases?

Like

f(x) = 3x+2

If i'm given this and someone asks if it's injective

[stone rapids]

I told you the starting point

Assume f(x)=f(y)

[toxic aspen]

then they are actually asking if for any x in the domain gives one unique output (y) in the codomain

[stone rapids]

f(2) and f(3) don't do anything

[ruby tree]

Yeah you can't just use two example values and say that it works, that's not a proof

[stone rapids]

I don't know why you keep jumping on f(2) and f(3) when I told you the starting point is f(x)=f(y), which is correct and didn't contain any inaccuracy

Actually f(2) and f(3) do something, but you're not using it to do anything

[ruby tree]

"distinct inputs always produce distinct outputs" is logically equivalent to "the same output is always produced by the same input"

The latter is how you prove that stuff

[toxic aspen]

$f(x) = 3x+2$ If i have this function and i want to prove it's injective then i just have to prove that $f(x_1) \ne f(x_2)$?

[thorny flame]

Merineth

[ruby tree]

Yes but for every single pair {x1, x2} such that x1 =/= x2

which is not feasible

[stone rapids]

Actually it is feasible

[toxic aspen]

Right, because we can't test every single pair?

[ruby tree]

Well kinda, using other properties

You should use this instead: "the same output is always produced by the same input"

"the same output" -> y1 = y2

[stone rapids]

Assume (x \neq y). Then (f(x) = 3x + 2 \neq 3y + 2 \iff 3x \neq 3y \iff x \neq y) which is what we assumed hence ( x \neq y \implies f(x) \neq f(y)) for every (x \neq y)

[thorny flame]

fishwhale

[ruby tree]

You start with that, and find x1 and x2, proving that they are equal, hence "the same input"

[stone rapids]

This is the same proof as starting from f(x) = f(y) implies x = y

The symbols are just different

Since for every pair of real numbers x,y we either have x = y or x \neq y

[stone rapids]

All in all given this definition, your starting points are

1. assume a \neq b for inputs a,b, goal to show f(a) neq b
2. assume f(a) = f(b) for inputs a,b, goal to show a = b

[stone rapids]

@toxic aspen Does this make sense?

[toxic aspen]

Honestly, no

[stone rapids]

Ok

[stone rapids]

What don't you understand from these statements

[toxic aspen]

Everything just became a million times more confusing after i asked for help

The only thning i understand is this:

"Injective functions are functions where any unique input of x gives a unique ouput y"

[stone rapids]

You shouldn't use words

Write the algebra

I'd also want to ask

A simpler question

Which is, what is a proof?

This one you can answer with words

[toxic aspen]

Idk something that is either true or false

[stone rapids]

no

And that didn't go in the direction I wanted

An object X has property Y if for every X, we have [ALGEBRAIC STATEMENTS ON X]

[toxic aspen]

I can't write this in algebra

[stone rapids]

To prove an object X has property Y, then we must show that, given suitable starting points, we have [ALGEBRAIC STATEMENTS ON X] holding true.

[stone rapids]

I wrote it here

[toxic aspen]

I dont understand it

$f(x_1) \ne f(x_2)$

[thorny flame]

Merineth

[stone rapids]

[\begin{aligned}
&\text{$f$ is injective} \
\iff & \color{red}{\left(a \neq b \implies f(a) \neq f(b) \right)} \
\iff & \color{blue}{\left( f(a) = f(b) \implies a = b \right)}
\end{aligned}]

[toxic aspen]

What so if we prove that a = b we have proved that a \ne b?

[stone rapids]

I left out the a,b in dom f, but putting them under f means they should be in dom f anyway, if not then the inequalities don't make sense

[stone rapids]

no

The collection of statements are separated

[toxic aspen]

Can you explain it in words?

like

[stone rapids]

The words are not helping

[toxic aspen]

Neither is the algebra

Like i need a human being to explain it

It doesn't make sense to make lots of alebra and expect someone to get i

[thorny flame]

fishwhale

[stone rapids]

This isn't a lot of algebra

The red statements are all together

The blue statements are all together

[toxic aspen]

Yes i know

But it still doesn't make sense

[stone rapids]

What does not make sense

[toxic aspen]

Like i said

are they equal?

if i prove the blue does that mean i prove the red?

[stone rapids]

a = b is never a starting point for proving if a function is injective

That is my answer to you

a = b is a possible result of trying to prove if a function is injective

[stone rapids]

The red and blue are equivalent, that's what the double sided arrow means

The one sided arrow means FROM_STATEMENT RESULTS_IN TOWARDS_STATEMENT

[ruby tree]

f is injective iff:

• if the inputs are different, then the outputs are different
  x1 =/= x2 => f(x1) =/= f(x2)
• if the outputs are the same, then the inputs are the same
  f(x1) = f(x2) => x1 = x2

[stone rapids]

You're just writing it without TeX, which is worse

[toxic aspen]

not THAT makes more sense

[hoary burrow]

How long have yall been talking 😭

[toxic aspen]

So we are essentially proving that "if the inputs are different, then the outputs are different" by proving the contrapositive that "if the outputs are the same, then the inputs are the same" ?

[ruby tree]

TeX makes things take more space than necessary

[stone rapids]

@toxic aspen Can we move on to examples

[ruby tree]

Saying that it's worse to not use it is as bad as saying words don't help

[stone rapids]

Because the definitions are not helping

@toxic aspen Give me a non-injective function

Which is probably easier to see

If you don't I'll give you one soon

I claim that f(x)=x^2 is NOT injective

Can you prove it is not injective

[toxic aspen]

yes becuase i saw a video of it earlier

[stone rapids]

ok prove it

[toxic aspen]

$\pm x = \pm x$ isn't true for the reason that -1 and 1 aren't the same

[thorny flame]

Merineth

[stone rapids]

That is not a proof

[hoary burrow]

Can I try frfr

[toxic aspen]

Okay, how would you prove it?

(was asking fish)

[stone rapids]

Assume we have f(2) = f(-2). Then 2^2 = (-2)^2 = 4. But 2 is not -2. Hence x^2 is not injective.

[hoary burrow]

Woah crazy wow

[stone rapids]

It's always the same steps

[ruby tree]

That's oddly phrased

(I'm guessing you're not a native english speaker but still)

[stone rapids]

For this to be a positive injective proof, we would have
Assume we have f(2)=f(b). Then 2^2 = b^2 = 4. Since b = 2. Hence x^2 is injective

[stone rapids]

It will be good if you can see what makes this 'proof' different from above. And what would make this a real proof

[ruby tree]

That's even more oddly phrased

[hoary burrow]

Where a is any real number not equal to 0
a ~= -a
(a)^2=(-a)^2
f(a) = f(-a)
So f(x) is not injective

Idk does that work

[stone rapids]

x^2 is not injective, I'm not sure what you're trying to do

[hoary burrow]

Mb cuh

Forgot the not

[stone rapids]

Yes it works now