help-19

[spice fiber]

For 102

Im fixing any 1

[civic citrus]

m

[spice fiber]

As 1

To 101

So my q becomes

Summation of case of

I+j=1 ,2....101

But permuting these would be extremely problematic xD

[civic citrus]

yall are way too smart tbh

[spice fiber]

The only reason this q is problematic is because u want a general solution xD

I cant simply multiply the cases bt 3 because we will overcount every single number divisible by 3 twice

And every odd number n-1/2 times

Oh dear

A diff approach then

[static totem]

@civic citrus if this is for a program, we can solve it

[civic citrus]

mh, is for a desmos calculation

nothing urgent though, just some fooling around

[static totem]

well nobody knows

like it looks impossible

[civic citrus]

that's fair

it's probably a lost cause then

thanks so much for trying though

have a nice day!

.close

[odd edge]

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[late dust]

Ok yeah I misread

.reopen

[odd edge]

✅

[late dust]

sum_{i=0}^{101} (min(101,N-i)+1-max(N-i-101,0)) works if you can input a sum

@civic citrus

[civic citrus]

oh? ofc

sums are cool, give me a second!

[late dust]

Also your number of pairs here is flawed

Unless the bounds are actually 0 and 100, inclusive

[civic citrus]

ah! misswrite

ye, I meant to write 101

[late dust]

Right so you can just put that in a sum and use N-i

as long as i < N

[civic citrus]

would this be the solution for between 101 and 202?
or would this just generally work?

[late dust]

Hold on

[civic citrus]

cus it starts giving negative results somewhere along the way, outside of that range
which def seems weird

[late dust]

So if i,j,k are in [0,B], then it works when N is in [B-2,2B+2]

Yeah you wanted it to work until 3B huh

[spice fiber]

For rest of the cases normal beggars works lol

From 2b+2 to 3b

Its 3b-n+2c2

For before b-2 its n+2c2

[late dust]

(conceptually it should be [B,2B] but it also works for the boundaries)

[spice fiber]

These work till n =B

[late dust]

(3*101-N+2) choose 2, yeah

Parentheses are kind of important here

[spice fiber]

Ig

[civic citrus]

I'm really new to this stuff, so I was also kind of confused just now

[spice fiber]

I get this from pnc

You elements must sum to N

So i+j+k=n

All solution of this

Is a triplet

Basically for n from 202 to 303

[spice fiber]

This is the number

If you want to know how i get this i can tell

[civic citrus]

tysm yall! you're wizards

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[lost folio]

help me please

J ai essayé d'utiliser la récurrence mais j'ai été bloqué

j ai essayé de faire disjonction des cas

SOS ¨PLEASE

m tired

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[round shore]

How do i solve this

[round shore]

Do i use distributive

[cold sage]

mhm

[round shore]

is it like this

5x + 15 _< 2x - 24

[ruby nacelle]

yea

[round shore]

then 3x _< -39

[ruby nacelle]

✅

[round shore]

so x is -13?

[ruby nacelle]

❌

[round shore]

x _< -13

Oh

Then what

Oh

Ok

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[weary jolt]

I need help with this I dont understand graphing

[ruby nacelle]

expand the function

and express in y=mx+c

if m>0 answer is C
if m<0 answer is A/B
then u need to look at y-intercept

[weary jolt]

I dont understand

I don't understand anything comeing from graphing in stuff

[odd edge]

@weary jolt Has your question been resolved?

[odd edge]

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[weary jolt]

.reopen

[odd edge]

✅

[spiral raven]

?

anything not understand?

[spiral raven]

for this lazy way

do x= 0 and get y-intercept

for this part try to remember what y=x and y=-x looklike

[weary jolt]

i dont get it

[spiral raven]

u know what y-intercept is?

when x = 0 u got y-intercept from question
when y =0 u also got x-intercept

then u dot the point on the graph

y-intercept should be (0,x)

x-intercept (x,0)

[weary jolt]

this hurts my head i dont understand

[spiral raven]

then did u understand equation

dont mind graph

[weary jolt]

no

i have an iep or smth like dat

[spiral raven]

its education plan or english program?

[weary jolt]

oh

[spiral raven]

iep mean two thing

[odd edge]

@weary jolt Has your question been resolved?

[odd edge]

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[mystic saffron]

In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that an adult over 40 years of age is diagnosed as having cancer?

[mystic saffron]

from what i would think to do, i would use the total probability

$$P(A) = \sum_{i=1}^{k} P(B_i)P(A|B_i)$$

[clever fjord]

Quasar

[mystic saffron]

i dont know how to use it here

.close

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[ornate summit]

so they gave us this definition

[ornate summit]

and this

why is [0] +[0] = [0]

[cold urchin]

Because 0 = 0 in mod4?

In fact 0=0 in any mod

[south plume]

what else would it be

also [0] is the identity

[ornate summit]

what exactly is Zm

[south plume]

integers mod m

have you heard of modular arithmetic

[ornate summit]

[1]

would be the

every element which remainder is 1?

[1]m is a set itself?

is this an equivalence class or do i mix stuff up?

[cold urchin]

Yes you put a relation on integers

[ornate summit]

how would this set be defined

[cold urchin]

[1] = {…,-3,1,5,9,…}

a = b modn iff n | b-a

[ornate summit]

hm why

[cold urchin]

Is the equiv relation

On Z x Z

[cold urchin]

They belong to the same equivalent class which we’ll call [1]

[ornate summit]

what is iff?

ah

[cold urchin]

if and only if

[ornate summit]

so basically every a-b which divided by n

gives the rest 1

is in [1]

[cold urchin]

Yeah

[ornate summit]

why you define it as subtraction from 2 elements

and not just one

why cant is say

n | a

[cold urchin]

By this you mean every integer a,b s.t b-a is divisible by n and 1 is in here.

Then it’s in the set 1

[ornate summit]

so when i have [2]_3

[2, ,5, 8, 11 ,...]

right?

[cold urchin]

I mean then what you’d just partition integers into 2 sets

One set contains all multiple of n

The other set contains all other number

How’d u define this relation anyway?

[ornate summit]

ah ok

[cold urchin]

Right

[ornate summit]

you kinda need 2 numbers because youll define a relation on them

[cold urchin]

and negative integers

Yes

[ornate summit]

a ~ b := n | a-b

[cold urchin]

Mhm

[ornate summit]

[a] = {z ∈ Z| m|(z − a)} = {z ∈ Z|∃k ∈ Z : z − a = k · m}

given that

m | (z-a)

m divides (z-a)

[cold urchin]

Correct

[ornate summit]

z − a = k · m

this just means

[cold urchin]

z-a is a multiple of m

[ornate summit]

jeah

so if m is 2

if i can divide z -a by 2

i can divide it by 4

aswell

so when i have [2]_3
[2, ,5, 8, 11 ,...]

here for example

i can do 2x3 where k =2 m=3

and 6 | 8 with rest 2

remainder 2

[cold urchin]

I’m not following

What are you doing ?

[ornate summit]

just brainstorming what this means: {z ∈ Z|∃k ∈ Z : z − a = k · m} given {z ∈ Z| m|(z − a)}

[cold urchin]

They’re equivalent statement

If m divides z-a

Then z-a can be written as a multiple of m

ie z-a = mk for some integer k

[ornate summit]

oh makes sense

and

when i have 2 of those equivalence classes

the addition is defined like that

but would i take every elements of one equivalence class and add it to the element of the other equivalence classß

e.g.

Z4

[0]_4 = {0, 4, 8, 16, ...}

[cold urchin]

It would but really when you think about operation in modulo you don’t really think in terms of equivalent classes

In the same way that when we think about integers we don’t think about the quotient of natural numbers

[ornate summit]

[1]_4 = {1, 5, 9, 17}

[cold urchin]

Adding this two will give you [1]

[ornate summit]

why?

[cold urchin]

Just try adding them

pairwise but for every element

[ornate summit]

[0]_4 + [1]_4 = {1, 9, 17, 33}

33:4

32 r 1

jeah true

lol

so [0]_4 is the neutral element in this group?

[cold urchin]

No we’re adding every element of [0] and [1]

S you’d get 1+4 = 5 in there as well

[ornate summit]

cartesian produkt

?

[cold urchin]

It’ll literally gives back [1]

Well no exactly but u can think of it that way

Take one element in [0] add to every element in [1]

Repeat for every element in [0]

[ornate summit]

{0, 4, 8, 16,] + {1, 5, 9, 17} = {0+1, 4+5, 8+9, 16+17}

not like this right

[cold urchin]

Well this will still give you [1]

[ornate summit]

{0+1, 0+5, 0+9, 0+17, 4+1, 4+5, 4+9, 4+17, 8+1,.....}

like this?

how should i know out of this definition

[cold urchin]

But if I recall correctly we can just define [x] + [y] to be we pick any element in x and y add them and see what mod it is and call it [c]

[cold urchin]

m| s - a and m| t-b means m | s+t-a-b

So s+t ~ (a+b)

ie. s+t belongs in [a+b]

[ornate summit]

ok

think i kinda got it

thanks mate

[cold urchin]

No worries

[odd edge]

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[vale horizon]

i know the dr/dt is -1 cm/min, and the dSA/dt is 8 pi r dr/dt, but im not sure where to go from there

[odd edge]

@vale horizon Has your question been resolved?

[vale horizon]

pls ping me when reply, i dont use discord often and i only get notified when pinged

[odd edge]

@vale horizon Has your question been resolved?

[vale horizon]

<@&286206848099549185> sry for the mention but it said i could ping after 15 mins

[topaz salmon]

what the problem ?

[vale horizon]

right here

just unsure where to go with my current info

[topaz salmon]

okie

gimme a few mins ?

[vale horizon]

tysm

[topaz salmon]

okay got it

D<50/pi

[vale horizon]

can you show how you did it pls

[topaz salmon]

okay

can i send ss on dms or here is ok?

[vale horizon]

here is okay, thanks

[topaz salmon]

okay

[vale horizon]

also brb sorry gime like 5 mins

[topaz salmon]

dw

thats finding the upper bound

so any value of D less than that will yeild a rate less than 200cm^2/min

theres the second half of the working

hope this helps @vale horizon

[vale horizon]

ohhhhh okay thank you so much

lemme look through this and check it out

[vale horizon]

whatd you do over here?

[topaz salmon]

oh thats js putting area interms of diamter

cuz r = d/2

if u plug d/2 inplace of r

then simplify

u get pid^2

[vale horizon]

ohh right

[topaz salmon]

yep

[vale horizon]

can u show me what you did here

i dont really understand where -1/2 came from

[topaz salmon]

cuz remember dD/dt was -2

if u js invert that dt/dD = -1/2

[vale horizon]

sry for asking so much but how did you realize that you needed to invert to find dA/dD

just trying to make sense of it all

[topaz salmon]

dw ask as much as u want

i was like that when i did rates

[vale horizon]

yeah its confusing 😭

[topaz salmon]

i was finding a way to link dA and dD

[vale horizon]

so you knew dA/dD was 2piD because its the derivative of the surface area function?

[topaz salmon]

yes

and the first half is to find the value of D which can satisfy the condition

then the second half is now using that to find the rate of change of area

therefor dA/dt

and if u break taht down using the info we have we can say dA/dt=dA/dD times dD/dt

[vale horizon]

OHHHHHHHHHHHHHHH I GET IT

[topaz salmon]

perfect

[vale horizon]

so rather than inverting dD/dt

you can also divide it from dA/dt

[topaz salmon]

in the start yea, but u get used to it 😄

[vale horizon]

to move it from the other side

[topaz salmon]

the inversion in the first half was to find the value of d

[topaz salmon]

then yea u move it to the other side

to find dA/dt

so we found that dA/dt = -4piD

[vale horizon]

derivative of SA with respect to time would be SA = piD^2 --> dSA/dt = 2 piD dD/dt

then move dD/dt to the other side of the equation

[topaz salmon]

so when we plug the initial value of D we found, we can say that that is the upper limit

[topaz salmon]

yes that is correct

[vale horizon]

okay i understand now

thank you so much 🙏

[topaz salmon]

ur wlcm man

[vale horizon]

oh oops

.close

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[jagged barn]

i still need help with it

[odd edge]

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

[jagged barn]

1

1. I don't know where to begin

[odd edge]

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[sweet pendant]

hi

[sweet pendant]

i am in need of help for a fourier series question

i finished and solved question 1)

im now working on 2)

and i solved the

an

and the bn

my question is, i have two different values for the odd and even, how would i represent it for the f(x)

[desert marlin]

!filetype

[odd edge]

Please post images (such as PNGs or JPGs) of the question rather than other filetypes such as PDFs which have to be downloaded. Non-image downloads can potentially contain viruses or other security risks.

[sweet pendant]

oh

ok

this is the question

i got An as this

i got Bn as this

my ao is this

my goal is to finally finsih and solve the question asked

which is what is the fourier series

how would i represent it as an equation

so far my professor has only given us cases where for example, the "even" bn is 0

so i never have to account for it

but in this case, they both have values, would it change the way I represent my final answer?

[desert marlin]

you have the formula here

[sweet pendant]

yes

[desert marlin]

in the case where bn is 0 you just ignore the sine term

but now, don't ignore it

that's all

[sweet pendant]

well

what i mean is

i never seen a case where it isnt ignored

so i dont even know how to write it out

like bn as a whole isnt 0

its like when n is even

or smth

[desert marlin]

Well you write it out in the exact same way as you've written out the an terms in the past

but now you use the values for bn, and you write sin instead of cosine

[sweet pendant]

hm, are u able to give me an example?

u dont need to use same values

just a diff thing

in the past, bn would have n even = 0

but the n is odd

still exists

meaning we would reach to a value

but in this case where i have bn has a value for both n even and n odd

what should i do?

wait im stupid

this has been a dumb question

🤦

well actually

so my prof keeps it "general"

but then further simplifies it

using the odd/even idea

[odd edge]

@sweet pendant Has your question been resolved?

[odd edge]

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[pale arrow]

how do you do this

[quasi sparrow]

!msgdel

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[viscid flint]

(it was someone else who deleted the msg; you should still grab a new channel anyway @pale arrow

[pale arrow]

thz

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[warm pivot]

I'm trying to solve this and I think I have an answer but I'm not sure how to check if it's right or not

[cold urchin]

whats ur answer

[warm pivot]

since there's no way i know of how to fact check it and it's past due so it won't tell me if it's right or wrong

i got 10

by finding the first few i get 3, 9/10, 27/100, ...

[cold urchin]

that doesnt seem right

[warm pivot]

plugging into S_n = 1(1 - 9/10^n-1)/(1 - 9/10)

9/10^n-1 is infinitely close to 0 so 1/(1 - 9/10)

[cold urchin]

the ratio is acutally 3/10

[cold urchin]

look at your sequence here

[warm pivot]

...i confused what that number meant

[cold urchin]

common ratio

[warm pivot]

i got it into my head that number was simply the 2nd of the sequence and not the ratio

ty

[cold urchin]

lol np

[warm pivot]

1/(1 - 3/10)
1/7/3
3/7?

[cold urchin]

should be 30/7

[warm pivot]

hm... what am i missing then

[cold urchin]

the first term is not 1

as you said 3,.....

[warm pivot]

ohh

the first 1 in 1(1 - 3/10^n) is a_1 isnt it

so it should be 3(1 - 3/10^n)

[cold urchin]

yes

[warm pivot]

thank you so much

.close

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[cold urchin]

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[stiff kettle]

,rotate

[stiff kettle]

How do I solve this limit

[south plume]

i don't think this will work here

[stiff kettle]

how so

[south plume]

both limits diverge

[stiff kettle]

yes but my teacher wants me to prove

it

showing that there is an infinity

i know it diverges

because of the discontinuity

he wants me to express them as limits

and solve

[south plume]

well

okay

[ember oak]

This is an improper integral

[south plume]

ln |a-3| will diverge

[ember oak]

In that context, what you have so far is correct, just cancel out the ln|a-3| terms

[stiff kettle]

wait what is Ln(0^-)

its two seperate limits so i cant cancel them out no

?

[ember oak]

Not necessarily. You are still fine to.

[south plume]

technically because you have an absolute value inside then 3+ and 3- won't matter here

[stiff kettle]

ok but can u explain to me why this here gives me -infinity

i know the other limit gives me +inf

but i dont get why this one is -inf

[south plume]

oh wait the integral converges right?

[stiff kettle]

diverge

[south plume]

if those ln|x-3| 's cancel out

hold up i'm being stupid

[stiff kettle]

hmmm

[south plume]

,w integral 0 to 5 of 1/(x-3) dx

[clever fjord]

Results provided by WolframAlpha

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

Requested by arte_m_etra

[south plume]

oh.

welp

[quasi sparrow]

yea this is essentially why

[stiff kettle]

yeah

Ln(0^+) is +inf

but what is ln(0^-)

is it -inf

like automatically since its the opposite

[south plume]

no

[stiff kettle]

oops

sorry

-inf

[south plume]

both are -inf

[cold urchin]

Well ln(0-) is undefined but in your case u have the absolute value

So it’s still ln(0+)

[stiff kettle]

ahhhh

ok i get it

ty

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[nocturne belfry]

I am kind of consfused about what my professor says

like yea okay the "order" of the bottom is higher, but doesn't this oscillate?

this limit doesnt exist does it?

like, the plot cosx = -xsinx has infinitely many intersections so we never run out of asymptotes

[tall veldt]

yeah the limit doesn't exist

[nocturne belfry]

is there some kind of good asymptotic limit here

[tall veldt]

i dont know anything about asymptotics sorry

[nocturne belfry]

alright, thanks

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[rocky frost]

Would if be true to say that if m does not divide a - b then m * n does not divide a - b

[blissful musk]

[tranquil granite]

if m is not a factor of a-b

why would m*n be

[outer wadi]

I was thinking that, logically it makes sense but im trying to prove it and im kinda stuck

how would you prove it

[rocky frost]

see my question is part of a bigger proof

trying to prove that if m divides a - b then n*m divides a - b

trying to prove this by contrapositive

[tranquil granite]

this is untrue

m=2, a-b=4, n=4000

2 divides 4 but 8000 obviously does not divide 4

unless if you are proving that n*m divides a-b ==> m divides a-b

[rocky frost]

yea oops

that

I put them wrong way

ok I think I got it

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[earnest moon]

How does the solution work in getting the value of k

I only saw the solution online so I don't know the logic behind it

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[graceful hatch]

How to solve?

[odd edge]

@graceful hatch Has your question been resolved?

[graceful hatch]

<@&286206848099549185>

This what I got now, but I only guessed

Oh wait nvm

Yeah it’s right

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[dim cliff]

im unsure about why there is a must in square rooting here.

[dim cliff]

<@&286206848099549185>

[sour gyro]

All terms in the expression must be multiplied by the exponent (3).

If the exponent is inside the parenthesis, it only applies to the term which it is raised to.

[dim cliff]

in simpler terms?

[sour gyro]

Such as (3+1)^3 means 3^3 + 1^3

Now if it was (3^3 + 1) it would be 3^3 + 1

For your problem, the exponent applies to the all terms as it is outside the parenthesis. Meaning all terms inside the parenthesis are raised to the third power.

[dim cliff]

okay that helped

thanks for your time

[sour gyro]

Does that fully make sense? It’s a bit hard to explain through text.

[dim cliff]

i mean i still dont understand why it must be raised and why the ^3 is kept

[sour gyro]

https://m.youtube.com/watch?v=LkhPRz7Hocg

Have a look at this tutorial.

It might be easier to understand.

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[waxen blade]

How to 3(c) I've done 3a and 3b

[waxen blade]

@fair token

[odd edge]

@waxen blade Has your question been resolved?

[zealous fossil]

don't ping people

[waxen blade]

Mb

But can u help me

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@waxen blade Has your question been resolved?

[waxen blade]

No

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[toxic sigil]

Define mappings f: R - > R by f(x) = 4x^(2)+1 and g: R - > R by g(x) = e^(3x).
A) Find the composite mapping fg.
B) Is fg injective? Give a proof or counterexample" Pretty sure I got A down, we do f((x)) = f(e^(3x)) = 4(e^(3x))^2+1 = 4e^(6x)+1. (edited)
I'm struggling with B.
How does one prove injectivity? How does one prove surjectivity?

[brittle beacon]

Generally you'd prove injectivity by showing that if $f(x_1) = f(x_2)$, then you must have $x_1 = x_2$

[clever fjord]

@brittle beacon

[brittle beacon]

There are shortcuts you could use, but as they say you should prove it, probably best to do it directly

[toxic sigil]

I think I got it down. So I set 4e^(6x)+1 = 4e^(6y)+1, subtract on both sides, isolate for e^6x and e^6y, apply natural logs, get 6x=6y, divide by 6, we get x = y, and since x and y are injective, fg is injective. right?

What about surjectivity? How can I determine if its surjective?

[brittle beacon]

Yep

[brittle beacon]

For any real number of your choosing, can you find an input that'll get you that? To reverse the question, are there any real numbers that aren't in the image of the function?

[toxic sigil]

Would I suppose f(g(a)=b an element in R, So 4e^(6x)+1 = -1 if I chose b = -1 E R?

[brittle beacon]

Well, that's kind of the idea, does any [real] x such that it's true exist?

[toxic sigil]

Getting a wild answer, so from here I'd just solve for x, right?

[brittle beacon]

You would try to - what's the answer you're getting?

[toxic sigil]

(ln(-1/2))/6

do you get same thiing

[brittle beacon]

hmmm

Are you allowed to take logs of negative numbers?

[toxic sigil]

no

undefined

[brittle beacon]

yep, you can't, so there's no solution to that

Another way to see it is that you know $e^p$ is strictly positive for any real $p$, so $4e^{6x} + 1 > 1$ for all real $x$

[clever fjord]

@brittle beacon

[brittle beacon]

So it's impossible to attain any values that are 1 or less

[toxic sigil]

So its not surjective? It is not surjective because there are some real numbers, like the one we just derived, is not covered. So it's not surjective over the entire set of real numbers R.

[brittle beacon]

Yep, correct

[toxic sigil]

It's making more sense. So for injectivity for functions means that if fg(x) = fg(y) then x = y. For surjectivity, we want to find any values that doesn't belong in the function?

[brittle beacon]

For [lack of] injectivity for a function h, you want to show that if h(x) = h(y), you must have x=y [or find two distinct x,y such that h(x) = h(y)]

For [lack of] surjectivity of h: A -> B, you want to show that for any choice of b in B, there's an a in A such that h(a) = b [or show that you can choose some b in B such that h(a) = b has no solutions in A]

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[dusk spire]

Whats with the parallel projection and perpendicular?

for i) i did

,tex p = $(F_1 \cdot \hat{F_2}) \hat{F_2}$

[clever fjord]

suds

[dusk spire]

which was -7.49i + 25.68j

i dont know what component is acting parallel/perpendicular and as they are both worth 2 marks, i dont think that it is just the i and the j values as that would mean that ii would be 1 mark

[forest sky]

all vectors can be expressed as the sum of two perpendicular component vectors. Normally those component vectors are in the x- and y- directions, but we can also consider directions that are parallel and perpendicular to some other vector

[dusk spire]

ok?

so the i value is the x- direction, therefore the parallel?

[forest sky]

we can split F1 into components in two different ways:

1. into F1x and F1y (the x- and y- components)
2. into F1 (parallel to F2) and F2(perpendicular to F2)
   In both cases, F1 is the sum of these two vectors (which are perpendicular to each other), but the two sets of components are distinct

we already found the parallel component by vector projection

[dusk spire]

i dont understand how you split f1 into parallel to f2 and perpendicular to f2

so F1 is -31.25i + 18.75j as the x- y- components you mean?

[forest sky]

the component of f1 parallel to f2 is the same as the vector projection of f1 onto f2

[dusk spire]

ok

wait so vector rejection is perpendicular?

ohhhh

i got it

[forest sky]

if we know the magnitude of vector F1 and one of its components, we can then solve for the magnitude of the other component

[dusk spire]

yes

so the parallel to F2 is basically F1 flattened to F2

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[jagged rover]

uh

[odd edge]

Please don't occupy multiple help channels.

[jagged rover]

what is the logarithms law/properties again

[noble forge]

DO you want to know in general, all properties?

[jagged rover]

yes

for the evaluating of them i guess

i remember one of them is like the change base

[jagged rover]

Wow perect

Thanks

[royal herald]

Homez

U legit coulda searched that in google

That’s what beard just did

Lol

[jagged rover]

i dont trust myself i feel like theres too much out there

i swear too much

[royal herald]

Nah all of them just saying the same thing

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@jagged rover Has your question been resolved?

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[dusk spire]

Consider the proposition

[dusk spire]

"For all integers j and k, if jk is even then j or k must be even"

State the contrapositive and hence prove the original statement by contraposition.

I think my proof isnt enough but im not sure

so "For integers j and k, if jk is not even then neither j or k are even."

If the contrapositive is true, the original statement is true.

2jk - 1 = odd for any integer j or k

we know that any multiple of 2 is even. if jk = n then 2n - 1 is odd

let j be 2n - 1

(2n - 1)k = 2nk - k

for any integer where k is odd, 2nk -k is odd. If 2n is even, 2nk is even

Assume that this is also true for j

(2n-1)(2n-1) = 4n^2 - 4n + 1

where 4n^2 and -4n are even, the plus one makes the equation odd

therefore for all integers j and k, if jk is odd, neither j or k is even

the contrapositive of this statement is true

[leaden widget]

angle multiple?

[dusk spire]

sorry?

[leaden widget]

"we know that angle multiple of 2 is even"

what does that mean?

is that a typo for every multiple?

[dusk spire]

sorry any

yes

[leaden widget]

oh ok

so

when you say "assume that this is also true for j"

Assume that this is also true for j
(2n-1)(2n-1) = 4n^2 - 4n + 1

[dusk spire]

i mean like (2n-1)k

[leaden widget]

youre using the same value of j for k?

[dusk spire]

so k could also be (2n-1)

[leaden widget]

j = k = 2n-1 ?

[dusk spire]

oh true

ok if i did 2n-1 and 2m-1

it would be 4nm - 2n - 2n + 1

that works?

[leaden widget]

yeah, because j is not necessarily equal to k

[leaden widget]

yup, that shows jk is odd

[dusk spire]

ok thanks

so that proof is good?

[leaden widget]

i think so, it's a bit confusing to read i guess but i think you got the gist of it

im not sure why you wrote this: "If the contrapositive is true, the original statement is true."

[dusk spire]

idk ive barely done any proofs

but its true that if the contrapositive is true then the original statement is true

do i not have to finish it like that when proving by contrapositive?

@leaden widget

I also have another proof i just did

Consider the proposition

"The sum of a rational and an irrational number must be irrational"

Prove the statement by contradiction

Assume this statement is false.. therefore teh sum of a rational and an irrational number must be rational

in order for a number to be rational, it can be expressed as a simplified fraction a/b where b is not 0

x,y,b,a,c are rational

a/b + sqrtc = x/y

sqrt c = x/y - a/b

sqrt c = (xb - ya)/yb

this is not true

[leaden widget]

why are you writing sqrt c

[dusk spire]

irrational numbers cannot be exporessed as simple fractions so the statement is false therefore the sum of a rational and an irrational number must be irrational

[dusk spire]

square root, i cant do that with typing

[leaden widget]

why not just write c

and say c is irrational

[dusk spire]

,tex $\sqrt{c} = x/y - a/b$

[clever fjord]

suds

[dusk spire]

because it more properly defines an irrational number

i could also do it with just c

[leaden widget]

and what is c, c is a rational number?

[dusk spire]

c is a rational number

so like sqrt 2

[leaden widget]

well not all irrational numbers can be expressed as roots of rational numbers

i.e. transcendental numbers, which are also irrational

[dusk spire]

ok so if i just do c, it shows for all irrational numbers?

true and also c could be like 4 which is rational

thanks for that

other than that?

[leaden widget]

yeah you could write $c\notin\mathds{Q}$

[clever fjord]

Soosh
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[dusk spire]

i just wronte where c is any irrational number

[leaden widget]

yea

[dusk spire]

i dont wanna potentially mess up the notation in the exam

its tomorrow

[leaden widget]

ya thats fine

[dusk spire]

so at the end of the first proof, i shouldnt write the contraposition is true so the original statement is true

[leaden widget]

and well the sum \ difference of any rational numbers is also rational

[dusk spire]

?

[leaden widget]

so writing c = x/y - a/b implies c is rational

which is a contradiction

[dusk spire]

in my actual writing i did =/ (not equal sign)

and if i say this statement is false isnt that valid?

how else do i write it?

[leaden widget]

yes just say it contradicts the premise that c is irrational

[dusk spire]

but when showing the working?

[leaden widget]

so the original assumption that the statement is false...must be false

[dusk spire]

oh

but if i wrong c = x/y + a/b

i know that c does not equal that

but i cant write =/ before proving it

[leaden widget]

im confused by what you are asking

at this point

the proof is valid

[dusk spire]

yes i know but you said

[dusk spire]

.

[dusk spire]

.

it does imply it but it also proves that it is false

it shows that it doesnt work

[leaden widget]

what implies what?

what proves what is false?

i dont understand what you are asking...

[dusk spire]

you said that writing c = x/y - a/b implies c is rational

which is a contradiction

i thought you were saying an error in my writing?

[leaden widget]

no

[leaden widget]

contradiction of this

[dusk spire]

yes it is, its supposed to be

because you prove the contradiction being wrong to prove the original statement

[leaden widget]

"The sum of a rational and an irrational number must be irrational"
-assume this is false (this is the assumption)
then for some irrational c and rationals a, b, we can write
c + a = b
c = b - a
but the sum of two rationals is also rational
therefore c is rational, but we assumed c is irrational
(this is the contradiction, both of those can't be true at same time)
contradiction implies our assumption on 2nd line is wrong
proof done

[dusk spire]

alright

so i say that if the equation is true, c must be rational

thanks

[leaden widget]

right then you get to the point that c is both irrational and rational

which makes things explode

[dusk spire]

... which it is not

boom proof

huzzah

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[real fox]

Hello, Sorry for disturbing.. Can anyone help me, Make a graph of F(x) = -x^2 - x + 12.

Thank you

[tepid hornet]

Factor it

It will be easier

To draw the graph

[odd edge]

@real fox Has your question been resolved?

[real fox]

Okaaay, THANKSS

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[real fox]

sorry i forgot to say; Thank youuu, Thankss!!

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[digital matrix]

u got a calculator?

[dull gulch]

yeah

[digital matrix]

convert to exponential form

[dull gulch]

1/2 = x^(-1/4) ?

[digital matrix]

yes

[dull gulch]

yeah im stuck what to do after that

[digital matrix]

get rid of the -¼

[dull gulch]

how do i root a number by a negative exponent

[digital matrix]

raise to -4

[dull gulch]

so reciprocal?

[digital matrix]

yeah

[dull gulch]

alr ty

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[tough rapids]

distribute the 1/3 on the left hand side, if you already not have done so

Can you show your work?

[mystic saffron]

@mystic saffron show your work please

[odd edge]

@mystic saffron Has your question been resolved?

[mystic saffron]

ignore the first thing can you explain this

i don't understand how it became 9 = 3/4x

@tough rapids ?

[tough rapids]

sure

$x=\frac{4}{4}x$

[clever fjord]

Ab

[tough rapids]

subtract that from both sides

and you have

$9=\frac{7}{4}x-\frac{4}{4}x$

[clever fjord]

Ab

[mystic saffron]

but why did we do this

how do i know when to do that

[tough rapids]

you want the x'es on one side and the numbers on the other

[mystic saffron]

right

[tough rapids]

well part of it is based on what people have seen before. but if you go back to the original equation, after distributing the 1/3, you have

$1/3x-2=4/9x-2$

[clever fjord]

Ab

[tough rapids]

you have something like

$ax-b=cx-d$

[clever fjord]

Ab

[tough rapids]

so some quantities multiplied by x'es and some that stand alone

then you just plus and minus to get the right thing

(another way of doing it)

and the plus and minus you can do is because:

it's like a balance (if you think of weight). taking away weight from both sides or adding does not change the overall equation

[mystic saffron]

I see hmm

[tough rapids]

and you have this unknown x quantity

[mystic saffron]

Yeah

[tough rapids]

any further questions?

[mystic saffron]

No not right now

thank you

[tough rapids]

alr then : ). you can type .close to close the channel

[mystic saffron]

.close

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[radiant crescent]

This is my exercise, though I get a different result from the solution, can anyone tell me where I went wrong?
I did:
alpha = arccos(c/x) = 31,94
h = c * sine(alpha) = 2,24

[steady tide]

cos = adjacent/hypotenuse, not the other way around

[radiant crescent]

sorry thats what I did

switched them when typing it down 🗿

[frigid yew]

Hi, I have a problem with this exercise, ∀n ∈ N* : (1 + 1/√2 + 1/√3 + ... + 1/√n) ≥ √n , plz can somone help me ?

[odd edge]

@radiant crescent Has your question been resolved?

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[spring knot]

im trying to get the eigen vectors for eigen value=-9 in matrix A, but im getting a zero vector... this is in the process of getting the form of a quadric

[echo ginkgo]

you don't have y=0

0y=0 that's what you have

no constraint on y whatsoever

[spring knot]

oh i see

bruh

thanks 🙏

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[spring knot]

.reopen

[odd edge]

✅

[echo ginkgo]

rip

@spring knot what happened

[spring knot]

i got too excited nyhaha

it's -9y = 0

[echo ginkgo]

no it should be 0 = 0

read your matrix again

[spring knot]

the equation is Au = lamda u right?

lambda = -9

[echo ginkgo]

yeah

you already incorporated the lambda in the matrix

[[11 0 ...] ... ]

[spring knot]

i see

got it, thanks

cause i used to start with (A-Iu)u = 0 before got stuff mixed up

thanks again 🙏

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[gleaming mantle]

is the number folllowing the x term a power or constant

[odd edge]

Please don't occupy multiple help channels.

[viral frost]

3/2 is a constant, 2 is an exponent.

[gleaming mantle]

okay thank you

close

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[lean night]

HELP ME

[odd edge]

Please don't occupy multiple help channels.

[reef sandal]

Jesus Christ someone is you narcoleptic egg yolk

Go back to your channel

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[ancient patrol]

Did i get the first question correct

[ancient patrol]

and how should i approach the second question

not sure how to start

can i do the same by multiplying the top and bottom by sqroot(x^2+2x) + x ?

[cursive charm]

Yeah. Though you could've done it by using the definition of a derivative, if you've studied that of course.

[cursive charm]

Yeah sounds good.

[ancient patrol]

alright thanks for the help

[ancient patrol]

🤔 not sure what that is

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[cursive charm]

Actually you can do it really quickly using limit laws.

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[native agate]

Going to find the anti derivative and just need someone to look after my method please

[native agate]

$$\frac{1}{x^2+1}\cdot 2x$$

[clever fjord]

Totalani

[native agate]

I start with designating x^2+1 as u, and du=2x

$$\frac{1}{u}\cdot du$$

[clever fjord]

Totalani

[native agate]

$$\ln(u)\cdot \frac{2x^2}{2}$$

[clever fjord]

Totalani

[native agate]

so far so good?

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[finite birch]

Write the force F as a cartesian vector

[finite birch]

I don't understand how to do this

The answers are in the southern corner at the right but I don't understand the process to getting there

[mystic saffron]

translation please

[finite birch]

Write the force F as a cartesian vector

[zealous fossil]

well

$F_z=F*cos(60)$

[clever fjord]

$Pure2$

[zealous fossil]

$F_x=F_y=\frac{F*sin(60)}{sqrt2}$

[clever fjord]

$Pure2$

[odd edge]

@finite birch Has your question been resolved?

[prisma cairn]

Mate this is wrong

Fx ≠ Fy

[prisma cairn]

This is correct

Angle of F with -x axis is 60 so Fx= -Fcos(60)

And find angle with y axis using cos²a+ cos²b + cos²c = 1 where a,b,c are angles with x,y,z axis

And when you find c Fz = F cos c

Btw force along a line= F*cos(angle of F with line)

[finite birch]

why do you use cos

isn't it sin?

[prisma cairn]

No it's cos

Sin would give you Force perpendicular to the line

Cos gives force along the line

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[finite birch]

and how do you calculate 354j then? I did 500 * sin(45) and got the answer, but i don't understand why it is sin(45)

[prisma cairn]

Uhh sin45 and cos 45 have the same value

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[finite birch]

right but why is it 45 degrees specifically

[prisma cairn]

Because of this

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[mystic saffron]

.reopen

grrr

[finite birch]

where a,b,c are angles with x,y,z axis
We miss the y angle though. If I do cos²(60) + cos²(60) - 1 = cos²(c) (c for the angle for y axis), I get -0.5

[prisma cairn]

No you get -cos²c = -0.5

[zenith tartan]

(reopen the channel)

[finite birch]

ah right right

cos²c = 0.5

[prisma cairn]

Idek who's channel is it lmao

[finite birch]

Do i then do cos(c) = sqrt(0.5)?

and then for the y angle i do 500 times sqrt(0.5)?

[prisma cairn]

-sqrt(0.5) because you can see that from the figure that it's -ve

[finite birch]

ah right

thank you

[prisma cairn]

Yea

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[finite birch]

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[frozen stream]

Sequence (b_n)_n for n in N converges to b.
Can somebody explain this?
I understand that |b_n - b| < |b|/2 bcs we can choose ε freely . But why is |b_n| > |b|/2?

[frozen stream]

I actually want to prove this, but I need to understand that Inequality first

[brittle beacon]

Well reverse triangle inequality gives you that $\abs{ \abs{b_n} - \abs{b} } \leq \abs{b_n - b} < \frac{\abs{b}}{2}$, which gives that $-\frac{\abs{b}}{2} < \abs{b_n} - \abs{b} < \frac{\abs{b}}{2}$, and then $\frac{\abs{b}}{2} < \abs{b_n} < \frac{3\abs{b}}{2}$

[clever fjord]

@brittle beacon

[brittle beacon]

Last by adding |b| to all sides and simplifying, and the "lower half" of that is what you want